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http://unapologetic.wordpress.com/2007/06/05/full-and-faithful-functors/?like=1&source=post_flair&_wpnonce=ae4d6bb289
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# The Unapologetic Mathematician
## Full and Faithful Functors
We could try to adapt the definitions of epics and monics to functors, but it turns out they’re not really the most useful. Really what we’re most concerned with in transforming one category into another is their morphisms. The more useful notions are best phrased specifically with regard to the hom-sets of the categories involved.
Remember that a functor $F:\mathcal{C}\rightarrow\mathcal{D}$ between two (locally small) categories defines a function for each hom set: $F_{X,Y}:\hom_\mathcal{C}(X,Y)\rightarrow\hom_\mathcal{D}(F(X),F(Y))$. We now apply our standard definitions to these functions.
We say that a functor is “full” if for every pair of objects $X$ and $Y$ in $\mathcal{C}$ the function $F_{X,Y}$ is surjective. Similarly, we say that a functor is “faithful” if each function $F_{X,Y}$ is injective. When we apply these definitions to monoids (considered as categories with one object) we recover monomorphisms and epimorphisms of monoids, so we know we’re on the right track.
Now a subcategory $\mathcal{S}$ of a category $\mathcal{C}$ consists of a subclass of the objects of $\mathcal{C}$ and a subclass of the morphisms of $\mathcal{C}$ so that
• For each arrow in $\mathcal{S}$ both its source and target objects are in $\mathcal{S}$.
• For each object in $\mathcal{S}$ its identity arrow is in $\mathcal{S}$.
• For each pair of composable arrows in $\mathcal{S}$ their composite is in $\mathcal{S}$.
These conditions are enough to ensure that $\mathcal{S}$ is a category in its own right. There is an inclusion functor $I:\mathcal{S}\rightarrow\mathcal{C}$ sending each object and morphism of $\mathcal{S}$ to itself in $\mathcal{C}$. Clearly it’s faithful, since any two distinct arrows in $\mathcal{S}$ are distinct in $\mathcal{C}$.
What if this inclusion functor is also full? Then given any two objects $X$ and $Y$ in $\mathcal{S}$, every morphism $f:X\rightarrow Y$ in $\mathcal{C}$ is included in $\mathcal{S}$. We say that $\mathcal{S}$ is a “full subcategory” of $\mathcal{C}$. In practice, this means that we’ve restricted which objects of $\mathcal{C}$ we care about, but we keep all the morphisms between the objects we keep. For example, the category $\mathbf{Ab}$ of abelian groups is a full subcategory of $\mathbf{Grp}$.
Note that we have not given any conditions on the map of objects induced by a functor. In fact, the (unique) functor from the category $\mathbf{2}$ to the category $\mathbf{1}$ is faithful, even though it appears to lose a lot of information. Indeed, though we identify both objects from $\mathbf{2}$ in the image, the function on each hom-set is injective. Similarly, either functor from $\mathbf{1}$ to $\mathbf{2}$ is full. One condition on the object map that’s useful is that a functor be “essentially surjective”. A functor $F$ has this property if each object $D$ in the target category is isomorphic to an object $F(C)$ in the image of $F$.
Now, let functors $F:\mathcal{C}\rightarrow\mathcal{D}$ and $G:\mathcal{D}\rightarrow\mathcal{C}$ furnish an equivalence of categories. I say that $F$ (and, by symmetry, $G$) is full, faithful, and essentially surjective. Indeed, since $F\circ G\cong1_\mathcal{D}$ we have an isomorphism $F(G(D))\cong D$ for each object $D\in\mathcal{D}$, so $F$ is essentially surjective. The natural isomorphism $\eta:G\circ F\rightarrow 1$ says that for every arrow $f:C_1\rightarrow C_2$ we have $G(F(f))=\eta_{C_2}^{-1}\circ f\circ\eta_{C_1}$. The right side of this equation is a bijection from $\hom_\mathcal{C}(C_1,C_2)$ to $\hom_\mathcal{C}(G(F(C_1)),G(F(C_2)))$, and the left side is the composite $G_{F(C_1),F(C_2)}\circ F_{C_1,C_2}$. If $G$ were not full then the left side could not be always surjective, while if $F$ were not faithful then the right side could not be always injective. Since the right side is always a bijection, $G$ must be full and $F$ must be faithful. By symmetry $G$ is essentially surjective and faithful, and $F$ is full.
On the other hand, if a functor $F:\mathcal{C}\rightarrow\mathcal{D}$ is full, faithful, and essentially surjective, then it is one side of an equivalence of categories. The proof is a bit more involved, but basically it proceeds by building a functor $G$. Since each object of $\mathcal{D}$ is isomorphic to some object in the image of $F$ we pick such an object $F(C)\cong D$ for each $D$ and set $G(D)=C$. Then we have to find mappings of the hom-sets of $\mathcal{D}$ in a coherent way so that the resulting composites $F\circ G$ and $G\circ F$ are naturally isomorphic to the identity functors. I won’t fill in all the details here, but suffice it to say it can be done.
This, then, characterizes equivalences of categories the way injectivity and surjectivity of functions characterize isomorphisms of sets.
### Like this:
Posted by John Armstrong | Category theory
## 10 Comments »
1. [...] now for the coup de grâce: the Yoneda embedding is fully faithful! Firstly, this means that the homomorphism is injective — its kernel is trivial — so [...]
Pingback by | June 8, 2007 | Reply
2. It seems to me that the exposition in the third-from-last paragraph is a bit mixed up. The isomorphism F\circ G\cong1_\mathcal{D} is what makes F essentially surjective.
\eta is reversed from what it usually is, and the positions of C_1 and C_2 need to be swapped in G(F(f))=\eta_{C_1}\circ f\circ\eta_{C_2}^{-1}; so I’d suggest reformulating this to G(F(f))=\eta_{C_2}\circ f\circ\eta_{C_1}^{-1}. Then all goes fine afaics.
Comment by Avery Andrews | September 22, 2007 | Reply
3. Avery, you’re right that I added the phrase about essential surjectivity to the wrong sentence, but I reversed the usual direction of $\eta$ (and the ensuing alterations to compensate) for a reason: this makes it clearer that an equivalence is actually a type of adjunction. I can flip the directions of these natural isomorphisms because they’re isomorphisms, but once I weaken them to mere natural transformations I no longer have that freedom.
Comment by | September 24, 2007 | Reply
4. Right, but what about \eta_{C_2}\circ f\circ\eta_{C_1}^{-1} vs \eta_{C_1}\circ f\circ\eta_{C_2}^{-1}; the latter is still looking wrong to me. Too bad categorists don’t use some sort of postfix notation for function-like things.
Comment by Avery Andrews | September 24, 2007 | Reply
5. Yes, that composition was also backwards.. A postfix notation would be useful, but unfortunately everyone else uses this ordering, and if I started out using the other then everyone reading my stuff would be confused when they try to read other material.
Comment by | September 24, 2007 | Reply
6. oops, \eta_{C_2}^{-1}\circ f\circ\eta_{C_1}, if \eta is from GF to 1_{\mathcal C}.
Comment by Avery Andrews | September 24, 2007 | Reply
7. Incidentally, Avery, try wrapping your LaTeX like this:
::dollar-sign::latex foo::dollar sign::
It comes up: $foo$. And it works on all WordPress-hosted weblogs!
Comment by | September 24, 2007 | Reply
8. cool. Maybe a `guide for commenters’ would be a useful addition?
Comment by Avery Andrews | September 24, 2007 | Reply
9. [...] show this, we must show that it is full, faithful, and essentially surjective. The first two conditions say that given [...]
Pingback by | June 24, 2008 | Reply
10. [...] , but with only one object. So we should really be thinking about the category of algebras as a full sub-2-category of the 2-category of categories enriched over [...]
Pingback by | November 18, 2008 | Reply
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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http://mathoverflow.net/questions/5865?sort=votes
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## Classical Calculi as Universal Quotients
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
As is well known, every differential calculus $(\Omega,d)$ over an algebra $A$ is a quotient of the universal calculus $(\Omega_A,d)$, by some ideal $I$. In the classical case, when $A$ is the coordinate ring of a variety $V(J)$ (for some ideal of polynomials $J$), and $(\Omega,d)$ is its ordinary calculus, how is $I$ related to $J$?
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## 1 Answer
In the classical case, if $\Omega(A)$ is the kernel of the multiplication map $m:A\otimes A\to A$, then—since $A$ is commutative, so that $m$ is not only a map of $A$-bimodules but also a morphism of $k$-algebras,—it turns out that $\Omega(A)$ is an ideal of $A\otimes A$, not only a sub-$A$-bimodule. In particular, you can compute its square $(\Omega(A))^2$. Then the classical module of Kähler differentials $\Omega^1_{A/k}$ is the quotient $\Omega(A)/\Omega(A)^2$.
(This is the construction used by Grothendieck in EGA IV, for example)
-
One way to think of this is that $\omega(A)$ is universal for derivations into arbitrary $A$-bimodules; the "classical" Kähler module $\omega(A)/\omeg(A)^2$ is universal for derivations into symmetric (a.k.a. commutative) $A$-bimodules. So the change of category means one works with a different notion of "universal derivation". – Yemon Choi Nov 18 2009 at 4:08
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http://math.stackexchange.com/questions/tagged/examples-counterexamples?page=4&sort=newest&pagesize=15
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# Tagged Questions
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http://en.wikipedia.org/wiki/Zero-sum
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# Zero–sum game
(Redirected from Zero-sum)
Not to be confused with Empty sum.
For other uses, see Zero sum (disambiguation).
This article needs attention from an expert in Game theory. Please add a reason or a talk parameter to this template to explain the issue with the article. WikiProject Game theory (or its Portal) may be able to help recruit an expert. (March 2011)
In game theory and economic theory, a zero–sum game is a mathematical representation of a situation in which a participant's gain (or loss) of utility is exactly balanced by the losses (or gains) of the utility of the other participant(s). If the total gains of the participants are added up, and the total losses are subtracted, they will sum to zero. Thus cutting a cake, where taking a larger piece reduces the amount of cake available for others, is a zero–sum game if all participants value each unit of cake equally (see marginal utility). In contrast, non-zero–sum describes a situation in which the interacting parties' aggregate gains and losses are either less than or more than zero. A zero–sum game is also called a strictly competitive game while non-zero–sum games can be either competitive or non-competitive. Zero–sum games are most often solved with the minimax theorem which is closely related to linear programming duality,[1] or with Nash equilibrium.
In common parlance, the expression zero-sum game is often used in a pejorative sense. It is especially used to point out cases where both sides may gain by an interaction (such as a business transaction), contrary to an assumption that the event must be competitive (zero-sum).
## Definition
| | | |
|-----------------------|----------|----------|
| | Choice 1 | Choice 2 |
| Choice 1 | –A, A | B, –B |
| Choice 2 | C, –C | –D, D |
| Generic zero-sum game | | |
The zero-sum property (if one gains, another loses) means that any result of a zero-sum situation is Pareto optimal (generally, any game where all strategies are Pareto optimal is called a conflict game).[2]
Zero–sum games are a specific example of constant sum games where the sum of each outcome is always zero. Such games are distributive, not integrative; the pie cannot be enlarged by good negotiation.
Situations where participants can all gain or suffer together are referred to as non-zero–sum. Thus, a country with an excess of bananas trading with another country for their excess of apples, where both benefit from the transaction, is in a non-zero–sum situation. Other non-zero–sum games are games in which the sum of gains and losses by the players are sometimes more or less than what they began with.
## Solution
For 2-player finite zero-sum games, the different game theoretic solution concepts of Nash equilibrium, minimax, and maximin all give the same solution. In the solution, players play a mixed strategy.
### Example
| A | B | C |
|---------|---------|---------|
| | | |
| 30, -30 | -10, 10 | 20, -20 |
| 10, -10 | 20, -20 | -20, 20 |
A game's payoff matrix is a convenient representation. Consider for example the two-player zero–sum game pictured at right.
The order of play proceeds as follows: The first player (red) chooses in secret one of the two actions 1 or 2; the second player (blue), unaware of the first player's choice, chooses in secret one of the three actions A, B or C. Then, the choices are revealed and each player's points total is affected according to the payoff for those choices.
Example: Red chooses action 2 and Blue chooses action B. When the payoff is allocated, Red gains 20 points and Blue loses 20 points.
Now, in this example game both players know the payoff matrix and attempt to maximize the number of their points. What should they do?
Red could reason as follows: "With action 2, I could lose up to 20 points and can win only 20, while with action 1 I can lose only 10 but can win up to 30, so action 1 looks a lot better." With similar reasoning, Blue would choose action C. If both players take these actions, Red will win 20 points. But what happens if Blue anticipates Red's reasoning and choice of action 1, and goes for action B, so as to win 10 points? Or if Red in turn anticipates this devious trick and goes for action 2, so as to win 20 points after all?
Émile Borel and John von Neumann had the fundamental and surprising insight that probability provides a way out of this conundrum. Instead of deciding on a definite action to take, the two players assign probabilities to their respective actions, and then use a random device which, according to these probabilities, chooses an action for them. Each player computes the probabilities so as to minimize the maximum expected point-loss independent of the opponent's strategy. This leads to a linear programming problem with the optimal strategies for each player. This minimax method can compute provably optimal strategies for all two-player zero–sum games.
For the example given above, it turns out that Red should choose action 1 with probability 4/7 and action 2 with probability 3/7, while Blue should assign the probabilities 0, 4/7, and 3/7 to the three actions A, B, and C. Red will then win 20/7 points on average per game.
### Solving
The Nash equilibrium for a two-player, zero–sum game can be found by solving a linear programming problem. Suppose a zero–sum game has a payoff matrix $M$ where element $M_{i,j}$ is the payoff obtained when the minimizing player chooses pure strategy $i$ and the maximizing player chooses pure strategy $j$ (i.e. the player trying to minimize the payoff chooses the row and the player trying to maximize the payoff chooses the column). Assume every element of $M$ is positive. The game will have at least one Nash equilibrium. The Nash equilibrium can be found (see ref. [2], page 740) by solving the following linear program to find a vector $u$:
Minimize:
$\sum_{i} u_i$
Subject to the constraints:
$u$ ≥ 0
$M u$ ≥ 1.
The first constraint says each element of the $u$ vector must be nonnegative, and the second constraint says each element of the $M u$ vector must be at least 1. For the resulting $u$ vector, the inverse of the sum of its elements is the value of the game. Multiplying $u$ by that value gives a probability vector, giving the probability that the maximizing player will choose each of the possible pure strategies.
If the game matrix does not have all positive elements, simply add a constant to every element that is large enough to make them all positive. That will increase the value of the game by that constant, and will have no effect on the equilibrium mixed strategies for the equilibrium.
The equilibrium mixed strategy for the minimizing player can be found by solving the dual of the given linear program. Or, it can be found by using the above procedure to solve a modified payoff matrix which is the transpose and negation of $M$ (adding a constant so it's positive), then solving the resulting game.
If all the solutions to the linear program are found, they will constitute all the Nash equilibria for the game. Conversely, any linear program can be converted into a two-player, zero-sum game by using a change of variables that puts it in the form of the above equations. So such games are equivalent to linear programs, in general.[citation needed]
## Non-zero–sum
### Economics
Many economic situations are not zero-sum, since valuable goods and services can be created, destroyed, or badly allocated in a number of ways, and any of these will create a net gain or loss of utility to numerous stakeholders. Specifically, all trade is by definition positive sum, because when two parties agree to an exchange each party must consider the goods it is receiving to be more valuable than the goods it is delivering. In fact, all economic exchanges must benefit both parties to the point that each party can overcome its transaction costs, or the transaction would simply not take place.
There is some semantic confusion in addressing exchanges under coercion. If we assume that "Trade X", in which Adam trades Good A to Brian for Good B, does not benefit Adam sufficiently, Adam will ignore Trade X (and trade his Good A for something else in a different positive-sum transaction, or keep it). However, if Brian uses force to ensure that Adam will exchange Good A for Good B, then this says nothing about the original Trade X. Trade X was not, and still is not, positive-sum (in fact, this non-occurring transaction may be zero-sum, if Brian's net gain of utility coincidentally offsets Adam's net loss of utility). What has in fact happened is that a new trade has been proposed, "Trade Y", where Adam exchanges Good A for two things: Good B and escaping the punishment imposed by Brian for refusing the trade. Trade Y is positive-sum, because if Adam wanted to refuse the trade, he theoretically has that option (although it is likely now a much worse option), but he has determined that his position is better served in at least temporarily putting up with the coercion. Under coercion, the coerced party is still doing the best they can under their unfortunate circumstances, and any exchanges they make are positive-sum.
There is additional confusion under asymmetric information. Although many economic theories assume perfect information, economic participants with imperfect or even no information can always avoid making trades that they feel are not in their best interest. Considering transaction costs, then, no zero-sum exchange would ever take place, although asymmetric information can reduce the number of positive-sum exchanges, as occurs in The Market for Lemons.
See also:
### Psychology
The most common or simple example from the subfield of Social Psychology is the concept of "Social Traps". In some cases we can enhance our collective well-being by pursuing our personal interests — or parties can pursue mutually destructive behavior as they choose their own ends.
### Complexity
It has been theorized by Robert Wright in his book Nonzero: The Logic of Human Destiny, that society becomes increasingly non-zero-sum as it becomes more complex, specialized, and interdependent. As former US President Bill Clinton states:
The more complex societies get and the more complex the networks of interdependence within and beyond community and national borders get, the more people are forced in their own interests to find non-zero-sum solutions. That is, win–win solutions instead of win–lose solutions.... Because we find as our interdependence increases that, on the whole, we do better when other people do better as well — so we have to find ways that we can all win, we have to accommodate each other....
—Bill Clinton, Wired interview, December 2000.[3]
## Extensions
In 1944 John von Neumann and Oskar Morgenstern proved that any zero-sum game involving n players is in fact a generalized form of a zero-sum game for two players, and that any non-zero-sum game for n players can be reduced to a zero-sum game for n + 1 players; the (n + 1) player representing the global profit or loss.[4]
## Misunderstandings
Zero–sum games and particularly their solutions are commonly misunderstood by critics of game theory, usually with respect to the independence and rationality of the players, as well as to the interpretation of utility functions. Furthermore, the word "game" does not imply the model is valid only for recreational games.[1]
## References
1. ^ a b Ken Binmore (2007). Playing for real: a text on game theory. Oxford University Press US. ISBN 978-0-19-530057-4. , chapters 1 & 7
2. Bowles, Samuel (2004). Microeconomics: Behavior, Institutions, and Evolution. Princeton University Press. pp. 33–36. ISBN 0-691-09163-3.
3. "Wired 8.12: Bill Clinton". Wired.com. 2009-01-04. Retrieved 2010-06-17.
4. "Theory of Games and Economic Behavior". Princeton University Press (1953). (Digital publication date)2005-06-25. Retrieved 2010-11-11.
## Further reading
• Created by Tony Kornheiser and Michael Wilbon. Performance by William Simmons (2010-09-23). "Misstating the Concept of Zero-Sum Games within the Context of Professional Sports Trading Strategies". Pardon the Interruption. ESPN.
• Raghavan, T. E. S. (1994). "Zero-sum two-person games". In Aumann; Hart. Handbook of Game Theory 2. Amsterdam: Elsevier. pp. 735–759. ISBN 0-444-89427-6.
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http://mathoverflow.net/questions/15486/how-strict-can-i-be-in-the-definition-of-2-group
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## How strict can I be in the definition of “2-group”?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Recall that a group is an associative, unital monoid $G$ such that the map $(p_1,m) : G \times G \to G\times G$ is an isomorphism of sets. Here $p_1$ is the first projection and $m$ is the multiplication, so the map is $(g_1,g_2) \mapsto (g_1,g_1g_2)$.
My question is a basic one concerning the definition of "2-group".
Recall that a monoidal category is a category $\mathcal G$ along with a functors $m : \mathcal G \times \mathcal G \to \mathcal G$ and $e: 1 \to \mathcal G$, where $1$ is the category with one object and only identity morphisms, such that certain diagrams commute up to natural isomorphism and those natural isomorphisms satisfy some axioms of their own (the natural isomorphisms are part of the data of the monoidal category).
Then a 2-group is a monoidal category $\mathcal G$ such that the functor $(p_1,m): \mathcal G \times \mathcal G \to \mathcal G \times \mathcal G$ is an equivalence of categories. I.e. there exists a functor $b: \mathcal G \times \mathcal G \to \mathcal G \times \mathcal G$ such that $b\circ (p_1,m)$ and $(p_1,m) \circ b$ are naturally isomorphic to the identity. Note that $b$ is determined only up to natural isomorphism of functors.
Question: Can I necessarily find such a functor $b$ of the form $b = (p_1,d)$, where $d : \mathcal G \times \mathcal G \to \mathcal G$ is some functor (called $d$ for "division")? If so, can I necessarily find $d = m\circ(i \times \text{id})$, where $i: \mathcal G \to \mathcal G$ is some functor (called $i$ for "inverse")?
In any case, the natural follow-up question is to ask all these at the level of 3-groups, etc.
-
## 2 Answers
You can always do this. Take any $b$ and define $d = p_2 b$. Then $b' = (p_1, d)$ is equivalent to the original $b$. To see this note that
$$(p_1, m) \circ b = (p_1b, m \circ (p_1 b, d)) \simeq id = (p_1, p_2)$$
The first component shows $p_1 b \simeq p_1$. We use this transformation $\times id$ to show that $b \simeq b' = (p_1, d)$.
Now we consider the equivalence $(p_1, m) \circ b' \simeq id$. Here we have,
$$(p_1, m) \circ (p_1, d) = (p_1, m \circ (p_1, d)) \simeq id = (p_1, p_2)$$
restricting to $G = G \times { 1} \subseteq G \times G$, this gives a natural isomorphism $m(x, d(x, 1)) \simeq 1$. You can take $i(x) = d(x,1)$, and we have $x i(x) \cong 1$.
We also have
$$(p_1, d) \circ (p_1, m) = (p_1, d \circ (p_1, m)) \simeq (p_1, p_2)$$
which gives a natural isomorphism, $d(x, xy) \simeq y$ (writing $m(x,y) = xy$). Thus we have,
$$d(x,y) \simeq d(x,1 y) \simeq d(x, x i(x) y) \simeq i(x) y,$$
which is the formula you were after. So we can replace $d(x,y)$ with $m(i(x), y)$ to get a third inverse functor b''. Note that this doesn't mean that we have a strict 2-group, just that we can define the inverse functors and difference functors you asked about.
Notice also that we didn't really use anything about G being a 1-category as opposed to an n-category (except the associator and unitors) so this argument generalizes to the n-group setting basically verbatim.
-
As an additional remark, it is also possible to choose isomorphisms $i(x) \cdot x \cong 1$ in such a way as to be compatible with the isomorphism $x \cdot i(x) \cong 1$. This takes a little more work, but the argument is explained in the paper by Baez and Lauda on 2-groups. – Chris Schommer-Pries Feb 16 2010 at 21:28
The Baez-Lauda paper is available here: arxiv.org/abs/math/0307200 – Chris Schommer-Pries Feb 17 2010 at 3:47
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
I think the answer is no, but cannot give a counterexample off hand. What is true is that there will be an equivalent monoidal category such that the corresponding functor $b$, does have the property. Any 2-group is equivalent to a strict 2-group (which can be assumed to come from a crossed module). What would be an interesting subsidiary question would be can the structure be `deformed' to one which is strict. (Deformed in the sense of a deformation theory.)
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http://sbseminar.wordpress.com/2007/10/16/freedman-on-distinguishing-manifolds-with-quantum-topology/
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## Freedman on distinguishing manifolds with quantum topology October 16, 2007
Posted by Noah Snyder in Category Theory, low-dimensional topology, quantum algebra, talks, tqft, Uncategorized.
trackback
This week Mike Freedman was in Berkeley for the annual series of three Bowen lectures. The first two were about topological quantum computation and the fractional quantum Hall effect. Since I missed one of those because of closing arguments in the case I was on the jury for, I’ll instead only discuss his third talk which dove-tails nicely with my last post.
His motivating question was why do we look for topological quantum computation in 3 space-time dimensions (that is, by restricting to a system stuck in a plane) rather than the full 4 space-time dimensions? His explanation was the following: In 3-dimensions it seems that unitary TQFT can distinguish all manifolds, however in 4-dimensions they can not.
Let’s begin by recalling the definition of a unitary TQFT. This is a unitary monoidal functor from the category of formal d+1-dimensional bordisms to the category of finite dimensional Hilbert spaces. Let’s unpack this definition a bit.
What is the category of formal d+1-dimensional bordisms? Its objects are closed d-manifolds. Given two d-manifolds A and B, the morphisms $\mathrm{Hom}(A,B)$ are linear combinations of d+1-manifolds $X_i$ where the boundary of $X_i$ is $A \cup \bar{B}$ (where $\bar{B}$ denotes the same manifold with the opposite orientation). Extending bar by complex conjugation on the scalars gives a bar structure on the category. The tensor product is just disjoint union extended linearly. Saying that the functor is unitary says that it sends bar to bar. Saying that it is monoidal means that it sends disjoint union to tensor product.
We want to know if there are any formal bordisms which are killed by all unitary TQFT. How might we find such a formal cobordism? Well, the bar structure on formal bordisms gives the space $\mathrm{Hom}(\emptyset,B)$ a Hermitian form. Under any unitary TQFT this Hermition form is mapped to a positive definite inner product. Hence, if $\mathrm{Hom}(\emptyset,B)$ has any isotropic (aka light-like) vectors, they must automatically be killed by any unitary TQFT.
(The converse is not automatically true, that is just because the Hermitian form is positive definite, that doesn’t necessarily imply that its whole structure can be captured by unitary TQFTs.)
What is published so far is that this Hermitian form on formal bordisms is positive definite for d+1 = 0, 1, 2, and has isotropic vectors for d+1 = 4, 5, etc. The d+1 = 3 case is going to be dealt with by Freedman and some of his coauthors in a work in progress that makes heavy use of large amounts of 3-manifold theory including details of Perelman’s work.
In 4-dimensions the construction of isotropic vectors is originally due to Freedman and his coauthors, but there is a nifty argument due to our own Peter Teichner. I’ll sketch that very rapidly. Using some intersection form theory one notes that the theory of even unimodular lattices can be embedded inside of manifold theory. That is, there are manifolds whose intersection forms exhibit an arbitrary unimodular lattice, and that if they have the same lattice then they’re the same manifold. However, the theory of indefinite lattices is easy: all you care about is parity and rank. So if you take the difference of the bordisms (just cut a ball out of the manifold) corresponding to two different definite unimodular lattices, you can easily check that its inner product with itself is (2-2)=0 times a single indefinite lattice.
In low dimensions the idea of the proof is to show “diagonal dominance.” That is you find a way of ranking how complicated manifolds are, and you show that if you glue a bunch of bordisms to each other pairwise that the most complicated ones must appear on the diagonal. On Scott’s request I’ll make this a bit more precise. Let C be our complexity function (it should take values in some poset), we want that for any pair of bordisms M and N that either $C(M \cup \bar{N}) < C(M \cup \bar{M})$ or $C(M \cup \bar{N}) < C(N \cup \bar{N})$.
Now, if you have $\langle\sum_i a_i M_i, \sum_j b_j N_j \rangle$ you see that the glued manifold $M_i \bar{N_j}$ has its largest complexity somewhere on the diagonal. Since the diagonal terms $a_i \bar{a_i} M_i \bar{N_j}$ all have positive coefficients, this means that the most complex term in the inner product cannot cancel. Hence the product is nonzero.
For example, in d+1=1 dimension the bordisms are just arcs and the complicatedness function is just number of connected components. When you glue two of these diagrams together, the number of components of the glued thing is smaller, and it’s strictly smaller unless you’re gluing to your mirror image. (This is a fun exercise, let me know in the comments if I should give you more details).
Freedman also left us all with a fascinating open question (which I think Vaughan asked about during questions). Is this inner product non-degenerate for any dimension? Or could you have a radical?
## Comments»
1. Scott Morrison - October 16, 2007
Perhaps it’s worth saying that the forthcoming result (Freedman, Walker, D. Calegari) is “positive” in d+1=3, that is, there are no isotropic vectors.
This leaves open the possibility that unitary TQFTs detect all 3-manifolds, although this is far from proven.
Their proof actually makes some use of unitary TQFTs along the way. Part of the complexity function has to “control” the fundamental groups of the manifolds. They take advantadge of the fact that $\pi_1(M^3)$ is residually finite (everything turns up in some finite group quotient), and consider the collection of all finite group TQFTs.
Perhaps someone will want to look at the proof, and see if more of the complexity function can actually be written in terms of other known TQFTs. This might be a first step towards proving the big result: that every (formal linear combination of) 3-manifold with boundary is detected by some unitary TQFT.
2. Greg Kuperberg - October 16, 2007
This leaves open the possibility that unitary TQFTs detect all 3-manifolds, although this is far from proven.
You could call it far from proven, but I would say that it is closer to proven than it is a complete mystery. Geometrization is true, so we know that fundamental groups of 3-manifolds are residually finite. This means that even the old-school invariants from finite groups, which are unitary TQFTs, carry a great deal of information.
3. Danny Calegari - October 27, 2007
The finite group TQFT’s are used at an early (but crucial) stage
to deal with compressing disks, which give rise to essential
spheres in the closed manifolds, and contribute to the sphere
decomposition. This is a very “crude” aspect of 3-manifold
topology, and it was therefore surprising to us that we needed
a (relatively) sophisticated tool to take care of it.
Other terms in the complexity function include -vol for the
hyperbolic pieces (in the geometric decomposition). The proof
that this term is diagonal dominant uses a significant amount of
analysis; I think it is plausible that this term could be recovered
by TQFT information (e.g. after Kashaev’s conjecture) but I would
guess that any proof that it satisfies the desired inequality would
be dramatically different from our proof.
4. TQFTs via Planar Algebras « Secret Blogging Seminar - December 6, 2008
[...] of the 3-manifold. Thus for any 3D TQFT we get 3-manifold invariants. In fact, as we discussed before, it might be that these invariants distinguish all 3-manifolds. Exciting [...]
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http://mathhelpforum.com/algebra/58898-algebric-inequality.html
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# Thread:
1. ## An algebric inequality
For $x,y,z>0$, prove
$x^{5}+y^{5}+z^{5}\leq x^{5}\sqrt{\frac{x^{2}}{yz}}+y^{5}\sqrt{\frac{y^{2 }}{xz}}+z^{5}\sqrt{\frac{z^{2}}{xy}}.$
thanks.
2. Originally Posted by bkarpuz
For $x,y,z>0$, prove
$x^{5}+y^{5}+z^{5}\geq x^{5}\sqrt{\frac{x^{2}}{yz}}+y^{5}\sqrt{\frac{y^{2 }}{xz}}+z^{5}\sqrt{\frac{z^{2}}{xy}}.$
thanks.
the inequality in this form is not true for all positive reals x, y, z, because it fails if you put z = 1, x = y, and then let x go to infinity! are you sure it's not $\leq$ instead of $\geq$ ?
3. Originally Posted by NonCommAlg
the inequality in this form is not true for all positive reals x, y, z, because it fails if you put z = 1, x = y, and then let x go to infinity! are you sure it's not $\leq$ instead of $\geq$ ?
I correct it now.
4. Originally Posted by bkarpuz
For $x,y,z>0$, prove
$x^{5}+y^{5}+z^{5}\leq x^{5}\sqrt{\frac{x^{2}}{yz}}+y^{5}\sqrt{\frac{y^{2 }}{xz}}+z^{5}\sqrt{\frac{z^{2}}{xy}}.$
thanks.
because of symmetry, we may assume that $x \leq y \leq z.$ then obviously: $\sqrt{\frac{x^{2}}{yz}} \leq \sqrt{\frac{y^{2}}{xz}} \leq \sqrt{\frac{z^{2}}{xy}}.$ hence by Chebyshev's inequality we'll have:
$x^{5}\sqrt{\frac{x^{2}}{yz}}+y^{5}\sqrt{\frac{y^{2 }}{xz}}+z^{5}\sqrt{\frac{z^{2}}{xy}} \geq \frac{1}{3}(x^5 + y^5 + z^5) \left(\sqrt{\frac{x^{2}}{yz}}+\sqrt{\frac{y^{2}}{x z}}+\sqrt{\frac{z^{2}}{xy}} \right).$ but by AM-GM we have: $\sqrt{\frac{x^{2}}{yz}}+\sqrt{\frac{y^{2}}{xz}}+\s qrt{\frac{z^{2}}{xy}} \geq 3. \ \ \ \ \ \Box$
imagine how many inequalities you can create using only Chebyshev's inequality + AM-GM inequality! i guess i'm back to high school again!
5. Originally Posted by NonCommAlg
because of symmetry, we may assume that $x \leq y \leq z.$ then obviously: $\sqrt{\frac{x^{2}}{yz}} \leq \sqrt{\frac{y^{2}}{xz}} \leq \sqrt{\frac{z^{2}}{xy}}.$ hence by Chebyshev's inequality we'll have:
$x^{5}\sqrt{\frac{x^{2}}{yz}}+y^{5}\sqrt{\frac{y^{2 }}{xz}}+z^{5}\sqrt{\frac{z^{2}}{xy}} \geq \frac{1}{3}(x^5 + y^5 + z^5) \left(\sqrt{\frac{x^{2}}{yz}}+\sqrt{\frac{y^{2}}{x z}}+\sqrt{\frac{z^{2}}{xy}} \right).$ but by AM-GM we have: $\sqrt{\frac{x^{2}}{yz}}+\sqrt{\frac{y^{2}}{xz}}+\s qrt{\frac{z^{2}}{xy}} \geq 3. \ \ \ \ \ \Box$
imagine how many inequalities you can create using only Chebyshev's inequality + AM-GM inequality! i guess i'm back to high school again!
I was thinking of a different solution.
I know that this inequality can be shown by special inequalities, the first one dropped in my mind was Muirhead's inequality (again because of the symmetry), which you may find in Muirhead's inequality - Wikipedia, the free encyclopedia.
If there is any other solutions, I would like to see.
Thanks NonCommAlg.
6. Originally Posted by bkarpuz
I was thinking of a different solution.
I know that this inequality can be shown by special inequalities, the first one dropped in my mind was Muirhead's inequality (again because of the symmetry), which you may find in Muirhead's inequality - Wikipedia, the free encyclopedia.
If there is any other solutions, I would like to see.
Thanks NonCommAlg.
when i was a high school kid, we were trained to avoid using Muirhead's inequality as much as we can! i never knew why? i guess i'm still under the infulence! lol
anyway, yes, Muirhead's inequality will also solve the problem because if you multiply both sides of the inequality by $xyz$ you'll get: $\sum_{sym}x^6yz \leq \sum_{sym}x^7 \sqrt{y}\sqrt{z},$
which is true because $7, \frac{1}{2}, \frac{1}{2}$ majorizes $6, 1, 1.$
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http://crypto.stackexchange.com/questions/1844/how-to-attack-a-general-polyalphabetic-cipher?answertab=oldest
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# How to attack a general polyalphabetic cipher?
I am able to decrypt vigenere cipher text using the index of coincidence and chi test.
However out of interested how do you go about attacking ciphertext that was encrypted using a mix alphabet shifted 26 times?
Also what about a ciphertext that has been encrypted with 26 random alphabets? so each line in the tableau is random.
Googling seems to bring up the basic vigenere, either an example of a link to a resource would be good.
Thanks
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## 2 Answers
A general polyalphabetical cipher is just a combination of several general monoalphabetical ciphers, each applied on every $n$-th letter of the message.
So the first thing is to find out what $n$ is (i.e. the key length). For this we can use the index of coincidence just like for Vigenere.
Then we can split the message into $n$ parts (columns), and try to break each of it as an individual monoalphabetic cipher, starting with frequency analysis. (We'll have to correlate information from the individual columns to do bigram frequency analysis, too.)
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This is what i expected, however if you have n splits of the ciphertexts which were encrypted with a general monoalphabetic cipher, carrying standard frequency analysis on each section seems like a right pain as even when you got the correct key it would have to be tested with the original ciphertext. Is there no simpler solution? as if you had a key lengt of 17 say, that would be 17 random pieces of ciphertext to decrypt, how would you ever know you got the right answer? unless you got a majority of them right? – Lunar Feb 13 '12 at 18:19
@Lunar: If you've got enough ciphertext, you can mostly guess each column of the tableau just by looking at single letter frequencies. The rest will then just be fine tuning. – Ilmari Karonen Feb 17 '12 at 17:08
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Also what about a ciphertext that has been encrypted with 26 random alphabets? so each line > in the tableau is random.
If you mean that you use a repeating sequence of 26 random unique characters as the key, then you break it in much the same way as any other Vigenere cipher - n, in this case, is 26.
If you mean a random key at least as long as the message using 26 random characters, then you won't be able to differentiate successful and unsuccessful decryptions as the message is information-theoretically secure.
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http://physics.stackexchange.com/questions/27012/does-entropy-measure-extractable-work
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# Does entropy measure extractable work?
Entropy has two definitions, which come from two different branches of science: thermodynamics and information theory. Yet, they both are thought to agree. Is it true?
Entropy, as seen from information theory, measures our ignorance about a system. Entropy, as seen from thermodynamics, measures the amount of extractable work. One is epistemological, the other is ontological. Yet, they both can agree, if entropy really measures the amount of extractable work, given one's knowledge about the system.
But, is this true? Does Shannon's entropy computed on the probability distribution of a physical system express the amount of work we can obtain from it? How do increases in our knowledge increase the amount of extractable work? Is Landauer principle powerful enough to establish the connection?
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I find the dramatically different answers to this question a bit baffling. I also thought the two kinds of entropy were different, but I liked Joe's answer and upvoted it. Would it be possible for someone to reconcile the different answers, or point out why one (or more) of them is wrong? – Aaron Sterling Oct 5 '11 at 18:55
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One reconciliation is that "work" has multiple meanings. Typically the first meaning learned is thermodynamic, but (as Dirac demonstrated) a more general meaning amounts to: Given a class of dynamical processes, 'work' is any mathematically natural measure of what those processes accomplish – John Sidles Oct 5 '11 at 21:21
You probably want to say that the thermodynamic entropy tells you how much of the energy is not available for work! I'll also go with Matt's answer than this only makes sense if the state is in thermal equilibrium, whereas Shannon entropy is more general. – Earl Oct 5 '11 at 21:25
@Aaron: Actually, the reconciliation is that I was answering yes to whether or not the two entropy definitions agree. I didn't mean that entropy measures extractable work. – Joe Fitzsimons Oct 6 '11 at 4:08
## 3 Answers
UPDATE: Below I am answering yes to the first question in the post (are the two kinds of entropy the same up to a constant). This has led to some confusion as both Matt and John gave answers saying "the answer is no", however I believe they are referring to the title "Does entropy measure extractable work?". Although the author uses the two interchangeably, the question itself contains a false premise: namely that physical entropy is a measure of extractable work ("Entropy, as seen from thermodynamics, measures the amount of extractable work"). This is simply not true in the general case, though can be true for certain special cases, as Matt's counter example shows. A concrete case of this is a ball placed anywhere on a level surface. If the ball is placed randomly, no more work can be extracted than if its location is known.
The answer is yes, the two kinds of entropy are the same up to a constant factor of $k_B \log 2$ (which is also the origin of Landauer's principle). On way to see this is via Maxwell's demon. In particular, via the Szilard engine, an idealised heat engine that uses a single particle gas. You then introduce a partition, which effectively partitions the gas into two regions, only one of which contains the particle. Now, if you knew which side of the partition the particle was on, you could use the pressure difference to extract work, and if you don't you can't since you don't know which way it will push.
Now the connection with information theory comes in when we measure which side of the partition the particle is on. From this we are gaining a certain amount of entropy (and hence information) in the register which holds our measurement result. But having this information decreases the entropy of the gas. And hence you can go back and forth between information entropy and physical entropy.
There is a fairly extensive literature on the matter, so instead of trying to give you a list, I'll point you to a review article on Maxwell's Demon and information theory from a few years ago: arXiv:0707.3400,
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Yes, I see. But still with the Szilard engine example, I'd like to go further. If I know in which side the particle si, I can extract kT ln(2) of work. If I don't know where the particle is, I can get no work. But, what if I have partial knowledge? Is still Shannon's entropy the answer? – Javier Rodriguez Laguna Sep 16 '11 at 14:14
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@Javier: Yes, you can use exactly the same trick, but use a measurement that returns 0 if the particle is in the left most compartment, and 1 with probability $p$ if the particle is in the rightmost compartment and 0 otherwise, and proceed from there. Have a look at the paper, there's lots of good stuff in it. – Joe Fitzsimons Sep 16 '11 at 14:30
More recent results here arxiv.org/abs/0908.0424 and here arxiv.org/abs/1009.1630 – Marco Sep 29 '11 at 7:20
The answer is no. Consider a system that has a degenerate ground state, such that the density matrix is a mixture of two ground state eigenstates. This has nonzero Shannon entropy, but you can't extract any work from it. More generally, thermodynamic entropy is not really well-defined for nonequilibrium systems, but Shannon entropy is.
My own take is that thermodynamic entropy and Shannon entropy are two conceptually distinct things. They happen to coincide in a wide variety of circumstances, but not always. It is not even clear to me whether the cases in which they happen to coincide in classical and quantum theory are necessary coincidences. It may be possible to come up with a well-defined physical theory in which they never coincide, e.g. in the convex set famework for generalized probabilistic theories studied in the quantum foundations community.
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A neat counterexample Matt! I was thinking a similar thing when reading the question. – Earl Oct 5 '11 at 21:20
@Matt: Good answer, and you get +1 from me, but I do not understand the claim that thermodynamic entropy is not well defined for non-equilibrium systems. It's given by $-k_B \sum_i p_i\log p_i$ where $p_i$ is the probability of the $i^{th}$ microstate. This is obviously defined for non-equilibrium systems and equilibrium systems alike. – Joe Fitzsimons Oct 6 '11 at 4:27
Actually I guess the issue is clouded by the existence of alternative non-equivalent definitions of entropy. – Joe Fitzsimons Oct 6 '11 at 5:55
Because the word "work" has multiple meanings, the answer in general is "no".
Typically the first meaning taught to students is thermodynamic, but (as Dirac demonstrated) a generalized meaning (which includes thermodynamic work as a particular case) amounts to:
Given a class of dynamical processes, 'work' is any potential function that naturally describes what that class accomplishes.
Specifically, in the context of isotope separation, Dirac established that the work potential $\mathcal{V}_c$ that is naturally associated to an isotope concentration $c$ is given by
$\displaystyle\qquad \mathcal{V}_c(c) = (2 c' - 1) \log\left[\frac{c'}{1-c'}\right]$
or equivalently for spin polarization $p = 2 c - 1$ the Dirac value function $\mathcal{V}_p$ associated to separative transport of spin polarization is
$\displaystyle\qquad\mathcal{V}_p(p') = \mathcal{V}_c(c')\big|_{c' = (1+p')/2} = p' \log\left[\frac{1+p'}{1-p'}\right]$
The key point is Dirac's value function is not proportional to an entropy difference (as is evident because $0\le \mathcal{V}_p(p') \lt \infty$ while per-mole entropy ranges over a finite range).
Moreover, the Dirac work associated to separation cannot be reversed to return mechanical work, since the separation process is entropically irreversible. Nonetheless, Dirac work has substantial economic value, and in fact defines the unit of value of a global market in separative work units (SWUs, pronounced "swooz").
For a derivation of the Dirac work function, see Dirac's own (unpublished) technical note "Theory of the separation of isotopes by statistical methods" (circa 1940), which appears in his Collected Works, or alternatively Donald Olander's survey article "Technical Basis of the Gas Centrifuge" (1972), or in general any textbook on isotope separation.
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http://math.stackexchange.com/questions/285520/where-exactly-are-complex-numbers-used-in-the-real-world/285534
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# “Where” exactly are complex numbers used “in the real world”?
I've always enjoyed solving problems in the complex world during my undergrad. However, I've always wondered where are they used and for what? In my domain (computer science) I've rarely seen it be used/applied and hence am curious.
So what practical applications of complex numbers exist and what are the ways in which complex transformation helps address the problem that wasn't immediately addressable?
Way back in undergrad when I asked my professor this he mentioned that "the folks in mechanical and aerospace engineering use it a lot" but for what? (Don't other domains use it too?). I'm well aware of its use in Fourier analysis but that's the farthest I got to a 'real world application'. I'm sure that's not it.
PS: I'm not looking for the ability to make one problem easier to solve, but a bigger picture where the result of the complex analysis is used for something meaningful in the real world. A naive analogy is deciding the height of tower based on trigonometry. That's going from paper to the real world. Similarly, what is it that is analyzed in the complex world and the result is used in the real world without imaginaries clouding the problem?
The question: Interesting results easily achieved using complex numbers is nice but covers a more mathematical perspective on interim results that make solving a problem easier. It covers different ground IMHO.
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I am pretty sure this has been asked... – Mariano Suárez-Alvarez♦ Jan 24 at 2:45
– Trevor Wilson Jan 24 at 2:48
– Rahul Narain Jan 24 at 2:51
1
– Rahul Narain Jan 24 at 2:52
## 3 Answers
Complex numbers are used in electrical engineering all the time, because Fourier transforms are used in understanding oscillations that occur both in alternating current and in signals modulated by electromagnetic waves.
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An example would go a long way for those who don't know this :) – PhD Jan 24 at 2:42
Complex numbers in a sense embody certain aspects of trigonometry. Therefore it is not unexpected for them to arise in situations involving trigonometric functions, such as waves and oscillations mentioned by Michael Hardy. A concrete example of their use is in phasors for example. – EuYu Jan 24 at 2:46
"one begin the current" . . . I presume "one being the current" was meant. – Michael Hardy Jan 24 at 2:52
The letter $j$ rather than $i$ is used in electrical engineering for the imaginary unit. The expression $t\mapsto e^{j\omega t}$ occurs frequently. The real and imaginary parts of that are $\cos(\omega t)$ and $\sin(\omega t)$. Frequency is $\omega$ and $t$ is time. – Michael Hardy Jan 24 at 2:55
Fourier transforms are also used in studying time series in statistics, including the stock market and biorhythms. – Michael Hardy Jan 24 at 2:56
show 1 more comment
Electrical engineering with signals.
For example
http://scipp.ucsc.edu/~johnson/phys160/ComplexNumbers.pdf
Regards
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+1 Excellent example! What would we do without $i$!? – amWhy May 7 at 0:15
@amWhy: I am actually very surprised how complex math has been used to prove some things, when there were thoughts that there was no connection. There is one famous theorem in this regard, bad sadly my memory fails me to as the name. Thx! – Amzoti May 7 at 0:31
Two-dimensional problems involving Laplace's equation (e.g. heat flow, fluid flow, electrostatics) are often solved using complex analysis, in particular conformal mapping.
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http://mathematica.stackexchange.com/questions/4330/using-contourplot3d-on-a-region-in-mathbbr3?answertab=oldest
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Using ContourPlot3D on a region in $\mathbb{R}^3$
I have a fundamental region in $\mathbb{R}^3$ defined solely by inequalities (i.e. the region is the intersection of 5 half-spaces and is a kind of square pyramid), and a function which is only well-defined inside that region.
I would like to plot that function's level surfaces inside the region using `ContourPlot3D`. However there is no easy way to stipulate the region simply in the form of `{a,a_min,a_max}`-type declarations as Mathematica seems to require. Does anyone know how to do this please?
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1 Answer
You can use the `RegionFunction` option with `ContourPlot3D` as follows:
```` ContourPlot3D[x^2 + y^2 + z^2, {x, -3, 3}, {y, -3, 3}, {z, -3, 3},
Contours -> 4,
RegionFunction -> Function[{x, y, z}, -4 < x + y < 4 && -4 < y + 2 z < 4 && -4 < x + z < 3],
ContourStyle -> {Red, Green, Yellow, Orange}, Mesh -> None]
````
which gives
Or, for a specific contour with the same region function, you can use
```` ContourPlot3D[x^2 + y^2 + z^2 == 10, {x, -3, 3}, {y, -3, 3}, {z, -3, 3},
RegionFunction -> Function[{x, y, z}, -4 < x + y < 4 && -4 < y + 2 z < 4 && -4 < x + z < 3],
ContourStyle -> {Red, Green, Yellow, Orange}, Mesh -> None]
````
to get
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Wow - thanks a lot! – gazza Apr 17 '12 at 11:14
@gazza, my pleasure. – kguler Apr 17 '12 at 11:22
lang-mma
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http://mathoverflow.net/questions/94503/domain-of-holomorphy
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## Domain of Holomorphy
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
How to show that $D={ |z_1|<1} \cup { |z_2|<1} \subset \mathbb{C}^2$ is not a domain of holomorphy. What is the smallest domain of holomorphy $S\supset D$ ?
I think we cannot just add the distinguished boundary (the torus $|z_1|=1, |z_2|=1$) to $D$ because it would violate analiticity. Should we use logarithmically convexity argument to extend our domain? Can anyone help?
Thank you!
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You can use the same arguments as for a Hartogs' figure to show that any holomophic function on $D$ extends to $\mathbb{C}^2$. Also beware that in general a smallest domain of holomorphy $S\subset \mathbb{C}^2$ containing $D$ need not exist. – Jérôme Poineau Apr 19 2012 at 9:14
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http://mathoverflow.net/questions/44165?sort=oldest
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## Reference request: lattice operations on the class of finitely presented groups
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
In my research, I work with certain finitely presented quotients of Coxeter groups. These are the automorphism groups of abstract polytopes, which are combinatorial generalizations of "usual" polytopes. (Essentially, an abstract polytope is an incidence complex.) Now, in this context, there is a useful combinatorial operation that has a nice effect on the automorphism groups. In fact, it's easy to generalize the operation on groups, so I'm curious whether any work has been done with this.
Let $G = \langle X \mid R \rangle$ and $H = \langle X \mid S \rangle$ be finitely presented groups. (I'm not sure that the finiteness of the presentation is essential, but let's assume it for now.) In other words, we have that $G = F(X) / \overline{R}$ and $H = F(X) / \overline{S}$, where $F(X)$ is the free group on $X$ and $\overline{R}$ is the normal closure of $R$ in $F(X)$. Then if $K$ naturally covers $G$ and $H$ (that is, if the identity map on $X$ extends to (surjective) homomorphisms from $K$ to $G$ and $K$ to $H$), we have that $K$ covers the group $F(X) / (\overline{R} \cap \overline{S})$. Similarly, if $G$ and $H$ naturally cover $K$, then the group $F(X) / (\overline{R} \overline{S})$ with presentation $\langle X \mid R \cup S \rangle$ naturally covers $K$ as well.
Therefore, the group $F(X) / (\overline{R} \cap \overline{S})$ is the minimal natural cover of $G$ and $H$, and $F(X) / (\overline{R} \overline{S})$ is the maximal natural quotient of $G$ and $H$. The first group is the fibre product of $G$ and $H$ over the second group.
These seem like such natural operations that I would guess they have been studied before, but I am having trouble finding anything. Any references would be greatly appreciated.
EDIT: Let me expand a little bit. As Mark Sapir points out, I'm essentially looking at the lattice of normal subgroups of F(X). I've looked a little at the general theory of subgroup lattices, but I'm really only interested in normal subgroups of F(X) or other finitely presented groups. Also, I find it difficult to work with the normal subgroups of F(X) directly, whereas it's not too hard to work with the quotients by these normal subgroups via presentations. So I'm hoping to find something that's somewhat more specific than just subgroup lattices; ideally, something that works with presentations.
Here are some examples of the type of questions I'm interested in:
1. Is there a simple way to write down a presentation for $F(X) / (\overline{R} \cap \overline{S})$ without changing the generators?
2. Given G and H, what are some conditions on the relations under which $\overline{R} \overline{S} = F(X)$? (This obviously won't be all-inclusive, but some instructive examples would be nice.)
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I have updated my answer. – Mark Sapir Nov 5 2010 at 11:38
## 1 Answer
What you are studying is the lattice of normal subgroups of the free group $F(X)$. The normal subgroup lattices of groups have been studied a lot. For example, this lattice is complete and modular. See the references here.
Update. About your newer, more concrete, questions. Even if $R$ and $S$ are finite, the normal subgroup $\bar R\cap \bar S$ may not be finitely generated. If the membership problem for $\bar R$ and $\bar S$ is decidable, then in principle one can decide the membership problem for the intersection, so you will find a generating set of the intersection - the whole intersection. But I am sure that the problem whether the intersection is finitely generated is undecidable (although I did not think about a proof).
The question whether $\bar R\bar S=F_k$ is the triviality problem which is undecidable (by a theorem of Adian-Rabin) even when $R$ is empty. For example, take any presentation of the trivial group, call a subset of it $R$ and the complement $S$. You get the equality $\bar R\bar S=F_k$. Another way: take a presentation of a finite group of exponent $n$, say, $S_3$ has exponent $6$ and presentation $\langle a,b \mid a^2=b^3=1, aba=b^{-1}\rangle$, and add relations that "kill" the generators, say $a^5=1, b^7=1$. The first set of relations is $R$, the second is $S$.
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Thanks for your updated answer. Regarding your last paragraph: I know there's no general algorithm to determine whether $\overline{R} \overline{S} = F_k$. What I'm interested in is finding simple conditions on R and S that are sufficient to guarantee that $\overline{R} \overline{S} = F_k$. But maybe that question is too broad to have a good answer. Mostly, I was just hoping to find a treatment of the normal subgroup lattice of F(X) from the point of view of presentations of the quotient groups. – Gabe Cunningham Nov 5 2010 at 14:55
@Gabe: Indeed, the question is too broad. I do not think one can say anything about the case $\bar R\bar S=F_k$ except something which would be clearly equivalent to the definition. If you have any specific information about $R$ or $S$, the answer could be more specific as well. For example, $R$ and $S$ may have some geometric meaning, then you can search for a geometric answer. – Mark Sapir Nov 5 2010 at 15:57
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http://mathoverflow.net/questions/68899?sort=oldest
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## How does one view the De Rham spectral sequence as a Grothendieck spectral sequence?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I was rereading basic results on de Rham cohomology, and this led me inevitably to the fact that $H^q(X,\Omega^p)$ converges to $H^*(X)$ for any smooth proper variety (over any field). How does one view this spectral sequence "maturely" as a Grothendieck spectral sequence?
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A naive comment: find the resolution, and this will give you the derived functor. – David Roberts Jun 27 2011 at 3:00
As far as I remember. the de Rham compleX is the hyper-resolution of the constant sheaf in the infinitesimal topology; see webcache.googleusercontent.com/… – Mikhail Bondarko Jun 27 2011 at 5:13
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In addition to what Mariano and Torsten have said, the Hodge-de Rham spectral sequence starts at $E_1$ and the Grothendieck spectral sequence starts at $E_2$. Hence it's unlikely that the former would be a special case of the latter. – Dan Petersen Jun 27 2011 at 6:18
Dan's point is excellent, I do not know of any nice interpretation of the $E_2$-term of the Hodge-de Rham spectral sequence which would come out of a composed functor interpretation (hence acting as an argument against such an interpretation). – Torsten Ekedahl Jun 27 2011 at 6:44
## 2 Answers
If by "Grothendieck spectral sequence" you mean the spectral sequence associated to the composite of functors (fulfilling the Grothendieck condition) then I am skeptical as to whether this is possible. Also I do not see that there would be any particular point in being able to view it in that light (unless the functors involved would be interesting in themselves). As you seem to be looking for some general principle that would give the dRss I would like to a mention one such which seem to give most spectral sequences in use very quickly. Usually one constructs the dRss by taking a Cartan-Eilenberg resolution of the de Rham complex and then consider the spectral sequence associated to a double complex. However, one can instead use the Massey exact couple construction; once one has an exact couple one automatically gets a spectral sequence. One systematic way of constructing such exact couples is to start with a triangulated category $T$, a sequence of morphisms $\cdots\to X_{i-1}\to X_i\to X_{i+1}\to\cdots$ and a homological functor $H$ on $T$. This gives a spectral sequence starting with $H^i(Y_j)$ and going towards $\varinjlim_iH^\ast(Y_i)$. (Convergence is not assured but is OK for instance of $Y_i$ is $0$ for small $i$ and equal if $i$ is large.) Starting with the naive truncations of de Rham complex gives the dRss (and starting with canonical truncations in the algebraic case gives the conjugate spectral sequence).
This setup works very generally, for instance it is how Adams first constructed the Adams spectral sequence.
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
You probably want one (the second one...) of the two hypercohomology spectral sequences which compute $\mathbb H^\bullet(X,\Omega^\bullet)$, the hypercohomology of the de Rham complex. A reference for this is Weibel's book.
I doubt you can view this as a Grothendieck spectral sequence, but it has sufficiently much hyper in it to be considered mature, I guess.
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3
In fact "hyper" is so mature that it seems to be no longer used... – Torsten Ekedahl Jun 27 2011 at 6:46
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Kids these days... What can you expect?! :) – Mariano Suárez-Alvarez Jun 27 2011 at 7:05
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http://en.wikipedia.org/wiki/Inductive_definition
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# Recursive definition
(Redirected from Inductive definition)
Four stages in the construction of a Koch snowflake. As with many other fractals, the stages are obtained via a recursive definition.
In mathematical logic and computer science, a recursive definition (or inductive definition) is used to define an object in terms of itself (Aczel 1978:740ff).
A recursive definition of a function defines values of the functions for some inputs in terms of the values of the same function for other inputs. For example, the factorial function n! is defined by the rules
0! = 1.
(n+1)! = (n+1)·n!.
This definition is valid for all n, because the recursion eventually reaches the base case of 0. The definition may also be thought of as giving a procedure describing how to construct the function n!, starting from n = 0 and proceeding onwards with n = 1, n = 2, n = 3 etc..
That such a definition indeed defines a function can be proved by induction.
An inductive definition of a set describes the elements in a set in terms of other elements in the set. For example, one definition of the set N of natural numbers is:
1. 1 is in N.
2. If an element n is in N then n+1 is in N.
3. N is the smallest set satisfying (1) and (2).
There are many sets that satisfy (1) and (2) - for example, the set {1, 1.649, 2, 2.649, 3, 3.649, ...} satisfies the definition. However, condition (3) specifies the set of natural numbers by removing the sets with extraneous members.
Properties of recursively defined functions and sets can often be proved by an induction principle that follows the recursive definition. For example, the definition of the natural numbers presented here directly implies the principle of mathematical induction for natural numbers: if a property holds of the natural number 0, and the property holds of n+1 whenever it holds of n, then the property holds of all natural numbers (Aczel 1978:742).
## Form of recursive definitions
Most recursive definition have three foundations: a base case (basis), an inductive clause, and an extremal clause.
The difference between a circular definition and a recursive definition is that a recursive definition must always have base cases, cases that satisfy the definition without being defined in terms of the definition itself, and all other cases comprising the definition must be "smaller" (closer to those base cases that terminate the recursion) in some sense. In contrast, a circular definition may have no base case, and define the value of a function in terms of that value itself, rather than on other values of the function. Such a situation would lead to an infinite regress.
## Proof that recursion defines a function
Suppose one has a pair of functions f and g on the natural numbers such that f(0)=0 and f(n+1)=g(f(n)) for all n. We want to construct, for any natural number n, a subset F(n) of N^2 such that:
(1) for all natural numbers a between 0 and n: there exists a unique natural number b such that (a, b) is in F(n);
(2) (0, f(0)) is in F(n);
(3) for all natural numbers a, where a is between 0 and n-1: if (a, f(a)) is in F then (a+1, f(f(a))) is in F(n).
Clearly (0, f(0)) satisfies (2) and (3).
Suppose these properties hold for n. Consider F(n)U{g(n)). Then this set satisfies all 3 properties (1), (2), (3). By induction we have such a set for any natural number n. Let F be the set theoretic union of all the F(n), where the union is taken over all natural numbers n.
## Examples of recursive definitions
### Elementary functions
Addition is defined recursively based on counting
$0+a=a$
$(1+n)+a=1+(n+a)$
Multiplication is defined recursively
$0 a=0$
$(1+n)a=a+na$
Exponentiation is defined recursively
$a^0=1$
$a^{1+n}=a a^n$
Binomial coefficients are defined recursively
$\binom{a}{0}=1$
$\binom{1+a}{1+n}=\frac{(1+a)\binom{a}{n}}{1+n}$
### Prime numbers
The set of prime numbers can be defined as the unique set of positive integers satisfying
• 1 is not a prime number
• any other positive integer is a prime number if and only if it is not divisible by any prime number smaller than itself
The primality of the integer 1 is the base case; checking the primality of any larger integer X by this definition requires knowing the primality of every integer between 1 and X, which is well defined by this definition. That last point can proved by induction on X, for which it is essential that the second clause says "if and only if"; if it had said just "if" the primality of for instance 4 would not be clear, and the further application of the second clause would be impossible.
### Non-negative even numbers
The even numbers can be defined as consisting of
• 0 is in the set E of non-negative evens (basis clause)
• For any element x in the set E, x+2 is in E (inductive clause)
• Nothing is in E unless it is obtained from the basis and inductive clauses (extremal clause).
### Well formed formulas
It is chiefly in logic or computer programming that recursive definitions are found. For example, a well formed formula (wff) can be defined as:
1. a symbol which stands for a proposition - like p means "Connor is a lawyer."
2. The negation symbol, followed by a wff - like Np means "It is not true that Connor is a lawyer."
3. Any of the four binary connectives (C, A, K, or E) followed by two wffs. The symbol K means "both are true", so Kpq may mean "Connor is a lawyer and Mary likes music."
The value of such a recursive definition is that it can be used to determine whether any particular string of symbols is "well formed".
• Kpq is well formed, because it's K followed by the atomic wffs p and q.
• NKpq is well formed, because it's N followed by Kpq, which is in turn a wff.
• KNpNq is K followed by Np and Nq; and Np is a wff, etc.
## References
• P. Aczel (1977), "An introduction to inductive definitions", Handbook of Mathematical Logic, J. Barwise (ed.), ISBN 0-444-86388-5
• James L. Hein (2009), Discrete Structures, Logic, and Computability. ISBN 0-7637-7206-2
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http://math.stackexchange.com/questions/43861/relationship-between-continuum-hypothesis-and-special-aleph-hypothesis-under-zf
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Relationship between Continuum Hypothesis and Special Aleph Hypothesis under ZF
Special Aleph Hypothesis AH(0) is the claim $2^{\aleph_0}=\aleph_1$, i.e. there is a bijection from $2^{\aleph_0}$ to $\aleph_1$.
Continuum Hypothesis CH is the claim $\aleph_0 \leq \mathfrak{a}< 2^{\aleph_0} \Rightarrow \mathfrak{a}=\aleph_0$, i.e. if there is an injection from $\aleph_0$ to $\mathfrak{a}$, an injection from $\mathfrak{a}$ to $2^{\aleph_0}$ and no injection from $2^{\aleph_0}$ to $\mathfrak{a}$, then there is a bijection from $\aleph_0$ to $\mathfrak{a}$.
AH(0) and CH are known to be equivalent under ZFC. What if we don't assume the Axiom of Choice? Under ZF, is it known that AH(0) $\nRightarrow$ CH or CH $\nRightarrow$ AH(0)?
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I would use something other than $\mathfrak b$, as it has a meaning in PCF theory (dominating number). It sure got me confused for a moment! – Asaf Karagila Jun 7 '11 at 13:38
Done. I thought $\mathfrak{c}$ would be an even worse choice, so $\mathfrak{a}$ it is. – LostInMath Jun 7 '11 at 14:55
2 Answers
January 25, 2012: I have found a mistake in one of the arguments, it is not an important one for the answer (in fact it is flat out irrelevant) but I should rewrite this answer anyway.
The short answer is that $AH(0)$ implies $CH$ under $ZF$, but is unprovable from $CH$ under $ZF$ alone.
$\boxed{ZF\vdash AH(0)\rightarrow CH}$:
Suppose $2^{\aleph_0} = \aleph_1$, suppose $\frak a$ is a set whose cardinality is between $\aleph_0$ and $2^{\aleph_0}=\aleph_1$. $\frak a$ can be well ordered (it can be injected into an ordinal) and therefore has the cardinality of some aleph number. If it is not $\aleph_1$, then it has to be $\aleph_0$.
So in $ZF$ you have that $AH(0)\rightarrow CH$.
From this we have that it is not consistent with $ZF$ that $CH\rightarrow\lnot AH(0)$. We also have that it is consistent relatively to $ZF$ that $AH(0)\land CH$ is true.
$\boxed{ZF\nvdash CH\rightarrow AH(0)}$:
We exhibit a model in which $CH$ is true, but $AH(0)$ is false. Note that this implies that $\aleph_1\nleq2^{\aleph_0}$. Such result can be achieved either when the continuum to be a countable union of countable sets or the presence of an inaccessible cardinal (which is a slight increase in consistency strength).
Surprisingly enough, in the Feferman-Levy model in which the continuum is a countable union of countable sets $CH$ fails. There exists a set of real numbers which is uncountable but there is no injection from $2^\omega$ into the set [1, Remark 3.4].
