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58,870 |
<p>I am teaching a introductory course on differentiable manifolds next term. The course is aimed at fourth year US undergraduate students and first year US graduate students who have done basic coursework in
point-set topology and multivariable calculus, but may not know the definition of differentiable manifold. I am following the textbook <a href="http://rads.stackoverflow.com/amzn/click/0132126052">Differential Topology</a> by
Guillemin and Pollack, supplemented by Milnor's <a href="http://rads.stackoverflow.com/amzn/click/0691048339">book</a>.</p>
<p>My question is: <strong>What are good topics to cover that are not in assigned textbooks?</strong> </p>
|
Paul Siegel
| 4,362 |
<p>I think there are two ways to approach a first course on manifolds: one can focus on either their geometry or their topology.</p>
<p>If you want to focus on geometry, then I think Anton Petrunin's suggestion is the end of the story. I'm a fourth year graduate student, and practically every time I find myself confused about something in differential geometry I realize that the root cause of my confusion is that I never properly learned surfaces. And I've taken <em>lots</em> of geometry courses.</p>
<p>If you want to focus on topology, I really think it makes a lot of sense to teach some Morse theory. It's rather elementary, it's extremely powerful and virtually ubiquitous in differential topology, and most of all it really feels like <em>topology</em> in a way that differential forms don't.</p>
<p>Finally, from looking at only the two books you mentioned in your question, I would be a little worried that your students won't have a lot of examples to work with. What about introducing Lie groups?</p>
|
58,870 |
<p>I am teaching a introductory course on differentiable manifolds next term. The course is aimed at fourth year US undergraduate students and first year US graduate students who have done basic coursework in
point-set topology and multivariable calculus, but may not know the definition of differentiable manifold. I am following the textbook <a href="http://rads.stackoverflow.com/amzn/click/0132126052">Differential Topology</a> by
Guillemin and Pollack, supplemented by Milnor's <a href="http://rads.stackoverflow.com/amzn/click/0691048339">book</a>.</p>
<p>My question is: <strong>What are good topics to cover that are not in assigned textbooks?</strong> </p>
|
Charlie Frohman
| 4,304 |
<p>The problem will be that the students do not have a firm grasp of multivariable calculus. </p>
<p>You should probably start with a rigorous review of multivariable calculus including the definition of the differentiable, C^1 implies differentiable on open sets, mixed partials are equal, inverse function theorem, local immersion theorem, local submersion theorem.
That will allow you to segue into the definition of smooth manifold as a parametrized subset of R^n as in Guilleman and Pollack. </p>
<p>Guillemann and Pollack is a softening of Milnor's "Topology from a Differentiable Viewpoint" and as
such is about the lowest level approach you can take to introducing the students to the "stuff" of topology. The exercises are good. I like to have the students divide up the long guided exercise sections to present at the board. I like to supplement the book by proving the Morse Lemma, having a discussion
of linking number, and proving the the Hopf fibration is not homotopic to a constant map using linking numbers. I also like touching on complex variables by proving the argument principle. Finally, I like proving
that two maps from a closed oriented n-manifold to the n-sphere are homotopic if and only if they have the same degree. I don't do all of these in any one year as there is not time. I generally key off of what seems to interest the particular group of students in the class that year.</p>
<p>Be careful in the section on integration, they leave out (or left out in an earlier edition) that you need to be using orientation preserving parametrizations to define the integral.</p>
<p>After teaching such a course for about 15 years, I changed directions and started teaching the foundations of smooth manifolds in the place of the Guilleman and Pollack course, so that students could learn a more mature definition of smooth manifold, and introduce vector bundles,
tensors, and Lie Groups. I have used both the books by Jack Lee and by Boothby. Each has its strong points and weak points (at least in use with graduate students at Iowa.) This turned out to be better for the graduate program as a whole because kids who wanted to do representation theory or PDE could get exposed to the ideas they would see in their research. It also allowed the Differential Geometry sequence to run more regularly. If you decided to go that route, it would still be wise to start with multivariable calculus, as really, very few kids going to graduate school in math have a sufficient background in the calculus.</p>
<p>However, the students are much less happy about taking the foundations of smooth manifolds, because it does not offer the immediate gratification of studying degree and winding number. In fact, when I teach the course as foundations of smooth manifolds, there will always be a block of 3 or 4 students who resent having taken the class. When I teach out of Guilleman and Pollack, even the students who never develop a clue, still enjoy the experience.</p>
|
294,519 |
<p>The problem I am working on is:</p>
<p>Translate these statements into English, where C(x) is “x is a comedian” and F(x) is “x is funny” and the domain consists of all people.</p>
<p>a) $∀x(C(x)→F(x))$ </p>
<p>b)$∀x(C(x)∧F(x))$</p>
<p>c) $∃x(C(x)→F(x))$ </p>
<p>d)$∃x(C(x)∧F(x))$</p>
<h2>-----------------------------------------------------------------------------------------</h2>
<p>Here are my answers:</p>
<p>For a): For every person, if they are a comedian, then they are funny.</p>
<p>For b): For every person, they are both a comedian and funny.</p>
<p>For c): There exists a person who, if he is funny, is a comedian</p>
<p>For d): There exists a person who is funny and is a comedian.</p>
<p>Here are the books answers:</p>
<p>a)Every comedian is funny.
b)Every person is a funny comedian.
c)There exists a person such that if she or he is a comedian, then she or he is funny.
d)Some comedians are funny. </p>
<p>Does the meaning of my answers seem to be in harmony with the meaning of the answers given in the solution manual? The reason I ask is because part a), for instance, is a implication, and "Every comedian is funny," does not appear to be an implication.</p>
|
Andreas Blass
| 48,510 |
<p>Your answers to parts a), b), and d) are OK; in part c) you've reversed the roles of "comedian" and "funny". The book's answers are correct and, in a couple of cases, closer to how people would ordinarily express these statements. One can make other equivalent statements that sound (to me) even more like ordinary usage, for example in d) I'd say "there is a funny comedian" and for b) I'd say "everybody is a funny comedian". </p>
<p>About the last sentence in your question, "Every comedian is funny" is, despite how it looks to you, an implication --- it says that being a comedian implies being funny. More generally, consider any restricted universal quantification, i.e., a situation where you assert some property $P(x)$ not for absolutely all $x$'s but only for those that satisfy some restriction $R(x)$. Such a restricted universal quantification is expressed by a universally quantified implication, $(\forall x)\,(R(x)\to P(x))$. I suspect that part of the purpose of part a) was to lead you to observe this connection between restricted universal quantification and implication.</p>
<p>Similarly, I suspect the purpose of part d) was to lead you to observe the connection between restricted existential quantification and conjunction. And the purpose of parts b) and c) is to show that if you mix these things up, using implication with existential quantification or using conjunction with universal quantification, you get some rather unnatural-sounding statements, and in any case you do not get the effect of just restricting the quantifier.</p>
|
1,877,558 |
<p>For instance, let $(\mathbb{R}, \mathfrak{T})$ be $\mathbb{R}$ with the usual topology. </p>
<p>Why is that $\mathfrak{T} \times \mathfrak{T}$ is a basis on $\mathbb{R} \times \mathbb{R}$ instead of topology?</p>
<p>It seems that people just take $\mathfrak{T} \times \mathfrak{T}$ as a basis by definition. There must be some open sets in the product that cannot be represented by Cartesian product of $\mathfrak{T} \times \mathfrak{T}$, but I don't have any examples handy. </p>
<p>Can someone please instruct?</p>
<p><em>I am still not convinced because the examples so far are not "constructive" :(</em></p>
|
Arkady
| 23,522 |
<p>Hint: If I take the union of two elements of $\mathcal I\times\mathcal I$ (boxes), it is very unlikely that their union will be a box. So $I\times I$ will not be topology).</p>
<p>Since you want an explicit construction, take the union of the sets $A=(0,1)\times(0,1)$ and $B=(1,2)\times(1,2)$. This is not of the form $I\times J$ for any open sets $I$ and $J$ in $\mathbb R$ because if there were, then $I$ would have to contain $1/2$ and $J$ would have to contain $3/2$ but the point $(1/2,3/2)$ is not in $A\cup B$.</p>
|
4,374,391 |
<blockquote>
<p>Find prime number <span class="math-container">$p$</span> such that <span class="math-container">$19p+1$</span> is a square number.</p>
</blockquote>
<p>Now, I have found out, what I think is the correct answer using this method.<br />
Square numbers can end with - <span class="math-container">$1, 4, 9, 6, 5, 0$</span>.<br />
So, <span class="math-container">$19p+1$</span> also ends with these digits.<br />
Thus, <span class="math-container">$19p$</span> ends with - <span class="math-container">$0, 3, 8, 5, 4, 9$</span>.<br />
As <span class="math-container">$p$</span> is either odd or <span class="math-container">$2$</span>. So, <span class="math-container">$19p$</span> is either odd or ends with <span class="math-container">$8$</span>.<br />
So, we can say that <span class="math-container">$19p$</span> ends with - <span class="math-container">$3, 8, 5, 9$</span>.<br />
So, <span class="math-container">$p$</span> ends with - <span class="math-container">$7, 2, 5, 1$</span>.<br />
As <span class="math-container">$19*7$</span> end with <span class="math-container">$3$</span>, <span class="math-container">$19*2$</span> ends with <span class="math-container">$8$</span> etc.<br />
Thus, possible one digit values of <span class="math-container">$p$</span> - <span class="math-container">$2, 7, 5$</span>.
<span class="math-container">$$19*2+1=39$$</span>
<span class="math-container">$$19*7+1=134$$</span>
<span class="math-container">$$19*5+1=96$$</span>
None of these are square numbers.<br />
Possible two digit values of <span class="math-container">$p$</span> - <span class="math-container">$11, 17$</span>.
<span class="math-container">$$19*11+1=210$$</span>
<span class="math-container">$$19*17+1=324=18^2$$</span>
Thus, <span class="math-container">$p=17$</span>.</p>
<p>But, I am not satisfied with the solution as it is basically trial and error. Is there a better way to do this?</p>
|
angryavian
| 43,949 |
<ul>
<li>What is the cardinality of <span class="math-container">$\{0,1\}^n$</span> (in terms of <span class="math-container">$n$</span>)?</li>
<li>Given that the cardinality of <span class="math-container">$2^A$</span> is <span class="math-container">$2^{|A|}$</span>, does this suggest some necessary condition on <span class="math-container">$A$</span> in order for <span class="math-container">$|\{0,1\}^n| = |2^A|$</span> to hold?</li>
<li>Try to let <span class="math-container">$0$</span> and <span class="math-container">$1$</span> encode membership in subsets of <span class="math-container">$A$</span>.</li>
</ul>
|
947,358 |
<p>Okay $g(x)= \sqrt{x^2-9}$</p>
<p>thus, $x^2 -9 \ge 0$</p>
<p>equals $x \ge +3$ and $x \ge -3$</p>
<p>thus the domains should be $[3,+\infty) \cup [-3,\infty)$ how come the answer key in my book is stating $(−\infty, −3] \cup[3,\infty)$. </p>
|
taninamdar
| 66,212 |
<p>$$\begin{align}\sqrt{x^2 - 9} \text{ exists} &\implies x^2-9 \ge 0\\&\implies (x+3)(x-3) \ge 0 \\&\implies (x-3), (x+3) \text{ both are non-negative OR } (x-3),(x+3) \text{both are non-positive} \end{align}$$</p>
<p><strong>case 1</strong>
Now, $(x-3)(x+3)$ both are non-negative when $x-3 \ge 0$ and $x+3 \ge 0 \implies x \ge -3 $ and $x \ge 3 \implies x \ge 3$.</p>
<p><strong>case 2</strong>
Now, $(x-3)(x+3)$ both are non-positive when $x-3 \le 0$ and $x+3 \le 0 \implies x \le -3 $ and $x \le 3 \implies x \le -3$.</p>
|
605,155 |
<p>$\newcommand{\ker}{\operatorname{ker}}$</p>
<p>Proof that: $\ker AB\subseteq\ker A+\ker B$</p>
<p>my solution:</p>
<p>$x\in \ker AB\to ABx=0\to \begin{cases} Ax=0\to x\in \ker A\\Bx=0\to x\in \ker B\end{cases}$</p>
<p>$\to x\in \ker A+\ker B\to \ker AB\subseteq \ker A+\ker B$</p>
<p>Question: Do it right? if false, the new right to do so?</p>
|
Ben Grossmann
| 81,360 |
<p>Note that we can have $x \in \ker(AB)$ with $x \not \in \ker(A)$ and $x \not \in \ker B$. For example, take
$$
A = \pmatrix{1&0\\0&0};\quad B=\pmatrix{0&1\\1&0}; \quad x=\pmatrix{1\\0}
$$
Note that $Ax \neq 0, Bx \neq 0,$ but $ABx=0$. Also, there are no vectors in the kernels of $A$ and $B$ whose sum is $x$.</p>
<p>I assume what you are trying to prove is that
$$
\dim \ker AB \leq \dim \ker A + \dim \ker B
$$
However, you will have to try a different approach.</p>
<hr>
<p>Assuming I have the correct definition, the fact that
$$
\dim(\ker A + \ker B) \leq \dim \ker A + \dim \ker B
$$
follows directly from the fact that $\ker A, \ker B$ are subspaces of some vector space $V$. So, we must prove that
$$
\ker AB \leq \dim(\ker A + \ker B)
$$
My hint for this is that we may write
$$
\dim \ker(AB) =
\dim(\ker(B) \cup (\ker(A) \cap \operatorname{Image}(B)))
= \dim\ker(B)+ \dim(\ker(A) \cap \operatorname{Image}(B))
$$
Why is this the case, and what does this statement allow you to conclude?</p>
|
1,910,983 |
<p>What conditions are equivalent to <strong>singularity</strong> of matrix $A\in \mathbb{R}^{n,n}$.<br>
<strong>a.</strong> $\dim(ker A) \ge 0$<br>
<strong>b.</strong> There is exist vector $b$ such that $Ax=b$ is contradictory.<br>
<strong>c.</strong> $rank(A^T) < n$ </p>
<p><strong>a.</strong> is true for each matrix, in other words $\dim$ can't be negative.<br>
<strong>c.</strong> $rank(A^T) = rank(A)$. Singularity means that some vector (row) is linearly <strong>dependent</strong> on other vectors. Then, we may using elementary operations on rows, reset this row -> so it is true that $rank(A) < n$</p>
<p>Is it correct ?<br>
When it comes to <strong>b.</strong> I suppose that it is true, however I can't prove it. </p>
|
H. H. Rugh
| 355,946 |
<p>The sequence verifies $a_{10+k} = a_{10 + (k \; {\rm mod}\; 4)}$ for $k\geq 0$ (and there are no other relations). So look for $\ell=10+k$, $k\geq 0$ so that
$a_{10+k}=a_{20+2k}=a_{10+(10+2k)}$. And this is equivalent to $k\geq 0$ and
$$ k \equiv 10+2 k \ {\rm mod} \ 4 $$
or $k \equiv 2 \ {\rm mod} \ 4$.</p>
|
178,028 |
<p>I am given $G = \{x + y \sqrt7 \mid x^2 - 7y^2 = 1; x,y \in \mathbb Q\}$ and the task is to determine the nature of $(G, \cdot)$, where $\cdot$ is multiplication. I'm having trouble finding the inverse element (I have found the neutral and proven the associative rule.</p>
|
user29999
| 29,999 |
<p>\begin{eqnarray}
\dfrac{1}{x+y\sqrt{7}} &=& \dfrac{x-y\sqrt{7}}{x^2-7y^2}\\
& =& x-y\sqrt{7}
\end{eqnarray}</p>
|
855,570 |
<p>I am having trouble with what seems like it should be a simple problem. I am trying to find intersections of connections between multiple people but I want to include any intersection of connections found between any two of the sets.</p>
<p>For example, Let</p>
<p>$A$ = {January, February, March, April, May, August}</p>
<p>$R$ = {January, February, March, April, September, October, November, December}</p>
<p>$Y$ = {January, February, May, July}</p>
<p>Now, $A \cap R \cap Y$ = {January, February},</p>
<p>but $A \cap R$ = {January, February, March, April}, </p>
<p>$A \cap Y$ = {January, February, May}, </p>
<p>and $R \cap Y$ = {January, February}</p>
<p>So what I really want is {January, February, March, April, May}</p>
<p>Now, I may not just have $A, R, $ and $Y$, I may have a lot more to sift through. Is there a more simple principle that I am missing to group all of these intersections together as a subset?</p>
<p>One thing I thought of was creating a graph, but I am rusty on my graph algorithms. Any suggestions though would be great.</p>
<p>Thanks!</p>
|
Ross Millikan
| 1,827 |
<p>It looks like you want any month that is in more than one subset. You can go through all the months and count how many subsets each one is in. If it is greater than one, put it in your final list.</p>
|
812,778 |
<p>Prove that $(4/5)^{\frac{4}{5}}$ is irrational.</p>
<p><strong>My proof so far:</strong></p>
<p>Suppose for contradiction that $(4/5)^{\frac{4}{5}}$ is rational.</p>
<p>Then $(4/5)^{\frac{4}{5}}$=$\dfrac{p}{q}$, where $p$,$q$ are integers.</p>
<p>Then $\dfrac{4^4}{5^4}=\dfrac{p^5}{q^5}$</p>
<p>$\therefore$ $4^4q^5=5^4p^5$</p>
<p>I've got to this point and now I don't know where to go from here.</p>
|
Aryan
| 153,656 |
<p>A slightly modified approach.</p>
<p>$(\frac{4}{5})^\frac{4}{5} = (\frac{4}{5})^{1-\frac{1}{5}} = (\frac{4}{5})^1 \times(\frac{5}{4})^\frac{1}{5}$</p>
<p>Since the first factor is rational we need only show that the latter term $(\frac{5}{4})^\frac{1}{5}$ is irrational.</p>
<p>Let us assume that the latter is rational and write (where $p$ and $q$ are coprime)
$\frac{p}{q}=(\frac{5}{4})^\frac{1}{5} \implies 4\times p^5 = 5 \times q^5$</p>
<p>The L.H.S is even and this implies that $q$ is even (sub $q=2^5\times n$ in above); $q^5$ then must have a factor of $2^5$. This would then require $p$ to be even and thus violates the coprime assumption and gives the required contradiction.</p>
|
1,304,529 |
<p>I have come across these while studying the limsup & liminf of sequence of subset of a set. In order to understand that, I have to understand what least upper bound & greatest lower bound of a sequence of subset mean. I would be grateful if anyone helps me comprehend this concept intuitively as I am new & novice to this topic.</p>
|
Mauro ALLEGRANZA
| 108,274 |
<p>Ref also to <a href="https://math.stackexchange.com/questions/1305004/what-is-meant-by-ordering-of-set-by-inclusion">your subsequent question</a>.</p>
<p>A <a href="http://en.wikipedia.org/wiki/Partially_ordered_set" rel="nofollow noreferrer">partially ordered set</a> is a very "simple" mathematical structure :</p>
<blockquote>
<p>A pooset [Partially Ordered SET] consists of a set together with a binary relation that indicates that, for certain pairs of elements in the set, one of the elements precedes the other. Such a relation is called a partial order to reflect the fact that not every pair of elements need be related: for some pairs, it may be that neither element precedes the other in the poset. </p>
</blockquote>
<p>An easy example is the set of human males with the "order relation" : "$x$ is father of $y$".</p>
<p>Obviously, not two males $a$ and $b$ whatever are in the relation : "$a$ is father of $b$" or "$b$ is father of $a$", but some are; this means that the order is <em>partial</em>.</p>
<p>Usually, the order relation is symbolized with $<$, because it is a "generalization" of the "less than" relation between numbers [that, by the way, is a partial order which is <em>total</em>].</p>
<p>Consider now the set $\mathcal P(X)$ of <em>subsets</em> of a set $X$; $\mathcal P(X)$ is <em>partially orderes</em> by the "inclusion" relation $\subseteq$.</p>
<p>Cosnider $X = \{a,b \}$; we have that the set $\mathcal P(X)$ containing all its subsets is $\{ \emptyset, \{a \}, \{ b \}, \{ a, b \} \}$.</p>
<p>As you can easily check, $\mathcal P(X)$ is <em>partially ordered</em> by $\subseteq$ :</p>
<blockquote>
<p>$\emptyset \subseteq \{a \}, \{ b \}, \{ a, b \}$</p>
<p>$\{a \}, \{ b \} \subseteq \{ a, b \}$</p>
</blockquote>
<p>but $\{a \} \nsubseteq \{ b \}$ and $\{b \} \nsubseteq \{ a \}$.</p>
<hr>
<p>Thus :</p>
<blockquote>
<blockquote>
<p>what <em>least upper bound</em> and <em>greatest lower bound</em> of a sequence of subset does mean ?</p>
</blockquote>
</blockquote>
<p>See <a href="http://en.wikipedia.org/wiki/Infimum_and_supremum" rel="nofollow noreferrer">here</a> :</p>
<blockquote>
<p>In mathematics, the <em>infimum</em> of a subset $S$ of a partially ordered set $T$ is the greatest element of $T$ that is less than or equal to all elements of $S$. Consequently the term <em>greatest lower bound</em> is also commonly used.</p>
<p>The definition of <em>greatest lower bounds</em> easily generalizes to subsets of arbitrary partially ordered sets and as such plays a vital role in order theory. </p>
<p>The dual concept of infimum is given by the notion of a supremum or <em>least upper bound</em>.</p>
<p>The <em>least upper bound</em> of a subset $S$ of $(\mathcal P(X), \subseteq)$, where $\mathcal P(X)$ is the power set of some set $X$, is the supremum with respect to [the relation of inclusion] $\subseteq$, and is the union of the elements of $S$. </p>
</blockquote>
<hr>
<p>Regarding <a href="http://en.wikipedia.org/wiki/Limit_superior_and_limit_inferior" rel="nofollow noreferrer">Limsup</a> of a sequence $\{ X_n \}$ of subset of the set $X$ (trivial example : $X=[0,1]$ and $X_n=[0,1/n]$ ) :</p>
<blockquote>
<p>consider the infimum, or <em>greatest lower bound</em>, of a sequence of sets. In the case of a sequence of sets, the sequence constituents "meet" at a set that is somehow smaller than each constituent set. Set inclusion provides an ordering that allows set intersection to generate a greatest lower bound $\bigcap X_n$ of sets in the sequence $\{ X_n \}$. Similarly, the supremum, or <em>least upper bound</em>, of a sequence of sets is the union $\bigcup X_n$ of sets in the sequence $\{ X_n \}$. </p>
</blockquote>
<p>Thus :</p>
<blockquote>
<p>If $\{ X_n \}$ is a sequence of subsets of $X$ [i.e. $X_n \subseteq X$, for every $n$], then:</p>
<blockquote>
<p>$\text {lim sup} \ X_n$ consists of elements of $X$ which belong to $X_n$ for <em>infinitely many</em> $n$. That is, $x \in \text {lim sup} X_n$ if and only if there exists a subsequence $\{ X_{n_k} \}$ of $\{ X_n \}$ such that $x \in X_{n_k}$ for all $k$.</p>
<p>$\text {lim inf} \ X_n$ consists of elements of $X$ which belong to $X_n$ for <em>all but finitely many</em> $n$. That is, $x \in \text {lim inf} X_n$ if and only if there exists some $m > 0$ such that $x \in X_n$ for all $n > m$.</p>
</blockquote>
<p>So the inferior limit acts like a version of the standard infimum that is unaffected by the set of elements that occur only finitely many times. That is, the <em>infimum limit</em> is a subset (i.e. a lower bound) for <em>all but finitely many</em> elements [of the sequence $\{ X_n \}$ ].</p>
</blockquote>
|
2,565,802 |
<p>Calculate the volume of the region bounded by $z=0, z=1,$, and $(z+1)\sqrt{x^2+y^2}=1$</p>
<p>The integral is $\int_{B}z\text{ dV}$</p>
<p>The area is like the thing between the top two green places. The first place is $z=1$, second is $z=0$</p>
<p>Clearly we have $0\leq z\leq 1$, but I'm not sure what to bound next? Should I be using cylindrical? </p>
<p>Would it be correct in saying $0\leq r\leq \dfrac{1}{z+1}$
<a href="https://i.stack.imgur.com/THRhl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/THRhl.png" alt="enter image description here"></a></p>
|
N. F. Taussig
| 173,070 |
<p>A circle can be formed by joining the ends of a row together. We solve the problem for a row, then remove those solutions in which two of the selected people are at the ends of the row.</p>
<p>We will arrange $96$ blue balls and $4$ green balls in a row so that the green balls are separated. Line up $96$ blue balls in a row, leaving spaces in between them and at the ends of the row in which to insert a green ball. There are $95$ spaces between successive blue balls and two at the ends of the row for a total of $97$ spaces. To separate the green balls, we choose four of these spaces in which to insert a green ball, which we can do in
$$\binom{97}{4}$$
ways. Now number the balls from left to right, the numbers on the green balls represent the positions of the selected people.</p>
<p>However, those selections in which there are green balls at both ends of the row are inadmissible since that means the people in those positions will be in adjacent seats when the ends of the row are joined. If two green balls are placed at the ends of the row, the other two green balls must be placed in two of the $95$ spaces between successive blue balls. Hence, there are
$$\binom{95}{2}$$
placements of green balls in which both ends of the row are filled by green balls. As these placements are inadmissible, the number of ways four of the $100$ people at the circular table may be selected so that no two are adjacent is
$$\binom{97}{4} - \binom{95}{2}$$ </p>
|
2,231,388 |
<blockquote>
<p>Consider a ring map $B \rightarrow A$. Consider the map $f:A \otimes_{B}A \rightarrow A$, where $x \otimes y$ goes to $xy$. Let $I$ be the kernel of $f$. Why is it true that $I/I^2$ is isomorphic to $I \otimes_{A \otimes_{B}A} A$?</p>
</blockquote>
<p>This is what I've been able to prove till now:</p>
<p>Let $R=A \otimes_{B}A$. Now, consider the $R$-module homomorphism $\phi$from $I$ to $I\otimes (R/I)$, $\phi(a)=a\otimes 1$. It is easy to see that the kernel of $\phi$ contains $I^2$. How do I prove it is exactly $I^2$?</p>
|
Andrew
| 154,986 |
<p>This can be made slightly more general: Let $R$ and $I$ be as you say, and let $M$ be any $R$-module. There is an isomorphism
$$
M\otimes_R R/I \cong M/IM.
$$
Here's how to see it: start with the exact sequence
$$
0 \to I \to R \overset{f}{\to} R/I \to 0
$$
(note that $f$ is the same as your $f$).
Then tensor with $M$:
$$
M\otimes_R I \to M\otimes_R R \overset{f\otimes 1}\to M\otimes_R (R/I) \to 0
$$
This is exact by the right-exactness of $\otimes$. Using the natural isomorphism $M \otimes_R R \cong M{}{}{}$, this sequence is exactly
$$
M\otimes_R I \overset{\phi}{\to} M \overset{\tilde f}\to M\otimes_R (R/I) \to 0, \qquad (*)
$$
where the map $\phi$ takes a pure tensor $m\otimes x$ to $xm$ and $\tilde{f}$ takes $m$ to $m\otimes 1$. It should be clear that the image of $\phi$ is equal to $IM\subset M$. </p>
<p>The nontrivial step that proves the direction you need is the exactness of $(*)$ at the term $M$, which gives
$$
IM = \text{image}(\phi) = \ker(M\to M\otimes_R(R/I)).
$$
So, in the case that $I=M$, we have that
$$
I^2 = \ker(I\to I\otimes_R(R/I)).
$$</p>
<p>Unfortunately, I do not know of a good way to prove this by hand. This seems to rely on the right exactness of $\otimes$, and you can see some discussion <a href="https://math.stackexchange.com/questions/81640/proving-that-the-tensor-product-is-right-exact">here</a> that discusses how this is proved.</p>
|
752,517 |
<p>From Wikipedia</p>
<blockquote>
<p>...the free group $F_{S}$ over a given set $S$ consists of all expressions (a.k.a. words, or terms) that can be built from members of $S$, considering two expressions different unless their equality follows from the group axioms (e.g. $st = suu^{−1}t$, but $s ≠ t$ for $s,t,u \in S$). The members of $S$ are called generators of $F_{S}$.</p>
</blockquote>
<p>I don't understand the distinction between " <strong>$a$ and $b$ freely generate a group</strong>" , "$a$ and $b$ <strong>generate a free group</strong>" and just checking if a group is free. For example I thought that if $a$ and $b$ freely generate a group then that group is free and vice versa. However I have seen statements like " $a$ and $b$ freely generate a free subgroup of..." quite a few times and so there seems to be some distinction between the terms. If so could you please provide an example where one of the conditions hold bit the other doesn't?</p>
<p>If, for example, $S=\{a,b\}$, but $a^{3}=1$ and $b^{2}=1$ is it the case that $S$ cannot generate a free group even if any word written using only the letters $a$,$a^2$ and $b$ (excluding powers of these) has a unique representation?
If we say that we are considering words in $\{a,b\}$ does that mean that we consider words using $a$,$b$, $a^{-1}$ and $b^{-1}$ as letters and identifying $b^2=1$ and $a^3=1$ in the process of reducing the letter or that we use any integer power of and b and only reduce identities that hold by group axioms independent of the actual group structure we are considering?</p>
<p>I apologize if the question is confusing but I am quite confused myself. If you think there is any way of clarifying the question please feel free to suggest it.</p>
<p>Thank you.</p>
|
Community
| -1 |
<p>The unit circle is a subspace of the plane; thus if you already know the plane is Hausdorff, that carries over to the circle too.</p>
<p>Showing that</p>
<p>$$ \mathrm{R} \to (-\pi, \pi) \to \text{everything but $(-1,0)$} $$</p>
<p>is surjective and $f(\infty) = (-1,0)$ would be enough to show $f$ is surjective.</p>
<p>Sometimes, though, it's easier to just find the <em>inverse</em> of your function.</p>
|
412,642 |
<p>Let $E=\mathcal{C}[0,1]$. How to prove that if $f_n\rightarrow f$ with the norm $\displaystyle{\|\cdot\|_\infty=\sup_{t\in[0, 1]}f(t)}$ then $f_n\rightarrow f$ with the norm $\displaystyle{\|\cdot\|_p=\left(\int_{0}^{1}|f(t)|^p\,dt\right)^{1/p}}$. </p>
<p>Give an example showing that the converse is not true</p>
|
Mhenni Benghorbal
| 35,472 |
<p>Here is a start for the first part
$$ \displaystyle{\|f_n-f\|^p_p= \int_{0}^{1}|f_n(t)-f(t)|^p\,dt }\leq \int_{0}^{1}\sup|f_n(t)-f(t)|^p\,dt \leq \dots. $$</p>
|
1,538,496 |
<p>I came across this riddle during a job interview and thought it was worth sharing with the community as I thought it was clever:</p>
<blockquote>
<p>Suppose you are sitting at a perfectly round table with an adversary about to play a game. Next to each of you is an infinitely large bag of pennies. The goal of the game is to be the player who is able to put the last penny on the table. Pennies cannot be moved once placed and cannot be stacked on top of each other; also, players place 1 penny per turn. There is a strategy to win this game every time. Do you move first or second, and what is your strategy?</p>
</blockquote>
<p>JMoravitz has provided the answer (hidden in spoilers) below in case you are frustrated!</p>
|
JMoravitz
| 179,297 |
<p>Yes, I've seen this one before. Assuming exactly one penny is allowed to be placed per turn:</p>
<blockquote class="spoiler">
<p> Go first and place a penny in the dead center of the table. From then on, any move your opponent makes, place a penny in the mirror opposite location (i.e. rotated 180 degrees). It stands to reason that if your opponent's move was valid, yours will be too. Hence, you will always have an available move if your opponent does. Since the table is only finitely large, there can only be finitely many turns, hence you will eventually win.</p>
</blockquote>
<p>A more complete proof:</p>
<p>Suppose the table is described using polar coordinates with the center of the table as the origin ($r=0$).</p>
<blockquote class="spoiler">
<p> My first move is to place at $r=0$. When my opponent makes a legal move at $(r,\theta)$ I attempt to place a coin at $(r,\theta+180^\circ)$.</p>
</blockquote>
<p>Claim: I am always allowed to do so and such a move will always be valid.</p>
<blockquote class="spoiler">
<p>Proof: Suppose otherwise. Then that implies that either the target location is not on the table (in which case my opponent's previous move will also have not been on the table and therefore was also invalid), or that target location would have a coin overlap with another previously placed coin. As it could <em>not</em> have been the coin that my opponent has just placed on his last turn (as it is $180^\circ$ away), that implies that those coins must have been placed previously. However... since my moves are always playing $180^\circ$ away from my opponent, that implies that there should be the same situation on the other side of the table and that my opponents coin <em>also</em> is overlapping the corresponding mirrored coins and therefore my opponents move was invalid. Either way, we reach a contradiction implying that if my opponents move <em>was</em> valid that my move is also guaranteed to be valid too.</p>
</blockquote>
|
1,335,842 |
<p>The smallest solution to the above equation for various primes are:</p>
<p>$(p=2)$ $3^2 = 2*2^2 +1$</p>
<p>$(p=3)$ $2^2 = 2*1^2 +1$</p>
<p>$(p=5)$ $9^2 = 5*4^2 +1$</p>
<p>$(p=7)$ $8^2 = 7*3^2 +1$</p>
<p>Is there at least one solution for each prime?
If there is one solution, there are infinite.</p>
|
Travis Willse
| 155,629 |
<p>If one permits a general positive integer $n$ in place of the prime $p$, the resulting equation (in integers $x, y$) is called <em><a href="https://en.wikipedia.org/wiki/Pell's_equation" rel="nofollow">Pell's equation</a></em>, and is usually rearranged to read
$$x^2 - n y^2 = 1.$$
Lagrange showed that there are solutions for all nonsquare $n$, and so in particular it holds for all prime $n$. Note that even for modest $n$ the fundamental solution (i.e., the one minimizing positive $x$) can be large. For $n = 61$, for example, the fundamental solution is $(1\,766\,319\,049, 226\,153\,980)$.</p>
|
3,124,158 |
<p>So what I want to prove is
<span class="math-container">$$ |xy+xz+yz- 2(x+y+z) + 3| \leq |x^2+y^2+z^2-2(x+y+z)+3| $$</span>
for <span class="math-container">$x,y,z\in \mathbb{R}$</span>, and I'm aware that the RHS is just <span class="math-container">$|(x-1)^2+(y-1)^2+(z-1)^2|$</span>.</p>
<p>Now I'm able to prove that <span class="math-container">$ x^2+y^2+z^2 \geq xy+xz+yz $</span> as this just follows from the AM-GM inequality. So I know that the statement <em>without</em> the absolute values must be true, i.e.
<span class="math-container">$$ xy+xz+yz- 2(x+y+z) + 3 \leq x^2+y^2+z^2-2(x+y+z)+3 $$</span>
But I can't see why I'm safe to just put absolute values on both sides here. Because I'm not sure why the LHS is guaranteed to be smaller in magnitude than the RHS?</p>
<p>(I thought about Cauchy-Schwarz being hidden here but then I realised that I could not see how.)</p>
<p>Edit: Alternatively I also understand that
<span class="math-container">$$ |xy+xz+yz| \leq |xy|+|xz|+|yz| \leq |x|^2+|y|^2+|z|^2 = x^2+y^2+z^2 = |x^2+y^2+z^2| $$</span>
but then if I try to adapt this path, the <span class="math-container">$-2(x+y+z) $</span> bit throws me off</p>
|
David Holden
| 79,543 |
<p>have you tried evaluating:
<span class="math-container">$$
(x-1)(y-1) + (y-1)(z-1) +(z-1)(x-1)
$$</span>
?</p>
|
2,007,373 |
<p>At some point in your life you were explained how to understand the dimensions of a line, a point, a plane, and a n-dimensional object. </p>
<p>For me the first instance that comes to memory was in 7th grade in a inner city USA school district. </p>
<p>Getting to the point, my geometry teacher taught,</p>
<p>"a point has no length width or depth in any dimensions, if you take a string of points and line them up for "x" distance you have a line, the line has "x" length and zero height, when you stack the lines on top of each other for "y" distance you get a plane"</p>
<p>Meanwhile I'm experiencing cognitive dissonance, how can anything with zero length or width be stacked on top of itself and build itself into something with width of length?</p>
<p>I quit math. </p>
<p>Cut to a few years after high school, I'm deep in the math's. </p>
<p>I rationalized geometry with my own theory which didn't conflict with any of geometry or trigonometry. </p>
<p>I theorized that a point in space was infinitely small space in every dimension such that you can add them together to get a line, or add the lines to get a plane. </p>
<p>Now you can say that the line has infinitely small height approaching zero but not zero.</p>
<p>What really triggered me is a Linear Algebra professor at my school said that lines have zero height and didn't listen to my argument. . . </p>
<p>I don't know if my intuition is any better than hers . . . if I'm wrong, if she's wrong . . . </p>
<p>I would very much appreciate some advice on how to deal with these sorts of things. </p>
|
hmakholm left over Monica
| 14,366 |
<p>The viewpoint you're groping towards is not crazy, and at least historically you're in excellent company -- e.g., Leibniz had similar ideas when he viewed an integral as a sum of the heights of infinitely many lines, weighted by their infinitesimal width, and this intuition is still the background for the notation $\int f(x)\,dx$ for integrals. However, from a modern perspective this viewpoint carries a large risk of accidentally concluding nonsense by stretching it beyond what it can do. Therefore it is not really in favor anymore.</p>
<p>A <strong>better answer</strong> to your misgivings (in the sense of being closer to the mainstream presentation) is probably simply to jump in with both feet and declare that <strong>a line is not really made of points</strong>.</p>
<p>Decide to think of a line as <em>something that is in principle a different kind of thing from a bunch of points glued together</em>. You can do this and still acknowledge that points <em>exist</em> and some of them are on the line while others are not.</p>
<p>It then turns out that all of the properties of a line segment (or a smooth curve in general) can be <em>recovered</em> from knowing which points lie on it and which don't. This doesn't necessarily mean that the points <em>make up</em> the line, but merely that the points tell us <em>enough</em> about the line.</p>
<p>It is <em>technically convenient</em>, then, to speak about the <em>set of points on the line</em> as a placeholder for the line itself, when we're formalizing our reasoning -- for the pragmatic reason that we have a well-developed common machinery and notation for speaking of sets of things, which means that we don't need to introduce a new formalism for an entirely different kind of things.</p>
<p>Some people are so comfortable with this representation that they happily declare that the line IS its set of points -- but nobody says you have to think of it that way. As long as you agree that the set of points <em>determine</em> the line, you can still communicate with people who prefer the other idea.</p>
|
2,755,733 |
<p>Why in this <a href="https://math.stackexchange.com/questions/625112/if-tossing-a-coin-400-times-we-count-the-heads-what-is-the-probability-that-t">If, tossing a coin 400 times, we count the heads, what is the probability that the number of heads is [160,190]?</a> question heropup's asnwer is like that? </p>
<p>I don't understand the blue text, I think it should be <strong>190</strong> instead of $\color{blue}{200}.$</p>
<p>And why to do this step $\color{blue}{\Pr[159.5 \le X \le 200.5] }?$ when you can pass directly to standarization.</p>
<p><strong>This is his/her answer:</strong></p>
<p>With $n = 400$ trials, the exact probability distribution for the number of heads $X$ observed is given by $X \sim {\rm Binomial}(n = 400, p = 1/2)$, assuming the coin is fair. Since calculating $\Pr[160 \le X \le \color{blue}{200}]$ requires a computer, and $n$ is large, we can approximate the distribution of $X$ as ${\rm Normal}(\mu = np = 200, \sigma^2 = np(1-p) = 100)$. Thus $$\begin{align*} \Pr[160 \le X \le 200] &\approx \color{blue}{\Pr[159.5 \le X \le 200.5] }\\ &= \Pr\left[\frac{159.5 - 200}{\sqrt{100}} \le \frac{X - \mu}{\sigma} \le \frac{200.5 - 200}{\sqrt{100}} \right] \\ &= \Pr[-4.05 \le Z \le 0.05] \\ &= \Phi(0.05) - \Phi(-4.05) \\ &\approx 0.519913. \end{align*}$$ Note that we employed continuity correction for this calculation. The exact probability is $0.5199104479\ldots$.</p>
<p>A similar calculation applies for $\Pr[160 \le X \le 190]$. Using the normal approximation to the binomial, you would get an approximate value of $0.171031$. Using the exact distribution, the probability is $0.17103699497659\ldots$.</p>
|
BruceET
| 221,800 |
<p>As mentioned in the linked post, if $X \sim \mathsf{Binom}(400, .5),$ then
$$P(160 \le X \le 200) = P(159 < X < 201) = P(159.5 < X < 200.5) = 0.5199$$
to four places. [The computation is from R statistical software, in which
<code>pbinom</code> is a binomial CDF.]</p>
<pre><code>pbinom(200, 400, .5) - pbinom(159, 400, .5)
[1] 0.5199104
</code></pre>
<p>Computing with the normal approximation, it is best to use the form
$P(159.5 < X < 201.5)$ because that gives the best fit for using the <em>continuous</em>
normal distribution to get probabilities for the <em>discrete</em> binomial distribution. The approximating normal distribution is $\mathsf{Norm}(\mu=200,\, \sigma = 10).$</p>
<p>Using endpoints 160 and 200 gives the normal approximation 0.49997, which
is a little too small.</p>
<pre><code>diff(pnorm(c(160,200), 200, 10))
[1] 0.4999683
</code></pre>
<p>Using the endpoints 159 and 201, we get 0.53980 (a little too large). </p>
<pre><code>diff(pnorm(c(159,201), 200, 10))
[1] 0.5398072
</code></pre>
<p>With the endpoints 159.5 and 200.5, we get 0.5199, which is very nearly the correct answer. [All three normal approximations are done using
software, which gives slightly more accurate values than are often possible from
printed normal tables of the standard normal distribution.] </p>
<pre><code>diff(pnorm(c(159.5,200.5), 200, 10))
[1] 0.5199132
</code></pre>
<p>The plot below shows the binomial distribution (vertical bars) and
the approximating normal distribution (blue curve). Vertical broken lines
enclose the desired probability.</p>
<p><a href="https://i.stack.imgur.com/8tBCJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8tBCJ.png" alt="enter image description here"></a></p>
<p><em>Note1:</em> (1) Notice that the binomial probability $P(X = 200)$ is
approximated by the normal probability $P(199.5 < X < 200.5),$ so if
you use 200 as the upper boundary, you are losing about half of the
probability of this one binomial value. </p>
<p>(2) Here the normal approximation (with continuity correction) is
accurate to four places because (a) $n=400$ is relatively large and (b) $p = 1/2,$ for
a symmetrical distribution. If $n$ is smaller or $p$ is far from $1/2,$
it is typical for the normal approximation to give only about two
places of accuracy. For example, if $Y \sim \mathsf{Binom}(36, 1/3),$
then $P(5 \le Y \le 9) = 0.1875,$ but the normal approximation with
continuity correction is $0.1844$ (or something like $0.1854,$ rounding
to use printed normal tables).</p>
<pre><code>pbinom(9, 36, 1/3) - pbinom(4, 36, 1/3)
[1] 0.1874903
diff(pnorm(c(4.5, 9.5), 12, sqrt(8)))
[1] 0.1843746
</code></pre>
|
189,014 |
<p>Ok, I know the simple answer is to set some form of Hold attribute to the function but bear with me for a bit while I explain my motivation and why that is not quite what I want.</p>
<hr>
<p>I have a collection of data that is naturally grouped together and some functions that operate on that data. To me, the obvious way to represent this is a struct-like data-type and an <a href="https://reference.wolfram.com/language/ref/Association.html" rel="noreferrer">Association</a> seemed perfect for the job. There are a couple of questions and answers describing this approach on this site. </p>
<p>Great, now I have an association as follows: (Just an example)</p>
<pre><code><|
"Atom Names" -> {"N","C","O","C","H","H"},
"Atom Nr" -> {56,23,117,81,211,5},
"Resname" -> {ALA,ALA,TYR,LEU,GLY,GLY},
"Bias Value" -> {1,5,1,5,1,1}
"getRandomAtom" :> RandomChoice[{56,23,117,81,211,5}],
"atomExists" :> (MemberQ[{"N","C","O","C","H","H"},#]&)
|>
</code></pre>
<p>I have some other functions that operate on this collection of data. </p>
<pre><code>f1[a_Association, args__] := a["getRandomAtom"] + Total[a["Atom Nr"]]
f2[a_Association, atom_String] := If[a[atomExists][atom], a["Bias Value"] + 1]
(*etc*)
</code></pre>
<p>Ideally these function would not just accept <em>any</em> <code>Association</code> but only the 'correct' type of <code>Association</code>. I can ensure this in a couple of ways. Eg. create a helper function and use <code>Condition</code> or <code>PatternTest</code> to check if the <code>Association</code> has all the correct keys or much more simply, I can just wrap the entire <code>Association</code> with a inert head.</p>
<pre><code>Protect[atomData]
atomData[<|
"Atom Names" -> {"N","C","O","C","H","H"},
"Atom Nr" -> {56,23,117,81,211,5},
"Resname" -> {ALA,ALA,TYR,LEU,GLY,GLY},
"Bias Value" -> {1,5,1,5,1,1}
"getRandomAtom" :> RandomChoice[{56,23,117,81,211,5}],
"atomExists" :> (MemberQ[{"N","C","O","C","H","H"},#]&)
|>]
</code></pre>
<p>And in my functions, I can just check the heads. </p>
<pre><code>f3[a_atomData,args__] := "Do Stuff"
</code></pre>
<p>But I'd like to retain the functionality of <code>Association</code> transparently. We can somewhat achieve this through the use of upvalues.</p>
<pre><code>atomData[a_Association][key_] := a[key];
atomData /: h_[atomData[a_Association],args___] := h[a,args]
(*And some additional stuff*)
atomData /: ToString[atomData[a_Association]] := a["Atom Names"]
atomData[a_Association][] := a
</code></pre>
<p>As <a href="https://mathematica.stackexchange.com/questions/109657/">this</a> question notes, this will not work for all functions (eg. Lookup) as they have the <code>HoldAllComplete</code> attribute. (Coincidentally, my motivation is almost exactly the same as the OP of that question)</p>
<hr>
<p>Here comes the problem. Since I defined these upvalues, my functions which check for the <code>Head</code> <code>atomData</code> won't work anymore. </p>
<pre><code>f3[myatomdata,3,4] (* myatomdata has head atomData *)
</code></pre>
<p>The upvalues associated with atomData will be applied first and this will result in <code>f3[<|...(*underlying association*)...|>,3,4]</code>, which will not be evaluated as the first argument nolonger has the <code>Head</code> <code>atomData</code>.</p>
<pre><code>SetAttributes[f3,HoldFirst]
</code></pre>
<p>won't help either, as in the function call <code>f3[myatomdata,3,4]</code>, the evaluator will leave <code>myatomdata</code> alone, which means that it will have <code>Head</code> <code>Symbol</code>, and once again f3 will not be evaluated.</p>
<hr>
<p>It seems that I have defeated my entire purpose by setting these upvalues. Is there a better way to do what I want? </p>
<p>I can think of 2 ways, both of which seem quite ugly. </p>
<ol>
<li><p>Modify the upvalue definition to exclude certain functions. Something like </p>
<pre><code> atomData /:
(h : Except[f1|f2|f3])[atomData[a_Association], args___] := h[a, args]
</code></pre></li>
</ol>
<p>This seems particularly inelegant, as I'd have to modify this every time I add another function that will use <code>atomData</code>.</p>
<ol start="2">
<li><p>Do the head checking yourself. </p>
<pre><code> SetAttributes[f3,HoldFirst]
atomData /: (h : Except[Head])[atomData[a_Association], args___] := h[a, args]
f3[a_ /; Head[a]===atomData] := "Do Stuff"
</code></pre></li>
</ol>
<p>The ideal solution would be a way to prevent <em>just</em> the upvalue from evaluating and leaving all others (OwnValues,DownValues etc.) alone.</p>
<hr>
<p>PS. I'm also open to the idea that this whole approach is rubbish if someone can suggest a better way. I come from a background of C++, Java, and Python; Thinking of everything in terms of objects has been ingrained in me. Apologies for the long-winded explanation. </p>
|
Jason B.