Consider now the Solovay model of $ZF$, we start with an inaccessible cardinal and end up with a model in which every subset of reals is Lebesgue measurable. It is also a model of the assertion "Every uncountable set of reals has a perfect set", where perfect sets always contain a copy of the Cantor set and therefore have cardinality continuum. In fact, Truss proved in [2] that repeating the Solovay construction from any limit cardinal results in a model where the perfect set property holds, and $AH(0)$ fails, so the inaccessible is redundant for this proof.
Every set of reals, in such model, is either countable or of cardinality continuum. In particular $CH$ holds, but $AH(0)$ not.
Therefore in a model of $ZF$ (without choice) exactly one of the options holds:
1. $CH\land AH(0)$,
2. $CH\land\lnot AH(0)$ (Solovay's model, and Truss models),
3. $\lnot CH\land\lnot AH(0)$ (Feferman-Levy model).
This shows that $CH$ cannot prove or disprove $AH(0)$. Note, by the way, that the first and third options can be found in models of choice, such as Godel's constructible universe and Cohen's construction where the continuum hypothesis fails; the latter can be even shown in models like Cohen's first model where there is a dense Dedekind-finite set of real numbers. However we see more here: the assertion $\aleph_1\leq2^{\aleph_0}$ is unprovable from $ZF$.
Bibliography:
1. Miller, A. A Dedekind Finite Borel Set. Arch. Math. Logic 50 (2011), no. 1-2, 1--17.
2. Truss, J. Models of set theory containing many perfect sets. Ann. Math. Logic 7 (1974), 197–219.
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2
Thank you for your answer. I'm aware of the fact ZFC does not prove the existence of an inaccessible, but still I can't see how Kanamori's Thm. 11.6 implies that ZF does not prove "CH implies AH(0)". Could you elaborate a bit on "This means that from ZF alone you cannot prove CH $\Longrightarrow$ AH(0)"? – LostInMath Jun 7 '11 at 15:00
@LostInMath: I have added a sketch of a proof, it was slightly harder that I thought, so it was very good of you to point that out. I hope that you understand the idea behind the proof in the addendum. – Asaf Karagila Jun 7 '11 at 16:13
Thank you for the clarification. But again, I have to ask because I could not figure it out by myself. Although this is probably obvious to everyone but me. This is how I understand your proof: The aim is to prove that CH $\Longrightarrow$ AH(0) is not a theorem of ZF. So suppose it is. Then the theory ZF+$\omega_1\nleq 2^{\aleph_0}$+"every subset of reals has the perfect set property" is inconsistent. Then Thm. 11.6 implies that the theory $ZFC+\exists$ inaccessible is inconsistent too. But from this we can't deduce that (ZFC and) ZF are inconsistent and reach a contradiction, can we? – LostInMath Jun 8 '11 at 13:04
No. What I proved is that $CH$ cannot prove from $ZF$ alone either $AH(0)$ or its negation. I.e. $AH(0)$ is independent from $ZF+CH$, and we cannot prove anything further without assuming more (e.g. assuming $AC$ they are in fact equivalent). Nothing in my addendum was pointing out an inconsistency. I will try to clarify things further a bit more. – Asaf Karagila Jun 8 '11 at 13:26
I think I understand it now. What confused me was that when demonstrating the consistency of $ZF+CH+\neg AH(0)$ you assumed the existence of a certain model. (In comparison, when demonstrating the consistency of, for example, $ZFC+\neg CH$, usually the model is constructed by extending an existing model of $ZFC$.) But now I see that the justification behind this assumption is that we have assumed the existence of an inaccessible cardinal. Is it possible to prove the consistency of $ZF+CH+\neg AH(0)$ without any extra assumptions? – LostInMath Jun 8 '11 at 14:35
show 1 more comment
It might also be worth mentioning also that it is known to be relatively consistent with $ZF+\neg AC$ that there are infinite Dedekind finite sets of reals, that is, a set $A\subset\mathbb{R}$ that is infinite, but which has no countable subset. Such a set is uncountable, with cardinality strictly less than the continuum, but is incomparable in cardinality with $\aleph_0$. In other words, it is relatively consistent with $ZF+\neg AC$ that there is a cardinality $\frak{a}$ with $\frak{a}\lt 2^{\aleph_0}$ and $\frak{a}$ is not finite (and even $\frak{a}$ is uncountable), yet $\aleph_0\not\leq\frak{a}$. In other words, just knowing a set $A$ is uncountable, one cannot conclude in ZF (unless inconsistent) that $\aleph_0\leq |A|$. Many would regard the existence of such cardinalities as even worse than the kind of counterexamples for which your question is asking.
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Joel: In the formulation of CH given in the question there might be other infinite cardinalities below the continuum, but they cannot be comparable with $\aleph_0$. Despite being "even worse than the kind of counterexamples" they still don't serve as counterexamples. – Asaf Karagila Jun 8 '11 at 7:31
Yes, I agree with that, and I intended this answer only as an interesting aside, rather than as a counterexample to the precise formulation of the question that was asked. Nevertheless, the formulation of CH that the OP gives may not be exactly what is desired in the $\neg AC$ context. For example, an alternative formulation would assert that every uncountable set of reals is bijective with $\mathbb{R}$, and the infinite Dedekind finite sets violate this. – JDH Jun 8 '11 at 10:11
Of course, I have some trouble in settling this in my mind prior to writing my answer on whether or not this is the "correct" way to write CH. I actually deleted my original answer (which turned into the comment on the original question) in which I suggested that CH will be formulated as "If $A$ is an infinite set whose cardinality is less than the continuum, then $A$ is countable", similarly to the formulation you gave here. – Asaf Karagila Jun 8 '11 at 10:39
Indeed, I guess there are a large variety of statements, and presumably many of the them are independent over ZF. It would be an interesting project to sort them all out into a big hierarchy. – JDH Jun 8 '11 at 11:01
I agree, it can be interesting. I'll put it on my crowded to-do list. – Asaf Karagila Jun 8 '11 at 12:36
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http://mathhelpforum.com/calculus/159682-complex-domains-paths.html
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# Thread:
1. ## Complex Domains and Paths
Find all possible values of
$\int_{\gamma } \frac{1}{1+z^2}dz$ where $\gamma$ is some path joining 0 to 1 which is in the domain of $f(z) = \frac{1}{1+z^2}$.
Any help would be great. Thanks!
2. Would the idea of independence of path help in this situation?
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http://mathhelpforum.com/advanced-algebra/203422-proving-sylow-p-subgroups-abelian.html
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1Thanks
• 1 Post By Deveno
# Thread:
1. ## proving the sylow p-subgroups are abelian
Assume that $p$ is an odd prime and $G$ is a finite simple group with exactly $2p+1$ Sylow $p$-subgroups. Prove that the Sylow $p$-subgroups are abelian.
I haven't made much progress: It's easy if $p^2$ doesn't divide the order of $G$, since then the Sylow $p$-subgroups have order $p$ and are therefore cyclic.
Thanks,
Hollywood
2. ## Re: proving the sylow p-subgroups are abelian
i figured it out:
let G act on the sylow p-subgroups by conjugation.
this gives a homomorphism of G into S2p+1.
since G is simple, and the action is transitive, the kernel of this action must be trivial.
therefore |G| divides (2p+1)!.
now we can ask, what is the highest possible power of p that divides (2p+1)! ?
well, the only integers 2 ≤ n ≤ 2p+1 that p divides are p and 2p. this means that the highest possible power of p that divides (2p+1)! is p2.
thus the sylow p-subgroups of G have either order p, or p2, and are abelian in either case.
Brilliant!!
Thanks,
Hollywood
4. ## Re: proving the sylow p-subgroups are abelian
Deveno - well done!
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http://mathhelpforum.com/advanced-algebra/169487-conjugacy-class.html
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# Thread:
1. ## conjugacy class
what is conjugacy class?
2. Two members of a group, x and y, are said to be "conjugate" if there exists a member of the group, z, such that $zxz^{-1}= y$. That, of course, is the same as saying that $zx= yz$. It is easy to prove that that is an equivalence relation (refexive: x is conjugate to itself by taking z equal to the group identity. symmetric: if x is conjugate to y, so that $zxz^{-1}= y$, then $x= z^{-1}yz= uyu^{-1}$ with $u= z^{-1}$. transitive: if x is conjugate to y and y is conjugate to z then $uxu^{-1}= y$ and $vyv^{-1}= z$. Then $v(uxu^{-1})v^{-1}= (vu)x(vu)^{-1}= z$ so x is conjugate to x).
A "conjugacy class" is the equivalence class using "conjugate" as the equivalence relation.
Notice that if the group is Abelian (commutative) then zx= xz so that if x and y are conjuate, zx= xz= yz and then x= y. That, is, in Abelian groups, a member of the group is conjugate only to itself and the conjugacy classes consist only of the single elements.
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http://en.wikipedia.org/wiki/Lebesgue's_density_theorem
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# Lebesgue's density theorem
In mathematics, Lebesgue's density theorem states that for any Lebesgue measurable set A, the "density" of A is 1 at almost every point in A. Intuitively, this means that the "edge" of A, the set of points in A whose "neighborhood" is partially in A and partially outside of A, is negligible.
Let μ be the Lebesgue measure on the Euclidean space Rn and A be a Lebesgue measurable subset of Rn. Define the approximate density of A in a ε-neighborhood of a point x in Rn as
$d_\varepsilon(x)=\frac{\mu(A\cap B_\varepsilon(x))}{\mu(B_\varepsilon(x))}$
where Bε denotes the closed ball of radius ε centered at x.
Lebesgue's density theorem asserts that for almost every point x of A the density
$d(x)=\lim_{\varepsilon\to 0} d_{\varepsilon}(x)$
exists and is equal to 1.
In other words, for every measurable set A, the density of A is 0 or 1 almost everywhere in Rn.[1] However, it is a curious fact that if μ(A) > 0 and μ(Rn \ A) > 0, then there are always points of Rn where the density is neither 0 nor 1.
For example, given a square in the plane, the density at every point inside the square is 1, on the edges is 1/2, and at the corners is 1/4. The set of points in the plane at which the density is neither 0 nor 1 is non-empty (the square boundary), but it is negligible.
The Lebesgue density theorem is a particular case of the Lebesgue differentiation theorem.
## References
1. Pertti, Mattila (1995). Geometry of Sets and Measures in Euclidean Spaces: Fractals and rectifiability, Corollary 2.14(1). Cambridge University Press, Cambridge. ISBN 0-521-65595-1.
• Hallard T. Croft. Three lattice-point problems of Steinhaus. Quart. J. Math. Oxford (2), 33:71-83, 1982.
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http://mathhelpforum.com/advanced-algebra/88635-deducing-matrix-minimum-polynomial.html
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# Thread:
1. ## Deducing matrix from minimum polynomial
Suppose V is a 3 dimensional vector space over $Z_2$ (the field of congruence classes mod 2) and that the characteristic polynomial of T is $(x-1)^3$. Then clearly the minimum polynomial is one of $(x-1)$, $(x-1)^2$, or $(x-1)^3$. I cannot see how we can use this to answer the final part of the question:
By considering the possible dimensions of eigenspaces, show that the matrix of T with respect to an arbitrary basis is exactly one of:
$\left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{ar ray}\right)$, $\left(\begin{array}{ccc}1&0&0\\0&1&1\\0&0&1\end{ar ray}\right)$, or $\left(\begin{array}{ccc}1&1&0\\0&1&1\\0&0&1\end{ar ray}\right)$
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http://mathhelpforum.com/advanced-algebra/74870-group-homomorphism.html
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# Thread:
1. ## group homomorphism
If $a:G\to H$ is a group homomorphism between G and H, then we have the following definition:
for all $g,k\in G$ we have $a(g*k)=a(g)*a(k)$
According to wikipedia, this means that $a(1_G)=1_H$, where $1_G,1_H$, are the identity elements of G and H respectively, and also that $a(g^{-1})=a(g)^{-1}$ for all $g\in G$.
Can someone show me how these conditions are true?
Thanks
2. Hi
One way is to write $a(1_G)=a(1_G1_G)=a(1_G)a(1_G)$ thus $a(1_G)=1_H.$
Now that we know that, $a(g^{-1})a(g)=a(g^{-1}g)=a(1_G)=1_H$ implies that $a(g^{-1})=(a(g))^{-1}.$
3. Thanks, but I don't understand how your first argument shows that $a(1_G)=1_H$
4. Ok, that's because a group law is strong enough: let $A$ be a group, $e$ its identity element and $a\in A,$ then $a^2=a\Rightarrow a=e$
Indeed, $a^2=a\Rightarrow a=a^{-1}a^2=a^{-1}a=e$
That may be a particular case of: $\forall a,b\in A,\ \exists !x\ \text{s.t.}\ ax=b$
Indeed if $b=a,$ since $e$ is a solution, by unicity, $aa=a\Rightarrow a=e$
5. Thanks, I got it now.
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http://physics.stackexchange.com/questions/51352/how-does-one-explain-this-pattern-generated-by-earthquake-waving-driving-a-pendu/51417
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How does one explain this pattern generated by earthquake waving driving a pendulum?
How to explain this pattern generated by earthquake wave driving a pendulum? Specially, there are three groups of curves that look categorically different:
1) The group of curves outside the olive-shaped envelope
2) The group of intertwined curves at the center
3) The group of curves inside the envelope but outside (2).
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2 Answers
This seems to be some new kind of seismogram to me. Looks like I am right. Robbed directly from wiki,
The principle can be shown by an early special purpose seismometer. This consisted of a large stationary pendulum, with a stylus on the bottom. As the earth starts to move, the heavy mass of the pendulum has the inertia to stay still in the non-earth frame of reference. The result is that the stylus scratches a pattern corresponding with the Earth's movement.
So, I'll start in the some sorted order.
The group of curves outside the olive-shaped envelope
I think this is the starting point of the shock. The analogy is something like this: The pendulum is still in motion with the last past earth's frame and it has not synced itself again. So, when it does - the pattern starts to change. During the tremors, the pattern is disturbed a lot. One more thing to notice (your image doesn't show the scale): They have used a very small pendulum (a sand tracing one) for the design (sensitivity maybe) and allowed to move on sand (looks like some muddy colloid so that it shouldn't be affected so much by friction, $\mu$ is very low)
The smooth ellipses were originally formed before the earthquake. How I tell this? We know the pattern for a pendulum. Don't we? It has started from the top of the pattern, curves somewhere near the middle, then the ellipse continues to incline about an angle (which is due to the earth's motion).
The group of curves inside the envelope but outside (2)
As the quake starts to show up, the pendulum notes down every fractional increase of the it's magnitude. And so, the inclination of the ellipses totally curve out (perpendicularly) thereby forming new ellipses at right angles to the previous ones. Now you might ask me a question...
Why are the perpendicular ellipses confined to a small region and do not spread out?
As you can see in the image, each and every fringe in the larger ellipses are equidistant (somewhat) from each other. As the magnitude increases, the fringes begin to compress which could be noticed in the small ellipses. This shows that the quakes weren't too smooth. As the pendulum starts the ellipse, the quake forces it to wiggle in the exactly opposite direction. For this reason,
The group of intertwined curves at the center
This is very very simple than the others. The earthquake has increased to its utmost magnitude. Now, the ground has shaken in every direction which has confused the pendulum to oscillate everywhere. Luckily, it has also made a rose by its random twist & twirl...
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This is just a plain blind suggestion. If you have any contradiction, feel free to comment it below. Feel free to correct me if I'm wrong :-) – Ϛѓăʑɏ βµԂԃϔ Jan 16 at 17:28
The link in the question explains how it was formed.
The outer concentric elipses were formed before the earthquake by someone just setting it in motion.
The inner elipses were formed by the earthquake itself, these gradually approaching circles over time to overwrite the earlier elipses, finally coming to rest at the centre.
The second position of rest suggests there was natural resettlement of the ground after the earthquake, and this gave rise to the strange rose pattern overwriting the earlier pattern.
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http://mathoverflow.net/questions/41042/unipotent-groups-their-forms-and-representations/41071
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## unipotent groups, their forms and representations
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
For simplicity fix a base field $k$ of characteristic zero, and consider smooth affine algebraic $k$-groups. (It is understood that unipotent groups in positive characteristic are more complicated, as one might have interesting non-smooth ones.)
Question 1: forms of unipotent groups
If $k$ is algebraically closed, then it is clear that every connected unipotent $k$-group is a successive extension of $\mathbb{G}_\mathrm{a}$'s. Then what about the case $k$ not algebraically closed? Is there non-trivial $K$-forms of, say, the upper trangular unipotent $k$-group of $GL_n$, and of the unipotent radicals of Levi $k$-subgroups of $GL_n$, etc?
If $K$ is a finite Galois extension of $k$ of Galois group $\Gamma$, then a $K$-form of the split $k$-torus is the same as a $\Gamma$-module structure on the group of characters $\mathbb{Z}^d$. Is there analogous results for $K$-forms of unipotent $k$-groups?
For example, with $K$ a finite extension field of $k$. the scalar restriction $$Res_{K/k}\mathbb{G}_{\mathrm{m}}$$
is not split as $k$-torus, but $$Res_{K/k}\mathbb{G}_\mathrm{a}$$
splits into a direct sum of $\mathbb{G}_\mathrm{a}$, becasue $K$ is a finite vector space over $k$ viewed additively. It is from this example that I want to know if there are interesting examples of forms of unipotent groups.
Question 2: representations of unipotent groups
If one has the one dimensional unipotent group $U=\mathbb{G}_\mathrm{a}$, then an algebraic representation of $U$ on $V$ a finite dimensional $k$-vector space is the same as a unipotent operator on $V$. One can then extend this description naturally to obtain the Tannakian category of finite dimensional algebraic representations of $U$.
And what about general unipotent $k$-group $U$? By the theorem of Lie-Engel, we know that such a representation of $U$ on $V$ is upper-triangular: it stabilizes a full flag of $V$, and acts trivially on the successive quotients (because of unipotence). Is there more precise information one can find about these representations so as the determine the Tannakian category of representations of $U$?
Again, let $K$ be a finite Galois extension of $k$ with Galois group $\Gamma$, and $U$, $W$ two connected unipotent $k$-groups that are isomorphic over $K$. Then how can one distinghuish the representations of the two groups by some "action" of $\Gamma$ one the representations spaces, in the spirit one finds in representations of the $k$-torus $$Res_{K/k}\mathbb{G}_\mathrm{m}$$
thanks!
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3
In characteristic zero unipotent groups and nilpotent Lie algebras correspond to each other using the Campbell-Baker-Haussdorf formulas. – Torsten Ekedahl Oct 4 2010 at 18:25
## 1 Answer
As Torsten points out, unipotent groups correspond naturally to nilpotent Lie algebras in characteristic 0. This is dealt with nicely on the scheme level, for example, in IV.2.4 of Demazure-Gabriel Groupes algebriques. They also treat in Chapter IV some questions about prime characteristic, which get quite tricky outside the commutative case. Both of your questions are more conveniently studied in the Lie algebra framework, I think, where standard Lie algebra methods for discussing forms in are available and where there is quite a bit of literature on structure, representations, and (in small dimensions) classification in characteristic 0. See for example Jacobson's 1962 book Lie Algebras.
Representation theory is potentially very complicated for nilpotent Lie algebras (say over the complex or real field), even in the finite dimensional situation: unlike the semisimple case, there is no nice general structure based on highest weights, etc. Dixmier and others have studied infinite dimensional representations extensively in connection with Lie groups. Classification of nilpotent Lie algebras is just about impossible in general, but up to dimension 7 or so there are lists. Anyway, there is a lot of literature out there. (Tori are on the other hand also studied a lot over fields of interest in number theory. They have at least the advantage of being commmutative.)
[ADDED] An older seminar write-up might be worth consulting, especially in prime characteristic, along with the relatively sparse literature published since then and best searched through MathSciNet: Unipotent Algebraic Groups by T. Kambayashi, M. Miyanishi, M. Takeuchi, Springer Lecture Notes in Math. 414 (1974). But as their treatment suggests, the main research challenges have occurred in prime characteristic. Over finite fields, there has been quite a bit of recent activity in studying the characters of finite unipotent groups related to the unipotent radical of a Borel subgroup.
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http://en.wikipedia.org/wiki/Cosine_transform
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# Sine and cosine transforms
(Redirected from Cosine transform)
In mathematics, the Fourier sine and cosine transforms are forms of the Fourier integral transform that do not use complex numbers. They are the forms originally used by Joseph Fourier and are still preferred in some applications, such as signal processing or statistics.
## Definition
The Fourier sine transform of $f (t)$, sometimes denoted by either ${\hat f}^s$ or ${\mathcal F}_s (f)$, is
$2 \int\limits_{-\infty}^\infty f(t)\sin\,{2\pi \nu t} \,dt.$
If $t$ means time, then $\nu$ is frequency in cycles per unit time, but in the abstract, they can be any pair of variables which are dual to each other.
This transform is necessarily an odd function of frequency, i.e.,
${\hat f}^s(\nu) = - {\hat f}^s(-\nu)$ for all $\nu$.
The numerical factors in the Fourier transforms are defined uniquely only by their product. Here, in order that the Fourier inversion formula not have any numerical factor, the factor of 2 appears because the sine function has $L^2$ norm of $\frac 1 {\sqrt2}$.
The Fourier cosine transform of $f (t)$, sometimes denoted by either ${\hat f}^c$ or ${\mathcal F}_c (f)$, is
$2 \int\limits_{-\infty}^\infty f(t)\cos\,{2\pi \nu t} \,dt.$
It is necessarily an even function of $\nu$, i.e., ${\hat f}^s(\nu) = {\hat f}^s(-\nu)$ for all $\nu$.
Some authors[1] only define the cosine transform for even functions of $t$, in which case its sine transform is zero. Since cosine is also even, a simpler formula can be used, $4 \int\limits_0^\infty f(t)\cos\,{2\pi \nu t} \,dt.$ Similarly, if $f$ is an odd function, then the cosine transform is zero and the sine transform can be simplified to $4 \int\limits_0^\infty f(t)\sin\,{2\pi \nu t} \,dt.$
## Fourier inversion
The original function $f(t)$ can be recovered from its transforms under the usual hypotheses, that $f$ and both of its transforms should be absolutely integrable. For more details on the different hypotheses, see Fourier inversion theorem.
The inversion formula is[2]
$f(t) = \int _0^\infty {\hat f}^c \cos (2\pi \nu t) d\nu + \int _0^\infty {\hat f}^s \sin (2\pi \nu t) d\nu,$
which has the advantage that all frequencies are positive and all quantities are real. If the numerical factor 2 is left out of the definitions of the transforms, then the inversion formula is usually written as an integral over both negative and positive frequencies.
Using the addition formula for cosine, this is sometimes rewritten as
$\frac\pi2 (f(x+0)+f(x-0)) = \int _0^\infty \int_{-\infty}^\infty \cos \omega (t-x) f(t) dt d\omega,$
where $f(x+0)$ denotes the one-sided limit of $f$ as $x$ approaches zero from above, and $f(x-0)$ denotes the one-sided limit of $f$ as $x$ approaches zero from below.
If the original function $f$ is an even function, then the sine transform is zero; if $f$ is an odd function, then the cosine transform is zero. In either case, the inversion formula simplifies.
## Relation with complex exponentials
The form of the Fourier transform used more often today is
$\hat f(\nu) = \int\limits_{-\infty}^\infty f(t) e^{-2\pi i\nu t}\,dt.$
Expanding the integrand by means of Euler's formula results in
$= \int\limits_{-\infty}^\infty f(t)(\cos\,{2\pi\nu t} - i\,\sin{2\pi\nu t})\,dt,$
which may be written as the sum of two integrals
$= \int\limits_{-\infty}^\infty f(t)\cos\,{2\pi \nu t} \,dt - i \int\limits_{-\infty}^\infty f(t)\sin\,{2\pi \nu t}\,dt,$
$= \frac 12 {\hat f}^c (\nu) - \frac i2 {\hat f}^s (\nu).$
## References
• Whittaker, Edmund, and James Watson, A Course in Modern Analysis, Fourth Edition, Cambridge Univ. Press, 1927, pp. 189, 211
1. Mary L. Boas, , 2nd Ed, John Wiley & Sons Inc, 1983. ISBN 0-471-04409-1
2. Poincaré, Henri (1895). Theorie analytique de la propagation de chaleur. Paris: G. Carré. pp. pp. 108ff.
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http://physics.stackexchange.com/questions/19724/how-to-represent-the-effect-of-linking-rigid-bodies-together/19727
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# how to represent the effect of linking rigid-bodies together?
I have 2 rigid-bodies (b1,b2) if i linked one to the other (as if they are conjoined together) , how to represent b1 effect on b2 and b2 effect on b1
Is there any LAW that affect the position/orientation of the other body ?
notes :
• i am using Quaternions for orientations
• i don't want to treat them as one body
• i have only primitive shapes (box,sphere,..) to link.
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Why don't you just treat them as a single, large rigid body? – yohBS Jan 19 '12 at 22:35
the idea that i am making a physics Engine that treats bodies as primitives (box,cone,sphere,..) . so if i treat them as a single rigid body ,i must add a new collision resolving logic for this new complex body – MhdSyrwan Jan 19 '12 at 22:49
## 3 Answers
The open-source physics engine ODE allows you to connect two bodies using any of a number of different joints. One of those joints is the "Fixed" joint. It's much more stable, in the physics engine, to represent the two bodies as a single body but maintain two separate geometries for collision purposes. However, ODE probably handles collision detection/resolution differently from what you have in mind. It only detects collision after one frame of interpenetration and then constrains the relative velocity of the colliding bodies in such a was as to force them apart on the next time step. That type of constraint is much easier to satisfy for a single rigid body than two, but perhaps you're actually preventing penetration and so need a different technique.
The fixed joint simply constrains the two bodies to have zero relative angular velocity and zero relative linear velocity (and also has an error correction term to eliminate small numerical drift). After that, the LCP solver handles the rest.
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The laws you are looking for are conservation of momentum and conservation of angular momentum. If you stick the two bodies together both laws still must be fulfilled (inelastic effects neglected). In the end you will have a compound single object. With the parallel axis theorem (wikipedia) you can calculate the mass moment of inertia and together with the centre of mass the whole motion of your compound object. You do not have to change the collision logic completely, create a sphere/box that includes your whole object and use that to test for collisions.
An alternative to connect the two rigid bodies is via springs as dmckee pointed out and this approach is quite successful in a lot of physics engines (bridge building games, World of Goo). Even liquids can be modeled with a few hard drops connected via springs.
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i am not talking about collision detection, i am talking about collision resolution that depends on Contact Normal which will be changed , am i right ? – MhdSyrwan Jan 20 '12 at 22:36
@MhdSyrwan: The contact normal will be changed, regardless of how you connect the bodies. For the calculation of the collision outcome conservation of momentum and angular momentum are good starting points. – Alexander Jan 21 '12 at 20:15
i want to start form my working calculation logic for my primitives. – MhdSyrwan Jan 21 '12 at 20:51
In Real Life (tm) there are no Rigid Bodies (tm), and to first order (i.e. under low strain {*}) all solids act as a springs. The rule you are looking for is Hooke's Law: $$F = -k \Delta x .$$
Of course, that leaves the matter of choosing the right spring constants (and noting that everything doesn't ring for ever) the right damping as well.
You're almost certainly better figuring out how to treat compound bodies and single units than trying to do a finite element analysis (even drastically simplified) at every step.
{*} Note that materials life clay which deform easily are only under low strain for exceedingly small pressures, so this rule does not apply to them in a practical sense even if it is arguably good in the low strain limit.
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http://www.nag.com/numeric/CL/nagdoc_cl23/html/G01/g01erc.html
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# NAG Library Function Documentnag_prob_von_mises (g01erc)
## 1 Purpose
nag_prob_von_mises (g01erc) returns the probability associated with the lower tail of the von Mises distribution between $-\pi $ and $\pi $ .
## 2 Specification
#include <nag.h>
#include <nagg01.h>
double nag_prob_von_mises (double t, double vk, NagError *fail)
## 3 Description
The von Mises distribution is a symmetric distribution used in the analysis of circular data. The lower tail area of this distribution on the circle with mean direction ${\mu }_{0}=0$ and concentration argument kappa, $\kappa $, can be written as
$PrΘ≤θ:κ=12πI0κ ∫-πθeκcosΘdΘ,$
where $\theta $ is reduced modulo $2\pi $ so that $-\pi \le \theta <\pi $ and $\kappa \ge 0$. Note that if $\theta =\pi $ then nag_prob_von_mises (g01erc) returns a probability of $1$. For very small $\kappa $ the distribution is almost the uniform distribution, whereas for $\kappa \to \infty $ all the probability is concentrated at one point.
The method of calculation for small $\kappa $ involves backwards recursion through a series expansion in terms of modified Bessel functions, while for large $\kappa $ an asymptotic Normal approximation is used.
In the case of small $\kappa $ the series expansion of Pr($\Theta \le \theta $: $\kappa $) can be expressed as
$PrΘ≤θ:κ=12+θ 2π +1πI0κ ∑n=1∞n-1Inκsinnθ,$
where ${I}_{n}\left(\kappa \right)$ is the modified Bessel function. This series expansion can be represented as a nested expression of terms involving the modified Bessel function ratio ${R}_{n}$,
$Rnκ=Inκ In-1κ , n=1,2,3,…,$
which is calculated using backwards recursion.
For large values of $\kappa $ (see Section 7) an asymptotic Normal approximation is used. The angle $\Theta $ is transformed to the nearly Normally distributed variate $Z$,
$Z=bκsinΘ2,$
where
$bκ=2π eκ I0κ$
and $b\left(\kappa \right)$ is computed from a continued fraction approximation. An approximation to order ${\kappa }^{-4}$ of the asymptotic normalizing series for $z$ is then used. Finally the Normal probability integral is evaluated.
For a more detailed analysis of the methods used see Hill (1977).
## 4 References
Hill G W (1977) Algorithm 518: Incomplete Bessel function ${I}_{0}$: The Von Mises distribution ACM Trans. Math. Software 3 279–284
Mardia K V (1972) Statistics of Directional Data Academic Press
## 5 Arguments
1: t – doubleInput
On entry: $\theta $, the observed von Mises statistic measured in radians.
2: vk – doubleInput
On entry: the concentration parameter $\kappa $, of the von Mises distribution.
Constraint: ${\mathbf{vk}}\ge 0.0$.
3: fail – NagError *Input/Output
The NAG error argument (see Section 3.6 in the Essential Introduction).
## 6 Error Indicators and Warnings
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
NE_REAL
On entry, ${\mathbf{vk}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{vk}}\ge 0.0$.
## 7 Accuracy
nag_prob_von_mises (g01erc) uses one of two sets of constants depending on the value of machine precision. One set gives an accuracy of six digits and uses the Normal approximation when ${\mathbf{vk}}\ge 6.5$, the other gives an accuracy of $12$ digits and uses the Normal approximation when ${\mathbf{vk}}\ge 50.0$.
## 8 Further Comments
Using the series expansion for small $\kappa $ the time taken by nag_prob_von_mises (g01erc) increases linearly with $\kappa $; for larger $\kappa $, for which the asymptotic Normal approximation is used, the time taken is much less.
If angles outside the region $-\pi \le \theta <\pi $ are used care has to be taken in evaluating the probability of being in a region ${\theta }_{1}\le \theta \le {\theta }_{2}$ if the region contains an odd multiple of $\pi $, $\left(2n+1\right)\pi $. The value of $F\left({\theta }_{2}\text{;}\kappa \right)-F\left({\theta }_{1}\text{;}\kappa \right)$ will be negative and the correct probability should then be obtained by adding one to the value.
## 9 Example
This example inputs four values from the von Mises distribution along with the values of the argument $\kappa $. The probabilities are computed and printed.
### 9.1 Program Text
Program Text (g01erce.c)
### 9.2 Program Data
Program Data (g01erce.d)
### 9.3 Program Results
Program Results (g01erce.r)
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http://mathoverflow.net/questions/62818/the-sets-in-mathematical-logic/62838
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## The sets in mathematical logic
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It is well known that intuitive set theory (or naive set theory) is characterized by having paradoxes, e.g. Russell's paradox, Cantor's paradox, etc. To avoid these and any other discovered or undiscovered potential paradoxes, the ZFC axioms impose constraints on the existense of a set. But ZFC set theory is build on mathematical logic, i.e., first-order language. For example, the axiom of extensionality is the wff $\forall A B(\forall x(x\in A\leftrightarrow x\in B)\rightarrow A=B)$. But mathematical logic also uses the concept of sets, e.g. the set of alphabet, the set of variables, the set of formulas, the set of terms, as well as functions and relations that are in essence sets. However, I found these sets are used freely without worrying about the existence or paradoxes that occur in intuitive set theory. That is to say, mathematical logic is using intuitive set theory. So, is there any paradox in mathematical logic? If no, why not? and by what reasoning can we exclude this possibility? This reasoning should not be ZFC (or any other analogue) and should lie beyond current mathematical logic because otherwise, ZFC depends on mathematical logic while mathematical logic depends on ZFC, constituting a circle reasoning. If yes, what we should do? since we cannot tolerate paradoxes in the intuitive set theory, neither should we tolerate paradoxes in mathematical logic, which is considered as the very foundation of the whole mathematics. Of course we have the third answer: We do not know yes or no, until one day a genius found a paradox in the intuitive set theory used at will in mathematical logic and then the entire edifice of math collapse. This problem puzzled me for a long time, and I will appreciate any answer that can dissipate my apprehension, Thanks!
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If yes, what we should do? Read the introductory material in a basic textbook on mathematical logic? – Gerald Edgar Apr 24 2011 at 12:09
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I have read serveral basic textbook on mathematical logic, but none of them answers my question successfully. – zzzhhh Apr 24 2011 at 12:36
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You mean "circularity", not "paradox". These are distinct notions. – Harry Altman Apr 24 2011 at 13:12
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@Gerald Edgar: Did you have any particular book in mind? If yes, why not reveal it? – Sergey Melikhov Apr 24 2011 at 14:34
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@Thierry, in set theory, everything is a set, so whether we're quantifying over sets, or over sets of sets, or sets of sets of sets, etc. these are all first-order quantifications. – Amit Kumar Gupta Apr 24 2011 at 18:06
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## 10 Answers
I have been asked this question several times in my logic or set theory classes. The conclusion that I have arrived at is that you need to assume that we know how to deal with finite strings over a finite alphabet. This is enough to code the countably many variables we usually use in first order logic (and finitely or countably many constant, relation, and function symbols).
So basically you have to assume that you can write down things. You have to start somewhere, and this is, I guess, a starting point that most mathematicians would be happy with. Do you fear any contradictions showing up when manipulating finite strings over a finite alphabet?
What mathematical logic does is analyzing the concept of proof using mathematical methods. So, we have some intuitive understanding of how to do maths, and then we develop mathematical logic and return and consider what we are actually doing when doing mathematics. This is the hermeneutic circle that we have to go through since we cannot build something from nothing.
We strongly believe that if there were any serious problems with the foundations of mathematics (more substantial than just assuming a too strong collection of axioms), the problems would show up in the logical analysis of mathematics described above.
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Finite set is OK, but there are also many infinite sets used freely in mathematical logic, how to transfer from finite set to infinite set safely without fearing any paradox? And, may I ask what is your exact opinion: 1)No paradox (why?), 2)There is paradoxes, but we have successfully excluded them (how?), 3)Not clear, but only hold a belief that there is or there is not any paradoxes? – zzzhhh Apr 24 2011 at 14:07
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Actually, I wouldn't say we are using any axioms of set theory. What we are using is some intuitive understanding of how to manipulate finite sequences of characters from a finite alphabet. If you want to formalise this, some fragment of number theory will be enough (as finite sequences can be coded as natural numbers). – Stefan Geschke Apr 24 2011 at 21:18
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+1 Stefan: "what we are using is some intuitive understanding of how to manipulate finite sequences of characters from a finite alphabet." I find it helpful to consider a computer which is programmed to detect whether a formal proof in a formal theory is valid. The proof is expressed in finitely many symbols, and there is no background set theory sitting inside the computer on which the proof-checking depends. The pattern of flow of electrons through the logic gates is an extra-linguistic entity, and this breaks the circularity. – Todd Trimble Apr 24 2011 at 23:40
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Stephan, your answer is GOLD. Yes, one must start somewhere, and what better place than finite strings and their manipulations? I, for one, do not believe in either sets or (platonic) numbers, but I have no problems with manipulating strings (my laptop does not have any issue either, and "it" does not know a thing of numbers and sets, unless by numbers ones means terms and their verifiable equivalence under the rules of arithmetics and term rewriting). The funny (and great) thing is: one can "play" ZFC without believing that anything there has any substance beyond syntactic games... – Mirco Mannucci Jun 14 2011 at 18:56
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@LeeMosher: One can write a huge novel about hobbits and orcs without believing in their existence in any deep sense. – Michael Greinecker Sep 26 at 14:20
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
The short answer is that there is no way to be absolutely certain that mathematics is free from contradiction.
To start with an extreme case, we all take for granted a certain amount of stability in our conscious experience. Take the equation $7\times 8 = 56$. I believe that I know what this means, and that if I choose to ponder it for a while, my mind will not somehow find a way to conclude firmly that $7\times 8 \ne 56$. This may sound silly, but it is not a totally trivial assumption, because I've had dreams in which I have found myself unable to count a small number of objects and come up with a consistent answer. Is there any way to rule out definitively the possibility that the world will somehow reach a consensus that $7\times 8 = 56$ and $7\times 8 \ne 56$ simultaneously? I would say no. We take some things for granted and there's no way to rule out the possibility that those assumptions are fundamentally flawed.
Suppose we grant that, and back off to a slightly less extreme case. Say we accept finitary mathematical reasoning without question. People might disagree about the precise definition of "finitary," but a commonly accepted standard is primitive recursive arithmetic (PRA). In PRA, we accept certain kinds of elementary reasoning about integers. (If you're suspicious about integers, then you can replace PRA with some kind of system for reasoning about symbols and strings, e.g., Quine's system of "protosyntax"; it comes to more or less the same thing.) Now we can rephrase your question as follows: can we prove, on the basis of PRA, that ZFC is consistent?
This, in essence, was Hilbert's program. If we could prove by finitary means that all that complicated reasoning about infinite sets would never lead to a contradiction, then we could use such infinitary reasoning "safely." Sounds like what you're asking for, doesn't it?
Unfortunately, Goedel's theorems showed that Hilbert's program cannot be carried out in its envisaged form. Even if we allow not just PRA, but all of ZFC, we still cannot prove that ZFC is consistent. Thus it's not just that we've all been too stupid so far to figure out how to show that ZFC doesn't lead to contradictions. There is an intrinsic obstacle here that is insurmountable.
So your scenario that someone may one day find a contradiction in ZFC cannot be ruled out, even if we take "ordinary mathematical reasoning" for granted. This is not as bad as it might seem, however. ZFC is not the only possible system on which mathematics can be based. There are many other systems of weaker logical strength. If a contradiction were found in ZFC, we would just scale back to some weaker system. For more discussion of this point, see this MO question.
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I should restate my question as follows: "Will inconsistency in meta-theory go into the theory built on it?" Here the meta-theory is naive set theory and the theory built is mathematical logic. I should express my thanks to kakaz and other people who answers and comments in this thread. Only after interaction with them can I formulate my question in a clear form. Now I wonder if the answer is still unknown. It seems that treating infinite sets as a proper class as suggested by Stefan Geschke works. Is that method will possibly produce paradox too? – zzzhhh Apr 24 2011 at 20:47
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Your restated question is easy enough to answer: Yes, inconsistent meta-reasoning could produce an inconsistent formal theory. – Timothy Chow Apr 25 2011 at 14:14
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A fine point: "There are two main approaches for carrying out consistency proofs. One is to introduce a model of the system in question in the Meta-theory. However, it seems to be implausible to assume that one can prove this way the consistency of a theory by using finitary methods, since such methods do not allow the use of sets. Hilbert realized this and suggested therefore that one should instead analyse proofs and show this way directly that it is not possible to derive in the formal system in question a contradiction." - From ehess.fr/revue-msh/pdf/N165R959.pdf – Sergey Melikhov Apr 25 2011 at 20:33
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"... Such theories will then be a substitute for Hilbert's finitary methods, we can call them extended finitary methods. The up to now most successful theories used for this purpose seem to be extensions of Martin-Löf type theory, ..." (from the same paper) – Sergey Melikhov Apr 25 2011 at 20:49
There are a few problems you seem to be having. First of all, the statement "mathematical logic depends on ZFC" doesn't make sense.
As mathematical logicians, when we study formal systems, we should imagine placing that formal system in a box. The box is full of formulas and deductions in the object language. For example, ZFC is a first-order theory with one binary predicate symbol and a bunch of axioms. It holds a privileged position since we tend to think of it as 'the' formal set theory, but there's no reason we couldn't instead use MK or NF or other set theories for the same purposes.
Mathematical logic is the act of studying formal systems using mathematical (not necessarily formal) methods, and ZFC is just one particular formal system. The important point is that mathematical logic is not a formal system, and although the statement "ZFC is consistent" is well formed, the statement "mathematical logic is consistent" is not. To claim a theory is inconsistent is to claim that there is a formal proof of false in some formal system. Russel's paradox, for example, can be cast as a formal proof of false in ZFC with unrestricted comprehension.
Without the context of first-order logic, and the collections of variables and symbols that are required to write down formulas and formal proofs, the statement "_ is (in)consistent" is not meangingful. The blank must be filled in with a first-order theory, or more generally, some formal system with a notion of formal proof. You can ask 'are we justified in forming and manipulating these collections?' But that's an informal question. As other users have pointed out, it has very good informal answers, for example, the fact that computers work gives us confidence that we shouldn't worry about doing arithmetic and manipulating strings informally.
In order to answer the question 'is the informal set theory we used to formulate first order logic consistent' either affirmitively or negatively, we must define the notion of consistency, and in doing so use the informal set theory in question. The point, again, is that consistency is only defined in the context of first order logic, where we take these collections as primitive and define consistency from there. In the same way we cannot speak of a simple group ouside the context of groups, we cannot speak about formal consistency outside the context of formal theories.
In short: One cannot provide a formal proof of anything without first defining formal proof! Hence, we must start somewhere and take the collections of symbols in first order logic as primitive, or convince ourselves informally that we are justified in forming such collections.
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I'm not a logician, but I consider myself a "formalist" as for the foundation of mathematics and logic, and I don't agree on your sentence: «mathematical logic is not a formal system». When you form and manipulate such strings of symbols, you should not be allowed to use (formal or informal) reasoning involving e.g. infinite sets, induction etc. Once you place yourself in such a restriction, I think in principle you can do formal mathematical logic (i.e. "blind" manipulation of symbols) without problems. Do you agree? – Qfwfq Sep 26 at 16:42
The computable strings on a finite alphabet-which include many infinite strings as well as well as all the finite strings-should suffice to encode all the naive sets that you need for a full mathematical description of a set theory such as Zf. I do not think that any paradoxes will turn up if you use only these computable strings. – Garabed Gulbenkian Sep 26 at 19:44
That is to say, mathematical logic is using intuitive set theory. So, is there any paradox in mathematical logic?
Yes, in set theory whose logic is based upon naive set theory there is Berry's paradox.
Consider the expression:
"The smallest positive integer not definable in under eleven words".
Suppose it defines a positive integer $n$. Then $n$ has been defined by the ten words between the quotation marks. But by its definition, $n$ is not definable in under eleven words. This is a contradiction.
A version of Berry's paradox is called Richard's paradox. I don't see an essential difference between the two paradoxes (except that Richard's paradox was brought out by Poincaré, who was concerned with impredicative definitions and Berry's paradox by Russell, who was concerned with types). But the two Wikipedia articles offer very different explanations.
The explanation of Berry's paradox is essentially by type theory (even though it's not named); and the more specific explanation given by recursion theory also seems to fit in the framework of constructive type theory.
Richard's paradox is resolved in the framework of ZFC (and more importantly, of first-order logic). Roughly, the resolution is that not all of what makes sense in set theory whose logic is based on naive set theory should make sense in ZFC; hence in some sense the paradox is only truly resolved in "the metatheory used to formalize ZFC". I don't know any textbook on such a meta-ZFC, which makes me feel uncomfortable about this resolution; but so did Poincaré and Russell, as I understand from their writings.
For this particular purpose I think a second-order ZFC would work just as well as a "meta-ZFC". But then a higher version of the same paradox would only be truly explained by a third-order ZFC, etc. But then again I don't know any textbook on third-order ZFC. Luckily, there are quite a few textbooks on higher-order logic/type theory, and as I understand there are no problems of this kind (Richard/Berry-type paradoxes) in a modern type theory, because it serves as its own logic in essence.
On the other hand, a type theory is said to still require a meta-theory. Does it mean that there must be deeper paradoxes that would not be resolved by type theory?
Added later. I'm a bit amused by this thread, which keeps growing with "orthodox" answers and with their lively discussion in the comments, whereas the "heretical" answer you're reading now has not been challenged (nor even downvoted) as yet.
Perhaps I should summarize my points so that it's easier to object if anybody cares.
1) The OP's question might be a bit informal or vague, but it does make sense as Berry and Richard paradoxes unambiguously demonstrate. One may hold different views of these paradoxes and their resolution, but one possible view (which I believe I picked from Poincare's writings) is that they are indeed caused by the factual mutual dependence between set theory and logic referred to in the OP. ("Set theory" and "logic" meant in colloquial sense here, not referring to a specific formalization.) You cannot just ignore these paradoxes, can you?
2) Type theories may have a lot of disadvantages compared to ZFC with first order logic (including the lack of a canonical formulation and of an equivalent of Bourbaki) but they do break the circularity between set theory and logic and thus do adequately resolve the paradoxes. ZFC is perfectly able to defend itself against these paradoxes but it does not attempt to explain them; for that it sends you to the meta-level, which then ends up with informal speculations about computers and one's experience with finite strings of symbols. (There are of course issues with such speculations, including: exactly which strings are finite, halting problems for idealized computers and finite memory of physical computers, not to mention potential finiteness of information in the physical universe and one's proof-checking software potentially involving higher-order logic already.) So in this case I see type theory as providing a mathematical solution, and ZFC, at best, a metaphysical one.
3) What I don't understand is whether there might exist a kind of type theory that would not need any metatheory (i.e. would serve as its own metatheory). Perhaps someone saying "You cannot get anything out of nothing" or "You cannot have any theory without metatheory" or "we must start somewhere" or "You have to start somewhere" could as well explain why it is impossible? If indeed it is impossible, can this impossibility be witnessed by a specific paradox to completely clarify matters?
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Thank you. Do you mean the modern type theory is sure to include no paradox inherent in the naive set theory, such as Russell's or Richard's or any other paradox, discovered or undiscovered yet? – zzzhhh Apr 24 2011 at 15:42
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Modern type theory avoids Russel's and Richard's paradoxes by design, while one's belief that ZFC avoids e.g. Russel-type paradoxes is based largely on experience. As to paradoxes that are not as yet discovered you'd better ask Voevodsky; he has strong opinions on such matters mathoverflow.net/questions/40920. – Sergey Melikhov Apr 24 2011 at 16:03
You cannot get anything out of nothing :-) But do not worry.
Mathematics existed long before ZFC was formulated, and well before “formal reasoning” rose to a kind of religion. Mathematically, there is nothing more formal in a “formal reasoning” than in any other “logically justified” (i.e. commonly accepted) reasoning. The true reason of encoding math in a single theory is to gather all doubts in a single place, earn confidence that our new theory is consistent (as long as the foundations are consistent), and help communicating with other mathematicians.
Moving back to your question. Let me distinguish between four cases – according to your terminology - a theory can be:
• naïve and formal – this means that by using formal reasoning we may show that there is an inconsistency in the theory (for example: ZFC with unrestricted comprehension)
• naïve and informal – this means that we see that there is an inconsistency in reasoning within the theory, but a proof of this fact is outside our math (the same example)
• non-naïve and formal – this means that we believe that the theory is consistent, and (sometimes) can “formally” prove its consistency relatively to another (formal or informal) commonly accepted theory
• non-naïve and informal – just like above with the last part of the sentence skipped.
So, as you may see, being formal cannot make a theory consistent/inconsistent, but can provide additional arguments for/against the theory – simply – correctness is invariant under changes of formality. For most situations the picture of formality looks like follows:
• there is an informal concept like “first-order logic”
• there is a formal theory of sets expressible in first-order logic
If we would like to investigate foundations themselves than we could extend the picture by introducing one (or more) additional level:
• there is an informal theory of sets (meta-theory; it has to be a bit stronger than the "inner" set theory to show that the "inner" set theory is consistent, but it may be far weaker to express the "inner" theory)
• there is a formal first-order logic expressible in the meta-theory
• there is a formal theory of sets expressible in first-order logic
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I'm a bit confused. Please let me rephrase some notions as following: The intuitive set theory containing Russell's paradox is naive and informal, the current mathematical logic is naive but formal, right? So my question is, for any theory that involves naive notion of sets like current mathematical logic, is there always any paradox as we encountered in the naive set theory (like Russell's paradox)? – zzzhhh Apr 24 2011 at 15:14
"The intuitive set theory containing Russell's paradox is naive and informal, the current mathematical logic is naive but formal, right?" I do not understand your term "intuitive". Formerly, you used it as a synonym for "naive". If this is correct, then "yes" - "intuitive set theory" is a "naive set theory", and as I said it may be formal or informal. "So my question is, for any theory that involves naive notion of sets like current mathematical logic" Mathematical logic does not involve "naive notion of sets". – Michal R. Przybylek Apr 24 2011 at 15:57
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The question is not formalization, but consistency(paradox), so let's temporarily forget ZFC and consider only the consistency problem. First we need the concept of set in mathematical logic, for example, the justification of proof by induction on structure of wffs relies on the concept of set. Second, we are using the concept of set in mathematical logic freely as we do in naive set theory. Third, there exists paradox in naive set theory in meta-universe, so by analogy I'm afraid there are also paradoxes in the naive set theory used in mathematical logic. This is why I post this thread. – zzzhhh Apr 24 2011 at 19:16
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To restate my question: Will inconsistency in meta-theory go into the theory built on it? Here the meta-theory is naive set theory and the theory built is mathematical logic. No formalization is involved in my question. – zzzhhh Apr 24 2011 at 20:23
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<<It now seems that every theory is inconsistent, that is, "there is no way to be absolutely certain that mathematics is free from contradiction." from Timothy Chow's answer.>> There is a giant difference between knowing non-something and not knowing something. – Michal R. Przybylek Apr 24 2011 at 23:17
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This is an answer to another question that unfortunately has been closed just before I could post it and that essentially revolves around the same, or very similar, points as the question of the above OP.
Even if, from the point of view of a logician (which I'm not), the question is elementary, I think it's worth trying to give an answer, so to clarify things (also to myself) and settle some doubts that many non-logicians like me often have about the foundations model theory and logic.
Forget for a moment about $ZFC$. When investigating any (let's say first order) formal theory (such as Peano Arithmetic $PA$, or Algebraically Closed Fields $ACF$) logicians tacitly assume to work within the framework of a metatheory $T$, which must be some kind of "set theory" (where the term is intended loosely, possibly including certain systems of second order arithmetic like $ACA_0$ or of category theory like $ETCS$ or of class theory like $NBG$, according to personal taste) otherwise they wouldn't be able to talk (in principle with rigour, i.e. formally) of all the classical syntactic concepts, such as: languages, signatures, sets of formulas, finite strings of symbols and concatenation thereof; but also of all the classical semantic concepts, like structures, interpretations, models, isomorphisms. Or at least they wouldn't be able to codify -in principle- all the above concepts in the metatheory. The metatheory $T$ is hence momentarily assumed to embody "ontological" concepts, i.e. to formalize the concept of the "real universe of sets" (as opposed to specific concepts of sets arising from theories that are under scrutiny within $T$).
As far as I understand, logicians (even if they usually don't like to explicitely spell out the metatheory they're tacitly using) are perfectly content in assuming the metatheory is the standard set theory $ZFC$ (after all, Logic is a part of Mathematics like any other).
Anyway, I presume much less is needed to "formalize Logic"; for example I think $ZF$ would be enough, or weaker theories such as $ACA_0$ would be ok, and I think (but some professional logician should say if my guess is correct) essentially nothing would be changed in Logic in switching from one suitable metatheory to another: that's why they usually don't bother themselves specifying the metatheory.
How does this formalization work, roughly? Let's consider for example the first order formal theory of groups (call it $GT$). It must be dealt with inside a metatheory (which is a "set theory") which we call $T$. So here we have sentences of $T$ talking about the "objects" $T$ is apt talking about (it could be sets, or sets of natural numbers...), and we have some way to codify the syntactic and the semantic concepts of Logic within $T$: for example some sentences of $T$ will talk about some particular sets that we (informally) interpret as being strings of symbols of $GT$, others as being elements of the language of $GT$, and so on. Of course there will also be sentences of $T$ talking about sets that are structures for the signature of $GT$ (that is, magmas equipped with an arbitrary element $1$ that need not be a unit, and maybe with an arbitrary unary function $x\mapsto (x)^{-1}$ which need not be the inverse), and sets that are models of $GT$ (that is, plain groups).
Now let's come back to $ZFC$. As in the case of $GT$, we must deal with the first order formal theory $ZFC$ within a metatheory ("set theory") $T$. Your question is: what about if $T$ is Zermelo Frenkel set theory with Choice?
For the syntactic aspect, nothing special happens.
You have sentences of $T$ talking about sets that we take to codify various syntactic notions of $ZFC$. Example: a certain sentence of $T$ will define a unique "set" (in the sense of "object of which $T$ is apt talking about" - in this case it means "set in the sense of Zermelo Frenkel set theory with Choice") which stands for the following finite concatenation of symbols in the language of $ZFC$:
$$\forall\forall\in\in\in\to\in x \in ((y(())\in\in = = \in,$$
others will be more meaningful, e.g. displaying an axiom, a theorem (or a conjecture) of $ZFC$.
For the semantic aspect, indeed, there are some subtleties.
In the case of $GT$ we consider models which are sets of $T$ equipped with some structure: $(G,\mu,\iota,1)$. Note that, in the language of $GT$, variables stand for (i.e. are interpreted as) elements of a group, and (the support of) a model is the set of all such elements. If we were asked to prove the existence of models of $GT$, we would have no problem: we would simply exhibit any group, for example $(\mathbb{Z}/ 2 \mathbb{Z},+,-,0)$ (or even the trivial group, for what it matters!).
In the case of $ZFC$, what would be a model? A model $(X,E)$, properly, would be a set $X$ of $T$ equipped with a relation $E \subseteq X \times X$ satisfying the axioms of $ZFC$ when we interpret (via $T$) the relation symbol "$\in$" of $ZFC$ as meaning $E$. Note that variables, in the language of $ZFC$, stand for sets, and the support $X$ of the model must be interpreted as the collection of all such elements, that is $X$ has to be interpreted as the class of all sets.
So far, nothing problematic: set theorists consider models (in the framework of Zermelo Frankel set theory with Choice) of $ZFC$ all the time. For example, if $\xi$ is an inaccessible cardinal, then $(V_{\xi},\in)$ will be a model of $ZFC$ (where $V_{\xi}$ is the slice of the cumulative hierarchy defined by $\xi$).
What about proving the existence of models of $ZFC$? This is the subtlety: when we take $T$ to be Zermelo Frenkel set theory with Choice, we cannot prove in $T$ that $ZFC$ has models, because otherwise we would contradict Goedel's second incompleteness theorem! For example, we don't know, on the grounds of $T$ alone, whether there exists an inaccessible cardinal. So, seen from the point of view of the metatheory $T$, the whole "ontologic" problem of existence of models of $ZFC$ is somehow "suspended". In order to be granted the existence of those models, we have to assume stronger axioms than the ones of $T$; for example, some large cardinal axioms. The existence of an inaccessible would be sufficient for having the mere existence of a model $(X,E)$ (in which, in this case, $E$ is the restriction of the "ambient" membership relation $\in$ itself) without further requirements.
There's another subtlety. In $T$, via some Russel like paradox, one can easily prove that there is no set $V$ such that every set belongs to $V$. This implies that, if for some reason we are handled a model $(X,E)$ of $ZFC$, then either the relation $E$ is not the restriction to $X$ of the "ambient" membership relation $\in$ of $T$ (those are the so called nonstandard models of set theory), or there are properties of some set $S$ defined in $ZFC$ that are not true "ontologically" about its interpretation in the model (i.e. the corresponding sentences of $T$ are false when applied to the interpretation $S^{\mathrm{int}}$ of $S$ in $(X,E)$). For example, if $(X,\in)$ is a countable (standard) model, then of course the notion of cardinality cannot be transferred literally from $ZFC$ to $(X,E)$, because $ZFC$ proves there are uncountable sets, yet the interpretation of any set in $(X,\in)$ will be "ontologically" countable just because $X$ is. To be more concrete, define $\mathbb{R}$ in $ZFC$ in one of the usual ways, then of course $ZFC$ proves « $\mathbb{R}$ is uncountable». In a countable standard model $(X,\in)$ there is a countable set $\mathbb{R}^{\mathrm{int}}$ which is the interpretation of $\mathbb{R}$. Being $X$ a model, the sentence «$\mathbb{R}^{\mathrm{int}}$ is $($ uncountable $)^{\mathrm{int}}$ » of $T$ is true, where «$($ uncountable $)^{\mathrm{int}}$» is the interpretation of the notion of uncountability in the model, but -this is the appearent paradox, which is called Skolem paradox- the sentence of $T$ «$\mathbb{R}^{\mathrm{int}}$ is uncountable» is false. The appearent confusion comes from mixing up the notion of uncountability in the theory and in the metatheory.
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You cannot have any theory without metatheory. Opposite statement is not true and is simple kind of ideology in the fundamentals of mathematics. Of course You may use the same language for both of them! Then You would build mathematics in language which is its own metalanguage - just like natural language, but it is not very useful for strict analysis regarding paradoxes and circularities. But is is how mathematics works in everyday practice;-)
Of course If You try to be strict and analise certain dilemma in fundamentals You have to build more or less strict metahteory for Your theory - here logic. So You have to separate theory and its metatheory. Assuming You agree with that, You question is: "is it possible that metalgic is inconsistent?" The answer is: Yes, it is!
Most of the logic may be performed in finite sets, but not all. The former part, which require infinite sets and quantifications over sets of symbols etc. requires to have its metatheory in order top be sure it may be consistent (and even for writing statements of a theory - mathematics is created for the people and by the people). So situation is as follows - You metatheory:
1. it ma be related to naive set theory ( natural language) which obviously may be inconsistent. It does not mean that logics have to be inconsistent! It ma be, but there is not a must!
2. Of course You may use ZFC as a metatheory! Then Your problem is transformed into question about consistency of ZFC. It may be also not related directly to logic!
Interesting question here is: may be theory consistent even if its metatheory is not? It seems that it may be one of the possible the cases...
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To avoid divergence, please let me rephrase part of your idea as follows: Any theory must be built on meta-theory. We should separate them. Here the meta-theory is "naive set theory" and the theory built is "mathematical logic". If so, my question is: "Will inconsistency in meta-theory go into the theory built on it?" I offered three answers to this question in my original post, the item 1 of your answer denies my second answer "if yes". Do you mean "if no" is also not certain, so there is not a must, as my third answer in my post indicates -- the answer is unknown, maybe yes, maybe no? – zzzhhh Apr 24 2011 at 19:52
"Any theory must be built on meta-theory." - are You sure? I am not. But is sometimes usable method. Theory-metatheory-metametatheory and so on hierarchy model resolves only certain problems. There are problems which in this type of reasoning are not resolved at all. For example claim beginning from the sentence: "for every level of theorem-metatheorem hierarchy .... " is not a part of any level of it, but still may be good, provable theorem... – kakaz Apr 24 2011 at 22:16
I happen to realize the same apparent circularity in mathematical logic. If it purports to establish a foundation for mathematics, it seems to me its methods need to be different from mathematical methods. However, I have the uneasy feeling that one still is doing mathematics when studying mathematical logic. In spite of all this ambivalence, I think it presents a rather satisfying foundation for other branches of mathematics such as abstract algebra. It then seems to be a pretension to think that mathematical logic offers a foundation for "all" mathematics; maybe, philosophy can help with this matter.
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The old (Hilbert) idea was to have a foundation of mathematics on elementary concepts that have a "itself evidence", and there was a intuitive "halph-real" (platonic?) idea about mathematics, it live in a pseudo-realty, then cannot have antinomy (a thing cannot be a book and a no-book). Of course there was the hope of find a proof of consistency of mathematics (necessarily inside mathematics itself). After K.GOdel all these hopes crush down, mathematics is building on the set theory rock, but this rock isnt "absolute", or "itself evidence natural", but is juast a formal logical theory. Inside "set theory (in some form) you can make mathematical logic. Then before make the "Set theory" (ZFC for example) what is the mathematical foundation of the logic? ANd why logics use the concept of set (before make the Set thory)?
The paradox is evident, you cannot found mathematics from itself: 1) Is a phylosophical contadiction, for make a first step on bulding mathematics is absurd supposing that you just have a "mathematical ground"..
2) FOr GOdel theorem: cannot have a first foundation step that is its own foundation or for a faith on its natural itself-evidence (or proof of consistency in itself).
THen MAthematics could have a foundation only from something that is external. For "mathematics" we mean of course the article, the books, the ZF axioms , the various literature about mathematical fields, but is also what make all these thgis, this is the human think, cannot think to mathematics without remember that doing-mathematics (thought) is what produce mathematics (in its formalism, articles ecc.).
Then the question become phylosophical (we have just see above that cannot have a logical formal resolution..): what is the sense of thinking mathematics? What is the sence of elementary concept used for buldind mathematics before its formal logical axiomatization?
When we think that "a element belong to a set" we mean a primitive concept that has its mean before a formalizated set theory and its axioms? (I think yes), Can we found mathematics on category theory ground, without have obligation about ZF when we say "this morphisms is a element of this set, or collection..."?
(excuse mine bad English)
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1
You don't need a theory of sets prior to a theory of categories, any more than you need a theory of sets prior to a theory of sets. – Todd Trimble Feb 17 2012 at 17:28
I think this problem shows itself at many stages of the human thinking. So we can form the expression "This expression is wrong" and immediately the liar paradox appears. So really this has nothing to do with mathematics proper, but with logic itself. And the solution is multistaged itself: at the basic level we can use Russell's (naive) type theory as a remedy. In fact when we talk about mathematics we already make sure we don't break certain rules on the metalanguage level. Next when we want to do set theory, we use the axioms of Zermelo-Fraenkel. Problem solved. Of course we can formalize the metalanguage and we can use again a ZF set theory in this task instead of type theory, but in doing so, we use an informal meta-meta-language for which no way to ignore type theory. So, type theory is the ultimate building block of "human logic" and not just "mathematical logic".
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http://mathoverflow.net/questions/7903/conjugate-gradient-iteration/75704
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## conjugate gradient iteration
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I'm having problems understanding why the conjugate gradient method breaks down for singular matrices. I've read a good introduction to intuitively understanding the CG method through visualizing the quadratic form, but I'm not really understanding mathematically why this seems to be. Does anyone have a good mathematical explanation or any sources that explain why conjugate gradient iteration doesn't work for singular matrices?
Thanks in advance.
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In what sense would you want it to work? Typically, a system won't have any solution at all if the coefficient matrix is singular. – Darsh Ranjan Dec 5 2009 at 22:04
My question was geared more towards any mathematical explanation of why singular matrices don't work. My understanding, from the previously referenced article (pg 5), is that singular matrices have multiple solutions rather than a single solution. – john Dec 5 2009 at 22:30
## 4 Answers
My question was geared more towards any mathematical explanation of why singular matrices don't work. My understanding, from the previously referenced article (pg 5), is that singular matrices have multiple solutions rather than a single solution.
Unfortunately, that's not so. If $A$ is a singular matrix, then for most $b$, the equation $Ax=b$ has no solution. It has a solution if and only if $b$ is in the column space of $A$ (or orthogonal to the kernel of $A$, if $A$ is symmetric), in which case there are infinitely many solutions. This is basic linear algebra.
Let's say $A$ is symmetric positive semidefinite, so we can at least entertain the thought of conjugate gradients. Conjugate gradients can be interpreted as maximizing the quadratic function $$f(x) = b^Tx-\frac12 x^TAx.$$ If $A$ is singular, then this function typically has no maximum value, as I will demonstrate. Since $A$ is singular, there are vectors $v$ such that $Av = 0$. If there is any such vector $v$ that is not orthogonal to $b$, consider substituting $x=tv$ into $f(x)$: $$f(tv) = t(b^Tv) - \frac12t^2(v^TAv) = t(b^Tv) = ct,$$ where $c$ is the nonzero real number $b^Tv$. Obviously, this is unbounded as a function of $t$, so we see that there is no maximum value if $b$ is not orthogonal to the kernel of $A$.
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
If the matrix is singular then then the images of some vectors may be zero and that could mean some of the steps may end up with division by zero.
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A search for "conjugate gradient singular matrix" took me to this question. While the answer is obviously given by the responses, the question can be refined: Can CG still give a working algorithm if the matrix is singular, but behaves as a symmetric positive definite form on a (large) subspace?
A standard example is given by the finite element discretization of the Neumann problem on a simply connected domain. The constant functions are both the kernel and the cokernel of the Laplacian. On functions with vanishing mean, the Laplacian is still a positive definite symmetric operator, and we would like to leverage this structure.
This is non-trivial and best our numerical method is derived from a fully analytic setting, because this might provide us the convergence analysis as well. --- This appraoch is for example elaborated in
````On the Finite Element Solution of the Pure Neumann Problem
Pavel Bochev and R. B. Lehoucq
SIAM Review
Vol. 47, No. 1 (Mar., 2005), pp. 50-66
Published by: Society for Industrial and Applied Mathematics
Article Stable URL: http://www.jstor.org/stable/20453601
````
Apart from this canon standard example for the Laplacian, system matrices with larger kernel appear in numerical methods for the de-Rham-complex, in particular if the domain is topologically non-trivial (Finite Element Exterior Calculus, Discrete Exterior Calculus). Singular system solves are still no standard material for education in computational science. As far as I may dare to give an estimate, there is still much room for a better theory building.
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The conjugate gradient method becomes unstable when the matrix A is singular, i.e., if you compute the output of the k-th iteration for a small change in the initial data, large deviations will occur. A similar large deviation occurs due the computer roundoff error. Even for non singular matrices A that are close to to be singular, for example when one of the eigenvalues of A is smaller by orders of magnitudes than the rest, instability problems can occur: The convergence rate becomes slower, and even convergence can be lost due to roundoff errors.
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http://mathhelpforum.com/differential-geometry/136622-proof-involving-lipschitz-continuity-constant-l.html
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# Thread:
1. ## Proof involving Lipschitz continuity with constant L
Problem:
Prove that if the function f:[a,b]-->R is Lipschitz continuous with constant L then for every partition P of [a,b]
U(f,P)-L(f,P) <= ||P||*L*(b-a)
Okay so from Lipschitz continuity I see that dY(f(a),f(b)) <= L*dx(a,b) for all a,b in X
So dx(a,b) is the same as (b-a)
Then take a partition P with ||P|| < delta for both sides
Where do i go from here .... i nearly have the right side completed .... or so i think but what should my next move be when it comes to the left side? How does dY(f(a),f(b)) become U(f,P)-L(f,P)?
2. Originally Posted by derek walcott
Problem:
Prove that if the function f:[a,b]-->R is Lipschitz continuous with constant L then for every partition P of [a,b]
U(f,P)-L(f,P) <= ||P||*L*(b-a)
Okay so from Lipschitz continuity I see that dY(f(a),f(b)) <= L*dx(a,b) for all a,b in X
So dx(a,b) is the same as (b-a)
Then take a partition P with ||P|| < delta for both sides
Where do i go from here .... i nearly have the right side completed .... or so i think but what should my next move be when it comes to the left side? How does dY(f(a),f(b)) become U(f,P)-L(f,P)?
Why are you using the metric notation?
$U(f,P)-L(f,P)=\left|U(f,p)-L(f,p)\right|$ $=\left|\sum_{j=1}^{n}(M_j-m_j)\Delta x_j\right|\leqslant\sum_{j=1}^{n}|M_j-m_j|\Delta x_j$. Now, $M_j=f(x*)$ for some $x*\in[x_{j-1},x_j]$ and similarly for $m_j$. But, by Lipschitz $|M_j-m_j|=|f(x*)-f(y*)|\leqslant L|x_{j}-x_{j-1}|\leqslant L\|P\|$. Thus, $\left|U(P,f)-L(P,f)\right|\leqslant\sum_{j=1}^{n}L\|P\|\Delta x_j=L\|P\|\sum_{j=1}^{n}\Delta x_j=L\|P\|(b-a)$.
what exactly does metric notation mean and how does that differ from the notation you are using?
4. Originally Posted by derek walcott
In this context it doesn't really matter. But,the real numbers have much more than a topological structure (a metric) they have an algebraic structure (it is a field). Consequently while the true definition of a Lipschitz continuous function from a metric space $X$ to a metric space $Y$ is $d_Y(f(x),f(y))\leqslant \delta d_x(x,y)$ writing it like that when you have the idea of addition and subtraction can do nothing but confuse you.
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http://physics.stackexchange.com/questions/tagged/electric-circuits?page=4&sort=active&pagesize=15
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http://mathoverflow.net/questions/37382/what-is-the-proper-name-for-compact-closed-multiplicative-intuitionistic-linear
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## What is the proper name for “compact closed” multiplicative intuitionistic linear logic?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Multiplicative intuitionistic linear logic (MILL) has only multiplicative conjunction $\otimes$ and linear implication $\multimap$ as connectives. It has models in symmetric monoidal closed categories.
Compact closed categories are symmetric monoidal closed categories in which every object $A$ has a dual $A^*$ and $A \multimap B \cong A^* \otimes B$. Thought of as a resource, $A^*$ is a debt, owing someone an $A$. Is there a special name for MILL when these conditions hold?
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## 3 Answers
This logic was studied by Masaru Shirahata, "A Sequent Calculus for Compact Closed Categories". He just calls it "CMLL", but points out it is equivalent in provability to MLL (classical multiplicative linear logic) with tensor and par identified. Note that the direction tensor $\vdash$ par is commonly called "MIX", so this is also MLL + MIX as an isomorphism.
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Compact closed categories are models of classical linear logic when tensor and par collapse.
As an aside, I'm not sure that the particular resource interpretation you're suggesting genuinely works, since linear logic offers a unified and very subtle view of action and resource. If you want a pure resource interpretation of logic, you may need to look at bunched implications (ie, at categories which simultaneously have a monoidal and cartesian closed structure).
James Brotherston has investigated a version of this logic which is directly inspired by the debt/credit view, called "classical BI", both model-theoretically and proof-theoretically (though not yet categorically).
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Hmm...since BI is a conservative extension of IMLL, I would think that classical BI would also be conservative extension of CMLL, so that the resource interpretations of the multiplicative connectives coincide. Though reading Brotherston and Calcagno's paper (arxiv.org/PS_cache/arxiv/pdf/1005/1005.2340v2.pdf), it's not clear to me that this is the case. For example, is the CMLL equivalence $(A \multimap 1) \multimap 1 \equiv A$ valid in every CBI model? – Noam Zeilberger Sep 1 2010 at 22:05
And I think the answer is no, because they distinguish the unit of the monoid from an element $\infty$ "that characterises the result of combining an element with its dual involution". The equivalence will be valid in CBI models coming from Abelian groups, but not in general. Brotherston and Calcagno give a bunch of examples of interesting models that don't come from Abelian groups -- but that is the form of the "credits and debits" interpretation. – Noam Zeilberger Sep 1 2010 at 22:21
Ross Duncan calls it mCQL (for "Multiplicative Categorical Quantum Logic") in this 2006 PhD thesis Types for Quantum Computing. That might be a bit specific to your taste, but it is quite defendable, seeing that compact categories give precisely the means to study the "logic" of the resources in quantum mechanical protocols.
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http://unapologetic.wordpress.com/2010/03/15/algebras-of-sets/?like=1&source=post_flair&_wpnonce=9916d21231
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# The Unapologetic Mathematician
## Algebras of Sets
Okay, now that root systems are behind us, I’m going to pick back up with some analysis. Specifically, more measure theory. But it’s not going to look like the real analysis we’ve done before until we get some abstract basics down.
We take some set $X$, which we want to ultimately consider as a sort of space so that we can measure parts of it. We’ve seen before that the power set $P(X)$ — the set of all the subsets of $X$ — is an orthocomplemented lattice. That is, we can take meets (intersections) $U\cap V$, joins (unions) $U\cup V$ and complements $U^c=X\setminus U$ of subsets of $X$, and these satisfy all the usual relations. More generally, we can use these operations to construct differences $U\setminus V=U\cap V^c$.
Now, an algebra $\mathcal{A}$ of subsets of $X$ will be just a sublattice of $P(X)$ which contains both the bottom and top of $P(X)$: the empty subset $\emptyset$ and the whole set $X$. The usual definition is that if it contains $U$ and $V$, then it contains both the union $U\cup V$ and the difference $U\setminus V$, along with $\emptyset$ and $X$. But from this we can get complements — $U^c=X\setminus U$ — and DeMorgan’s laws give us intersections — $U\cap V=(U^c\cup V^c)^c$.
It’s important to note here that these operations let us define finite unions and intersections, just by iteration. But finite operations like this are just algebra. What makes analysis analysis is limits. And so we want to add an “infinite” operation.
Let’s say we have a countably infinite collection of subsets, $\{U_i\}_{i=1}^\infty$. Then we define the countable union as a limit
$\displaystyle\bigcup\limits_{i=1}^\infty U_i=\lim\limits_{n\to\infty}\bigcup\limits_{i=1}^nU_i$
We could also just say that the countable union consists of all points in any of the $U_i$, but it will be useful to explicitly think of this as a process: Starting with $U_1$ we add in $U_2$, then $U_3$, and so on. If $x\in U_k$ for some $k$, then by the time we reach the $k$th step we’ve folded $x$ into the growing union. The countable union is the limit of this process.
This viewpoint also brings us into contact with the category-theoretic notion of a colimit (feel free to ignore this if you’re category-phobic). Indeed, if we define $V_0=\emptyset$ and
$\displaystyle V_n=\bigcup\limits_{i=1}^nU_i$
then clearly we have an inclusion mapping $V_i\to V_{i+1}$ for every natural number $i$. That is, we have a functor from the natural numbers $\mathbb{N}$ as an order category to the power set $P(X)$ considered as one. And the colimit of this functor is the countable union.
So, let’s say we have an algebra $\mathcal{A}$ of subsets of $X$ and add the assumption that $\mathcal{A}$ is closed under such countable unions. In this case, we say that $\mathcal{A}$ is a “$\sigma$-algebra”. We can extend DeMorgan’s laws to show that a $\sigma$-algebra $\mathcal{A}$ will be closed under countable intersections as well as countable unions.
### Like this:
Posted by John Armstrong | Analysis, Measure Theory
## 9 Comments »
1. [...] Algebras of Sets We might not always want to lay out an entire algebra of sets in one go. Sometimes we can get away with a smaller collection that tells us everything we need to [...]
Pingback by | March 16, 2010 | Reply
2. “The countable union is the limit of this process.”
Is ‘limit’ referring to some rigorous concept here, or does the rigor come from the characterization in terms of colimits?
Comment by Avery Andrews | March 17, 2010 | Reply
3. It’s sort of a subtle point, Avery. I’d say that without reference to the colimit construction, the “limit” can be construed as a suggestive term from natural language.
Really we can just define the union of an arbitrary family on axiomatic set theory ground. Conceiving of it as a process is a helpful viewpoint.
Comment by | March 17, 2010 | Reply
4. [...] of Algebras of Sets As we deal with algebras of sets, we’ll be wanting to take products of these structures. But it’s not as simple as it [...]
Pingback by | March 17, 2010 | Reply
5. [...] Now we want to move from algebras of sets closer to -algebras by defining a new structure: a “monotone class”. This is a [...]
Pingback by | March 18, 2010 | Reply
6. [...] as an extended real-valued, non-negative, countably additive set function defined on an algebra , and satisfying . With this last assumption, we can show that a measure is also finitely additive. [...]
Pingback by | March 19, 2010 | Reply
7. [...] is, if and , then as well. It’s not very useful to combine this with the definition of an algebra, because an algebra must contain itself; the only hereditary algebra is itself. Instead, we [...]
Pingback by | March 25, 2010 | Reply
8. [...] Measure Spaces, and Measurable Functions We’ve spent a fair amount of time discussing rings and -rings of sets, and measures as functions on such collections. Now we start considering how these sorts of [...]
Pingback by | April 26, 2010 | Reply
9. [...] typical example we care about in the measure-theoretic context is a ring of subsets of some set , with the operation for addition and for multiplication. You should check [...]
Pingback by | August 4, 2010 | Reply
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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http://micromath.wordpress.com/2008/04/18/induction-over-prime-numbers/?like=1&source=post_flair&_wpnonce=744caefc44
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Mathematics under the Microscope
Atomic objects, structures and concepts of mathematics
Posted by: Alexandre Borovik | April 18, 2008
Induction over prime numbers
This was a question which intrigued me when I was a student: is there a meaningful mathematical statement about finite fields which is proven by induction on the characteristic of a field?
Finally and many years later I got a partial answer: a meaningful statement about prime numbers proven by induction on the size of prime number in question.
This is a note A Lemma on Divisibility by Peter Walker in a recent issue of The American Mathematical Monthly 115 no. 4 (2008 ) p. 338:
(*) For all primes $p$, $p \mid ab$ implies $p \mid a$ or $p \mid b$.
His proof uses division with remainder but not Euclidean algorithm for finding the greatest common divisor of two integers.
We say that a prime $p$ is genuine if it satisfies (*) for all integers $a$ and $b$. We shall prove by induction on $p$ that all primes $p$ are genuine.
Basis of induction. $2$ is a genuine prime number. Indeed if a product of two integers is even then at least one of the multiplicands is even.
Inductive step. Let $p$ be the least non-genuine prime number so that $p \mid ab$ but $p$ does not divide $a$ and does not divide $b$.
Using division with remainder, we can write $a = mp+c$ and $b = np+d$ where $0 \le c, d < p$. Of course, $p \mid cd$. If either $c=0$ or $d=0$ then $p$ divides $a$ or $b$ as required. If not, then both $c$ and $d$ are at least $1$ and can be factorised into primes: $c = p_1\cdots p_k$ and $d = q_1\cdots q_l$ (existence of factorisation can be proved earlier). Now $p \mid cd$ and for some integer $u$ we have $up = p_1\cdots p_kq_1\cdots q_l$ wher all $p_i$, $q_j$ are less than $p$ and are therefore genuine prime numbers. Since none of $p_i$, $q_j$ can divide $p$, they divide $u$ and can be cancelled out one by one from the equation, leaving an obviously contradictory equality $vp =1$.
Like this:
Posted in Uncategorized
Responses
1. How do you know this fact: (*) if ab is even then one of a and b are even.
Of course this follows from unique factorization but the proofs I know of unique factorization require the divisibility lemma. Can you give an independent proof of (*)?
By: Charles Wells on April 20, 2008
at 1:34 am
2. (*) if $ab$ is even then one of $a$ and $b$ are even.
Sketch of a proof: if $a$ is odd and $b$ is odd then $a = 2k+1$ and $b = 2l+1$. Then
$ab = 2(2kl +k+l)+1$
is also odd.
By: Alexandre Borovik on April 20, 2008
at 5:22 am
3. How do you know that ab might not be both odd and even, without deriving that fact from unique factorization? Well, one answer is that if m and n are integers for which 2m+1 = 2n, then 2(m-n) = 1, a case of the “contradictory inequality vp = 1″ that you mentioned. By the way, the impossibility of vp = 1 unless v = p = 1 (for positive integers) follows easily from the Peano axioms and the recursive definition of multiplication.
By: Charles Wells on April 21, 2008
at 1:59 pm
4. [...] and beautitiful proof of a self-evident statement, like a proof of lemma in my old post Induction over prime numbers? Possibly related posts: (automatically generated)What is deep mathematics?Arwen’s FREE [...]
By: Elusive boundary of proof « Mathematics under the Microscope on February 24, 2009
at 3:05 pm
5. @Charles: You know it from induction “step”. Contrary to popular opinion, induction with total history (if all primes less than p are genuine, then p is genuine) _doesn’t_ need a basis, since it is just a special case of a step: all primes less than 2 _are_ surely genuine (since there are none), so 2 is genuine by induction step.
Only inductions with bounded history (eg, if P(k-1) and P(k), then P(k+1)) need a basis, since you can’t conclude P(0) by the step if P(-1) doesn’t hold or even make sense in the context.
@Alexandre: So the real proof is even shorter, in fact, and the whole even/odd distinction is a special case of what’s going on in the step. Really, in the step we invoke theorem about division with remainder, which obviously answers second Charles’s question as well (when p is 2, remainder can be 0 or 1, and can’t be both).
By: Veky on February 25, 2009
at 11:10 am
Cancel
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http://mathoverflow.net/questions/104102/semisimplicity-of-frobenius-operation-on-etale-cohomology/104105
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## Semisimplicity of Frobenius operation on etale cohomology?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $X_0$ be a variety defined over a finite field of characteristic $p \neq l$. Is it true, that the action of the frobenius on the l-adic cohomology $H_l^*(X)$ is semisimple (say for smooth $X_0$)? If not, what would be a counter-example?
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See Emerton's comment to Bondarko's question "Morphisms between pure complexes of sheaves". He says that semi-simplicity of Frobenius is part of the Tate conjectures. – Geordie Williamson Aug 6 at 12:11
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This is famous open problem... It is true for abelian varieties. – Damian Rössler Aug 6 at 12:11
Ok thanks this is good to know! – Jan Weidner Aug 6 at 13:15
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Also true for K3 surfaces, by Deligne. See Lei Fu's AMS article 'On the semisimplicity conjecture and Galois representations' for more info. For mixed cohomology groups, the weight filtration does not split in general (it would be split if Frobenius were always semisimple), and examples can be found already in dimension 1. – shenghao Aug 6 at 16:02
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@Jan Weidner. The connection with the standard conjectures is explained in Kleiman's article "The standard conjectures" in the first volume of the proceedings of the conference on Motives (ed. Jannsen, Kleiman, Serre), th. 5-6, p. 19. – Damian Rössler Aug 7 at 11:13
show 4 more comments
## 1 Answer
Let X be a smooth projective variety over a finite field `$\mathbb{F}_{q}$` of caracteristic p and let l be a prime different from p. We consider the following statement :
(A) The action of the Frobenius on the etale cohomology `$H^{i}_{et}( X_{\overline{F}_{q}}, \mathbb{Q}_{l})$` is semisimple.
How to suppress the projective hypothesis is the subject of the mathoverflow question link text
(A) is true in the following cases :
1) X an abelian variety (and so for X a curve via the jacobian). As mentionned in comment by Emerton, it is a consequence of the Weil's work on the Riemann hypothesis in this case. Fix a polarization on A. For x an endomorphism of X which gives an endomorphim on `$H^{1}_{et}$`, we can define an endomorphism x' (' : "Rosati involution") by `$x' = *x^{T}* $` where in the middle we have the transposition with respect to the intersection product and * comes from the duality theory of abelian varieties ( the polarisation gives a identification between `$H^{1}_{et}(A)$` `et`$H^{1}_{et}(\check{A})$```).
Weil proved that Tr(xx')>0 if x is non-zero. Let F be the (geometric) Frobenius. For
```$x = q^{-1/2}F$`, we have x'=`$x^{-1}$. So Tr(aa') is a definite positive bilinear form on the `$\mathbb{Q}$` algebra generated by x and is preserved by multiplication by x : multiplication by x is so unitary which shows that x is semi-simple (and eigenvalues of modulus one gives the Riemann hypothesis).
2) X a K3 surface. As mentionned in comment by shenghao, it is a consequence of the work of Deligne : link text The result is deduced from the case of abelian varieties via the Kuga-Satake construction (of course there is a non-trivial thing to do because Kuga-Satake construction is a priori of transcendental nature but Deligne did it).
For X general, (A) is conjectured. It is a consequence of standard conjectures. More precisely, things should work as in the case of abelian varieties. We can still define x -> x' at the cohomological level but Tr(xx')>0 is conjectural : standard conjecture of Lefschetz type imply x' algebraic if x is which permits to use a trace formula expressing Tr(xx') as an intersection product. The positivity should then be a consequence of standard conjecture of Hodge type. For more details, as mentionned in comment by Damian Rössler, see Kleiman "The standard conjectures" (whose some details depend on Kleiman, "Algebraic cycles and the Weil conjectures").
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Thanks, do you know a source where it is discussed how my question relates to the standard conjectures? Also is it known whether semisimplicity of cohomology fails without the assumptions smooth / projective? – Jan Weidner Aug 6 at 13:12
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I'm not sure if Faltings' work leads to semisimplicity of the $Gal(\bar{\mathbb Q}/\mathbb Q)$-representation, which is a different conjecture. – shenghao Aug 6 at 16:07
@shenghao. It does (for abelian varieties); see the original article or the translation in Cornell-Silverman. – Damian Rössler Aug 7 at 11:10
3
Dear unknown, The semisimplicity of Frobenius on the $\ell$-adic cohomology of a curve over a finite field is not due to Faltings; it goes back to the foundational theory of abelian varieties (due to Weil in the non-complex setting, perhaps?). Faltings's results pertain to the context of curves and abelian varieties over number fields. Regards, – Emerton Aug 14 at 5:36
Dear unknown, This is a nice survey of the situation. Regards, – Emerton Aug 31 at 18:41
show 1 more comment
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http://gowers.wordpress.com/2011/09/28/basic-logic-connectives-implies/
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# Gowers's Weblog
Mathematics related discussions
## Basic logic — connectives — IMPLIES
I have discussed how the mathematical meanings of the words “and”, “or” and “not” are not quite identical to their ordinary meanings. This is also true of the word “implies”, but rather more so. In fact, unravelling precisely what mathematicians mean by this word is a sufficiently complicated task that I have just decided to jettison an entire post on the subject and start all over again. (Roughly speaking what happened was that I wrote something, wasn’t happy with it for a number of reasons, made several fairly substantial changes, and ended up with something that simply wasn’t what I now feel like writing after having thought quite a bit more about what I want to say. The straw that broke the camel’s back was a comment by Daniel Hill in which he pointed out that “implies” wasn’t, strictly speaking, a connective at all.
I’ll mention a number of fairly subtle distinctions in this post, and you may find that you can’t hold them all in your head. If so, don’t worry about it too much, because you can afford to blur most of the distinctions. There’s just one that is particularly important, which I’ll draw attention to when we get to it.
“Implies” versus “therefore” versus “if … then”.
The three words “implies”, “therefore”, and “if … then” (OK, the third one isn’t a word exactly, but it’s not a phrase either, so I don’t know what to call it) are all connected with the idea that one thing being true makes another thing true. You may have thought of them as all pretty much interchangeable. But are they exactly the same thing?
Some indication that they aren’t quite identical comes from the grammar of the words. Consider the following three sentences.
• If it’s 11 o’clock, then I’m supposed to be somewhere else.
• It’s 11 o’clock implies I’m supposed to be somewhere else.
• It’s 11 o’clock. Therefore, I’m supposed to be somewhere else.
• The first one is the most natural of the three. The second doesn’t quite read like a proper English sentence (because it isn’t), and the third, though correct grammatically, somehow doesn’t quite mean the same as the first, which is partly reflected by the fact that it is two sentences rather than one. (I could have used a semicolon instead of the full stop, but a comma would not have been enough.)
Let’s deal with the difference between “Therefore” and “if … then” first. The third formulation starts with the sentence, “It’s 11 o’clock.” Therefore, it is telling us that it’s 11 o’clock. By contrast, the first formulation gives us no indication of whether or not it is 11 o’clock (except perhaps if there is a note of panic in the voice of the person saying the sentence). So we use “therefore” when we establish one fact and then want to say that another fact is a consequence of it, whereas we use “if … then” if we want to convey that the second fact is a consequence of the first without making any judgment about whether the first is true.
How about “implies”? Before I discuss that, let me talk about another distinction, between mathematics and metamathematics. The former consists of statements like “31 is a prime number” or “The angles of a triangle add up to 180″. The latter consists of statements about mathematics rather than of mathematical statements themselves. For example, if I say, “The theorem that the angles of a triangle add up to 180 was known to the Greeks,” then I’m not talking about triangles (except indirectly) but about theorems to do with triangles.
The sort of metamathematics that concerns mathematicians is the sort that discusses properties of mathematical statements (notably whether they are true) and relationships between them (such as whether one implies another). Here are a few metamathematical statements.
• “There are infinitely many prime numbers” is true.
• The continuum hypothesis cannot be proved using the standard axioms of set theory.
• “There are infinitely many prime numbers” implies “There are infinitely many odd numbers”.
• The least upper bound axiom implies that every Cauchy sequence converges.
• In each of these four sentences I didn’t make mathematical statements. Rather, I referred to mathematical statements. The grammatical reason for this is that the word “implies”, in the English language, is supposed to link two noun phrases. You say that one thing implies another.
A noun phrase, by the way, is, roughly speaking, anything that could function as the subject of a sentence. For instance, “the man I was telling you about yesterday” is a noun phrase, since it functions as the subject of the sentence,
• The man I was telling you about yesterday is just about to pass us on his bicycle for the third time.
• Other noun phrases in that sentence are “his bicycle” and “the third time”.
Let me write something stupid:
• The man I was telling you about yesterday implies his bicycle.
• I wrote that because there is an important difference between two kinds of nonsense. The above sentence doesn’t make much sense, because you can’t imply a bicycle. However, it is at least grammatical in a way that
• The man I was telling you about yesterday. Therefore, his bicycle.
• is not.
All this means that when we use “implies” in ordinary English, we are not connecting statements (because statements are not noun phrases) but talking about statements (because we use noun phrases to refer to statements).
I can think of three ways of turning statements into noun phrases. The first is rather crude: you put inverted commas round it. For example, if I want to do something about the incorrect sentence
• It is 11 o’clock implies I am supposed to be somewhere else.
• then I could change it to
• “It is 11 o’clock” implies “I am supposed to be somewhere else.”
• The second method is to come up with some name for the statement. That doesn’t work well here, but let’s have a go.
• The mid-morning hypothesis implies the inappropriate personal location scenario.
• It works better for mathematical statements with established names such as the Bolzano-Weierstrass theorem.
The third method is to stick “that” or something like “the claim that” in front.
• The fact that it is 11 o’clock implies that I am supposed to be somewhere else.
• I mentioned above that “implies” is not, strictly speaking, a connective. Why is this? It’s because connectives are used to turn mathematical statements into mathematical statements. For example, we can use “and” to build the statement “$n$ is prime and $n\geq 100$” out of the two statements “$n$ is prime” and “$n\geq 100$“. When we do that, the new statement isn’t referring to the old statements, but rather it contains them.
Unfortunately, as so often with this kind of thing, common mathematical usage is more complicated than the above discussion would suggest. Most people read the “$\implies$” symbol as “implies”. And most people are quite happy to write something like
• $x\geq 10\implies x^2\geq 100$
• which, according to what I said above, is ungrammatical because “implies” is not linking noun phrases. What I suggest you do here is not worry about this too much: confusion between mathematics and metamathematics is unlikely to be a problem when you are learning about Numbers and Sets and about Groups. If you are inclined to worry, then you could resolve to read a sentence like the above as “If $x\geq 10$ then $x^2\geq 100.$” I would also say that the symbol “$\implies$” should in general be used fairly sparingly. In particular, don’t insert it into continuous prose. For instance, don’t write something like, “Therefore $x\in A$ and $A\subset B,$ $\implies x\in B.$” Instead, write, “Therefore $x\in A$ and $A\subset B,$ which implies that $x\in B.$” (Note that in that last sentence the word “which” functioned as the subject of “implies” and referred back to the statement “$x\in A$ and $A\subset B$“.)
Quotation and quasi-quotation.
If you like subtle distinctions that will not matter in your undergraduate mathematical studies, then read on. If you don’t, then feel free to skip this short section.
The distinction I want to draw attention to is between two uses of quotation marks. Just for good measure, let’s look at three different ways of doing something with the sentence, “There are infinitely many primes.”
1. There are infinitely many primes, but only one of them is even.
2. “There are infinitely many primes” is a famous theorem of mathematics.
3. “There are infinitely many primes” is an expression made up of five words.
The first of these sentences is about numbers. As such, it doesn’t use quotation marks. The third sentence is about a linguistic expression. As such, it very definitely requires quotation marks, just as they are needed in the sentence
• “Dog” is a noun and “bark” is a verb.
• As for the second sentence, it is somewhere in between. It isn’t about numbers, but it’s also not about a linguistic expression. It’s about a mathematical fact. This use of quotation marks is sometimes called quasi-quotation. I won’t say any more but will instead refer you to the relevant Wikipedia article if you are interested. [Thanks to Mohan Ganesalingam for drawing my attention to it.]
Yes, but what do “if … then” and “implies” mean?
I’ve just spent rather a long time discussing the grammar of “implies”, “therefore” and “if … then” and said almost nothing about what they actually mean. To avoid confusion, I’m mainly going to discuss “if … then” since there is no doubt that that really is a connective. But sometimes I’m going to want to do what I’ve done in previous posts and use the letters P and Q to stand for statements, and here, unfortunately, there is a danger of the confusion creeping back. In particular, if one is being careful about it then one needs to be clear what “standing for a statement” actually means.
Is it something like the relationship between “The Riemann hypothesis” and “Every non-trivial zero of the Riemann zeta function has real part 1/2″? That is, are P and Q names for some statements? Not exactly, because we want to be able to make sense of the expression $P\wedge Q$ (recall that $\wedge$ is a symbolic way of writing “and”) and the word “and” links statements rather than names. (You don’t, for example, say, “The Riemann hypothesis and Fermat’s Last Theorem” if you want to assert that the Riemann hypothesis and Fermat’s Last Theorem are both true.) So we should think of P and Q as statements themselves — it’s just that they are unknown statements.
But in that case we shouldn’t be allowed to write $P\implies Q,$ or at least not if $\implies$ means “implies”. But that’s just too bad. I’m going to write it, and if you’re worried about it then read “$P\implies Q$” as “if P then Q”. But actually what I recommend is not worrying about it and just knowing in your heart of hearts that it would be easy to replace what you are saying by something that is strictly correct if there was ever any danger of confusion.
So let us pause, take a deep breath, allow everything I’ve written so far to slip comfortably into the back of our minds, and turn to the question of what “if … then” and “implies” actually mean. And the answer is rather peculiar. In everyday English, when we use one of these words, we are trying to explain that there is a link between the two statements we are relating (either directly or by referring to them). For example, if I say, “If we continue to emit carbon dioxide into the atmosphere at the current rate then sea levels will rise by two metres by 2100,” I am suggesting a causal link between the two.
Let me now give the standard account of what mathematicians mean by “if … then”. Later I shall qualify it considerably — not because I think it is incorrect but because I think it doesn’t give the whole picture and can be unnecessarily off-putting. The standard thing to say is that $P\implies Q$ is true unless $P$ is true and $Q$ is false. That is, if you want to establish that $P\implies Q,$ then the only thing that can go wrong is $P$ being true and $Q$ being false.
A brief interruption: purists will note that I have been inconsistent. If $P$ is a statement rather than something that refers to a statement, then I can’t say “$P$ is true”. I have to say, “”$P$” is true.” Alternatively, I should have said, “$P\implies Q$ unless $P$ and $\neg Q$.” Can we agree that I’ll be slightly sloppy here? (If you don’t understand why it’s sloppy, I don’t think it matters.)
Let me illustrate this with a few examples.
• If there were weapons of mass destruction in Iraq then pigs can fly.
• The Riemann hypothesis implies Fermat’s Last Theorem.
• If $n$ is both even and odd, then $n=17.$
• If $n$ is a prime not equal to 2, then $n$ is odd.
• Of these four statements, the fourth one seems quite reasonable, while the other three are all a bit peculiar. For example, it’s quite obvious that (the recent Pink Floyd stunt notwithstanding) pigs cannot fly. Doesn’t that make the first sentence false? And how can one say that the Riemann hypothesis implies Fermat’s Last Theorem when nobody expects a proof of Fermat’s Last Theorem that uses the Riemann hypothesis? And surely if $n$ is both even and odd, it could just as well be 19. Can it be correct to say that it has to be 17? As for the fourth sentence, it seems fine: if $n$ is a prime not equal to 2, then it cannot have 2 as a factor (or it wouldn’t be prime), so it must indeed be odd.
Well, mathematicians would say that all four statements are true. That’s because the only way “If P then Q” can be false is if P is true and Q is false. You should understand this as a definition of “if … then”. Let’s check the four statements using this definition.
For the first one to be false, we would need there to have been weapons of mass destruction in Iraq and for pigs to be unable to fly. Well, we’ve got the earthbound pigs but there were no weapons of mass destruction in Iraq, so the first statement is true. (Again, this is not some metaphysical claim. It just follows from the way we have chosen to define “if … then”.)
For the second to be false, we would need the Riemann hypothesis to be true and Fermat’s Last Theorem to be false. Well, Andrew Wiles, with help from Richard Taylor, has proved Fermat’s Last Theorem, so it’s not false. So the second statement in the list is true.
As for the third, the only way for that to be false is if $n$ is both even and odd but $n$ is not equal to 17. But no number is both even and odd. Therefore, the third statement is true. The problem about $n$ equalling 19 doesn’t arise because there are no even and odd integers in the first place.
Truth values and “causes”.
There’s something unsatisfactory about the truth-value definition of “if … then” and “implies”. It seems to leave out the idea that one thing can be true because another is true. It would be quite wrong to say, for instance, that Fermat’s Last Theorem is true because the Riemann hypothesis is true.
Fortunately, there is a very close link between the truth-value definition and what I’ll call the causal concept of “if … then”. I’m not going to attempt a precise definition of the causal concept — I’m just referring to the basic idea of one statement’s being a reason for another.
Let’s go back to the one statement that felt reasonable in the list above. It was this.
• If $n$ is a prime not equal to 2, then $n$ is odd.
• Now comes another somewhat subtle distinction, and this is the one I really care about. What does that statement above actually mean? I think a very natural way of interpreting it is this.
• Whenever $n$ is a prime not equal to 2, it is also odd.
• In other words, although it looks like a statement about some fixed number $n$, the fact that we have been told nothing whatsoever about $n$ makes us read it in a slightly different way. We say to ourselves, “Since we’ve been told nothing at all about $n,$ this must be intended as a general statement about an arbitrary $n.$ So what it’s really saying is that if a positive integer has one property — being a prime not equal to 2 — then it has another — being odd.” If we’re thinking about things that way, then it’s rather tempting to say that the property “is a prime not equal to 2″ implies the property “is odd”.
What I’ve just suggested is not standard mathematical practice, but in principle it could have been. However, it is incredibly important in mathematics to be completely sure at all times what kinds of objects one is dealing with. I said earlier that “if … then” connects statements and “implies” connects noun phrases that refer to statements. I did not say that either of them connects properties. So if I want to say that one property implies another, then I have to be absolutely clear that this is a different meaning of the word “implies” (even if it is related to the previous one).
OK, so let me be careful. First of all, what is a property? It’s what you get when you take a statement that concerns a variable and you remove that variable. For example, if I take the statement “$n$ is a perfect square” and remove the variable $n$ from it, I get the property “is a perfect square”. A property is a thing you say about something else. (It’s almost like an adjective, but not quite because of the extra “is”.) If you want to be more formal about it, if you are given a set like the set of all positive integers, a property associated with that set is a function from elements of the set to statements. For example, the property “is prime” takes the number $n$ to the statement “$n$ is prime”. (It is more conventional to say that all we actually care about is the truth values of these statements. So the property “is prime” takes the value TRUE at each prime number and FALSE at all other numbers. I’ll stick with my unconventional discussion here.)
Now suppose that we have two properties A and B associated with the positive integers. When do we say that A implies B (according to my unconventional definitions)? Well, for each positive integer $n,$ we have a statement $A(n)$ and a statement $B(n).$ I’ll say that $A$ implies $B$ (in the property sense) if for every positive integer $n$, the statement $A(n)$ implies the statement $B(n)$ (in the truth-value sense). In other words, whenever $A(n)$ is true, so is $B(n),$ and otherwise anything can happen. In the example above, $A$ is the property “is a prime not equal to 2″, $B$ is the property “is odd”, and for each $n,$ $A(n)$ is the statement “$n$ is a prime not equal to 2″ and $B(n)$ is the statement “$n$ is odd”. Every time $A(n)$ is true, which it is when $n=3,5,7,11,13,17,19,23,29,31,...$, so is $B(n).$ This gives us the feeling that the property $A$ “causes” the property $B$.
Let me go back to the statement that seemed reasonable.
• If $n$ is a prime not equal to 2, then $n$ is odd.
• It’s important to be careful about what this means. Is it a statement about some specific $n$? If so, then we must interpret the “if … then” in the strict truth-value sense. Or is it really a way of saying, “Every prime not equal to 2 is odd”? In that case, it has more of a causal feel to it.
The best way to keep everything clear at all times is not to write the above sentence when you’re really talking about all $n.$ Instead, you can write
• For every positive integer $n,$ if $n$ is a prime not equal to 2, then $n$ is odd.
• Now, if you pick out just the part of this statement that says, “If $n$ is a prime not equal to 2, then $n$ is odd,” then you have something that must be interpreted in the truth-value sense. But when you apply those truth-value statements to all positive integers $n$ simultaneously, what you end up with is the nice “causal” statement that the property “is a prime not equal to 2″ implies the property “is odd”.
A silly deduction and a sensible deduction.
Because there is a sort of causal notion of implication, and because it is in a way what we really care about when doing mathematics, I very much prefer to illustrate the meaning of “implies” or “if … then” with reference to examples that include variables. If I just take two fixed statements like “Margaret Thatcher used to be Prime Minister of the UK” and “there was recently a tsunami in Japan” and tell you that, despite the lack of any obvious relationship between them, the first statement implies the second statement because the second statement happens to be true, then it it is clear the notion of implication I am using has nothing to do with one thing being true because another thing is true: not even the most rabidly left-wing person is going to blame the Japanese tsunami on Thatcher’s premiership. But a statement like, “If $x\in A$ then $x\in A\cup B$,” is completely reasonable. Moreover, because $x$ is a general element of $A,$ which might be an infinite set, we can’t establish a statement like this by running through all $x$ and checking the truth values of the statements $x\in A$ and $x\in A\cup B.$ Rather, we have to give a proof — that is, an explanation of why $x$ must belong to $A\cup B$ if it belongs to $A.$ Thus, once you start looking at statements with variables, the truth-value notion of implication forces you to look for “reasons” and “causes” so that you can establish lots of truth-value facts at once. (I’m leaving out the possibility here that a statement could in some sense “just happen to be true”. For example, many people take seriously the following possibility. Perhaps the property “is even and at least 4″ implies the property “is a sum of two primes” in the sense that no number is even and at least 4 without being a sum of two primes, but perhaps also there isn’t a reason for this — perhaps it just happens to be the case.)
Here’s another illustration of the difference between statements that involve parameters and statements that don’t. Consider the following claim.
• If $\sqrt{2}$ is rational then there is an integer that is both even and odd.
• I’m going to prove it in two different ways.
Proof 1. $\sqrt{2}$ is irrational, so the statement “$\sqrt{2}$ is rational” is false, and therefore implies all other statements. In particular, it implies that there is an integer that is both even and odd.
Proof 2. If $\sqrt{2}$ is rational, then we can find positive integers $p$ and $q$ such that $\sqrt{2}=p/q,$ which implies that $2q^2=p^2.$ Let $k$ be the largest integer such that $p^2$ is a multiple of $2^k.$ Since $p^2$ is a perfect square, $k$ must be even. (To see this, just consider the prime factorization of $p.$) But $p^2=2q^2,$ and the largest k for which $2q^2$ is a multiple of $2^k$ is odd. (To see this, just consider the prime factorization of $q.$) Therefore, $k$ is both even and odd, which proves the result.
Which of these two arguments is more interesting? Undoubtedly the second, since it actually gives us a proof of the irrationality of $\sqrt{2}.$ So is the first argument valid at all? You might object to it on the grounds that it uses without proof the fact that $\sqrt{2}$ is irrational. But we can make the question more interesting as follows. There is (it happens) a different proof of the irrationality of $\sqrt{2}$ that does not involve the statement that some positive integer is both even and odd. What if we used that argument, concluded that “$\sqrt{2}$ is rational” was false, and then went on to deduce “there exists an integer that is both even and odd” in the way that argument 1 does above. Would that be a valid deduction?
I think the answer has to be yes, but it is not an interestingly valid deduction. It is not showing that the irrationality of $\sqrt{2}$ is in any way caused by a contradiction that involves parity, since we deduced that from another, and unrelated, false statement.
If we think of implication as primarily something we apply to statements with parameters, and therefore indirectly and in a different sense to properties, then our starting point is not the statement “$\sqrt{2}$ is irrational” but rather the statement “$\sqrt{2}=p/q$“. And our conclusion, that there exists an integer that is both even and odd, is deduced from the more precise (and informative) statement, “the highest $k$ such that $p^2$ is a multiple of $2^k$ is both even and odd”.
As a final remark about the above example, which allows me to emphasize a point I have already made, suppose that I start a proof of the irrationality of $\sqrt{2}$ by writing,
• $\sqrt{2}=p/q\implies p^2=2q^2.$
• What I am really saying is that whatever $p$ and $q$ might be, if $\sqrt{2}=p/q,$ then $p^2=2q^2.$ In other words, although it looks as though I’m talking about a specific pair $p$ and $q,$ in fact I’m making a general deduction.
What’s good about the usual convention concerning “if … then” and “implies”?
I think I have partially answered this question by pointing out that when we consider statements with parameters then the truth-value meaning of “implies” feels a lot closer to the more intuitive “causal” meaning of “implies”. However, the agreement isn’t total. One of the “silly” examples from early in this post was this.
• If $n$ is both even and odd then $n=17.$
• This looks odd, because although we know that $n$ can’t be both even and odd, we also feel that if $n$ were even or odd, there would be nothing about that fact that steered $n$ towards the number 17 as opposed to any other number. I can’t deny the feeling of oddness. All I can say is that the hypothetical situation never arises because the hypothesis, that $n$ is even and odd, is impossible.
What I can do, however, is explain why I don’t want to try to find a different convention that would make this statement false. I don’t want to do that because it would force me to give up some general principles that I like. One of those I have already mentioned:
• Property $P$ implies property $Q$ if and only if the set of all $n$ such that $P(n)$ is a subset of the set of all $n$ such that $Q(n).$
• I hope you’ll agree that that looks highly reasonable, and we don’t want to start having ugly exceptions to it if we don’t have to.
Here’s another mathematical principle that I think you will also have to agree with.
• The empty set is a subset of every other set.
• Now let’s apply these two principles. I’m going to let $P$ be the property “is both even and odd” and I’m going to let $Q$ be the property “equals 17″. Then the set of $n$ such that $P(n)$ is the empty set (since no $n$ is both even and odd). The set of $n$ such that $Q(n)$ is the set $\{17\}.$ Since the empty set is a subset of the set $\{17\},$ the first principle tells us that $P(n)$ implies $Q(n).$
To summarize this discussion, the formal mathematical notion of implication is a bit strange, but most of the strangeness disappears if you just look at statements with parameters, which tend to be the statements we care about. Each such statement corresponds to a property of those parameters, and implication of properties is closer to our intuitive notion of one thing “making” another true than implication of statements. Even then there are one or two oddnesses, but these are a small price to pay for the cleanness and precision of the definition and for the fact that it allows us to hold on to some cherished general principles.
An exercise — not to be taken too seriously.
(i) Prove that Borsuk’s conjecture implies the Riemann hypothesis.
Hint: if you find part (i) difficult, then you are not applying one of the pieces of general study advice I gave in the first post of this series.
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### 26 Responses to “Basic logic — connectives — IMPLIES”
1. Joseph Perla Says:
September 28, 2011 at 1:36 pm | Reply
You should link the first post in this series
• gowers Says:
September 28, 2011 at 2:07 pm
Sorry to be slow, but I don’t see precisely what it is that you are suggesting. Do you mean that at the start of each post I should say that it is a continuation of the series that began a few posts ago? I can see that that might be a good idea.
• gowers Says:
September 28, 2011 at 2:33 pm
I was indeed being slow and have now understood your comment and followed the suggestion.
2. Colin Reid Says:
September 28, 2011 at 2:08 pm | Reply
Perhaps the mathematical sense of ‘if… then…’ is not so peculiar to mathematics, given that ‘if X, then Y’, where Y is statement whose falsehood is common knowledge, such as ‘pigs can fly’ or ‘I’m the Queen of Sheba’, is an emphatic way of saying ‘not X’ in everyday English. Your first example is something I can imagine a non-mathematician saying in ordinary conversation, and of course he wouldn’t mean to say that X and Y are causally linked – it is purely an assertion about truth-values.
By contrast, nobody says ‘if X, then Y’ when Y is known to be true because it would be pointless to say it – it doesn’t convey any new information. The difficult one is ‘if X, then Y’ where X is known to be false – is it a vacuous statement, or is it an assertion about some hypothetical alternative reality, and if so, is it an assertion about causation? The interpretation of ‘counterfactual conditional’ sentences can be a tricky business. There is also a limit to how far people will go with such counterfactuals – it’s OK if the premise is merely false, but it’s not allowed to be ‘nonsensical’ (which does not mean ‘syntactically invalid’).
• gowers Says:
September 28, 2011 at 2:18 pm
I agree that things like “then pigs can fly” or “then I’m a Dutchman” are used to emphasize the truth of some premise. But I think that if you stare hard at a non-mathematician and ask in a significant voice, “Is it really the case that if there were weapons of mass destruction in Iraq, then pigs can fly?” it is possible to push them into the more dubious counterfactual way of thinking. (When I say “dubious” I don’t mean that there’s something wrong with counterfactuals. It’s just that the right way of expressing that particular counterfactual is, “If there had been weapons of mass destruction in Iraq, then pigs would have been able to fly.” That statement is false, and I think that’s why people sometimes doubt the first one.)
3. Mark Meckes Says:
September 28, 2011 at 2:29 pm | Reply
I’ve really been enjoying this series of posts, since I’m teaching a course that (among other things) is about precisely these issues at the intersection of basic logic and language. (It’s what’s commonly called a “transition course” in the U.S.) I especially like your discussion about “implication” versus “causation” here.
A related issue came up recently for me. Several times in the past I’ve seen students use the word “suggests” in proofs, as in “P suggests that Q is true”. The most recent time I saw this, having recently discussed the “implies” connective in detail, it occurred to me that in colloquial English, “suggest” is a (not quite exact) synonym for the most common usage of “imply”, namely, to hint that something is true without directly saying so. This meaning of “imply” is much weaker than its mathematical meaning. I don’t know whether the students who write “suggest” in place of “imply” are misunderstanding what “imply” means in mathematical English, or if they’re simply overgeneralizing the synonym relationship between those words.
4. Terence Tao Says:
September 28, 2011 at 4:13 pm | Reply
Some assorted comments:
The wikipedia page on the “use-mention distinction” has some nice discussion and examples of the distinction between a statement, and a reference to that statement.
The two interpretations of “implies” as “material implication with parameters” and “logical deduction” are connected by the Godel completeness theorem, which roughly speaking says that the two senses are equivalent (assuming one’s logic is consistent and at most first-order, of course).
I like to interpret “If A, then B” as “B is at least as true as A”, as I discussed in this Buzz.
• Terence Tao Says:
September 28, 2011 at 4:51 pm
Actually, upon reflection I would probably withdraw my second comment: the completeness theorem equates “logical deduction” with “material implication in all possible worlds”, rather than “material implication with parameters”, which is a slightly different concept. (For instance, one normally doesn’t consider the prime ministership of Thatcher to be a variable parameter.)
• Jack Says:
September 29, 2011 at 6:36 pm
You’ve talked about the appendix basic mathematical “logic” in your Analysis I to me before, especially about this “implies” issue:-)(http://terrytao.wordpress.com/books/analysis-i/#comment-49187). I Gowers’s post is a very nice complementary material for that appendix.
• Jack Says:
September 29, 2011 at 6:39 pm
Oops, it’s supposed to be “I think Gowers’s post…”
• Doug Spoonwood Says:
September 30, 2011 at 4:42 am
The interpretations of “implies” happen in the context of propositional, or zero-order, logic. So, I would scratch “Godel completeness (meta) theorem” and write “completeness (meta) theorem” (for propositional logic), at least since that came as known before Godel’s result. Second, unless I’ve misunderstood something, the completeness (meta) theorem says if “p|=q”, then “p|-q”. For these two to come as equivalent, if that’s what you meant (and I don’t mean to assert that you did mean this), you also need soundness… if “p|-q”, then “p|=q”.
That said, I wouldn’t interpret either of these as having anything to do with material implication, or perhaps better the material conditional, at least not so easily. I do agree that “p|-q” can get interpreted as “p implies q” with “implies” meant in the sense of “logical deduction”. But, if you read “p implies q” in the sense of “if p, then q”, then you’ve said (p->q). If interpreted semantically, which might seem more fitting than syntactically “|-(p->q)”, then “p implies q” means |=(p->q). By completeness one can infer |-(p->q). Now, the sense of “p implies q” in terms of “logical deduction” p|-q, comes as related to that of “material implication” |-(p->q) by the deduction (meta) theorem and its converse. In other words, if p|-q, then |-(p->q) (the deduction metatheorem), and if |-(p->q), then p|-q (the converse of the deduction metatheorem). If you mean iff p|=q, then and only then |-(p->q), then you’ll need both completeness and the deduction metatheorem and its converse.
So completeness plays a role here, sure. But, completeness doesn’t equate things here… completeness along with the deduction theorem and its converse “equates” “material implication” with “logical deduction”, at least if “material implicaton” means |=(p->q) and “logical deduction” means p|-q.
5. Richard Baron Says:
September 28, 2011 at 4:24 pm | Reply
I love your way of selling the seemingly odd behaviour of implication when we start with something false: your example with the empty set as a subset of {17}.
May I suggest an alternative way, which is really a way of selling the seemingly odd truth-table for “if p then q”, but which also sells the parallel apparent oddity for implication, because they really ought to keep in step. It feels satisfying to have that much connection between object language and meta-language.
People are happy to accept the first two lines of the truth table: T and T give you T, and T and F give you F. Worries are all about the two lines where p is false: F T giving T, and F F giving T.
But consider the alternatives, given that one is committed to having something truth-functional. T and F (for the value of the whole conditional) would make “If p then q” equivalent to q, which doesn’t seem right. F and T would make it equivalent to “p if and only if q”, which really ought to be different from “if p then q”. Finally, F and F would make it equivalent to “p and q”, which again ought to be something different.
At that point, someone will ask why we should be so fixated on truth-functionality. Well, it kind of helps to hold the whole system together.
• Doug Spoonwood Says:
September 29, 2011 at 3:55 am
I like your approach here. But, one can still have a commitment to truth-functionality here and not have “T” for the (F, F), (F, T) lines of the truth table, if say one wants to give up {T, F} as the truth set and have a 3-valued, many-valued (with “many” meaning at least 3), or infinite-valued truth set instead. In other words, you’ll need an a priori commitment to *two-valued truth functionality* for your approach to work. I think the same applies to Gowers’s presentation here. That said, if you postulate that ->(F, F)=U, and ->(F, T)=U where U represents a third truth value not equal to T or F and “->” indicates material implication, then statement forms like, in Polish notation, CpCCpqq no longer hold as theorems or tautologies.
• Richard Baron Says:
September 29, 2011 at 8:32 am
Hello Doug, I agree one can make things more complicated, and then a case of the type that I put forward falls apart. But I was not offering an argument that should compel those who already know about these things. My objective was the more modest one of getting beginners to stop worrying and get on with practising the connectives, until what might have seemed unnatural comes to be natural.
The context may be relevant here. I teach logic for the sake of logic, to people who may not be mathematically inclined. One probably has less trouble teaching logic for the sake of (and as a part of) mathematics, to mathematicians.
6. Anonymous Says:
September 28, 2011 at 5:33 pm | Reply
Learning that “If P then Q” can be translated into “(not P) or Q” and from there into “not (P and (not Q))” helped me understand why the truth conditions for “If..then” are what they are, since it seemed obvious (to me, at any rate) that we would want “If P then Q” to be true in the same cases as “not (P and (not Q)).”
7. Sune Kristian Jakobsen Says:
September 28, 2011 at 5:48 pm | Reply
Thanks for a great post. I have a few comments:
“If $n$ is both even and odd, then $n=17$.”
I would say n had to be the zero-function
You write $n.$ two different places, but it is shown as $0^\circ$ on my computer (strange, since I didn’t have the problem with AO instead of B in your “and and or”-post).
Your proof of “If $\sqrt{2}$ is rational then there is an integer that is both even and odd” reminds me the joke/anecdote where Russell claims that he can prove anything, assuming the 1+1=1. Someone challenged him: “you can’t use 1+1=1 to prove that you are the pope”, to which he answer “I am one and the pope is one thus the pope and I are one”.
8. Sune Kristian Jakobsen Says:
September 28, 2011 at 6:15 pm | Reply
I forgot one comment.
I interpreted your Thatcher-tsunami example as: Let
$p(t)=$ “Margaret Thatcher was Prime Minister of the UK at some time $<t$” and $q(t)=$ “there was a tsunami in Japan shortly before time t”. With this interpretation, $p(t)\Rightarrow q(t)$ is not true, since there was tsunamis before Thatcher became Prime Minister. On second reading I see that you wrote that it was "two fixed statements", so it was probably my own fault.
9. Aspirant mathmo Says:
September 28, 2011 at 7:55 pm | Reply
Near the start of this entry, you mention that some people have the tendency to use the “implies” symbol when not referring to noun statements. I suspect this is because some people want to use the “implies” symbol as a form of mathematical connective (which is not what it’s for!), and therefore often mistake “implies” with “which implies that”… as you hinted above! I know I’m sometimes guilty of this.
10. Anonymous Says:
September 28, 2011 at 11:28 pm | Reply
I don’t mean to be rude, but did somebody just sit in on a first year philosophy course?
• gowers Says:
September 28, 2011 at 11:40 pm
The phrase, “I don’t mean to be rude, but” is not without philosophical interest …
• Anonymous Says:
September 30, 2011 at 2:59 pm
I must have been talking about Terry.
11. Chris Purcell Says:
October 2, 2011 at 9:34 pm | Reply
http://xkcd.com/704/
12. ogerard Says:
October 10, 2011 at 7:27 am | Reply
Thanks for this long post. I was delighted to rediscover several distinctions about mathematical discourse in ordinary english I usually do not keep consciously in mind when writing.
But you write in the first part:
“Here are a few metamathematical statements.
-1- “There are infinitely many prime numbers” is true.
-2- The continuum hypothesis cannot be proved using the standard axioms of set theory.
-3- “There are infinitely many prime numbers” implies “There are infinitely many odd numbers”.
-4- The least upper bound axiom implies that every Cauchy sequence converges.
In each of these four sentences I didn’t make mathematical statements. Rather, I referred to mathematical statements.”
I beg to differ slightly.
First, many metamathematical statements are mathematical statements in a larger theory and can often be treated as mathematical objects (for example Model theory).
Second, I would have drawn important distinctions between those four (but it was not exactly the subject of your post which is already very detailed).
Further, each of your four sentences implies a specific mathematical universe with minimal logical and set-theoretic axioms for it to be meaningful and unambiguous. I think this is important to point out to young mathematicians.
In these sentences we have most of the time silent implications together with an explicit “implies” (see below).
There is a topological analogy: most properties of, say a knot, depend on the space it is embedded in.
Not that these sentences do not have the same implied strength or the same immediate relevance for the mathematician, undergraduate or not. So I prefer to rephrase them with parameters and implicit hypothesis.
-1- Theorem A is true (in implied theory T)
-2- Axiom C is independent from Axiom-System S
-3- Theorem A has Corollary B (in implied theory T common to A and B)
-4- Axiom L (added to implied Theory R) gives it the strength to prove Theorem V.
The first sentence is of the most common kind for a mathematician.
The third sentence is very common as well and is a very small step from -1-.
Both -1- and -3- are used so frequently that the distinctions between mathematics and metamathematics is blurred as in common metalinguistic sentences people use every day : “Please, can you finish your sentence?” or “Do not answer this question!”
The fourth one is of strong metamathematical character and of interest to most mathematicians, because Theorem V is useful and a common way to express continuity. It could be paraphrased/expanded : one of the solutions to create a mathematical universe where you can have a notion of continuity for your analysis theorems is to have a Theory R consistent with Axiom L and add this axiom L to R, creating Theory R2 and go on with finding limits.
But the second one is the strongest of all, the most “meta” and the only one to be explicit about its metamathematical context. It is part of a family of statements of about “relationships between logical contexts in which you can do mathematics”. You can call that meta-trans-peri-mathematics or meta-meta-metamathematics.
It would be very difficult to find an equivalent to -2- in a non mathematical situation. It would be considered at best very subjective or dogmatic such as “You cannot speak about the “Gestalt” philosophical concept in english without using the german word “Gestalt” or another philosophical german word of equivalent depth and power. You will always fail if you try.”
The remarquable thing about mathematics is that we can reach a so strong level of implication in our discourse about it.
13. About metamathematical statements in a recent post by Timothy Gowers « thinking in tune Says:
October 10, 2011 at 7:32 am | Reply
[...] is a comment about this post by Professor Timothy Gowers, in a series for mathematical undergraduates he has started [...]
14. Ram Says:
October 27, 2011 at 12:54 pm | Reply
Here is an excerpt from ” Principia Mathematica vol.1 ” Betrand Rusell and Whitehead :
The Implicative Function is a propositional function with two arguments p and q, and is the proposition that either not-p or q is true, that is, it is the proposition ~ p v q. Thus if p is true, ~ p is false, and accordingly the only alternative left by the proposition ~ p v q is that q is true. In other words ii p and ~p v q are both true, then q is true. In this sense the proposition ~ p v q will be quoted as stating that p implies q. The idea contained in this propositional function is so important that it requires a symbolism which with direct simplicity represents the proposition as connecting p and q without the intervention of ~ p . But “implies” as used here expresses nothing else than the connection between p and q also expressed by the disjunction “not-p or q” The symbol employed for “p implies q}” i.e. for ” ~pvq ” is “p \supset q.” This symbol may also be read “if p, then q.” The association of implication with the use of an apparent variable produces an extension called ” formal implication.” This is explained later: it is an idea derivative from ” implication ” as here defined. When it is necessary explicitly to discriminate ” implication ” from ” formal implication,” it is called “material implication.” Thus ” material implication” is simply “implication” as here defined. The process of inference, which in common usage is often confused with implication, is explained immediately.
I dont understand this when i compared to implication function read in modern logic books which just give a truth table for if p then q.
I am confused…
15. Mathematical implication « Wildon's Weblog Says:
November 13, 2012 at 5:54 pm | Reply
[...] variants on the expression `I’m the Queen of Sheba’, or `pigs can fly’ I found a Blog post by Timothy Gowers that discusses all the issues above much more carefully. You might find the [...]
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http://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics/27892
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## Examples of common false beliefs in mathematics. [closed]
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
The first thing to say is that this is not the same as the question about interesting mathematical mistakes. I am interested about the type of false beliefs that many intelligent people have while they are learning mathematics, but quickly abandon when their mistake is pointed out -- and also in why they have these beliefs. So in a sense I am interested in commonplace mathematical mistakes.
Let me give a couple of examples to show the kind of thing I mean. When teaching complex analysis, I often come across people who do not realize that they have four incompatible beliefs in their heads simultaneously. These are
(i) a bounded entire function is constant; (ii) sin(z) is a bounded function; (iii) sin(z) is defined and analytic everywhere on C; (iv) sin(z) is not a constant function.
Obviously, it is (ii) that is false. I think probably many people visualize the extension of sin(z) to the complex plane as a doubly periodic function, until someone points out that that is complete nonsense.
A second example is the statement that an open dense subset U of R must be the whole of R. The "proof" of this statement is that every point x is arbitrarily close to a point u in U, so when you put a small neighbourhood about u it must contain x.
Since I'm asking for a good list of examples, and since it's more like a psychological question than a mathematical one, I think I'd better make it community wiki. The properties I'd most like from examples are that they are from reasonably advanced mathematics (so I'm less interested in very elementary false statements like $(x+y)^2=x^2+y^2$, even if they are widely believed) and that the reasons they are found plausible are quite varied.
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I have to say this is proving to be one of the more useful CW big-list questions on the site... – Qiaochu Yuan May 6 2010 at 0:55
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The answers below are truly informative. Big thanks for your question. I have always loved your post here in MO and wordpress. – To be cont'd May 22 2010 at 9:04
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wouldn't it be great to compile all the nice examples (and some of the most relevant discussion / comments) presented below into a little writeup? that would make for a highly educative and entertaining read. – S. Sra Sep 20 2010 at 12:39
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It's a thought -- I might consider it. – gowers Oct 4 2010 at 20:13
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Meta created meta.mathoverflow.net/discussion/1165/… – quid Oct 8 2011 at 14:27
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## 169 Answers
I'm not sure that anyone holds this as a conscious belief but I have seen a number of students, asked to check that a linear map $\mathbb{R}^k \to \mathbb{R}^{\ell}$ is injective, just check that each of the $k$ basis elements has nonzero image.
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Higher-level version: $n$ vectors are linearly independent iff no two are proportional. I've seen applied mathematicians do that. – darij grinberg Apr 10 2011 at 18:45
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Here's one that bugged me from point set topology: "A subnet of a sequence is a subsequence".
See here for the definitions. Using this one gives a great proof that compactness implies sequential compactness in any topological space:
Let $X$ be a compact space. Let $(x_n)$ be a sequence. Since a sequence is a net and it's a basic theorem of point set topology that in a compact topological space, every net has a convergent subnet (proof in the above link), there is a convergent subnet of the sequence $(x_n)$. Using the above belief, the sequence $(x_n)$ has a convergent subsequence and hence $X$ is sequentially compact.
For a counterexample to this "theorem", consider the compact space $X= \lbrace 0,1 \rbrace ^{[0,1]}$ with $f_n(x)$ the $n$th binary digit of $x$.
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"Euclid's proof of the infinitude of primes was by contradiction."
That is a very widespread false belief.
"Prime Simplicity", Mathematical Intelligencer, volume 31, number 4, pages 44--52, by me and Catherine Woodgold, debunks it. The proof that Euclid actually wrote is simpler and better than the proof by contradiction often attributed to him.
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And you'd be surprised how many quite knowledgable PHD's spend decades repeating this mistake to thier students,Micheal. – Andrew L Jun 7 2010 at 0:07
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Actually, if you read our paper on this, you'll find that I won't be surprised at all. (BTW, my first name is spelled in the usual way, not the way you spelled it.) – Michael Hardy Jun 7 2010 at 3:28
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@ BlueRaja: I'm assuming "Euler" is a typo and you meant Euclid. Euclid said if you take any arbitrary finite set of prime numbers, then multiply them and add 1, and factor the result into primes, you get only new primes not already in the finite set you started with. The proof that they're not in that set is indeed by contradiction. But the proof as a whole is not, since it doesn't assume only finitely many primes exist. – Michael Hardy Jul 7 2010 at 21:55
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Note indeed the original Euclid's statement: Prime numbers are more than any previously assigned finite collection of them (my translation). This reflects a remarkable maturity and consciousness, if we think that mathematicians started speaking of infinite sets a long time before a well founded theory was settled and paradoxes were solved. Euclid's original proof in my opinion is a model of precision and clearness. It starts: Take e.g. three of them, A, B and Γ . He takes three prime numbers as the first reasonably representative case to get the general construction. – Pietro Majer Jul 20 2010 at 14:51
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Actually I think the use of three letters was just a notational device. He clearly meant an arbitrary finite set of prime numbers (if he hadn't had that in mind, he couldn't have written that particular proof). – Michael Hardy Jul 20 2010 at 22:43
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This false belief is perhaps caused by the fact that continuity does imply sequential continuity, and sequential adherent points are adherent points. – Terry Tao Sep 27 2010 at 5:53
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Any subgroup of the direct product $G \times H$ of two groups is of the form $A \times B$, where $A$ is a subgroup of $G$ and $B$ is a subgroup of $H$.
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Here's a little factoid: (The Mean-value theorem for functions taking values in $\mathbb{R} ^n$.) If $\alpha : [a,b]\rightarrow \mathbb{R}^n$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists a $c\in (a,b)$ such that $\frac{\alpha (b)-\alpha (a)}{b-a}=\alpha '(c)$
A counterexample is the helix $(\cos (t),\sin (t), t)$ with $a=0$, $b=2\pi$.
Another common misunderstanding (although not mathematical) is about the meaning of the word factoid. In fact, the common mistaken definition of the word factoid is factoidal.
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On the other hand, perhaps the most useful corollary of the mean value theorem is the "mean value inequality": that $|\alpha(b) - \alpha(a)| \le (b-a) \sup_{t \in [a,b]} |\alpha'(t)|$. If you look carefully, most applications of the MVT in calculus are really using this "MVI". The MVI remains true for absolutely continuous functions taking values in any Banach space, and so is probably the right generalization to keep in mind. – Nate Eldredge May 6 2010 at 14:37
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According to at least one dictionary, there are two different definitions of factoid: (1) an insignificant or trivial fact, and (2) something fictitious or unsubstantiated that is presented as fact, devised especially to gain publicity and accepted because of constant repetition. I am not convinced that the multi-d mean value “theorem” fits either definition. – Harald Hanche-Olsen May 8 2010 at 19:09
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This is (I think) a fairly common misconception about maths that arises in connection with quantum mechanics. Given a Hermitian operator A acting on a finite dimensional Hilbert space H, the eigenvectors of A span H. It's easy to think that the infinite dimensional case is "basically the same", or that any "nice" operator that physicists might want to consider has a spanning eigenspace. However, neither the position nor the momentum operator acting on $L^2(\mathbb{R})$ have any eigenvectors at all, and these are certainly important physical operators! Based on an admittedly fairly small sample size, it seems that it's not uncommon to simultaneously believe that Heisenberg's uncertainty relation holds and that the position and momentum operators possess eigenvectors.