| 9,490 |
<p>You should definitely go the route of using an inert wrapper for the association, <code>atomData[Association[...]]</code>. </p>
<p>You can make a general subvalue like what you have,</p>
<pre><code>atomData[a_Association][key_] := a[key];
</code></pre>
<p>and you can make more specific definitions like</p>
<pre><code>atomData[a_Association]["getRandomAtom"] :=
RandomChoice[a["Atom Names"]
</code></pre>
<p>and </p>
<pre><code>atomData[a_Association]["atomExists", s_] := MemberQ[a["Atom Names"], s]
</code></pre>
<p>so that you don't need to waste space in your <code>atomData</code> for computed properties. This definition however,</p>
<pre><code>atomData /: h_[atomData[a_Association],args___] := h[a,args]
</code></pre>
<p>is, as you say, <s>pure rubbish</s> suboptimal. This upvalue is so general that it renders your object useless for any other purpose. Just try <code>{atomData[<|a->b|>],2}</code>.</p>
<p>I use <code>UpValues</code> so seldom, that I tend to see it as code smell, an indication you should rethink things (except <a href="https://reference.wolfram.com/language/ref/Nothing.html" rel="nofollow noreferrer"><code>Nothing</code></a>, I use <code>Nothing</code> in a lot of cases). In the above, the definition should be on <code>h</code> and not on <code>atomData</code>.</p>
<p>In your workarounds, you try to implement a list of special heads that shouldn't trigger an upvalue on <code>atomData</code>. Instead, it should be in the definition of the other functions to access the data in <code>atomData</code>:</p>
<pre><code>f1[a:atomData[assoc_Association], args_] := f[ assoc, args]
f1[a_Association, args_] := ....
</code></pre>
<p>Write a definition like the above for every function you want to just work with the raw data directly. But the general behavior is for <code>atomData</code> head to <strong>not disappear</strong> whenever it's in some other expression.</p>
<p>A couple of other points:</p>
<p>You should use a more specific pattern than <code>_Association</code>. Even <code>_?AssociationQ</code> is better, since it rules out things like <code>Association[1]</code>. But in your case you could define a pattern:</p>
<pre><code>$atomDataPattern = KeyValuePattern[
{"Atom Names" -> {__String}, "Atom Nr" -> {__Integer} }
]
</code></pre>
<p>and so then you can use this in function definitions:</p>
<pre><code>f[atomData[ad:$atomDataPattern], args] := f[ad,args]
</code></pre>
<p>You can then make a nice formatting for <code>atomData</code> using the functions described in <a href="https://mathematica.stackexchange.com/a/79891/9490">this post</a>.</p>
<p>If you want to program in Mathematica but keep to OOP methods you are used to, you might benefit from the package described <a href="https://mathematica.stackexchange.com/a/165486/9490">in this post</a>.</p>
|
3,765,555 |
<p>Let <span class="math-container">$\triangle ABC$</span> be an isosceles triangle with base <span class="math-container">$a$</span> and altitude to the base <span class="math-container">$b.$</span> I am trying to find the sides of the rectangle inscribed in <span class="math-container">$\triangle ABC$</span> if its diagonals are parallel to the triangle legs.</p>
<p>Does an inscribed rectangle exist in every isosceles triangle? How are we to construct that rectangle?</p>
<p>Thank you in advance!</p>
|
Narasimham
| 95,860 |
<p>Yes, any number of triangles can be constructed that way. Draw a line parallel to one leg. See where it cuts the altitude. Reflect this parallel parallel line about altitude to be parallel to the other leg. Draw the inscribed rectangle as shown including cutting points on both the slant sides of the isosceles triangle.</p>
<p>The slant parallel lines <em>cannot be called diagonals</em> in general. They help to locate intersection point.. of concurrency ( slant leg, breadth and height of rectangle. )</p>
<p><a href="https://i.stack.imgur.com/JD3IE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JD3IE.png" alt="enter image description here" /></a></p>
<p>The cutting point <em>can be even outside</em> the given isosceles triangle <span class="math-container">$ABC$</span> ( base <span class="math-container">$a$</span>, height <span class="math-container">$b$</span>). That is, <span class="math-container">$h>b$</span> possible as well as <span class="math-container">$h<0$</span> is possible.</p>
<p>Equation of slant leg</p>
<p><span class="math-container">$$ \frac{2x}{a}+\frac{y}{b}=1 $$</span></p>
<p>If you plug in <span class="math-container">$ y=h$</span> then the <span class="math-container">$x-$</span> coordinate of the corner of rectangle is:</p>
<p><span class="math-container">$$ x1=(1-\frac{h}{b}) \frac{a}{2}$$</span></p>
<p>When <span class="math-container">$ h>b, x1 $</span> goes negative, to the left of base center.</p>
|
4,594 |
<p>I would like to open an Excel file and manipulate it as a COM object. While I'm able to open an instance of excel with</p>
<pre><code>Needs["NETLink`"]
InstallNET[]
excel = CreateCOMObject["Excel.Application"]
</code></pre>
<p>This doesn't work for me:</p>
<pre><code> wb = excel@Workbooks@Open["D:\\prices.csv"]
</code></pre>
<p>Producing these errors:</p>
<pre>NET::netexcptn: A .NET exception occurred: System.Runtime.InteropServices.COMException (0x80028018): Old format or invalid type library. (Exception from HRESULT: 0x80028018 (TYPE_E_INVDATAREAD))
at Microsoft.Office.Interop.Excel.Workbooks.Open(String Filename, Object UpdateLinks, Object ReadOnly, Object Format, Object Password, Object WriteResPassword, Object IgnoreReadOnlyRecommended, Object Origin, Object Delimiter, Object Editable, Object Notify, Object Converter, Object AddToMru, Object Local, Object CorruptLoad).</pre>
<p>Is this a known problem? I would very much appreciate any ideas on how to open an excel file with Mathematica as a COM object.</p>
|
Szabolcs
| 12 |
<p>Per request, I'm posting this as an answer:</p>
<p>The same problem is mentioned in the following support article:</p>
<ul>
<li><a href="http://support.microsoft.com/kb/320369">http://support.microsoft.com/kb/320369</a></li>
</ul>
<p>The problem appears if the language of Excel differs from <a href="http://windows.microsoft.com/en-US/windows-vista/Change-the-system-locale">the locale setting of the operating system</a>. One workaround is to set the system locale to match with the language of Excel (probably US English for most users).</p>
<p>Another is executing the code described in the support article. I was unable to translate that code from VB/.NET to .NET/Link. If anyone can do the translation, please post a new answer!</p>
|
3,545,250 |
<p>Being new to calculus, I'm trying to understand Part 1 of the Fundamental Theorem of Calculus. </p>
<p>Ordinarily, this first part is stated using an " area function" <em>F</em> mapping every <em>x</em> in the domain of <em>f</em> to the number " integral from a to x of f(t)dt". </p>
<p>However, I encounter <strong><em>difficulties to understand what is the status of this area function, being apparently neither an indefinite integral , nor a definite integral</em></strong>( for, I think, a definite integral is a number, not a function); if this " area function" is not an " integral " ( of some sort), I do not understand in which way asserting that <em>F'=f</em> amounts to saying " integration and differentiation are inverse processes" as it is said informally. </p>
<p>Hence my question : is there an easier to understand version of FTC Part 1 that does not make use of the area function concept? </p>
<p>Note : I think I understand in which way the area function is a function and what it " does". What I do not understand is the role it plays in proving that " integration and differentiation a reverse processes" ( being given this function is neither a definite integral, nor an indefinite integral, as MSE answers I got previously tend to show). </p>
|
José Carlos Santos
| 446,262 |
<p>Yes, <span class="math-container">$\int_a^bf(t)\,\mathrm dt$</span> is a number. But if you change <span class="math-container">$a$</span> or <span class="math-container">$b$</span> (or both), you usually get a different number. So, <span class="math-container">$(a,b)\mapsto\int_a^bf(t)\,\mathrm dt$</span> is a <em>function</em> of <span class="math-container">$a$</span> and <span class="math-container">$b$</span> (and <span class="math-container">$f$</span>). And, in particular, for <span class="math-container">$a$</span> (and <span class="math-container">$f$</span>) fixed, <span class="math-container">$x\mapsto\int_a^xf(t)\,\mathrm dt$</span> is a function. And the Fundamental Theorem of Calculus states that, if <span class="math-container">$f$</span> is continuous, then <span class="math-container">$F$</span> is differentiable and <span class="math-container">$F'=f$</span>.</p>
|
3,820,465 |
<p>I'm working on the following problem but I'm having a hard time figuring out how to do it:</p>
<p>Q: Let A and B be two arbitrary events in a sample space S. Prove or provide a counterexample:</p>
<p>If <span class="math-container">$P(A^c) = P(B) - P(A \cap B)$</span> then <span class="math-container">$P(B) = 1$</span></p>
<p>Drawing Venn diagrams I can see how this is true, as <span class="math-container">$A \subset B$</span>, but I'm not sure how to formally prove this. Any help would be great!</p>
|
Graham Kemp
| 135,106 |
<blockquote>
<p>Drawing Venn diagrams I can see how this is true, as <span class="math-container">$A \subset B$</span>, but I'm not sure how to formally prove this. Any help would be great!</p>
</blockquote>
<p>Nope. (Unless you have a typo.)</p>
<p><span class="math-container">$$\begin{align}\mathsf P(A^{\small\complement})&= \mathsf P(B)-\mathsf P(A\cap B) \\[1ex]&=\mathsf P(B)-\mathsf P(B)\,\mathsf P(A\mid B)\\[1ex]&=\mathsf P(B)\,\big(1-\mathsf P(A\mid B)\big)\\[1ex]&=\mathsf P(B)\,\mathsf P(A^{\small\complement}\mid B)\\[1ex]&=\mathsf P(A^{\small\complement}\cap B)\end{align}$$</span> Therefore the statement indicates that <span class="math-container">$A^{\small\complement}\cap B^{\small\complement}$</span> is a null set (which is not <em>quite</em> that <span class="math-container">$A^{\small\complement}\subseteq B$</span> ).</p>
<p>It would only entail that <span class="math-container">$\mathsf P(B)=1~~$</span> <em>when</em> <span class="math-container">$~~0<\mathsf P(A^{\small\complement})=\mathsf P(A^{\small\complement}\mid B)$</span>, which requires … .</p>
|
1,121,205 |
<p>Can we find an bijective continuous map $f:X\to Y$ from a disconnected topological space $X$ to a connected topological space $Y$?</p>
<p>It seems counter intuitive for me, but I am not able to prove that $f(X)$ will be disconnected. I cannot think of any counterexample either. Can someone help?</p>
|
Seth
| 31,659 |
<p>Take the interval $[0,1)$ disjoint union with $[2,3]$ mapping in the obvious way to $[0,2]$.</p>
|
132,862 |
<p>Is it true that given a matrix $A_{m\times n}$, $A$ is regular / invertible if and only if $m=n$ and $A$ is a basis in $\mathbb{R}^n$?</p>
<p>Seems so to me, but I haven't seen anything in my book yet that says it directly.</p>
|
penartur
| 25,016 |
<p>You are trying to deduce something about the importance of 5th axiom only from the axioms itself, which is impossible.
The axioms by itself do not carry any significance. What is important is that the set defined by the axioms is (a) unique and (b) isomorphic to some real-world object (real-world natural numbers, as in "two apples" and "three oranges").</p>
<p>Giving away the fifth axiom means that:</p>
<p>a) Peano axioms no longer <strong>define</strong> a set of natural numbers, as there could be two non-isomorphic (and even not of the same cardinality) sets, both compatible with Peano axioms; rather they define a class of sets, each containing a subset, isomorphic to $\{0, 1, 2, 3, \ldots\}$. For example, let us consider the set $\mathbb N$ in its usual sense, and the set $\mathbb C \setminus {\mathbb N}^+$. Both sets fulfill the Peano axioms (except for the fifth one), but they are quite different in itself.</p>
<p>b) Of course, the second set from the previous paragraph has nothing in common with a real=world object called "the natural numbers".</p>
|
3,366,064 |
<p>I have a baking recipe that calls for 1/2 tsp of vanilla extract, but I only have a 1 tsp measuring spoon available, since the dishwasher is running. The measuring spoon is very nearly a perfect hemisphere. </p>
<p>My question is, to what depth (as a percentage of hemisphere radius) must I fill my teaspoon with vanilla such that it contains precisely 1/2 tsp of vanilla? Due to the shape, I obviously have to fill it more than halfway, but how much more?</p>
<p>(I nearly posted this in the Cooking forum, but I have a feeling the answer will involve more math knowledge than baking knowledge.)</p>
|
Rafi
| 307,853 |
<p>Note about eyeballing: Your eye's reference is the surface of the spoon, so when you eyeball you may actually be measuring along the arc from the bottom of the spoon to its top edge. </p>
<p>That is, your eye may be watching the red curve, not the blue line:</p>
<p><a href="https://i.stack.imgur.com/H1lzK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/H1lzK.png" alt="half full spoon"></a></p>
<p>Using the 65.27% from other answers, the depth measured along the red curve is <span class="math-container">$$ \frac{\arccos(1 - 0.6527)} {90\deg }\approx 77.42\%$$</span></p>
<p>So to the eye, the "depth" of a half-full spoon may look like more like three quarters than two thirds.</p>
|
496,255 |
<p>Let $u$ be an integer of the form $4n+3$, where $n$ is a positive integer. Can we find integers $a$ and $b$ such that $u = a^2 + b^2$? If not, how to establish this for a fact? </p>
|
triple_sec
| 87,778 |
<p><strong>Lemma 1</strong>: $a$ is odd $\Longrightarrow$ $a^2\equiv 1(\operatorname{mod} 4)$.</p>
<p><em>Proof</em>: $a^2-1=(a-1)(a+1)$. Since $a$ is odd, both $a-1$ and $a+1$ are even, so that $a^2-1$ is divisible by $4$. $\blacksquare$</p>
<p><strong>Lemma 2</strong>: $a$ is even $\Longrightarrow$ $a^2\equiv 0(\operatorname{mod} 4)$.</p>
<p><em>Proof</em>: Trivial. $\blacksquare$</p>
<p>Now, suppose that $u=a^2+b^2$.</p>
<p>(1) If both $a$ and $b$ are even, then $u$ is divisible by four by lemma 2.</p>
<p>(2) If both $a$ and $b$ are odd, then $u\equiv 2(\operatorname{mod}4)$ by lemma 1.</p>
<p>(3) If $a$ is even and $b$ is odd (wlog), then $u\equiv1(\operatorname{mod}4)$ by lemmas 1 and 2.</p>
<p>That is, it is never the case that $u\equiv 3(\operatorname{mod} 4)$.</p>
|
496,255 |
<p>Let $u$ be an integer of the form $4n+3$, where $n$ is a positive integer. Can we find integers $a$ and $b$ such that $u = a^2 + b^2$? If not, how to establish this for a fact? </p>
|
user66733
| 66,733 |
<p>I'll write another argument with more group theoretic flavor in my opinion. Suppose that $p=4k+3$ is a prime number and you can write $p=x^2+y^2$. then $x^2+y^2 \equiv 0 \pmod{p} \iff x^2 \equiv -y^2 \pmod{p} \iff (xy^{-1})^2 \equiv -1 \pmod{p}$. Therefore $t=xy^{-1}$ is a solution of $x^2 \equiv -1 \pmod{p}$.</p>
<p>Now consider the group $\mathbb{Z}^*_p$ which consists of all non-zero residues in mod $p$ under multiplication of residues. $|G|=(4k+3)-1=4k+2$. Therefore, by a group theory result (you can also use a weaker theorem in number theory called Fermat's little theorem), for any $a \in \mathbb{Z}^*_p: a^{|G|}=1$, i.e. $a^{4k+2}=1$.</p>
<p>We know that there exists $x=t$ in $\mathbb{Z}^*_p$ such that $x^2 = -1$, hence, $x^4 = 1$. But this means that $\operatorname{ord}(x) \mid |G| \implies 4 \mid 4k+2$. But $4 \mid 4k$ and therefore $4 \mid 4k+2 - 4k = 2$ which is absurd. This contradiction means that it's not possible to write $p=x^2+y^2$ for $x,y \in \mathbb{Z}$.</p>
<p>EDIT: I should also add that any integer of the form $4k+3$ will have a prime factor of the form $4k+3$. The reason is, if none of its factors are of this form, then all of its prime factors must be of the form $4k+1$. But you can easily check that $(4k+1)(4k'+1)=4k''+1$ which leads us to a contradiction. This is how you can generalize what I said to the case when $n=4k+3$ is any natural number.</p>
|
1,012,895 |
<p>I am stuck with my revision for the upcoming test.</p>
<p>The question asks"</p>
<p>An implementation of insertion sort spent 1 second to sort a list of ${10^6}$ records. How many seconds it will spend to sort ${10^7}$ records?</p>
<p>By using $\frac{T(x)}{T(1)}$ = $\frac{10^7}{10^6}$ I thought the answer was $10$ seconds but the actual answer says it's $100$ seconds.</p>
<p>Can someone please help me out? :(</p>
|
Irvan
| 172,851 |
<p><strong>Hint</strong>: what is the <a href="http://en.wikipedia.org/wiki/Time_complexity" rel="nofollow">Time Complexity</a> of insertion sort?</p>
|
1,634,725 |
<p>It's bee a long time since I've worked with sums and series, so even simple examples like this one are giving me trouble:</p>
<p>$\sum_{i=4}^N \left(5\right)^i$</p>
<p>Can I get some guidance on series like this? I'm finding different methods online but not sure which to use. I know that starting at a non-zero number also changes things.</p>
<p>My original thought was to do (sum from 0 to N of 5^i) - (sum from 0 to 3 of 5^i) but I'm not sure that's right.</p>
|
stackoverflowuser2010
| 9,177 |
<p>The problem asks for a closed-form solution to:</p>
<p><span class="math-container">$$\sum_{i=4}^{N} 5^i = 5^4 + 5^5 + ... + 5^N$$</span></p>
<p>The OP's original intuition was correct:
<span class="math-container">$$\sum_{i=4}^{N} = \sum_{i=0}^{N} 5^i - \sum_{i=0}^{3} 5^i$$</span></p>
<p>More generally, for summing a geometric series starting at an arbitrary index <span class="math-container">$m$</span>:
<span class="math-container">$$
\sum_{i=m}^{N} r^i = \sum_{i=0}^{N} r^i - \sum_{i=0}^{m-1} r^i \\
$$</span></p>
<p>To get a closed form for the above expression, let's start with the closed-form equation for a geometric series:
<span class="math-container">$$
\sum_{i=0}^{N} r^i = \frac{r^{N+1}-1}{r-1}
$$</span></p>
<p>So:</p>
<p><span class="math-container">$$
\begin{align*}
\sum_{i=m}^{N} r^i &= \sum_{i=0}^{N} r^i - \sum_{i=0}^{m-1} r^i \\
&= (\frac{r^{N+1}-1}{r-1}) - (\frac{r^{m-1+1}-1}{r-1}) \\
&= \frac{r^{N+1} - r^m}{r-1}
\end{align*}
$$</span></p>
<p>An alternative and equivalent form can be found if we multiply the top and bottom by <span class="math-container">$-1$</span>:
<span class="math-container">$$
\sum_{i=m}^{N} r^i = \frac{r^m - r^{N+1}}{1-r}
$$</span></p>
|
3,892,246 |
<p>I am dealing with sets of vectors <span class="math-container">$\big\{x_1, x_2, x_3, \dotsc, x_m \big\}$</span> from some abstract vector space <span class="math-container">$\mathcal{V}$</span>. <strong>Occasionally</strong>, <span class="math-container">$\mathcal{V}=\mathbb{R}^n$</span> an I need to address the elements these vectors e.g. sum over all elements of the vector <span class="math-container">$x_j$</span>. However, the subindex is already used to denote a specific vector.</p>
<p>Is there a common notation address the elements of a vector? E.g. <span class="math-container">$x_j[i]$</span> or <span class="math-container">$x_j^i$</span> or <span class="math-container">$x_j^{(i)}$</span>?</p>
|
Wuestenfux
| 417,848 |
<p>Well, there is no common notation here. The policy is to keep the notation as simple as possible.</p>
<p>In your case, I would write <span class="math-container">$x_{ij}$</span> for the <span class="math-container">$j$</span>th component of vector <span class="math-container">$x_i$</span>.</p>
|
4,486,594 |
<p>Let <span class="math-container">$X$</span> be the Riemann surface of <span class="math-container">$w^{2} \ =\text{sin} \ z$</span> in <span class="math-container">$ \mathbb{C}^{2}$</span>, i.e. let <span class="math-container">$X = \{(z,w): w^2 = \text{sin} \ z\}$</span>.</p>
<p>The Riemann surface structure on <span class="math-container">$X$</span> is obtained by paramertizing by <span class="math-container">$z$</span> at all places where <span class="math-container">$w \ne 0$</span> and parametrizing by <span class="math-container">$w$</span> at places where <span class="math-container">$w=0$</span>.</p>
<p>The question is to show is not the interior of a compact surface-with-boundary.</p>
<hr />
<p>My attempt:</p>
<p>Consider the holomorphic function <span class="math-container">$\displaystyle F:X\rightarrow \mathbb{C}$</span> given by <span class="math-container">$\displaystyle F:( z,w)\rightarrow w$</span>.
Now assume there is a compact surface-with-boundary <span class="math-container">$\displaystyle \tilde{X}$</span> such that interior of <span class="math-container">$\displaystyle \tilde{X}$</span> is <span class="math-container">$\displaystyle X$</span>.
It is natural to expect that <span class="math-container">$\displaystyle F$</span> extends to an holomorphic function <span class="math-container">$\displaystyle \tilde{F} :\tilde{X} \ \rightarrow \mathbb{C} P^{1}$</span>.</p>
<p>Now observe <span class="math-container">$\displaystyle \tilde{F}^{-1}( 0)$</span> has a limit point as it contains an infinite sequence <span class="math-container">$\displaystyle \{( n\pi ,0)\}_{n=1}^{\infty }$</span> in <span class="math-container">$\displaystyle \tilde{X}$</span>. Holomorphicity of <span class="math-container">$\displaystyle \tilde{F}$</span> forces <span class="math-container">$\displaystyle \tilde{F} $</span> to be a constant.
<br />
This is a contradiction as <span class="math-container">$\displaystyle F$</span> is not constant.</p>
<p>The issue with this proof is that there is no reason to believe that <span class="math-container">$\displaystyle F$</span> should have an extension <span class="math-container">$\displaystyle \tilde{F}$</span>.</p>
|
onriv
| 195,865 |
<p>I am trying to use some other way to prove that the Riemann surface <span class="math-container">$X=\{(z,w):w^2=\sin z\}$</span> has infinite genus, as shown in the answer of Kohan's. The main result I wanna to show is that:</p>
<blockquote>
<p>For any positive integer <span class="math-container">$n$</span>, there is closed, compactly supported, smooth 1-forms <span class="math-container">$\theta_1, \theta_2, \cdots, \theta_n$</span> (not required to be holomorphic) that is linear independent over <span class="math-container">$\Bbb{R}$</span>, i.e. if <span class="math-container">$a_1 \theta_1 + \cdots + a_n \theta_n = \mathrm{d}u$</span> for some smooth function <span class="math-container">$u$</span>, then <span class="math-container">$a_i=0$</span> for all <span class="math-container">$i$</span>.</p>
</blockquote>
<p>And the main proposition to use is that(It's Proposition 15 in Chapter 5 in Donaldson's <em>Riemann Surfaces</em>, Section II.3.3 in Farkas&Kra's <em>Riemann Surfaces</em>, and Formula 6.71 in Page 289 in Kodaira's <em>Complex analysis</em>):</p>
<blockquote>
<p><strong>Proposition 15</strong> For any loop <span class="math-container">$\gamma$</span> in <span class="math-container">$X$</span>, there is a compactly supported 1-form <span class="math-container">$\theta$</span> such that for any closed smooth compactly supported 1-form <span class="math-container">$\phi$</span>,</p>
<p><span class="math-container">$$\int_\gamma \phi = \int_X \theta \wedge\phi
$$</span></p>
</blockquote>
<p>All references explain this result in details. Mainly it's using partition of unity to construct a smooth function <span class="math-container">$\rho_\gamma^+$</span> on <span class="math-container">$X-\gamma$</span> and taking <span class="math-container">$\mathrm{d} \rho_\gamma^+$</span> as <span class="math-container">$\theta$</span>.</p>
<p>Furthermore, for this 1-form <span class="math-container">$\theta$</span> associated to the curve <span class="math-container">$\gamma$</span>, if <span class="math-container">$\delta$</span> is some other loop does not intersect the support of <span class="math-container">$\theta$</span> (the support of <span class="math-container">$\theta$</span> is near <span class="math-container">$\gamma$</span>, it can be shown from the construction of <span class="math-container">$\theta$</span>), then <span class="math-container">$\int_\delta\theta = 0$</span>. And if <span class="math-container">$\delta$</span> intersects <span class="math-container">$\gamma$</span> once, then <span class="math-container">$\int_\delta\theta = 1$</span> (from one direction, from the other direction it's <span class="math-container">$-1$</span>).</p>
<p>With this proposition, using the loops as follow: (sorry for the poor drawing, the black loops are in the <span class="math-container">$u-x$</span> plane, but the red loops are in the <span class="math-container">$v-y$</span> plane, where <span class="math-container">$w=u+iv$</span> and <span class="math-container">$z=x+iy$</span>. For the loop <span class="math-container">$\gamma_0$</span>, imagine that <span class="math-container">$z$</span> goes from <span class="math-container">$0$</span> to <span class="math-container">$\pi$</span>, then <span class="math-container">$w$</span> goes from <span class="math-container">$0$</span> to <span class="math-container">$1$</span> and back to <span class="math-container">$0$</span>, and the opposite direction is similar. I attached the picture that projecting <span class="math-container">$z=\mathrm{arcsin}(w^2)$</span> to the <span class="math-container">$u$</span> coordinate to help the illustration.)
<a href="https://i.stack.imgur.com/narJt.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/narJt.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/3N2QJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3N2QJ.jpg" alt="enter image description here" /></a></p>
<p>Then for each <span class="math-container">$\gamma_n$</span>, there is an associated closed smooth compactly supported 1-form <span class="math-container">$\theta_n$</span>. The set <span class="math-container">$\{\theta_n\}$</span> is linear independent over <span class="math-container">$\Bbb{R}$</span>:</p>
<p>Suppose <span class="math-container">$a_1 \theta_1 + \cdots a_n \theta_n = 0$</span>, applying the loop <span class="math-container">$\delta_0$</span> to it (integrating it on <span class="math-container">$\delta_0$</span>), we get <span class="math-container">$a_1 = 0$</span> since <span class="math-container">$\delta_0$</span> only intersects <span class="math-container">$\gamma_1$</span>. Similarlly <span class="math-container">$a_i = 0$</span> for all other <span class="math-container">$i$</span>.</p>
<p>Hence <span class="math-container">$X$</span> cannot be the interior of a compact surface-with-boundary.(It's not required to be a Riemman surface. The argument doesn't use any condition of being holomorphic)</p>
|
4,321,604 |
<p>I have come across an expression like this,</p>
<p><span class="math-container">$$ \frac{f(x) + f(a)}{2\sqrt{f(x)f(a)}}\,\delta(x-a), $$</span></p>
<p>where I expected to find just <span class="math-container">$\delta(x-a)$</span>. When I thought about it, though, I realised maybe... they are identical? Because both yield <span class="math-container">$1$</span> when integrating over a domain that contains <span class="math-container">$a$</span> and <span class="math-container">$0$</span> otherwise, so both distributions behave identically.</p>
<p>So, can I say
<span class="math-container">$$ \frac{f(x) + f(a)}{2\sqrt{f(x)f(a)}}\,\delta(x-a) = \delta(x-a) $$</span>
or can I not?</p>
|
Mostafa Ayaz
| 518,023 |
<p>The famous property of sampling of the delta function yields
<span class="math-container">$$
f(x)\delta(x-a)=f(a)\delta(x-a).
$$</span>
However, subtlety lies in this property. It should be mentioned that the validity of the above equation is the domain of <span class="math-container">$f(x)$</span>, which in this case, is the set <span class="math-container">$\{x:f(x)f(a)>0\}$</span>. Other than this, no more considerations are required.</p>
|
1,617,698 |
<p>While I was trying to find the formula of something by my own means I came across this sum which I need to solve, however I don't know if there is a solution for it, maybe it doesn't mean anything and I made a mistake. However if there's an equation which can replace this sum I will appreciate it a lot if you show me which one and how did you find the answer!</p>
|
sinbadh
| 277,566 |
<p>Let $z=e^{i\theta}=\cos\theta+i\sin\theta$. Then $z^j=e^{ij\theta}=\cos j\theta+i\sin j\theta$ for all $j\in\mathbb{N}$.</p>
<p>Thus</p>
<p>$\begin{eqnarray}
1+\sin\theta+\sin2\theta+...+\sin n\theta&=&Im(z^0+z^1+z^2+...+z^n)\\
&=&Im\left(\frac{z^{m+1}-1}{z-1}\right)
\end{eqnarray}$</p>
<p>Putting $\theta=\frac{\pi}{2n}$ you get an implicit form for your sum (of course, you must continue simplifying the RHS)</p>
|
1,562,010 |
<p>Let $f(x) \in \mathbb Z[x]$ be an irreducible monic polynomial such that $|f(0)|$ is not a perfect square . Then is $f(x^2)$ also irreducible in $\mathbb Z[x]$ ?</p>
<p>( It is supposed to have an elementary solution , without using any field-extension etc. )</p>
|
Eric Wofsey
| 86,856 |
<p>Sophisticated answer: Let us suppose $f(x)\in \mathbb{Z}[x]$ is monic and irreducible of degree $n$ but $g(x)=f(x^2)$ is reducible. Let $K\supset\mathbb{Q}$ be a splitting field of $g$ and let $a_1,\dots,a_n\in K$ be the roots of $f$, so $\pm\sqrt{a_1}\dots,\pm\sqrt{a_n}$ are the roots of $g$. Since $f$ is irreducible, $G=Gal(K/\mathbb{Q})$ acts transitively on the set $\{a_1,\dots,a_n\}$. It follows that the $G$-orbit of any root of $g$ must contain at least one of the square roots of $a_i$ for each $i$. Since $g$ is reducible, $G$ does not act transitively on the roots of $g$, so there must be exactly two orbits of roots of $g$, with each one containing exactly one of the square roots of each $a_i$. This means that $g$ has two irreducible factors $h_1$ and $h_2$ of degree $n$, and the roots of $h_1$ are the negatives of the roots of $h_2$. Thus the constant term of $h_1$ is $(-1)^n$ times the constant term of $h_2$ (if we take $h_1$ and $h_2$ to be monic). We conclude that the constant term of $g$ is $(-1)^n$ times the square of the constant term of $h_1$. It follows that $|f(0)|$ is a square.</p>
<p>(Strictly speaking, there is a special case when $a_i=0$ for some $i$, in which case $g$ has a double root and the argument above doesn't work as stated. But if that happens, then $|f(0)|=0$ is trivially a square. Also, the only $f$ for which this happens is $f(x)=x$)</p>
<hr>
<p>Less sophisticated answer: Note that since $g(x)=g(-x)$, if $h(x)$ is an irreducible factor of $g(x)$, then so is $h(-x)$. So unless $h(x)=\pm h(-x)$ for some irreducible factor $h$ of $g$, the irreducible factors of $g$ come in pairs which have (up to sign) the same constant term, and so multiplying them all together we get that the constant term of $g$ is a square (up to sign).</p>
<p>So we just have to rule out $h(x)=\pm h(-x)$ for an irreducible monic factor $h$ of $g$. If $h(x)=-h(-x)$, then every term of $h$ has odd degree, and so $h$ is divisible by $x$. Thus $g(x)$ is divisible by $x$, and in particular its constant term is $0$, which is a square. If $h(x)=h(-x)$, then every term of $h$ has even degree, and we can write $h(x)=h'(x^2)$ for some $h'$. Writing $g(x)=h(x)k(x)$, we see that $k(x)$ also satisfies $k(x)=k(-x)$ (since $g$ and $h$ do), so $k(x)=k'(x^2)$ for some $k'$. But then $f(x)=h'(x)k'(x)$ is a factorization of $f$. This means that actually $k'(x)=1$, so $h'=f$ and $h=g$, so in this case $g$ is irreducible.</p>
|
313,030 |
<p>I often find myself writing a definition which requires a proof. You are defining a term and, contextually, need to prove that the definition makes sense. </p>
<p>How can you express that? What about a definition with a proof?</p>
<p>Sometime one can write the definition and then the theorem. But often happens that many definition which should stay together need to be split
because a theorem is required in between.</p>
<p>A tentative example:</p>
<p><strong>Definition</strong> (rational numbers)
Let <span class="math-container">$\sim$</span> be the equivalence relation on <span class="math-container">$\mathbb Z^*\times \mathbb Z$</span> given
by
<span class="math-container">$$
(q,p) \sim (q',p') \iff pq' = p'q.
$$</span>
We define <span class="math-container">$\mathbb Q= (\mathbb Z^*\times \mathbb Z)/\sim$</span>.
On <span class="math-container">$\mathbb Q$</span> we define addition and multiplication as follows
<span class="math-container">$$
[(q,p)] + [(q',p')] = [(qq',pq'+p'q)] \\
[(q,p)] \cdot[(q',p')] = [(qq',pp')]
$$</span>
With these operations and choosing
<span class="math-container">$0_\mathbb Q=[(1,0)]$</span>, and <span class="math-container">$1_\mathbb Q=[(1,1)]$</span>
turns out that <span class="math-container">$\mathbb Q$</span> is a field.</p>
<p><strong>Proof.</strong>
We are going to prove that <span class="math-container">$\sim$</span> is indeed an equivalence relation,
that addition and multiplication are well defined and that the resulting
set is a field. [...]</p>
|
usul
| 29,697 |
<p>One approach always available is to decompose your problem into a series of definitions and theorems each of which is formally correct and relies only on the previous ones. It may require defining and naming sub-objects. For example: (1) Definition of ~. (2) Theorem: ~ is an equivalence relation. Proof. (3) Definition of the set Q. (4) Definition of the + operation. (5) Definition of the x operation. (6) Theorem: (Q,+,x) is a field. Proof.[0]</p>
<p>This decomposition must be possible, otherwise what you are trying to do is not formally correct. But I admit, it feels inelegant. We are taking up more space and time than we "need". We are introducing apparently "global" definitions ~,+,x but we know these are really "local" to the definition of the field Q.</p>
<p>So I think one just has to decide stylistically whether it sacrifices clarity to compress this into a single more concise definition. This is harder if you are introducing a new definition nobody has seen before. In this case you can consider wrapping the whole construction into a subsection of the paper with a summary/overview. ("The goal of this part is to define Q, but to do so we need some intermediate objects...").</p>
<hr>
<p>[0] A similar common example is to define a function <span class="math-container">$f: A \to B$</span> in some way and prove it is well-defined. Formally this can be broken into two steps, e.g. (1) define the <em>relation</em> <span class="math-container">$f$</span>, (2) Theorem: <span class="math-container">$f$</span> is a function. In both cases you want to define a _____, but really you are first defining an object then proving it is a _____.</p>
|
1,751,196 |
<p>I have an interesting question that I certainly don't know how to solve it. I've already read many topics on probability, eg: <a href="https://math.stackexchange.com/questions/1041325/probability-that-someone-will-pick-a-red-ball-first">Probability that someone will pick a red ball first?</a> and <a href="https://math.stackexchange.com/questions/1043245/comparing-probabilities-of-drawing-balls-of-certain-color-with-and-without-repl?rq=1">Comparing probabilities of drawing balls of certain color, with and without replacement</a> etc. But unfortunately, I can't apply the same methodology in this case and get the right answer from the given ones (it seems I'm really silly one). So here is the question:</p>
<p>There are 5 balls in a bucket: green, blue, red, orange and black. Each turn you take a random ball from the bucket. What is a probability that at 2nd turn you will pick blue ball? The answers:</p>
<ul>
<li>1/2</li>
<li>2/3</li>
<li>1/3</li>
<li>2/5</li>
<li>1/5</li>
</ul>
<p>The first way I thought is to add probability of each turn like this: $\frac{1}{5} + \frac{1}{4}$ - 1/4 because at 2nd turn we have only four balls. However, the answer become $\frac{9}{20}$ which is not correct.</p>
<p>I know there is something to do with either factorial or combination (just my assumption).</p>
|
barak manos
| 131,263 |
<p>If you choose with replacement:</p>
<p>$$\frac15$$</p>
<hr>
<p>If you choose without replacement:</p>
<p>$$\left(1-\frac15\right)\cdot\frac14$$</p>
|
1,751,196 |
<p>I have an interesting question that I certainly don't know how to solve it. I've already read many topics on probability, eg: <a href="https://math.stackexchange.com/questions/1041325/probability-that-someone-will-pick-a-red-ball-first">Probability that someone will pick a red ball first?</a> and <a href="https://math.stackexchange.com/questions/1043245/comparing-probabilities-of-drawing-balls-of-certain-color-with-and-without-repl?rq=1">Comparing probabilities of drawing balls of certain color, with and without replacement</a> etc. But unfortunately, I can't apply the same methodology in this case and get the right answer from the given ones (it seems I'm really silly one). So here is the question:</p>
<p>There are 5 balls in a bucket: green, blue, red, orange and black. Each turn you take a random ball from the bucket. What is a probability that at 2nd turn you will pick blue ball? The answers:</p>
<ul>
<li>1/2</li>
<li>2/3</li>
<li>1/3</li>
<li>2/5</li>
<li>1/5</li>
</ul>
<p>The first way I thought is to add probability of each turn like this: $\frac{1}{5} + \frac{1}{4}$ - 1/4 because at 2nd turn we have only four balls. However, the answer become $\frac{9}{20}$ which is not correct.</p>
<p>I know there is something to do with either factorial or combination (just my assumption).</p>
|
Fabich
| 320,938 |
<p>you need the blue ball to be in the bucket after first pick : probability $\frac{4}{5}$</p>
<p>you need to pick the blue ball among the 4 remaining balls : probability $\frac{1}{4}$</p>
<p>Global probability : $$\frac{1}{4}\times \frac{4}{5} = \frac{1}{5}$$</p>
<hr>
<p>You can also see it this way : The 5 balls are exactly the same so the probability to choose any ball at 2nd turn is $\frac{1}{5}$ for each ball.</p>
|
2,781,153 |
<p>I've a right triangle that is inscribed in a circle with radius $r$ the hypotunese of the triangle is equal to the diameter of the circle and the two other sides of the triangle are equal to eachother.</p>
<blockquote>
<p>Prove that when you divide the area of the circle by the area of the triangle that you will get $\pi$.</p>
</blockquote>
<p>This is what I did:</p>
<p>The area of a triangle is $\frac{height\times width}{2}$ and the area of a circle is $\pi r^2$. Now I do not know how to continue.</p>
|
The Integrator
| 538,397 |
<p>Consider the triangle $\triangle ABC$ with angles $45^\circ-90^\circ -45^\circ$
and hypotenuse $BC = 2r$</p>
<p>$AB = BC\cdot \cos(45^\circ) = \sqrt2r$</p>
<p>$AC = BC\cdot \sin(45^\circ) = \sqrt2r$</p>
<p>Area of triangle = $\frac12\cdot AB\cdot AC = r^2$</p>
<p>ratio = $\frac{\pi r^2}{r^2} = \pi$</p>
|
227,833 |
<p>In the documentation article for <code>Polygon</code> in Mathematica 12, there is an example with the input:</p>
<pre><code>pol = Polygon[{{1, 0}, {0, Sqrt[3]}, {-1, 0}}]
</code></pre>
<p>In the documentation article the output is displayed as:</p>
<blockquote>
<pre><code>Polygon[{{1, 0}, {0, Sqrt[3]}, {-1, 0}}]
</code></pre>
</blockquote>
<p>But when I evaluate the code I get a different output with some information about the polygon, the number of points, the dimension, and more. It looks like this:</p>
<p><a href="https://i.stack.imgur.com/iA16Z.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iA16Z.png" alt="polygon" /></a></p>
<p>Is there a way to control what is obtained as output?</p>
|
m_goldberg
| 3,066 |
<p>The examples in the documentation article for <code>Polygon</code> should have been re-evaluated before Mathematica 12 was released, but it is evident that they weren't. This is true even for V12.1.1. If you manually evaluate the examples shown in the documentation they will show the new iconized argument form. This is a documentation bug,</p>
<p>That said, you can get the old output form like this:</p>
<pre><code>pol = Defer @ Polygon[{{1, 0}, {0, Sqrt[3]}, {-1, 0}}]
</code></pre>
<blockquote>
<pre><code>Polygon[{{1, 0}, {0, Sqrt[3]}, {-1, 0}}]
</code></pre>
</blockquote>
|
1,137,079 |
<p>I'm new to the concept of complex plane. I found this exercise:</p>
<blockquote>
<p>Let $z,z_1,z_2\in\mathbb C$ such that $z=z_1/z_2$. Show that the length of $z$ is the quotient of the length of $z_1$ and $z_2$.</p>
</blockquote>
<p>If $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$ then $|z_1|=\sqrt{x_1^2+y_1^2}$ and $|z_2|=\sqrt{x_2^2+y_2^2}$, which yields $|z_1|/|z_2|=\sqrt{\dfrac{x_1^2+y_1^2}{x_2^2+y_2^2}}$.</p>
<p>Now, $z=\dfrac{z_1}{z_2}=\dfrac{x_1+iy_1}{x_2+iy_2} $. The first issue is to try and separate the imaginary part from the real one. I did this by:
$$\dfrac{x_1+iy_1}{x_2+iy_2}=\dfrac{x_1+iy_1}{x_2+iy_2}\times\frac{x_2-iy_2}{x_2-iy_2}=\frac{x_1x_2-ix_1y_2+ix_2y_1+y_1y_2}{x_2^2+y_2^2}\\ =\frac{x_1x_2+y_1y_2}{x_2^2+y_2^2}+i\frac{x_2y_1-x_1y_2}{x_2^2+y_2^2}.$$