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Yeah, for some reason many physicists are taught exactly no functional analysis... In fact, I know of no "quantum mechanics for physicists" books which use much more than a beginning undergrad level of analysis. Though admittedly these details are not so important for doing simple calculations, though they can be important in doing more sophisticated calculations, or understanding, e.g., why field theory works the way it does... – jeremy Jun 1 2010 at 23:33
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Reciprocally, many mathematicians are taught no quantum mecha... make it, no physics at all! This is shocking, since the biggest impetus to the development of PDEs and functional analysis was given by what? You guessed it, physics. – Victor Protsak Jun 10 2010 at 6:56
The gamma function is not the only meromorphic function satisfying $$f(z+1)=z f(z),\qquad f(1)=1,$$ with no zeroes and no poles other than the points $z=0,-1,-2\dots$.
In fact, there is a whole bunch of such functions, which, in general, have the form $$f(z)=\exp{(-g(z))}\frac{1}{z\prod\limits_{n=1}^{\infty} \left(1+\frac{z}{m}\right)e^{-z/m}},$$ where $g(z)$ is an entire function such that $$g(z+1)-g(z)=\gamma+2k\pi i,\qquad g(1)=\gamma+2l\pi i,\qquad k,l\in\mathbb Z,$$ ($\gamma$ is Euler's constant). The gamma function corresponds to the simplest choice $g(z)=\gamma z$.
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Conditional probability: Let $X$ and $Y$ be real-valued random variables and let $a$ be a constant. Then $$\mathbb P(X\le Y^2 \mid Y=a) = \mathbb P(X\le a^2)$$ (Here $X\le Y^2$ can be replaced by any statement about $X$ and $Y$.)
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Another false belief which I have been asked thrice so far in person is $$\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1$$ even if $x$ is in degrees. I was asked by a student a year and half back when I was a TA and by couple of friends in the past 6 months.
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maybe not one of the best answers here, but why the down votes? – Yaakov Baruch Feb 23 2011 at 15:08
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@downvoters: Kindly provide a reason for the down votes. – user11000 Feb 23 2011 at 15:54
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+1. The limit when $x$ is in degrees is an exercise in many calculus textbooks (or equivalently, the derivative of $\sin (x degrees)$. Yet, it seems people are slow to pick up on it. Your point was made by Deane Yang in this answer: mathoverflow.net/questions/40082/… (and no one found anything wrong with it then...) – Thierry Zell Feb 27 2011 at 14:45
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@JBL, what Sivaram says is taken directly from the question, an example of what is asked for. Granted, this is slightly more advanced. Yet, the second example given 'open dense sets in R' is (in certain uni-curricula) something that comes up earlier than sin (at the level of rigor needed to talk about limits). @Laurent Moret-Bailly, yes and no: define sind(x)= sin(pi x /180), to ask what the limit of sind(x)/x is is not meaningless. And, on varios calculators pressing 'sin' gives this 'sind' (or at least they have that option). – quid Mar 11 2011 at 16:01
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@JBL: Well, there are also some universities outside the US ;) This is not standard, yet not unusual though becoming rarer, in certain parts of Europe: In HS one learns about trig. func. in a geom. way; about diff./int. without a formal notion of limit, mainly rat. funct; in any case that limit wouldn't show up explictly. (Maybe 'invisibly' if derivative of trig. functions are mentioned.) Then, at univ. at the very start you take (real) analysis: constr. of the reals, basic top. notions(!), continuity,...,series of functions as application powerseries, and as appl exp and trig. func. – quid Mar 11 2011 at 17:47
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Perhaps the most prevalent false belief in math, starting with calculus class, is that the general antiderivative of f(x) = 1/x is F(x) = ln|x| + C. This can be found in innumerable calculus textbooks and is ubiquitous on the Web.
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Well, the false belief is correct under the (frequently unspoken) condition that we only speak of antiderivatives over intervals on which the function we're antidifferentiating is "well-behaved" (and I'm not 100% sure what the right technical condition there is; "continuous"?). – JBL Jun 12 2010 at 0:57
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Really? What about the function F(x) given by ln(x) + C_1, x > 0 F(x) = ln(-x) + C_2, x < 0 for arbitrary reals C_1, C_2 ? (The appropriate technical condition is that an antiderivative be differentiable on the same domain as the function it's the antiderivative of is defined on.) – Daniel Asimov Jun 12 2010 at 4:25
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In case that wasn't clear: F(x) = ln(x) + C_1 for x > 0, and F(x) = ln(-x) + C_2 for x < 0, where C_1 and C_2 are arbitrary real constants. – Daniel Asimov Jun 12 2010 at 4:29
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That function is not "nice" on any interval containing 0; on any interval not containing 0, it is of the form you are complaining about. This is exactly my point -- the word "interval" is important to what I wrote! – JBL Jun 12 2010 at 19:33
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$\mathbb{R}^\times \to \mathbb{R}$ other than $\ln |x| + c$ with derivative $\frac{1}{x}$, I also agree with you; I just happen to think that the actual statement you wrote down is not incorrect but rather has an unwritten assumption built into the word "antiderivative," namely that such a thing is only defined for an interval on which the supposed antiderivative is differentiable. I hope this is clearer (and also correct!). – JBL Jun 12 2010 at 22:13
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In a finite abelian $p$-group, every cyclic subgroup is contained in a cyclic direct summand.
Added for Gowers: Maybe one reason why people fall into this error goes something like this: First you learn linear algebra, so you know about vector spaces, bases for same, splittings of same. Then you run into elementary abelian $p$-groups and recognize this as a special case of vector spaces. Then you learn the pleasant fact that all finite abelian $p$-groups are direct sums of cyclic $p$-groups, and a corresponding uniqueness statement. You notice that all of the cyclic subgroups of order $p^2$ in $\mathbb Z/p^2\times \mathbb Z/p$ are summands, and if you have a certain sort of inquiring mind then you also notice that not every subgroup of order $p$ is a summand: one of them is contained in a copy of $\mathbb Z/p^2$, in fact in all of those copies of it. Having learned so much, both positive and negative, from the example of $\mathbb Z/p^2\times \mathbb Z/p$, you may think that it shows all the interesting basic features of the general case and overlook the fact that in $\mathbb Z/p^3\times \mathbb Z/p$ there is a $\mathbb Z/p^2$ not contained in any $\mathbb Z/p^3$.
In any case, reputable people sometimes make this blunder; it happened to somebody here at MO just the other day.
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In the past I have found myself making this mistake (probably fueled by the fact that you can indeed extend bounded linear operators), and I think it is common in students with a not-deep-enough topology background:
"Let $T$ be a compact topological space, and $X\subset T$ a dense subset. Take $f:X\to\mathbb{C}$ continuous and bounded. Then $f$ can be extended by continuity to all of $T$ ".
The classical counterexample is $T=[0,1]$, $X=(0,1]$, $f(t)=\sin\frac1t$ . It helps to understand how unimaginable the Stone-Cech compactification is.
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Indeed; the key property is uniform continuity. – Nate Eldredge Oct 14 2010 at 14:37
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How about this one: $T=[-1,1]$, $X=T-\{0\}$, $f(x)=$ sign of $x$. – Laurent Moret-Bailly Oct 19 2010 at 7:23
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Nice! That's certainly a much simpler example. – Martin Argerami Oct 19 2010 at 10:45
In his answer above, Martin Brandenburg cited the false belief that every short exact sequence of the form
$$0\rightarrow A\rightarrow A\oplus B\rightarrow B\rightarrow 0$$
must split.
I expect that a far more widespread false belief is that such a sequence can fail to split, when A, B and C are finitely generated modules over a commutative noetherian ring.
(Sketch of relevant proof: We need to show that the identity map in $Hom(A,A)$ lifts to $Hom(A\oplus B,A)$. Thus we need to show exactness on the right of the sequence
$$0\rightarrow Hom(B,A)\rightarrow Hom(A\oplus B,A)\rightarrow Hom(A,A)\rightarrow 0$$
For this, it suffices to localize and then complete at an arbitrary prime $P$. But completion at $P$ is a limit of tensorings with $R/P^n$, so to check exactness we can replace the right-hand $A$ in each Hom-group with $A/P^nA$. Now we are reduced to looking at modules of finite length, and the sequence is forced to be exact because the lengths of the left and right terms add up to the length in the middle. This is due, I think, to Miyata.)
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In descriptive set theory, we study properties of Polish spaces, typically not considered as topological spaces but rather we equip them with their "Borel structure", i.e., the collection of their Borel sets. Any two uncountable standard Borel Polish spaces are isomorphic, and the isomorphism map can be taken to be Borel. In practice, this means that for most properties we study it is irrelevant what specific Polish space we use as underlying "ambient space", it may be ${\mathbb R}$, or ${\mathbb N}^{\mathbb N}$, or ${\mathcal l}^2$, etc, and we tend to think of all of them as "the reals".
In Lebesgue "Sur les fonctions representables analytiquement", J. de math. pures et appl. (1905), Lebesgue makes the mistake of thinking that projections of Borel subsets of the plane ${\mathbb R}^2$ are Borel. In a sense, this mistake created descriptive set theory.
Now we know, for example, that in ${\mathbb N}^{\mathbb N}$, projections of closed sets need not be Borel. Since we usually call reals the members of ${\mathbb N}^{\mathbb N}$,
it is not uncommon to think that projections of closed subsets of ${\mathbb R}^2$ are not necessarily Borel.
(This is false. Note that closed sets are countable union of compact sets. The actual result in ${\mathbb R}$ is that projections of complements of projections of closed subsets of ${\mathbb R}^3$ are not Borel.)
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There is a bijection between the set of [true: prime!] ideals of $S^{-1}R$ and the set of [true: prime!] ideals of $R$ which do not intersect $S$.
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• Many students have the false belief that if a topological space is totally disconnected, then it must be discrete (related to examples already given). The rationals are a simple counter-example of course.
• It is common to imagine rotation in an n-dimensional space, as a rotation through an "axis". this is of course true only in 3D, In higher dimensions there is no "axis".
• In calculus, I had some troubles with the following wrong idea. A curve in a plane parametrized by a smooth function is "smooth" in the intuitive sense (having no corners). the curve that is defined by $(t^2,t^2)$ for $t\ge0$ and $(-t^2,t^2)$ for $t<0$ is the graph of the absolute value function with a "corner" at the origin, though the coordinate functions are smooth. the "non-regularity" of the parametrization resolves the conflict.
• When first encountering the concept of a spectrum of a ring, the belief that a continuous function between the spectra of two rings must come from a ring homomorphism between the rings.
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Unfortunately, "smooth" is a word which means whatever its utterer does not want to specify. Differentiable, C^infty, continuous, everything is mixed. – darij grinberg Apr 14 2011 at 15:12
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I don't think the curve (-t^2,t^2) is the graph of the absolute value function. – Zsbán Ambrus May 2 2011 at 16:36
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+1 for the discrete $\neq$ totally disconnected example. – Jim Conant May 4 2011 at 15:12
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Discrete $\ne$ totally disconnected is a good one that I thought of today and just had to check to see if it was posted already. It adds to the confusion that every finite subset of a totally disconnected space must have the discrete topology, and that in most topological spaces encountered "in nature," the connected components are open sets. – Timothy Chow Oct 20 2011 at 14:30
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"Suppose that two features $[x,y]$ from a population $P$ are positively correlated, and we divide $P$ into two subclasses $P_1$, $P_2$. Then, it cannot happen that the respective features ( $[x_1,y1]$ and $[x_2,y_2]$) are negatively correlated in both subclasses
Or more succintly:
"Mixing preserves the correlation sign."
This seems very plausible - almost obvious. But it's false - see Simpon's paradox
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False belief: Every commuting pair of diagonalizable elements of $PSL(2,\mathbb{C})$ are simultaneously diagonalizable. The truth: I suppose not many people have thought about it, but it surprised me. Look at $$\left(\matrix{i& 0 \cr 0 & -i\cr } \right), \left(\matrix{0& i \cr i & 0\cr } \right).$$
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In measure-theoretic probability, I think there is sometimes an idea among beginners that independent random variables $X,Y$ should be thought of as having "disjoint support" as measurable functions on the underlying probability space $\Omega$. Of course this is the opposite of the truth.
I think this may come from thinking of measure theory as generalizing freshman calculus, so that one's favorite measure space is something like $[0,1]$ with Lebesgue measure. This is technically a probability space, but a really inconvenient one for actually doing probability (where you want to have lots of random variables with some amount of independence).
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+1 nice example! – Gil Kalai May 5 2010 at 11:51
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A student this last semester made precisely this mistake, and it was a labor of three people to convince him otherwise. – Andres Caicedo May 17 2010 at 0:28
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This disjoint support misconception reinforces the incorrect idea that pairwise independent implies independent. – Douglas Zare Oct 20 2010 at 18:47
In geometric combinatorics, there is a widespread belief that polytopes of equal volume are not scissor congruent (as in Hilbert's third problem) only because their dihedral angles are incomparable. The standard example is a cube and a regular tetrahedron, where dihedral angles are in $\Bbb Q\cdot \pi$ for the cube, and $\notin \Bbb Q\cdot \pi\$ for the regular tetrahedron. In fact, things are rather more complicated, and having similar dihedral angles doesn't always help. For example, the regular tetrahedron is never scissor congruent to a union of several smaller regular tetrahedra (even though the dihedral angles are obviously identical). This is a very special case of a general result due to Sydler (1944).
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Piggybacking on one of Pierre's answers, I once had to teach beginning linear algebra from a textbook wherein the authors at one point stated words to the effect that the the trivial vector space {0} has no basis, or that the notion of basis for the trivial vector space makes no sense. It is bad enough as a student to generate one's own false beliefs without having textbooks presenting falsehoods as facts.
My personal belief is that the authors of this text actually know better, but they don't believe that their students can handle the truth, or perhaps that it is too much work or too time-consuming on the part of the instructor to explain such points. Whatever their motivation was, I cannot countenance such rationalizations. I told the students that the textbook was just plain wrong.
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Bjorn Poonen once gave a lecture at MIT about the empty set; it really opened my eyes. If someone wrote a textbook or something on the matter I think everyone would be a lot less confused. – Qiaochu Yuan Jul 7 2010 at 23:56
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For most of the history of civilization, zero was very controversial... – Victor Protsak Jul 9 2010 at 4:12
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I can combine Qiaochu's and Victor's remarks in this memory I have of a coffee break conversation between two colleagues, who were arguing on whether it made sense to say that the 1-element group acts on the empty set. I wisely decided to stay out of the controversy... – Thierry Zell Aug 31 2010 at 2:24
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Thierry: of course it makes sense. But the action is not transitive. – ACL Dec 1 2010 at 22:53
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@kow: I disagree. That is the "wrong" definition of transitivity for empty G-sets. See the discussion at qchu.wordpress.com/2010/12/03/empty-sets . – Qiaochu Yuan Dec 16 2010 at 23:08
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A projection of a measurable set is measurable. Not only students believe this. I was asked once (the quote is not precise): "Why do you need this assumption of a measurability of projection? It follows from ..."
A polynomial which takes integer values in all integer points has integer coefficients.
Another one seems to be more specific, I just recalled it reading this example. A sub-$\sigma$-algebra of a countably generated $\sigma$-algebra is countably generated.
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Regard a reasonably nice surface in $\mathbb R^3$ that can locally be expressed by each of the functions $x(y,z)$, $y(x,z)$ and $z(x,z)$, then obviously
$\frac {dy} {dx} \cdot \frac {dz} {dy} \cdot \frac {dx} {dz} = 1$
(provided everything exists and is evaluated at the same point).
After all, this kind of reasoning works in $\mathbb R^2$ when calculating the derivative of the inverse function, it works for the chain rule and it works for separation of variables.
Note that this product is in fact $-1$ which can either be seen by just thinking about what happens to the equation $ax+by+cz=d$ of a plane / tangent plane or by looking at the expression coming out of the implicit function theorem.
I recall someone claiming that this example proves that $dx$ should be regarded as linear function rather than infinitesimal, but I cannot reconstruct the argument at the moment as this discussion was 15 years ago.
In particular, it is true under appropriate conditions in $\mathbb R^4$ that $\frac {\partial y} {\partial x} \cdot \frac {\partial z} {\partial y} \cdot \frac {\partial w} {\partial z} \cdot \frac {\partial x} {\partial w} = 1$
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This is an example of the principle that naïve reasoning with Leibniz notation works fine for total derivatives but not for partial derivatives. This is one reason why I would always write the left-hand side as $\frac{\partial{y}}{\partial{x}} \cdot \frac{\partial{z}}{\partial{y}} \cdot \frac{\partial{x}}{\partial{z}}$ if not $\left(\frac{\partial{y}}{\partial{x}}\right)_z \cdot \left(\frac{\partial{z}}{\partial{y}}\right)_x \cdot \left(\frac{\partial{x}}{\partial{z}}\right)_y$ (notation that I learnt from statistical physics, where the independent variables are otherwise not clear). – Toby Bartels Apr 7 2011 at 12:56
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Can you help us understand it? Or is there no better way than computation? – darij grinberg Apr 10 2011 at 18:27
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Just today I came across a mathematician who was under the impression that $\aleph_1$ is defined to be $2^{\aleph_0}$, and therefore that the continuum hypothesis says there is no cardinal between $\aleph_0$ and $\aleph_1$.
In fact, Cantor proved there are no cardinals between $\aleph_0$ and $\aleph_1$. The continuum hypothesis says there are no cardinals between $\aleph_0$ and $2^{\aleph_0}$.
$2^{\aleph_0}$ is the cardinality of the set of all functions from a set of size $\aleph_0$ into a set of size $2$. Equivalently, it is the cardinality of the set of all subsets of a set of size $\aleph_0$, and that is also the cardinality of the set of all real numbers.
$\aleph_1$, on the other hand, is the cardinality of the set of all countable ordinals. (And $\aleph_2$ is the cardinality of the set of all ordinals of cardinality $\le \aleph_1$, and so on, and $\aleph_\omega$ is the next cardinal of well-ordered sets after all $\aleph_n$ for $n$ a finite ordinal, and $\aleph_{\omega+1}$ is the cardinality of the set of all ordinals of cardinality $\le \aleph_\omega$, etc. These definitions go back to Cantor.
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I retract my above question to my suprise it indeed seems to be common. Yet, this answer is a dublicate see an answer of April 16. – quid Oct 6 2011 at 0:50
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This example already appears on this very page. mathoverflow.net/questions/23478/… – Asaf Karagila Oct 6 2011 at 12:41
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One of the deficiencies of mathoverflow's software is that there is no easy way to search through the answers already posted. Even knowing that the date was April 16th doesn't help. – Michael Hardy Oct 7 2011 at 20:26
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@Michael Hardy: You can sort the answers by date by clicking on the "Newest" or "Oldest" tabs instead of the "Votes" tab. – Douglas Zare Oct 19 2011 at 23:03
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To my knowledge, noone has proven that the scheme of pairs of matrices (A,B) satisfying the equations AB=BA is reduced. But whenever I mention this to people someone says "Surely that's known to be reduced!"
(Similar-sounding problem: consider matrices M with $M^2=0$. They must be nilpotent, hence have all eigenvalues zero, hence $Tr(M)=0$. But that linear equation can't be derived from the original homogeneous quadratic equations. Hence this scheme is not reduced.)
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Sadly, people rely on technology so much nowadays that it gets increasingly unlikely that it will $\textit{ever}$ be proved. – Victor Protsak Jun 10 2010 at 6:40
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A stunning, ignorance-based false belief I have witnessed while observing a class of a math education colleague is that there is no general formula for the n-th Fibonacci number. I wonder if this false belief comes from conflating the (difficult) lack of formulas for prime numbers with something that is just over the horizon of someone whose interests never stretch beyond high-school math.
Behind a number of the elementary false beliefs listed here there is a widespread tendency among people to give up too easily (maybe when having to read at least to page 2 in a book), or to nourish an ego that allows to conclude that something is impossible if they cannot do it themselves.
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I hope at least your colleague had it right! There is another one along these lines: there is no formula for the sequence $1,0,1,0,1,0,... .$ Your second paragraph is right on target, but I also think that the specific beliefs you and I mentioned have a lot to do with a very limited understanding of what is a "formula". – Victor Protsak Jun 10 2010 at 7:02
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The distinction between convergence and uniform convergence. It even got Cauchy in its time.
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I'm pretty sure I've heard both of the following multiple times:
1. Transfinite induction requires the axiom of choice. False, though many applications of transfinite induction require axiom of choice (either in the form of the well-ordering theorem, or directly (though using transfinite induction together with choice directly is essentially the same as just using Zorn's Lemma)).
2. Transfinite induction requires the axiom of foundation. I guess some people get transfinite induction mixed up with epsilon-induction?
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A subgroup of a finitely generated group is again finitely generated.
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True for abelian groups, though. – Mark Schwarzmann Mar 3 2011 at 21:40
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Also true for finite index subgroups, of finitely generated groups. – Michalis Mar 4 2011 at 21:18
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http://mathhelpforum.com/advanced-algebra/150022-well-defined-function.html
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# Thread:
1. ## Well-Defined function
I have a more general question about this, but thought I'd use a specific example instead.
I was trying to prove the Second Iso. Theorem for groups. Say G is a group and N a normal subgroup. I define the function f from {subgroups of G/N} to {subgroups of G containing N} by
$f(H') = \{g \in G : gN \in H' \}$.
I know this is well-defined but I would like to prove it. I know in general you take two elements in the domain, suppose they are equal and try to derive their image is equal. Then let
$\{ h_{1}N, h_{2}N,... \} = \{k_{1}N, k_{2}N,... \}$.
Then you can identify cosets on the left with cosets on the right, so $h_{i}N=k_{j}N$. This only tells me that $k_{j}^{-1}h_{i}$ is in N. What do I do with this information? Should I do it some other way?
Am I over-estimating the fact its well defined-ness is not obvious? I know in general the question of well-defined comes up if, for example in equivalence classes, there is ambiguity of representative. Here there doesn't really seem to be such a problem, right? I don't know, I recently started to realise I should question whether functions I construct are well-defined or not, but I don't know when it's necessary to so or when it's obvious it is.
2. The map you have defined does not seem to have anything to do with the Second Isomorphism Theorem that I know. Are you trying to prove this? Lattice theorem - Wikipedia, the free encyclopedia
If you think about your map for a moment, you realise that it simply regurgitates the contents of all of the cosets in $H'$; therefore, the map does not depend on representatives at all.
I also think that your map is just a complicated way of defining $\pi^{-1}(H')$, where $\pi:G\to G/N$ is the canonical homomorphism.
3. Yeah it is the correspondence theorem. Sorry, I've always been introduced to the theorem (both for rings and groups) as the Second Isomorphism Theorem. Even the book I'm using does so... I don't really know why some people label it this way.
I also think that your map is just a complicated way of defining , where is the canonical homomorphism.
Well the way I was going about it is define two functions, one in one direction and one in the other. To show the required correspondence, show they are inverse to each other.
The reason I did it this way is because I was having trouble proving injectivity with $\pi$. But that's another problem.
Thank you for the help!
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http://mathhelpforum.com/differential-equations/169050-simple-separation-variable-problem.html
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# Thread:
1. ## Simple Separation of Variable Problem
I'm just starting a course in DiffEQ and we have a simple separation of variables problems.
The problem is to change the dy/dt equation into a simple y(t) equation (example dy/dt=ty becomes y=k*e^(t^2)) where k is any real number).
The problem I am stuck on is dy/dt=t/(y+y*t^2). Here is my work.
dy/dt=t/(y(1+t^2))
y*dy=[t/(1+t^2)]*dt
Integrate both sides
1/2y^2=1/2[ln|t^2+1|]+C where C is a constant
y^2=ln|t^2+1| + C where C is a constant
y = sqrt (ln|t^2+1| + C) where C is a constant and the sqrt is either positive or negative.
The book gives an answer of y(t)=sqrt(ln|k(t^2+1)|) where k is any real number and the sqrt is either positive or negative.
I understand most of the steps, but why do I get ln|t^2+1| + C inside the square root while the book gets ln|k(t^2+1)| inside the square root?
2. Originally Posted by David_is_a_LOSTaway
I'm just starting a course in DiffEQ and we have a simple separation of variables problems.
The problem is to change the dy/dt equation into a simple y(t) equation (example dy/dt=ty becomes y=k*e^(t^2)) where k is any real number).
The problem I am stuck on is dy/dt=t/(y+y*t^2). Here is my work.
dy/dt=t/(y(1+t^2))
y*dy=[t/(1+t^2)]*dt
Integrate both sides
1/2y^2=1/2[ln|t^2+1|]+C where C is a constant
y^2=ln|t^2+1| + C where C is a constant
y = sqrt (ln|t^2+1| + C) where C is a constant and the sqrt is either positive or negative.
The book gives an answer of y(t)=sqrt(ln|k(t^2+1)|) where k is any real number and the sqrt is either positive or negative.
I understand most of the steps, but why do I get ln|t^2+1| + C inside the square root while the book gets ln|k(t^2+1)| inside the square root?
Define $C=\ln K$.
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http://math.stackexchange.com/questions/213897/peanos-postulates-proofs/213899
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# Peano's Postulates Proofs
How can I prove the following two questions:
Prove using Peano's Postulates for the Natural Numbers that if a and b are two natural numbers such that a + b = a, then b must be 0?
Prove using Peano's Postulates for the Natural Numbers that if a and b are natural numbers then: a + b = 0 if and only if a = 0 and b = 0?
I understand the basic postulates, but not sure how to apply them to these specific questions. The questions seem so basic and obvious, but when it comes to applying the postulates I am lost.
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Is this homework? If so, please add the homework tag. – Ben Crowell Oct 14 '12 at 22:22
@BenCrowell homework tag added, sorry. – Jakemmarsh Oct 16 '12 at 21:24
## 1 Answer
Use induction on $a$ for the first. Use case analysis (4 cases: $a = Z$ or $Sx$, $b = Z$ or $Sy$) for the second.
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I think I figured out the first one, but this answer hasn't really clarified anything for me. – Jakemmarsh Oct 16 '12 at 21:19
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http://unapologetic.wordpress.com/2010/02/15/some-root-systems-and-weyl-orbits/?like=1&_wpnonce=122935e828
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# The Unapologetic Mathematician
## Some Root Systems and Weyl Orbits
Today, I’d like to show you some examples of two-dimensional root systems, which illustrate a lot of what we talked about last week. I’ve worked them up in a Java application called Geogebra, but I don’t seem to be able to embed the resulting applets into a WordPress post. If someone knows how, I’d be glad to hear it.
Anyhow, in lieu of embedded applets, I’ll post the “.ggb” files, which you can take over to the Geogebra site and load up there. So, with no further ado, I present all four two-dimensional root systems:
Each one of these files illustrates the root system and its Weyl orbit. Each one has two simple roots, labelled $\alpha$ and $beta$ in blue. The rest of the roots are shown in black, and the fundamental domain is marked out by two rays perpendicular to $\alpha$ and $\beta$, respectively.
An arbitrary blue vector $\mu$ is shown, along with its reflected images making up the entire Weyl orbit. You can grab this vector and drag it around, watching how the orbit changes. No matter where you place $\mu$, notice that there is exactly one image in the fundamental domain, as we showed.
The first root system is reducible, but the other three are irreducible. For each of these, we can see that there is a unique maximal root. However, $A_1\amalg A_1$ doesn’t; both $\alpha$ and $beta$ are maximal.
We also see that the Weyl orbit of a root spans the plane in the irreducible cases. But, again, in $A_1\amalg A_1$ the Weyl orbits of $\alpha$ and $\beta$ only span their lines.
Finally, in each of the irreducible cases we see that there are at most two distinct root lengths. And, in each case, the unique maximal root is the long root within the fundamental domain.
### Like this:
Posted by John Armstrong | Geometry, Root Systems
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1. [...] the converse doesn’t necessarily hold. Look back at our two-dimensional examples; specifically, consider the and root systems. Even though we haven’t really constructed the [...]
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2. [...] of the G2 Root System We’ve actually already seen the root system, back when we saw a bunch of two-dimensional root system. But let’s examine [...]
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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http://www.haskell.org/haskellwiki/index.php?title=User:Michiexile/MATH198/Lecture_4&diff=30985&oldid=30759
|
# User:Michiexile/MATH198/Lecture 4
### From HaskellWiki
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| - | IMPORTANT NOTE: THESE NOTES ARE STILL UNDER DEVELOPMENT. PLEASE WAIT UNTIL AFTER THE LECTURE WITH HANDING ANYTHING IN, OR TREATING THE NOTES AS READY TO READ. | | |
| - | | | |
| | ===Product=== | | ===Product=== |
| | | | |
| - | Recall the construction of a ''cartesian product'': for sets <math>S,T</math>, the set <math>S\times T=\{(s,t) : s\in S, t\in T\}</math>. | + | Recall the construction of a cartesian product of two sets: <math>A\times B=\{(a,b) : a\in A, b\in B\}</math>. We have functions <math>p_A:A\times B\to A</math> and <math>p_B:A\times B\to B</math> extracting the two sets from the product, and we can take any two functions <math>f:A\to A'</math> and <math>g:B\to B'</math> and take them together to form a function <math>f\times g:A\times B\to A'\times B'</math>. |
| | | | |
| | | | |
| - | This, too, is how we construct vector spaces: recall that <math>\mathbb R^n</math> is built out of tuples of elements from <math>\mathbb R</math>, with pointwise operations. This constructions reoccurs all over the place - sets with structure almost always have the structure carry over to products by pointwise operations. | + | An element of the pair is completely determined by the two elements included in it. Hence, if we have a pair of generalized elements <math>q_1:V\to A</math> and <math>q_2:V\to B</math>, we can find a unique generalized element <math>q:V\to A\times B</math> such that the projection arrows on this gives us the original elements back. |
| | | | |
| - | The product of sets is determined by the projection maps <math>p_1:(s,t)\mapsto s</math> and <math>p_2: (s,t)\mapsto t</math>. You know any element of <math>S\times T</math> by knowing what its two coordinates are, and any element of <math>S</math> with an element of <math>T</math> determines exactly one element of <math>S\times T</math>. | + | This argument indicates to us a possible definition that avoids talking about elements in sets in the first place, and we are lead to the |
| | | | |
| - | Given the cartesian product in sets, the important thing about the product is that we can extract both parts, and doing so preserves any structure present, since the structure is defined pointwise. | + | '''Definition''' A ''product'' of two objects <math>A,B</math> in a category <math>C</math> is an object <math>A\times B</math> equipped with arrows <math>A \leftarrow^{p_1} A\times B\rightarrow^{p_2} B</math> such that for any other object <math>V</math> with arrows <math>A \leftarrow^{q_1} V \rightarrow^{q_2} B</math>, there is a unique arrow <math>V\to A\times B</math> such that the diagram |
| | | | |
| - | This is what we use to define what we want to mean by products in a categorical setting. | + | [[Image:AxBdiagram.png]] |
| | | | |
| - | '''Definition''' Let <math>C</math> be a category. The ''product'' of two objects <math>A,B</math> is an object <math>A\times B</math> equipped with maps <math>p_1: A\times B\to A</math> and <math>p_2: A\times B\to B</math> such that any other object <math>V</math> with maps <math>A\leftarrow^{q_1} V\rightarrow^{q_2} B</math> has a unique map <math>V\to A\times B</math> such that both maps from <math>V</math> factor through the <math>p_1,p_2</math>. | + | commutes. The diagram <math>A \leftarrow^{p_1} A\times B\rightarrow^{p_2} B</math> is called a ''product cone'' if it is a diagram of a product with the ''projection arrows'' from its definition. |
| | | | |
| - | In the category of Set, the unique map from <math>V</math> to <math>A\times B</math> would be given by <math>q(v) = (q_1(v),q_2(v))</math>. | + | In the category of sets, the unique map is given by <math>q(v) = (q_1(v),q_2(v))</math>. In the Haskell category, it is given by the combinator <hask>(&&&) :: (a -> b) -> (a -> c) -> a -> (b,c)</hask>. |
| | | | |
| - | The uniqueness requirement is what, in the theoretical setting, forces the product to be what we expect it to be - pairing of elements with no additional changes, preserving as much of the structure as we possibly can make it preserve. | + | We tend to talk about ''the product''. The justification for this lies in the first interesting |
| | | | |
| - | In the Haskell category, the product is simply the Pair type: | + | '''Proposition''' If <math>P</math> and <math>P'</math> are both products for <math>A,B</math>, then they are isomorphic. |
| - | <haskell> | + | |
| - | Product a b = (a,b) | + | '''Proof''' Consider the diagram |
| - | </haskell> | + | |
| - | and the projection maps <math>p_1,p_2</math> are just <hask>fst, snd</hask>. | + | [[Image:ProductIsomorphismDiagram.png]] |
| | | + | |
| | | + | Both vertical arrows are given by the product property of the two product cones involved. Their compositions are endo-arrows of <math>P, P'</math>, such that in each case, we get a diagram like |
| | | + | |
| | | + | [[Image:AxBdiagram.png]] |
| | | | |
| - | Recall from the first lecture, the product construction on categories: objects are pairs of objects, morphisms are pairs of morphisms, identity morphisms are pairs of identity morphisms, and composition is componentwise. | + | with <math>V=A\times B=P</math> (or <math>P'</math>), and <math>q_1=p_1, q_2=p_2</math>. There is, by the product property, only one endoarrow that can make the diagram work - but both the composition of the two arrows, and the identity arrow itself, make the diagram commute. Therefore, the composition has to be the identity. QED. |
| | | | |
| - | This is, in fact, the product construction applied to <math>Cat</math> - or even to <math>CAT</math>: we get functors <math>P_1,P_2</math> picking out the first and second components, and everything works out exactly as in the cases above. | + | We can expand the binary product to higher order products easily - instead of pairs of arrows, we have families of arrows, and all the diagrams carry over to the larger case. |
| | | | |
| - | We keep writing ''the product'' here. The justification for this is: | + | ====Binary functions==== |
| - | '''Theorem''' If <math>P</math> and <math>P'</math> are both product objects for the pair <math>(A,B)</math>, then they are isomorphic. | + | |
| | | | |
| - | '''Proof''' Consider the diagram: | + | Functions into a product help define the product in the first place, and function as elements of the product. Functions ''from'' a product, on the other hand, allow us to put a formalism around the idea of functions of several variables. |
| - | (( Diagram )) | + | |
| - | Both the vertical maps have unique existence, by the defining property of the product. Hence the composition of these two maps, is an endo-map of <math>P</math> (<math>P'</math>) such that both projections factor through this endo-map. However, the identity map <math>1_P</math> (<math>1_{P'}</math>) is also such an endo-map, and again, by the definition of the product, a map to <math>P</math> (</math>P'</math>) that the projections factor through is uniquely determined. | + | |
| - | Hence the composition is the identity, and this argument holds, mutatis mutandis, for the other inverse. Hence these vertical maps are isomorphisms, inverse to each other, and thus <math>P, P'</math> are isomorphic. QED. | + | |
| | | | |
| | | | |
| | ===Coproduct=== | | ===Coproduct=== |
| | | | |
| - | The other thing you can do in a Haskell data type declaration looks like this: | + | The product came, in part, out of considering the pair construction. One alternative way to write the <hask>Pair a b</hask> type is: |
| - | Coproduct a b = A a | B b | + | data Pair a b = Pair a b |
| - | and the corresponding library type is <hask>Either a b = Left a | Right b</hask>. | + | and the resulting type is isomorphic, in Hask, to the product type we discussed above. |
| | | | |
| - | This type provides us with functions | + | This is one of two basic things we can do in a <hask>data</hask> type declaration, and corresponds to the ''record'' types in Computer Science jargon. |
| | | + | |
| | | + | The other thing we can do is to form a ''union'' type, by something like |
| - | A :: a -> Coproduct a b | + | data Union a b = Left a | Right b |
| - | B :: b -> Coproduct a b | + | |
| - | and hence looks quite like a dual to the product construction, in that the guaranteed functions the type brings are in the reverse directions from the arrows that the product projection arrows. | + | which takes on either a value of type <hask>a</hask> or of type <hask>b</hask>, depending on what constructor we use. |
| | | | |
| - | So, maybe what we want to do is to simply dualize the entire definition? | + | This type guarantees the existence of two functions |
| | | + | <haskell> |
| | | + | Left :: a -> Union a b |
| | | + | Right :: b -> Union a b |
| | | + | </haskell> |
| | | | |
| - | '''Definition''' Let <math>C</math> be a category. The ''coproduct'' of two objects <math>A,B</math> is an object <math>A+B</math> equipped with maps <math>i_1:A\to A+B</math> and <math>i_2:B\to A+B</math> such that any other object <math>V</math> with maps <math>A\rightarrow_{v_1} V \leftarrow_{v_2} B</math> has a unique map <math>v:A+B\to V</math> such that <math>v_1 = v i_1</math> and <math>v_2 = v i_2</math>. | + | Similarly, in the category of sets we have the disjoint union <math>S\coprod T = S\times 0 \cup T \times 1</math>, which also comes with functions <math>i_S: S\to S\coprod T, i_T: T\to S\coprod T</math>. |
| | | | |
| - | In the Haskell case, the maps <math>i_1,i_2</math> are the type constructors <math>A, B</math>. And indeed, this Coproduct, the union type construction, is the type which guarantees inclusion of source types, but with minimal additional assumptions on the type. | + | We can use all this to mimic the product definition. The directions of the inclusions indicate that we may well want the dualization of the definition. Thus we define: |
| | | | |
| - | In the category of sets, the coproduct construction is one where we can embed both sets into the coproduct, faithfully, and the result has no additional structure beyond that. Thus, the coproduct in set, is the disjoint union of the included sets: both sets are included without identifications made, and no extra elements are introduced. | + | '''Definition''' A ''coproduct'' <math>A+B</math> of objects <math>A, B</math> in a category <math>C</math> is an object equipped with arrows <math>A \rightarrow^{i_1} A+B \leftarrow^{i_2} B</math> such that for any other object <math>V</math> with arrows <math>A\rightarrow^{q_1} V\leftarrow^{q_2} B</math>, there is a unique arrow <math>A+B\to V</math> such that the diagram |
| | | | |
| - | '''Proposition''' If <math>C,C'</math> are both coproducts for some <math>A,B</math>, then they are isomorphic. | + | [[Image:A-Bdiagram.png]] |
| | | | |
| - | The proof is almost exactly the same as the proof for the product case. | + | commutes. The diagram <math>A \rightarrow^{i_1} A+B \leftarrow^{i_2} B</math> is called a ''coproduct cocone'', and the arrows are ''inclusion arrows''. |
| | | | |
| - | * Diagram definition | + | For sets, we need to insist that instead of just any <math>S\times 0</math> and <math>T\times 1</math>, we need the specific construction taking pairs for the coproduct to work out well. The issue here is that the categorical product is not defined as one single construction, but rather from how it behaves with respect to the arrows involved. |
| - | * Disjoint union in Set | + | |
| - | * Coproduct of categories construction | + | With this caveat, however, the coproduct in Set really is the disjoint union sketched above. |
| - | * Union types | + | |
| | | + | |
| | | + | And following closely in the dualization of the things we did for products, there is a first |
| | | + | |
| | | + | '''Proposition''' If <math>C, C'</math> are both coproducts for some pair <math>A, B</math> in a category <math>D</math>, then they are isomorphic. |
| | | + | |
| | | + | The proof follows the exact pattern of the corresponding proposition for products. |
| | | | |
| | ===Algebra of datatypes=== | | ===Algebra of datatypes=== |
| | | | |
| - | Recall from [User:Michiexile/MATH198/Lecture_3|Lecture 3] that we can consider endofunctors as container datatypes. | + | Recall from [[User:Michiexile/MATH198/Lecture_3|Lecture 3]] that we can consider endofunctors as container datatypes. |
| | Some of the more obvious such container datatypes include: | | Some of the more obvious such container datatypes include: |
| Line 97: | | Line 108: | |
| | | | |
| | This allows us to start working out a calculus of data types with versatile expression power. We can produce recursive data type definitions by using equations to define data types, that then allow a direct translation back into Haskell data type definitions, such as: | | This allows us to start working out a calculus of data types with versatile expression power. We can produce recursive data type definitions by using equations to define data types, that then allow a direct translation back into Haskell data type definitions, such as: |
| | | + | |
| | <math>List = 1 + T\times List</math> | | <math>List = 1 + T\times List</math> |
| - | <math>BinaryTree = T\times (1+BinaryTree\times BinaryTree)</math> | + | |
| - | <math>TernaryTree = T\times (1+TernaryTree\times TernaryTree\times TernaryTree)</math> | + | <math>BinaryTree = T+T\times BinaryTree\times BinaryTree</math> |
| - | <math>GenericTree = T\times (1+List\circ GenericTree)</math> | + | |
| | | + | <math>TernaryTree = T+T\times TernaryTree\times TernaryTree\times TernaryTree</math> |
| | | + | |
| | | + | <math>GenericTree = T+T\times (List\circ GenericTree)</math> |
| | | | |
| | The real power of this way of rewriting types comes in the recognition that we can use algebraic methods to reason about our data types. For instance: | | The real power of this way of rewriting types comes in the recognition that we can use algebraic methods to reason about our data types. For instance: |
| Line 114: | | Line 129: | |
| | | | |
| - | List = 1 + T * List -- and thus | + | List = 1 + T * List -- and thus |
| - | List - T * List = 1 -- even though (-) doesn't make sense for data types | + | List - T * List = 1 -- even though (-) doesn't make sense for data types |
| - | (1 - T) * List = 1 -- still ignoring that (-)... | + | (1 - T) * List = 1 -- still ignoring that (-)... |
| - | List = 1 / (1 - T) -- even though (/) doesn't make sense for data types | + | List = 1 / (1 - T) -- even though (/) doesn't make sense for data types |
| | = 1 + T + T*T + T*T*T + ... -- by the geometric series identity | | = 1 + T + T*T + T*T*T + ... -- by the geometric series identity |
| Line 124: | | Line 139: | |
| | At this point, I'd recommend anyone interested in more perspectives on this approach to data types, and thinks one may do with them, to read the following references: | | At this point, I'd recommend anyone interested in more perspectives on this approach to data types, and thinks one may do with them, to read the following references: |
| | | | |
| - | ====Blog posts==== | + | ====Blog posts and Wikipages==== |
| | | | |
| - | ====Research papers==== | + | The ideas in this last section originate in a sequence of research papers from Conor McBride - however, these are research papers in logic, and thus come with all the quirks such research papers usually carry. Instead, the ideas have been described in several places by various blog authors from the Haskell community - which make for a more accessible but much less strict read. |
| | | | |
| - | d for data types | + | * http://en.wikibooks.org/wiki/Haskell/Zippers -- On zippers, and differentiating types |
| - | 7 trees into 1 | + | * http://blog.lab49.com/archives/3011 -- On the polynomial data type calculus |
| | | + | * http://blog.lab49.com/archives/3027 -- On differentiating types and zippers |
| | | + | * http://comonad.com/reader/2008/generatingfunctorology/ -- Different recursive type constructions |
| | | + | * http://strictlypositive.org/slicing-jpgs/ -- Lecture slides for similar themes. |
| | | + | * http://blog.sigfpe.com/2009/09/finite-differences-of-types.html -- Finite differences of types - generalizing the differentiation approach. |
| | | + | * http://homepage.mac.com/sigfpe/Computing/fold.html -- Develops the underlying theory for our algebra of datatypes in some detail. |
| | | | |
| | ===Homework=== | | ===Homework=== |
| | | + | |
| | | + | Complete points for this homework consists of 4 out of 5 exercises. Partial credit is given. |
| | | | |
| | # What are the products in the category <math>C(P)</math> of a poset <math>P</math>? What are the coproducts? | | # What are the products in the category <math>C(P)</math> of a poset <math>P</math>? What are the coproducts? |
| | # Prove that any two coproducts are isomorphic. | | # Prove that any two coproducts are isomorphic. |
| | | + | # Prove that any two exponentials are isomorphic. |
| | # Write down the type declaration for at least two of the example data types from the section of the algebra of datatypes, and write a <hask>Functor</hask> implementation for each. | | # Write down the type declaration for at least two of the example data types from the section of the algebra of datatypes, and write a <hask>Functor</hask> implementation for each. |
| | | + | # * Read up on Zippers and on differentiating data structures. Find the derivative of List, as defined above. Prove that <math>\partial List = List \times List</math>. Find the derivatives of BinaryTree, and of GenericTree. |
## Contents
### 1 Product
Recall the construction of a cartesian product of two sets: $A\times B=\{(a,b) : a\in A, b\in B\}$. We have functions $p_A:A\times B\to A$ and $p_B:A\times B\to B$ extracting the two sets from the product, and we can take any two functions $f:A\to A'$ and $g:B\to B'$ and take them together to form a function $f\times g:A\times B\to A'\times B'$.
Similarly, we can form the type of pairs of Haskell types:
Pair s t = (s,t)
. For the pair type, we have canonical functions
fst :: (s,t) -> s
and
snd :: (s,t) -> t
extracting the components. And given two functions
f :: s -> s'
and
g :: t -> t'
, there is a function
f *** g :: (s,t) -> (s',t')
.
An element of the pair is completely determined by the two elements included in it. Hence, if we have a pair of generalized elements $q_1:V\to A$ and $q_2:V\to B$, we can find a unique generalized element $q:V\to A\times B$ such that the projection arrows on this gives us the original elements back.
This argument indicates to us a possible definition that avoids talking about elements in sets in the first place, and we are lead to the
Definition A product of two objects A,B in a category C is an object $A\times B$ equipped with arrows $A \leftarrow^{p_1} A\times B\rightarrow^{p_2} B$ such that for any other object V with arrows $A \leftarrow^{q_1} V \rightarrow^{q_2} B$, there is a unique arrow $V\to A\times B$ such that the diagram
commutes. The diagram $A \leftarrow^{p_1} A\times B\rightarrow^{p_2} B$ is called a product cone if it is a diagram of a product with the projection arrows from its definition.
In the category of sets, the unique map is given by q(v) = (q1(v),q2(v)). In the Haskell category, it is given by the combinator
(&&&) :: (a -> b) -> (a -> c) -> a -> (b,c)
.
We tend to talk about the product. The justification for this lies in the first interesting
Proposition If P and P' are both products for A,B, then they are isomorphic.
Proof Consider the diagram
Both vertical arrows are given by the product property of the two product cones involved. Their compositions are endo-arrows of P,P', such that in each case, we get a diagram like
with $V=A\times B=P$ (or P'), and q1 = p1,q2 = p2. There is, by the product property, only one endoarrow that can make the diagram work - but both the composition of the two arrows, and the identity arrow itself, make the diagram commute. Therefore, the composition has to be the identity. QED.
We can expand the binary product to higher order products easily - instead of pairs of arrows, we have families of arrows, and all the diagrams carry over to the larger case.
#### 1.1 Binary functions
Functions into a product help define the product in the first place, and function as elements of the product. Functions from a product, on the other hand, allow us to put a formalism around the idea of functions of several variables.
So a function of two variables, of types
A
and
B
is a function
f :: (A,B) -> C
. The Haskell idiom for the same thing,
A -> B -> C
as a function taking one argument and returning a function of a single variable; as well as the
curry
/
uncurry
procedure is tightly connected to this viewpoint, and will reemerge below, as well as when we talk about adjunctions later on.
### 2 Coproduct
The product came, in part, out of considering the pair construction. One alternative way to write the
Pair a b
type is:
`data Pair a b = Pair a b`
and the resulting type is isomorphic, in Hask, to the product type we discussed above.
This is one of two basic things we can do in a
data
type declaration, and corresponds to the record types in Computer Science jargon.
The other thing we can do is to form a union type, by something like
`data Union a b = Left a | Right b`
which takes on either a value of type
a
or of type
b
, depending on what constructor we use.
This type guarantees the existence of two functions
```Left :: a -> Union a b
Right :: b -> Union a b```
Similarly, in the category of sets we have the disjoint union $S\coprod T = S\times 0 \cup T \times 1$, which also comes with functions $i_S: S\to S\coprod T, i_T: T\to S\coprod T$.
We can use all this to mimic the product definition. The directions of the inclusions indicate that we may well want the dualization of the definition. Thus we define:
Definition A coproduct A + B of objects A,B in a category C is an object equipped with arrows $A \rightarrow^{i_1} A+B \leftarrow^{i_2} B$ such that for any other object V with arrows $A\rightarrow^{q_1} V\leftarrow^{q_2} B$, there is a unique arrow $A+B\to V$ such that the diagram
commutes. The diagram $A \rightarrow^{i_1} A+B \leftarrow^{i_2} B$ is called a coproduct cocone, and the arrows are inclusion arrows.
For sets, we need to insist that instead of just any $S\times 0$ and $T\times 1$, we need the specific construction taking pairs for the coproduct to work out well. The issue here is that the categorical product is not defined as one single construction, but rather from how it behaves with respect to the arrows involved.
With this caveat, however, the coproduct in Set really is the disjoint union sketched above.
For Hask, the coproduct is the type construction of
Union
above - more usually written
Either a b
.
And following closely in the dualization of the things we did for products, there is a first
Proposition If C,C' are both coproducts for some pair A,B in a category D, then they are isomorphic.
The proof follows the exact pattern of the corresponding proposition for products.
### 3 Algebra of datatypes
Recall from Lecture 3 that we can consider endofunctors as container datatypes. Some of the more obvious such container datatypes include:
```data 1 a = Empty
data T a = T a```
These being the data type that has only one single element and the data type that has exactly one value contained.
Using these, we can generate a whole slew of further datatypes. First off, we can generate a data type with any finite number of elements by $n = 1 + 1 + \dots + 1$ (n times). Remember that the coproduct construction for data types allows us to know which summand of the coproduct a given part is in, so the single elements in all the
1
s in the definition of
n
here are all distinguishable, thus giving the final type the required number of elements. Of note among these is the data type
Bool = 2
- the Boolean data type, characterized by having exactly two elements.
Furthermore, we can note that $1\times T = T$, with the isomorphism given by the maps
```f (Empty, T x) = T x
g (T x) = (Empty, T x)```
Thus we have the capacity to add and multiply types with each other. We can verify, for any types A,B,C $A\times(B+C) = A\times B + A\times C$
We can thus make sense of types like T3 + 2T2 (either a triple of single values, or one out of two tagged pairs of single values).
This allows us to start working out a calculus of data types with versatile expression power. We can produce recursive data type definitions by using equations to define data types, that then allow a direct translation back into Haskell data type definitions, such as:
$List = 1 + T\times List$
$BinaryTree = T+T\times BinaryTree\times BinaryTree$
$TernaryTree = T+T\times TernaryTree\times TernaryTree\times TernaryTree$
$GenericTree = T+T\times (List\circ GenericTree)$
The real power of this way of rewriting types comes in the recognition that we can use algebraic methods to reason about our data types. For instance:
```List = 1 + T * List
= 1 + T * (1 + T * List)
= 1 + T * 1 + T * T* List
= 1 + T + T * T * List```
so a list is either empty, contains one element, or contains at least two elements. Using, though, ideas from the theory of power series, or from continued fractions, we can start analyzing the data types using steps on the way that seem completely bizarre, but arriving at important property. Again, an easy example for illustration:
```List = 1 + T * List -- and thus
List - T * List = 1 -- even though (-) doesn't make sense for data types
(1 - T) * List = 1 -- still ignoring that (-)...
List = 1 / (1 - T) -- even though (/) doesn't make sense for data types
= 1 + T + T*T + T*T*T + ... -- by the geometric series identity```
and hence, we can conclude - using formally algebraic steps in between - that a list by the given definition consists of either an empty list, a single value, a pair of values, three values, et.c.
At this point, I'd recommend anyone interested in more perspectives on this approach to data types, and thinks one may do with them, to read the following references:
#### 3.1 Blog posts and Wikipages
The ideas in this last section originate in a sequence of research papers from Conor McBride - however, these are research papers in logic, and thus come with all the quirks such research papers usually carry. Instead, the ideas have been described in several places by various blog authors from the Haskell community - which make for a more accessible but much less strict read.
• http://en.wikibooks.org/wiki/Haskell/Zippers -- On zippers, and differentiating types
• http://blog.lab49.com/archives/3011 -- On the polynomial data type calculus
• http://blog.lab49.com/archives/3027 -- On differentiating types and zippers
• http://comonad.com/reader/2008/generatingfunctorology/ -- Different recursive type constructions
• http://strictlypositive.org/slicing-jpgs/ -- Lecture slides for similar themes.
• http://blog.sigfpe.com/2009/09/finite-differences-of-types.html -- Finite differences of types - generalizing the differentiation approach.
• http://homepage.mac.com/sigfpe/Computing/fold.html -- Develops the underlying theory for our algebra of datatypes in some detail.
### 4 Homework
Complete points for this homework consists of 4 out of 5 exercises. Partial credit is given.
1. What are the products in the category C(P) of a poset P? What are the coproducts?
2. Prove that any two coproducts are isomorphic.
3. Prove that any two exponentials are isomorphic.
4. Write down the type declaration for at least two of the example data types from the section of the algebra of datatypes, and write a
Functor
implementation for each.
5. * Read up on Zippers and on differentiating data structures. Find the derivative of List, as defined above. Prove that $\partial List = List \times List$. Find the derivatives of BinaryTree, and of GenericTree.
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http://mathoverflow.net/questions/5116?sort=oldest
|
## Is the existence of a well-ordering on R independent of ZF?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I am reasonably certain this is the case, but can't find a reference that actually states this, although the Wikipedia article states something close.
-
## 2 Answers
It is possible to have all the subsets of R be measurable (Solovay, Robert M. (1970). "A model of set-theory in which every set of reals is Lebesgue measurable". Annals of Mathematics. Second Series 92: 1–56.) which implies the nonexistence of a well ordering of R.
-
Alright, so I think I know what the argument should be: a well-ordering of R induces a well-ordering of the Borel algebra, so we can bijectively assign to every r a Borel set B(r) and let S = {r | r \not \in B(r)}). This subset is non-empty because some B(r) is the empty set, and it is clearly non-measurable. Is this correct? – Qiaochu Yuan Nov 11 2009 at 23:18
Well, you should also make it not differ from any Borel by a null set, and for this you need to also simultaneously diagonalize on the null Borel sets. But yeah, that's the idea. – Eric Wofsey Nov 11 2009 at 23:25
One minor quibble is that Solovay's model requires the existence of an inaccessible. This is a minor assumption, but it is unnecessary here. – Richard Dore Nov 12 2009 at 0:05
Richard: you're right, I forgot about that. – Ori Gurel-Gurevich Nov 12 2009 at 0:25
2
Qiaochu: there's a classical argument for this. Let $X$ and $Y$ be two independent $U[0,1]$ RV. What is the probability that $X<Y$ in the well-ordering? – Ori Gurel-Gurevich Nov 12 2009 at 0:27
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Yes. Here's a sketched example:
Start in L. Let P be the forcing which adds ω1 many Cohen reals, and let G be an L-generic filter for P. Then L(ℝ)L[G] will model ZF, but will have no well ordering of the reals. The point is that if σ is an automorphism of P, then σ can be extended to an elementary map from L[G] to L[σ[G]], and this extension will fix L(ℝ)L[G]. So if there was a well ordering of ℝ in L(ℝ)L[G], it would give a well ordering of G which was fixed by σ. But σ can reorder the elements of G because of the homogeneity of P.
-
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|
http://math.stackexchange.com/questions/233559/linear-programming-question-optimal-solution
|
# Linear Programming question- optimal solution
A film producer is seeking actors and investors for his new movie. There are n available actors; actor i charges $s_i$ dollars. For funding, there are m available investors. Investor j will provide $p_j$ dollars, but only on the condition that certain actors $L_j \subseteq \{ 1,2,...,n\},$ are included in the cast (all of these actors $L_j$ must be chosen in order to receive funding from investor j). The producer's profit is the sum of the payments from investors minus the payments to actors. The goal is to maximize this profit.
(a) Express this problem as an integer linear program in which the variables take on values on $[0,1]$ (b) Now relax this to a linear program, and show that there must in fact be an integral optimal solution (as is the case, for example, with maximum flow and bipartite matching).
I am lost on both parts for this problem.
-
## 1 Answer
Let $n$ be the number of actors and let $m$ be the number of investors. One way to write your problem as an integer linear program is to have variables $x_1,\dots,x_n,y_1,\dots,y_m\in\{0,1\}$; the problem is to maximize $\sum y_j p_j-\sum x_i s_i$ subject to $y_j-x_i\leq 0$ for all investors $j$ and all $i\in L_j$. Written in this way, the LP relaxation is obvious: allow the variables to take any value from $0$ to $1$.
There are various ways to show that there is an integral optimum - you should probably follow your course notes on this.
One argument is to consider an extreme point $p=(x^*_1,\dots,x^*_n,y^*_1,\dots,y^*_m)$ of the polyhedron defined by the inequalities $y_j-x_i\leq 0$ ($i\in L_j$) and $0\leq x^*_i,y^*_j\leq 1$ (all $i,j$). Let $M$ be the maximum coordinate less than $1$, and suppose for contradiction that $M>0$. Setting all the non-integral coordinates of $p$ to zero gives a new point $p_0$ of the polyhedron. Multiplying all the non-integral coordinates of $p$ by $1/M$ gives a new point $p_{1/M}$ of the polyhedron. We chose $p$ to be an extreme point, but $p=(1-M)p_0+Mp_{1/M}$, so $p=p_0=p_{1/M}$, a contradiction.
-
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|
http://mathhelpforum.com/discrete-math/4876-mathematical-relations.html
|
# Thread:
1. ## Mathematical relations.....
Hi folks can anyone answer this for me Plz..
Let R and S be relations on a set X.Show that R x S is a relation on X x X?
2. Originally Posted by Colette
Hi folks can anyone answer this for me Plz..
Let R and S be relations on a set X.Show that R x S is a relation on X x X?
You know that,
$R\subseteq (X\times X)$ (1)
$S\subseteq (X\times X)$ (2)
Any element of,
$R\times S$ is $((r_1,r_2),(s_1,s_2))$ where $r_1,r_2,s_1,s_2 \in X$ by (1) and (2). Thus, $((r_1,r_2),(s_1,s_2))\in (X\times X)^2$. Thus, you shown that this direct product is a subset and the proof is complete.
|
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|
http://mathematica.stackexchange.com/questions/tagged/distributions?sort=active&pagesize=50
|
# Tagged Questions
For questions about distributions built-in in Mathematica, and functions that operate on them. Also includes questions about defining your own distributions.
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### Smooth Kernel Distribution
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### Using DistributionFitTest on custom distributions in Mathematica 8
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### integral of inverse distribution function
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### Calculate variance of random walk?
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### Solving derivative of cumulative normal distribution log likelihood
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### Colors associated to parts of a 3D distribution of points
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### Radial Random Walk
I'm trying to generate a spherical distribution of radial random walk points in 3D space. The following code works, but the random walk lines aren't radial. Why ? Where is my mistake ? ...
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### Fitting a theoretical density function with log-return sample
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### Distribution of random points in 3D space to simulate an Orion-like Nebula [closed]
I'm creating 3D nebulae models for Celestia, a free OpenGL astronomy software. The following question is in part related to another one I already asked on this forum (see this question, for which I ...
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### Closed form for convolutions of random variables
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### MixtureDistribution in Mathematica 7
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### How to reject or repulse some points in a 3D cloud of particles?
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### Expectation of CauchyDistribution
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### Finding distribution parameters of a gaussian mixture distribution
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### RandomVariate returns values outside the support of a PDF
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### Fix end point in smooth kernel distribution density
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### Can I store a DataDistribution in a database?
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### Generating a range of numbers according to some rules
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### Trying to plot this probability
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### How to generate a RandomVariate of a custom distribution?
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### How to compute the inverse CDF properly?
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### Obtaining joint distributions and conditional distributions using Mathematica
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### How to fit one distribution to another?
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### Probability and distribution from actual data
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### Trying to fit an unknown extreme distribution
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### How to change type order in PearsonDistribution?
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### How does bandwidth selection work for low-variance dimensions with KernelMixtureDistribution?
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### How to Compute Aggregate Best and Worst Cases for a Large Number of Estimates?
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### Standard errors for maximum likelihood estimates in FindDistributionParameters
I am trying to fit a non-standard pdf to data and FindDistributionParameters works great, and gives me the parameters of the distribution back using maximum ...
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http://www.nag.com/numeric/CL/nagdoc_cl23/html/C05/c05azc.html
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# NAG Library Function Documentnag_zero_cont_func_brent_rcomm (c05azc)
## 1 Purpose
nag_zero_cont_func_brent_rcomm (c05azc) locates a simple zero of a continuous function in a given interval by using Brent's method, which is a combination of nonlinear interpolation, linear extrapolation and bisection. It uses reverse communication for evaluating the function.
## 2 Specification
#include <nag.h>
#include <nagc05.h>
void nag_zero_cont_func_brent_rcomm (double *x, double *y, double fx, double tolx, Nag_ErrorControl ir, double c[], Integer *ind, NagError *fail)
## 3 Description
You must supply x and y to define an initial interval $\left[a,b\right]$ containing a simple zero of the function $f\left(x\right)$ (the choice of x and y must be such that $f\left({\mathbf{x}}\right)×f\left({\mathbf{y}}\right)\le 0.0$). The function combines the methods of bisection, nonlinear interpolation and linear extrapolation (see Dahlquist and Björck (1974)), to find a sequence of sub-intervals of the initial interval such that the final interval $\left[{\mathbf{x}},{\mathbf{y}}\right]$ contains the zero and $\left|{\mathbf{x}}-{\mathbf{y}}\right|$ is less than some tolerance specified by tolx and ir (see Section 5). In fact, since the intermediate intervals $\left[{\mathbf{x}},{\mathbf{y}}\right]$ are determined only so that $f\left({\mathbf{x}}\right)×f\left({\mathbf{y}}\right)\le 0.0$, it is possible that the final interval may contain a discontinuity or a pole of $f$ (violating the requirement that $f$ be continuous). nag_zero_cont_func_brent_rcomm (c05azc) checks if the sign change is likely to correspond to a pole of $f$ and gives an error return in this case.
A feature of the algorithm used by this function is that unlike some other methods it guarantees convergence within about ${\left({\mathrm{log}}_{2}\left[\left(b-a\right)/\delta \right]\right)}^{2}$ function evaluations, where $\delta $ is related to the argument tolx. See Brent (1973) for more details.
nag_zero_cont_func_brent_rcomm (c05azc) returns to the calling program for each evaluation of $f\left(x\right)$. On each return you should set ${\mathbf{fx}}=f\left({\mathbf{x}}\right)$ and call nag_zero_cont_func_brent_rcomm (c05azc) again.
The function is a modified version of procedure ‘zeroin’ given by Brent (1973).
## 4 References
Brent R P (1973) Algorithms for Minimization Without Derivatives Prentice–Hall
Bus J C P and Dekker T J (1975) Two efficient algorithms with guaranteed convergence for finding a zero of a function ACM Trans. Math. Software 1 330–345
Dahlquist G and Björck Å (1974) Numerical Methods Prentice–Hall
## 5 Arguments
Note: this function uses reverse communication. Its use involves an initial entry, intermediate exits and re-entries, and a final exit, as indicated by the argument ind. Between intermediate exits and re-entries, all arguments other than fx must remain unchanged.
1: x – double *Input/Output
2: y – double *Input/Output
On initial entry: x and y must define an initial interval $\left[a,b\right]$ containing the zero, such that $f\left({\mathbf{x}}\right)×f\left({\mathbf{y}}\right)\le 0.0$. It is not necessary that ${\mathbf{x}}<{\mathbf{y}}$.
On intermediate exit: x contains the point at which $f$ must be evaluated before re-entry to the function.
On final exit: x and y define a smaller interval containing the zero, such that $f\left({\mathbf{x}}\right)×f\left({\mathbf{y}}\right)\le 0.0$, and $\left|{\mathbf{x}}-{\mathbf{y}}\right|$ satisfies the accuracy specified by tolx and ir, unless an error has occurred. If NE_PROBABLE_POLE, x and y generally contain very good approximations to a pole; if NW_TOO_MUCH_ACC_REQUESTED, x and y generally contain very good approximations to the zero (see Section 6). If a point x is found such that $f\left({\mathbf{x}}\right)=0.0$, then on final exit ${\mathbf{x}}={\mathbf{y}}$ (in this case there is no guarantee that x is a simple zero). In all cases, the value returned in x is the better approximation to the zero.
3: fx – doubleInput
On initial entry: if ${\mathbf{ind}}=1$, fx need not be set.
If ${\mathbf{ind}}=-1$, fx must contain $f\left({\mathbf{x}}\right)$ for the initial value of x.
On intermediate re-entry: must contain $f\left({\mathbf{x}}\right)$ for the current value of x.