Hence $|z|=\sqrt{\left(\dfrac{x_1x_2+y_1y_2}{x_2^2+y_2^2}\right)^2+\left(\dfrac{x_2y_1-x_1y_2}{x_2^2+y_2^2}\right)^2}=\sqrt{\dfrac{(x_1x_2+y_1y_2)^2+(x_2y_1-x_1y_2)^2}{(x_2^2+y_2^2)^2}}$. Continuing I get
$$\sqrt{\frac{(x_1x_2)^2+(y_1y_2)^2+(x_2y_1)^2+(x_1y_2)^2}{(x_2^2+y_2^2)^2}},$$
but I can't see any future here. Is there any mistake? How can I simplify the expression for $|z|$? I appreciate your help.</p>
|
Tim Raczkowski
| 192,581 |
<p>Since $|z|^2=z\overline{z}$,</p>
<p>$$\left|{z_1\over z_2}\right|^2={z_1\over z_2}{\overline{z_1}\over\overline{z_2}}={|z_1|^2\over|z_2|^2}.$$</p>
<p>Take square root of both sides to get your result.</p>
|
2,768,187 |
<blockquote>
<p>Let $w$ and $z$ be complex numbers such that $w=\frac{1}{1-z}$, and
$|z|^2=1$. Find the real part of $w$.</p>
</blockquote>
<p>The answer is $\frac{1}{2}$ but I don't know how to get to it.</p>
<hr>
<p>My attempt</p>
<p>as $|z|^2=1$</p>
<p>$z\bar z = 1$</p>
<p>If $z = x+yi$</p>
<p>$z=\frac{1}{\bar z} = \frac{1}{x-yi}$</p>
<p>$\therefore w= \frac{1}{1-\frac{1}{x-yi}}=\frac{x-yi}{(x-1)-yi}$</p>
<p>$=\frac{(x-yi)((x-1)+yi)}{((x-1)-yi)((x-1)+yi)}$</p>
<p>$=\frac{x^2-x+yi+y^2}{x^2-x+1+y^2}$</p>
<p>$=\frac{x^2-x+y^2}{x^2-x+1+y^2}+\frac{y}{x^2-x+1+y^2}i$</p>
<p>Hence $Re(w)=\frac{x^2-x+y^2}{x^2-x+1+y^2}$</p>
<p>$=\frac{x^2+y^2-x}{x^2+y^2-x+1}$</p>
<p>$=\frac{1-x}{1-x+1}$</p>
<p>$=\frac{1-x}{2-x}$</p>
<p>$=\frac{-1}{2-x}+\frac{2-x}{2-x}$</p>
<p>$=1-\frac{1}{2-x}$</p>
<p>$=1+\frac{1}{x-2}$ ?????</p>
|
nonuser
| 463,553 |
<p>Pure geometric solution. </p>
<p>All points $z$ satisfying $|z|=1$ are on circle $\mathcal{C}$ with radius $r=1$ and center at $0$. </p>
<p>Now transformation $z\mapsto 1-z$ is reflection across $0$ followed by translation for $1$. So $\mathcal{C}$ ''moves'' to the right for $1$ and let this new circle be $\mathcal{C}'$. </p>
<p>Transformation $z\mapsto {1\over z}$ is inversion with center at $0$ with radius $1$ (i.e. circle $\mathcal{C}$). Since $0\in \mathcal{C}'$ and $\mathcal{C}'$ touches imaginary axis, $\mathcal{C}'$ maps to line $\ell$ which is parallel to imaginary axis. Finally, since points of intersection of $\mathcal{C}$ and $\mathcal{C}'$, which have clearly real part ${1\over 2}$, are invariant, the line $\ell$ has equation $z={1\over 2}$ and thus conclusion.</p>
|
2,921,439 |
<p>I got this summation from the book <a href="https://rads.stackoverflow.com/amzn/click/0201558025" rel="nofollow noreferrer">Concrete Mathematics</a> which I didn't exactly understand:</p>
<p>$$
\begin{align}
Sn &= \sum_{1 \leqslant k \leqslant n} \sum_{1 \leqslant j \lt k} {\frac{1}{k-j}} \\
&= \sum_{1 \leqslant k \leqslant n} \sum_{1 \leqslant k-j \lt k} {\frac{1}{j}} \\
&= \sum_{1 \leqslant k \leqslant n} \sum_{0 \lt j \leqslant k-1} {\frac{1}{j}} \\
\end{align}
$$</p>
<p>I didn't understant why $1 \leqslant j \lt k$ became $1 \leqslant k-j \lt k$ in the second line and why $1 \leqslant k-j \lt k$ became $0 \lt j \leqslant k-1$ in the third line.</p>
<p>Can you guys help me understanding that?</p>
|
user
| 505,767 |
<p>We have indeed that</p>
<p>$$\frac{x+1}{-x^3-x^2+4x+4}=\frac{x+1}{-x^2(x+1)+4(x+1)}=-\frac{1}{x^2-4}=\frac14 \frac1{x+2}-\frac14 \frac1{x-2}$$</p>
|
4,608,805 |
<p>Suppose that I have a class of 35 students whose average grade is 90. I randomly picked 5 students whose average came out to be 85. Assume their grades are i.i.d and of normal <span class="math-container">$N(\mu, \sigma^2)$</span>. From the example I have seen, <span class="math-container">$\mu$</span> is usually called the population mean and should be equal to <span class="math-container">$90$</span>. The sample me is usually referred to as <span class="math-container">$\frac{\sum{X_i}}{5}$</span>. When we do hypothesis testing we can ask whether the sample mean is equal to <span class="math-container">$90$</span>.</p>
<ol>
<li><p>Is the sample mean <span class="math-container">$\frac{\sum X_i}{5}$</span> or <span class="math-container">$85$</span>?</p>
</li>
<li><p>The population mean mathematically should be <span class="math-container">$\mu$</span>, but I think people also say that <span class="math-container">$90$</span> is the population mean. This does not make sense to mean since it is not obvious to me that why <span class="math-container">$90 = \mu$</span>. <span class="math-container">$90$</span> is calculated through the sum of the grades divided by 35, whereas <span class="math-container">$\mu$</span> is equal to some integral. I just do not see how they can be equal to each other.</p>
</li>
</ol>
|
JonathanZ supports MonicaC
| 275,313 |
<p>You have a valid point here. In fact, the population has 35 members, so its distribution <em>must</em> be discrete, while a normal distribution is continuous. The sentence starting "Assume ...." is telling you that we are going to approximate that discrete distribution with a continuous, normal one.</p>
<p>Your other (implicit) point, "Look, there are just 35 values, and apparently we already know their average (though we're not told how we know that!). Surely we know all the scores, then. So why isn't the whole class our sample?" also has a point. This one is mostly explained by the fact that this is an artificial problem, set up so that you can solve it with your current tool set, which I'm betting you're just starting to build.</p>
<p>A real problem might have a much larger population, or individual data points might be a lot harder or more expensive to gather than just asking a student their test score. As for somehow magically knowing the population mean, that's also unlikely in reality.</p>
<p>To answer your two numbered questions:</p>
<ul>
<li><p>Yes, the sample mean is <span class="math-container">$\Sigma_{i=1}^5 x_i /5$</span>, which you are told equals 85.</p>
</li>
<li><p>You are told to model/approximate the (discrete) distribution of the 35 grades with the a (continuous) <span class="math-container">$N(\mu, \sigma)$</span> distribution. Honestly, you've confused me a bit here. In hypothesis testing, the population/model mean is usually unknown, and its exact value is never determined. Maybe if you're just learning how to do the calculations required for hypothesis testing, you might be provided with a value of <span class="math-container">$\mu$</span> to use, but in general you won't be setting <span class="math-container">$\mu$</span> to any value. I can't really see of what use the class's average score is. And without seeing an actual question, I'm having trouble coming up with a reason why it would have been provided to you. Maybe you've combined some parts of two or more different questions?</p>
</li>
</ul>
<hr />
<p>Some other observations: The class of 35 students is kind of immaterial here. All we care about is that we are going to model each student's score as if it's drawn from i.i.d. normal distributions. This might be a nationally administered test, with millions of scores, and it wouldn't make much of a difference.</p>
<p>The other point is that you say we can use a hypothesis test to ask whether the <strong>sample mean</strong> is 90. That's not right. We know the sample mean, and we know it's not equal to 90. A hypothesis test makes an assertion about the <strong>unknown population mean</strong> (which is the <span class="math-container">$\mu$</span> we have in our model), and we use our observed (known) sample mean to make statements about how probable (or improbable) that assertion is.</p>
|
2,261,410 |
<blockquote>
<p>The generating function for a Bessel equation is:</p>
<p>$$g(x,t) = e^{(x/2)(t-1/t))}$$</p>
<p>Using the product $g(x,t)\cdot g(x,-t)$ show that:</p>
<p>a) $$[J_0(x)]^2 + 2[J_1(x)]^2 + 2[J_2(x)]^2 + \cdots = 1$$</p>
<p>and consequently:</p>
<p>b)</p>
<p>$$|J_0(x)|\le 1, \forall x$$</p>
<p>c) $$|J_n(x)| \le \frac{1}{\sqrt{2}}; n=1,2,3,\cdots$$</p>
</blockquote>
<p>For a) I tried the product:</p>
<p>$$e^{(x/2)(t-1/t))}\cdot e^{(x/2)(-t+1/t))} = 1$$</p>
<p>I at least arrived at the right side of the equation. Since this generates the bessel functions, I should arrive at something related to $J_n$ in the left side. I know that</p>
<p>$$e^{(x/2)(t-1/t))} = \sum_{n=0\infty}^{\infty} J_n(x)t^n$$</p>
<p>$$e^{(x/2)(-t+1/t))} = \sum_{n=0\infty}^{\infty} J_n(x)(-t)^n$$</p>
<p>but it's not just a matter of multiplying coefficients from the two infinite series, right?</p>
<p>For $b$, I tried to use <a href="https://math.stackexchange.com/questions/2261372/proving-bessel-equation-j-0uv-j-0u-cdot-j-0v2-sum-s-1-inftyj-s">Proving Bessel equation $J_{0}(u+v) = J_0(u)\cdot J_0(v)+2\sum_{s=1}^{\infty}J_s(u)\cdot J_{-s}(v)$</a> but it's not as obvious</p>
<p>I have no idea how to deal with $c$, could somebody help me?</p>
|
Michael Seifert
| 248,639 |
<p>For part (a), note that you also have
$$
g(x,-t) = g(x,1/t) = \sum_{n = 0}^\infty J_n(x) \left( \frac{1}{t} \right)^n.
$$
You can then multiply this with the original series to get
$$
\left[ \sum_{n = -\infty}^\infty J_n(x) t^{-n} \right] \left[ \sum_{n = -\infty}^\infty J_n(x) t^n \right] = 1.
$$
By multiplying these two series together and matching coefficients, you should be able to prove the given identity.</p>
<p>Parts (b) and (c) follow from the identity in part (a). Note that if you have a set of quantities $a_i$ such that $a_i > 0$ for all $i$, then $a_i < \sum_j a_j$ for all $i$.</p>
|
3,948,418 |
<p>Ok so on doing a whole lot of Geometry Problems, since I am weak at Trigonometry, I am now focused on <span class="math-container">$2$</span> main questions :-</p>
<p><span class="math-container">$1)$</span> <strong>How to calculate the <span class="math-container">$\sin,\cos,\tan$</span> of any angle?</strong></p>
<p><strong>Some Information</strong> :- This site :- <a href="https://www.intmath.com/blog/mathematics/how-do-you-find-exact-values-for-the-sine-of-all-angles-6212" rel="noreferrer">https://www.intmath.com/blog/mathematics/how-do-you-find-exact-values-for-the-sine-of-all-angles-6212</a> , produces a clear understanding and a detailed approach of finding the <span class="math-container">$\sin$</span> of any angle from <span class="math-container">$1$</span> to <span class="math-container">$90^\circ$</span> , and I found it very interesting. But now the Questions arise :-</p>
<p>Can you find the <span class="math-container">$\sin$</span>, <span class="math-container">$\cos$</span> or <span class="math-container">$\tan$</span> of any fraction angles, like <span class="math-container">$39.67$</span>? <br/>
Can you find the <span class="math-container">$\sin$</span>, <span class="math-container">$\cos$</span> or <span class="math-container">$\tan$</span> of recurring fractions like <span class="math-container">$\frac{47}{9}$</span>? <br/>
Can you find the <span class="math-container">$\sin$</span>, <span class="math-container">$\cos$</span> or <span class="math-container">$\tan$</span> of irrationals, like <span class="math-container">$\sqrt{2}?$</span></p>
<p>Since I am a bit new to Trigonometry, I will be asking if there is a formula to find the <span class="math-container">$\sin$</span> of fractions, or even recurring fractions. I can use the calculator to find them obviously, but I have another Question :-</p>
<p><span class="math-container">$2)$</span> <strong>How to calculate the trigonometric ratios of every angle in fractional form?</strong></p>
<p>We all know <span class="math-container">$\sin 45^\circ = \frac{1}{\sqrt{2}}$</span> , but what will be <span class="math-container">$\sin 46^\circ$</span> in fractions? I can use a calculator to calculate the decimal of it, but it is hard to deduce the fraction out of the value, especially because the decimal will be irrational. I know how to convert recurring decimals to fractions, but this is not the case. Right now I am focused on a particular problem, which asks me to find the <span class="math-container">$\sin$</span> of a recurring fraction, in a fraction form. I am struggling to do this unless I clear up the ideas.</p>
<p><strong>Edit</strong>: My problem is to find the <span class="math-container">$\sin$</span> of <span class="math-container">$\frac{143}{3}^\circ$</span> . I do not have any specific formula to find this, and I am mainly stuck here. I need a formula which shows how this can be done.</p>
<p>Can anyone help me? Thank You.</p>
|
WindSoul
| 715,008 |
<p>For any value angle I would turn it into radians in order to use the power series of sine.</p>
<p><span class="math-container">$sin(x)=\sum_{k=0}^{\infty}{(-1)^k\frac{x^{2k+1}}{(2k+1)!}}$</span>, <span class="math-container">$x\epsilon\mathbb R$</span></p>
<p><span class="math-container">$\theta=\frac{143^o}{3}\Rightarrow x=\frac{\pi}{180}\cdot \theta$</span></p>
<p>The result will always be a rational number that is an approximation of a transcendental number since both x and sin(x) are transcendental numbers related to <span class="math-container">$\pi$</span>.</p>
|
3,948,418 |
<p>Ok so on doing a whole lot of Geometry Problems, since I am weak at Trigonometry, I am now focused on <span class="math-container">$2$</span> main questions :-</p>
<p><span class="math-container">$1)$</span> <strong>How to calculate the <span class="math-container">$\sin,\cos,\tan$</span> of any angle?</strong></p>
<p><strong>Some Information</strong> :- This site :- <a href="https://www.intmath.com/blog/mathematics/how-do-you-find-exact-values-for-the-sine-of-all-angles-6212" rel="noreferrer">https://www.intmath.com/blog/mathematics/how-do-you-find-exact-values-for-the-sine-of-all-angles-6212</a> , produces a clear understanding and a detailed approach of finding the <span class="math-container">$\sin$</span> of any angle from <span class="math-container">$1$</span> to <span class="math-container">$90^\circ$</span> , and I found it very interesting. But now the Questions arise :-</p>
<p>Can you find the <span class="math-container">$\sin$</span>, <span class="math-container">$\cos$</span> or <span class="math-container">$\tan$</span> of any fraction angles, like <span class="math-container">$39.67$</span>? <br/>
Can you find the <span class="math-container">$\sin$</span>, <span class="math-container">$\cos$</span> or <span class="math-container">$\tan$</span> of recurring fractions like <span class="math-container">$\frac{47}{9}$</span>? <br/>
Can you find the <span class="math-container">$\sin$</span>, <span class="math-container">$\cos$</span> or <span class="math-container">$\tan$</span> of irrationals, like <span class="math-container">$\sqrt{2}?$</span></p>
<p>Since I am a bit new to Trigonometry, I will be asking if there is a formula to find the <span class="math-container">$\sin$</span> of fractions, or even recurring fractions. I can use the calculator to find them obviously, but I have another Question :-</p>
<p><span class="math-container">$2)$</span> <strong>How to calculate the trigonometric ratios of every angle in fractional form?</strong></p>
<p>We all know <span class="math-container">$\sin 45^\circ = \frac{1}{\sqrt{2}}$</span> , but what will be <span class="math-container">$\sin 46^\circ$</span> in fractions? I can use a calculator to calculate the decimal of it, but it is hard to deduce the fraction out of the value, especially because the decimal will be irrational. I know how to convert recurring decimals to fractions, but this is not the case. Right now I am focused on a particular problem, which asks me to find the <span class="math-container">$\sin$</span> of a recurring fraction, in a fraction form. I am struggling to do this unless I clear up the ideas.</p>
<p><strong>Edit</strong>: My problem is to find the <span class="math-container">$\sin$</span> of <span class="math-container">$\frac{143}{3}^\circ$</span> . I do not have any specific formula to find this, and I am mainly stuck here. I need a formula which shows how this can be done.</p>
<p>Can anyone help me? Thank You.</p>
|
G Cab
| 317,234 |
<p>I do not catch exactly what is your problem, so I am proposing some considerations which might be useful,
at least to substantiate what you need.</p>
<p>a) The sine of an angle will be rational when the angle corresponds to that of an integral triangle (i.e. a Pythagorean triple).<br />
Among the angles which are rational multiples of <span class="math-container">$\pi$</span>, only <span class="math-container">$0, \pi /6 , \pi /2 (+ k \pi)$</span> provide a rational value of the sine.<br />
see <a href="https://mathpages.com/home/kmath460/kmath460.htm" rel="nofollow noreferrer">this reference</a>.<br />
And if the angle is a rational multiple of <span class="math-container">$\pi$</span> it is as well a rational multiple of <span class="math-container">$180^{\circ}$</span> and thus a rational value in degrees.</p>
<blockquote>
<p>Therefore the sines you are looking for are irrational.</p>
</blockquote>
<p>b) The methods for finding a rational approximation of a (irrational) <span class="math-container">$\sin x$</span> are various, and are the subject
of the hard work of many scholars of the past, when the trigonometric tables were much of need
and the computer was far to come.<br />
An example is the <a href="https://en.wikipedia.org/wiki/Bhaskara_I%27s_sine_approximation_formula" rel="nofollow noreferrer">Bhaskara sine approximation</a>.
<span class="math-container">$$
\sin x^ \circ \approx
{{4x\left( {180 - x} \right)} \over {40500 - x\left( {180 - x} \right)}}
$$</span>
which for one of the angles you cite as example will give
<span class="math-container">$$
\eqalign{
& x = \left( {{{143} \over 3}} \right)^{\, \circ }
= \left( {{{143} \over {540}}} \right)\pi \;rad\quad \Rightarrow \cr
& \Rightarrow \quad \sin x \approx {{227084} \over {307729}} \to err = 0.17\,\% \cr}
$$</span></p>
<p>And when the computer arrived, also there has been much work to find suitable algorithms to express the trig functions which are summarized in <a href="https://en.wikipedia.org/wiki/Trigonometric_tables" rel="nofollow noreferrer">this Wikipedia article</a> .<br />
There it is stated that the sine of rational multiples of <span class="math-container">$\pi$</span> are in fact algebraic numbers (generally of degree <span class="math-container">$2$</span> and higher).</p>
<p>A <a href="https://mae.ufl.edu/%7Euhk/IEEETrigpaper8.pdf" rel="nofollow noreferrer">recent paper</a> illustrates a rational approximation for the <span class="math-container">$\tan$</span> and for <span class="math-container">$\sin , \cos$</span> functions.</p>
<p>c) However most of the algorithms developed for being implemented on computers relies on the angle expressed in radians,
and therefore they will provide a rational approximation of the sine when the angle is expressed as a rational multiple of a radian.<br />
The problem that you pose concerns instead the sine of an angle expressed as a fraction of degrees, which is irrational in radians.<br />
Aside from the Bhaskara's formula above, with the other available methods you cannot avoid to introduce
a rational approximation for <span class="math-container">$\pi$</span>, which is going to fix the threshold on the precision that can be achieved.</p>
<p>To this end we would better fix some lower/upper couple of bounds on <span class="math-container">$\pi$</span> such as
<span class="math-container">$$
\left( {{{25} \over 8},{{22} \over 7}} \right),\left( {{{91} \over {29}},{{22} \over 7}} \right),
\cdots ,\left( {{{688} \over {219}},{{355} \over {113}}} \right), \cdots ,
\left( {{{9918} \over {3157}},{{355} \over {113}}} \right), \cdots
$$</span>
obtainable by a Stern-Brocot approximation.</p>
<p>Then if you are using the Taylor series, for instance, you have
<span class="math-container">$$
x - {1 \over 6}x^{\,3} < \sin x < x - {1 \over 6}x^{\,3} + {1 \over {120}}x^{\,5}
$$</span>
so that for the angle already considered
<span class="math-container">$$
x = \left( {{{143} \over 3}} \right)^{\, \circ } = \left( {{{143} \over {540}}} \right)\pi \;rad
$$</span>
and using for <span class="math-container">$\pi$</span> the second couple of values above, you get
<span class="math-container">$$
\left( {{{143} \over {540}}{{91} \over {29}}} \right)
- {1 \over 6}\left( {{{143} \over {540}}{{91} \over {29}}} \right)^{\,3}
< \sin x <
\left( {{{143} \over {540}}{{22} \over 7}} \right)
- {1 \over 6}\left( {{{143} \over {540}}{{22} \over 7}} \right)^{\,3}
+ {1 \over {120}}\left( {{{143} \over {540}}{{22} \over 7}} \right)^{\,5}
$$</span>
i.e.
<span class="math-container">$$
{{16943907583603} \over {23042336976000}} < \sin x < {{2140128005530465093} \over {2893944959388000000}}
$$</span>
which put into decimal is
<span class="math-container">$$
0.735338 \ldots < 0.739239 \ldots < 0.7395132 \ldots
$$</span>
corresponding to a relative error of
<span class="math-container">$$
- \,0.5\,\% \,,\; + 0.04\,\%
$$</span></p>
<p>Of course , depending on the accuracy required, it is possible to increase the precision on <span class="math-container">$\pi$</span>, increase the degree of the
series, or also to change to Padè approximants or other techniques.</p>
|
109,423 |
<p>Let $f$ be an isometry (<em>i.e</em> a diffeomorphism which preserves the Riemannian metrics) between Riemannian manifolds $(M,g)$ and $(N,h).$ </p>
<p>One can argue that $f$ also preserves the induced metrics $d_1, d_2$ on $M, N$ from $g, h$ resp. that is, $d_1(x,y)=d_2(f(x),f(y))$ for $x,y \in M.$ Then, it's easy to show that $f$ sends geodesics on $M$ to geodesics on $N,$ using the length minimizing property of geodesics and that $f$ is distance-preserving. </p>
<p>My question, </p>
<blockquote>
<p>Is it possible to derive the result without using the distance-preserving property of isometries, by merely the definition?</p>
</blockquote>
<p>What I have found so far;</p>
<p>Let $\gamma : I \to M$ be a geodesic on $M$, i.e. $\frac{D}{dt}(\frac{d\gamma}{dt})=0,$ where $\frac{D}{dt}$ is the covariant derivative and $ \frac{d\gamma}{dt}:=d\gamma(\frac{d}{dt}).$ Let $t_0 \in I,$ we have to show that $\frac{D}{dt}(\frac{d(f \circ \gamma)}{dt})=0$ at $t=t_0,$ or $\frac{D}{dt}(df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0}))=0.$</p>
<p>We also know that </p>
<p>$$ \langle \frac{d \gamma}{dt}|_{t=t_0},\frac{d \gamma}{dt}|_{t=t_0}\rangle_{\gamma(t_0)}= \langle df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0}), df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0})\rangle_{f(\gamma(t_0))}.$$</p>
<p>Since $\frac{d}{dt} \langle \frac{d \gamma}{dt}|_{t=t_0},\frac{d \gamma}{dt}|_{t=t_0}\rangle_{\gamma(t_0)}=2 \langle \frac{D}{dt}(\frac{d \gamma}{dt}|_{t=t_0}),\frac{d \gamma}{dt}|_{t=t_0}\rangle_{\gamma(t_0)}=0,$ therefore</p>
<p>$$\frac{d}{dt} \langle df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0}), df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0})\rangle_{f(\gamma(t_0))}=$$</p>
<p>$$2\langle \frac{D}{dt}(df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0})), df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0})\rangle _{f(\gamma(t_0))}=0.$$</p>
<p>How can I conclude from $\langle\frac{D}{dt}(df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0})), df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0}) \rangle _{f(\gamma(t_0))}=0$ that $\frac{D}{dt}(df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0}))=0?$</p>
|
H-H
| 528,586 |
<p>It is actually an exercise in the Lee's book, I try to do it by following the hint. First, you have to understand the naturality of Riemannian connection, then everything will be clear. I like using $\nabla_{\frac{d}{dt}}$ instead of $D_t$ here.</p>
<p>First, Define an operator $\varphi^*\tilde{\nabla}_{\frac{d}{dt}}:\mathcal{J}(\gamma)\rightarrow\mathcal{J}(\gamma)$ by $(\varphi^*\tilde{\nabla}_{\frac{d}{dt}})V=\varphi^*(\tilde{\nabla}_{\frac{d}{dt}}(\varphi_*V))$. Suppose $V$ is a vector field along $\gamma.$ Show this operator satisfies three properties in the Lemma 4.9. in Lee's Riemannian Geometry.</p>
<p>(1) Linearity over $\mathcal{R}$:$$(\varphi^*\tilde{\nabla}_{\frac{d}{dt}})(aV+bW)=a(\varphi^*\tilde{\nabla}_{\frac{d}{dt}})V+b(\varphi^*\tilde{\nabla}_{\frac{d}{dt}})W$$ (2) Product rule: $$(\varphi^*\tilde{\nabla}_{\frac{d}{dt}})(fV)=\frac{df}{dt}V+f(\varphi^*\tilde{\nabla}_{\frac{d}{dt}})V.$$
(3) If $V$ is extendible, then for any extension $\tilde{V}$ of $V$,
$$(\varphi^*\tilde{\nabla}_{\frac{d}{dt}})V=(\varphi^*\tilde{\nabla}_{\dot{\gamma}(t)})\tilde{V}.$$
Then by uniqueness, the operator we defined above is just the unique operator $\nabla_{\frac{d}{dt}}$. that is
$$\nabla_{\frac{d}{dt}}V=\varphi^*(\tilde{\nabla}_{\frac{d}{dt}}(\varphi_*V)) \ \text{or} \ \ \varphi_*(\nabla_{\frac{d}{dt}}V)=\tilde{\nabla}_{\frac{d}{dt}}(\varphi_*V).$$</p>
<p>Now If $\gamma$ is the geodesic in $M$ with initial $p$ and initial velocity $V$, i.e., $\nabla_{\frac{d}{dt}}\dot{\gamma}(t)=0$. Obviously $\varphi\circ\gamma$ is a curve in $\tilde{M}$ with initial point $\varphi(p)$ and initial velocity $\varphi_*V$, moreover, it is also the geodesic since
$$\tilde{\nabla}_{\frac{d}{dt}}\dot{\varphi}(\gamma(t))=\tilde{\nabla}_{\frac{d}{dt}}\varphi_*\dot{\gamma}(t)=\varphi_*(\nabla_{\frac{d}{dt}}\dot{\gamma}(t))=0.$$</p>
|
30,718 |
<p>As we all know, questions lacking context are strongly discouraged on this site. This includes mainly "homework questions" that look a bit like:</p>
<blockquote>
<p>Prove that <span class="math-container">$\lim_{x\to0}x^2=0$</span> using <span class="math-container">$\epsilon$</span>-<span class="math-container">$\delta$</span> definition of the limit</p>
</blockquote>
<p>and contain nothing else in the question body. It is for this reason that the close reason of "lacking context and/or other details" exists, to ensure that we are not spammed with and overwhelmed by such questions which display no effort on the part of the OP.</p>
<p>I used to think this was quite clear: if the OP showed effort, even if it might have led to little to no progress, then it has context and should be allowed and not closed. But recently, I encountered <a href="https://math.stackexchange.com/questions/3366064/how-deep-is-the-liquid-in-a-half-full-hemisphere">this question</a> about how deep the liquid in a half-full hemisphere should fill it up to. Seeing essentially no mathematical effort by the OP, I was immediately tempted to downvote and close the question as off-topic due to lack of context. This is especially since the question is something that could come up in any introductory course on calculus, just phrased differently. But at the same time, the OP did provide some sort of "context", albeit a non-mathematical one---that they were trying to measure the exact amount of vanilla extract to use in a cooking recipe. This makes the motivation clear in some sense, but the "context" provided isn't what one would usually expect, and certainly not one I would have considered prior to coming across this question.</p>
<p>So, my question is: <strong>How exactly should we define "context" in general, and in this particular case, should the question be regarded as lacking context?</strong></p>
|
quid
| 85,306 |
<p>For the specific question, the context seems completely clear. The question reads:</p>
<blockquote>
<p><strong>How deep is the liquid in a half-full hemisphere?</strong></p>
<p>I have a baking recipe that calls for 1/2 tsp of vanilla extract, but I only have a 1 tsp measuring spoon available, since the dishwasher is running. The measuring spoon is very nearly a perfect hemisphere.</p>
<p>My question is, to what depth (as a percentage of hemisphere radius) must I fill my teaspoon with vanilla such that it contains precisely 1/2 tsp of vanilla? Due to the shape, I obviously have to fill it more than halfway, but how much more?</p>
<p>(I nearly posted this in the Cooking forum, but I have a feeling the answer will involve more math knowledge than baking knowledge.)</p>
</blockquote>
<p>The way the question is presented it seems clear that is an actual practical question (as opposed to a word-problem in an educational context).</p>
<p>This context is relevant for at least the following reasons:</p>
<ul>
<li><p>The main point is the actual answer, the methods are basically irrelevant.</p>
</li>
<li><p>An approximate answer is sufficient. Indeed, an approximate numerical answer might be more helpful than an exact answer whose numerical value is not apparent.</p>
</li>
</ul>
<p>Now, we can still strive to explain the answer and give an exact answer etc. But, still the context gives a clear indication what is expected from an answer: a ready to use number, and likely something that lends some credibility to the claim it is the correct answer, everything else is a bonus.</p>
<p><strong>On the general problem.</strong></p>
<p>The "work" and "effort" aspect is often over-stressed. Historically, it is to a large part a compromise-solution. What I, and at least some others, actually would want is that users asking explain the bigger picture in which the question arises and pinpoint a specific issue.</p>
<p>It was then claimed by others (and lets admit that at least for the sake of argument but there is at least some truth to it, I think) that this is not a realistic expectation for users below a relatively elevated level of mathematical or more broadly intelecttual sophistication.</p>
<p>Thus, came the compromise, they could at least explain what they were/are trying to do.</p>
<p>While users should put some effort into the question, via presenting it well and thinking first about it, it is not necessary to present attempts if the specific concern of the question is otherwise clear.</p>
<p>Contrary to some believes that is often not the case for what goes under PSQ. One could ask many questions related to:</p>
<blockquote>
<p>Prove that <span class="math-container">$\lim_{x\to0}x^2=0$</span> using <span class="math-container">$\epsilon$</span>-<span class="math-container">$\delta$</span> definition of the limit.</p>
</blockquote>
<p>One of them being: What does a typical instructor expect and accept as satisfactory solution to this?</p>
<p>That often may be the intended question, but sometimes it might not be it, and in any case <em>it should at least be made explicit.</em> Frankly, personally, I might even accept <em>that</em> as context.</p>
|
4,376,076 |
<p>i) the Matrix P has only real elements</p>
<p>ii) 2+i is an eigenvalue of Matrix P</p>
<p>I got that the zeros has to be <span class="math-container">$(x-(2+i))(x-(2-i))$</span> which is equal to <span class="math-container">$(x-2)^2 -i^2$</span> so the characteristical polynom is equal to <span class="math-container">$x^2-4x+4+1 =x^2-4x+5$</span> how can I find the matrix of this characteristical polynom?</p>
|
Martin Argerami
| 22,857 |
<p>If you find <span class="math-container">$A$</span> with eigenvalues <span class="math-container">$\pm i$</span>, then <span class="math-container">$\alpha I+A$</span> has eigenvalues <span class="math-container">$\alpha \pm i$</span>. Indeed,
<span class="math-container">$$ \det\bigl(( \alpha I+A )-\lambda I \bigr)=\det\bigl(A-( \lambda-\alpha)\,I\bigr).
$$</span>
That is, <span class="math-container">$\lambda$</span> is an eigenvalue for <span class="math-container">$\alpha I+A$</span> if and only if <span class="math-container">$\lambda-\alpha=i$</span>.</p>
|
603,986 |
<p>Show that in a finite field $F$ there exists $p(x)\in F[X]$ s.t $p(f)\neq 0\;\;\forall f\in F$</p>
<p>Any ideas how to prove it?</p>
|
Community
| -1 |
<p>Take some element $\alpha_1\in F$</p>
<p>Then consider $f_1(x)=(x-\alpha_1)+1$.. What would be $f_1(\alpha_1)$?</p>
<p>Soon you will see that $f(\alpha_1)$ is non zero but may probably for some $\alpha_2$ we have $f_1(\alpha_2)=0$</p>
<p>Because of this i would now try to include $(x-\alpha_2)$ in $f_1(x)$ to make it </p>
<p>$f_2(x)=(x-\alpha_1)(x-\alpha_2)+1$.. What would be $f_1(\alpha_1),f
_2(\alpha_2)$?</p>
<p>keep doing this until you believe that resultant function does not have a root in $F$.</p>
<p>You have two simple questions :</p>
<ul>
<li>will the resultant be a polynomial in general if you repeat this steps.</li>
<li>How do you make sure that no element in field is a root of resultant</li>
</ul>
|
2,265,782 |
<p>Number of twenty one digit numbers such that Product of the digits is divisible by $21$</p>
<p>Since product is divisible by $21$ the number should contain the digits $3,6,7,9$ But i am unable to decide how to proceed...can i have any hint</p>
|
Mark Fischler
| 150,362 |
<p>We will massage
$$(p\wedge q) \rightarrow p$$ into $\mbox{TRUE}$
showing justifications for each step.</p>
<p>By definition of the "implies" relation,
$$(p\wedge q) \rightarrow p = \lnot (p\wedge q) \lor q$$ </p>
<p>By negation of an and relationship
$$(p\wedge q) \rightarrow p = \lnot (p\wedge q) \lor p = ( (\lnot p) \lor (\lnot q) \lor p $$
By associativity and commutativity of the or relationship</p>
<p>$$(p\wedge q) \rightarrow p =(\lnot p) \lor ((\lnot q) \lor p )
= (\lnot p) \lor (p \lor (\lnot q) ) = ((\lnot p) \lor p) \lor (\lnot q) $$ </p>
<p>By the "law of the excluded middle"
$$(p\wedge q) \rightarrow p = ((\lnot p) \lor p) \lor (\lnot q) = (\mbox{TRUE}) \lor q$$ </p>
<p>By the axiom that $\forall s: \mbox{TRUE} \lor s = \mbox{TRUE} $
$$(p\wedge q) \rightarrow p = (\mbox{TRUE}) \lor q = \mbox{TRUE}$$ </p>
|
1,959,949 |
<blockquote>
<p>We introduce new variables as
$\begin{cases}\xi:=x+ct\\\eta:=x-ct\end{cases}
$
which implies that
$
\begin{cases} \partial_ x=\partial_\xi+\partial_\eta\\\partial_t=c\partial_\xi+c\partial_\eta
\end{cases}
$</p>
</blockquote>
<p>This is from page 34 of <em>Partial Differential Equation --- an Introduction</em> (2nd edition) by Strauss. I don't understand the implication here. </p>
<p>How does this proceeds? Thanks a lot.</p>
|
Community
| -1 |
<p>There is a typo in Strauss's book. One should have
$$
\partial_t=c\partial_\xi-c\partial_\eta.
$$
Note that both $\xi$ and $\eta$ are functions of $x$ and $t$:
$$
\xi=h(x,t);\quad \eta=g(x,t).
$$</p>
<p>Now, suppose you have a function $u=u(\xi,\eta)$. Then you have
$$
u=u(h(x,t),g(x,t)).
$$
Can you write down by chain rule, what is $u_x$ and $u_t$?</p>
|
1,237,077 |
<p>For a periodic function we have: $$\int_{b}^{b+a}f(t)dt = \int_{b}^{na}f(t)dt+\int_{na}^{b+a}f(t)dt = \int_{b+a}^{(n+1)a}f(t)dt+\int_{an}^{b+a}f(t)dt = \int_{na}^{(n+1)a}f(t)dt = \int_{0}^{a}f(t)dt.$$ , but I don't understand how we obtain $\int _{b+a}^{\left(n+1\right)a}\:f\left(t\right)\:dt=\int _b^{na}\:f\left(t\right)dt$ in our equality?</p>
|
Idris Addou
| 192,045 |
<p>Let $F$ be a primitive of $f$. Then, by the fundamental theorem of calculus, one has</p>
<p>$ \frac{d}{db} \int_{b}^{b+a}f(t)dt= \frac{d}{db} [F(b+a)-F(b)]=F^ \prime (b+a)-F^ \prime (b)=f(b+a)-f(b)=0.$ </p>
<p>since $f$ is $a$-periodic. Then, the integral $\int_{b}^{b+a}f(t)dt$ is independent of $b$, so one can take $b=0$.</p>
|
786,086 |
<p>For my research I am working with approximations to functions which I then integrate or differentiate and I am wondering how this affects the order of approximation.</p>
<p>Consider as a minimal example the case of $e^x$ for which integration and differentiation doesn't change anything. Now if I would approximate this function with a second order taylor series I get:
$$e^x\approx 1+x+\frac{x^2}{2}+O(x^3) \tag{1}$$</p>
<p>If I were to integrate this function I get:
$$ \int 1+x+\frac{x^2}{2}+O(x^3) dx = C+x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^{?4?}) \tag{2}$$</p>
<p>I wrote $O^{?4?}$ because that is what my question is about: <strong>do I indeed get a higher order approximation when I do this integration or is it appropriate to cut-off the solution to the integral at $O(x^3)$, thus removing the $\frac{x^3}{6}$ term?</strong></p>
<p>And what about differentiation? In that case I seem to lose an order of accuracy, is that indeed the case?</p>
|
Claude Leibovici
| 82,404 |
<p>This is indeed a very interesting question about the use of truncated series for integration and differentiation.</p>
<p>Taking you example of $e^x$, if we write $$f(x) =1+x+\frac{x^2}{2}+O\left(x^3\right)$$ from a formal point of view (at least, in my opinion), the derivative should write $$f'(x) =1+x+O\left(x^2\right)$$ and the antiderivative should write $$\int f(x) dx=C+x+\frac{x^2}{2}+\frac{x^3}{6}+O\left(x^4\right)$$</p>
|
737,835 |
<p>Why is $[0,1]$ not homeomorphic to $[0,1]^2$? It seems that the easiest way to show this is to find some inconsistency between the open set structures of the two. It is clear that the two share the same cardinality. Both are compact. Both are normal since they are metric spaces. However, where to find the open set structure that is not shared by the two? Any hint, please?</p>
|
user126154
| 126,154 |
<p>As comments and other questions point out, this case is easily solved by a connectedness argument.</p>
<p>It may be worth to point out that the general proof that $\mathbb R^n$ is not homeomorphic to $\mathbb R^m$ for $m\neq n$ is not trivial.</p>
<p>There is a nice important result, called the invariance of domain, (see <a href="http://en.wikipedia.org/wiki/Invariance_of_domain" rel="nofollow">http://en.wikipedia.org/wiki/Invariance_of_domain</a>) that states that if a map $f:U\subset \mathbb R^n\to \mathbb R^n$ is continuous and injective, then it is open. This implies that there is no continuous injective map from $\mathbb R^{n+k}$ to $\mathbb R^n$.</p>
|
229,606 |
<p>I need little help in proving the following result :</p>
<p>Consider the ring $R:=\mathbb{F}_q[X]/(X^n-1)$, where $\mathbb{F}_q$ is a finite field of cardinality $q$ and $n\in\mathbb{N}$. Then any ideal $I$ of $R$ is principle and can be written as $I=(g(X))$, such that $g(X)|(X^n-1)$.</p>
|
Lior B-S
| 26,713 |
<p>If $\varphi\colon S\to T$ is an epimorphism of rings with $1$, and if $I$ is an ideal of $T$, then $J=\varphi^{-1}(I)$ is an ideal of $S$. </p>
<p>Now if $S$ is <a href="http://en.wikipedia.org/wiki/Principal_ideal_domain" rel="nofollow">PID</a>, that is, if every ideal is generated by a single element, then $J=xS$, for some $x\in S$. So $I=\varphi(J)=\varphi(s)I$, is also principle. </p>
<p>Apply the above to $\varphi\colon \mathbb{F}_q[X]\to R$ and you'll get that every ideal of $R$ is principle. </p>
|
1,109,552 |
<p>So the Norm for an element $\alpha = a + b\sqrt{-5}$ in $\mathbb{Z}[\sqrt{-5}]$ is defined as $N(\alpha) = a^2 + 5b^2$ and so i argue by contradiction assume there exists $\alpha$ such that $N(\alpha) = 2$ and so $a^2+5b^2 = 2$ , however, since $b^2$ and $a^2$ are both positive integers then $b=0$ and $a=\sqrt{2}$ however $a$ must be an integer and so no such $\alpha$ exists, same goes for $3$.</p>
<p>I already proved that </p>
<ol>
<li>$N(\alpha\beta) = N(\alpha)N(\beta)$ for all $\alpha,\beta\in\mathbb{Z}[\sqrt{-5}]$.</li>
<li>if $\alpha\mid\beta$ in $\mathbb{Z}[\sqrt{-5}]$, then $N(\alpha)\mid N(\beta)$ in $\mathbb{Z}$.</li>
<li>$\alpha\in\mathbb{Z}[\sqrt{-5}]$ is a unit if and only if $N(\alpha)=1$.</li>
<li>Show that there are no elements in $\mathbb{Z}[\sqrt{-5}]$ with $N(\alpha)=2$ or $N(\alpha)=3$. (I proved it above)</li>
</ol>
<p>Now I need to prove that $2$, $3$, $1+ \sqrt{-5}$, and $1-\sqrt{-5}$ are irreducible.</p>
<p>So I also argue by contradiction, assume $1 + \sqrt{-5}$ is reducible then there must exists Non unit elements $\alpha,\beta \in \mathbb{Z}[\sqrt{-5}]$ such that $\alpha\beta = 1 + \sqrt{-5} $ and so $N(\alpha\beta) =N(\alpha)N(\beta)= N(1 + \sqrt{-5}) = 6$ but we already know that $N(\alpha) \neq 2$ or $3$ and so $N(\alpha) = 6$ and $N(\beta) = 1$ or vice verse , in any case this contradicts the fact that both $\alpha$ and $\beta$ are both non units.I just want to make sure i am on the right track here. And how can i prove that $2$, $3$, $1+ \sqrt{-5}$, and $1-\sqrt{-5}$ are not associate to each other.</p>
|
Community
| -1 |
<p>I've seen these same arguments in quite a few books, it all looks pretty standard issue to me.</p>
<p>But what you seem to be missing is the <em>motivation</em> for all of this, the why do we care. That motivation is this famous fact: $$6 = 2 \times 3 = (1 - \sqrt{-5})(1 + \sqrt{-5}).$$ It's also true that $$6 = -2 \times -3 = (-1 - \sqrt{-5})(-1 + \sqrt{-5}).$$ We've got four factorizations here, but how many of them are distinct factorizations into irreducibles?</p>
<p>Well, $6$ has a norm of $36$. Per the familiar arguments you've already quoted, $2, 3, 1 - \sqrt{-5}, 1 + \sqrt{-5}$ are irreducible with norms of $4, 9, 6, 6$ respectively, and $36 = 4 \times 9 = 6 \times 6$.</p>
<p>I suppose you also need a good definition of associates: in a commutative ring, $x$ and $y$ are associates <em>if and only if</em> there is a unit $u$ such that $x = uy$ (see <a href="https://www.proofwiki.org/wiki/Definition:Associate/Commutative_Ring">https://www.proofwiki.org/wiki/Definition:Associate/Commutative_Ring</a>). In an imgainary quadratic ring like $\mathbb{Z}[\sqrt{-5}]$, if $u$ is a unit and $x$ and $y$ are associates, then $N(u) = 1$ and $N(x) = N(y)$.</p>
<p>Clearly $2$ is not an associate of $3$ nor of $1 - \sqrt{-5}$ nor of $1 + \sqrt{-5}$. It's possible that $1 - \sqrt{-5}$ and $1 + \sqrt{-5}$ are associates of each other. But don't jump to the conclusion that $N(x) = N(y)$ automatically means $x$ and $y$ are associates, the definition explicitly requires a unit $u$ such that $x = uy$. There are only two units in $\mathbb{Z}[\sqrt{-5}]$, and those are $-1$ and $1$. As it turns out, $(-1)(1 - \sqrt{-5}) = -1 + \sqrt{-5}$, not $1 + \sqrt{-5}$, so this absence of a suitable unit means that $1 - \sqrt{-5}$ and $1 + \sqrt{-5}$ are <em>not</em> associates of each other.</p>
<p>Although we've shown four factorizations of $6$ in $\mathbb{Z}[\sqrt{-5}]$, only two of these are distinct because the other two consist of associates of numbers in the first two.</p>
|
526,820 |
<p>How do I integrate the inner integral on 2nd line? </p>
<p><img src="https://i.stack.imgur.com/uIxQX.png" alt="enter image description here"></p>
<hr>
<p>$$\int^\infty_{-\infty} x \exp\{ -\frac{1}{2(1-\rho^2)} (x-y\rho)^2 \} \, dx$$</p>
<p>I know I can use integration by substitution, let $u = \frac{x-y\rho}{\sqrt{1-\rho^2}}$ resulting in</p>
<p>$$\sqrt{1-\rho^2}\int^{\infty}_{-\infty} [u\sqrt{1-\rho^2} + y\rho] e^{-u^2/2} \; du$$</p>
<p>Thats the 3rd line in the image, but how do I proceed? </p>
|
martini
| 15,379 |
<p>Then, recalling that $u \mapsto (2\pi)^{-1/2}\exp(-u^2/2)$ is the density of a standard normal distribution, we have that
$$ \int_{\mathbb R} (2\pi)^{-1/2} \exp(-u^2/2)\, du = 1 \iff \int_{\mathbb R} \exp(-u^2/2)\, du = (2\pi)^{1/2} $$
and $\int_{\mathbb R} (2\pi)^{-1/2}u\exp(-u^2/2)\, du$ is the expectation of a standard normal, hence equal to 0 (you can also argue that this holds as the integrand is odd), so
$$ \int_{\mathbb R} (2\pi)^{-1/2} u \exp(-u^2/2)\, du = 0 $$</p>
|
3,058,139 |
<p>Let us consider the statement <span class="math-container">$\exists x P(x)$</span> - translated into English, "there exists an <span class="math-container">$x$</span> in our universe of discourse such that <span class="math-container">$P(x)$</span> is true." In writing the negation of this, we are taught to switch quantifiers (<span class="math-container">$\exists \leftrightarrow \forall$</span>) and to negate that statement <span class="math-container">$P(x)$</span>.</p>
<p>Thus, </p>
<p><span class="math-container">$$\neg (\exists x P(x)) = \forall x (\neg P(x))$$</span></p>
<p>However, humor me for a second. Let's consider what negation is - it is the "logical complement." The negation of a statement is always false when the statement is true, and vice versa. In that light, why would we not say the following is also the negation?</p>
<p><span class="math-container">$$\neg (\exists x P(x)) = \not \exists x P(x)$$</span></p>
<p>Or, taking this a bit further, why would we not write this as well?</p>
<p><span class="math-container">$$\forall x (\neg P(x)) = \not \exists x P(x)$$</span></p>
<p>Both seem to imply the same thing: there does not exist an <span class="math-container">$x$</span> such that <span class="math-container">$P(x)$</span> is true (and thus for all <span class="math-container">$x$</span>, <span class="math-container">$P(x)$</span> is false, or, rather, <span class="math-container">$\neg P(x)$</span> is true). </p>
<p>So is there some underlying reason why we don't do negations in this way? As far as I can tell, they mean the same thing, yet I always have seen the <span class="math-container">$\forall$</span> version as above. Looking around MSE, I've only seen some posts which have <span class="math-container">$\neg \exists$</span> (basically the same as <span class="math-container">$\not \exists$</span>), but they're only in the context of simplifying a logical expression. </p>
<p>So I guess, if indeed these are logically equivalent, my follow-up question would be - why is <span class="math-container">$\forall$</span> considered a simplification of <span class="math-container">$\neg \exists$</span> or <span class="math-container">$\not \exists$</span>?</p>
<p>My only guess is that "<span class="math-container">$\not \exists$</span>" isn't a standard notation, or so I recall from some notes my complex analysis professor gave us last semester. Or perhaps to say "for all <span class="math-container">$x$</span>, this is false" more immediately is understood (or a more "direct" way of saying it) than "there does not exist <span class="math-container">$x$</span> such this is true?"</p>
<p><em>(Footnote of note: I haven't had much education in predicate logic and such. We went over it for a little while in one of my classes so I understand some basics like the above but we never went into much detail. So I apologize if this question is poorly framed or worded.)</em></p>