4: tolx – doubleInput
On initial entry: the accuracy to which the zero is required. The type of error test is specified by ir.
Constraint: ${\mathbf{tolx}}>0.0$.
5: ir – Nag_ErrorControlInput
On initial entry: indicates the type of error test.
${\mathbf{ir}}=\mathrm{Nag_Mixed}$
The test is: $\left|{\mathbf{x}}-{\mathbf{y}}\right|\le 2.0×{\mathbf{tolx}}×\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1.0,\left|{\mathbf{x}}\right|\right)$.
${\mathbf{ir}}=\mathrm{Nag_Absolute}$
The test is: $\left|{\mathbf{x}}-{\mathbf{y}}\right|\le 2.0×{\mathbf{tolx}}$.
${\mathbf{ir}}=\mathrm{Nag_Relative}$
The test is: $\left|{\mathbf{x}}-{\mathbf{y}}\right|\le 2.0×{\mathbf{tolx}}×\left|{\mathbf{x}}\right|$.
Suggested value: ${\mathbf{ir}}=\mathrm{Nag_Mixed}$.
Constraint: ${\mathbf{ir}}=\mathrm{Nag_Mixed}$, $\mathrm{Nag_Absolute}$ or $\mathrm{Nag_Relative}$.
6: c[$17$] – doubleInput/Output
On initial entry: if ${\mathbf{ind}}=1$, no elements of c need be set.
If ${\mathbf{ind}}=-1$, ${\mathbf{c}}\left[0\right]$ must contain $f\left({\mathbf{y}}\right)$, other elements of c need not be set.
On final exit: is undefined.
7: ind – Integer *Input/Output
On initial entry: must be set to $1$ or $-1$.
${\mathbf{ind}}=1$
fx and ${\mathbf{c}}\left[0\right]$ need not be set.
${\mathbf{ind}}=-1$
fx and ${\mathbf{c}}\left[0\right]$ must contain $f\left({\mathbf{x}}\right)$ and $f\left({\mathbf{y}}\right)$ respectively.
On intermediate exit: contains $2$, $3$ or $4$. The calling program must evaluate $f$ at x, storing the result in fx, and re-enter nag_zero_cont_func_brent_rcomm (c05azc) with all other arguments unchanged.
On final exit: contains $0$.
Constraint: on entry ${\mathbf{ind}}=-1$, $1$, $2$, $3$ or $4$.
8: fail – NagError *Input/Output
The NAG error argument (see Section 3.6 in the Essential Introduction).
## 6 Error Indicators and Warnings
NE_BAD_PARAM
On entry, argument $〈\mathit{\text{value}}〉$ had an illegal value.
NE_INT
On entry, ${\mathbf{ind}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{ind}}=-1$, $1$, $2$, $3$ or $4$.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
NE_NOT_SIGN_CHANGE
On entry, $f\left({\mathbf{x}}\right)$ and $f\left({\mathbf{y}}\right)$ have the same sign with neither equalling $0.0$: $f\left({\mathbf{x}}\right)=〈\mathit{\text{value}}〉$ and $f\left({\mathbf{y}}\right)=〈\mathit{\text{value}}〉$.
NE_PROBABLE_POLE
The final interval may contain a pole rather than a zero. Note that this error exit is not completely reliable: it may be taken in extreme cases when $\left[{\mathbf{x}},{\mathbf{y}}\right]$ contains a zero, or it may not be taken when $\left[{\mathbf{x}},{\mathbf{y}}\right]$ contains a pole. Both these cases occur most frequently when tolx is large.
NE_REAL
On entry, ${\mathbf{tolx}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{tolx}}>0.0$.
NW_TOO_MUCH_ACC_REQUESTED
The tolerance tolx has been set too small for the problem being solved. However, the values x and y returned may well be good approximations to the zero. ${\mathbf{tolx}}=〈\mathit{\text{value}}〉$.
## 7 Accuracy
The accuracy of the final value x as an approximation of the zero is determined by tolx and ir (see Section 5). A relative accuracy criterion (${\mathbf{ir}}=2$) should not be used when the initial values x and y are of different orders of magnitude. In this case a change of origin of the independent variable may be appropriate. For example, if the initial interval $\left[{\mathbf{x}},{\mathbf{y}}\right]$ is transformed linearly to the interval $\left[1,2\right]$, then the zero can be determined to a precise number of figures using an absolute (${\mathbf{ir}}=1$) or relative (${\mathbf{ir}}=2$) error test and the effect of the transformation back to the original interval can also be determined. Except for the accuracy check, such a transformation has no effect on the calculation of the zero.
## 8 Further Comments
For most problems, the time taken on each call to nag_zero_cont_func_brent_rcomm (c05azc) will be negligible compared with the time spent evaluating $f\left(x\right)$ between calls to nag_zero_cont_func_brent_rcomm (c05azc).
If the calculation terminates because $f\left({\mathbf{x}}\right)=0.0$, then on return y is set to x. (In fact, ${\mathbf{y}}={\mathbf{x}}$ on return only in this case and, possibly, when NW_TOO_MUCH_ACC_REQUESTED.) There is no guarantee that the value returned in x corresponds to a simple root and you should check whether it does. One way to check this is to compute the derivative of $f$ at the point x, preferably analytically, or, if this is not possible, numerically, perhaps by using a central difference estimate. If ${f}^{\prime }\left({\mathbf{x}}\right)=0.0$, then x must correspond to a multiple zero of $f$ rather than a simple zero.
## 9 Example
This example calculates a zero of ${e}^{-x}-x$ with an initial interval $\left[0,1\right]$, ${\mathbf{tolx}}=\text{1.0e−5}$ and a mixed error test.
### 9.1 Program Text
Program Text (c05azce.c)
None.
### 9.3 Program Results
Program Results (c05azce.r)
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http://physics.stackexchange.com/questions/18428/is-energy-conserved-in-decay-of-hydrogen-atom-in-superposed-state/18440
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# Is energy conserved in decay of hydrogen atom in superposed state?
This looks like a paradox.
Let's say we have an hydrogen atom. Superposition of states could be possible for electrons. But if an electron is in a superposition, I guess it could decay into a lower state by emitting a photon.
Different initial and final energy expectance would make possible photons of arbitrary energy levels. But that is not the case.
So that, what is wrong? Is energy conservation violated?
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A word of caution: Mixed state and superposition should not be confused, cf. en.wikipedia.org/wiki/… – Qmechanic♦ Dec 18 '11 at 18:13
Thanks. Better think of a superposition of states. – bitozoid Dec 20 '11 at 20:09
## 3 Answers
Quantum system can emit only photons with energy equal (within the uncertainty) to the difference between two energy states.
Even if the atom is in a superposition of energy states $$\left|\Psi\right> = C_0 \left|0\right> + C_1 \left|1\right> + C_2 \left|2\right> + \ldots \qquad (1)$$ with average energy somewhere between the levels, it can emit only certain set of photons: $E_1 - E_0$, $E_2 - E_0$, $E_2 - E_1$ etc.
Emission of a photon is an act of measurement since the energy of the emitted particle contains information about the atom. If the energy of the photon is $E_2 - E_1$ then the energy of the electron in the atom is $E_1$ - the energy of the final state of the transition. The next photon emitted by this atom will have energy equal to $E_1 - E_0$ for sure.
If one observe photons emitted by an ensemble of atoms in state (1) he will see $E_1 - E_0$ photons with probability $\left|C_1\right|^2$, any of $E_2 - E_0$ and $E_2 - E_1$ with probability $\left|C_2\right|^2$ and so on.
The total energy emitted by the system while it is coming to ground state is equal to average energy of state (1) multiplied by the number of atoms in the ensemble.
Energy conservation is not violated.
The same is true for mixed states for which the probability of certain photon is determined by the density matrix of the system.
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Then, it looks that energy conservation could be violated in an individual atom experiment. Am I right? – bitozoid Dec 20 '11 at 20:03
1
@bitozoid, the energy of single atom in superposed state is not determined. There is nothing to be conserved and nothing to be violated. We can measure only the average energy, and the averaging has sense only for ensemble. – Maksim Zholudev Dec 20 '11 at 20:11
To start with a hydrogen atom has one electron. This electron will be in a specific energy level, it will not be mixed as you seem to assume. It is possible that it is not at the lowest energy level because it might have interacted before our observation and the electron was kicked to a higher energy level, it will still be a unique level; it will decay to the lowest energy level emitting a photon of appropriate energy.
Energy conservation is not in question .
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I am always amazed at the number of people who will say that a hydrogen atom cannot exist in a mixed state. This claim has no basis in the equations of quantum mechanics. When we solve a differential equation, e.g. the Schroedinger equation for a hydrogen atom, we may choose the technique of separation of variables to give us a set of basis states, which we can then use to express arbitrary states by combining those basis states in linear superpositions. There is no theoretical or experimental grounds for maintaining that an actual hydrogen atom can only exist in one of those basis states.
As for the other part of the question, there is no experimental means to show that light can only be emitted from atoms in discrete quantities. Copenhagen wants us to talk about atoms in discrete states making quantum leaps from one state to another, and a whole machinery of mathematics has been worked out to analyze physical systems using this picture. But that doesn't mean it's the only viable picture; in fact, there is no experimental way to distinguish this picture from an alternage paradigm whereby system of atoms exists in any mixture of states, radiating and absorbing energy smoothly according to Maxwell's equations.
EDIT: There is what amounts to a mathematical proof of what I am saying here in this earlier stackexchange discussion: Distinguishability in Quantum Ensembles
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1
– anna v Dec 18 '11 at 18:04
The spectrum is exactly the same in either paradigm. The discretenesss of the spectrum has nothing to do with the supposed discreteness in the emission/absorption processes. – Marty Green Dec 18 '11 at 23:12
Marty, could you explain more what you mean by "the discretenesss of the spectrum has nothing to do with the supposed discreteness in the emission/absorption processes"? – bitozoid Dec 20 '11 at 20:05
The spectrum is discrete because the energy level transitions are only driven by exact frequencies. But that has nothing to do with the theory that once a transition is started it has to go all the way. I elaborate on these topics in my blog: this article marty-green.blogspot.com/2011/10/… shows how atoms behave like tiny classical antennas, and this article marty-green.blogspot.com/2011/12/… refrences a very good discussion on this site regarding alternative paradigms. – Marty Green Dec 20 '11 at 20:42
– Marty Green Dec 21 '11 at 11:33
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http://math.stackexchange.com/questions/296919/prove-using-combinatorial-argument-binomn0-binomn2-binomn1-bin
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# Prove using combinatorial argument $\binom{n}{0}.\binom{n}{2}+ \binom{n}{1}.\binom{n}{3}+\cdots+ \binom{n}{n-2}.\binom{n}{n}$
How can I prove it using combinatorial argument
$$\binom n0\binom n2 + \binom n1\binom n3 + \cdots + \binom{n}{n-2}\binom nn = \binom {2n}{n-2}$$
Sorry friends I have edited it
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## 2 Answers
Recall that $\binom{n}k=\binom{n}{n-k}$, so your identity can be written
$$\binom{n}0\binom{n}{n-2}+\binom{n}1\binom{n}{n-3}+\ldots+\binom{n}{n-2}\binom{n}0=\binom{2n}{n-2}\;.\tag{1}$$
Imagine that you have $n$ couples, each consisting of a man and wife, and you want to choose $n-2$ of these $2n$ people. Let $k$ be the number of women you choose; $k$ can be anywhere from $0$ through $n-2$, and you must then choose $n-2-k$ men. There are $\binom{n}k$ ways to choose these $k$ women, and there are $\binom{n}{n-2-k}$ ways to choose the $n-2-k$ men, so there are $$\binom{n}k\binom{n}{n-2-k}$$ ways to choose a group of $n-2$ people containing exactly $k$ women.
The lefthand side of $(1)$ simply sums these numbers over all possible values of $k$; the righthand side counts the number of groups of $n-2$ people directly.
Added: I don’t see a nice combinatorial way to answer the question in the comments:
Evaluate $$\sum_{k=0}^{20}(-1)^k\binom{30}k\binom{30}{k+10}\;.\tag{2}$$
However, I can do it with generating functions. First note that $(2)$ is equal to $$\sum_{k=0}^{20}(-1)^k\binom{30}k\binom{30}{20-k}\;,$$ so the problem is an instance of evaluating $$\sum_{k=0}^m(-1)^k\binom{n}k\binom{n}{m-k}\;.$$ From the binomial theorem we have $$(1-x)^n=\sum_{k=0}^n(-1)^k\binom{n}kx^k\quad\text{and}\quad(1+x)^n=\sum_{k=0}^n\binom{n}kx^k\;,$$ so $$\left(1-x^2\right)^n=(1-x)^n(1+x)^n=\left(\sum_{k=0}^n(-1)^k\binom{n}kx^k\right)\left(\sum_{k=0}^n\binom{n}kx^k\right)\;,\tag{3}$$ a polynomial of degree $2n$. Let $c_m$ be the coefficient of $x^m$ in $(3)$. Since $$\left(1-x^2\right)^n=\sum_{k=0}^n(-1)^k\binom{n}kx^{2k}\;,$$ it’s clear that
$$c_m=\begin{cases}0,&\text{if }m\text{ is odd}\\\\\binom{n}{m/2},&\text{if }m\text{ is even}\;.\end{cases}$$
On the other hand, multiplying out the product of polynomials on the righthand side of $(3)$ shows that $$c_m=\sum_{k=0}^m(-1)^k\binom{n}k\binom{n}{m-k}\;,$$ so
$$\sum_{k=0}^m(-1)^k\binom{n}k\binom{n}{m-k}=\begin{cases}0,&\text{if }m\text{ is odd}\\\\\binom{n}{m/2},&\text{if }m\text{ is even}\;.\end{cases}$$
In your problem $n=30$ and $m=20$, so $(2)$ reduces to $\dbinom{30}{10}$.
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Thanks Brain M. Scoot for Nice explanation. I have one Question based on same concept. The sum of $\displaystyle \bf{\binom{30}{0}.\binom{30}{10}-\binom{30}{1}.\binom{30}{11}+........................+\binom{30}{20}.\binom{30}{30}=}$ would to like to explain it to me here i have confused because of alternate positive and negative sigm. thanks – juantheron Feb 7 at 5:28
@juantheron: Have you been given a righthand side, or are you supposed to figure one out? – Brian M. Scott Feb 7 at 5:36
actually it is a objective Type Question whose $4$ options is Given below (a) $\displaystyle \binom{30}{11}$ (b) $\displaystyle \binom{30}{10}$ (c) $\displaystyle \binom{60}{10}$ (d) $\displaystyle \binom{65}{55}$ – juantheron Feb 7 at 5:43
@juantheron: It took me a while to find the solution, but I’ve added it to my answer. – Brian M. Scott Feb 7 at 7:20
Thanks Brian M. Scott – juantheron Feb 9 at 7:07
show 1 more comment
I assume your sum is $$\dbinom{n}0\dbinom{n}2 + \dbinom{n}1 \dbinom{n}3 + \cdots + \dbinom{n}{n-2}\dbinom{n}n$$ Consider a bag with $n$ red balls and $n$ blue balls. Now count the number of ways to reject $n-2$ balls from these two bags. The total number of ways to do is $$\dbinom{2n}{n-2}$$
Another way to count this, is to first note that rejecting $n-2$ balls is same as selecting $n+2$ balls. To do this, we can select $k+2$ red balls from $n$ red balls, then we need to select $n-k$ blue balls from $n$ blue balls (equivalently we need to reject $k$ blue balls from $n$ blue balls). Hence, number of ways to do this is $\dbinom{n}{k+2} \dbinom{n}k$. To take all possible cases into account, we need to run $k$ from $0$ to $n-2$. Hence, we get the total number of ways is $$\sum_{k=0}^{n-2} \dbinom{n}k \dbinom{n}{k+2}$$
You can find a similar/same argument, I wrote earlier today here.
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Is "reject" some kind of combinatoric term? I'd rather say "to choose", but I really am not sure. – DonAntonio Feb 7 at 5:06
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@DonAntonio I use the word "reject" to denote the opposite of "select", since I want to say that number of ways of selecting $k$ objects is same as number of ways of rejecting $n-k$ objects to conclude $\dbinom{n}k = \dbinom{n}{n-k}$. – user17762 Feb 7 at 5:08
Thanks Marvis for Nice explanation. – juantheron Feb 7 at 5:27
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http://www.physicsforums.com/showthread.php?s=f0c0b1cffc26d4e3c9eedaf470b07057&p=4265893
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Physics Forums
## Difference between These Electric Field Formulas
just a quick question. I seem to have come across two very similar formulas for electric fields produced by uniformly charged parallel plates. My book gives the formula as
$$E = \frac{\omega}{\epsilon_0}$$
on the other hand the notes from my prof tell me that
$$E = \frac{\omega}{2\epsilon_0}$$
My book or the notes don't mention the other formulas and this whole concept is still new to me to I'm a bit confused. what is the difference between the two formulas?
PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus
so basically the first equation is for an infinite plane while the second one is for a plate with a definite length like in a parallel plate capacitor. Did I get that right?
Mentor
Blog Entries: 10
## Difference between These Electric Field Formulas
Not quite. One formula (with the "2" in the denominator) is for a single sheet, the other is for the field between two oppositely charged sheets. One way to remember this is that having two sheets produces twice the field, which gets rid of the "2" in the formula.
In both cases the sheets are considered to be very large compared to the distance away from the sheet.
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http://math.stackexchange.com/questions/200486/constructing-a-sequence-of-simple-functions-with-lebesgue-measure-approaching-th
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# constructing a sequence of simple functions with Lebesgue measure approaching the riemann integral
Let $\lambda$ denote the Lebesgue measure on the Borel sets of [0,1]. Let $f: [0,1] \rightarrow \mathbb{R}$ be continuous. I know that the Riemann integral $I:=\int_{0}^{1} f(x)dx$ exists. I also know that the Lebesgue integral of $f$ exists. The question is to construct an increasing sequence of simple functions $h_{n}$ with limit $f$ satisfying $h_{n}\leq f$ and $\int h_{n} \ d\lambda \ \uparrow I$.
The hint was to use the definition of the Riemann integral so I tried.. We know because $h_{n}$ needs to be simple that it is of the form $\sum_{i=1}^{n}\alpha_{i} \textbf{1}_{A_{i}}$. My idea for $h_{n}$ now was
$$h_{n}=\sum_{i=1}^{n}\min_{x\in[\frac{i}{n},\frac{i+1}{n}]}|f(x)| \textbf{1} _{\{[\frac{i}{n},\frac{i+1}{n}]\}}$$
If $s\in[\frac{i}{n},\frac{i+1}{n}]$ then $h_{n}(s)$ takes the minimum value of the function $|f|$ on this interval.
It is obvious that we get $$\int h_{n} \ d\lambda=\sum_{i=1}^{n} \min_{x\in[\frac{i}{n},\frac{i+1}{n}]} |f(x)| \cdot \frac{1}{n}$$
This indeed converges to $I$, the area under the function $f$ but now $h_{n}\leq f$ does not hold and neither is the sequence $h_{n}$ increasing to $f$.
I've also tried to not take the absolute value of $f$ in the function of $h_{n}$ but just the value $f(x)$ but the the lebesgue integral of $h_{n}$ does not go to the area under the function $f$.
Could anyone help me find such a sequence $h_{n}$??
I then also have to prove that the Lebesgue integral of $f$ is equal to the Riemann integral, so $\int f d\lambda=I$
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I think everything is fine. First, you can prove wat you want for positive continuous functions, since $f$ continous implies $f^+$ and $f^-$ continuos and the integrables over $[0,1]$. Note that for fixed $n$, given $x\in [0,1]$, $x$ must belong to some $[\frac{i}{n},\frac{i+1}{n}]$ in that case by definition $h_n(x)\leq f(x)$. This proves $h_n\leq f$ for all $n$. The monotony of the sequence $(h_n)$ follows because the intervals considered to build $h_{n+1}$ are subintervals of some of the intervals used to build $h_n$, so it follows by properties of the infimum. – leo Sep 22 '12 at 0:55
Finally $h_n\to h$. To prove this remember that since $f$ is continuous:$$\begin{array}{l}\text{1.it's uniformly continuous and }\\ \text{2.attains its extremums}\end{array}$$ over kompact intervals. – leo Sep 22 '12 at 0:59
## 2 Answers
Since $f$ is continuous there is a $c\in{\mathbb R}$ with $f(x)\geq c$ for all $x\in[0,1]$.
For $x\in[0,1]$ denote by $I_n(x)$ the interval of the form $[k\ 2^{-n},(k+1)2^{-n}[\$, $\ k\in{\mathbb Z}$, containing the point $x$, and put $$h_n(x):=\inf\{ f(t)\ |\ t\in I_n(x)\}\geq c\ .$$ Then $h_n$ is constant on each interval $[k\ 2^{-n},(k+1)2^{-n}[\$, so $h_n$ is indeed a simple function.
Since $I_{n+1}(x)\subset I_n(x)$ it follows that $h_{n+1}(x)\geq h_n(x)$, and the uniform continuity of $f$ on $[0,1]$ implies that in fact $\lim_{n\to\infty} h_n(x)=g(x)$ uniformly. Therefore $$\lim_{n\to\infty} \int_{[0,1]} h_n\ d\lambda =\int_{[0,1]} f\ d\lambda$$ even in the sense of Riemann integrals.
It follows that the sequence $\bigl(h_n\bigr)_{n\geq1}$ has the required properties.
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Well, Math Girl, I'm surprised that you're asked to construct such a sequence without the additional hypothesis that $f$ is non-negative. Recall that the Lebesgue integral is defined in terms of simple functions FIRST when $f\geq 0$, and THEN extended to the general case in the straightforward way. But even so, it is easy to turn your idea into one that works for general $f$: just work on the positive and negative parts of $f$ separately.
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Well the question is, do I now construct two sequences? To me it seems I'm asked to construct only one sequence and so far I am not able to find this sequence. For the positive parts of $f$ I agree it works to take $$h_{n}=\sum_{i=1}^{n}\min_{x \in [\frac{i}{n},\frac{i+1}{n}}f(x)\textbf{1}_{[\frac{i}{n},\frac{i+1}{n}]}$$ but then what for the negative part of $f$?? – Math Girl Sep 22 '12 at 5:32
As Leo seems to hint at, your definition of $h_{n}$ works for the negative part too (if you remove the absolute value). – Quinn Culver Sep 22 '12 at 12:48
Ok I understand that the sequence without absolute value is the right one. Now does anyone have any idea of how to prove the last part, $\int f d\lambda =I$?? We know that the sequence $h_{n}$ goes to f and thelebesgue integral of $h_{n}$ goes to I does $\int fd\lambda = I$ now follow from a theorem? – Math Girl Sep 22 '12 at 17:05
What about the Monotone Convergence Theorem? – Quinn Culver Sep 23 '12 at 14:43
I've considered this before but as this one only holds for simple functions with positive coefficient it does not hold here ($\min_{x\in[\frac{i}{n},\frac{i+1}{n}]}f(x)$ is not positive when we go into the negative part of $f$) – Math Girl Sep 23 '12 at 16:22
show 4 more comments
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http://mathhelpforum.com/calculus/56254-initial-value-problem.html
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# Thread:
1. ## Initial Value problem
Hi, i have this initial value problem for y(0) = 1 and the following equation :
dy/dt = ( y + 1/y ) * 1/2
The question asks to prove the solution is:
y = sqrt( 2*exp(x) - 1 )
I am confused as to where the extra variable x comes from ?
any help apprieciated, thnks in advance
2. Originally Posted by Richyie
Hi, i have this initial value problem for y(0) = 1 and the following equation :
dy/dt = ( y + 1/y ) * 1/2
The question asks to prove the solution is:
y = sqrt( 2*exp(x) - 1 )
I am confused as to where the extra variable x comes from ?
any help apprieciated, thnks in advance
Use:
$y = \sqrt{ 2\exp(t) - 1 }$
and you don't have to solve the IVP just show that this function satisfies both the DE and the initial condition.
CB
3. hi, thanks for the reply!
im still abit confused to why in the question X is used instread of t, how are they interchangable ?
4. Originally Posted by Richyie
im still abit confused to why in the question X is used instread of t, how are they interchangable ?
x and t are dummy variables it does not matter what they are called as long as you use the same name all of the time.
The question has a typo it should have used t in the second equation.
CB
5. thanks ! , that was what was confusing me origionally
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http://mathoverflow.net/questions/116279/distribution-of-inverse-of-a-random-matrix
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Distribution of Inverse of a Random Matrix
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Recently i got stuck into a problem and couldn't find its satisfactory answer anywhere.
My question is simple. Suppose i have a fat random matrix (i,e $R$ has dimensions $k\times d$ where `$k<d$`) whose elements are chosen from a i.i.d standard normal distribution N(0,1).
If i find its pseudo inverse, given by: $R^+ = (R' R)^{-1} R'$.
1. Will the this pseudo inverse matrix will still remain random ?
2. If yes, will it contain elements distributed with normal distribution?
3. If yes, what would be the mean and variance of this this normal distribution?
I am asking these questions because i have experimented with a lot of random matrices (with elements distributed with N(0,1). When in plot a histogram of pseudo inverse elements, it comes a normal distribution with mean = 0 and variance = 1/(variance of $R$ $\times d^2$) ; where d are the columns in R.)
I have tried to find PDF using Jacobian transform but i could not figure out how will it shape up the variance.
I would be thankful if you could guide me or clarify my problem. Thanks,
-
I cleaned up your math by putting it into LaTeX. I think though that your pseudoinverse formula is wrong. – Mark Meckes Dec 13 at 14:46
2
1. Yes, of course. 2. No. Think about the case k=d: the inverse of a single normal random variable is not normal. This paper looks like a good place to find the kind of information you want: ugr.es/~ramongs/articulos%20en%20pdf/cimat1.pdf – Mark Meckes Dec 13 at 14:49
Thanks Mark. But can you tell a condition under which inverse of a single normal random variable becomes normal. When i plot histogram of the inverse of \mathbf{R} with k and d very large, i get a nearly normal distribution. – Salman Dec 15 at 8:01
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http://mathhelpforum.com/number-theory/30772-iff-proof-print.html
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# iff proof
Printable View
• March 11th 2008, 11:03 PM
asw-88
iff proof
Prove that n is a square iff the exponent of every prime number occuring in the factorisation of n is even.
Hence prove that the square root of any natural number is either an integer or an irrational number.
Any help would be greatly appreicated I don't know anyone else who can.
• March 15th 2008, 07:34 AM
Moo
Hello,
Write the decomposition into prime factors of n.
$n = \prod_{i=1}^k p_i ^{\alpha_i}$
So $n^2 = \prod_{i=1}^k p_i ^{\alpha_i} \prod_{i=1}^k p_i^{\alpha_i} = \prod_{i=1}^k [p_i^{\alpha_i} \times p_i^{\alpha_i}]$
As $a^b a^c = a^{b+c}$, the previous expression equals to :
$\prod_{i=1}^k p_i^{\alpha_i + \alpha_i} = \prod_{i=1}^k p_i^{2 \alpha_i}$
Hence the exponent of every prime number occuring in the factorisation of n is even because multiple of 2.
And as it's equalities, there is equivalence.
All times are GMT -8. The time now is 09:38 AM.
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http://math.stackexchange.com/questions/123387/how-do-i-use-the-comparison-test-when-the-geometric-function-is-divergent
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# How do I use the Comparison Test when the geometric function is divergent?
I have the following problem
$$A=\sum_{n=1}^{\infty}\frac{2n+7^n}{2n+6^n}$$
... and I'm trying to figure out if it is convergent or divergent using the Comparison Test.
A similar problem:
$$D = \sum_{n=1}^{\infty}\frac{4n+5^n}{4n+8^n}$$
... is solved by using the Comparison Test like so:
$$\frac{4n+5^n}{4n+8^n}<\frac{5^n+5^n}{4n+8^n}<\frac{2\cdot 5^n}{8^n}$$
and since:
$$2\sum _{n=1}^{\infty }\frac{5^n}{8^n}$$
... is a convergent Geometric Series, $D$ converges too.
If I try to do something similar to the original problem, $A$, I get the following:
$$B=\sum _{n=1}^{\infty }\frac{7^n}{6^n}$$
... is a divergent geometric series.
Therefore, I believe I need to use the comparison test to show that:
$$A>B$$
... this time, and I will be able to conclude that $A$ is divergent as well. If I try doing that, though, I run into a problem:
$$\frac{2n+7^n}{2n+6^n}>\frac{7^n}{2n+6^n}<\frac{7^n}{6^n}$$
Am I comparing it to the wrong function? Can I not use the Comparison Test on this problem and instead need to use the Integral Comparison Test?
-
The terms of the series do not approach $0$, so you have automatic divergence. It is important to take a good informal look at a series first. – André Nicolas Mar 22 '12 at 19:16
In other words, you can do comparison $$\frac{2n+7^n}{2n+6^n}>1$$ and get divergence. – GEdgar Mar 22 '12 at 19:20
## 3 Answers
As Andre Nicolas said your terms do not approach zero, and hence your series will diverge.
If however you want to use the comparison test you are pretty close $$\frac{2n+7^n}{2n+6^n}>\frac{7^n}{2n+6^n}>\frac{7^n}{6^n+6^n}=\frac{1}{2}\cdot\frac{7^n}{6^n}$$ And so your since $\frac{7}{6}>1$, you can say the series of the right diverge, and hence your series is larger than a divergent, and hence divergent.
-
If you really want to do a comparison of the type you described, note that $2n+6^n<6^n+6^n$. So the terms of your series are greater than $$\frac{7^n}{2\cdot 6^n}.$$
Remark: But this is too much work. Look at the original series. The top is always bigger than the bottom.
-
I am sorry to be nitpicky-but we would be sending wrong signals by using top and bottom. Why don't we stick to numerator and denominator? – user21436 Mar 22 '12 at 19:41
@KannappanSampath: The use of "top" and "bottom" is admittedly very casual. The idea is to stress, through the use of casual language, that we need to look first at what's happenin'. Students, even very good ones, all too often attack a problem by looking through the screwdrivers in the toolbox. But I will demote the last sentence of the post to a remark. – André Nicolas Mar 22 '12 at 19:59
Once again I apologise, if it came across to you wrongly. I understand your perspective too. : ) – user21436 Mar 22 '12 at 20:00
One relation… $$\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$$
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http://pubget.com/paper/pgtmp_13051338/Parametrization_of_Extremal_Trajectories_in_Sub_Riemannian_Problem_on___Group_of_Motions_of_Pseudo_Euclidean_Plane
|
Advanced search×
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## Parametrization of Extremal Trajectories in Sub-Riemannian Problem on
Group of Motions of Pseudo Euclidean Plane
arXiv:1305.1338 [math.OC] 6 May 2013
We consider the sub-Riemannian length optimization problem on the group of motions of hyperbolic plane i.e. the special hyperbolic group SH(2). The system comprises of left invariant vector fields with 2 dimensional linear control input and energy cost functional. We prove the global controllability of control distribution and use Pontryagin Maximum Principle to obtain the extremal control input and sub-Riemannian geodesics. The abnormal and normal extremal trajectories of the system are analyzed qualitatively and investigated for strict abnormality. A change of coordinates transforms the vertical subssystem of the normal Hamiltonian system into mathematical pendulum. In suitable elliptic coordinates the vertical and horizontal subsystems are integrated such that the resulting extremal trajectories are parametrized by Jacobi elliptic functions.
Version: za2963e q8zaa q8zb1 q8zc2 q8zdf q8ze4 q8zfa q8zgb
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http://torus.math.uiuc.edu/cal/math/cal?year=2012&month=09&day=20&interval=day
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Seminar Calendar
for events the day of Thursday, September 20, 2012.
.
events for the
events containing
Questions regarding events or the calendar should be directed to Tori Corkery.
``` August 2012 September 2012 October 2012
Su Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa
1 2 3 4 1 1 2 3 4 5 6
5 6 7 8 9 10 11 2 3 4 5 6 7 8 7 8 9 10 11 12 13
12 13 14 15 16 17 18 9 10 11 12 13 14 15 14 15 16 17 18 19 20
19 20 21 22 23 24 25 16 17 18 19 20 21 22 21 22 23 24 25 26 27
26 27 28 29 30 31 23 24 25 26 27 28 29 28 29 30 31
30
```
Thursday, September 20, 2012
Number Theory Seminar
11:00 am in 241 Altgeld Hall, Thursday, September 20, 2012
Del Edit Copy
Submitted by ford.
Harold Diamond (UIUC Math)Chebyshev bounds for Beurling generalized numbersAbstract: This is a semi-expository talk. It begins with a survey of Beurling generalized numbers, a structure that is similar to rational integers, except for having only multiplicative structure. We seek conditions on the counting function of g-numbers that allow us to deduce analogs of the Chebyshev upper and lower prime bounds. An early conjecture of the speaker is shown to be inadequate, and further conditions are given for which the bounds hold. The results are proved to be optimal in their class.
Math/Theoretical Physics Seminar
12:00 pm in 464 Loomis Laboratory, Thursday, September 20, 2012
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Submitted by katz.
Sheldon Katz (Illinois Math)Supermanifolds and Berezin IntegrationAbstract: This is an expository talk on the mathematical formalism of supermanifolds, which provides a rigorous way to analyze the "anticommuting coordinates" of physics used in the study of fermions. I will also introduce Berezin integration, a mathematical formalism which gives a precise meaning to fermionic integration as it arises in physics. I would like to say something as well about supersymmetry but I won't promise due to time constraints.
Graduate Geometry Topology Seminar
2:00 pm in 241 Altgeld Hall, Thursday, September 20, 2012
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Submitted by collier3.
Mike DiPasquale (UIUC Math)Bezout, Cayley-Bacharach, and PascalAbstract: We introduce some basic constructions of algebraic geometry in the process of exploring the geometry of curves in the complex projective plane. In particular we will discuss Bezout's theorem and the Cayley-Bacharach theorem for plane cubics, pointing out the special case of Pascal's 'mystic hexagon.' The object is to communicate the power of algebraic machinery in proving some beautiful geometric theorems.
Women's Seminar
3:00 pm in 147 Altgeld Hall, Thursday, September 20, 2012
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Submitted by funk3.
Sarah Loeb and Sogol Jahanbekam (UIUC Math)Combinatorics at MightyAbstract: Results in Chromatic-Paintability and the Paintability of Complete Bipartite Graphs by Sarah Loeb Introduced independently by Schauz and by Zhu, the Marker-Remover game is an on-line version of list coloring. The game is played on a graph $G$ with a token assignment $f$ giving each $v \in V(G)$ a nonnegative number of tokens. On each round Marker marks a subset $M$ of the remaining vertices, which uses up a token on each vertex in $M$. Remover deletes from the graph an independent subset of vertices in $M$. Marker wins by marking a vertex that has no tokens. Remover wins if the entire graph is removed. The paint number, or paintability, of a graph $G$ is the least $k$ such that Remover has a winning strategy when $f(v) = k$ for all $v \in V(G)$. We show that if $G$ is $k$-paintable and $|V(G)| \le \frac{t}{t-1} k$, then the join of $G$ with $\overline{K}_t$ is $(k+1)$-paintable. As a corollary, the paint number of $G$ equals to its chromatic number when $|V(G)| \le \chi(G) + 2 \sqrt{\chi(G)-1}$. This strengthens a result of Ohba. We also explore the paintability of complete bipartite graphs. Extending a result of Erd\H{o}s, Rubin, and Taylor, $K_{k,r}$ is $k$-paintable if and only if $r < k^k$. For $j \ge 1$ we provide an upper bound on the least $r$ such that $K_{k+j,r}$ is not $k$-paintable. 1,2,3-Conjecture and 1,2-Conjecture for sparse graphs by Sogol Jahanbekam We apply the Discharging Method to prove the 1, 2, 3-Conjecture and the 1, 2-Conjecture for graphs with maximum average degree less than 8 3. As a result, the conjectures hold for planar graphs with girth at least 8.
Commutative Ring Theory
3:00 pm in 243 Altgeld Hall, Thursday, September 20, 2012
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Submitted by s-dutta.
Javid Validashti (UIUC Math)On Syzygies and Singularities of Tensor Product SurfacesAbstract: On Syzygies and Singularities of Tensor Product Surfaces. Let $U \subseteq H^0({\mathcal{O}_{\mathbb{P}^1 \times \mathbb{P}^1}}(2,1))$ be a basepoint free four-dimensional vector space. We study the associated bigraded ideal $I_U \subseteq \textsf{k}[s,t;u,v]$ from the standpoint of commutative algebra, proving that there are exactly six numerical types of possible bigraded minimal free resolution. These resolutions play a key role in determining the implicit equation for the image of the projective surface in $\mathbb{P}^3$ parametrized by generators of $U$ over $\mathbb{P}^1 \times \mathbb{P}^1$. This problem arises from a real world application in geometric modeling, where one would like to understand the implicit equation and singular locus of a parametric surface. This talk is based on a joint work with H. Schenck and A. Seceleanu.
Fall Department Faculty Meeting
4:00 pm in 245 Altgeld Hall, Thursday, September 20, 2012
Del Edit Copy
Submitted by seminar.
Fall Department Faculty Meeting
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http://mathoverflow.net/questions/57079/special-killing-vector-fields/57170
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## Special Killing Vector Fields
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Consider $(M^{n},g)$ to be a Riemannian manifold and suppose that $X$ is a smooth non-trivial Killing vector field on $M$. Away from the zeros of $X$ we have a natural distribution $D$ of $(n-1)$-planes defined so that $D_p$ is orthogonal to $X_p$. If the distribution $D$ is (completely) integrable then it is straightforward to verify that the one form $\omega$ defined by $$\omega(\cdot )=\frac{1}{g(X,X)} g(X, \cdot).$$ is closed (away from $\lbrace X=0\rbrace$). Moreover, the converse also holds.
Examples in $\mathbb{R}^n$ with the euclidean metric include the the translations along the $x_i$-axis, $T_i$ and rotations around the $x_i$-axis, $R_i$. The Killing fields $T_i+R_i$ are non-examples.
My question is whether this concept already has a name and where it might appear in the literature.
-
2
Not a complete answer, but such Killing vectors are *twist-free*; although this condition seems stronger. Recall that a Killing vector $X$ is twist-free if $X^\flat \wedge dX^\flat = 0$, where $X^\flat = g(X,\cdot)$ in your notation. – José Figueroa-O'Farrill Mar 2 2011 at 13:00
## 2 Answers
I now think that my comment might indeed be the complete answer in the case when $X$ has no zeroes.
Guiseppe's answer has been a sort of Socratic catalyst.
Indeed, in that case the distribution $D$ defined by $\omega$ and $X^\flat$ agree. So $D$ is integrable if and only if the ideal generated by either $\omega$ or $X^\flat$ is differentiably closed, hence $dX^\flat = \alpha \wedge X^\flat$ for some one-form $\alpha$. In turn this is equivalent to $X^\flat \wedge dX^\flat = 0$, which is precisely the condition that $X$ be twist-free.
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Great! I figured there was a more convenient way to think about things. – Rbega Mar 2 2011 at 20:45
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Mine is not an answer but a question. I'll delete it if it is improper.
Why, as the questioner says, if $X$ were Killing and $D$ integrable then $\frac{1}{X^\flat(X)}.X^\flat$ should be closed on $M$? Could someone explain me the reason for this?
Here follows what I have understood:
Given a smooth non-singular vector field $X$ on a Riemannian manifold $(M,g)$, we get the smooth distribution $D$ on $M$ globally generated by the smooth non-vanishing 1-form $X^\flat=g(X,\cdot)$. By Frobenius' Theorem, $D$ is integrable iff ${X^\flat}\wedge{d{X^\flat}}=0$ on $M$. This integrability condition is at the same time necessary and sufficient for the local existence of integrating factors for $X^\flat$: i.e. for any point $p$ of $M$, there exists a function $f$ such that $f.X^\flat$ is closed in a neighborhood of $p$.
-
As it ended up answering the question your approach makes the most sense. The way I thought about things was: Fix a point $p$ so that $X(p)\neq 0$ Let $U$ be a small neighborhood of $p$ so that in $p$ there is a smooth surface $\Sigma$ through $p$ and normal to $X$. Let $\phi_s$ be the family of isometries generated by $X$ clearly for $U$ small enough $\phi_s(\Sigma)$ foliate $U$ and all the leaves are orthogonal to $X$. This allows one to think of $s$ as a function on $U$. One has $ds=\frac{X}{|X|^2}$. – Rbega Mar 2 2011 at 20:50
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http://mathoverflow.net/revisions/98898/list
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## Return to Answer
2 added 7 characters in body
Each connected component of an algebraic group has the same dimension. Thus if it has a connected subvariety whose complement has a lower dimension, it is connected. (If it had two connected components, only one could contain the connected subvariety, and so it the other would have a lower dimension.)
The subvariety where $I_n+A$ is invertible is birational to $\mathbb A^{\frac{n(n-1)}{2}}$ and thus has dimension $\frac{n(n-1)}{2}$.
The complement is where $A$ has a $-1$ eigenvalue. But since every eigenvalue to an orthogonal matrix must have its inverse also an eigenvalue, the determinant is the product of all the eigenvalues which are their own inverse, which are just $-1$ and $1$. Thus, if the determinant is $1$, the number of $-1$ eigenvalues is even, so is at least $2$. We can split the matrix into a $2$-dimensional $-1$-eigenspace and a matrix in $SO(n-2)$. These things are paramaterized by a $SO_{n-2}$-bundle on $G_{2}^{n-2}$, whose dimension is $2(n-2)+\frac{(n-2)(n-3)}{2}=\frac{n(n-1)}{2}-1$. If the $-1$-eigenspace is more than two-dimensional there is more than one way to express a matrix in this way, but that can only decrease the dimension.
1
Each connected component of an algebraic group has the same dimension. Thus if it has a connected subvariety whose complement has a lower dimension, it is connected. (If it had two connected components, only one could contain the connected subvariety, and so it would have a lower dimension.)
The subvariety where $I_n+A$ is invertible is birational to $\mathbb A^{\frac{n(n-1)}{2}}$ and thus has dimension $\frac{n(n-1)}{2}$.
The complement is where $A$ has a $-1$ eigenvalue. But since every eigenvalue to an orthogonal matrix must have its inverse also an eigenvalue, the determinant is the product of all the eigenvalues which are their own inverse, which are just $-1$ and $1$. Thus, if the determinant is $1$, the number of $-1$ eigenvalues is even, so is at least $2$. We can split the matrix into a $2$-dimensional $-1$-eigenspace and a matrix in $SO(n-2)$. These things are paramaterized by a $SO_{n-2}$-bundle on $G_{2}^{n-2}$, whose dimension is $2(n-2)+\frac{(n-2)(n-3)}{2}=\frac{n(n-1)}{2}-1$. If the $-1$-eigenspace is more than two-dimensional there is more than one way to express a matrix in this way, but that can only decrease the dimension.
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http://mathhelpforum.com/geometry/27028-rectangular-box.html
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# Thread:
1. ## Rectangular box
A rectangular box is to be made from a rectangular piece of shee metal 20in wide and 30in long by cutting a square piece of equal size from each of the corners and bending up the sides so that they make a 90 degree angle with the base. The base of the finished box will be 200 square inches. About how many inches high will the box be?
2. Originally Posted by sarahh
A rectangular box is to be made from a rectangular piece of shee metal 20in wide and 30in long by cutting a square piece of equal size from each of the corners and bending up the sides so that they make a 90 degree angle with the base. The base of the finished box will be 200 square inches. About how many inches high will the box be?
Let x be the length of one side of one of the squares you have to cut off.
Then you get:
$(20-2x)(30-2x) = 200$ ......... Expand the brackets. You'll get aquadratic equation:
$4x^2-100x+400 = 0$ ......... Solve this equation for x. Only one of the two solutions is valid with your problem.
(For confirmation only: x = 5)
3. Why do we reject the 20 answer again?
4. Originally Posted by sarahh
Why do we reject the 20 answer again?
Plug in the solution into the original equation:
$\begin{array}{ccc}(20-2 \cdot 20)(30-2 \cdot 20)& = &200\\\\ (-20)(-10)&=&200 \end{array}$
That means you cut off more than there exists. In my opinion quite impossible but the modern science ...
5. Ahh that does make sense. Many thanks earboth!
Sarah H.
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http://www.all-science-fair-projects.com/science_fair_projects_encyclopedia/Trigonometric_identity
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# All Science Fair Projects
## Science Fair Project Encyclopedia for Schools!
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# Science Fair Project Encyclopedia
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# Trigonometric identity
In mathematics, trigonometric identities are equations involving trigonometric functions that are true for all values of the occurring variables. These identities are useful whenever expressions involving trigonometric functions need to be simplified. An important application is the integration of non-trigonometric functions: a common trick involves first using the substitution rule with a trigonometric function, and then simplifying the resulting integral with a trigonometric identity.
Contents
## Notation
The following notations hold for all six trigonometric functions: sine (sin), cosine (cos), tangent (tan), cotangent (cot), secant (sec), and cosecant (csc). For brevity, only the sine case is given in the table.
Notation Reading Description Definition
sin2(x) "sine squared [of] x" the square of sine; sine to the second power sin2(x) = (sin(x))2
arcsin(x) "arcsine [of] x" the inverse function for sine arcsin(x) = y if and only if sin(y) = x and $-{\pi \over 2} \le y \le {\pi \over 2}$
(sin(x))−1 "sine [of] x, to the negative-one power" the reciprocal of sine; the multiplicative inverse of sine (sin(x))−1 = 1 / sin(x)
arcsin(x) can also be written sin−1(x)
## Definitions
$\tan (x) = \frac {\sin (x)} {\cos(x)} \qquad \operatorname{cot}(x) = \frac {\cos (x)} {\sin(x)} = \frac{1} {\tan(x)}$
$\operatorname{sec}(x) = \frac{1} {\cos(x)} \qquad \operatorname{csc}(x) = \frac{1} {\sin(x)}$
For more information, including definitions based on the sides of a right triangle, see Trigonometric functions.
## Periodicity, symmetry, and shifts
These are most easily shown from the unit circle:
$\sin(x) = \sin(x + 2\pi) \qquad \sin(x) = \cos\left(\frac{\pi}{2} - x\right)$
$\cos(x) = \cos(x + 2\pi) \qquad \cos(x) = \sin\left(\frac{\pi}{2}-x\right)$
$\tan(x) = \tan(x + \pi) \qquad \tan(x) = \cot\left(\frac{\pi}{2} - x\right)$
$\sin(-x) = -\sin(x) \qquad\; \cos(-x) =\; \cos(x)$
$\tan(-x) = -\tan(x) \qquad \cot(-x) = -\cot(x)$
For some purposes it is important to know that any linear combination of sine waves of the same period but different phase shifts is also a sine wave with the same period, but a different phase shift. In other words, we have
$a\sin x+b\cos x=\sqrt{a^2+b^2}\cdot\sin(x+\varphi)$
where
$\varphi= \left\{ \begin{matrix} {\rm arctan}(b/a),&&\mbox{if }a\ge0; \; \\ \pi+{\rm arctan}(b/a),&&\mbox{if }a<0. \; \end{matrix} \right. \;$
## Pythagorean identities
These identities are based on the Pythagorean theorem. The first is sometimes simply called the Pythagorean trigonometric identity.
$\sin^2(x) + \cos^2(x) = 1 \;$
$\tan^2(x) + 1 = \sec^2(x) \;$
$1 + \cot^2(x) = \csc^2(x) \;$
Note that the second equation is obtained from the first by dividing both sides by cos²(x). To get the third equation, divide the first by sin²(x) instead.
## Angle sum and difference identities
These are also known as the addition and subtraction theorems or formulas. The quickest way to prove these is Euler's formula. The tangent formula follows from the other two. A geometric proof of the sin(x + y) identity is given at the end of this article.
$\sin(x \pm y) = \sin(x) \cos(y) \pm \cos(x) \sin(y)\,$
$\cos(x \pm y) = \cos(x) \cos(y) \mp \sin(x) \sin(y)\,$
$\tan(x \pm y) = \frac{\tan(x) \pm \tan(y)}{1 \mp \tan(x)\tan(y)}$
${\rm cis}(x+y)={\rm cis}(x)\,{\rm cis}(y)$
${\rm cis}(x-y)={{\rm cis}(x)\over{\rm cis}(y)}$
where
${\rm cis}(x)=e^{ix}=\cos(x)+i\sin(x)\,$
and
$i=\sqrt{-1}.\,$
See also Ptolemaios' theorem.
## Double-angle formulas
These can be shown by substituting x = y in the addition theorems, and using the Pythagorean formula for the latter two. Or use de Moivre's formula with n = 2.
$\sin(2x) = 2 \sin (x) \cos(x) \,$
$\cos(2x) = \cos^2(x) - \sin^2(x) = 2 \cos^2(x) - 1 = 1 - 2 \sin^2(x) \,$
$\tan(2x) = \frac{2 \tan (x)} {1 - \tan^2(x)}$
The double-angle formulas can also be used to find Pythagorean triples. If (a, b, c) are the lengths of the sides of a right triangle, then (a2 − b2, 2ab, c2) also form a right triangle, where angle B is the angle being doubled. If a2 − b2 is negative, take its opposite and use the supplement of B.
## Multiple-angle formulas
If Tn is the nth Chebyshev polynomial then
$\cos(nx)=T_n(\cos(x)). \,$
De Moivre's formula:
$\cos(nx)+i\sin(nx)=(\cos(x)+i\sin(x))^n \,$
The Dirichlet kernel Dn(x) is the function occurring on both sides of the next identity:
$1+2\cos(x)+2\cos(2x)+2\cos(3x)+\cdots+2\cos(nx) \;$
$= \frac{ \sin\left(\left(n+\frac{1}{2}\right)x\right) } { \sin(x/2) } \;$
The convolution of any integrable function of period 2π with the Dirichlet kernel coincides with the function's nth-degree Fourier approximation. The same holds for any measure or generalized function.
## Power-reduction formulas
Solve the second and third versions of the cosine double-angle formula for cos2(x) and sin2(x), respectively.
$\cos^2(x) = {1 + \cos(2x) \over 2}$
$\sin^2(x) = {1 - \cos(2x) \over 2}$
$\sin^2(x) \cos^2(x) = {1 - \cos(4x) \over 8}$
## Half-angle formulas
Sometimes the formulas in the previous section are called half-angle formulas. To see why, substitute x/2 for x in the power reduction formulas, then solve for cos(x/2) and sin(x/2) to get:
$\cos\left(\frac{x}{2}\right) = \pm\, \sqrt{\frac{1 + \cos(x)}{2}}$
$\sin\left(\frac{x}{2}\right) = \pm\, \sqrt{\frac{1 - \cos(x)}{2}}$
These may also be called the half-angle formulas. Then
$\tan\left(\frac{x}{2}\right) = {\sin (x/2) \over \cos (x/2)} = \pm\, \sqrt{1 - \cos x \over 1 + \cos x}. \qquad \qquad (1)$
Multiply both numerator and denominator inside the radical by 1 + cos x, then simplify (using a Pythagorean identity):
$\tan\left(\frac{x}{2}\right) = \pm\, \sqrt{(1 - \cos x) (1 + \cos x) \over (1 + \cos x) (1 + \cos x)} = \pm\, \sqrt{1 - \cos^2 x \over (1 + \cos x)^2}$
$= {\sin x \over 1 + \cos x}.$
Likewise, multiplying both numerator and denominator inside the radical — in equation (1) — by
1 − cos x, then simplifying:
$\tan\left(\frac{x}{2}\right) = \pm\, \sqrt{(1 - \cos x) (1 - \cos x) \over (1 + \cos x) (1 - \cos x)} = \pm\, \sqrt{(1 - \cos x)^2 \over (1 - \cos^2 x)}$
$= {1 - \cos x \over \sin x}.$
Thus, the pair of half-angle formulas for the tangent are:
$\tan\left(\frac{x}{2}\right) = \frac{\sin(x)}{1 + \cos(x)} = \frac{1-\cos(x)}{\sin(x)}.$
If we set
$t = \tan\left(\frac{x}{2}\right),$
then
$\sin(x) = \frac{2t}{1 + t^2}$ and $\cos(x) = \frac{1 - t^2}{1 + t^2}$ and $e^{i x} = \frac{1 + i t}{1 - i t}.$
This substitution of t for tan(x/2), with the consequent replacement of sin(x) by 2t/(1 + t2) and cos(x) by (1 − t2)/(1 + t2) is useful in calculus for converting rational functions in sin(x) and cos(x) to functions of t in order to find their antiderivatives. For more information see tangent half-angle formula.
## Products-to-sum identities
These can be proven by expanding their right-hand-sides using the addition theorems.
$\cos(x) \cos(y) = {\cos(x + y) + \cos(x - y) \over 2} \;$
$\sin(x) \sin(y) = {\cos(x - y) - \cos(x + y) \over 2} \;$
$\sin(x) \cos(y) = {\sin(x + y) + \sin(x - y) \over 2} \;$
## Sum-to-product identities
Replace x by (x + y) / 2 and y by (x – y) / 2 in the product-to-sum formulas.
$\sin(x) + \sin(y) = 2 \sin\left( \frac{x + y}{2} \right) \cos\left( \frac{x - y}{2} \right) \;$
$\cos(x) + \cos(y) = 2 \cos\left( \frac{x + y}{2} \right) \cos\left( \frac{x - y}{2} \right) \;$
## Inverse trigonometric functions
$\arcsin(x)+\arccos(x)=\pi/2\;$
$\arctan(x)+\arccot(x)=\pi/2.\;$
$\arctan(x)+\arctan(1/x)=\left\{\begin{matrix} \pi/2, & \mbox{if }x > 0 \\ -\pi/2, & \mbox{if }x < 0 \end{matrix}\right..$
$\arctan(x)+\arctan(y)=\arctan\left(\frac{x+y}{1-xy}\right) \;$
$\sin(\arccos(x))=\sqrt{1-x^2} \,$
$\cos(\arcsin(x))=\sqrt{1-x^2} \,$
$\sin(\arctan(x))=\frac{x}{\sqrt{1+x^2}}$
$\cos(\arctan(x))=\frac{1}{\sqrt{1+x^2}}$
Every trigonometric function can be related directly to every other trigonometric function. Such relations can be expressed by means of inverse trigonometric functions as follows: let φ and ψ represent a pair of trigonometric functions, and let arcψ be the inverse of ψ, such that ψ(arcψ(x))=x. Then φ(arcψ(x)) can be expressed as an algebraic formula in terms of x. Such formulas are shown in the table below: φ can be made equal to the head of one of the rows, and ψ can be equated to the head of a column:
| | sin
| cos
| tan
| csc
| sec
| cot
|
|-----------------------------------------------------------------------------|-----------------------------------------------------------------------------|-----------------------------------------------------------------------------|-----------------------------------------------------------------------------|-----------------------------------------------------------------------------|-----------------------------------------------------------------------------|---------|
| φ / ψ | sin | cos | tan | csc | sec | cot |
| $x\$ | $\sqrt{1 - x^2}$ | ${x \over \sqrt{1 + x^2}}$ | ${1 \over x}$ | ${\sqrt{x^2 - 1} \over x}$ | ${1 \over \sqrt{1 + x^2}}$ | |
| $\sqrt{1 - x^2}$ | $x\$ | ${1 \over \sqrt{1 + x^2}}$ | ${\sqrt{x^2 - 1} \over x}$ | ${1 \over x}$ | ${x \over \sqrt{1 + x^2}}$ | |
| ${x \over \sqrt{1 - x^2}}$ | ${\sqrt{1 - x^2} \over x}$ | $x\$ | ${1 \over \sqrt{x^2 - 1}}$ | $\sqrt{x^2 - 1}$ | ${1 \over x}$ | |
| ${1 \over x}$ | ${1 \over \sqrt{1 - x^2}}$ | ${\sqrt{1 + x^2} \over x}$ | $x\$ | ${x \over \sqrt{x^2 - 1}}$ | $\sqrt{1 + x^2}$ | |
| ${1 \over \sqrt{1 - x^2}}$ | ${1 \over x}$ | $\sqrt{1 + x^2}$ | ${x \over \sqrt{x^2 - 1}}$ | $x\$ | ${\sqrt{1 + x^2} \over x}$ | |
| ${\sqrt{1 - x^2} \over x}$ | ${x \over \sqrt{1 - x^2}}$ | ${1 \over x}$ | $\sqrt{x^2 - 1}$ | ${1 \over \sqrt{x^2 - 1}}$ | $x\$ | |
One procedure that can be used to obtain the elements of this table is as follows:
Given trigonometric functions φ and ψ, what is φ(arcψ(x)) equal to?
1. Find an equation that relates φ(u) and ψ(u) to each other:
$f(\varphi(u), \psi(u)) = 0 \$
2. Let u = arc ψ(x), so that:
$f(\varphi({\rm arc}\psi(x)),\psi({\rm arc}\psi(x)) = 0 \$
$f(\varphi({\rm arc}\psi(x)),x) = 0 \$
3. Solve the last equation for φ(arcψ(x)).
Example. What is cot(arccsc(x)) equal to? First, find an equation which relations the functions cot and csc to each other, such as
$\cot^2 u + 1 = \csc^2 u \$.
Second, let u = arccsc(x):
$\cot^2(\arccsc(x)) + 1 = \csc^2(\arccsc(x)) \$,
$\cot^2(\arccsc(x)) + 1 = x^2 \$.
Third, solve this equation for cot(arccsc(x)):
$\cot^2(\arccsc(x)) = x^2 - 1, \$
$\cot(\arccsc(x)) = \pm\sqrt{x^2 - 1},$
and this is the formula which shows up in the sixth row and fourth column of the table.
## Exponential forms
$\cos(x) = \frac{e^{ix} + e^{-ix}}{2} \;$
$\sin(x) = \frac{e^{ix} - e^{-ix}}{2i} \;$
where $i^{2}=-1.\,$
## The Gudermannian function
The Gudermannian function relates the circular and hyperbolic trigonometric functions without resorting to complex numbers -- see that article for details.
## Identities without variables
Richard Feynman is reputed to have learned as a boy, and always remembered, the following curious identity:
$\cos 20^\circ\cdot\cos 40^\circ\cdot\cos 80^\circ=\frac{1}{8}.$
However, this is a special case of an identity that contains one variable:
$\prod_{j=0}^{k-1}\cos(2^j x)=\frac{\sin(2^k x)}{2^k\sin(x)}.$
The following are perhaps not as readily generalized to identities containing variables:
$\cos 36^\circ+\cos 108^\circ=\frac{1}{2}$
$\cos 24^\circ+\cos 48^\circ+\cos 96^\circ+\cos 168^\circ=\frac{1}{2}.$
Degree measure ceases to be more felicitous than radian measure when we consider this identity with 21 in the denominators:
$\cos\left( \frac{2\pi}{21}\right) \,+\, \cos\left(2\cdot\frac{2\pi}{21}\right) \,+\, \cos\left(4\cdot\frac{2\pi}{21}\right)$
$\,+\, \cos\left( 5\cdot\frac{2\pi}{21}\right) \,+\, \cos\left( 8\cdot\frac{2\pi}{21}\right) \,+\, \cos\left(10\cdot\frac{2\pi}{21}\right)=\frac{1}{2}.$
The factors 1, 2, 4, 5, 8, 10 may start to make the pattern clear: they are those integers less than 21/2 that have no prime factors in common with 21. The last several examples are corollaries of a basic fact about the irreducible cyclotomic polynomials: the cosines are the real parts of the zeroes of those polynomials; the sum of the zeroes is the Möbius function evaluated at (in the very last case above) 21; only half of the zeroes are present above. The two identities preceding this last one arise in the same fashion with 21 replaced by 10 and 15, respectively.
An efficient way to compute π is based on the following identity without variables, due to Machin:
$\frac{\pi}{4} = 4 \arctan\frac{1}{5} - \arctan\frac{1}{239}$
or, alternatively, by using Euler's formula:
$\frac{\pi}{4} = 5 \arctan\frac{1}{7} + 2 \arctan\frac{3}{79}.$
Under this heading, there are also "special values" of trigonometric functions, including the ones that every student of trigonometry learns:
$\sin 0=0=\cos 90^\circ,\,$
$\sin 30^\circ=1/2=\cos 60^\circ,\,$
$\sin 45^\circ=\sqrt{2}/2=\cos 45^\circ,\,$
$\sin 60^\circ=\sqrt{3}/2=\cos 30^\circ,\,$
$\sin 90^\circ=1=\cos 0.\,$
Some are less well-known:
$\cos 36^\circ=\cos(\pi/5)={1+\sqrt{5} \over 4}\,$
(this one is related to the golden ratio).
$\sin{\frac{\pi}{7}}=\frac{\sqrt{7}}{6}- \frac{\sqrt{7}}{189} \sum_{j=0}^{\infty} \frac{(3j+1)!}{189^j j!\,(2j+2)!} \!$
$\sin{\frac{\pi}{18}}= \frac{1}{6} \sum_{j=0}^{\infty} \frac{(3j)!}{27^j j!\,(2j+1)!} \!$
## Calculus
In calculus the relations stated below require angles to be measured in radians; the relations would become more complicated if angles were measured in another unit such as degrees. If the trigonometric functions are defined in terms of geometry, then their derivatives can be found by verifying two limits. The first is:
$\lim_{x\rightarrow 0}\frac{\sin(x)}{x}=1,$
verified using the unit circle and squeeze theorem. It may be tempting to propose to use L'Hôpital's rule to establish this limit. However, if one uses this limit in order to prove that the derivative of the sine is the cosine, and then uses the fact that the derivative of the sine is the cosine in applying L'Hôpital's rule, one is reasoning circularly—a logical fallacy. The second limit is:
$\lim_{x\rightarrow 0}\frac{1-\cos(x)}{x}=0,$
verified using the identity tan(x/2) = (1 − cos(x))/sin(x). Having established these two limits, one can use the limit definition of the derivative and the addition theorems to show that sin′(x) = cos(x) and cos′(x) = −sin(x). If the sine and cosine functions are defined by their Taylor series, then the derivatives can be found by differentiating the power series term-by-term.
${d \over dx}\sin(x) = \cos(x)$
The rest of the trigonometric functions can be differentiated using the above identities and the rules of differentiation. We have:
${d \over dx}\cos(x) = -\sin(x)$
${d \over dx}\tan(x) = \sec^2(x)$
${d \over dx}\cot(x) = -\csc^2(x)$
${d \over dx}\sec(x) = \sec(x) \tan(x)$
${d \over dx}\csc(x) = - \csc(x)\cot(x)$
${d \over dx}\arcsin(x)=\frac{1}{\sqrt{1-x^2}}$
${d \over dx}\arctan(x)=\frac{1}{1+x^2}$
The integral identities can be found in Wikipedia's table of integrals.
## Proofs using a differential equation
Consider this differential equation:
$y^{\prime\prime} + y=0$
Using Euler's formula and the method for solving linear differential equations combined with the uniqueness theorem and the existence theorem we can define sine and cosine as the following:
cos(x) is the unique solution of
$y^{\prime\prime} + y = 0$ subject to the initial conditions of $y(0)=1 \;$ and $y^\prime(0) = 0$
sin(x) is the unique solution of
$y^{\prime\prime} + y = 0$ subject to the initial conditions of $y(0)=0 \;$ and $y^\prime(0) = 1$
Now, let's prove that
$\frac{d}{dx}\sin(x) = \cos(x)$
First let
$T(x) = \sin^\prime(x)$
Now we find the first and second derivatives of T(x)
$T^\prime(x) = \sin^{\prime\prime}(x)$ but since sin(x) is a solution of $y^{\prime\prime} + y = 0$ we can say $\sin^{\prime\prime}(x) + \sin(x) = 0$ so $\sin^{\prime\prime}(x)=-\sin(x)$
Therefore
$T^\prime(x) = -\sin(x)$
$T^{\prime\prime}(x) = -\sin^\prime(x) = -T(x)$
Therefore we can say
$T^{\prime\prime}(x) + T(x) = 0$
Once again according to our technique of solving linear differential equations and Euler's formula the solution to $T^{\prime\prime}(x) + T(x) = 0$ must be a linear combination of sin(x) and cos(x), therefore
$T(x) = A\sin(x) + B\cos(x) \;$
Now we solve for B by plugging in 0 for x
$T(0) = 0 + B \;$
but according to our initial values $T(0) = \sin^\prime(0) = 1$, therefore
$B=1 \;$
To solve for A we take the derivative of T(x) and plug in 0 for x
$T^\prime(x) = A\sin^\prime(x) + B\cos^\prime(x)$
$T^\prime(0) = A\sin^\prime(0) + B\cos^\prime(0)$
Using our initial values and since $T^\prime(x) = -\sin(x)$
$-\sin(0)=A{1}+B{0} \;$
$A=0 \;$
Plugging A and B back into our original equation for T(x) we get
$T(x)=\cos(x) \;$
But since T(x) was defined as $\sin^\prime(x)$ we get
$\sin^\prime(x) = \cos(x)$
or
$\frac{d}{dx}\sin(x) = \cos(x)$
Q.E.D.