|
N. S.
| 9,176 |
<p>The two statements are equivalent, the reason why we write it that way it is because it is easier to deal/prove it. It is irrelevant which way it is easier to say it, the important thing is being able to use.</p>
<p>Just consider the statement <span class="math-container">$\exists x \in \mathbb Z, x^2+(x+1)^2 \mbox{ is even}$</span>. This is not true, now try to show that this is false.</p>
<p>Try to prove separately each of the following two statements:</p>
<ul>
<li><span class="math-container">$\not\exists x \in \mathbb Z, x^2+(x+1)^2 \mbox{ is even}$</span></li>
<li><span class="math-container">$\forall x \in \mathbb Z, x^2+(x+1)^2 \mbox{ is odd}$</span></li>
</ul>
<p>It is easier to deal with the second one.</p>
|
2,117,420 |
<p>We all know that particular solution of $A_{n} = A_{(n-1)} + f(n)$</p>
<p>where $f(n)=n^c$ , c is a random positive integer.</p>
<p>Can be set to $(n^c+n^{(c-1)}+.....+1)$</p>
<p>But what about when $c\lt0$?</p>
<p>How do we find a particular solution of the form:</p>
<blockquote>
<p>$A_{n} = A_{(n-1)} + f(n)$</p>
<p>where $f(n)=n^{(c)}$ , c is a random negative integer.</p>
</blockquote>
<p>For example. What's the particular solution of $T(n)=T(n-1)+{1\over n}$</p>
<p>(P.S I know it's Harmonic series and we can use Integral test to prove that it's diverges by comparing its sum with an improper integral. But it doesn't matter. )</p>
<p>I can't find any clue in "Discrete Mathematics, 7th Edition". </p>
<p>Does anyone know the answer?</p>
|
MatheMagic
| 397,530 |
<p>We can write $M=I+N$ where </p>
<p>$$
I = \begin{pmatrix}1&0&0 \\ 0&1&0 \\0&0&1 \end{pmatrix}
\qquad
N = \begin{pmatrix}0&1&0 \\ 0&0&1 \\0&0&0 \end{pmatrix}
$$</p>
<p>Since $I$ is the identity, $N$ and $I$ commute; hence $e^{I+N} = e^I e^N$ and $N$ is nilpotent with index $3$. So, $N^3=0$</p>
<p>$$ e^I e^N = eI\cdot(I+N+\frac{1}{2}N^2) = e\begin{pmatrix}1&1&\frac{1}{2} \\ 0&1&1 \\0&0&1 \end{pmatrix}$$</p>
<p>Thus the sum of the entries of the matrix $M$ is $\displaystyle 5.5e$.</p>
|
921,893 |
<p>If we have: $f(x)=\frac { 1+x }{ 1+{ e }^{ x } } $</p>
<p>I am told to determine if $f(x)=x$ has multiple roots on $\left[ 0;+\infty \right] $</p>
<p>I tried to manually solve this equation, but I don't understand the result:</p>
<p>$f(x)-x=0\rightarrow \frac { 1+x }{ 1+{ e }^{ x } } =0\rightarrow { e }^{ x }(x+1)=0$</p>
<p>which would mean that -1 is the only solution. So there shouldn't be any solution on $\left[ 0;+\infty \right] $.</p>
<p>But when I check on Wolfram, I see that there is a unique solution, and that this solution is 0.56.</p>
<p>Can anybody help please ?</p>
<p>1) How can I find the nb of solutions to this equation ?
2) How can we explain the result I get and the one from wolfram ?</p>
|
Sheheryar Zaidi
| 131,709 |
<p>$$f(x) - x = \frac{1+x}{1+e^x} - x = \frac{1+x}{1+e^x} - \frac{x(1+e^x)}{1+e^x} = \frac{1-xe^x}{1+e^x}$$</p>
|
788,995 |
<p>I need to prove that function $\mathbb R × \mathbb R → \mathbb R $ : $f(x,y) = \frac{|x-y|}{1 + |x-y|}$ is a metric on $\mathbb R$. First two axioms are trivial; it's the triangle inequality which is pain. $\frac{|x-y|}{1 + |x-y|}$ + $\frac{|y-z|}{1 + |y-z|} ≥ \frac{|x-z|}{1 + |x-z|} ⇒ \frac{|x-y| + |y-z| + 2|(x-y)(y-z)|}{1 + |x-y| + |y-z| + |(x-y)(y-z)|}≥ \frac{|x-z|}{1 + |x-z|}$, but then I am stuck. Can somebody show me way out of this?</p>
|
Tom
| 103,715 |
<p><b>Hint</b>: For $a,b \geq 0$
$$\frac{a+b}{1+a+b} = \frac{a}{1+a+b} + \frac{b}{1+a+b} \leq \frac{a}{1+a}+\frac{b}{1+b}$$
Next, made judicious choices for $a$ and $b$. </p>
|
788,995 |
<p>I need to prove that function $\mathbb R × \mathbb R → \mathbb R $ : $f(x,y) = \frac{|x-y|}{1 + |x-y|}$ is a metric on $\mathbb R$. First two axioms are trivial; it's the triangle inequality which is pain. $\frac{|x-y|}{1 + |x-y|}$ + $\frac{|y-z|}{1 + |y-z|} ≥ \frac{|x-z|}{1 + |x-z|} ⇒ \frac{|x-y| + |y-z| + 2|(x-y)(y-z)|}{1 + |x-y| + |y-z| + |(x-y)(y-z)|}≥ \frac{|x-z|}{1 + |x-z|}$, but then I am stuck. Can somebody show me way out of this?</p>
|
Santosh Linkha
| 2,199 |
<p>Since $1 + |x-y| + |y -z| \ge 1 + |x-z|$, </p>
<p>$$1 - \frac{1}{1 + |x-z|} \le 1 - \frac{1}{1 + |x-y| + |y -z|} \\
\implies \frac{|x-z|}{1 +|x-z|} \le \frac{|x-y| + |y-z|}{1 + |x-y| + |y-z|} \le \frac{|x-y|}{1 + |x-y|} + \frac{|y-z|}{1 + |y-z|}$$</p>
|
1,586,286 |
<p>There are $30$ red balls and $50$ white balls. Sam and Jane take turns drawing balls until they have drawn them all. Sam goes first. Let $N$ be the number of times Jane draws the same color ball as Sam. Find $E[N].$</p>
<p>I have been proceeding with indicators...</p>
<p>$$ I_{j} =
\begin{cases}
1, & \text{if the $i^{th}$ pick is the same as the $(i-1)^{th}$ pick.} \\
0, & \text{otherwise}
\end{cases}
$$</p>
<p>But I am having problems with this. Because when I proceed, I am find all the instances where the same ball is drawn on every draw, not just Jane's. I know the answer is $\frac{9}{19}$ but I can't get there. How do I change the indicators, or the probability?</p>
|
MichaelChirico
| 205,203 |
<h1>Hint:</h1>
<p>Here's the simulated distribution (300,000 repetitions)</p>
<p><a href="https://i.stack.imgur.com/1GpDR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1GpDR.png" alt="enter image description here" /></a></p>
<p>(let me know if you want the code, it's pretty straightforward though)</p>
|
2,227,027 |
<p>If $f(x)$ is defined everywhere except at $x=x_0$, would $f'(x_0)$ be undefined at $x=x_0$ as well?</p>
<p>One example is: $$f(x)=\ln(x)\rightarrow f'(x)=\frac{1}{x}$$</p>
<p>In this particular case, both $f(x)$ and $f'(x)$ are undefined at $x=0$. I wonder if this always holds true.</p>
<p>Thank you.</p>
|
Ross Millikan
| 1,827 |
<p>The usual definition, <span class="math-container">$f'(x)=\lim_{h \to 0}\frac {f(x+h)-f(x)}h$</span> clearly requires that we be able to evaluate <span class="math-container">$f(x)$</span> in a neighborhood of <span class="math-container">$x$</span>. A challenging case is where there is a removable discontinuity. For example, take <span class="math-container">$f(x)=\frac {x(x-1)}x$</span>. It cannot be evaluated at <span class="math-container">$x=0$</span>, but otherwise is the same as <span class="math-container">$g(x)=x-1$</span>. For a function which is differentiable at <span class="math-container">$x$</span>, another expression is <span class="math-container">$f'(x)=\lim_{h \to 0}\frac {f(x+h)-f(x-h)}{2h}$</span>. This one avoids the problem of evaluating <span class="math-container">$f(x)$</span> at the point in question, so will give the sensible answer of <span class="math-container">$1$</span> for <span class="math-container">$f'(0)$</span>. Should <span class="math-container">$f(x)$</span> have a derivative at <span class="math-container">$0$</span>? If you think so, you can propose the second as a replacement definition. The fact that you can give a sensible answer to another type of problem is a point in favor. That is how definitions get changed-people find the new one more useful than the old one.</p>
|
3,273,756 |
<blockquote>
<p>I am supposed to give a 9-dimensional irreducible representation of <span class="math-container">$\mathfrak{so}(4)$</span>.</p>
</blockquote>
<p>I know that <span class="math-container">$\mathfrak{so}(4)\cong\mathfrak{so}(3)\oplus\mathfrak{so}(3)$</span> and hence I have a 6-dimensional reducible representation of <span class="math-container">$\mathfrak{so}(4)$</span>. But how do I get a irreducible 9-dimensional representation?
This may be a stupid question but I couldn't find anything.</p>
|
Torsten Schoeneberg
| 96,384 |
<p>Almost every treatment of representations of real semisimple Lie algebras explicitly treats the basic case <span class="math-container">$\mathfrak{su}_2$</span>; the standard result is that for each positive integer <span class="math-container">$n$</span>, there is up to equivalence exactly one irreducible <span class="math-container">$n$</span>-dimensional <span class="math-container">$\mathfrak{su}_2$</span>-representation, call it <span class="math-container">$\rho_n$</span>.</p>
<p>Now if you already know <span class="math-container">$\mathfrak{so}_4 \simeq \mathfrak{so}_3 \oplus \mathfrak{so}_3$</span>, and further know <span class="math-container">$\mathfrak{so}_3 \simeq \mathfrak{su}_2$</span>, then let</p>
<p><span class="math-container">$pr_1: \mathfrak{so}_4 \twoheadrightarrow \mathfrak{su}_2, \qquad pr_2: \mathfrak{so}_4 \twoheadrightarrow \mathfrak{su}_2$</span></p>
<p>be the projections on the respective factors. Then <span class="math-container">$\rho_9 \circ pr_i$</span> give two inequivalent (why?), irreducible (why?), <span class="math-container">$9$</span>-dimensional representations of <span class="math-container">$\mathfrak{so}_4$</span>.</p>
|
1,358,735 |
<p>I'm sorry to sound like a dummy, but I've had trouble with Algebra all my life. I'm studying online with Khan Academy and one of the questions is: </p>
<p>Point $E$'s $y$-coordinate is $0$, but its $x$-coordinate is not $0$.
Where could point $E$ be located on the coordinate plane?</p>
<p>There is not graph or nothing, just a multiple choice of </p>
<ul>
<li>Quadrant $I$ </li>
<li>Quadrant $II$</li>
<li>Quadrant $III$</li>
<li>Quadrant $IV$ </li>
<li>$x$-axis </li>
<li>$y$-axis</li>
</ul>
<p>What does the $E$ mean? I understand plotting numbers and points, etc, but what is $E$? </p>
|
Tyler Hilton
| 287 |
<p>Each point is represented by its coordinates on the plane. So your point $E$ can be represented by coordinates $$(x, y)$$</p>
<p>You are already told that the $y$-coordinate is 0, so the point $E$ has the form $$(x, 0)$$</p>
<p>Since no information is given for $x$ (ie, no restrictions for $x$), plug in some values for $x$ and draw the points on the plane. Consider $x = 3$ so plot the point $(3, 0)$. Or consider $x = -5$, and plot $(5, 0)$. It turns out that all possible points, no matter which value of $x$ you pick, will lie on the $x$-axis. </p>
<p>Also, an additional note: Here $x$ was free to be whatever value you want (and thus covers the entire x-axis) but some questions may even restrict what $x$ could be. A simple example would be the above + the additional restriction that $x > 0$. Plot a few points using this, and see what you get. </p>
|
1,666,615 |
<blockquote>
<p>For the series $\sum_{k=1}^{\infty}a_k$, suppose that there is a number $r$ with $0\leq r<1$ and a natural number $N$ such that $$|a_k|^{1/k}<r\qquad\text{for all indices $k\geq N$}$$ Prove that $\sum_{k=1}^{\infty}a_k$ converges absolutely.</p>
</blockquote>
<p>Proof:</p>
<p>For a given $r\in\mathbb{R}$ with $0\leq r<1$ and $N\in\mathbb{N}$ satisfy $|a_k|^{1/k}<r$ for all indices $k\geq N$, that gives $|a_k|<r^k$. Now, define $s_n=\sum_{k=1}^{n}|a_k|$ be a sequence of partial sum of $\sum_{k=1}^{\infty}|a_k|$. Since $\sum_{k=1}^{n}r^k$ converges to $(1-r^{n+1})/(1-r)$, for all $\epsilon>0$, this gives $$\left|\sum_{k=1}^{n}r^k-\frac{1-r^{n+1}}{1-r}\right|<\frac{\epsilon}{2}\qquad\text{for all $k\geq N$}$$ Then for all $j,k\geq N$, we have
\begin{align*}
\left|\sum_{j=1}^{n}a_j-\sum_{k=1}^{n}a_k\right|<\left|\sum_{j=1}^{n}r^j-\sum_{k=1}^{n}r^k\right|&=\left|\sum_{j=1}^{n}r^j-\frac{1-r^{n+1}}{1-r}+\frac{1-r^{n+1}}{1-r}-\sum_{k=1}^{n}r^k\right|\\
&\leq\left|\sum_{j=1}^{n}r^j-\frac{1-r^{n+1}}{1-r}\right|+\left|\frac{1-r^{n+1}}{1-r}-\sum_{k=1}^{n}r^k\right|\\
&=\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon
\end{align*}
Hence, $\{s_n\}$ is a Cauchy sequence which implies $\{s_n\}$ is convergent, so there exists an $M\in\mathbb{R}$ such that $\sum_{k=1}^{n}a_k\leq M$. This inequality implies $\sum_{k=1}^{\infty}|a_k|$ is convergent; therefore, $\sum_{k=1}^{\infty}a_k$ converges absolutely.</p>
<hr>
<p>Does this solution valid? If not, can someone give me a hint or suggestion to receive the answer? Thanks.</p>
|
choco_addicted
| 310,026 |
<p>If you know the comparison test, then you can prove the proposition easily. You know $|a_k| \le r^k$ for all $k \ge N$. Since $0\le r < 1$, the geometric series $\sum_{n=1}^{\infty} r^n$ converges. Therefore, by comparison test, $\sum_{n=1}^{\infty} |a_n|$ converges.</p>
<p>The proposition you have to prove is called the 'root test'.</p>
|
4,351,794 |
<p>I am trying not exactly to solve equation, but just change it from what is on right side to what is on left side. But I didn't do any math for years and can't remember what to.</p>
<blockquote>
<p><span class="math-container">$$\frac{1}{2jw(1+jw)}=\frac{-j(1-jw)}{2w(1+w^2)}$$</span></p>
<p>Here, <span class="math-container">$j^2=-1$</span>.</p>
</blockquote>
<p>If I am right, I should do something like this.
<span class="math-container">$$\frac{1}{(2jw(1+jw))}=\frac{(2jw(1-jw))}{(2jw(1+jw))}$$</span>
But whatever I do I can't get what's shown above.</p>
<p>I will be thankful for any help.</p>
|
Azur
| 656,302 |
<p><em>So, correct me if I'm wrong, but this equation seems to come from some electrical physics context? The reason I'm saying that is because it seems that <span class="math-container">$j$</span> is the imaginary unit (such that <span class="math-container">$j^2 = -1$</span>), which is usually called <span class="math-container">$i$</span>, except for some areas of physics (to not confuse it with intensity)</em></p>
<p>If so, then a first step would be to nice that <span class="math-container">$\dfrac{1}{j} = -j$</span> (can you see why this is true?). And once you have that, you can do a common trick which is "multiplying by the conjugate".</p>
<p>Example: say you have an expression of the form <span class="math-container">$\dfrac{1}{a + bj}$</span>, and you want to get a real denominator. You can achieve this by multiplying both the top and bottom by the <strong>conjugate</strong> of <span class="math-container">$a + bj$</span>, which is defined as <span class="math-container">$a - bj$</span> (see <a href="https://en.wikipedia.org/wiki/Complex_conjugate" rel="nofollow noreferrer">complex conjugate</a>, keeping in mind that the article uses <span class="math-container">$i$</span> instead of <span class="math-container">$j$</span>). I will let you perform the calculations for yourself (because that will help you understand what's going on), but this should be enough for you to finish the computations :)</p>
|
947,730 |
<p>I'm trying to do this for practice but I'm just going nowhere with it, I'd love to see some work and answers on it.</p>
<p>Thanks :)</p>
<p>Find a polynomial that passes through the points (-2,-1), (-1,7), (2,-5), (3,-1). Present the answer in standard form.</p>
<p>What I've tried:</p>
<p><img src="https://i.stack.imgur.com/Wsvj9.jpg" alt="What I firstly tried but went nowhere"></p>
<p><img src="https://i.stack.imgur.com/R6USZ.jpg" alt="Another attempt to get somewhere"></p>
|
0xfee1dead
| 898,565 |
<p><strong>Step 1 :</strong> List the equations that define the function according to ax<sup>3</sup> + bx<sup>2</sup> + cx + d:<br></p>
<pre>-8a+ 4b - 2c + d = -1 (equation1)<br>
-1a + 1b + -1c + d = 7 (equation2) <br>
8a + 4b + 2c + d = -5 (equation3) <br>
27a + 9b + 3c + d = -1 (equation4)</pre>
<p><strong>Step 2 :</strong> Use matrix operations to simplify the system of equations<br><br>
for example: add/subtract multiples of equation1 from/to each of the equations below it to remove the a's from them.</p>
<p>Then add/subtract multiples of the <strong>new</strong> equation2 from/to the equations below it to remove the b's from them. .</p>
<p>Then add/subtract multiples of the <strong>new</strong> equation3 from/to the equations below it (which is only equation 4) to remove the c's from it.</p>
<p>After completing this process, the system of equations will be easy to solve since equation4 will end up having 1 variable (which will be d), which you can calculate and substitute in equation3 to get c and then substitute d and c into equation2 and so on</p>
|
83,336 |
<p>Sorry if this is a naive question-- I'm trying to learn this stuff (cross-posted from <a href="https://math.stackexchange.com/questions/89248/induced-representations-of-topological-groups">https://math.stackexchange.com/questions/89248/induced-representations-of-topological-groups</a>)</p>
<p>If $G$ is a group with subgroup $H$, then we have the restriction functor $\operatorname{Res}$ from $G-\operatorname{mod}$ to $H-\operatorname{mod}$. We also have this idea of induction, a functor $\operatorname{Ind}^G_H$ from $H-\operatorname{mod}$ to $G-\operatorname{mod}$. These are adjoints, which means (I think) that $\operatorname{Hom}_G(\operatorname{Ind}^G_H(V), U) \cong \operatorname{Hom}_H(V,\operatorname{Res}(U))$ naturally, for $G$-modules $U$ and $H$-modules $V$.</p>
<p>For locally compact groups, there is a theory worked out by MacKey and others. Actually, I have only read Rieffel's work on the subject (as I come from a functional Analysis background). For a locally compact $G$ and closed subgroup $H$, there is a very satisfactory notion of the functor $\operatorname{Ind}^G_H$ (where we consider "Hermitian modules", i.e. unitary representations on Hilbert spaces). What I don't see is how (or even if) this relates to the restriction functor?</p>
<blockquote>
<p>In the topological setting, are $\operatorname{Ind}^G_H$ and $\operatorname{Res}$ in any sense adjoints?</p>
</blockquote>
<p>A slightly vague rider-- if (as I suspect) the answer is "no", can we be more precise about <em>why</em> the answer is no?</p>
|
Jesse Peterson
| 6,460 |
<p>If $G$ is compact this is the Frobenius Reciprocity Theorem, see e.g., Section 6.2 in Folland's A Course in Abstract Harmonic Analysis for a proof. When $G$ is not compact then this fails already for $H$ the trivial group and $U$, and $V$ trivial representations. Indeed, in this case Ind$_H^G(1_H) = L^2G$ the left regular representation, and since constant functions are not square integrable we have Hom$_G( L^2G, 1_G ) = \{ 0 \}$, while Hom$_H(1_H, 1_H) \not= \{ 0 \}$.</p>
|
2,400,336 |
<p>My first try was to set the whole expression equal to $a$ and square both sides. $$\sqrt{6-\sqrt{20}}=a \Longleftrightarrow a^2=6-\sqrt{20}=6-\sqrt{4\cdot5}=6-2\sqrt{5}.$$</p>
<p>Multiplying by conjugate I get $$a^2=\frac{(6-2\sqrt{5})(6+2\sqrt{5})}{6+2\sqrt{5}}=\frac{16}{2+\sqrt{5}}.$$</p>
<p>But I still end up with an ugly radical expression.</p>
|
John Joy
| 140,156 |
<p>In this particular problem, you can pretty much guess the answer.</p>
<p>$$\sqrt{6-\sqrt{20}}=\sqrt{6-2\sqrt{5}}$$</p>
<p>Now, suppose that the $-2\sqrt{5}$ was the middle term of a perfect square trinomial, where $x = \sqrt{5}$. In other words, that middle term is $-2x$.</p>
<p>What would the first and last term look like? Obviously it would be $x^2$ and $1$ respectively.</p>
<p>$x^2 - 2x +1 = (x-1)^2$</p>
<p>Substituting $\sqrt{5}$ for $x$ we have...