Using these rigorous definitions of sine and cosine, you can prove all the other properties of sine and cosine using the same technique.
See also Rigorous Definition of Sine and Cosine (pdf).
## Geometric proofs
### sin(x + y) = sin(x) cos(y) + cos(x) sin(y)
In the figure the angle x is part of right angled triangle ABC, and the angle y part of right angled triangle ACD. Then construct DG perpendicular to AB and construct CE parallel to AB.
Angle x = Angle BAC = Angle ACE = Angle CDE.
EG = BC.
$\sin(x + y) \,$
$= DG / AD \,$
$= (EG + DE) / AD \,$
$= (BC + DE) / AD \,$
$= (BC / AD) + (DE / AD) \,$
$= \frac{BC}{AD} \cdot \frac{AC}{AC} + \frac{DE}{AD} \cdot \frac{CD}{CD} \,$
$= \frac{BC}{AC} \cdot \frac{AC}{AD} + \frac{DE}{CD} \cdot \frac{CD}{AD} \,$
$= \sin( x ) \cos( y ) + \cos( x ) \sin( y ). \,$
### cos(x + y) = cos(x) cos(y) − sin(x) sin(y)
Using the above figure:
$\cos(x + y) \,$
$= AG / AD \,$
$= (AB - GB) / AD \,$
$= (AB - EC) / AD \,$
$= (AB / AD) - (EC / AD) \,$
$= \frac{AB}{AD} \cdot \frac{AC}{AC} - \frac{EC}{AD} \cdot \frac{CD}{CD} \,$
$= \frac{AB}{AC} \cdot \frac{AC}{AD} - \frac{EC}{CD} \cdot \frac{CD}{AD} \,$
$= \cos( x ) \cos( y ) - \sin( x ) \sin( y ). \,$
## See also
03-10-2013 05:06:04
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http://math.stackexchange.com/questions/49649/which-steps-i-have-to-do-to-get-this-equation?answertab=oldest
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# Which steps I have to do to get this equation?
I don't know what to do to derive the right side from the left side: $$\frac{B}{1+r} = B - \frac{r B}{1+r}.$$
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1
Instead, derive the left side from the right side (with a least common denominator), then work backwards. – MartianInvader Jul 5 '11 at 17:55
## 2 Answers
Hint: Write the numerator of the left hand side as $B(1+r-r)$ and use the fact that $$\frac{a+c}{d}=\frac{a}{d}+\frac{c}{d}$$ for $d\neq 0.$
Added:
Observe that $\,\,$ $B=B(1+r-r)=B(1+r)-rB.$
So $$\begin{align*}\frac{B}{1+r}&=\frac{B(1+r)-rB}{1+r}\\ &= \frac{B(1+r)}{1+r}-\frac{rB}{1+r}\\ &= B-\frac{rB}{1+r}. \end{align*}$$
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Could you please show the whole transformation? – Georg Jul 5 '11 at 18:12
@Georg: I have added to my answer. – Nana Jul 5 '11 at 18:23
Great, thanks a lot! – Georg Jul 5 '11 at 18:27
@georg: please accept nana's answer – user9413 Jul 5 '11 at 19:22
HINT $\$ It's simply $\rm\displaystyle\ 1\ =\ \frac{1+r}{1+r}\ =\ \frac{1}{1+r}\ +\ \frac{r}{1+r}\$ rearranged, then scaled by $\rm\:B\:.$
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http://physics.stackexchange.com/questions/44120/what-are-wightman-fields-functions?answertab=votes
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What are Wightman fields/functions
Simple question: What are Wightman fields? What are Wightman functions? What are their uses? For example can I use them in operator product expansions? How about in scattering theory?
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1 Answer
In axiomatic approaches to quantum field theory, the basic field operators are usually realized as operator-valued distributions. That's what Wightman fields are: operator-valued distributions satisfying the Wightman Axioms.
Wightman functions are the correlation functions of Wightman fields, nothing more. There's a nice theorem that says if you have a bunch of functions that look like they're the Wightman functions of QFT, then you can actually reconstruct the Hilbert space and the algebra of Wightman fields from it.
Neither of these concepts is anything new physically. They're just more precise ways of speaking about things physicists already know. (Thinking of fields as operator-valued distributions instead of operator-valued functions lets you make precise what goes wrong when you multiply two fields at the same point.) You can talk about OPEs or scattering theory in this language, but it won't gain you anything, unless you're trying to publish in math journals.
If you're interested in the Wightman Axioms, they're explained nicely in the first of Kazhdan's lectures in the IAS QFT Year.
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Ok, so based on my reading, the field operators of a theory are understood to be distribution-valued, that is, to be integrated over a smooth function: e.g. $\hat\phi(f)\equiv\int dx f(x)\hat\phi(x)$. But if the smooth function has support over such a large region in space-time that if I try to compute time-ordered products, I run into some ambiguity. How do I define the time-ordering symbol for Wightman functions? – QuantumDot Nov 17 '12 at 22:23
@QuantumDot: Why not just ask this as another question instead of trying to have a discussion here in the comments? – user1504 Nov 18 '12 at 0:13
– QuantumDot Nov 18 '12 at 0:18
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http://math.stackexchange.com/questions/179526/prerequisites-for-studying-smooth-manifold-theory/179530
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# Prerequisites for studying smooth manifold theory?
I am attending first year graduate school in about three weeks and one of the courses I am taking is an introduction to smooth manifolds. Unfortunately, my topology knowledge is minimal, limited to self study. Besides some basic topological definitions, are there specific areas where I should become acquainted with? I have heard multivariable calculus is a 'prerequisite' for the study of smooth manifolds -- is this from an intuition perspective, and if so, what parts and to what degree of rigor are they referring?
Thank you. (My apologies if this is a repeat question, by the way -- I could not find it)
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3
You need multivariable calculus, because you need to understand what it means for a function $\mathbb{R}^m\to\mathbb{R}^n$ to be differentiable. – M Turgeon Aug 6 '12 at 15:38
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Open up a book on topological manifolds and see what topology they use. – Qiaochu Yuan Aug 6 '12 at 15:39
Well, I have and so I understand I need to understand several topological concepts: Hausdorff, bases, countability, etc., but I suppose I am wondering if there is anything else I need, perhaps only to help my intuition while learning, or more. – Three Aug 6 '12 at 15:43
## 3 Answers
You will need
• Calculus/real analysis of functions of one and several variables up to and including the implicit and inverse function theorem. This (the implicit function theorem) is the basic starting point for (smooth) manifold theory, you will not be able to get anywhere without it.
• sound knowledge of linear algebra (at least in/for real vector spaces)
• 1-d integration (Riemann integral will suffice in the beginning), preferably basic knowledge of an n-dimensional integration theory.
• basic theory of ODE will sooner or later become important, in order to be able to deal with, e.g, the flow of vector fields (actually one needs existence (Picard-Lindelöf) and rather soon smooth dependence of the solution on the initial values, but the latter may be stated and believed. It'll become difficult without existence of solutions).
Depending on your book of choice everything else will probably be developed in the course of said book, see the other answers for some suggestion. The basics in topology you'll need will likely be known to you, once you really covered the above list. Spivak's calculus on manifolds comes to my mind.
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Very helpful -- thank you. – Three Aug 6 '12 at 16:39
In my home University, the textbook used for the first graduate course on Manifold theory is Introduction to Smooth Manifolds, by John M. Lee. If you look it up at your library, you will see there is an Appendix with three sections: Topology, Linear Algebra, Calculus. This Appendix will give you some prerequisites you are looking for.
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They usually refer to using inverse function theorem, etc to construct smooth manifolds, and to my knowledge this is not included in Stewart calculus book. There are other topics usually included in a differentiable manifolds book but not covered in calculus too, like Lie derivative, homogeneous spaces, Poincare-Hopf theorem, etc.
The classical reference on smooth manifolds is Boothy's book, and you may read it online or grab a copy yourself. The other book by Lee others mentioned is also good for the purpose, and probably covered more topology than Boothy's book. I read them several years ago and do not remember the strucutre very well. You may be interested to read Munkres's Analysis on Manifolds as well.
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I am quite enjoying Boothy's book. It's a good introduction. Thank you. – Three Aug 6 '12 at 16:40
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http://unapologetic.wordpress.com/2008/12/31/the-determinant/?like=1&source=post_flair&_wpnonce=671450efa2
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# The Unapologetic Mathematician
## The Determinant
Let’s look at the dimensions of antisymmetric tensor spaces. We worked out that if $V$ has dimension $d$, then the space of antisymmetric tensors with $n$ tensorands has dimension
$\displaystyle\dim\left(A^n(V)\right)=\binom{d}{n}=\frac{d!}{n!(d-n)!}$
One thing should leap out about this: if $n$ is greater than $d$, then the dimension formula breaks down. This is connected with the fact that at that point we can’t pick any $n$-tuples without repetition from $d$ basis vectors.
So what happens right before everything breaks down? If $n=d$, then we find
$\displaystyle\dim\left(A^d(V)\right)=\binom{d}{d}=\frac{d!}{d!(d-d)!}=\frac{d!}{d!}=1$
There’s only one independent antisymmetric tensor of this type, and so we have a one-dimensional vector space. But remember that this isn’t just a vector space. The tensor power $V^{\otimes d}$ is both a representation of $\mathrm{GL}(V)$ and a representation of $S_d$, which actions commute with each other. Our antisymmetric tensors are the image of a certain action from the symmetric group, which is an intertwiner of the $\mathrm{GL}(V)$ action. Thus we have a one-dimensional representation of $\mathrm{GL}(V)$, which we call the determinant representation.
I want to pause here and point out something that’s extremely important. We’ve mentioned a basis for $V$ in the process of calculating the dimension of this space, but the space itself was defined without reference to such a basis. Similarly, the representation of any element of $\mathrm{GL}(V)$ is defined completely without reference to any basis of $V$. It needs only the abstract vector space itself to be defined. Calculating the determinant of a linear transformation, though, is a different story. We’ll use a basis to calculate it, but as we’ve just said the particular choice of a basis won’t matter in the slightest to the answer we get. We’d get the same answer no matter what basis we chose.
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## 7 Comments »
1. [...] the Determinant Today we’ll actually calculate the determinant representation of an element of for a finite-dimensional vector space . That is, what [...]
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2. [...] Determinant of a Noninvertible Transformation We’ve defined and calculated the determinant representation of for a finite-dimensional vector space . But we [...]
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3. [...] polynomial of an endomorphism on a vector space of finite dimension , then we can get its determinant from the constant [...]
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4. [...] always unapologetic John Armstrong is currently on an exposition of the determinant, starting here and now here . This must be about the fifth time I’m relearning the determinant, each time [...]
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5. [...] of some appropriate — we have a homomorphism to the multiplicative group of given by the determinant. We originally defined the determinant on itself, but we can easily restrict it to any subgroup. [...]
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6. [...] we can use this to get back to our original definition of the determinant of a linear transformation . Pick a orthonormal basis for and wedge them all [...]
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7. [...] these matrices are exactly the Jacobian matrices of the functions! And since the by definition, the determinant of the product of two matrices is the product of their determinants. That is, we [...]
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
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http://mathhelpforum.com/advanced-algebra/35984-fields-subfields.html
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# Thread:
1. ## fields and subfields
Assume Q[sqrt(2)] is a subfield of the Real field R, and prove that it is the smallest subfield of R that contains sqrt(2)
Q[sqrt(2)]= a + b(sqrt(2))
2. Originally Posted by JCIR
Assume Q[sqrt(2)] is a subfield of the Real field R, and prove that it is the smallest subfield of R that contains sqrt(2)
how is $Q[\sqrt{2}]$ defined?
3. Originally Posted by JCIR
Assume Q[sqrt(2)] is a subfield of the Real field R, and prove that it is the smallest subfield of R that contains sqrt(2)
Q[sqrt(2)]= a + b(sqrt(2))
The way $\mathbb{Q}(\sqrt{2})$ is usually defined is to be the smallest subfield containing $\mathbb{Q}$ and $\sqrt{2}$. Look Here.
So I will assume you are defining $\mathbb{Q}(\sqrt{2}) = \{ a+b\sqrt{2}|a,b\in \mathbb{Q}\}$ and you want to show it is the smallest such field. Let $\mathbb{Q}\subseteq K\subseteq \mathbb{R}$ be a field containing $\mathbb{Q}$ and $\sqrt{2}$. We need to show $\mathbb{Q}(\sqrt{2})\subseteq K$. Let $x\in \mathbb{Q}(\sqrt{2})$ then $x=a+b\sqrt{2}$ but then $a+b\sqrt{2}\in K$ because it is closed under sums and products, thus, $x\in K$.
4. This is JCIR you answered my question on fields and subfields, can you tell my why is it closed under sums and products ( i need to prove that too)
Because a field has to be closed under sums and products (that is basically the definition of "field"). Meaning if $a,b\in F$ then $a+b,ab\in F$.
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http://physics.stackexchange.com/questions/946/psf-measurements-in-fluorescence-imaging/990
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# PSF Measurements In Fluorescence Imaging
Quite a technical question!
I have measured the Point Spread Function of 100nm fluorescent breads with my Olympus scanning head. I'm two-photon exciting the beads with a wavelength of 800nm and focused in the sample with a 100x with N.A.: 1.4
The theory suggests me the resolution of such a system to be: $$d=\frac{0.7\cdot\lambda}{N.A.}\approx 400nm$$
Now that value should be equal (with at most a 18% correction) to the FWHM of the 2D gaussian I obtain on the image.
But from my analysis of the images I obtain a FWHM close to 2um. Now surely the formula is for an ideal optical setup but a factor five seems to me quite strange! Is that possible to obtain such a result, in the case of very not ideal optical elements, or should I look for some sort of problem with the acquisition sw that tells me the pixel dimension of the images?
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## 2 Answers
You have many things which can go wrong when you try to go down to the wavelength limit, without resorting to Very not ideal optical element : non-Gaussian beam, wrong wavelength, spherical aberration, etc. I fear there is no easy answer to your question.
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no well, if someone else tells me he obtained a result similar (factor 5) i will be satisfied.. cause i've never really take a measurements like this and i don't know if it is normal or i have some technical problem in the analysis.. thx – Steve Nov 17 '10 at 10:20
ok i've found the solution to my problem.. first of all i had a factor 2 discrepancy due to the pixel dimension and, more important from an optical point of view, the laser beam was underfilling the entrance pupil of the objective: i.e. the focusing was worse!
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http://mathoverflow.net/questions/93324/source-for-nbgequipollence-conservative-over-zfc
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## Source for NBG+Equipollence conservative over ZFC?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
The "conservative" class theory, NBG, proves no new theorems about sets (with respect to ZFC). The choice function used here is set choice, and it's not too hard to prove (if M is a ctm for ZFC, then D(M) is a ctm for NBG and has the same set universe).
However, if we add the axiom that there is a bijection (in the class universe) between $\mathbf{V}$ and $\mathbf{ON}$, the classes of sets and ordinals, respectively, this is apparently still conservative over ZFC (with a much stronger class choice axiom). However, I can't find a reference for this.
Apparently this fact is credited to Easton and Solovay, published by Easton in 1964, and apparently it uses class forcing, but I can't find any more specific information on this topic, or the paper itself. Does anyone have more specific information on this, or better search skills than I do?
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I don't have the reference handy, but the forcing consists of all set-sized partial injections from $\mathbf{ON}$ to $\mathbf{V}$, ordered by extension. Since this is $\kappa$-closed for all $\kappa$, no new sets are added in the forcing extension, and the generic is a bijection from $\mathbf{ON}$ to $\mathbf{V}$. – François G. Dorais♦ Apr 6 2012 at 15:50
This is the forcing I wanted to do, but unfortunately I lacked the background to prove (to my own satisfaction) that it actually works. – Richard Rast Apr 6 2012 at 16:12
## 1 Answer
A detailed proof of the conservativity can be found in: Ulrich Felgner, Comparison of the axioms of local and universal choice, Fundamenta Mathematicae 71 (1971), 43–62 (pdf).
The basic idea of the proof is pretty straightforward: You take the class of injective (set) functions whose domain is an initial segment of On as your forcing conditions, and then a generic filter gives you a bijection of On and V. (Felgner uses choice functions as forcing conditions, but that does not make much of a difference.)
As an aside: In the last chapter of my master thesis, which unfortunately only exists in Czech, I prove a generalization of the result to set theory without foundation ($\mathrm{ZFC}_-$). It turns out that in the absence of foundation, global choice has nontrivial consequences even for sets. $\mathrm{ZFC}_-$ (or $\mathrm{NBG}_-$) + global choice is a conservative extension of $\mathrm{ZF}_-$ extended by the following schema: $$\forall x\,\exists y\,\phi(x,y)\to\forall\alpha\in\mathrm{On}\,\exists f\colon\alpha\to\mathrm{V}\,\forall\beta< \alpha\,\phi(f\restriction\beta,f(\beta))$$ (a sort of class version of $\mathrm{DC}_\alpha$; an equivalent formulation: any class tree whose height is bounded by an ordinal has a maximal path).
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Thank you for the reference! – Richard Rast Apr 6 2012 at 16:16
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http://mathoverflow.net/revisions/49135/list
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## Return to Answer
2 added 310 characters in body
@David Hansen: Teichmuller modular forms are basically the natural analogue of Siegel modular forms when one considers sections of line bundles on $M_g$ instead of $A_g$. Search for papers by Ichikawa.
I don't know very much about Teichmuller modular forms but my advisor knows a bit about them and from what I can tell from conversations with him, pretty much nothing at all is known about this question. I can think of some plausible reasons why this is so. (This answer is probably a bit naive, my apologies.)
One is that the difference between Siegel modular forms and Teichmuller modular forms does not become visible until genus three. For genus one and two the map from $M_g$ to $A_g$ is an isomorphism and an open immersion respectively, and all Teichmuller modular forms are just pullbacks of forms from $A_1$ and $A_2$. In genus three the Torelli map is an open immersion on coarse moduli spaces but on the level of stacks it is a ramified double cover, which makes it plausible that there should be more functions on $M_3$ than on $A_3$. And indeed the ring of Teichmuller modular forms in genus three is obtained from the ring of Siegel modular forms by adjoining a square root of $\chi_{18}$, which vanishes exactly at the hyperelliptic locus i.e. the branch locus of the Torelli map. Anyway, the point is that already Siegel modular forms in higher genera are not so well understood compared to the rich theory we have in low genus. In particular we know extremely few examples of genuine Teichmuller modular forms.
Another more serious reason is that it is not clear how any of the standard tools for studying ordinary modular forms can be applied to the Teichmuller case. As a very basic example, it is for instance completely open if or how one can define a notion of Hecke operators acting on Teichmuller modular forms: double cosets in the mapping class group are a scary prospect, and there seems to be no analogue of the standard Hecke correspondences in terms of cyclic subgroups of order p. What's worse is that none of the standard automorphic/Langlands etc tools seem to be applicable to the study of Teichmuller modular forms. Unlike Siegel's upper half space, Teichmuller space is not a symmetric space. If we believe in the Langlands philosophy then there should be a reductive group somewhere that the Teichmuller modular forms "come from", but there is no natural reductive group anywhere in sight.
Anyway, to actually answer your question: there probably is some connection between Teichmuller modular forms and number theory. In particular it seems that there are $\ell$-adic Galois representations naturally attached to Teichmuller modular forms. There are just no tools to study these representations.
1
@David Hansen: Teichmuller modular forms are basically the natural analogue of Siegel modular forms when one considers sections of line bundles on $M_g$ instead of $A_g$. Search for papers by Ichikawa.
I don't know very much about Teichmuller modular forms but my advisor knows a bit about them and from what I can tell from conversations with him, pretty much nothing at all is known about this question. I can think of some plausible reasons why this is so. (This answer is probably a bit naive, my apologies.)
One is that the difference between Siegel modular forms and Teichmuller modular forms does not become visible until genus three. For genus one and two the map from $M_g$ to $A_g$ is an isomorphism and an open immersion respectively, and all Teichmuller modular forms are just pullbacks of forms from $A_1$ and $A_2$. In genus three the Torelli map is an open immersion on coarse moduli spaces but on the level of stacks it is a ramified double cover, which makes it plausible that there should be more functions on $M_3$ than on $A_3$. And indeed the ring of Teichmuller modular forms in genus three is obtained from the ring of Siegel modular forms by adjoining a square root of $\chi_{18}$, which vanishes exactly at the hyperelliptic locus i.e. the branch locus of the Torelli map. Anyway, the point is that already Siegel modular forms in higher genera are not so well understood compared to the rich theory we have in low genus. In particular we know extremely few examples of genuine Teichmuller modular forms.
Another more serious reason is that it is not clear how any of the standard tools for studying ordinary modular forms can be applied to the Teichmuller case. As a very basic example, it is for instance completely open if or how one can define a notion of Hecke operators acting on Teichmuller modular forms: double cosets in the mapping class group are a scary prospect, and there seems to be no analogue of the standard Hecke correspondences in terms of cyclic subgroups of order p. What's worse is that none of the standard automorphic/Langlands etc tools seem to be applicable to the study of Teichmuller modular forms. Unlike Siegel's upper half space, Teichmuller space is not a symmetric space. If we believe in the Langlands philosophy then there should be a reductive group somewhere that the Teichmuller modular forms "come from", but there is no natural reductive group anywhere in sight.
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http://mathhelpforum.com/number-theory/137704-holomorphic-extension-dirichlet-series.html
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# Thread:
1. ## Holomorphic extension of Dirichlet Series
Define $L : \{s \in \mathbb{C}| Re(s) > 0\} \rightarrow \mathbb{C}$ by $L(s) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s}$.
Note that $L(s) = \frac{1}{\Gamma(s)} \int_0^\infty \frac{t^{s-1}}{e^t+e^{-t}}dt$
Show that L can be extended to an entire function $L : \mathbb{C} \rightarrow \mathbb{C}$
2. Originally Posted by EinStone
Define $L : \{s \in \mathbb{C}| Re(s) > 0\} \rightarrow \mathbb{C}$ by $L(s) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s}$.
Note that $L(s) = \frac{1}{\Gamma(s)} \int_0^\infty \frac{t^{s-1}}{e^t+e^{-t}}dt$
Show that L can be extended to an entire function $L : \mathbb{C} \rightarrow \mathbb{C}$
Tell me more i.e. what else do you know?
Are you familiar with the derivation of the analytic continuation of $\zeta(s)$? Because this series has a similar continuation.
If not, that's ok because I don't think you're asking for the functional equation itself, just that it exists.
Before I make any head way on this problem, what are your thoughts on how to approach it?
3. What I know is how to show that the Gamma Function is meromorphic on $\mathbb{C}$. Then by evaluating $\Gamma(s) * \zeta(s)$ in a specific way, one can show that $\Gamma(s)*\zeta(s)$ is also meromorphic on $\mathbb{C}$, and therefore $\zeta(s)$ has to be meromorphic as well.
I actually thought that $\int_0^\infty \frac{t^{s-1}}{e^t+e^{-t}}dt$ can somehow be extended holomorphically to $\mathbb{C}$ similarly to the Gamma function. But thats all I can think of.
4. Originally Posted by EinStone
Note that $L(s) = \frac{1}{\Gamma(s)} \int_0^\infty \frac{t^{s-1}}{e^t+e^{-t}}dt$
Show that L can be extended to an entire function $L : \mathbb{C} \rightarrow \mathbb{C}$
I guess you know that $\frac{1}{\Gamma}$ is an entire function. Then all you have to prove is that the integral defines an entire function as well. There is a general theorem for proving that functions of the form $s\mapsto\int f(s,t)dt$ are analytic. If you know it (which is probably the case), then you know what left you have to do... and this shouldn't be too hard (very similar to Gamma). What did you already try this way?
5. Actually I don't remember the proof exactly, it was very strange. Also I don't know the theorem you are talking about . What can I do?
6. I still need help!
Meanwhile I asked myself why the integral converges in the first place for $Re(s) > 0$ ?
7. Originally Posted by EinStone
I still need help!
Actually I was wrong: the integral is only defined when ${\rm Re}(s)>0$... It is not simple to extend it to $\mathbb{C}$. Like Chiph588@ said, you should repeat arguments used for zeta.
Meanwhile I asked myself why the integral converges in the first place for $Re(s) > 0$ ?
We have $\frac{|t^{s-1}|}{e^t+e^{-t}}\leq |t^{s-1}e^{-t}|$, so if you know that the integral defining Gamma converges, then so does this one.
8. Ok, So to prove that Zeta is holomorphic everywhere, we used $\frac{t}{e^t-1} = \sum_{k=0}^\infty \frac{B_k}{k!} t^k$ where $B_k$ are the Bernoulli numbers.
So I think I need a similar expression for $\frac{t}{e^t+e^{-t}}$, but I don't see it.
9. Originally Posted by EinStone
Ok, So to prove that Zeta is holomorphic everywhere, we used $\frac{t}{e^t-1} = \sum_{k=0}^\infty \frac{B_k}{k!} t^k$ where $B_k$ are the Bernoulli numbers.
So I think I need a similar expression for $\frac{t}{e^t+e^{-t}}$, but I don't see it.
$\frac{t}{e^t+e^{-t}} = \frac12 t \,\text{sech}(t) = \sum_{n=0}^\infty \frac{E_{2n}t^{2n+1}}{2(2n)!} \;\; |t|<\frac\pi2$, where $E_k$ are the Euler numbers.
I don't know if this will help at all though, since $|t|<\frac\pi2$...
10. Originally Posted by EinStone
Ok, So to prove that Zeta is holomorphic everywhere, we used $\frac{t}{e^t-1} = \sum_{k=0}^\infty \frac{B_k}{k!} t^k$ where $B_k$ are the Bernoulli numbers.
What all was said to show $\zeta(s)$ is meromorphic everywhere?
11. Ok here is the proof for Zeta being meromorphic everywhere.
$\frac{t}{e^t-1} = \sum_{k=0}^\infty \frac{B_k}{k!} t^k = 1 - \frac{t}{2} + \frac{t^2}{12} + 0t^3 - \frac{t^4}{720} + \cdots$ where $B_k$ are the Bernoulli numbers.
Fix $n > 0$, $f_n(t):= \sum_{k=0}^n (-1)^k \frac{B_k}{k!} t^k = 1 + \frac{t}{2} + \frac{B_2}{2!}t^2 + \frac{B_4}{4!}t^4 + \cdots + \frac{B_n}{n!}t^n$ (all terms even)
Now for Re(s) > 1:
$\Gamma(s) * \zeta(s) = \int_0^\infty \frac{t e^t}{e^t-1} e^{-t} t^{s-2}dt = \int_0^\infty (\frac{t e^t}{e^t-1} - f_n(t))e^{-t} t^{s-2}dt$ $+ \int_0^\infty f_n(t) e^{-t} t^{s-2}dt = I_1(s) + I_2(s)$
$\frac{t e^t}{e^t-1}$ is holomorphic at t = 0, has Taylor expansion: $1 + \frac{t}{2} + \frac{B_2}{2!}t^2 + \frac{B_4}{4!}t^4 + \cdots = \frac{t}{e^t-1} +t$
hence $\frac{t e^t}{e^t-1}-f_n(t) = O(t^{n+1})$ near t = 0.
$\rightsquigarrow$ Integrand of $I_1$ near t = 0 is $\approx O(t^{n+1}t^{s-2}) = O(t^{n+s-1})$. So $I_1$ converges if Re(n+s-1) > -1, i.e. if Re(s) > -n.
For $I_2$: $I_2(s) = \int_0^\infty (1 + \frac{t}{2} + \sum_{k=2}^n \frac{B_k}{k!}t^{k}) e^{-t} t^{s-2}dt = \Gamma(s-1) + \frac{1}{2}\Gamma(s) + \sum_{k=2}^n \frac{B_k}{k!} \Gamma(s+k-1)$ , a finite sum of meromorphic functions on $\mathbb{C}$.
$\rightsquigarrow \zeta$ is meromorphic for Re(s) > -n, n arbitrary, hence $\zeta$ is meromorphic on $\mathbb{C}$.
12. I didn't know this proof; it is rather simple, and adaptable to many situations.
Notice that you only use the fact that there exists an expansion $\frac{t e^t}{e^t-1}=a_0+a_1t+\cdots+a_n t^n+O(t^{n+1})=f_n(t)+O(t^{n+1})$ when $t\to 0$, regardless of the value of $a_n$ (or of the convergence of the series expansion).
Since $t\mapsto\frac{t e^t}{e^t+e^{-t}}$ is $\mathcal{C}^\infty$ (even analytic), there is also such an expansion $\frac{t e^t}{e^t+e^{-t}}=a_0+a_1t+\cdots+a_n t^n+O(t^{n+1})$ when $t\to0$ (this is Taylor-Young's theorem). And you can transpose the proof seamlessly.
It will show you that $L$ is meromorphic on $\mathbb{C}$, so you also have to justify it has no poles... That may require the values of the coefficients $a_n$ ; I'll think about it.
Sidenote: there is a justification missing in your proof, but maybe it was an implicit reference to a proof you did for Gamma. An argument is indeed necessary to justify that $I_1$ is analytic; it is of the form $\int_0^\infty f(t,s)dt$ where, for all $t>0$, $s\mapsto f(t,s)$ is analytic, but this is not sufficient to conclude that the integral itself is analytic. A "domination" is needed, like for the continuity or differentiability theorems under the integral sign. But it is simple to apply here, so I guess you were told it is just like for Gamma.
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Addendum: in fact, the same method proves the analyticity on $\mathbb{C}$.
Indeed, we get an expression like $L(s)=\frac{1}{\Gamma(s)}{I_1(s)}+\sum_{k=0}^n a_k\frac{\Gamma(s+k-1)}{\Gamma(s)}$, for ${\rm Re}(s)>-n$, where $I_1(s)$ is analytic. We know that $\Gamma(s)\neq 0$, hence the first term $\frac{I_1(s)}{\Gamma(s)}$ defines an analytic function.
Furthermore, for $k\geq 1$, $\Gamma(s+k-1)=(s+k-2)(s+k-3)\cdots s \Gamma(s)$ (usual functional equation of $\Gamma$ used several times) hence $\frac{\Gamma(s+k-1)}{\Gamma(s)}=(s+k-2)(s+k-3)\cdots s$ is just a polynomial, hence it is analytic.
The only piece that is not analytic for $\zeta$ is the term $\frac{\Gamma(s-1)}{\Gamma(s)}=\frac{1}{s-1}$, which comes from $k=0$. (This proves that 1 is the only pole for $\zeta$; it is simple with residue 1).
However, for $L(s)$, the term $a_0$ is 0 because $\frac{te^t}{e^t+e^{-t}}$ is 0 at 0. Therefore, all terms are analytic. Thus $L$ is analytic on ${\rm Re}(s)>-n$ for all n, hence on $\mathbb{C}$. qed.
13. So I want to apply the same proof, but how do I find $I_1$ for L? Also to show that $I_2$ is meromorphic, don't I need to know the $a_n$?
14. Originally Posted by EinStone
So I want to apply the same proof, but how do I find $I_1$ for L? Also to show that $I_2$ is meromorphic, don't I need to know the $a_n$?
All you have to do is replace $\frac{t e^t}{e^t-1}$ by $\frac{t e^t}{e^t+e^{-t}}$, and of course $\zeta(s)$ by $L(s)$. Everything is the same, except that $a_0=0$ for $L$ (hence the analyticity), but you don't need the other values (neither for zeta of for L), just the fact that they exist. They are constants, so they matter in no way.
15. Now I get it, its just the existence (even for Zeta) that is needed, which follows from analyticity.
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http://mathoverflow.net/questions/86915/continuity-of-a-weight-on-its-definition-domain-in-a-von-neumann-algebra
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## Continuity of a weight on its definition domain in a von Neumann algebra
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Let $M$ be a von Neumann algebra and $\varphi$ be a normal weight on it, and let $A$ be its definition subalgebra. We still denote $\varphi$ the extension to $A$ as a linear positive functional. It is known that $\varphi$ is lower-ultraweakly-semicontinuous on $M^+$ (the positive elements of $M$).
Questions:
• Is $\varphi$ ultraweakly continuous on $A$ (as a linear positive functional), where $A$ has the induced ultraweak topology of $M$ ? (Clearly, if $A=M$ this assertion is classical, right ?)
• If we fix a Hilbert space representation of $M$ (so that $M\subseteq \mathcal{B}(H)$). Do we have that $\varphi$ is strongly or weakly continuous on A ?
(If it helps, one can suppose $\varphi$ to be a trace, semifinite and faithful).
• A third (somewhat related) question: Suppose that $B$ is a subalgebra of a von Neumann algebra, and that $f:B\to M$ is a positive linear map, such that $f$ is normal in the following sense : for any increasing net in $B^+$ with supremum in $B^+$, the image (by $f$) of this supremum is the supremum of the image of the net (the usual notion of normality, but with the hypothesis that the supremum lies in $B^+$). Do we have that $f$ is continuous (for the ultraweak topologies)?
These questions seem natural to me, but I haven't been able to locate any reference about them.
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## 1 Answer
1) if $A=M$ the assertion is indeed classical: normal states are exactly those ultraweakly continuous. But consider the case where $M=B(H)$ and $\varphi$ is the trace. Then the definition subalgebra $A$ is exactly the trace-class operators. Let `$\{p_k\}\subset A$` be a maximal net of pairwise orthogonal projections of trace 1 (i.e. `$\{e_{kk}\}$` for any choice of matrix units). Then $p_k\to0$ ultraweakly, but $\mbox{Tr}(p_k)=1$ for all $k$. So the trace is not ultraweakly continuous on the definition algebra, only ultraweakly lower-semicontinuous.
2) The example on 1) is already explicitly represented, so no.
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1
For (3), doesn't the same example work, considered as the map $T(H)\rightarrow B(H), x\mapsto \tr(x) 1$? I think this follows, as the question insists that the supremum of our increasing net in $B^+$ exists in $B^+$. – Matthew Daws Jan 29 2012 at 9:38
I first answered that same example to 3, but then I thought that maybe the question required $B$ to be a von Neumann subalgebra and the map to be defined everywhere. So I stopped to consider cases like when $B$ is a monotone complete C$^*$-subalgebra of $B(H)$. Maybe Oliver can clarify what he expects in 3). – Martin Argerami Jan 29 2012 at 13:28
I have just added a possible answer to 3). – Martin Argerami Jan 29 2012 at 14:02
1
Never mind. Let's wait for Oliver to clarify question 3. – Martin Argerami Jan 29 2012 at 14:46
1
Indeed, this answers also my question 3. Thanks! But I am actually also curious about the case you mentioned where $f$ is everywhere defined and $B$ is a von Neumann subalgebra. – Oliver Jan 29 2012 at 20:42
show 3 more comments
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http://crypto.stackexchange.com/questions/2388/how-much-bigger-does-a-precomputed-lookup-table-get-when-salt-is-added/3063
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How much bigger does a precomputed lookup table get when salt is added?
I am trying to wrap my head around the benefits of salt in cryptography.
http://en.wikipedia.org/wiki/Salt_(cryptography)
I understand that adding salt makes it harder to precompute a table. But exactly how much harder do things get with salt?
It seems to me, like when you add salt, the number of entries in your precomputed table would = number of common passwords to precompute x number of entries in the password table (ie number of different possibile salts).
So if you have a list of 100 common passwords, then without salt, you would have 100 hashed passwords. But if you have 10 users on a system, with 10 different salts, then you would now have 1000 different combinations to check.
So as the numbers of users or the size of common password list increases, the precomputed table gets so big that you can't pre-compute it easily (if at all)
Am I getting this? Do I have it right?
*Note*Cross posted on cs theory http://cstheory.stackexchange.com/questions/11118/how-much-bigger-does-a-precomputed-lookup-table-get-when-salt-is-added#comment30381_11118 CS Theory Users suggested I post here too
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I don't get why you're talking about the number of users. – CodesInChaos Jun 26 '12 at 20:16
3 Answers
Looks right to me, assuming you know what the 10 salts are.
If you don't know a priori what the salts are (which is typically the case when building a rainbow table), you'd have to compute for the entire salt space. This is where the big gain comes into play as for each additional salt bit, the storage requirement doubles. Thus you get exponential growth of the storage requirement.
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Well, you've got the main gist of it ("it tries to make precomputation take too long to be practical") correct, but you've gotten some of the details wrong. Here are the corrected details:
• When someone actually uses a precomputed table, an attacker with a big table of passwords and hashed versions of those passwords isn't the only thing we have to worry about. Another thing we need to worry about is if someone puts together a Rainbow table. In essence, a rainbow table is a clever way of compressing the table of passwords and hashes; a rainbow table that covers N passwords still takes $O(N)$ time to build, but it may take up only $O(N/k)$ space (at the cost of making a check to see if something is in the table take up $O(k)$ time; the attacker chooses $k$ when he builds the Rainbow table). What a salt does to a Rainbow table is force the table to cover various passwords and salts; if you have $M$ possible salts, this effectively increases the amount of time building the table to $O(NM)$.
Salt does have this limitation; it limits the amount of precomputation that the attacker can do, however if the attacker is interested in one specific password, then after he has recovered the salt/hash, then he can still spend $O(N)$ time to check to see if $N$ passwords are correct; this implies that huge values of $M$ (that is, really large salts) don't really add that much protection.
• As mikeazo mentioned, it doesn't really matter how many salts are currently active. If the attacker doesn't know the salt when he is precomputing, he'll need to cover all the salt values (or, at least, any salt value he doesn't cover will be a guaranteed miss when he actually gets the salt/hash combination).
• Salt also has another advantage (which has nothing to do with precomputation); it also disguises when two passwords are the same. If Alice and Bob happen to pick the same password (perhaps because they're really the same person), then without salt, their password will hash the same, and that will be apparent to someone just viewing the hash. However, if we add salt, the hashes will appear to be unrelated (unless we happen to pick the same salts). Of course, there are other ways to achieve this as well (say, include the username in the hash).
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when you add salt you multiply the number possible hashes by the number of possible salts.
The time to generate a rainbow table is proportional to the number of hashes.
Now the time to crack for a rainbow table is proportional to the square of the number of hashes divided by the square of size of the table.
So for every bit of salt you add, you make generating a table twice as long and using a table 4 times as slow, if the size of the table is limited.
You want to add salt until it is not feasible to create a table anymore (2^80 combinations of salt and passwords should be safe, but use 128 bits of salt anyway to be on the safe side)
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You don't even need salts that large. You don't need to make creating the table infeasible, it's enough to make multi-target attacks less efficient than single target attacks. For user password-hashes, 64 bits is plenty. – CodesInChaos Jun 26 '12 at 20:28
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http://crypto.stackexchange.com/questions/1490/creating-a-hash-of-xord-blocks?answertab=oldest
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# Creating a hash of XOR'd blocks
Suppose a message $m$ is divided into blocks of length $160$ bits: $m > = M_1 || M_2 || ... || M_l$ And define $h(m) = M_1 \oplus M_2 \oplus ... \oplus M_l$
Which of the three desirable properties of a good cryptographic hash function does h satisfy? Show that it does not satisfy the other two.
Three desirable properties of a cryptographic hash function http://en.wikipedia.org/wiki/Cryptographic_hash_function#Properties
I feel like if there is only one, like the question states, it has to be the first one, preimage resistance. However, given a hash $h = 000....0$ I can find a bunch of $m$'s that could give that particular hash. Maybe since there are lots of solutions you can't pin point the specific $m$ that gave that value?
What do you think?
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I first thought it might've been supposed to be $h(m)=H(M_1)\oplus H(M_2)\oplus\dotsb\oplus H(M_\ell)$, where $H$ is a PRF, but that's not preimage resistant either. (You need to hash about as many random blocks as there are bits in the output and do some linear algebra to get first preimages, but that's a negligible amount of work.) – Ilmari Karonen Dec 16 '11 at 19:07
## 1 Answer
I believe that this is a poorly written question: such an $h$ obviously doesn't have either preimage resistance, second preimage resistance or collision resistance.
The inability to rederive the specific value of $m$ based on its hash is not an interesting property; it's pretty much true of any function which generates an output shorter than its input.
I don't know where you're getting these questions from; based on this question, I'd suggest you look elsewhere.
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Old Exams, thanks for the confirmation of my worst fears! – Bobby S Dec 16 '11 at 17:43
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http://math.stackexchange.com/questions/177267/fast-algorithms-for-calculating-the-mobius-inversion
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# Fast algorithms for calculating the Möbius inversion
Recall the Möbius inversion formula: if we have two functions $f,g \colon \mathbf{N} \to \mathbf{N}$ such that $$g(n) = \sum_{k=1}^n f\left(\left\lfloor \frac{n}{k} \right\rfloor\right)$$ holds for every $n \in \mathbf{N}$, then the values of $f$ can be recovered as $$f(n) = \sum_{k=1}^n \mu(k) g\left(\left\lfloor \frac{n}{k} \right\rfloor\right),$$ where $\mu(k)$ is the Möbius function.
I am interested in fast algorithms for calculating a single value $f(n)$, assuming that $g$ can be calculated in $O(1)$ time.
The best algorithm I know of is as follows: There are $O(\sqrt{n})$ different values $\left\lfloor \frac{n}{k} \right\rfloor$, $1 \le k \le n$. If we have already calculated $f\left(\left\lfloor \frac{n}{k} \right\rfloor\right)$ for $2 \le k \le n$, then $$f(n) = g(n) - \sum_{k=2}^n f\left(\left\lfloor \frac{n}{k} \right\rfloor\right)$$ can be calculated in $O(\sqrt{n})$ time by counting the multiplicities of the terms in the sum. By calculating the $f\left(\left\lfloor \frac{n}{k} \right\rfloor\right)$ from bigger to lower $k$ the total time required will then be $O(n^{3/4})$ while using $O(\sqrt{n})$ space.
I'm also interested in possible improvements to the above algorithm, even if they reduce the required time just by a constant factor.
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## 1 Answer
You can improve the $O(n^{3/4})$ algorithm by a constant factor of roughly three by computing only the odd summands of $k$ by using the identity:
$$\begin{array}{lll} f \left( n \right) & = & g \left( n \right) - g \left( \frac{n}{2} \right) - \sum_{3 \leq k \leq n, k \:\mathrm{odd}} f \left( \left\lfloor \frac{n}{k} \right\rfloor \right) \end{array}$$
which can be derived by subtracting $g(\frac{n}{2})$ from $g(n)$. The odd-only approach is hinted at in [Lehman 1960] (search for "in practice").
You can further improve the time complexity to $O(n^{2/3+\epsilon})$ if you can compute $f(k)$ for each $k\in \left\{1,..,u \right\}$ in $O(u^{1+\epsilon})$, as is typical for arithmetic functions by using a (possibly segmented) sieve.
Two references for the $O(n^{2/3})$ approach are [Pawlewicz 2008] and [Deleglise 1996].
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Thank you for the good answer. – J. J. Aug 1 '12 at 16:31
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http://nrich.maths.org/5338/index
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nrich enriching mathematicsSkip over navigation
### Whole Number Dynamics I
The first of five articles concentrating on whole number dynamics, ideas of general dynamical systems are introduced and seen in concrete cases.
### Whole Number Dynamics II
This article extends the discussions in "Whole number dynamics I". Continuing the proof that, for all starting points, the Happy Number sequence goes into a loop or homes in on a fixed point.
### Whole Number Dynamics III
In this third of five articles we prove that whatever whole number we start with for the Happy Number sequence we will always end up with some set of numbers being repeated over and over again.
# Odd Stones
#### $27$ stones are distributed between $3$ circles
On a "move" a stone is removed from two of the circles and placed in the third circle.
So, in the illustration, if a stone is removed from the $4$ and the $10$ circles and added to the $13$ circle, the new distribution would be $3$ - $9$ - $15$
#### Check you can turn $2$ - $8$ - $17$ into $3$ - $9$ - $15$ in
two "moves"
Here are five of the ways that $27$ stones could be distributed between the three circles :
$6$ - $9$ - $12$
$3$ - $9$ - $15$
$4$ - $10$ - $13$
$4$ - $9$ - $14$
$2$ - $8$ - $17$
There is always some sequence of "moves" that will turn each distribution into any of the others - apart from one.
Identify the distribution that does not belong with the other four.
Can you be certain that this is actually impossible rather than just hard and so far unsuccessful?
The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.
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http://math.stackexchange.com/questions/114406/how-can-i-apply-youngs-inequality-here
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How can I apply Young's inequality here?
I'm reading Fourier Analysis and Nonlinear Partial Differential Equations by Rapha\"el Danchin et al. There are lines on page 9 reads:
Using Young's inequality for $\mathbb{Z}$ equipped with the counting measure, we may now deduce that $$I(f,g,h)\leqslant C\sum_{j,k,\ell \in \mathbb{Z}}c_kd_{\ell}2^{-\varepsilon |jq+k|-\varepsilon |jq+\ell|-\varepsilon |k-\ell|}$$ $$\leqslant \frac{C}{\varepsilon}\sum_{k,\ell \in \mathbb{Z}}c_kd_{\ell}2^{-\varepsilon |k-\ell|}$$ $$\leqslant\frac{C}{\varepsilon^2}\lVert(c_k)\rVert_{\ell^{p}}\lVert(d_{\ell})\rVert_{\ell^{p'}}$$ where: $$\frac{1}{p}+\frac{1}{q}=1+\frac{1}{r},\,\,\varepsilon\triangleq\frac{1}{4}\left(\frac{1}{p}-\frac{1}{r}\right),\,\,(p,q,r)\in (1,\infty)^3, c_k,d_{\ell}>0.$$
I really don't know how to obtain the last two "$\leqslant$". For the last "$\leqslant$" I can only get $\frac{C}{\varepsilon}$ but $\frac{C}{\varepsilon^2}$ because $2^{-\varepsilon |k-\ell|}$ are always no bigger than 1.
Anyone could help me? Any advice will be appreciated.
Edit:
Actually, this is to prove the Young's inequality for weak $L^q$ space, i.e., $$\lVert f*g\rVert_{L^r(G, \mu)}\leqslant C\lVert f\rVert_{L^p(G, \mu)}\lVert g\rVert_{L_w^q(G, \mu)}$$ by homogeneity, he assume $\lVert f\rVert_{L^p(G,\mu)}=\lVert g\rVert_{L_w^q(G,\mu)}=\lVert h\rVert_{L^{r'}(G,\mu)}=1$, and then prove $$I(f,g,h)=\int_{G^2}f(y)g(y^{-1}\cdot x)h(x)\,d\mu(x)\,d\mu(y)\leqslant C$$ instead.
So I think it's enough once we get (Juli\'{a}n's answer) $$I(f,g,h)\leqslant C\sum_{j,k,\ell \in \mathbb{Z}}c_kd_{\ell}2^{-\varepsilon |jq+k|-\varepsilon |jq+\ell|-\varepsilon |k-\ell|}$$ $$\leqslant 2^{\varepsilon2q+1}\frac{1}{1-2^{-\varepsilon2q}}\sum_{k,\ell \in \mathbb{Z}}c_kd_{\ell}2^{-\varepsilon |k-\ell|}$$ $$\leqslant 2^{\varepsilon2q+1}\frac{1}{1-2^{-\varepsilon2q}}\lVert(c_k)\rVert_{\ell^{p}}\lVert(d_{\ell})\rVert_{\ell^{p'}}$$ $$\leqslant 2^{\varepsilon2q+1}\frac{1}{1-2^{-\varepsilon2q}}\lVert(c_k)\rVert_{\ell^{p}}\lVert(d_{\ell})\rVert_{\ell^{r'}}$$ By the definition of $\varepsilon$ we can denotes $2^{\varepsilon2q+1}\frac{1}{1-2^{-\varepsilon2q}}$ by $C(q,r)$. And by the assumption of norm 1, the coefficients of atomic decompositions of $f$ and $h$ we know $\lVert(c_k)\rVert_{\ell^{p}}\leqslant 2, \lVert(d_{\ell})\rVert_{\ell^{r'}}\leqslant 2$. So we get $I\leqslant C(q,r)$. I don't know why the $\frac{1}{\varepsilon}$ and $\frac{1}{\varepsilon^2}$ appear. Or am I miss understand something?
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What are $c_k,d_l$ and $I$? – Ilya Feb 28 '12 at 12:21
@Ilya: $I=\int_{G^2}f(y)g(y^{-1}\cdot x)h(x)\,d\mu(x)\,d\mu(y)$, $G$ is a locally compact topological group and $\mu$ a left-invariant Haar measure. $f\in L^p$ and $g$ in weak $L^q$, $h\in L^{r'}$. $f=\sum_k c_kf_k,h=\sum_{\ell}d_{\ell}h_{\ell}$ are atomic decompositions. But I don't think these matters. – Y.Z Feb 28 '12 at 12:44
There is no $j$ index in the last summation, right? – Ilya Feb 28 '12 at 12:58
@Ilya: I'm so sorry! No $j$ index in the last summation! Corrected. – Y.Z Feb 28 '12 at 13:09
1 Answer
For the first inequality, since $|2\,q\,j+k+\ell|\le|j\,q+k|+|j\,q+\ell|$, we have $$\sum_{j\in\mathbb{Z}}2^{-\varepsilon |jq+k|-\varepsilon |jq+\ell|}\le\sum_{j\in\mathbb{Z}}2^{-\varepsilon|2qj+k+\ell|}\quad\forall k,\ell\in\mathbb{Z}.$$ For fixed $k$ and $\ell$ choose $j_0\in\mathbb{Z}$ such that $\Bigl|\dfrac{k+\ell}{2\,q}-j_0\Bigr|<1$. Then $$|2\,q\,j+k+\ell|=|2\,q(j+j_0)+k+\ell-2\,q\,j_0|\ge|2\,q(j+j_0)|-2\,q.$$ Then $$\begin{align*} \sum_{j\in\mathbb{Z}}2^{-\varepsilon|2qj+k+\ell|}&\le2^{\varepsilon2q}\sum_{j\in\mathbb{Z}}2^{-\varepsilon|2q(j+j_0)|}\\ &=2^{\varepsilon2q}\sum_{j\in\mathbb{Z}}2^{-\varepsilon|2qj|}\\ &\le2^{\varepsilon2q+1}\sum_{j=0}^\infty2^{-\varepsilon2qj}\\ &=2^{\varepsilon2q+1}\frac{1}{1-2^{-\varepsilon2q}}\\ &\le\frac{C_q}{\varepsilon}, \end{align*}$$ where $C_q$ is independent of $k$ and $\ell$.
The second inequality holds because $\varepsilon\le1$.
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Thanks for your help! I have a question about the $\frac{1}{\varepsilon}$ and I edit the post for that. Another question is where did we use the Young's inequality? I can just see I use H\"older's inequality after the last summation. Thanks! – Y.Z Feb 28 '12 at 16:27
Observe that $\bigl(1-2^{\varepsilon2q}\bigr)^{-1}\sim C/\epsilon$ as $\epsilon\to0$. As for the use of Young's inequality, I assumed it was used to bound $I$ im terms of the sum. – Julián Aguirre Feb 28 '12 at 18:36
Here may be a Young's inequality: $$\sum_{k,\ell \in \mathbb{Z}}c_kd_{\ell}2^{-\varepsilon |k-\ell|}\leqslant \lVert(c_k)\rVert_{\ell^{p}}\lVert(d_{\ell})\rVert_{\ell^{p'}}\lVert(2^{-\varepsilon |\cdot|})\rVert_{\ell^{\infty}} =\lVert(c_k)\rVert_{\ell^{p}}\lVert(d_{\ell})\rVert_{\ell^{p'}} .$$ But still I don't know why $\frac{1}{\varepsilon}$ becomes $\frac{1}{\varepsilon^2}$. – Y.Z Feb 29 '12 at 2:33
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http://www.absoluteastronomy.com/topics/Pierre_de_Fermat
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Pierre de Fermat
# Pierre de Fermat
Overview
Pierre de Fermat was a French
French people
The French are a nation that share a common French culture and speak the French language as a mother tongue. Historically, the French population are descended from peoples of Celtic, Latin and Germanic origin, and are today a mixture of several ethnic groups...
lawyer at the Parlement
Parlement
Parlements were regional legislative bodies in Ancien Régime France.The political institutions of the Parlement in Ancien Régime France developed out of the previous council of the king, the Conseil du roi or curia regis, and consequently had ancient and customary rights of consultation and...
of Toulouse
Toulouse
Toulouse is a city in the Haute-Garonne department in southwestern FranceIt lies on the banks of the River Garonne, 590 km away from Paris and half-way between the Atlantic Ocean and the Mediterranean Sea...
, France
France
The French Republic , The French Republic , The French Republic , (commonly known as France , is a unitary semi-presidential republic in Western Europe with several overseas territories and islands located on other continents and in the Indian, Pacific, and Atlantic oceans. Metropolitan France...
, and an amateur mathematician who is given credit for early developments that led to infinitesimal calculus
Infinitesimal calculus
Infinitesimal calculus is the part of mathematics concerned with finding slope of curves, areas under curves, minima and maxima, and other geometric and analytic problems. It was independently developed by Gottfried Leibniz and Isaac Newton starting in the 1660s...
, including his adequality
Adequality
In the history of infinitesimal calculus, adequality is a technique developed by Pierre de Fermat. Fermat said he borrowed the term from Diophantus. Adequality was a technique first used to find maxima for functions and then adapted to find tangent lines to curves...
. In particular, he is recognized for his discovery of an original method of finding the greatest and the smallest ordinate
Ordinate
In mathematics, ordinate refers to that element of an ordered pair which is plotted on the vertical axis of a two-dimensional Cartesian coordinate system, as opposed to the abscissa...
s of curved lines, which is analogous to that of the then unknown differential calculus
Differential calculus
In mathematics, differential calculus is a subfield of calculus concerned with the study of the rates at which quantities change. It is one of the two traditional divisions of calculus, the other being integral calculus....
, and his research into number theory
Number theory
Number theory is a branch of pure mathematics devoted primarily to the study of the integers. Number theorists study prime numbers as well...
. He made notable contributions to analytic geometry
Analytic geometry
Analytic geometry, or analytical geometry has two different meanings in mathematics. The modern and advanced meaning refers to the geometry of analytic varieties...
, probability
Probability
Probability is ordinarily used to describe an attitude of mind towards some proposition of whose truth we arenot certain. The proposition of interest is usually of the form "Will a specific event occur?" The attitude of mind is of the form "How certain are we that the event will occur?" The...
, and optics
Optics
Optics is the branch of physics which involves the behavior and properties of light, including its interactions with matter and the construction of instruments that use or detect it. Optics usually describes the behavior of visible, ultraviolet, and infrared light...
. He is best known for Fermat's Last Theorem
Fermat's Last Theorem
In number theory, Fermat's Last Theorem states that no three positive integers a, b, and c can satisfy the equation an + bn = cn for any integer value of n greater than two....
, which he described in a note at the margin of a copy of Diophantus
Diophantus
Diophantus of Alexandria , sometimes called "the father of algebra", was an Alexandrian Greek mathematician and the author of a series of books called Arithmetica. These texts deal with solving algebraic equations, many of which are now lost...
' Arithmetica
Arithmetica
Arithmetica is an ancient Greek text on mathematics written by the mathematician Diophantus in the 3rd century AD. It is a collection of 130 algebraic problems giving numerical solutions of determinate equations and indeterminate equations.Equations in the book are called Diophantine equations...
.
Discussion
Ask a question about 'Pierre de Fermat' Start a new discussion about 'Pierre de Fermat' Answer questions from other users Full Discussion Forum
Unanswered Questions
Recent Discussions
Quotations
a^p \equiv a \pmod\,\!
Also known as Fermat's little theorem|Fermat's little theorem, the formula above is true for any prime p where a
$a^p \equiv a \pmod\,\!$
Also known as Fermat's little theorem|Fermat's little theorem, the formula above is true for any prime $p$ where $a$
$a^p \equiv a \pmod\,\!$
Also known as Fermat's little theorem|Fermat's little theorem, the formula above is true for any prime $p$ where
Encyclopedia
Pierre de Fermat was a French
French people
The French are a nation that share a common French culture and speak the French language as a mother tongue. Historically, the French population are descended from peoples of Celtic, Latin and Germanic origin, and are today a mixture of several ethnic groups...
lawyer at the Parlement
Parlement
Parlements were regional legislative bodies in Ancien Régime France.The political institutions of the Parlement in Ancien Régime France developed out of the previous council of the king, the Conseil du roi or curia regis, and consequently had ancient and customary rights of consultation and...
of Toulouse
Toulouse
Toulouse is a city in the Haute-Garonne department in southwestern FranceIt lies on the banks of the River Garonne, 590 km away from Paris and half-way between the Atlantic Ocean and the Mediterranean Sea...
, France
France
The French Republic , The French Republic , The French Republic , (commonly known as France , is a unitary semi-presidential republic in Western Europe with several overseas territories and islands located on other continents and in the Indian, Pacific, and Atlantic oceans. Metropolitan France...
, and an amateur mathematician who is given credit for early developments that led to infinitesimal calculus
Infinitesimal calculus
Infinitesimal calculus is the part of mathematics concerned with finding slope of curves, areas under curves, minima and maxima, and other geometric and analytic problems. It was independently developed by Gottfried Leibniz and Isaac Newton starting in the 1660s...
, including his adequality
Adequality
In the history of infinitesimal calculus, adequality is a technique developed by Pierre de Fermat. Fermat said he borrowed the term from Diophantus. Adequality was a technique first used to find maxima for functions and then adapted to find tangent lines to curves...
. In particular, he is recognized for his discovery of an original method of finding the greatest and the smallest ordinate
Ordinate
In mathematics, ordinate refers to that element of an ordered pair which is plotted on the vertical axis of a two-dimensional Cartesian coordinate system, as opposed to the abscissa...
s of curved lines, which is analogous to that of the then unknown differential calculus
Differential calculus
In mathematics, differential calculus is a subfield of calculus concerned with the study of the rates at which quantities change. It is one of the two traditional divisions of calculus, the other being integral calculus....
, and his research into number theory
Number theory
Number theory is a branch of pure mathematics devoted primarily to the study of the integers. Number theorists study prime numbers as well...
. He made notable contributions to analytic geometry
Analytic geometry
Analytic geometry, or analytical geometry has two different meanings in mathematics. The modern and advanced meaning refers to the geometry of analytic varieties...
, probability
Probability
Probability is ordinarily used to describe an attitude of mind towards some proposition of whose truth we arenot certain. The proposition of interest is usually of the form "Will a specific event occur?" The attitude of mind is of the form "How certain are we that the event will occur?" The...
, and optics
Optics
Optics is the branch of physics which involves the behavior and properties of light, including its interactions with matter and the construction of instruments that use or detect it. Optics usually describes the behavior of visible, ultraviolet, and infrared light...
. He is best known for Fermat's Last Theorem
Fermat's Last Theorem
In number theory, Fermat's Last Theorem states that no three positive integers a, b, and c can satisfy the equation an + bn = cn for any integer value of n greater than two....
, which he described in a note at the margin of a copy of Diophantus
Diophantus
Diophantus of Alexandria , sometimes called "the father of algebra", was an Alexandrian Greek mathematician and the author of a series of books called Arithmetica. These texts deal with solving algebraic equations, many of which are now lost...
' Arithmetica
Arithmetica
Arithmetica is an ancient Greek text on mathematics written by the mathematician Diophantus in the 3rd century AD. It is a collection of 130 algebraic problems giving numerical solutions of determinate equations and indeterminate equations.Equations in the book are called Diophantine equations...
.
## Life and work
Fermat was born in Beaumont-de-Lomagne
Beaumont-de-Lomagne
Beaumont-de-Lomagne is a commune in the Tarn-et-Garonne department in the Midi-Pyrénées region in southern France.-History:Beaumont-de-Lomagne, bastide, was founded in 1276 following the act of coregency between the abbey of Grandselve and King Philip III of France - the King was represented by his...
, Tarn-et-Garonne
Tarn-et-Garonne
Tarn-et-Garonne is a French department in the southwest of France. It is traversed by the Rivers Tarn and Garonne, from which it takes its name.-History:...
, France; the late 15th century mansion where Fermat was born is now a museum. He was of Basque
Basque people
The Basques as an ethnic group, primarily inhabit an area traditionally known as the Basque Country , a region that is located around the western end of the Pyrenees on the coast of the Bay of Biscay and straddles parts of north-central Spain and south-western France.The Basques are known in the...
origin. Fermat's father was a wealthy leather merchant and second consul of Beaumont-de-Lomagne. Pierre had a brother and two sisters and was almost certainly brought up in the town of his birth. There is little evidence concerning his school education, but it may have been at the local Franciscan
Franciscan
Most Franciscans are members of Roman Catholic religious orders founded by Saint Francis of Assisi. Besides Roman Catholic communities, there are also Old Catholic, Anglican, Lutheran, ecumenical and Non-denominational Franciscan communities....
monastery.
He attended the University of Toulouse
University of Toulouse
The Université de Toulouse is a consortium of French universities, grandes écoles and other institutions of higher education and research, named after one of the earliest universities established in Europe in 1229, and including the successor universities to that earlier university...
before moving to Bordeaux
Bordeaux
Bordeaux is a port city on the Garonne River in the Gironde department in southwestern France.The Bordeaux-Arcachon-Libourne metropolitan area, has a population of 1,010,000 and constitutes the sixth-largest urban area in France. It is the capital of the Aquitaine region, as well as the prefecture...
in the second half of the 1620s. In Bordeaux he began his first serious mathematical researches and in 1629 he gave a copy of his restoration of Apollonius
Apollonius of Perga
Apollonius of Perga [Pergaeus] was a Greek geometer and astronomer noted for his writings on conic sections. His innovative methodology and terminology, especially in the field of conics, influenced many later scholars including Ptolemy, Francesco Maurolico, Isaac Newton, and René Descartes...
's De Locis Planis to one of the mathematicians there. Certainly in Bordeaux he was in contact with Beaugrand and during this time he produced important work on maxima and minima
Maxima and minima
In mathematics, the maximum and minimum of a function, known collectively as extrema , are the largest and smallest value that the function takes at a point either within a given neighborhood or on the function domain in its entirety .More generally, the...
which he gave to Étienne d'Espagnet who clearly shared mathematical interests with Fermat. There he became much influenced by the work of François Viète
François Viète
François Viète , Seigneur de la Bigotière, was a French mathematician whose work on new algebra was an important step towards modern algebra, due to its innovative use of letters as parameters in equations...
.
From Bordeaux, Fermat went to Orléans
Orléans
-Prehistory and Roman:Cenabum was a Gallic stronghold, one of the principal towns of the Carnutes tribe where the Druids held their annual assembly. It was conquered and destroyed by Julius Caesar in 52 BC, then rebuilt under the Roman Empire...
where he studied law at the University. He received a degree in civil law before, in 1631, receiving the title of councillor at the High Court of Judicature in Toulouse, which he held for the rest of his life. Due to the office he now held he became entitled to change his name from Pierre Fermat to Pierre de Fermat. Fluent in Latin, Basque, classical Greek, Italian, and Spanish, Fermat was praised for his written verse in several languages, and his advice was eagerly sought regarding the emendation of Greek texts.
He communicated most of his work in letters to friends, often with little or no proof of his theorems. This allowed him to preserve his status as an "amateur" while gaining the recognition he desired. This naturally led to priority disputes with contemporaries such as Descartes
René Descartes
René Descartes ; was a French philosopher and writer who spent most of his adult life in the Dutch Republic. He has been dubbed the 'Father of Modern Philosophy', and much subsequent Western philosophy is a response to his writings, which are studied closely to this day...
and Wallis. He developed a close relationship with Blaise Pascal
Blaise Pascal
Blaise Pascal , was a French mathematician, physicist, inventor, writer and Catholic philosopher. He was a child prodigy who was educated by his father, a tax collector in Rouen...
.
Anders Hald
Anders Hald
Anders Hald was a Danish statistician who made contributions to the history of statistics.He was a professor at the University of Copenhagen from 1960 to 1982.- Bibliography :...
writes that, "The basis of Fermat's mathematics was the classical Greek treatises combined with Vieta's new algebra
New algebra
The new algebra or symbolic analysis is a formalization of algebra promoted by François Viète in 1591 and by his successors...
ic methods."
### Work
Fermat's pioneering work in analytic geometry was circulated in manuscript form in 1636, predating the publication of Descartes' famous La géométrie. This manuscript was published posthumously in 1679 in "Varia opera mathematica", as Ad Locos Planos et Solidos Isagoge, ("Introduction to Plane and Solid Loci").
In Methodus ad disquirendam maximam et minima and in De tangentibus linearum curvarum, Fermat developed a method for determining maxima, minima, and tangents to various curves that was equivalent to differentiation. In these works, Fermat obtained a technique for finding the centers of gravity of various plane and solid figures, which led to his further work in quadrature
Numerical integration
In numerical analysis, numerical integration constitutes a broad family of algorithms for calculating the numerical value of a definite integral, and by extension, the term is also sometimes used to describe the numerical solution of differential equations. This article focuses on calculation of...
.
Fermat was the first person known to have evaluated the integral of general power functions. Using an ingenious trick, he was able to reduce this evaluation to the sum of geometric series. The resulting formula was helpful to Newton
Isaac Newton
Sir Isaac Newton PRS was an English physicist, mathematician, astronomer, natural philosopher, alchemist, and theologian, who has been "considered by many to be the greatest and most influential scientist who ever lived."...
, and then Leibniz, when they independently developed the fundamental theorem of calculus
Fundamental theorem of calculus
The first part of the theorem, sometimes called the first fundamental theorem of calculus, shows that an indefinite integration can be reversed by a differentiation...
.
In number theory, Fermat studied Pell's equation
Pell's equation
Pell's equation is any Diophantine equation of the formx^2-ny^2=1\,where n is a nonsquare integer. The word Diophantine means that integer values of x and y are sought. Trivially, x = 1 and y = 0 always solve this equation...
, perfect number
Perfect number
In number theory, a perfect number is a positive integer that is equal to the sum of its proper positive divisors, that is, the sum of its positive divisors excluding the number itself . Equivalently, a perfect number is a number that is half the sum of all of its positive divisors i.e...
s, amicable number
Amicable number
Amicable numbers are two different numbers so related that the sum of the proper divisors of each is equal to the other number. A pair of amicable numbers constitutes an aliquot sequence of period 2...
s and what would later become Fermat numbers. It was while researching perfect numbers that he discovered the little theorem
Fermat's little theorem
Fermat's little theorem states that if p is a prime number, then for any integer a, a p − a will be evenly divisible by p...
. He invented a factorization method—Fermat's factorization method
Fermat's factorization method
Fermat's factorization method, named after Pierre de Fermat, is based on the representation of an odd integer as the difference of two squares:N = a^2 - b^2.\...
—as well as the proof technique of infinite descent
Infinite descent
In mathematics, a proof by infinite descent is a particular kind of proof by contradiction which relies on the fact that the natural numbers are well ordered. One typical application is to show that a given equation has no solutions. Assuming a solution exists, one shows that another exists, that...
, which he used to prove Fermat's Last Theorem for the case n = 4. Fermat developed the two-square theorem, and the polygonal number theorem
Fermat polygonal number theorem
In additive number theory, the Fermat polygonal number theorem states that every positive integer is a sum of at most -gonal numbers. That is, every positive number can be written as the sum of three or fewer triangular numbers, and as the sum of four or fewer square numbers, and as the sum of...
, which states that each number is a sum of three triangular number
Triangular number
A triangular number or triangle number numbers the objects that can form an equilateral triangle, as in the diagram on the right. The nth triangle number is the number of dots in a triangle with n dots on a side; it is the sum of the n natural numbers from 1 to n...
s, four square numbers
Lagrange's four-square theorem
Lagrange's four-square theorem, also known as Bachet's conjecture, states that any natural number can be represented as the sum of four integer squaresp = a_0^2 + a_1^2 + a_2^2 + a_3^2\ where the four numbers are integers...
, five pentagonal number
Pentagonal number
A pentagonal number is a figurate number that extends the concept of triangular and square numbers to the pentagon, but, unlike the first two, the patterns involved in the construction of pentagonal numbers are not rotationally symmetrical...
s, and so on.
Although Fermat claimed to have proved all his arithmetic theorems, few records of his proofs have survived. Many mathematicians, including Gauss
Carl Friedrich Gauss
Johann Carl Friedrich Gauss was a German mathematician and scientist who contributed significantly to many fields, including number theory, statistics, analysis, differential geometry, geodesy, geophysics, electrostatics, astronomy and optics.Sometimes referred to as the Princeps mathematicorum...
, doubted several of his claims, especially given the difficulty of some of the problems and the limited mathematical tools available to Fermat. His famous Last Theorem
Fermat's Last Theorem
In number theory, Fermat's Last Theorem states that no three positive integers a, b, and c can satisfy the equation an + bn = cn for any integer value of n greater than two....
was first discovered by his son in the margin on his father's copy of an edition of Diophantus, and included the statement that the margin was too small to include the proof. He had not bothered to inform even Marin Mersenne
Marin Mersenne
Marin Mersenne, Marin Mersennus or le Père Mersenne was a French theologian, philosopher, mathematician and music theorist, often referred to as the "father of acoustics"...
of it. It was not proved until 1994, using techniques unavailable to Fermat.
Although he carefully studied, and drew inspiration from Diophantus, Fermat began a different tradition. Diophantus was content to find a single solution to his equations, even if it were an undesired fractional one. Fermat was interested only in integer solutions to his Diophantine equation
Diophantine equation
In mathematics, a Diophantine equation is an indeterminate polynomial equation that allows the variables to be integers only. Diophantine problems have fewer equations than unknown variables and involve finding integers that work correctly for all equations...
s, and he looked for all possible general solutions. He often proved that certain equations had no solution
Empty set
In mathematics, and more specifically set theory, the empty set is the unique set having no elements; its size or cardinality is zero. Some axiomatic set theories assure that the empty set exists by including an axiom of empty set; in other theories, its existence can be deduced...
, which usually baffled his contemporaries.
Through his correspondence with Pascal in 1654, Fermat and Pascal helped lay the fundamental groundwork for the theory of probability. From this brief but productive collaboration on the problem of points
Problem of points
The problem of points, also called the problem of division of the stakes, is a classical problem in probability theory. One of the famous problems that motivated the beginnings of modern probability theory in the 17th century, it led Blaise Pascal to the first explicit reasoning about what today is...
, they are now regarded as joint founders of probability theory
Probability theory
Probability theory is the branch of mathematics concerned with analysis of random phenomena. The central objects of probability theory are random variables, stochastic processes, and events: mathematical abstractions of non-deterministic events or measured quantities that may either be single...
. Fermat is credited with carrying out the first ever rigorous probability calculation. In it, he was asked by a professional gambler why if he bet on rolling at least one six in four throws of a die he won in the long term, whereas betting on throwing at least one double-six in 24 throws of two dice
Dice
A die is a small throwable object with multiple resting positions, used for generating random numbers...
resulted in him losing. Fermat subsequently proved why this was the case mathematically.
Fermat's principle of least time (which he used to derive Snell's law
Snell's law
In optics and physics, Snell's law is a formula used to describe the relationship between the angles of incidence and refraction, when referring to light or other waves passing through a boundary between two different isotropic media, such as water and glass...
in 1657) was the first variational principle
History of variational principles in physics
A variational principle in physics is an alternative method for determining the state or dynamics of a physical system, by identifying it as an extremum of a function or functional...
enunciated in physics since Hero of Alexandria
Hero of Alexandria
Hero of Alexandria was an ancient Greek mathematician and engineerEnc. Britannica 2007, "Heron of Alexandria" who was active in his native city of Alexandria, Roman Egypt...
described a principle of least distance in the first century CE. In this way, Fermat is recognized as a key figure in the historical development of the fundamental principle of least action
Principle of least action
In physics, the principle of least action – or, more accurately, the principle of stationary action – is a variational principle that, when applied to the action of a mechanical system, can be used to obtain the equations of motion for that system...
in physics. The terms Fermat's principle
Fermat's principle
In optics, Fermat's principle or the principle of least time is the principle that the path taken between two points by a ray of light is the path that can be traversed in the least time. This principle is sometimes taken as the definition of a ray of light...
and Fermat functional were named in recognition of this role.
### Death
He died at Castres
Castres
Castres is a commune, and arrondissement capital in the Tarn department and Midi-Pyrénées region in southern France. It lies in the former French province of Languedoc....
, Tarn. The oldest and most prestigious high school in Toulouse
Toulouse
Toulouse is a city in the Haute-Garonne department in southwestern FranceIt lies on the banks of the River Garonne, 590 km away from Paris and half-way between the Atlantic Ocean and the Mediterranean Sea...
is named after him: the Lycée Pierre de Fermat. French sculptor Théophile Barrau
Théophile Barrau
Théophile Barrau was a French sculptor.Barrau was born in Carcassonne. He was a student of Alexandre Falguière and started at the Salon in 1874. He received awards in 1879, 1880, 1889, and became a Chevalier of the Legion of Honor in 1892...
made a marble statue named Hommage à Pierre Fermat as tribute to Fermat, now at the Capitole of Toulouse.
## Assessment of his work
Together with René Descartes
René Descartes
René Descartes ; was a French philosopher and writer who spent most of his adult life in the Dutch Republic. He has been dubbed the 'Father of Modern Philosophy', and much subsequent Western philosophy is a response to his writings, which are studied closely to this day...
, Fermat was one of the two leading mathematicians of the first half of the 17th century. According to Peter L. Bernstein, in his book Against the Gods, Fermat "was a mathematician of rare power. He was an independent inventor of analytic geometry, he contributed to the early development of calculus, he did research on the weight of the earth, and he worked on light refraction and optics. In the course of what turned out to be an extended correspondence with Pascal, he made a significant contribution to the theory of probability. But Fermat's crowning achievement was in the theory of numbers."
Regarding Fermat's work in analysis, Isaac Newton
Isaac Newton
Sir Isaac Newton PRS was an English physicist, mathematician, astronomer, natural philosopher, alchemist, and theologian, who has been "considered by many to be the greatest and most influential scientist who ever lived."...
wrote that his own early ideas about calculus came directly from "Fermat's way of drawing tangents."
Of Fermat's number theoretic work, the great 20th-century mathematician André Weil
André Weil
André Weil was an influential mathematician of the 20th century, renowned for the breadth and quality of his research output, its influence on future work, and the elegance of his exposition. He is especially known for his foundational work in number theory and algebraic geometry...
wrote that "... what we possess of his methods for dealing with curves
Algebraic curve
In algebraic geometry, an algebraic curve is an algebraic variety of dimension one. The theory of these curves in general was quite fully developed in the nineteenth century, after many particular examples had been considered, starting with circles and other conic sections.- Plane algebraic curves...
of genus 1
Elliptic curve
In mathematics, an elliptic curve is a smooth, projective algebraic curve of genus one, on which there is a specified point O. An elliptic curve is in fact an abelian variety — that is, it has a multiplication defined algebraically with respect to which it is a group — and O serves as the identity...
is remarkably coherent; it is still the foundation for the modern theory of such curves. It naturally falls into two parts; the first one ... may conveniently be termed a method of ascent, in contrast with the descent
Infinite descent
In mathematics, a proof by infinite descent is a particular kind of proof by contradiction which relies on the fact that the natural numbers are well ordered. One typical application is to show that a given equation has no solutions. Assuming a solution exists, one shows that another exists, that...
which is rightly regarded as Fermat's own." Regarding Fermat's use of ascent, Weil continued "The novelty consisted in the vastly extended use which Fermat made of it, giving him at least a partial equivalent of what we would obtain by the systematic use of the group theoretical
Group theory
In mathematics and abstract algebra, group theory studies the algebraic structures known as groups.The concept of a group is central to abstract algebra: other well-known algebraic structures, such as rings, fields, and vector spaces can all be seen as groups endowed with additional operations and...
properties of the rational point
Rational point
In number theory, a K-rational point is a point on an algebraic variety where each coordinate of the point belongs to the field K. This means that, if the variety is given by a set of equationsthen the K-rational points are solutions ∈Kn of the equations...
s on a standard cubic." With his gift for number relations and his ability to find proofs for many of his theorems, Fermat essentially created the modern theory of numbers.
## External links
• Fermat's Achievements
• Fermat's Fallibility at MathPages
• History of Fermat's Last Theorem (French)
• The Life and times of Pierre de Fermat (1601 - 1665) from W. W. Rouse Ball's History of Mathematics
• The Mathematics of Fermat's Last Theorem
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|
http://mathhelpforum.com/math-challenge-problems/25061-find-speed.html
|
# Thread:
1. ## Find the speed...
Here's a problem that I cam across the other day, I can't seem to find the answer though at first glance the question looks relatively simple.
There is a 2mile stretch of road. I drive down the first mile with an average speed of 30mph, find what speed I must travel down the remaining mile for my overall average speed to be 60mph.
At first I thought time=distance/speed, so 2/(30+x)=2/60, but that yields the incorrect answer. Does it have something to do with the harmonic mean? Can someone help?
2. Originally Posted by free_to_fly
Here's a problem that I cam across the other day, I can't seem to find the answer though at first glance the question looks relatively simple.
There is a 2mile stretch of road. I drive down the first mile with an average speed of 30mph, find what speed I must travel down the remaining mile for my overall average speed to be 60mph.
At first I thought time=distance/speed, so 2/(30+x)=2/60, but that yields the incorrect answer. Does it have something to do with the harmonic mean? Can someone help?
I was stumped!
3. Originally Posted by free_to_fly
Here's a problem that I cam across the other day, I can't seem to find the answer though at first glance the question looks relatively simple.
There is a 2mile stretch of road. I drive down the first mile with an average speed of 30mph, find what speed I must travel down the remaining mile for my overall average speed to be 60mph.
At first I thought time=distance/speed, so 2/(30+x)=2/60, but that yields the incorrect answer. Does it have something to do with the harmonic mean? Can someone help?
To drive 2 miles at an average speed of 60mph takes 1/30 of an hour.
You have already driven 1 mile at 30mph, which took 1/30 hour
So you must drive the final mile at $\infty$mph
RonL
4. Originally Posted by free_to_fly
Here's a problem that I cam across the other day, I can't seem to find the answer though at first glance the question looks relatively simple.
There is a 2mile stretch of road. I drive down the first mile with an average speed of 30mph, find what speed I must travel down the remaining mile for my overall average speed to be 60mph.
At first I thought time=distance/speed, so 2/(30+x)=2/60, but that yields the incorrect answer. Does it have something to do with the harmonic mean? Can someone help?
This is a variation of a Mensa question. I love this one!
-Dan
5. Hello, free_to_fly!
As Dan pointed out, this is a classic problem . . .
There is a 2-mile stretch of road.
I drive down the first mile with an average speed of 30 mph.
What speed must I travel on the remaining mile
for my overall average speed to be 60 mph?
To average 60 mph over the 2-mile road,
. . you must cover the distance in: . $\frac{\text{2 miles}}{\text{60 mph}} \:=\:\frac{1}{30}\text{ hours} \:=\:2\text{ minutes}$
You have already driven 1 mile at 30 mph.
. . This took you: . $\frac{\text{1 mile}}{\text{30 mph}}\:=\:\frac{1}{30}\text{ hours} \:=\:2\text{ minutes}$
You have already used up all of the allotted time.
. . It is impossible to drive the last mile in 0 minutes.
|
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|
http://math.stackexchange.com/questions/tagged/computer-science?page=1&sort=unanswered&pagesize=50
|
# Tagged Questions
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0answers
126 views
### what is the relationship between the complexity class E(and EXP) and NP?
I want to know any relationship between the complexity class E(and EXP) and NP. I also would like to know whether there is any $DTIME$ formulation or relations of $NTIME(O(n^k))$ where n is the size ...
0answers
97 views
### How is the loop invarient obtained in this square root bound finding algorithm?
Consider the following algorithm. ...
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http://en.wikipedia.org/wiki/Quasi-Newton_method
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# Quasi-Newton method
In optimization, quasi-Newton methods (a special case of variable metric methods) are algorithms for finding local maxima and minima of functions. Quasi-Newton methods are based on Newton's method to find the stationary point of a function, where the gradient is 0. Newton's method assumes that the function can be locally approximated as a quadratic in the region around the optimum, and uses the first and second derivatives to find the stationary point. In higher dimensions, Newton's method uses the gradient and the Hessian matrix of second derivatives of the function to be minimized.
In quasi-Newton methods the Hessian matrix does not need to be computed. The Hessian is updated by analyzing successive gradient vectors instead. Quasi-Newton methods are a generalization of the secant method to find the root of the first derivative for multidimensional problems. In multi-dimensions the secant equation is under-determined, and quasi-Newton methods differ in how they constrain the solution, typically by adding a simple low-rank update to the current estimate of the Hessian.
The first quasi-Newton algorithm was proposed by W.C. Davidon, a physicist working at Argonne National Laboratory. He developed the first quasi-Newton algorithm in 1959: the DFP updating formula, which was later popularized by Fletcher and Powell in 1963, but is rarely used today. The most common quasi-Newton algorithms are currently the SR1 formula (for symmetric rank one), the BHHH method, the widespread BFGS method (suggested independently by Broyden, Fletcher, Goldfarb, and Shanno, in 1970), and its low-memory extension, L-BFGS. The Broyden's class is a linear combination of the DFP and BFGS methods.
The SR1 formula does not guarantee the update matrix to maintain positive-definiteness and can be used for indefinite problems. The Broyden's method does not require the update matrix to be symmetric and it is used to find the root of a general system of equations (rather than the gradient) by updating the Jacobian (rather than the Hessian).
One of the chief advantages of quasi-Newton methods over Newton's method is that the Hessian matrix (or, in the case of quasi-Newton methods, its approximation) $B$ does not need to be inverted. Newton's method, and its derivatives such as interior point methods, require the Hessian to be inverted, which is typically implemented by solving a system of linear equations and is often quite costly. In contrast, quasi-Newton methods usually generate an estimate of $B^{-1}$ directly.
## Description of the method
As in Newton's method, one uses a second order approximation to find the minimum of a function $f(x)$. The Taylor series of $f(x)$ around an iterate is:
$f(x_k+\Delta x) \approx f(x_k)+\nabla f(x_k)^T \Delta x+\frac{1}{2} \Delta x^T {B} \, \Delta x,$
where ($\nabla f$) is the gradient and $B$ an approximation to the Hessian matrix. The gradient of this approximation (with respect to $\Delta x$) is
$\nabla f(x_k+\Delta x) \approx \nabla f(x_k)+B \, \Delta x$
and setting this gradient to zero provides the Newton step:
$\Delta x=-B^{-1}\nabla f(x_k), \,$
The Hessian approximation $B$ is chosen to satisfy
$\nabla f(x_k+\Delta x)=\nabla f(x_k)+B \, \Delta x,$
which is called the secant equation (the Taylor series of the gradient itself). In more than one dimension $B$ is under determined. In one dimension, solving for $B$ and applying the Newton's step with the updated value is equivalent to the secant method. The various quasi-Newton methods differ in their choice of the solution to the secant equation (in one dimension, all the variants are equivalent). Most methods (but with exceptions, such as Broyden's method) seek a symmetric solution ($B^T=B$); furthermore, the variants listed below can be motivated by finding an update $B_{k+1}$ that is as close as possible to $B_{k}$ in some norm; that is, $B_{k+1} = \textrm{argmin}_B \|B-B_k\|_V$ where $V$ is some positive definite matrix matrix that defines the norm. An approximate initial value of $B_0=I$ is often sufficient to achieve rapid convergence. The unknown $x_k$ is updated applying the Newton's step calculated using the current approximate Hessian matrix $B_{k}$
• $\Delta x_k=- \alpha_k B_k^{-1}\nabla f(x_k)$, with $\alpha$ chosen to satisfy the Wolfe conditions;
• $x_{k+1}=x_{k}+\Delta x_k$;
• The gradient computed at the new point $\nabla f(x_{k+1})$, and
$y_k=\nabla f(x_{k+1})-\nabla f(x_k),$
is used to update the approximate Hessian $\displaystyle B_{k+1}$, or directly its inverse $\displaystyle H_{k+1}=B_{k+1}^{-1}$ using the Sherman-Morrison formula.
• A key property of the BFGS and DFP updates is that if $B_k$ is positive definite and $\alpha_k$ is chosen to satisfy the Wolfe conditions then $\displaystyle B_{k+1}$ is also positive definite.
The most popular update formulas are:
Method $\displaystyle B_{k+1}=$ $H_{k+1}=B_{k+1}^{-1}=$
DFP $\left (I-\frac {y_k \, \Delta x_k^T} {y_k^T \, \Delta x_k} \right ) B_k \left (I-\frac {\Delta x_k y_k^T} {y_k^T \, \Delta x_k} \right )+\frac{y_k y_k^T} {y_k^T \, \Delta x_k}$ $H_k + \frac {\Delta x_k \Delta x_k^T}{y_k^{T} \, \Delta x_k} - \frac {H_k y_k y_k^T H_k^T} {y_k^T H_k y_k}$
BFGS $B_k + \frac {y_k y_k^T}{y_k^{T} \Delta x_k} - \frac {B_k \Delta x_k (B_k \Delta x_k)^T} {\Delta x_k^{T} B_k \, \Delta x_k}$ $\left (I-\frac {y_k \Delta x_k^T} {y_k^T \Delta x_k} \right )^T H_k \left (I-\frac { y_k \Delta x_k^T} {y_k^T \Delta x_k} \right )+\frac {\Delta x_k \Delta x_k^T} {y_k^T \, \Delta x_k}$
Broyden $B_k+\frac {y_k-B_k \Delta x_k}{\Delta x_k^T \, \Delta x_k} \, \Delta x_k^T$ $H_{k}+\frac {(\Delta x_k-H_k y_k) \Delta x_k^T H_k}{\Delta x_k^T H_k \, y_k}$
Broyden family $(1-\varphi_k) B_{k+1}^{BFGS}+ \varphi_k B_{k+1}^{DFP}, \qquad \varphi\in[0,1]$
SR1 $B_{k}+\frac {(y_k-B_k \, \Delta x_k) (y_k-B_k \, \Delta x_k)^T}{(y_k-B_k \, \Delta x_k)^T \, \Delta x_k}$ $H_{k}+\frac {(\Delta x_k-H_k y_k) (\Delta x_k-H_k y_k)^T}{(\Delta x_k-H_k y_k)^T y_k}$
## Implementations
Owing to their success, there are implementations of quasi-Newton methods in almost all programming languages. The NAG Library contains several routines[1] for minimizing or maximizing a function[2] which use quasi-Newton algorithms.
In MATLAB's Optimization toolbox, the `fminunc` function uses (among other methods) the BFGS Quasi-Newton method. Many of the constrained methods of the Optimization toolbox use BFGS and the variant L-BFGS. Many user-contributed quasi-Newton routines are available on MATLAB's file exchange.
Mathematica includes quasi-Newton solvers. R's `optim` general-purpose optimizer routine uses the BFGS method by using `method="BFGS"`[1]. In the SciPy extension to Python, the `scipy.optimize.minimize` function includes, among other methods, a BFGS implementation.
## References
1. The Numerical Algorithms Group. "Keyword Index: Quasi-Newton". NAG Library Manual, Mark 23. Retrieved 2012-02-09.
2. The Numerical Algorithms Group. "E04 – Minimizing or Maximizing a Function". NAG Library Manual, Mark 23. Retrieved 2012-02-09.
## Further reading
• Bonnans, J. F., Gilbert, J.Ch., Lemaréchal, C. and Sagastizábal, C.A. (2006), Numerical optimization, theoretical and numerical aspects. Second edition. Springer. ISBN 978-3-540-35445-1.
• William C. Davidon, Variable Metric Method for Minimization, SIOPT Volume 1 Issue 1, Pages 1–17, 1991.
• Fletcher, Roger (1987), Practical methods of optimization (2nd ed.), New York: John Wiley & Sons, ISBN 978-0-471-91547-8 .
• Nocedal, Jorge & Wright, Stephen J. (1999). Numerical Optimization. Springer-Verlag. ISBN 0-387-98793-2.
• Press, WH; Teukolsky, SA; Vetterling, WT; Flannery, BP (2007). "Section 10.9. Quasi-Newton or Variable Metric Methods in Multidimensions". Numerical Recipes: The Art of Scientific Computing (3rd ed.). New York: Cambridge University Press. ISBN 978-0-521-88068-8.
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http://crypto.stackexchange.com/questions/2165/factoring-a-polynomial-over-a-gf/2176
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# Factoring a polynomial over a GF [closed]
I have the following question:
What polynomial, when factored over the field $GF (2^8)$ based on the irreducible polynomial that is used in Rijndael, will factor into all the polynomials in the field?
As I understand it, the polynomial should be an irreducible polynomial and this polynomial must factor all the polynomials. How can I get this polynomial? What steps should I follow?
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I closed this question as even after one month, it is not clear what you actually want to know. Feel free to edit it and comment if you want to have it reopened. – Paŭlo Ebermann♦ Apr 21 '12 at 16:24
## closed as not a real question by Paŭlo Ebermann♦Apr 21 '12 at 16:22
It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, see the FAQ.
## 2 Answers
There are several questions hidden in your question.
You'll have to distinguish between the polynomial ring $\mathbb Z_2[X]$, and the factor field $GF(2^8) = \mathbb Z_2[X]/F$ (where $F$ is any irreductible polynomial of degree 8).
In $\mathbb Z_2[X]$, the only polynomial which is a factor of all others is the trivial polynomial $1$. (In general, in a (unitary) ring the set of elements with this property are called units, and this ring has just the trivial unit one.)
In $GF(2^8)$ (like in any field), every non-zero element is a unit and thus a factor of every other element, but we normally don't call them "polynomials".
But the original question actually asks something else:
What polynomial when factored over the field $GF (2^8)$ based on the irreducible polynomial that is used in Rijndael will factor into all the polynomials in the field?
The expression "factors into" can be understood as "is a multiple of" ... and "when factored over $GF(2^8)$" could be understood as "when seen as a polynomial with coefficients in $GF(2^8)$ (i.e. an element of $GF(2^8)[X]$, and then factored (into its prime divisors)". But then the expression "into all the polynomials in the field" makes no sense, as a field element is either $0$ (which is not a factor other than for itself) or a unit (which is a factor for everything), and you likely don't want $0$ as the result.
Quite likely the original question was something else, but I have no idea, what.
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Even after revision, this question makes litle sense. The OP wants a polynomial with the property that
... this polynomial must factor all the polynomials
or possibly
... will factor into all the polynomials in the field
With regard to the first, a polynomial is not an operator or algorithm that can be used to factor a polynomial (or decide that the polynomial is irreducible) and so perhaps the word into was inadvertently omitted, and this question was supposed to be identical to the second.
With regard to the second, it is worth noting that there are no polynomials in the field GF$(2^8)$ except in the sense that all the field elements can be expressed as polynomials of degree $7$ or less with binary coefficients. There are $2^8$ such polynomials ranging from $0$ or $0+0x+0x^2+\cdots+0x^7$ to $1+1x+1x^2+\cdots+1x^7$, and if factor into means a divisor of, then the only polynomial that is a divisor of all these $2^8$ polynomials is $1$, the constant polynomial.
Other possibilities might be the question
What binary polynomials factor into a product of monic linear polynomials over GF$(2^8)$?
Here, monic linear polynomial means $(x-\alpha_i)$ where $\alpha_i \in \text{GF}(2^8)$. One answer would be
All the irreducible binary polynomials of degrees $1, 2, 4$, and $8$ factor into products of linear polynomials over GF$(2^8)$.
Some people would say that GF$(2^8)$ is the splitting field of irreducible binary polynomials of degree $8$. More generally, any polynomial that is a product of any number of these irreducible polynomials (including repeats so that $[f(x)]^i[g(x]^j[h(x)]^k\cdots$ is allowed) will also factor into a product of linear polynomials over GF$(2^8)$ but might have roots with multiplicity greater than $1$.
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http://unapologetic.wordpress.com/2008/12/08/the-zero-representation/?like=1&source=post_flair&_wpnonce=c1c9c39881
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# The Unapologetic Mathematician
## The Zero Representation
Okay, this is going to sound pretty silly and trivial, but I’ve been grading today. There is one representation we always have for any group, called the zero representation $\mathbf{0}$.
Pretty obviously this is built on the unique zero-dimensional vector space $\mathbf{0}$. It shouldn’t be hard to convince yourself that $\mathrm{GL}(\mathbf{0})$ is the trivial group, and so any group $G$ has a unique homomorphism to this group. Thus there is a unique representation of $G$ on the vector space $\mathbf{0}$.
We should immediately ask: is this representation a zero object? Suppose we have a representation $\rho:G\rightarrow\mathrm{GL}(V)$. Then there is a unique arrow $0:V\rightarrow\mathbf{0}$ sending every vector $v\in V$ to $0\in\mathbf{0}$. Similarly, there is a unique arrow $0:\mathbf{0}\rightarrow V$ sending the only vector $0\in\mathbf{0}$ to the zero vector $0\in V$. It’s straightforward to show that these linear maps are intertwinors, and thus that the zero representation is indeed a zero object for the category of representations of $G$.
This is all well and good for groups, but what about representing an algebra $A$? This can only make sense if we allow rings without unit, which I only really mentioned back when I first defined a ring. This is because there’s only one endomorphism of the zero-dimensional vector space at all! The endomorphism algebra will consist of just the element $0:\mathbf{0}\rightarrow\mathbf{0}$, and a representation of $A$ has to be an algebra homomorphism to this non-unital algebra. Given this allowance, we do have the zero representation, and it’s a zero object just as for groups. It’s sort of convenient, so we’ll tacitly allow this one non-unital algebra to float around just so we can have our zero representation, even if we allow no other algebras without units.
### Like this:
Posted by John Armstrong | Algebra, Representation Theory
## 12 Comments »
1. There is nothing in the definition of rings with unit that demands that the unit be distinct from 0. (Or if there is, it’s the wrong definition!) Indeed, the theory of unital rings is equational, and the condition that 1 not equal 0 is a non-equational condition; demanding it would be somewhat wrong-headed from the categorical and logical points of view.
Moral: the endomorphism algebra on the zero vector space may be considered unital.
Comment by | December 9, 2008 | Reply
2. I suppose there’s a point to that. I just see every time someone says “unital” they’re very emphatic that it’s not zero.
Comment by | December 9, 2008 | Reply
3. The only case I can remember such an explicit emphasis is for the case of fields, where the convention would be I think more defensible (since, e.g., already the theory of fields is non-equational in the sense we mean, i.e., Lawvere theories). If I saw it in the more general case of rings, I’d be tempted to write the author!
Comment by | December 9, 2008 | Reply
4. 0 is indeed a unital ring, but it isn’t a zero object in the category of unital rings.
Consider a morphism f with domain 0 and codomain an algebra A. f(0) = 0 and (if f is unital) f(1) = 1, and 0 = 1 in the zero ring; so 0 = 1 in A, and A must be the zero ring, otherwise no such morphism exists.
Whether that matters for your purposes, I don’t know.
Comment by Chad Groft | December 9, 2008 | Reply
5. It doesn’t. I’m making the vector space a zero object, not its endomorphism ring.
Comment by | December 9, 2008 | Reply
6. Todd, Chad kind of nailed it. The zero ring may be a ring with unit, but when you’re talking about rings with unit, you tend to require that the homomorphisms preserve the unit, which isn’t viable here.
Comment by | December 9, 2008 | Reply
7. Charles, no. The “here” in question is an algebra morphism going the other way: $A \to End(0)$. The unit is preserved.
It’s perfectly fine that there be no algebra homomorphism $\mathrm{End}(0) \to A$ unless $A$ is also the one-element ring. Looked at from the point of view of the opposite category, affine spectra, that’s like saying there’s no morphism $X \to 0$ to the empty spectrum, unless $X$ is itself empty. Which is a general feature of what are called distributive categories.
Comment by | December 9, 2008 | Reply
8. Hm, obviously the bit about the opposite category pertains to commutative unital rings. Not that that affects any of what we’re talking about here.
Incidentally, I like the convention of calling rings not assumed as carrying units “rngs”: rings without the i-dentity. I think the term is due to Jacobson.
Comment by | December 9, 2008 | Reply
9. Yes, Todd’s right. I’m not trying to make the homomorphism go back the other way. Remember, the homomorphism of rings is the representation, which doesn’t have to go back the other way.
And Chad’s right that the zero ring is not a zero object in the category of unital rings. But I’m not trying to find a zero object in that category.
As for rngs.. what happens if you throw away negatives as well? Say, the strictly positive natural numbers? Are they a “rg”?
Comment by | December 9, 2008 | Reply
10. Yes, there seems to be some scholarly controversy about that: whether those ancient Indian treatises should be rendered as the Rig Vedas or the Rg Vedas.
(Seriously, there are such things as rigs = rings without n-egatives = monoids in the monoidal category of commutative monoids, as John already knows I’m sure. I’ve never seen rgs mentioned though. How would you even pronounce that: “erg”?)
Comment by | December 9, 2008 | Reply
11. Well, yes, I know that there are rigs. That’s the point: take away identity and you have rng. Take away negatives and you have rig. Take away *both*?
Comment by | December 9, 2008 | Reply
12. Yes, I knew what you meant. But I’ve never seen them mentioned by anyone until now.
Comment by | December 9, 2008 | Reply
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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http://physics.stackexchange.com/questions/52135/confusion-about-time-shift-in-special-relativity
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Confusion about time shift in special relativity
I have never really found a way to comfortably comprehend the idea of time shift even though I know its not the hard part of relativity theory. In that light, can someone point out what is wrong or right about the following logic I'm thinking of here:
The scenario is:
There are four things in my universe: a wallclock, a flashlight, my wrist watch and myself. I shine the flashlight at my wrist which is in the same line of sight as the wallclock and see they two clocks are synchronized. Now my wristwatch, my flashlight and myself begin traveling away from the wallclock along the line of sight.
It seems to me that logically the light I'm running away from will take longer to reach my eye, so that means I would see the wall clock run slower, no?
But according to what I think I know, when a minute passes for me, more than a minute passes for the wall clock.
I just feel like I must be looking at it wrong. Any pointers are appreciated. This is just recreational thinking...
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1 Answer
This is the situation you describe seen from the perspective of the wall clock i.e. the wall clock is stationary and you are moving.
We'll choose time zero to be when you pass the wall clock i.e. the spacing between you and the wall clock can be taken to be zero. At this point you synchronise your watch with the clock. At this point both you and the clock agree about the time and your relative speed.
From the perspective of the wall clock your time is running slowly i.e. when the wall clock measures 1 hour your writwatch will have measured less than one hour.
However from your perspective you are stationary and it's the wall clock that's moving. So when you measure one hour on your wristwatch less than one hour will pass for the wall clock. This is the opposite of what you say in your question:
But according to what I think I know, when a minute passes for me, more than a minute passes for the wall clock.
The situation is symmetric. Both frames see time running more slowly in the other frame. It must be this way otherwise you could say which frame was moving and which was stationary, and this would contradict the basic principle of relativity.
It is very important to be clear that this is not due to the time the light from your flashlight takes to get to the clock and back. Because you know how fast the wall clock is moving (both frames agree the relative velocity is $v$, and you know how much time has elapsed since you separated) you can calculate the flight time of the light and correct for it. If you do this you will still find that time is running more slowly for the wall clock.
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http://mathoverflow.net/questions/74791?sort=newest
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## Do representations of Fuchsian groups have unitary deformations?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $G$ be $SL_2({\mathbb C})$ and for $a,b\in G$ let $[a,b]=aba^{-1}b^{-1}$ be the commutator bracket. Let $n$ be a natural number $\ge 2$ and let $X\subset G^{2n}$ be the set of all $g\in G^{2n}$ such that $$[g_1,g_2]\cdots[g_{2n-1},g_{2n}]=1.$$ The first question is, whether $X$ is connected. If not, can one give a list of the connected components? Finally, does the subset $SU(2)^{2n}\cap X$ meet every connected component?
If the last question has an affirmative answer, every $G$-valued representation of the fundamental group $\Gamma$ of a compact Riemann surface of genus $n$ can be deformed to a unitary one, which explains the title of my question.
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## 1 Answer
$X$ is the $SL_2(\mathbb{C})$--representation variety of the surface group, and, by Goldman's thesis, it is irreducible, and so connected.
See
Goldman, Topological components of spaces of representations. Invent. Math. 93 (1988), no. 3, 557–607.
If you take $G$ to be $PSL_2(\mathbb{C})$, then there are two components (see also Goldman), one for each Stiefel-Whitney class.
Edit: I should say that I recall that this is perhaps not so easy to find in Goldman's paper as he doesn't state it explicitly, but at some point he proves that the smooth locus of $X$ is connected, which gives the result.
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http://mathhelpforum.com/calculus/198895-convergence-integral-involving-hermite-polynomials.html
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2Thanks
• 1 Post By girdav
• 1 Post By girdav
# Thread:
1. ## Convergence of an integral involving Hermite polynomials
Dear All,
I would like to know if the following integral converges, and if so, what is the easiest way to prove it?
$I_n = \int_{- \infty}^{+ \infty} e^{- x^2 / 2} H_n(x) dx, \forall \ n \in \mathbb{N}^{+},$
where $H_n(x)$ is the physicists' Hermite polynomial:
$H_n(x) = (-1)^n e^{x^2} \frac{d^n e^{-x^2}}{d x^n}$.
Thank you.
Regards.
2. ## Re: Convergence of an integral involving Hermite polynomials
If $P$ is a polynomial of degree $d$, the integral $\int_{-\infty}^{+\infty}e^{-x^2/2}P(x)dx$ is convergent. To see that, write $e^t\geq \frac{t^{d+2}}{(d+2)!}$ for $t\geq 0$.
3. ## Re: Convergence of an integral involving Hermite polynomials
Dear girdav,
Thanks a lot for your reply. I appreciate it.
Would you mind detailing your suggestion that uses $e^{t} \geq \frac{t^{d+2}}{(d+2)!}$ for $t \geq 0$? Indeed, the latter uses $t$ positive but $-x^2/2$ is negative, and it seems to me easier to prove that the integral in question is finite by using $\leq$ rather than $\geq$.
Regards.
4. ## Re: Convergence of an integral involving Hermite polynomials
You get an upper bound after taking the inverse.
5. ## Re: Convergence of an integral involving Hermite polynomials
Thanks girdav. That makes sense.
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http://mathhelpforum.com/number-theory/141542-there-quicker-way-solve-congruence.html
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# Thread:
1. ## Is there a quicker way to solve this congruence
Is $2^{35} \equiv 1 \textrm{ mod } 561$?
In the question we were given maple commands to use so I assume we were meant to do it that way. Is there a way to solve this without a calculator?
I had to use one but just did it like...
$2^8 = 256$,
so $2^{16} = 450$ mod 561
and $2^{32} = (2^{16})^2 = 103$ mod 561...
Then $2^{35} = 2^{32} 2^3 = 103 \times 8$ which is not equal to 1 mod 561..
2. Originally Posted by Deadstar
Is $2^{35} \equiv 1 \textrm{ mod } 561$?
Since $561=3\cdot 11\cdot 17$ , we can do:
*** Arithmetic modulo $3$ : $2^{35}=(2^3)^{11}\cdot 2^2=2^{11}\cdot 2^2=(2^3)^4\cdot 2=2^4\cdot 2=2^3\cdot 2^2=2^3=2\!\!\!\pmod 3$
*** Arithmetic modulo $11$ : $2^{35}=(2^{11})^3\cdot 2^2=2^3\cdot 2^2=32=10=-1\!\!\!\pmod {11}$
*** Arithmetic modulo $17$ : $2^{35}=(2^{17})^2\cdot 2=2^2\cdot 2=8\!\!\!\pmod {17}$ .
Well, now use the Chinese Remainder Theorem to find an element $x\in\mathbb{Z}\,\,\,s.t.\,\,\,x=2\!\!\!\pmod 3\,,\,x=-1\!\!\!\pmod {11}\,,\,x=8\!\!\!\pmod{17}$ .
I found $x=-298=263\!\!\!\pmod{561}\Longrightarrow 2^{35}=263\!\!\!\pmod{561}$ , as you can easily check with any calculator.
Tonio
In the question we were given maple commands to use so I assume we were meant to do it that way. Is there a way to solve this without a calculator?
I had to use one but just did it like...
$2^8 = 256$,
so $2^{16} = 450$ mod 561
and $2^{32} = (2^{16})^2 = 103$ mod 561...
Then $2^{35} = 2^{32} 2^3 = 103 \times 8$ which is not equal to 1 mod 561..
.
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http://mathoverflow.net/questions/22592?sort=votes
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## Mean minimum distance for K random points on a N-dimensional (hyper-)cube
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Given K points in a N-dimensional (hyper-)cube with all edges length 1. What is the expected minimal distance between 2 points. I found the 1-dimensional case in this topic: http://mathoverflow.net/questions/1294/mean-minimum-distance-for-n-random-points-on-a-one-dimensional-line and I wonder if this can be generalized into multiple dimensions in general. I don't seem to succeed in extending the card analogy in the other topic. Does anyone have any hints as to proceed with this?
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1
If you just count the average distance between two points in an $N$-cube (which is some kind of continuous version of the probelm), this is known as box integrals. They are studied intensively during the last years by D.Bailey, J.Borwein, R.Crandall, and their collaborators. The $2N$-dimensional integral is easy to write down but hard to evaluate by means of "known functions". The "treatable" cases (in a certain sense and as far as I know) end at dimension $N=5$. For $N=2$ the answer is compact enough, while higher dimensions involve arctangents, logs and their repeated antiderivatives... – Wadim Zudilin Apr 26 2010 at 13:34
If it isn't manageable to evaluate the integrals, is possible to approximate them? I also don't really understand, assuming I know the average distance between 2 points (on that subject i found this interesting paper: jstor.org/stable/1427094?seq=1) , how it's possible to extend this to the average minimal distance on k points? – Ingdas Apr 26 2010 at 14:13
You may be interested in some of J. Philip's papers at <math.kth.se/~johanph/>;. – villemoes Jul 2 2010 at 6:59
The case $K=2$ is called hypercube line picking here: mathworld.wolfram.com/HypercubeLinePicking.html – Douglas Zare Mar 17 at 16:10
## 2 Answers
It depends on what kind of accuracy you are looking for, but you can get a crude bound of the order of $1/K^{1-1/N}$ by breaking the space into K regions and applying a balls and bins argument.
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The following paper:
Bhattacharyya, P., and B. K. Chakrabarti. The mean distance to the nth neighbour in a uniform distribution of random points: an application of probability theory. Eur J. Phys. 29, pp. 639-645.
Claims to provide exact, approximate, and handwaving estimates for the mean 'k'th nearest neighbor distance in a uniform distribution of points over a D-dimensional Euclidean space (or a D-dimensional hypersphere of unit volume) when one ignores certain boundary conditions.
However, Wadim's response is making me feel some concern that the exact problem is much more complex. Please see the paper for the full derivation (and approximate methods), but I'll write the exact expression they converge on using two different method of absolute probability and conditional probability.
Let $D$ be the dimension of the Euclidean space, let $N$ be the number of points randomly and uniformly distributed over the space, and let $MeanDist(D, N, k)$ be the mean distance to a given points $kth$ nearest-neighbor. This yields:
$MeanDist(D, N, k) = \frac{(\Gamma(\frac{D}{2}+1))^{\frac{1}{D}}}{\pi^{\frac{1}{2}}} \frac{(\Gamma(k + \frac{1}{D}))}{\Gamma(k)} \frac{\Gamma(N)}{\Gamma(N + \frac{1}{D})}$
Where $\Gamma(...)$ is the complete Gamma function.
Wadim - might it be possible for you to provide some feedback about the derivations here vs. the method of box integrals you described in your comment?
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http://mathoverflow.net/questions/61678/what-is-the-general-geometric-interpretation-of-modules-in-algebraic-geometry/61751
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## What is the general geometric interpretation of modules in algebraic geometry?
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Algebraic geometry is quite new for me, so this question may be too naive. therefore, I will also be happy to get answers explaining why this is a bad question.
I understand that the basic philosophy begins with considering an abstract commutative ring as a function space of a certain "geometric" object (the spectrum of the ring). I also understand that at least certain types of modules correspond to well known geometric constructions. For example, projective modules should be thought of as vector bundles over the spectrum (and that there are formal statements such as the Serre-Swan theorem which make this correspondence precise in certain categories). My question is, what is the general geometric counterpart of modules?
This is not a formal mathematical question, and I am not looking for the formal scheme-theoretic concept (of sheaves of certain type and so on), but for the geometric picture that I should keep in mind when working with modules.
I will appreciate any kind of insight or even just a particularly Enlightening example.
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I chose the answer that was the clearest for me personally, but all the answers were interesting and helpful. thanks a lot to everyone. – KotelKanim Apr 17 2011 at 11:15
## 5 Answers
Roughly a module can be thought of as a vector bundle on the spectrum, where the dimension of fibers may vary. Let me give some examples and facts:
• A free modules corresponds to trival vector bundles, or more generally projective modules correspond to vector bundles as you already pointed out.
• Let $R$ be the coordinate ring of a variety and $I$ a radical. Then the $R$ module $R/I$ corresponds to attaching a one dimensional vector space on each point of $Z(I)$ and the zero vector space everywhere else. For Example $R=k[x,y]$ and $I=(x,y)$ gives the skyscraper sheaf at the origin. $I=(x)$ gives the trivial one dimensional bundle on the y-axis etc. If your Ideal is not a radical, the situation is slightly more complicated. $R/I$ can be thought of as the trival bundle on an infinitesimal neighborhood of Z(I).
• Another nice example is a geometric explanation why the tensorproduct $\mathbb Z/p \otimes_{\mathbb Z} \mathbb Z/q$ for say $p,q$ without common divisor vanishes. Our space $spec(\mathbb Z)$ consists just of a point for each prime (and a generic point). Now with the above intuition in mind, our two modules are geometrically just one dimensional vectorspaces attached to infinitesimal neighborhoods of the prime divisors of $p,q$. Since $p$ and $q$ have no common divisor there is no point where both $\mathbb Z/p$ and $\mathbb Z/q$ have nonzero fiber. As with vectorbundles, tensor produduct of modules can be thought of geometrically as fiberwise tensor product ($i^*$ commutes with $\otimes$). But of course the fiberwise tensor product vanishes because there are no points where both modules have nonzero fibers, so $\mathbb Z/p \otimes \mathbb Z/q=0$ .
• Finally any finitely generated module (more generally a coherent sheaf on a noetherian scheme) is built up of vector bundles on subspaces in the following way: There exists a stratification of spec(R) such that the module pulled back to the strata is a vector bundle. This follows from Hartshorne Ex II.5.8.
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As you say, projective modules correspond in a highly moral way to vector bundles. Bundles pull back but do not push forward, in topologists' terms. This might be a good point at which to start. Sheaves, on the other hand, push forward easily. You don't get something for nothing (well, I've known mathematicians argue that you can elegantly do just that ...) and so there is a "picture deficit": you can think of the pullback of a fibre bundle easily in terms of the fibres you already had being attached to inverse image points. You don't have quite the same easy access to the pushforward of the sheaf of sections of a bundle.
Following up this line of thought, algebraic geometry as practised since Serre's FAC is full of pushforwards of sheaves that start off locally free. They don't end up locally free - really can't because you can't do that successfully just with vector bundles. This is more like a Grothendieck justification of what the general quasicoherent sheaves are doing in the theory.
Historically it seems that the concept of module evolved because "ideal theory", as commutative algebra was once called, needed examples that were less rigid than ideals of a ring.
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These somewhat naive remarks are motivated by the desire to think about the question myself.
Sections of bundles are natural examples of (locally) free modules and the standard examples are the tangent and cotangent bundles of smooth varieties. The need for more general modules is the fact that kernels and cokernels of maps of (locally) free modules need not be so. In addition to ideal sheaves of subvarieties, as noted above the kahler differentials are a standard example of not necessarily locally free sheaves.
These two remarks are connected by the standard exact sequence of a closed subscheme, which exhibits the sheaf of kahler differentials of a local complete intersection, as the cokernel of a map of locally free sheaves. I,.e, the kahler differentials on a closed subvariety of a smooth variety, are the quotient of the kahler differentials on the ambient variety by the conormal sheaf of the subvariety.
Thus, the standard examples of modules are ideals of subvarieties and kahler differentials, and at least for lci subvarieties these arise as either kernels or cokernels of maps of bundles. So geometrically one could say one wants to consider differentials and subschemes, and algebraically one wants to be able to take kernels and cokernels.
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Here is a (hopefully) enlightening example. Let $X$ be an irreducible variety over a perfect field, and consider its sheaf of Kähler differentials $\Omega_X$. If $X$ is smooth, then $\Omega_X$ is locally free, and corresponds to the cotangent bundle. If $X$ is not smooth, then the notion of "cotangent bundle" does not really make sense, but the sheaf $\Omega_X$ serves as a sort of substitute:
-The open subset of $X$ on which $\Omega_X$ is locally free consists precisely of the set of points at which $X$ is smooth.
-If $x \in X$ is a point with residue field $\Bbbk(x)$, then any reasonable definition of the cotangent space at $x$ will turn out to be equivalent to $\Omega_{X,x} \otimes \Bbbk(x)$. The singular points of $X$ are precisely those points at which the dimension of the cotangent space is "too large."
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Have a look at Serre's definition of a sheaf [FAC], namely via étale spaces. This gives a geometric picture of a sheaf. Over every point of our topological space $X$, there sits the fiber of the sheaf. Of course we make some compatibility conditions on these fibers, namely that they vary continuously. Topologists call this a bundle on $X$. Now if $X$ has some extra structure, it is reasonable to study the bundles with some appropriate extra structure. Namely, if $X$ is a ringed space, then the fibers should be modules. The corresponding sheafes are called module sheaves. If $X$ is a scheme, then we restrict to quasi-coherent sheaves in order to involve the local affine charts. In every case, you get a special type of a bundle over $X$.
In the same way as the structure of a ring $A$ may be studied by means of the modules over $A$, the structure of a scheme $X$ may be studied by means of quasi-coherent sheaves on $X$. Actually the Reconstruction Theorem by Rosenberg justifies this. Even in the affine case this helps to enlighten some concepts of module theory. An example is the support of a module.
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http://nrich.maths.org/4905
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nrich enriching mathematicsSkip over navigation
### Matchsticks
Reasoning about the number of matches needed to build squares that share their sides.
### Doplication
We can arrange dots in a similar way to the 5 on a dice and they usually sit quite well into a rectangular shape. How many altogether in this 3 by 5? What happens for other sizes?
### The Great Tiling Count
Compare the numbers of particular tiles in one or all of these three designs, inspired by the floor tiles of a church in Cambridge.
# Tables Without Tens
##### Stage: 2 Challenge Level:
You will need a piece of squared paper for this activity. If you have none you can get a sheet here.
Write the units digits of the numbers in the two times table from $1 \times 2$ up to $10 \times 2$ in a line. (Leave some room at the top of the paper, and some space to the left and right.)
It should look like this:
You might be able to see some patterns in these numbers.
Now, do the same thing with the multiples of three. Remember, just write the units digits, this time directly underneath the line of the two times table, like this:
Continue by writing in the four and five times tables in the same way. Again, just using the units digits.
Now look at the whole array of numbers you have created.
What patterns can you find?
Try to explain why the patterns occur.
What do you notice about these four sets of numbers?
Can you predict what would happen next if we wrote in the next times table?
***************
Well, why not add in the tables of sixes, sevens, eights, nines and finally tens?
After that, for the sake of completeness, we could put in the table of ones and zeros.
Do you have a grid that looks like this?
What patterns are there here?
What about repeats?
Can you predict what you will find?
How might you record the repeats that you find?
Each line could be written like this:
But you will probably find some other ways which are just as good.
You could try writing all the tables like that.
Are some tables the same or similar to others?
Does it matter which way the arrows go?
What can you discover about the pattern of repeats?
Can you predict what you will find out about 'pairs' of tables?
The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.
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http://mathoverflow.net/questions/122760?sort=oldest
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Linearization instability and singular points of algebraic varieties
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
In a well known 1973 paper, Fischer and Marsden pointed out (with similar, contemporary remarks made in the physics literature by Brill and Deser) that the space of solutions of some non-linear partial differential equations, like the Einstein or Yang-Mills equations, on a manifold with compact Cauchy slices cannot be manifold (of an appropriate infinite dimensional kind). This phenomenon was given then name linearization instability; I'll explain it in more detail below.
As far as I can tell, this terminology is only used in the literature that followed this observation. However, the actual mathematical phenomenon is very similar to the fact that algebraic varieties are also not manifolds, since they possess singular points. My question is this: what terminology is used in algebraic geometry for the various aspects of this phenomenon? I'll be more specific below.
Given a non-linear PDE, $E[u] = 0$, where $u$ is the unknown function, defined on some manifold $M$. Let its linearization about a background solution $E[U]=0$ be $E_U[u] = 0$, that is, $$E[U+\varepsilon u]=\varepsilon E_U[u] + O(\varepsilon^2).$$ Let $X$ denote the set of solutions of the full equation (exact solutions) and let $\ker E_U$ denote the set of all solutions of the linearized equation (linearized solutions). The background solution $U$ is said to be linearization stable if each linearized solution $u\in \ker E_U$ can be extended to a 1-parameter family $v_\varepsilon = U + \varepsilon u + O(\varepsilon^2)$ of exact solutions, $E[v_\varepsilon] = 0$. If $U$ is linearization stable, then one can show that the space $X$ of exact solutions that are close to $U$ is an (infinite dimensional) manifold, modeled on the space $\ker E_U$.
On the other hand, it is possible that there exists a function $Q[u]$ (typically involving some integral of a local expression over a submanifold of $M$) such that a linearized solution $u\in \ker E_U$ cannot be extended to a 1-parameter family $v_\varepsilon = U + \varepsilon u + O(\varepsilon^2)$ of exact solutions unless $Q(u)=0$. If such a non-trivial function $Q[u]$ exists, the background $U$ is said to be linearization unstable. In the examples of Einstein or Yang-Mills equations, $U$ happens to be linearization unstable if it possess some symmetries and the corresponding $Q(u)$ is quadratic. One can then show that the space $X$ of exact solutions of $E[v]=0$ around $v=U$ is not a manifold, since it has neighborhoods that are not homeomorphic to any vector space (in particular not $\ker E_U$) but rather to a conical subset of $\ker E_U$ defined by $Q[u]=0$.
Sometimes it is possible to identify a set of functions $S_i[u]$ such that the intersection of the zero set $S_i[u]=0$, for all $i$, with the space $X$ of exact solutions consists only of linearization unstable backgrounds. In the example of Einstein's equations, the $S_i[u]=0$ could consist of the Killing equations, which identify metrics with non-trivial symmetries.
As I've already mentioned, a linearization unstable point in $X$ is very similar to the singular point of an algebraic variety. Namely, consider the following analogy. Instead of a function on a manifold, $u$ is a finite dimensional vector. Instead of a PDE, $E[u]=0$ is a polynomial equation, which defines an algebraic variety $X$. A linearization unstable point $U\in X$ is, I believe, called a singular point of $X$. For instance, we could have $u=(x,y,z)$, $E[u]=x^2+y^2-z^2$, $U=(0,0,0)$, $E_U[u] = 0$, $Q[u] = x^2+y^2-z^2$, $S[u] = z$.
Could someone provide a translation dictionary into the appropriate terminology used in algebraic geometry (preserving grammatical structure if possible)?
• linearization sability/instability --- ?
• linearization stable/unstable point $U\in X$ --- ?
• linearized solutions $\ker E_U$ --- ?
• functions like $Q[u]$ --- ?
• the zero set $Q[u]=0$ in $\ker E_U$ --- ?
• functions like $S_i[u]$ --- ?
• the zero set $S_i[u]=0$ --- ?
Also, do functions like $S_i[u]$ play a significant role in blowing up singular points?
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1 Answer
Maybe the best answer I can give is ''learn about deformation theory''. But I will take a shot at writing the dictionary you request, in the case where $E(u)$ is a polynomial function
• linearization stability <--> smoothness / instability <--> singularity
• linearization stable/unstable point <--> singular / smooth point
• linearized solutions <--> Zariski tangent space
• functions like $Q[u]$ <--> obstructions
• the zero set $Q[u]=0$ <--> unobstructed deformations I guess? or the formal completion of $X$ near the given point
• functions like $S_i[u]$ <--> the functions $\partial E/\partial u$, or maybe it is better to say, a certain Fitting ideal of the sheaf of Kahler differentials on $X$
• the zero set $S_i[u] = 0$ <--> on $X$, you would I guess call it the singular locus; in the whole ambient space I don't know.
Regarding ''learn about deformation theory'', the point is that you started with some very explicit algebraic variety $X$ and then decided to think about its structure near a point. But deformation theory lets you think about the following thing: say you want to think about the space of all algebraic ''solutions'' of some sort near a given ''solution'', but to a less explicit problem like ''describe all holomorphic maps from a genus 5 curve into projective space'', which of course can also be written as a PDE. It may be difficult or worse to construct such a space globally, but still you can start to think about the local infinitesimal structure near a given solution using infinitesimal methods; and then there are very powerful algebraic tools (the Artin approximation theorem) which let you under good conditions construct an algebraic neighborhood of your given solution.
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Thanks, this is very helpful! What is a good (perhaps on the elementary side) reference for deformation theory? – Igor Khavkine Feb 24 at 10:29
unfortunately I'm not sure if that exists. You could try Hartshorne's notes math.berkeley.edu/~robin/math274root.pdf – Vivek Shende Feb 24 at 14:44
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http://math.stackexchange.com/questions/10527/constraints-on-sum-of-rows-and-columns-of-matrix?answertab=oldest
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# Constraints on sum of rows and columns of matrix
Suppose $r_i$, $1 \le i \le n$, and $c_j$, $1 \le j \le m$, are nonnegative integers. When does there exist an $n \times m$ matrix in $\text{Mat}_{n \times m} (\mathbb{Z}^+)$, i.e. nonnegative entries, such that $r_i$ is the sum of the entries in its $i$th row and $c_j$ is the sum of the entries in its $j$th columns?
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@user314: Why the algebra-precalculus tag? – Mike Spivey Nov 16 '10 at 5:50
## 1 Answer
Non-negative integer solutions exist whenever the (also necessary) "balance condition" is met:
$$\sum_i r_i = \sum_j c_j$$
which simply sums the total matrix entries in two ways.
There is a survey paper by Alexander Barvinok on this: http://www.math.lsa.umich.edu/~barvinok/linalg.pdf
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Yes, it's a very active area. One of my fellow graduate students is working on this for his thesis. – Timothy Wagner Nov 16 '10 at 5:33
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http://math.stackexchange.com/questions/174021/complex-equation-in-maxima
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# Complex equation in maxima
I rested on this tutorial.
After issuing the command with "solve" function:
````%i2 solve((a-b-sqrt(-c^2+2*c*y-y^2+r^2))^2+(d-y)^2=2*r^2*(1-cos(e)),y);
````
The output is:
Why there is unknown quantity "y" on the right side?
P.S.
There's no "Maxima" tag, what a pity! However, I was redirected here by a stackoverflow moderator, so I assume it is not offtopic.
Maxima is a computer algebra system based on a 1982 version of Macsyma. It is written in Common Lisp and runs on all POSIX platforms such as Mac OS X, Unix, BSD, and GNU/Linux as well as under Microsoft Windows. It is free software released under the terms of the GNU General Public License.
source: http://en.wikipedia.org/wiki/Maxima_%28software%29
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Clearly, Maxima was not entirely successful at isolating the variable you want... – J. M. Jul 23 '12 at 0:42
@J.M. Was it my fault? – 0x6B6F77616C74 Jul 23 '12 at 0:47
W|A spits out a pretty lengthy result that doesn't seem to have $y=$ (something with $y$ in it)... – The Chaz 2.0 Jul 23 '12 at 0:50
@TheChaz there are two "y" under the square root. Or this "W|A" shortcut which I don't understand (google also doesn't know anything about "W|A maxima") is significant. – 0x6B6F77616C74 Jul 23 '12 at 1:05
1
I copied your code "solve(...,y)" into www.wolframalpha.com – The Chaz 2.0 Jul 23 '12 at 1:16
show 1 more comment
## 1 Answer
The general solve command in Macsyma has limited capabilities for dealing with algebraic functions. You can work around this by using the rational function package as in the code below. There I define w to be the sqrt, then solve for w, square, then solve for y.
(algebraic:true, tellrat( w^2 = -c^2+2*c*y-y^2+r^2 ));
solve(rat(solve(rat((a-b-w)^2+(d-y)^2 = 2*r^2*(1-cos(e))),w)^2),y)
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This is useful. – copper.hat Jul 23 '12 at 3:01
The result is five lines long, like in the wolframalpha.com. – 0x6B6F77616C74 Jul 23 '12 at 14:53
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http://mathhelpforum.com/algebra/82705-remainder-factor-theorems.html
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# Thread:
1. ## The remainder and factor theorems
The remainder when x^2 - 3x + 1 is divided by (x + d) is 11. Find the possible values of d.
2. Originally Posted by Mr Rayon
The remainder when x^2 - 3x + 1 is divided by (x + d) is 11. Find the possible values of d.
Since the divisor is x + d, then the zero (when doing the synthetic division) is -d. So do the synthetic division:
Code:
```-d | 1 -3 1
| -d d^2+3d
+------------------
1 -d-3 d^2+3d+1```
Set the remainder equal to the given value:
. . . . . $d^2\, +\, 3d\, +\, 1\, =\, 11$
This rearranges as:
. . . . . $d^2\, +\, 3d\, -\, 10\, =\, 0$
Either factor the quadratic and then solve the factors for the values of d, or else plug the above into the Quadratic Formula.
3. Hello, Mr Rayon!
The remainder when $x^2 - 3x + 1$ is divided by $(x + d)$ is 11.
Find the possible values of $d.$
From the heading, I assume you're expected to know the Remainder Theorem.
. . If a polynomial $p(x)$ is divided by $(x-a)$, the remainder is $p(a).$
We have: . $p(x) \:=\:x^2-3x+1$
When $p(x)$ is divided by $(x+d)$, the remainder is $p(\text{-}d)$
We have: . $p(x) \:=\:x^2-3x+1$ divided by $(x-[\text{-}d])$
Hence: . $(\text{-}d)^2 - 3(\text{-}d) + 1 \:=\:1 \quad\Rightarrow\quad d^2 + 3d \:=\:0 \quad\Rightarrow\quad d(d+3) \:=\:0$
Therefore: . $d \;=\;0,\:\text{-}3$
4. Originally Posted by Soroban
Hence: . $(\text{-}d)^2 - 3(\text{-}d) + 1 \:=\:1$
...But what happened to the 11?
5. Originally Posted by Mr Rayon
...But what happened to the 11?
A small mistake by Soroban is what happened. But surely you can make the appropriate corrections .... Not that you need to do this, given Stapel's earlier post. Did you read it by the way?
6. Originally Posted by mr fantastic
A small mistake by Soroban is what happened. But surely you can make the appropriate corrections .... Not that you need to do this, given Stapel's earlier post. Did you read it by the way?
Yes...I never skip a post
d^2 + 3d - 10 = 0
Using the quadratic formula we get:
(-3 - 7)/2 = -5 or (-3 +7)/2 = 2
But are there any other ways to solve this apart from by factorising?
7. Originally Posted by Mr Rayon
But are there any other ways to solve this apart from by factorising?
Yes; you can solve the quadratic by using the Quadratic Formula, or you can complete the square.
8. Originally Posted by stapel
Yes; you can solve the quadratic by using the Quadratic Formula, or you can complete the square.
Yes...I tried to use completing the square but it got a bit messy down the end. I prefer using the quadratic formula.
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http://mathforum.org/mathimages/index.php?title=Transformations_and_Matrices&diff=27392&oldid=23039
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# Transformations and Matrices
### From Math Images
(Difference between revisions)
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- {{Image Description + {{Image Description Ready
- |ImageName=Transformations and Matrices + |ImageName=Transformations
- |Image=Tranformations2.png + |Image=Tranformations4.png
- |ImageIntro=placeholder + |ImageIntro= This picture shows an example of four basic transformations (where the original teapot is a red wire frame). On the top left is a translation, which is essentially the teapot being moved. On the top right is a scaling. The teapot has been squished or stretched in each of the three dimensions. On the bottom left is a rotation. In this case the teapot has been rotated around the x axis and the z axis (veritcal). On the bottom right is a shearing, creating a skewed look.
- |ImageDescElem=When an object undergoes a transformation, it can be represented as a matrix. Different transformations such as translations, rotations, scaling and shearing are represented mathematically in different ways. One matrix can also represent multiple transformations in sequence when the matrices are multiplied together. <br /><br /> + |ImageDescElem=When an object undergoes a transformation, the transformation can be represented as a [[Matrix|matrix]]. Different transformations such as translations, rotations, scaling and shearing are represented mathematically in different ways. One matrix can also represent multiple transformations in sequence when the matrices are multiplied together. <br />
+
+ ==Basic Transformations For Graphics==
+ Computer graphics works by representing objects in terms of simple [[Graphics Primitives|primitives]] that are manipulated with transformations that preserve some primitives’ essential properties. These properties may include angles, lengths, or basic shapes. Some of these transformations can work on primitives with vertices in standard 2D or 3D space, but some need to have vertices in homogeneous coordinates. The general graphics approach is to do everything in homogeneous coordinates, but we’ll talk about the primitives in terms of both kinds when we can. <br /><br />
+
+ The most fundamental kinds of transformations for graphics are rotation, scaling, and translation. There are also a few cases when you might want to use shear transformations, so we’ll talk about these as well. <br /><br />
+ |ImageDesc=
==Linear Transformations Are Matrices== ==Linear Transformations Are Matrices==
A linear transformation on 2D (or 3D) space is a function f from 2D (or 3D) space to itself that has the property that<br /> A linear transformation on 2D (or 3D) space is a function f from 2D (or 3D) space to itself that has the property that<br />
::<math> f(aA + bB) = af(A) + bf(B). </math> <br /> ::<math> f(aA + bB) = af(A) + bf(B). </math> <br />
- Since points in 2D or 3D space can be written as <math> P = xi + yj </math> or <math> P = xi + yj +zk </math> with i, j, and k the coordinate vectors, then we see that <math> f(P) = xf(i) + yf(j) </math> or <math> f(P) = xf(i) + yf(j) + zf(k) </math> <br /> + Since points in 2D or 3D space can be written as <math> P = xi + yj </math> or <math> P = xi + yj +zk </math> with <math>i</math>, <math>j</math>, and <math>k</math> the coordinate vectors, then we see that <math> f(P) = xf(i) + yf(j) </math> or <math> f(P) = xf(i) + yf(j) + zf(k) </math> <br />
This tells us that the linear transformation is completely determined by what it does to the coordinate vectors. <br /><br /> This tells us that the linear transformation is completely determined by what it does to the coordinate vectors. <br /><br />
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From this example, we see that the linear transformation is exactly determined by the matrix whose first column is <math>f(i)</math>, whose second column is <math>f(j)</math>, and whose third column is <math>f(k)</math>, and that applying the function f is exactly the same as multiplying by the matrix. So the linear transformation <b>is</b> the matrix multiplication, and we can use the concepts of linear transformation and matrix multiplication interchangeably.<br /> From this example, we see that the linear transformation is exactly determined by the matrix whose first column is <math>f(i)</math>, whose second column is <math>f(j)</math>, and whose third column is <math>f(k)</math>, and that applying the function f is exactly the same as multiplying by the matrix. So the linear transformation <b>is</b> the matrix multiplication, and we can use the concepts of linear transformation and matrix multiplication interchangeably.<br />
- ==Transformation Composition Is Matrix Multiplication==
- Transformations are usually not used by themselves, especially in graphics, so you need to have a way to compose transformations, as in <math>g(f(P))</math>. But if G is the matrix for the transformation g, and F is the matrix for the transformation f, then the matrix product G*F is the matrix for the composed functions gf.<br />
- ==Basic Transformations For Graphics==
- Computer graphics works by representing objects in terms of simple primitives (link to the graphics primitives page) that are manipulated with transformations that preserve some primitives’ essential properties. These properties may include angles, lengths, or basic shapes. Some of these transformations can work on primitives with vertices in standard 2D or 3D space, but some need to have vertices in homogeneous coordinates. The general graphics approach is to do everything in homogeneous coordinates, but we’ll talk about the primitives in terms of both kinds when we can. <br /><br />
-
- The most fundamental kinds of transformations for graphics are rotation, scaling, and translation. There are also a few cases when you might want to use shear transformations, so we’ll talk about these as well. <br /><br />
- |ImageDesc=
===Rotation=== ===Rotation===
{{hide|1= {{hide|1=
- [[Image:Rotationgraph.png|300px|right]] A 2D rotation transformation rotates everything in 2D space around the origin by a given angle. In order to see what it does, let’s take a look at what a rotation by a positive angle does to the coordinate axes. Now (x,y) is the result when you apply the transformation to (1,0), which means that<br /> + [[Image:Rotationgraph.png|300px|right]] A 2D rotation transformation rotates everything in 2D space around the origin by a given angle. In order to see what it does, let’s take a look at what a rotation by a positive angle <math>\theta </math> does to the coordinate axes. Now (x,y) is the result when you apply the transformation to (1,0), which means that<br />
::<math> \begin{align} x = cos(\theta ) \\ ::<math> \begin{align} x = cos(\theta ) \\
y = sin(\theta ) \end{align} </math><br /> y = sin(\theta ) \end{align} </math><br />
But (x’,y’) is the result when you apply the transformation to (0,1), or <br /> But (x’,y’) is the result when you apply the transformation to (0,1), or <br />
- ::<math> \begin{align} x' = cos(\theta +\frac{p}{2}) = cos(\theta )cos(\frac{p}{2}) - sin(\theta )sin(\frac{p}{2}) = -sin(\theta ) \\ + ::<math> \begin{align} x' &= cos(\theta +\frac{p}{2}) = cos(\theta )cos(\frac{p}{2}) - sin(\theta )sin(\frac{p}{2}) = -sin(\theta ) \\
- y' = sin(\theta +\frac{p}{2}) = sin(\theta )cos(\frac{p}{2}) + cos(\theta )sin(\frac{p}{2}) = cos(\theta ) \end{align} </math> <br /> + y' &= sin(\theta +\frac{p}{2}) = sin(\theta )cos(\frac{p}{2}) + cos(\theta )sin(\frac{p}{2}) = cos(\theta ) \end{align} </math> <br />
Then as we saw above, the rotation transformation must have the image of (1,0) as the first column and the image of (0,1) as the second column, or<br /> Then as we saw above, the rotation transformation must have the image of (1,0) as the first column and the image of (0,1) as the second column, or<br />
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sin(\theta ) & cos(\theta ) sin(\theta ) & cos(\theta )
\end{bmatrix} </math><br /> \end{bmatrix} </math><br />
- or, as XKCD (http://xkcd.com/184/) sees it (notice that the rotation is by -90° and <math>sin(-90\,^{\circ}) = -sin(90\,^{\circ})</math>, <br /><br + or, as XKCD (http://xkcd.com/184/) sees it (notice that the rotation is by -90° and <math>sin(-90\,^{\circ}) = -sin(90\,^{\circ})</math>, <br /><br />
The situation for 3D rotations is different because a rotation in 3D space must leave a fixed line through the origin. In fact we really only handle the special cases where the fixed line is one of the coordinate axes. Let’s start with the easiest one.<br /><br /> The situation for 3D rotations is different because a rotation in 3D space must leave a fixed line through the origin. In fact we really only handle the special cases where the fixed line is one of the coordinate axes. Let’s start with the easiest one.<br /><br />
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For rotations around the Y-axis, the view down the Y-axis looks different from the one down the Z-axis; it is <br /> For rotations around the Y-axis, the view down the Y-axis looks different from the one down the Z-axis; it is <br />
::[[Image:Xzplane.png| 200px]] <br /> ::[[Image:Xzplane.png| 200px]] <br />
- Here a positive-angle is from the X-axis towards the Z-axis, but <math> X \times Z = -Z \times X = -Y </math>, so the rotation axis dimension is pointing in the opposite direction from the Y-axis. Thus a the angle for the rotation is the negative of the angle we would see in the axes above, and since cos is an even function but sin is odd, we have the rotation matrix <br /> + Here a positive-angle is from the X-axis towards the Z-axis, but <math> X \times Z = -Z \times X = -Y </math>, so the rotation axis dimension is pointing in the opposite direction from the Y-axis. Thus a the angle for the rotation is the negative of the angle we would see in the axes above, so we use <math>-\theta </math> instead of <math>\theta</math>. Since cos is an even function but sin is odd, we can substitute in <math>cos(\theta )</math> for <math> cos(-\theta ) </math> and <math>-sin(\theta ) </math> for <math>sin(-\theta ) </math>. Thus we have the rotation matrix <br />
::<math> \begin{bmatrix} ::<math> \begin{bmatrix}
cos(-\theta ) & 0 & -sin(-\theta ) \\ cos(-\theta ) & 0 & -sin(-\theta ) \\
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* & * & * & 0 \\ [0.3em] * & * & * & 0 \\ [0.3em]
* & * & * & 0 \\ * & * & * & 0 \\
- 0 & 0 & 0 & 1 \end{bmatrix} </math> where the * terms are the terms from the 3D rotations above. + 0 & 0 & 0 & 1 \end{bmatrix} </math><br />
+ where the * terms are the terms from the 3D rotations above.
}} }}
===Scaling=== ===Scaling===
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f(0,1,0) = (0,3,0) \\ f(0,1,0) = (0,3,0) \\
f(0,0,1) = (0,0,4) \end{align} </math><br /> f(0,0,1) = (0,0,4) \end{align} </math><br />
- So the matrix for this transformation is <math> \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix} </math> and, in general, a scaling matrix looks like<br /> + So the matrix for this transformation is<br />
- ::<math> \begin{bmatrix} \sigma_x & 0 & 0 \\ 0 & \sigma_y & 0 \\ 0 & 0 & \sigma_z \end{bmatrix} </math> where the <math> \sigma_x , \sigma_y ,\text{ and }\sigma_z </math> are the scaling factors for x, y, and z respectively. <br /> + ::<math> \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix} </math><br />
+ and, in general, a scaling matrix looks like<br />
+ ::<math> \begin{bmatrix} \sigma_x & 0 & 0 \\ 0 & \sigma_y & 0 \\ 0 & 0 & \sigma_z \end{bmatrix} </math><br />
+ where the <math> \sigma_x , \sigma_y ,\text{ and }\sigma_z </math> are the scaling factors for x, y, and z respectively. <br />
In case of only 2D transformations, scaling simply scales down to two dimensions and we simply have<br /> In case of only 2D transformations, scaling simply scales down to two dimensions and we simply have<br />
- <math> \begin{bmatrix} \sigma_x & 0 \\ 0 & \sigma_y \end{bmatrix} </math><br /><br /> + ::<math> \begin{bmatrix} \sigma_x & 0 \\ 0 & \sigma_y \end{bmatrix} </math><br /><br />
- In case we are working with homogeneous coordinates, a scaling transformation only acts on the three primary components and leaves the homogeneous component alone, so we simply have the matrix <math> \begin{bmatrix} \sigma_x & 0 & 0 & 0 \\ 0 & \sigma_y & 0 & 0 \\ 0 & 0 & \sigma_z & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} </math> for the scaling transformation. + In case we are working with homogeneous coordinates, a scaling transformation only acts on the three primary components and leaves the homogeneous component alone, so we simply have the matrix<br />
+ ::<math> \begin{bmatrix} \sigma_x & 0 & 0 & 0 \\ 0 & \sigma_y & 0 & 0 \\ 0 & 0 & \sigma_z & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} </math> <br />
+ for the scaling transformation.
}} }}
===Translation=== ===Translation===
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::<math> \begin{bmatrix} 1 & 0 & T_x \\ 0 & 1 & T_y \\ 0 & 0 & 1 \end{bmatrix} \times ::<math> \begin{bmatrix} 1 & 0 & T_x \\ 0 & 1 & T_y \\ 0 & 0 & 1 \end{bmatrix} \times
\begin{bmatrix} x \\ y \\ 1 \end{bmatrix} = \begin{bmatrix} x \\ y \\ 1 \end{bmatrix} =
- \begin{bmatrix} x+T_x \\ y+T_y \\ 1 \end{bmatrix} </math> so that the matrix <math> \begin{bmatrix} 1 & 0 & T_x \\ 0 & 1 & T_y \\ 0 & 0 & 1 \end{bmatrix} </math> gives a 2D translation. <br /><br /> + \begin{bmatrix} x+T_x \\ y+T_y \\ 1 \end{bmatrix} </math> <br />
+ so that the matrix <br />
+ ::<math> \begin{bmatrix} 1 & 0 & T_x \\ 0 & 1 & T_y \\ 0 & 0 & 1 \end{bmatrix} </math> <br />
+ gives a 2D translation. <br /><br />
The 3D case is basically the same, and by the same argument we see that the 3D translation is given by <br /> The 3D case is basically the same, and by the same argument we see that the 3D translation is given by <br />
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The shear transformation is not widely used in computer graphics, but can be used for things like the oblique view in engineering drawings. The concept of a shear is to add a multiple of one coordinate to another coordinate of each point, or, for example,<br /> The shear transformation is not widely used in computer graphics, but can be used for things like the oblique view in engineering drawings. The concept of a shear is to add a multiple of one coordinate to another coordinate of each point, or, for example,<br />
::<math>shear(x,y,z) = (x+3y,y,z)</math><br /> ::<math>shear(x,y,z) = (x+3y,y,z)</math><br />
- The matrix for this shear transformation looks like <math> \begin{bmatrix} 1 & 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}</math>.<br /><br /> + The matrix for this shear transformation looks like <br />
+ ::<math> \begin{bmatrix} 1 & 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}</math>.<br /><br />
- In general, the matrix for a shear transformation will look like the identity matrix with one non-zero element A off the diagonal. If A is in row i, column j, then the matrix will add A times the jth coordinate of the vector to the ith coordinate. <br /><br /> + In general, the matrix for a shear transformation will look like the identity matrix with one non-zero element A off the diagonal. If A is in row <math>i</math>, column <math>j</math>, then the matrix will add A times the <math>j^{th}</math> coordinate of the vector to the <math>i^{th}</math> coordinate. <br /><br />
For the oblique view of engineering drawings, we look at the shear matrices that add a certain amount of the z-coordinate to each of the x- and y-coordinates. The matrices are<br /> For the oblique view of engineering drawings, we look at the shear matrices that add a certain amount of the z-coordinate to each of the x- and y-coordinates. The matrices are<br />
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In general, getting the inverse of a matrix can be difficult, but for the basic graphics transformations the inverses are easy because we can simply undo the geometric action of the original transformation.<br /> <br /> In general, getting the inverse of a matrix can be difficult, but for the basic graphics transformations the inverses are easy because we can simply undo the geometric action of the original transformation.<br /> <br />
- The inverse of the scaling matrix <math> \begin{bmatrix} \sigma_x & 0 & 0 \\ 0 & \sigma_y & 0 \\ 0 & 0 & \sigma_z \end{bmatrix} </math> is clearly + The inverse of the scaling matrix<br />
- <math> \begin{bmatrix} \frac{1}{\sigma_x } & 0 & 0 \\ 0 & \frac{1}{\sigma_y } & 0 \\ 0 & 0 & \frac{1}{\sigma_z } \end{bmatrix} </math><br /><br /> + ::<math> \begin{bmatrix} \sigma_x & 0 & 0 \\ 0 & \sigma_y & 0 \\ 0 & 0 & \sigma_z \end{bmatrix} </math><br />
+ is clearly <br />
+ ::<math> \begin{bmatrix} \frac{1}{\sigma_x } & 0 & 0 \\ 0 & \frac{1}{\sigma_y } & 0 \\ 0 & 0 & \frac{1}{\sigma_z } \end{bmatrix} </math><br /><br />
- The inverse of a rotation transformation by angle <math> \theta </math> is clearly the rotation around the same line by the angle <math>-\theta </math>.<br /><br /> + The inverse of a rotation transformation by angle <math> \theta </math> is clearly the rotation around the same line by the angle <math>-\theta </math>. For example, the rotation matrix
+ ::<math>\begin{bmatrix}cos(\theta ) & -sin(\theta ) & 0 \\ sin(\theta ) & cos(\theta ) & 0 \\ 0 & 0 & 1\end{bmatrix}</math><br />
+ has an inverse of <br />
+ ::<math>\begin{bmatrix}cos(\theta ) & sin(\theta ) & 0 \\ -sin(\theta ) & cos(\theta ) & 0 \\ 0 & 0 & 1\end{bmatrix}</math><br />
+ Note that <math>sin(-\theta ) = -sin(\theta )</math> and <math> cos(-\theta ) = cos(\theta) </math>.<br /><br />
- The inverse of the translation matrix <math> \begin{bmatrix} 1 & 0 & 0 & T_x \\ 0 & 1 & 0 & T_y \\ 0 & 0 & 1 & T_z \\ 0 & 0 & 0 & 1 \end{bmatrix} </math> is clearly <math> \begin{bmatrix} 1 & 0 & 0 & -T_x \\ 0 & 1 & 0 & -T_y \\ 0 & 0 & 1 & -T_z \\ 0 & 0 & 0 & 1 \end{bmatrix} </math><br /><br /> + The inverse of the translation matrix <br />
+ ::<math> \begin{bmatrix} 1 & 0 & 0 & T_x \\ 0 & 1 & 0 & T_y \\ 0 & 0 & 1 & T_z \\ 0 & 0 & 0 & 1 \end{bmatrix} </math> <br />
+ is clearly<br />
+ ::<math> \begin{bmatrix} 1 & 0 & 0 & -T_x \\ 0 & 1 & 0 & -T_y \\ 0 & 0 & 1 & -T_z \\ 0 & 0 & 0 & 1 \end{bmatrix} </math><br /><br />
The inverse of the simple shear transformation is also straightforward. Since a simple shear adds a multiple of one vector component to another component, the inverse only needs to subtract that multiple. So we have<br /><br /> The inverse of the simple shear transformation is also straightforward. Since a simple shear adds a multiple of one vector component to another component, the inverse only needs to subtract that multiple. So we have<br /><br />
- ::<math> \begin{bmatrix} 1 & A \\ 0 & 1 \end{bmatrix}^{-1} = \begin{bmatrix} 1 & -A \\ 0 & 1 \end{bmatrix} </math> and the 3D case is a simple extension of this. <br /><br /> + ::<math> \begin{bmatrix} 1 & A \\ 0 & 1 \end{bmatrix}^{-1} = \begin{bmatrix} 1 & -A \\ 0 & 1 \end{bmatrix} </math><br />
+ and the 3D case is a simple extension of this. <br /><br />
So we have a major observation: If any transformation is the product of basic graphics transformations, it is easy to find the inverse of its matrix (and hence its inverse transformation) as the product of the inverses of the components in reverse order. Or: <br /> So we have a major observation: If any transformation is the product of basic graphics transformations, it is easy to find the inverse of its matrix (and hence its inverse transformation) as the product of the inverses of the components in reverse order. Or: <br />
::<math>(A \times B \times C)^{-1} = C^{-1} \times B^{-1} \times A^{-1} </math> <br /> ::<math>(A \times B \times C)^{-1} = C^{-1} \times B^{-1} \times A^{-1} </math> <br />
}} }}
+
+ ==Transformation Composition Is Matrix Multiplication==
+ Transformations are usually not used by themselves, especially in graphics, so you need to have a way to compose transformations, as in <math>g(f(P))</math>. But if G is the matrix for the transformation g, and F is the matrix for the transformation f, then the matrix product G*F is the matrix for the composed functions gf.<br /><br />
+ For example, we have the translation represented by the matrix <br />
+ ::<math>\begin{bmatrix} 1 & 0 & 2 \\
+ 0 & 1 & 1 \\
+ 0 & 0 & 1
+ \end{bmatrix}</math> <br />
+ which represents a move two units in the x direction and one unit in the y direction. If we want to then rotate the same object with the matrix <br />
+ ::<math>\begin{bmatrix} -0.5 & 0.866 & 0 \\
+ -0.866 & -0.5 & 0 \\
+ 0 & 0 & 1
+ \end{bmatrix}</math><br />
+ we can represent the combination of the two actions with a single composed matrix. This matrix is found by multiplying the second action by the first action. <br />
+ ::<math>\begin{bmatrix} -0.5 & 0.866 & 0 \\
+ -0.866 & -0.5 & 0 \\
+ 0 & 0 & 1
+ \end{bmatrix} * \begin{bmatrix} 1 & 0 & 2 \\
+ 0 & 1 & 1 \\
+ 0 & 0 & 1
+ \end{bmatrix} = \begin{bmatrix} -0.5 & 0.866 & -0.134 \\
+ -0.866 & -0.5 & -2.23 \\
+ 0 & 0 & 1
+ \end{bmatrix}</math> <br />
+ So this matrix represents moving, then rotating an object in sequence.<br /><br />
+
+ In the example below, the teapot on the left has just been translated by the translation matrix above. The next image is just the rotation from the rotation images. The two images that follow are the translation then rotation and rotation then translation respectively. This demonstrates the combination of different transformations and how they must be done in the right order. <br />
+ <center>[[Image:Rotationimage.png|1200px]] </center>
+
==Transformations and Graphics Environments== ==Transformations and Graphics Environments==
+ {{hide|1=
Attention – this concept needs a bit of programming background; it involves stacks.<br /><br /> Attention – this concept needs a bit of programming background; it involves stacks.<br /><br />
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As an example, consider the simple picture of a bunny head, basically made up of several spheres. Each sphere is scaled (making it an ellipsoid of the right size), rotated into the right orientation, and then translated into the proper place. The tree next to the picture shows how this is organized.<br /><br /> As an example, consider the simple picture of a bunny head, basically made up of several spheres. Each sphere is scaled (making it an ellipsoid of the right size), rotated into the right orientation, and then translated into the proper place. The tree next to the picture shows how this is organized.<br /><br />
- [[Image:Bunnyhead.jpg|300px]] [[Image:Tranformationtree.png|800px|right]] + <center>[[Image:Bunnyhead.jpg|200px]] [[Image:Tranformationtree.png|400px]] </center>
<br /><br /> <br /><br />
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::sphere for main part ::sphere for main part
:pop :pop
+ :push
::translate ::translate
::scale ::scale
::sphere for left eye ::sphere for left eye
:pop :pop
+ :push
::Translate ::Translate
::Scale ::Scale
::sphere for right eye ::sphere for right eye
:pop :pop
+ :push
::Translate ::Translate
::Rotate ::Rotate
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::sphere for left ear ::sphere for left ear
:pop :pop
+ :push
::Translate ::Translate
::Rotate ::Rotate
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Notice something important: the transformations are written in the order they are applied, with the one closest to the geometry to be applied first. The right ear operations are really Translate(Rotate(Scale(sphere-for-right-ear)))<br /><br /> Notice something important: the transformations are written in the order they are applied, with the one closest to the geometry to be applied first. The right ear operations are really Translate(Rotate(Scale(sphere-for-right-ear)))<br /><br />
- If you are not familiar with stacks, this won’t make much sense, but this isn’t necessary to understand basic transformation concepts. A simple way to look at these stacks is to notice that a transformation is a 4x4 matrix or, equivalently, a 16-element array, so maintaining a stack is simply a matter of building an array<br /> + If you are not familiar with stacks, this won’t make much sense, but you don't need to understand this to understand basic transformation concepts. A simple way to look at these stacks is to notice that a transformation is a 4x4 matrix or, equivalently, a 16-element array, so maintaining a stack is simply a matter of building an array<br />
::<b>float transStack[N][16];</b><br /> ::<b>float transStack[N][16];</b><br />
or<br /> or<br />
::<b>float transStack[N][4][4];</b><br /> ::<b>float transStack[N][4][4];</b><br />
where N is the number of transformations one wants to save. where N is the number of transformations one wants to save.
+ }}
+
|other=stacks |other=stacks
- |AuthorName=Steve Cunningham + |AuthorName= Nordhr
|Field=Geometry |Field=Geometry
- |References=references + |Field2=Algebra
- |InProgress=Yes + |References=Page written by Steve Cunningham.
+ |Pre-K=No
+ |Elementary=No
+ |MiddleSchool=No
+ |HighSchool=Yes
+ |HigherEd=Yes
+ |InProgress=No
}} }}
+
+
+
+
+ =Messages to the Future=
+ When there is a page for 2D, 3D, 4D real spaces; affine spaces; homogeneous coordinates, this page should link to that page (to the homogeneous coordinates section).
## Current revision
Transformations
Fields: Geometry and Algebra
Image Created By: Nordhr
Transformations
This picture shows an example of four basic transformations (where the original teapot is a red wire frame). On the top left is a translation, which is essentially the teapot being moved. On the top right is a scaling. The teapot has been squished or stretched in each of the three dimensions. On the bottom left is a rotation. In this case the teapot has been rotated around the x axis and the z axis (veritcal). On the bottom right is a shearing, creating a skewed look.
# Basic Description
When an object undergoes a transformation, the transformation can be represented as a matrix. Different transformations such as translations, rotations, scaling and shearing are represented mathematically in different ways. One matrix can also represent multiple transformations in sequence when the matrices are multiplied together.
## Basic Transformations For Graphics
Computer graphics works by representing objects in terms of simple primitives that are manipulated with transformations that preserve some primitives’ essential properties. These properties may include angles, lengths, or basic shapes. Some of these transformations can work on primitives with vertices in standard 2D or 3D space, but some need to have vertices in homogeneous coordinates. The general graphics approach is to do everything in homogeneous coordinates, but we’ll talk about the primitives in terms of both kinds when we can.
The most fundamental kinds of transformations for graphics are rotation, scaling, and translation. There are also a few cases when you might want to use shear transformations, so we’ll talk about these as well.
# A More Mathematical Explanation
Note: understanding of this explanation requires: *stacks
[Click to view A More Mathematical Explanation]
## Linear Transformations Are Matrices
A linear transformation on 2D (or 3D) space is a function f f [...]
[Click to hide A More Mathematical Explanation]
## Linear Transformations Are Matrices
A linear transformation on 2D (or 3D) space is a function f from 2D (or 3D) space to itself that has the property that
$f(aA + bB) = af(A) + bf(B).$
Since points in 2D or 3D space can be written as $P = xi + yj$ or $P = xi + yj +zk$ with $i$, $j$, and $k$ the coordinate vectors, then we see that $f(P) = xf(i) + yf(j)$ or $f(P) = xf(i) + yf(j) + zf(k)$
This tells us that the linear transformation is completely determined by what it does to the coordinate vectors.
Let’s see an example of this: if the transformation has the following action on the coordinates:
$\begin{align}f(1,0,0) = f(i) = (2,-2,1) \\ f(0,1,0) = f(j) = (-1,3,2) \\ f(0,0,1) = f(k) = (4,3,-2) \end{align}$
then for any point we have:
$\begin{align}f(x,y,z) = (2x,-2x,x)+(-y,3y,2y)+(4z,3z,-2z) \\ = (2x-y+4z, -2x+3y+3z, x+2y-2z) \\ = \begin{bmatrix} 2x - y + 4z \\ -2x + 3y + 3z \\ x + 2y - 2z \end{bmatrix} = \begin{bmatrix} 2 & -1 & 4 \\ -2 & 3 & 3 \\ 1 & 2 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} \end{align}$
From this example, we see that the linear transformation is exactly determined by the matrix whose first column is $f(i)$, whose second column is $f(j)$, and whose third column is $f(k)$, and that applying the function f is exactly the same as multiplying by the matrix. So the linear transformation is the matrix multiplication, and we can use the concepts of linear transformation and matrix multiplication interchangeably.
### Rotation
A 2D rotation transformation rotates everything in 2D space around the origin by a given angle. In order to see what it does, let’s take a look at what a rotation by a positive angle $\theta$ does to the coordinate axes. Now (x,y) is the result when you apply the transformation to (1,0), which means that
$\begin{align} x = cos(\theta ) \\ y = sin(\theta ) \end{align}$
But (x’,y’) is the result when you apply the transformation to (0,1), or
$\begin{align} x' &= cos(\theta +\frac{p}{2}) = cos(\theta )cos(\frac{p}{2}) - sin(\theta )sin(\frac{p}{2}) = -sin(\theta ) \\ y' &= sin(\theta +\frac{p}{2}) = sin(\theta )cos(\frac{p}{2}) + cos(\theta )sin(\frac{p}{2}) = cos(\theta ) \end{align}$
Then as we saw above, the rotation transformation must have the image of (1,0) as the first column and the image of (0,1) as the second column, or
$rotate(\theta ) = \begin{bmatrix} cos(\theta ) & -sin(\theta ) \\ sin(\theta ) & cos(\theta ) \end{bmatrix}$
or, as XKCD (http://xkcd.com/184/) sees it (notice that the rotation is by -90° and $sin(-90\,^{\circ}) = -sin(90\,^{\circ})$,
The situation for 3D rotations is different because a rotation in 3D space must leave a fixed line through the origin. In fact we really only handle the special cases where the fixed line is one of the coordinate axes. Let’s start with the easiest one.
A rotation around the Z-axis is a 2D rotation as above with the third dimension fixed. So the matrix for this rotation is pretty clearly
$\begin{bmatrix} cos(\theta ) & -sin(\theta ) & 0 \\ sin(\theta ) & cos(\theta ) & 0 \\ 0 & 0 & 1 \end{bmatrix}$
A rotation around the X-axis is pretty similar. If we look down the X-axis, we see the following 2D coordinates:
with $Y \times Z = X$, the axis of rotation. This looks like an exact analogue of the XY-plane, and so we can see that the rotation matrix must leave X fixed and operate only on Y and Z as
$\begin{bmatrix} 1 & 0 & 0 \\ 1 & cos(\theta ) & -sin(\theta ) \\ 0 & sin(\theta ) & cos(\theta ) \end{bmatrix}$
For rotations around the Y-axis, the view down the Y-axis looks different from the one down the Z-axis; it is
Here a positive-angle is from the X-axis towards the Z-axis, but $X \times Z = -Z \times X = -Y$, so the rotation axis dimension is pointing in the opposite direction from the Y-axis. Thus a the angle for the rotation is the negative of the angle we would see in the axes above, so we use $-\theta$ instead of $\theta$. Since cos is an even function but sin is odd, we can substitute in $cos(\theta )$ for $cos(-\theta )$ and $-sin(\theta )$ for $sin(-\theta )$. Thus we have the rotation matrix
$\begin{bmatrix} cos(-\theta ) & 0 & -sin(-\theta ) \\ 0 & 1 & 0 \\ sin(-\theta ) & 0 & cos(-\theta ) \end{bmatrix} = \begin{bmatrix} cos(\theta ) & 0 & sin(\theta ) \\ 0 & 1 & 0 \\ sin(-\theta ) & 0 & cos(\theta ) \end{bmatrix}$
around the Y-axis.
When you want to get a rotation around a different line than a coordinate axis, the usual approach is to find two rotations that, when composed, take a coordinate line into the fixed line you want. You can then apply these two rotations, apply the rotation you want around the coordinate line, and apply the inverses of the two rotations (in inverse order) to construct the general rotation. The sequence goes like this:
apply a rotation $R_1$ around the Z-axis to move your fixed line into the YZ-plane
apply a rotation $R_2$ around the X-axis to move that line to the Y-axis
apply the rotation by your desired angle around the Y-axis
apply the inverse $R_2^{-1}$ to move your rotation line back into the YZ-plane
apply the inverse $R_1^{-1}$ to move your rotation line back to the original line.
Whew! This can all be put into a function – or you can simply keep everything in terms of rotations around X, Y, and Z.
If we are working in homogeneous coordinates, we see that all of the rotation operations take place in standard 3D space and so the fourth coordinate is not changed. Thus the general pattern for all the rotations in homogeneous coordinates is
$\begin{bmatrix} * & * & * & 0 \\ [0.3em] * & * & * & 0 \\ [0.3em] * & * & * & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$
where the * terms are the terms from the 3D rotations above.
### Scaling
Scaling is the action of multiplying each coordinate of a point by a constant amount. As an example, let $f(x,y,z) = (2x,3y,4z)$. Then
$f((x,y,z)+(a,b,c)) = f(x+a,y+b,z+c) = (2(x+a),3(y+b), 4(z+c)) = (2x,3y,4z)+(2a,3b,4c)$
So this is a linear transformation. If we look at what this transformation does to each of the coordinate vectors, we have
$\begin{align} f(1,0,0) = (2,0,0) \\ f(0,1,0) = (0,3,0) \\ f(0,0,1) = (0,0,4) \end{align}$
So the matrix for this transformation is
$\begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix}$
and, in general, a scaling matrix looks like
$\begin{bmatrix} \sigma_x & 0 & 0 \\ 0 & \sigma_y & 0 \\ 0 & 0 & \sigma_z \end{bmatrix}$
where the $\sigma_x , \sigma_y ,\text{ and }\sigma_z$ are the scaling factors for x, y, and z respectively.
In case of only 2D transformations, scaling simply scales down to two dimensions and we simply have
$\begin{bmatrix} \sigma_x & 0 \\ 0 & \sigma_y \end{bmatrix}$
In case we are working with homogeneous coordinates, a scaling transformation only acts on the three primary components and leaves the homogeneous component alone, so we simply have the matrix
$\begin{bmatrix} \sigma_x & 0 & 0 & 0 \\ 0 & \sigma_y & 0 & 0 \\ 0 & 0 & \sigma_z & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$
for the scaling transformation.
### Translation
Notice that a translation function cannot be a linear transformation on normal space because it does not take the origin to the origin. These are examples of affine transformations, transformations that are composed of a linear transformation, such as a rotation, scaling, or shear, and a translation. In order to write a translation matrix, we need to use homogeneous coordinates.
If we want to add $T_x$ to the X-coordinate and $T_y$ to the Y-coordinate of every point in 2D space, we see that
$\begin{bmatrix} 1 & 0 & T_x \\ 0 & 1 & T_y \\ 0 & 0 & 1 \end{bmatrix} \times \begin{bmatrix} x \\ y \\ 1 \end{bmatrix} = \begin{bmatrix} x+T_x \\ y+T_y \\ 1 \end{bmatrix}$
so that the matrix
$\begin{bmatrix} 1 & 0 & T_x \\ 0 & 1 & T_y \\ 0 & 0 & 1 \end{bmatrix}$
gives a 2D translation.
The 3D case is basically the same, and by the same argument we see that the 3D translation is given by
$\begin{bmatrix} 1 & 0 & 0 & T_x \\ 0 & 1 & 0 & T_y \\ 0 & 0 & 1 & T_z \\ 0 & 0 & 0 & 1 \end{bmatrix}$
hese are linear transformations in the space one degree higher than the geometry you are working with. In fact, the main reason for including homogeneous coordinates is the math for graphics is to be able to handle translations (and thus all basic transformations) as linear transformations represented by matrices.
### Shear
The shear transformation is not widely used in computer graphics, but can be used for things like the oblique view in engineering drawings. The concept of a shear is to add a multiple of one coordinate to another coordinate of each point, or, for example,
$shear(x,y,z) = (x+3y,y,z)$
The matrix for this shear transformation looks like
$\begin{bmatrix} 1 & 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
In general, the matrix for a shear transformation will look like the identity matrix with one non-zero element A off the diagonal. If A is in row $i$, column $j$, then the matrix will add A times the $j^{th}$ coordinate of the vector to the $i^{th}$ coordinate.
For the oblique view of engineering drawings, we look at the shear matrices that add a certain amount of the z-coordinate to each of the x- and y-coordinates. The matrices are
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ A & 0 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & B & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ A & B & 1 \end{bmatrix}$
that take $(x,y,z) \times (x+Az,y+Bz,z)$. The values of A and B are adjusted to give precisely the view that you want, and the z-term of the result is usually dropped to give the needed 2D view of the 3D object. An example is the classical cabinet view shown below:
To experiment with these transformations, we have two interactive applets. The first one lets you apply the 2D transformations to a 2D figure. Transformation Matrix
The second applet lets you apply the 3D transformations to a 3D figure.
(Currently Unavailable)
### Matrix Inverses
In general, getting the inverse of a matrix can be difficult, but for the basic graphics transformations the inverses are easy because we can simply undo the geometric action of the original transformation.
The inverse of the scaling matrix
$\begin{bmatrix} \sigma_x & 0 & 0 \\ 0 & \sigma_y & 0 \\ 0 & 0 & \sigma_z \end{bmatrix}$
is clearly
$\begin{bmatrix} \frac{1}{\sigma_x } & 0 & 0 \\ 0 & \frac{1}{\sigma_y } & 0 \\ 0 & 0 & \frac{1}{\sigma_z } \end{bmatrix}$
The inverse of a rotation transformation by angle $\theta$ is clearly the rotation around the same line by the angle $-\theta$. For example, the rotation matrix
$\begin{bmatrix}cos(\theta ) & -sin(\theta ) & 0 \\ sin(\theta ) & cos(\theta ) & 0 \\ 0 & 0 & 1\end{bmatrix}$
has an inverse of
$\begin{bmatrix}cos(\theta ) & sin(\theta ) & 0 \\ -sin(\theta ) & cos(\theta ) & 0 \\ 0 & 0 & 1\end{bmatrix}$
Note that $sin(-\theta ) = -sin(\theta )$ and $cos(-\theta ) = cos(\theta)$.
The inverse of the translation matrix
$\begin{bmatrix} 1 & 0 & 0 & T_x \\ 0 & 1 & 0 & T_y \\ 0 & 0 & 1 & T_z \\ 0 & 0 & 0 & 1 \end{bmatrix}$
is clearly
$\begin{bmatrix} 1 & 0 & 0 & -T_x \\ 0 & 1 & 0 & -T_y \\ 0 & 0 & 1 & -T_z \\ 0 & 0 & 0 & 1 \end{bmatrix}$
The inverse of the simple shear transformation is also straightforward. Since a simple shear adds a multiple of one vector component to another component, the inverse only needs to subtract that multiple. So we have
$\begin{bmatrix} 1 & A \\ 0 & 1 \end{bmatrix}^{-1} = \begin{bmatrix} 1 & -A \\ 0 & 1 \end{bmatrix}$
and the 3D case is a simple extension of this.
So we have a major observation: If any transformation is the product of basic graphics transformations, it is easy to find the inverse of its matrix (and hence its inverse transformation) as the product of the inverses of the components in reverse order. Or:
$(A \times B \times C)^{-1} = C^{-1} \times B^{-1} \times A^{-1}$
## Transformation Composition Is Matrix Multiplication
Transformations are usually not used by themselves, especially in graphics, so you need to have a way to compose transformations, as in $g(f(P))$. But if G is the matrix for the transformation g, and F is the matrix for the transformation f, then the matrix product G*F is the matrix for the composed functions gf.
For example, we have the translation represented by the matrix
$\begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$
which represents a move two units in the x direction and one unit in the y direction. If we want to then rotate the same object with the matrix
$\begin{bmatrix} -0.5 & 0.866 & 0 \\ -0.866 & -0.5 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
we can represent the combination of the two actions with a single composed matrix. This matrix is found by multiplying the second action by the first action.
$\begin{bmatrix} -0.5 & 0.866 & 0 \\ -0.866 & -0.5 & 0 \\ 0 & 0 & 1 \end{bmatrix} * \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} -0.5 & 0.866 & -0.134 \\ -0.866 & -0.5 & -2.23 \\ 0 & 0 & 1 \end{bmatrix}$
So this matrix represents moving, then rotating an object in sequence.
In the example below, the teapot on the left has just been translated by the translation matrix above. The next image is just the rotation from the rotation images. The two images that follow are the translation then rotation and rotation then translation respectively. This demonstrates the combination of different transformations and how they must be done in the right order.
## Transformations and Graphics Environments
Attention – this concept needs a bit of programming background; it involves stacks.
When you are defining the geometry for a graphics image, you will sometimes want to model your scene as a hierarchy of simpler objects. You might have a desk, for example, that is made up of several parts (legs, drawers, shelves); the drawers may have handles or other parts; you may want to put several things on top of the desk; and so on. It’s common to define general models for each simple part, and then to put the pieces together in a common space, called the “world space.”
Each simple part will be defined in its own “model space,” and then you can apply transformations that move all the parts into the right place in the more complex part. In turn, that whole more complex part may be moved into another position, and so on – you can build up quite complex models this way. One common technique for this kind of hierarchical modeling is to build a “scene graph” that shows how everything is assembled and the transformations that are used in the assembly.
As an example, consider the simple picture of a bunny head, basically made up of several spheres. Each sphere is scaled (making it an ellipsoid of the right size), rotated into the right orientation, and then translated into the proper place. The tree next to the picture shows how this is organized.
In order to make this work, you have to apply each set of transformations to its own sphere and then “forget” those transformations so you can apply the transformations for the next piece. You could, of course, use inverses to undo the transformations, but that’s slow and invites roundoff errors from many multiplications.
instead, it is common to maintain a “transformation stack” that holds the history of every place you want to get back to – all the transformations you have saved. This is a stack of 4x4 matrices that implement the transformations. You also have an active transformation to which you apply any new transformations by matrix multiplication.
To save a transformation to get back to later, you push a copy of the current active transformation (as a 4x4 matrix) onto the stack. Later, when you have applied whatever new transformations you need and want to get back to the last saved transformation, you pop the top matrix off the stack and make it the current active transformation. Presto – all the transformations you had used since the corresponding push operation are gone.
So let’s get back to the rabbit. We want to create the rabbit head, and we have whatever active transformation was in place when we wanted to draw the head. Then we have
push
scale
sphere for main part
pop
push
translate
scale
sphere for left eye
pop
push
Translate
Scale
sphere for right eye
pop
push
Translate
Rotate
Scale
sphere for left ear
pop
push
Translate
Rotate
Scale
sphere for right ear
pop
Notice something important: the transformations are written in the order they are applied, with the one closest to the geometry to be applied first. The right ear operations are really Translate(Rotate(Scale(sphere-for-right-ear)))
If you are not familiar with stacks, this won’t make much sense, but you don't need to understand this to understand basic transformation concepts. A simple way to look at these stacks is to notice that a transformation is a 4x4 matrix or, equivalently, a 16-element array, so maintaining a stack is simply a matter of building an array
float transStack[N][16];
or
float transStack[N][4][4];
where N is the number of transformations one wants to save.
# Teaching Materials
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# References
Page written by Steve Cunningham.
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# Messages to the Future
When there is a page for 2D, 3D, 4D real spaces; affine spaces; homogeneous coordinates, this page should link to that page (to the homogeneous coordinates section).
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http://rjlipton.wordpress.com/2010/01/27/programming-turing-machines-is-hard/
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## a personal view of the theory of computation
by
Two classic Turing Machine simulation theorems
Patrick Fischer is one of the founders of complexity theory. He played a critical role in starting the field, in proving some of the early results, and in making theory as vibrant an area as it is today. We owe him a great deal of thanks. The picture is of his wife, Charlotte Froese Fischer, who is a research professor of Computer Science—I could not find one of him.
Today I would like to talk about two central results of complexity theory. They are old, they are important, and they have not been improved for years. I believe that the time may be right to finally make progress on them.
One of the curious features of theory, especially results concerning Turing Machines (TM’s), is that their proofs often depend on the ability to program TM’s in clever ways. Programming a TM well is a skill that requires a programmer to understand how linear memory—Turing Tapes—can be manipulated. Linear memory is not as easy to use as random access memory, and this is why TM programmers get paid so well. Okay, you probably can make much more money programming iPhone apps, but being a master TM programmer is hard.
I recall one conference, not FOCS or STOC, where a speaker gave a talk on TM’s that did not seem correct to me, but I was not sure. Pat was also in the audience and said nothing during the talk. Immediately after the talk was over and a coffee break was in progress, Pat walked up to the speaker and said:
Son, what you said about Turing Machines is not right. I program them for a living, and you cannot do what you claim.
I have always liked the phrase “I program them for a living.” Pat has been doing that for almost half a century. His first paper that I can locate was On computability by certain classes of restricted Turing machines in the proceedings of the 1963 Fourth Annual Symposium on Switching Circuit Theory and Logical Design—the conference that became the conference that became FOCS.
Let’s turn to the two theorems.
The Two Theorems
The first is by Fred Hennie and Richard Stearns in 1966 and is titled “Two-Tape Simulation of Multitape Turing Machines.”
Theorem: Let ${M}$ be a multi-tape TM that runs in time ${t(n)}$. Then, there is a two-tape TM ${M^{*}}$ that accepts the same language as ${M}$ and runs in time ${O(t(n)\log t(n))}$.
For the next theorem we need to define what it means for a TM to be oblivious. A TM is called oblivious provided for all ${n}$ and for all inputs ${x}$ of length ${n}$, the heads of the TM move in exactly the same way. That is the tape motions do not depend on the input ${x}$—they only depend on its length.
The second is by Nicholas Pippenger and Michael Fischer in 1979 and is titled “Relations Among Complexity Measures.” They extend the Hennie-Stearns result to make the simulation oblivious. An aside: Michael and Pat are brothers. They probably are one of the first brothers that both worked in complexity theory.
Theorem: Let ${M}$ be a multi-tape TM that runs in time ${t(n)}$. Then, there is a two-tape TM ${M^{*}}$ that accepts the same language as ${M}$ and runs in time ${O(t(n)\log t(n))}$. Further, the machine ${M^{*}}$ is oblivious.
Applications
I have already discussed the second theorem here. The reason I wanted to raise the two theorems together is that they really show how programming TM’s is a skill that we are still learning how to do well.
The main application of the Hennie-Stearns theorem is to prove efficient deterministic time hierarchy theorems. The main application of the Pippenger-Fischer theorem is to prove efficient circuit simulations of languages computable by Turing Machines. Note, I also wish to fix an error that I made earlier: the order of the authors is “Pippenger-Fischer” not “Fischer-Pippenger.” I apologize to Nick and Mike.
The best known separation theorems for nondeterministic time are much stronger than those known for deterministic time, which are based on Hennie-Stearns. The best is still, I believe, due to Joel Seiferas, Michael Fischer, and Albert Meyer in Separating Nondeterministic Time Complexity Classes.
Theorem: Suppose that ${t(n)}$ and ${T(n)}$ are time-constructible functions such that ${t(n + 1) = o(T(n))}$, then
$\displaystyle \mathsf{NTIME}(t(n)) \subsetneq \mathsf{NTIME(T(n))}.$
One key reason is the beautiful simulation theorem of Ron Book that I discussed before.
Proofs Strategies of the Theorems
I will give the flavor of how these theorems are proved—as usual see the papers for the actual details.
What is interesting is that while most—many?—know the theorems, I believe that most do not know the proofs. I did know the proofs at one time, but have long forgotten the details.
Somehow the results are important, but their proofs are no longer well known. I think that this is a shame for two reasons. First, they are beautiful non-trivial theorems. Second, if we forget how they are proved, then they will never get improved. This would be too bad, since an improvement to either of the theorems would have important consequences.
The main idea both use is what we call today amortized complexity. Both proofs use amortization, but do not call it that. I am not a historian, but I believe that the notion of amortized complexity is usually associated with the work of Danny Sleator and Bob Tarjan in their famous work on splay-trees. For example, they won the prestigious Paris Kanellakis Award in 1999 for this brilliant work.
Both theorems, Hennie-Stearns and Pippenger-Fischer, essentially have to use a kind of amortized simulation. The critical issue is this: Suppose that you want to simulate multiple Turing tapes. The obvious method is to store their contents on one tape, and use your other tape for “working” storage. This is exactly what they both do.
The problem is that one linear device does not seem well suited to store two linear sequences. Suppose that you store the sequences ${a_{1},\dots,a_{m}}$ and ${b_{1},\dots,b_{m} \dots }$ as one sequence
$\displaystyle a_{1},b_{1}, \dots, \underbrace{a_{k},b_{k}}_{\bigtriangleup}, \dots a_{m},b_{m}$
Here ${\bigtriangleup}$ is the location of the tape head, and ${a_{k}}$ and ${b_{k}}$ are the contents of the two tapes that are being simulated.
The simulation is easy. Just look at the two symbols, decide what to do next, and do it. No problem. Wrong. The difficulty is that what if the simulation needs to keep one tape head stationary and move the next one to the right. Then, the next state should be:
$\displaystyle a_{1},b_{2}, \dots, \underbrace{a_{k},b_{k+1}}_{\bigtriangleup}, \dots a_{m},b_{m+1}$
The trouble is that to get to this global state is very expensive: it requires the simulator to move ${O(m)}$ symbols. Since ${m}$ can be huge, this would be a very inefficient simulation.
What to do? The details are complex, but the high level idea is to be conservative. Why move everything? The tape heads might go back to where they were previously at the very next step, or they might continue to diverge—move further and further apart. The clever idea is to think of the linear storage as a series of blocks that double in size. Then, as the two heads move the simulator only moves at first a block of size ${2}$, then a block of size ${4}$, and so on. Further, it is necessary to organize the tape so that there are “slack” or empty blocks. These allow content to be moved into the slack area from neighboring blocks. Also to get obliviousness, one can schedule in advance block moves that may turn out to be unnecessary—if they are found to be redundant when the time comes, the simulator writes dummy characters.
The insight is that there is no bound on the cost of any simulation step: some take constant time, and others take time ${2^{i}}$ where ${i}$ is the size of the block that they need to move. But, the total cost over the whole simulation is bounded by the “average” cost of each step times the number of steps, which is shown to be
$\displaystyle O(\log t(n)) \cdot t(n).$
Very neat. Again see the papers for the details.
Open Problems
The obvious open problems are to try and improve these classic theorems. I conjecture that it should be possible to improve them, but I have no clear idea how to proceed. I hope that you can find a trick and program TM’s better than I can.
A final comment. I wonder what would have happened if the proofs, not the results, of these early TM theorems had become well known. The proofs both used the key idea of amortization that today plays such an important role in the theory of algorithms.
A related question is: are there tricks in complexity theorems that could be used in non-complexity situations? More bluntly: is there some fundamental programming trick that is used in complexity theory toady that could be used in other parts of theory or even computer science in general? I wonder.
### Like this:
from → History, People, Proofs
15 Comments leave one →
1. January 27, 2010 11:21 am
The Pippenger-Fischer proof and its timing can be approached by considering a kind of sequence that I believe was studied by Axel Thue, before Turing was even born. Define:
Jag(1) = 1,0,-1,0
Jag(2) = 1,2,3,2,1,0,-1,-2,-3,-2,-1,0
Jag(3) = 1,2,3,4,5,6,7,6,5,4,3,2,1,0,-1,-2,-3,-4,-5,-6,-7,-6,-5,-4,-3,-2,-1,0
…and so on, with Jag(i) going out to +/-(2^i – 1).
Then the infinite sequence is:
Jag(1)Jag(2)Jag(1)Jag(3)Jag(1)Jag(2)Jag(1)Jag(4)Jag(1)Jag(2)…
The program of the 2-tape oblivious TM M’ can be written so that each “jag” simulates 1 step of the original k-tape TM M, with the jag describing the head motions on the first tape (which has 2k tracks, a data track and “slack track” for each tape of M). There are analogous motions on the second tape, and since the sequence is followed regardless of bit values (until the M’ sees that M has halted), M’ is oblivious. It remains to count the total sizes of the jags thru the t-th one, where we may assume t is a power of 2, t = 2^k, so the t-th jag is Jag(k+1). Then the contributions from all Jag(1), all Jag(2), all Jag(3).. are roughly equal, so the sum of steps is roughly 4t(k+1), which gives time O(t*log t) to simulate t steps of M.
Thue didn’t include the negative numbers, so I call the above the “two-way Thue sequence” . I’ve never written out the program of this M’ in full, and would love to know if someone has.
The circuits resulting from M’ embed the full binary tree, and so have exponential expansion. Whereas, the t^2-size circuits from the standard Savage space-time tableau simulation can be laid out on a square mesh, and so have only quadratic expansion. One of the improvements I’d be interested to see is obtaining t^{1+\epsilon} size with reasonable polynomial expansion.
2. January 27, 2010 12:30 pm
This is O(t log t) simulation is very clever. But it seems to me that what it really reveals is that the Turing machine is the wrong model of computation if you want to prove strong time hierarchy theorems. If you adopt a RAM model with a simple programming language, then a universal program can simulate other programs with only constant overhead. Jones, Sudborough and Zalcberg, and Ben-Amram have shown that in such a model—which is the one that students are used to working with in the real world—there are constant-factor time hierarchy theorems, and even (1+epsilon)-factor hierarchy theorems. Their results are not well-known in the theory community, but they should be.
So, is the Turing machine model of computation really the right one for teaching computational complexity? Many students get the wrong impression that the O(t log t) overhead is something fundamental about simulation and the hierarchy theorem, but it’s really just an artificial limitation of the constant-tape TM.
The TM is certainly important historically. But in this case it obscures the central aspect of the hierarchy theorem—namely, simulation and bounded diagonalization—rather than making it clear.
3. January 27, 2010 1:29 pm
Hi, Cris—
I actually devised a variant of the TM model that has constant-factor time overhead for getting k tapes down to one. It has random-access (actually “tree-access”) to the input tape, for four “fingers” labeled a,b,c,d. Its computations consist of a series of finite transductions that read from [a,...,b] and write output to [c,...,d]. The cost for each is |b – a| plus m(max{a,b,c,d}), where m(a) stands for the time to address memory location a. This and several other nice robustness theorems hold for functions m(a) = a^{1/d}, where I think of d as the implicit dimensionality of the memory. But the memory itself is linear, thus embodying the notion that “locality is one-dimensional”. My papers on this model are SIAM J. Comput. 25, 1996 and STACS’94; the latter uses the number of simple finite transductions as a parallel time measure to characterize the NC^k classes. I’d be curious to know whether this helps restore some of the “honour” of the Turing machine architecture in the face of your comment…
My “4t(k+1)” should be 2t(k+1)—though programming details might force each jag to be repeated once, a-la how “Jag(1)” is repeated in the standard linear-speedup-theorem proof, thus bringing back the “4″.
4. richde
January 28, 2010 11:25 am
I really liked this post. It’s been a long time since I’ve actually seen a TM argument — are they still used in class? I was at the conference presentation where Pat made his famous “I program these for a living” comment. My recollection is that it was one of those Johns Hopkins conferences although we were also all together with him at a Computer Science Conference — early, when it was all abstracts and 15 minute presentations — session that same year where he took off on someone who was messing up a multi-tape TM proof.
5. January 28, 2010 6:07 pm
Dick,
The proofs of these claims are not too lost. The recently published Arora-Barak complexity textbook proves the Hennie-Stearns theorem you cite (although, it does so in an appendix). It leaves the Pippenger-Fischer result as an exercise. Other books might mention these results, but I am not sure.
-Michael
• rjlipton *
January 28, 2010 8:05 pm
Thanks. I do wish we could improve these results. I really do not think they are best. But I am stuck.
• steve
February 2, 2010 10:54 am
Homer/Selman goes through this in blow-by-blow detail. Many students here have seen these proofs.
s.
• Cem Say
February 10, 2010 4:17 pm
Arora-Barak covers the first theorem in detail in Chapter 1, not in an appendix.
6. Jonathan Katz
February 14, 2010 9:34 pm
Programming non-deterministic Turing machines is that much harder…
In this line, you may be interested in a recent result showing that any language accepted in NTIME(f(n)) can be accepted by a non-deterministic machine running in time f(n) and space \sqrt(f(n)). See
7. Steven Twentyman
June 30, 2010 5:55 pm
Hello.
What if, in a general theory of everything kind of way, some singular human conscious had used simple Yes(I should feel guilty) or No(I do not feel guilty) answers to answer every moral question that he could remember that had ever occurred in his life. In this way he could have become completely pure. He would have truly asked himself all the questions that could be asked of him. He would have emotionally evolved back to when he was a child.
What if he then assigned an ‘It doesn’t matter as life is only made to make mistakes’ label on every decision that he had analysed.
This would not make him God or the devil, but just very still and very exhausted. Anybody can do this but just for the purpose of this experiment lets say I have done this. Which I have.
There are no fears in me and if I died today I could deal with that because who can know me better than myself? Neither God or the Devil. I am consciously judging myself on ‘their’ level.
To make this work, despite my many faults, take ME as the ONLY universal constant. In this sense I have killed God and the Devil external to myself.The only place that they could exist is if I chose to believe they are internal.
This is obviously more a matter for a shrink more than a mathematician, but that would only be the case depending on what you believed the base operating system of the universe to be. Math / Physics / morals or some other concept.
As long I agreed to do NOTHING, to never move or think a thought, humanity would have something to peg all understanding on. Each person could send a moral choice and I would simply send back only one philosophy. ‘ LIFE IS ONLY FOR MAKING MISTAKES’.
People, for the purpose of this experiment could disbelief their belief system knowing they could go back to it at any time. It would give them an opportunity to unburden themselves to someone pure. A new Pandora’s box. Once everyone had finished I would simply format my drive and always leave adequate space for each individual to send any more thoughts that they thought were impure. People would then eventually end up with clear minds and have to be judged only on their physical efforts. Either good or bad. It would get rid of a lot of maybes which would speed lives along..
If we then assume that there are a finite(or at some point in the future, given a lot of hard work, there will be a finite amount) amount of topics that can be conceived of then we can realise that there will come to a time when we, as a people, will NOT have to work again.
Once we reach that point we will only have the option of doing the things we love or doing the things we hate as society will be completely catered for in every possible scenario. People will find their true path in life which will make them infinitely more happy, forever.
In this system there is no point in accounting for time in any context.
If we consider this to be The Grand Unified Theory then we can set the parameters as we please.
This will be our common goals for humanity. As before everyone would have their say. This could be a computer database that was completely updated in any given moment when a unique individual thought of a new idea / direction that may or may not have been thought before.
All that would be required is that every person on the planet have a mobile phone or access to the web and a self organising weighting algorithm biased on an initial set of parameters that someone has set to push the operation in a random direction.
As I’m speaking first I say we concentrate on GRAINE.
Genetics – Robotics – Artificial Intelligence – Nanotechnology and Zero Point Energy.
I have chosen these as I think the subjects will cross breed information(word of mouth first) at the present day optimum rate to get us to our finishing point, complete and finished mastery of all possible information.
Surely mastery of information in this way will lead us to the bottom of a fractal??? What if one end of the universes fractal was me??? What could we start to build with that concept???
As parameters, we can assume that we live only in 3 dimensions. We can place this box around The Milky Way galaxy as this gives us plenty of scope for all kinds of discoveries.
In this new system we can make time obsolete as it only making us contemplate our things that cannot be solved because up to now, no one has been willing to stand around for long enough. It has been holding us back.
All watches should be banned so that we find a natural rhythm with each other, those in our immediate area and then further afield.
An old clock watch in this system is should only be used when you think of an event that you need to go to. It is a compass, a modern day direction of thought.
A digital clock can be adapted to let you know where you are in relation to this new milky way boxed system.(x,y,z).
With these two types of clocks used in combination we can safely start taking small steps around the box by using the digital to see exactly where you are and then using the old watch to direct you as your thoughts come to you.
We can even build as many assign atomic clocks as are needed to individually track stars. Each and every star in the Milky Way galaxy.
I supposed a good idea would be to assume that I was inside the centre of the super-massive black hole at the centre of the galaxy. That would stop us from being so Earth centric.
We could even go as far as to say that we are each an individual star and that we project all our energies onto the super-massive black hole at the centre of the galaxy.
You can assume that I have stabilized the black hole into a non rotating perfect cube. 6 outputs /f aces in which we all see ‘the universe and earth, friends’ etc. This acts like a block hole mirror finish. Once you look it is very hard to look away from.
The 6 face cube should make the maths easier to run as you could construct the inner of the black hole with solid beams of condensed light(1 unit long) arranged into right angled triangles with set squares in for strength.
Some people would naturally say that if the universe is essentially unknowable as the more things we ‘discover’ the more things there are to combine with and therefore talk about. This is extremely fractal in both directions. There can be no true answers because there is no grounding point. Nothing for the people who are interested, to start building there own personal concepts, theories and designs on.
Is it possible that with just one man becoming conscious of a highly improbable possibility that all of universes fractals might collapse into one wave function that would answer all of humanities questions in a day? Could this be possible?
Answers to why we are here? What the point of life really is et al?
Is it possible that the insignificant possibility could become an escalating one that would at some point reach 100%???
Could it be at that point that the math would figure itself out so naturally that we would barely need to think about it. We would instantly understand Quantum theory and all.
Can anybody run the numbers on that probability?
• colorlessSleepyIdeas
October 3, 2011 4:22 pm
I don’t think science or math could ever answer philosophical questions. Also, why center coordinate systems on the black hole? The only significance of that point in space is its extraordinary mass gradient.
Also, how high were you when you wrote this?
8. Joel Seiferas
October 21, 2010 9:06 am
Note that Wolfgang Paul reported in 1982 (Information and Control 53:1-8) that the 2-tape Hennie-Stearns simulation can be improved at best to time O(n(log n)^{1/3}). (See also Paturi et al., Information and Computation 88:88-104.)
• rjlipton *
October 21, 2010 12:35 pm
Joel,
Thanks for this pointer…dick
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http://math.stackexchange.com/questions/46042/why-does-multiplicatively-weighted-voronoi-diagram-mwvd-with-2-sites-create-a
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# why does multiplicatively weighted voronoi diagram (mwvd) with 2 sites create a circle?
I want to understand the structure of a `multiplicatively weighted voronoi diagram`. I found that the bisector between 2 sites is circle shaped, but couldn't formally see it. can someone explain?
Thanks, Ohad.
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## 1 Answer
About "multiplicatively weighted Voronoi diagrams" see here:
http://www.nirarebakun.com/voro/emwvoro.html
For the benefit of the readers I give a short description: You have a set of cities $p_i$ in the euclidean plane $E:={\mathbb R}^2$, and each of these cities has a given weight $w_i>0$. The "felt distance" $d(z,p_i)$ from an arbitrary point $z\in E$ to the city $p_i$ is defined to be $$d(z,p_i)\ :=\ {|z-p_i| \over w_i}\ ,$$ where $|z-z'|$ denotes the euclidean distance. This means that the "felt distance" to a city with large weight is comparatively small, and conversely, that the $d$-unit-disk of a city with large weight has a large euclidean radius.
Cosider now two cities $p$ and $p'$ with respective weights $w$ and $w'$. The $d$-Voronoi-boundary $\partial$ between these two cities consists of the points $z\in E$ which have the same $d$-distance to $p$ and to $p'$, i.e., it is defined by the equation $${|z-p|\over |z-p'|} \ =\ {w\over w'}\ .\qquad (*)$$ This says that $\partial$ is the locus of all points $z\in E$ for which the ratio $|z-p|/ |z-p'|$ has the a priori given value $\lambda:=w/w'$. It is a theorem of elementary geometry that such a locus is a circle, called the ${\it Apollonian\ circle}$ for the given data $p$, $p'$ and $\lambda$. The simplest proof is by choosing $p=(0,0)$, $p'=(c,0)$. Then the condition $(*)$ becomes $x^2+y^2=\lambda^2\bigl((x-c)^2 + y^2 \bigr)$, which can be simplified to the equation of a circle (or a line).
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http://mathhelpforum.com/algebra/39200-simplifying-expressions-not-sure-kinds.html
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# Thread:
1. ## Simplifying Expressions... Not sure which kinds
Ok, I'm going to post 3 so I can be sure of how to do each kind.
$<br /> \frac {6}{x^2-9x+20} * (5x-25)<br />$
$<br /> \frac {2x-6}{x^2+5x-24} * \frac {x^2+6x-16}{3x}<br />$
$<br /> \frac {(\frac {3x^2}{8x^5} * 2x^3)}{\frac {9x}{8x^2}}<br />$
2. Originally Posted by jscalalamoboy
Ok, I'm going to post 3 so I can be sure of how to do each kind.
$<br /> \frac {6}{x^2-9x+20} * (5x-25)<br />$
$\frac{6\cdot 5(x-5)}{(x-4)(x-5)}=\frac{30(x-5)}{(x-4)(x-5)}=\frac{30}{(x-4)}<br />$
---
$\frac{(3x^{2})(2x^{3})}{8x^{5}}\times \frac{8x^{2}}{9x}$
$\frac{3x(2x^{4})}{9x(8x^{3})}=\frac{2x^{4}}{3(8x^{ 3})}=\frac{2x^{4}}{24x^{3}}=\frac{x}{12}<br />$
3. Originally Posted by jscalalamoboy
Ok, I'm going to post 3 so I can be sure of how to do each kind.
$<br /> \frac {2x-6}{x^2+5x-24} * \frac {x^2+6x-16}{3x}<br />$
$\frac{2x-6}{x^2+5x-24} = \frac{2(x-3)}{(x+8)(x-3)} = \frac{2}{x+8}$
$\frac{x^2+6x-16}{3x} = \frac{(x+8)(x -2)}{3x}$
so the product is $\frac{2}{x+8}\cdot \frac{(x+8)(x-2)}{3x}$
$= \frac{2(x+8)(x-2)}{3x(x+8)} = \frac{2(x-2)}{3x}$
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http://mathoverflow.net/questions/80988/canonical-divisor-of-a-curve-base-point-free-if-g0
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## Canonical divisor of a curve base point free (if g>0)
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Is there a way to prove that the canonical divisor $W$ of an algebraic function field in one variable $F$ over a field $K$ (that is the function field of an algebraic curve) of genus $g>0$ is base point free, without using Clifford´s theorem?
Note that K is not necessarily algebraically closed! in that case I know how to solve the problem.
Equivalently, I have to show that for any place $P$ of $F/K$ there exists an holomorphic differential $\omega$ of $F/K$ such that $v_P(\omega)=0$, that is the support of the associated canonical divisor $(\omega)$ to $\omega$ does not contain the place $P$.
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1
Could you remind me (us) what Clifford's theorem says ? – Damian Rössler Nov 15 2011 at 15:47
if $A$ is a divisor of $F/K$ such that $0\leq\deg(A)\leq 2g-2$, then $$l(A)\leq 1+\deg(A)/2$$ where as usual $l(A)$ is the dimension over $K$ of the vector space $L(A)=\{x\in F|(x)+A\geq0\}$. – GiulioP Nov 15 2011 at 16:23
3
Surely the case where $K$ is not algebraically closed follows from the case where $K$ is algebraically closed? – Daniel Loughran Nov 15 2011 at 16:38
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In the algebraically closed case (that is enough for your purpose, as remarked above) it follows directly by Riemann-Roch, without using Clifford's theorem. – rita Nov 15 2011 at 16:42
exactly, that is what I meant... I used clifford´s theorem (as in the exercises of Stichtenoth book) to solve the problem, but I would like a simple proof which avoids that. Essentially is the fact that if $l(A)=\deg(A)+1$ in genus greater than zero then $A$ is principal, which I can´t prove. Is it possible to prove it without extending the base field to the algebraic closure, as it seems that you suggest? It seems that the result is contained in the book of Deuring, "Lectures in the theory of alf. func.", Lemma Lectures 10, §20, but I don´t understand the proof... – GiulioP Nov 15 2011 at 17:21
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