$$x^2 - 2x +1 = (x-1)^2$$
$$\sqrt{5}^2 - 2\sqrt{5} +1 = (\sqrt{5}-1)^2$$
$$5 - 2\sqrt{5} +1 = (\sqrt{5}-1)^2$$
$$6 - 2\sqrt{5} = (\sqrt{5}-1)^2$$
and taking the square root of both sides we have $$\sqrt{6-2\sqrt{5}} = \sqrt{5}-1$$</p>
<p>It's almost like somebody just make up that problem to work out cleanly like that. ;)</p>
|
144,375 |
<p>We know that every $2\times 2$ matrix in $PGL(2, \mathbb{Z})$ of order $3$ is conjugate to the matrix $$ \left( \begin{array}{cc} 1 & -1 \\ 1 & 0 \end{array} \right) $$. </p>
<p>I am interested in finding out to what extent this holds for $3\times 3$ integer invertible matrices.</p>
<p>In other words how many conjugacy classes of order 3 matrices in $PGL(3, \mathbb{Z})$ are there?</p>
|
Alex B.
| 35,416 |
<p>I will work in ${\rm GL}$ instead of ${\rm PGL}$.</p>
<p>The corresponding question over ${\rm GL}_3(\mathbb{Z})$ is essentially$^1$ equivalent to asking how many faithful $\mathbb{Z}[G]$-modules, free of rank 3 over $\mathbb{Z}$ there are up to isomorphism, where $G$ is the cyclic group of order 3. Any such module is a direct sum of indecomposable modules, and those have been classified in I. Reiner, Integral representations of cyclic groups of prime order, Proc. Amer. Math. Soc. 8 (1957), 142–146. There are three indecomposable $\mathbb{Z}[G]$-modules:</p>
<ul>
<li>the trivial module of $\mathbb{Z}$-rank 1, $\Gamma_1$,</li>
<li>the augmentation ideal of $\mathbb{Z}[G]$, which has $\mathbb{Z}$-rank 2, $\Gamma_2$,</li>
<li>the regular module $\Gamma_3=\mathbb{Z}[G]$ itself.</li>
</ul>
<p>So there are two isomorphism classes of faithful modules of rank 3:</p>
<ul>
<li>$\Gamma_1\oplus \Gamma_2$,</li>
<li>$\Gamma_3$.</li>
</ul>
<p>One should beware, that, in general, the Krull-Schmidt Theorem fails for $\mathbb{Z}[G]$-modules, but in this case it is easy to see that the two guys are not isomorphic, since e.g. in one of them the trivial isotypical component is a direct summand, and in the other one it isn't. Alternatively, the two are not isomorphic over $\mathbb{Z}_3$, and Krull-Schmidt <em>does</em> hold over local rings.</p>
<p>${}^1$ To remove the word "essentially", one needs to check that any matrix obtained in this way is conjugate to its inverse. This is true because each of the indecomposable modules listed above can be extended to a module under the symmetric group $S_3$, as is easy to check.</p>
|
3,375,375 |
<p>I noticed this issue was throwing off a more sophisticated problem I'm working on. When computing the indefinite integral </p>
<p><span class="math-container">$$ I(x) = \int \frac{dx}{1-x} = \log | 1-x | + C,$$</span></p>
<p>I realized I could equivalently write</p>
<p><span class="math-container">$$ I(x) = - \int \frac{dx}{x-1} = -\log|x-1| +C = \log \frac{1}{|1-x|} + C.$$</span></p>
<p>How are these two answers compatible? What am I missing here? </p>
|
Z Ahmed
| 671,540 |
<p><span class="math-container">$$I=\int \frac{dx}{1-x}=-\log (x-1), ~if~ x>1~~~(1)$$</span>
and
<span class="math-container">$$I=\int \frac{dx}{1-x}=-\log (1-x), ~if~ x<1~~~(2)$$</span></p>
<p>By the way <span class="math-container">$$I=\int \frac{dx}{1-x} = - \log |1-x|~~~~(3)$$</span>
will perform like (1) or (2) for the integrals
<span class="math-container">$$\int_{2}^{5} \frac{dx}{1-x}= -\log 4 ~and ~\int_{-3}^{-1} \frac{dx}{1-x}=\log 2$$</span></p>
<p>But when one limit is negative and the other one is positive (3) will yield
<span class="math-container">$$I=\int_{-2}^{3} \frac{dx}{1-x}= \log (3/2),$$</span> which is only the principal value of the integral.</p>
|
3,370,750 |
<p>Show that the moment if inertia of an elliptic area of mass
M
and semi-axis
a
and
b
about a semi-diameter
of length
r
is <span class="math-container">$$\frac{Ma^2b^2}{4r^2}$$</span>.
My attempt.</p>
<ol>
<li>I know that MI about ox is <span class="math-container">${Mb^2 \over 4}$</span>.</li>
<li>MI about oy axis is <span class="math-container">${Ma^2 \over 4}$</span>.
How to proceed further.... </li>
</ol>
|
achille hui
| 59,379 |
<p>Let <span class="math-container">$\theta$</span> be the angle between the axis you wish to compute MI and the <span class="math-container">$x$</span>-axis.</p>
<p>For any point <span class="math-container">$(x,y)$</span> in the plane, its distance to the axis equals to <span class="math-container">$|x\sin\theta - y\cos\theta|$</span>. The desired MI is given by following integral</p>
<p><span class="math-container">$${\rm MI}_\theta \stackrel{def}{=} \rho\int_{\Omega} (x\sin\theta - y\cos\theta)^2 dx dy$$</span>
where <span class="math-container">$\rho$</span> is mass density and <span class="math-container">$\Omega$</span> is the region for the ellipse.</p>
<p>Expand the integrand and notice under transform <span class="math-container">$y \mapsto -y$</span>, <span class="math-container">$\Omega$</span> remains invariant while the cross term proportional to <span class="math-container">$xy$</span> pickup a minus sign. The cross term will not contribute anything to the integral. This leads to</p>
<p><span class="math-container">$$MI_\theta = \left(\rho \int_{\Omega} x^2 dxdy\right) \sin^2\theta + \left(\rho\int_{\Omega} y^2 dx dy\right) \cos^2\theta$$</span>
The two coefficients inside the parentheses are nothing but the MI with respect to <span class="math-container">$y$</span> and <span class="math-container">$x$</span>-axes. This means</p>
<p><span class="math-container">$$MI_{\theta} = \frac{M}{4}((a\sin\theta)^2 + (b\cos\theta)^2)
= \frac{Ma^2b^2}{4}\left(\frac{\cos^2\theta}{a^2} + \frac{\sin^2\theta}{b^2}\right)
$$</span>
Since the axis has semi-diameter <span class="math-container">$r$</span>, when you start at origin and move forward for a distance <span class="math-container">$r$</span>, you will hit the circumference of ellipse. This implies</p>
<p><span class="math-container">$$\frac{(r\cos\theta)^2}{a^2} + \frac{(r\sin\theta)^2}{b^2} = 1 \quad\implies\quad
\frac{\cos^2\theta}{a^2} + \frac{\sin^2\theta}{b^2} = \frac{1}{r^2}$$</span></p>
<p>Substitute this back into above expression of <span class="math-container">${\rm MI}_\theta$</span> and you are done.</p>
|
1,261,504 |
<p>I am trying to proof $ab = \gcd(a,b)\mathrm{lcm}(a,b)$.</p>
<p>The definition of $\mathrm{lcm}(a,b)$ is as follows:</p>
<p>$t$ is the lowest common multiple of $a$ and $b$ if it satisfies the following:</p>
<p>i) $a | t$ and $b | t$ </p>
<p>ii) If $a | c$ and $b | c$, then $t | c$.</p>
<p>Similiarly for the $\gcd(a,b)$.</p>
<p>Here is my proof:</p>
<p>Case I: $\gcd(a,b)\neq 1$</p>
<p>Suppose $\gcd(a,b) = d$.</p>
<p>Then $ab = dq_1b = dbq_1 = d(dq_1q_2)$</p>
<p>Claim: $\mathrm{lcm}(a,b) = dq_1q_2$</p>
<p>$a = dq_1 | dq_1q_2$ </p>
<p>$b = dq_2 | dq_2q_1$.</p>
<p>Supppose $\mathrm{lcm}(a,b) = c$.
Hence $c \leq dq_1q_2$ .</p>
<p>To get the other inequality we have $dq_1 | a$ and $dq_2 | b$. Hence $dq_1 \leq a \leq c \leq dq_1q_2$ similarly for $dq_2$.</p>
<p>Suppose that c is strictly less than $dq_1q_2$, so we have $dq_1q_2 < cq_2$ and $dq_1q_2 < cq_1$.</p>
<p>So $dq_1q_2 < c < cq_2 < dq_2^2q_1$ and $dq_1q_2 < c < cq_2 < dq_1^2q_2$, but $dq_1^2q_2 > dq_1q_2$ so $c < dq_1q_2$ and </p>
<p>$c > dq_1q_2$ contradiction. Hence $c = dq_1q_2$ </p>
<p>Notice that the case where $\gcd(a,b) = 1$ we can just set $q_1 = a$ and $q_2 = b$, and the proof will be the same.</p>
|
callculus42
| 144,421 |
<blockquote>
<p>At the same time, my notes state this formula (with no explanation..)
that $z = \frac{x - \mu}{\sigma}$, where $z$ is a $z-score$. But what
is this and what does it have to even do with these problems?</p>
</blockquote>
<p>Suppose you have a random variable X, which is $ \mathcal N(\mu,\sigma^2 )$ distributed-with $\mu\neq 0$ and/or $\sigma^2\neq 1$. Now you want to calculate $P(X \leq x)$ and you have only a table of a standard normal distribution. Thus the table contains only values of a $\mathcal N(0,1 )$ distributed variable. The following equation shows a useful relation:</p>
<p>$P(X\leq x)=P(Z\leq z)=\Phi \left( \frac{x - \mu}{\sigma}\right)$</p>
<p>$\Phi \left( z\right)$ is the function of the standard normal distribution.</p>
<p>$Z\sim \mathcal N(0,1)$</p>
<p>Example:</p>
<p>Your variable X is $ \mathcal N(4, 9)$ distributed. And you want to calculate the probability, that X is smaller than 5. The equation becomes </p>
<p>$P(X\leq 5)=P(Z\leq \frac 13)=\Phi \left( \frac{5 - 4}{3 }\right)=\Phi (\frac 13)$</p>
<p>If you look at <a href="https://www.stat.tamu.edu/~lzhou/stat302/standardnormaltable.pdf" rel="nofollow">this table</a> you will see, that $P(X\leq 5)\approx \Phi (0.33)=0.6230$</p>
|
68,438 |
<p>I recall reading somewhere that if a conformal class contains an Einstein metric then that metric is the unique metric with constant scalar curvature in its conformal class, with the exception of the case of the round sphere. Does this sound right? If it is true: where can I find the proof of this result? </p>
|
Brian Clarke
| 2,063 |
<p>This is not the original reference, but the most general (i.e., applicable to the non-compact and pseudo-Riemannian cases) result I know is:</p>
<p><a href="http://www.ams.org/mathscinet-getitem?mr=1260173">Kühnel, W., & Rademacher, H. Conformal diffeomorphisms preserving the Ricci tensor. Proceedings of the American Mathematical Society, 123 (1995), no. 9, 2841–2848.</a></p>
<p>(The <a href="http://www.ams.org/journals/proc/1995-123-09/S0002-9939-1995-1260173-6/home.html">article</a> is publicly available for free.)</p>
<p>Theorem 1* in there gives uniqueness in any case other than simply-connected, constant curvature spaces, and certain warped products with Ricci-flat spaces.</p>
<p>Check out the references in that paper, too.</p>
|
3,401,260 |
<p>I am supposed to find the intersection of :
<span class="math-container">$$\begin{cases} 2^{x}=y \\ 31x+8y-94=0 \end{cases}$$</span>
When I substitute the first equation into the second one:
<span class="math-container">$$\frac{94-31x}{8}=2^{x}$$</span> and I do not know how to continue. </p>
<p>Can anyone help me?</p>
|
Dr. Sonnhard Graubner
| 175,066 |
<p>Hint: <span class="math-container">$x$</span> can only be <span class="math-container">$2$</span>, <span class="math-container">$2$</span> solves the equation.</p>
|
3,401,260 |
<p>I am supposed to find the intersection of :
<span class="math-container">$$\begin{cases} 2^{x}=y \\ 31x+8y-94=0 \end{cases}$$</span>
When I substitute the first equation into the second one:
<span class="math-container">$$\frac{94-31x}{8}=2^{x}$$</span> and I do not know how to continue. </p>
<p>Can anyone help me?</p>
|
Cornman
| 439,383 |
<p><span class="math-container">$$\dfrac{94-31x}{8}=2^x$$</span></p>
<p>is correct, you might proceed like this: It is <span class="math-container">$8=2^3$</span> and we get:</p>
<p><span class="math-container">$$94-31x=2^{x+3}$$</span></p>
<p>Note that the RHS is always positive, while the LHS is negative when <span class="math-container">$x>3$</span>, so there are not many values of <span class="math-container">$x$</span> to check.</p>
|
2,226,337 |
<p>How many multiples of $5$ are greater than $60,000,$ and can be made from the digits:
$$0, 1, 2, 3, 4, 5, 6$$ </p>
<p>if <strong>all</strong> digits have to be used and each can only be used once with no repeats.</p>
<p>Am I looking at this in the wrong way too simplistically or is it simply $2 \times 6!$</p>
<p>Many thanks</p>
<p>KM</p>
|
Kanwaljit Singh
| 401,635 |
<p>First digit can't be 0. So we have two cases.</p>
<p>Case 1 -</p>
<p>Number start with 5. Then it should be end with 0.</p>
<p>So we have,</p>
<p>$1 \times 5 \times 4 \times 3 \times 2 \times 1 \times 1$</p>
<p>Case 2 - </p>
<p>Number start with any digit except 5 or 0. Then it should be end with either 5 or 0.</p>
<p>So we have,</p>
<p>$5 \times 5 \times 4 \times 3 \times 2 \times 1 \times 2$</p>
|
850,390 |
<p>Let $f(x)$ be differentiable function from $\mathbb R$ to $\mathbb R$, If $f(x)$ is even, then $f'(0)=0$. Is it always true?</p>
|
gniourf_gniourf
| 51,488 |
<p>Let $f$ be an even function defined on a (symmetric) neighborhood of $0$ and differentiable at $0$. From the definition of differentiability:
$$f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}h.$$
By composition, we also have:
$$f'(0)=\lim_{h\to0}\frac{f(-h)-f(0)}{-h}$$
and using the fact that $f$ is even, this equality can be written as:
$$f'(0)=\lim_{h\to0}-\frac{f(h)-f(0)}h=-f'(0).$$
Hence $f'(0)=0$.</p>
|
2,403,404 |
<p>I would be thankful if anyone can answer my question. This is a very basic question. Let's say we wish to minimise the quantity</p>
<p>$$\hat{h}= \|h-h_i\|+\lambda\|h-u\|,$$</p>
<p>where:</p>
<p>$$h=[13,17,20, 17, 20, 14, 17, 18, 16, 15, 15, 12, 19, 13, 17, 13]^\top,\\ h_i=[18, 17, 14, 13, 17, 15, 17, 19, 12, 20, 15, 13, 16, 17, 20, 13]^\top, \\u = [16, 16,16,16,16,16,16,16,16,16,16,16,16,16,16,16]^\top,\\
\text{with }\lambda \in [0,100].$$</p>
<p>I know this is a very basic question, but please help me to understand. Also, please suggest me any book where I can start from zero to learn to solve these kinds of problems.</p>
|
Michael Rozenberg
| 190,319 |
<p>I think your way is not true. Why the minimum occurs for right-angled triangle? </p>
<p>I think we can solve your problem by the following way.
Let $CD$ be an altitude of the triangle and $BD=x$.</p>
<p>Thus,
$$CD=\frac{2\cdot12}{6}=4$$ and by Minkowski (triangle inequality) we obtain:
$$AC+BC=\sqrt{(6-x)^2+4^2}+\sqrt{x^2+4^2}\geq\sqrt{(6-x+x)^2+(4+4)^2}=10.$$
The equality occurs for $(6-x,4)||(x,4)$, which happens for $x=3$.</p>
<p>Id est, we got a minimal value.</p>
<p>Done!</p>
|
4,567,390 |
<p>Let</p>
<ul>
<li><span class="math-container">$X$</span> be a metric space,</li>
<li><span class="math-container">$\mathcal C_b(X)$</span> the space of real-valued bounded continuous functions,</li>
<li><span class="math-container">$\mathcal C_0(X)$</span> the space of real-valued continuous functions that vanish at infinity, and</li>
<li><span class="math-container">$\mathcal C_c(X)$</span> the space of real-valued continuous functions with compact supports.</li>
</ul>
<p>Then <span class="math-container">$\mathcal C_b(X)$</span> and <span class="math-container">$\mathcal C_0(X)$</span> are real Banach space with supremum norm <span class="math-container">$\|\cdot\|_\infty$</span>. It's mentioned <a href="https://math.stackexchange.com/questions/184766/when-is-c-0x-separable?rq=1">here</a> that <em>"if <span class="math-container">$E$</span> is locally compact and separable, then <span class="math-container">$\mathcal C_0 (E)$</span> is separable"</em>. Now I would like to prove a somehow reverse direction, i.e.,</p>
<blockquote>
<p><strong>Theorem:</strong> If <span class="math-container">$X$</span> is locally compact and <span class="math-container">$\mathcal C_c (X)$</span> is separable, then <span class="math-container">$X$</span> is separable.</p>
</blockquote>
<p>Could you have a check on my below attempt?</p>
<hr />
<p><strong>Proof:</strong> Because <span class="math-container">$\mathcal C_0(X)$</span> <a href="https://math.stackexchange.com/questions/4566771/mathcal-c-0x-is-the-closure-of-mathcal-c-cx-in-mathcal-c-bx-is-l">is the closure</a> of <span class="math-container">$\mathcal C_c(X)$</span> in <span class="math-container">$\mathcal C_b(X)$</span>, it suffices to show that</p>
<blockquote>
<p>If <span class="math-container">$X$</span> is locally compact and <span class="math-container">$\mathcal C_0 (X)$</span> is separable, then <span class="math-container">$X$</span> is separable.</p>
</blockquote>
<p>Let <span class="math-container">$\mathcal M(X)$</span> the space of finite signed Radon measures on <span class="math-container">$X$</span>. Let <span class="math-container">$[\cdot]$</span> be the total variation norm on <span class="math-container">$\mathcal M(X)$</span>. We define a map
<span class="math-container">$$
f:X \to \mathcal M(X), x \mapsto \delta_x.
$$</span></p>
<p>Notice that <span class="math-container">$f(X)$</span> is a subset of the closed unit ball of <span class="math-container">$\mathcal M (X)$</span>. Let <span class="math-container">$E := \mathcal C_0 (X)$</span>. By <a href="https://math.stackexchange.com/questions/4500358/3-versions-of-riesz-markov-kakutani-theorem">Riesz–Markov–Kakutani theorem</a>, <span class="math-container">$(\mathcal M(X), [\cdot])$</span> is isometrically isomorphic to <span class="math-container">$E^*$</span> though a canonical map <span class="math-container">$\Phi:\mathcal M(X) \to E^*$</span>. Let <span class="math-container">$B$</span> be the closed unit ball of <span class="math-container">$E^*$</span>. Clearly, <span class="math-container">$\Phi \circ f (X) \subset B$</span>.</p>
<p>By Banach-Alaoglu's theorem, <span class="math-container">$B$</span> is compact in the weak<span class="math-container">$^*$</span> topology <span class="math-container">$\sigma(E^*, E)$</span>. Because <span class="math-container">$E$</span> is separable, the subspace topology <span class="math-container">$\sigma_B(E^*, E)$</span> that <span class="math-container">$\sigma(E^*, E)$</span> induces on <span class="math-container">$B$</span> <a href="https://math.stackexchange.com/questions/4546596/let-e-be-a-separable-banach-space-then-e-star-is-metrizable-in-the-wea?rq=1">is metrizable</a>. It follows that <span class="math-container">$\sigma_B(E^*, E)$</span> is compact metrizable and thus separable.</p>
<p>If one proves that <span class="math-container">$\Phi \circ f$</span> is a <a href="https://math.stackexchange.com/questions/4567328/the-bijection-in-riesz-markov-kakutani-theorem-is-a-homeomorphism-w-r-t-both-no">homeomorphism</a> from <span class="math-container">$X$</span> (together with metric topology) onto <span class="math-container">$\Phi \circ f(X)$</span> (together with its subspace topology induced by <span class="math-container">$\sigma_B(E^*, E)$</span>), It would follow that <span class="math-container">$X$</span> is separable.</p>
|
Oliver Díaz
| 121,671 |
<p>To show that <span class="math-container">$\Phi\circ f: X\rightarrow\Phi(f(X))$</span> is a homeomorphism it seems that is enough to show that for any sequence <span class="math-container">$(x_m:m\in\mathbb{N})\subset X$</span>, <span class="math-container">$\delta_{x_m}\stackrel{v}{\longrightarrow}\delta_x$</span> iff <span class="math-container">$x_m\xrightarrow{m\rightarrow\infty} x$</span> in <span class="math-container">$X$</span>.</p>
<p>Sufficiency is obvious. As for necessity, I suggest the OP to consider a sequence of open and relatively compact neighborhoods <span class="math-container">$V_n$</span> around <span class="math-container">$x$</span> such that
<span class="math-container">$V_{n+1}\subset\overline{V_{n+1}}\subset V_n$</span>, and <span class="math-container">$\operatorname{diam}(V_n)\xrightarrow{n\rightarrow\infty}0$</span>.</p>
<p>Then define functions <span class="math-container">$f_n\in\mathcal{C}_{00}(X)$</span> with <span class="math-container">$0\leq f_n\leq 1$</span> such that <span class="math-container">$f_{n+1}=1$</span> on <span class="math-container">$\overline{V_{n+1}}$</span> and <span class="math-container">$f_n=0$</span> on <span class="math-container">$X\setminus V_n$</span>. Then
If <span class="math-container">$\delta_{x_m}\stackrel{v}{\rightarrow}\delta_x$</span>, for any <span class="math-container">$n$</span>,
<span class="math-container">$f_n(x_m)\xrightarrow{m\rightarrow\infty} f_n(x)=1$</span>. This means that for all <span class="math-container">$m$</span> large enough, the <span class="math-container">$x_m$</span> are close to <span class="math-container">$x$</span>, i.e. <span class="math-container">$x_m\xrightarrow{m\rightarrow\infty}x$</span>.</p>
|
3,637,526 |
<p>I'm trying to understand the proof of Theorem 16.10, Probability and Measure, Patrick Billingsley, I put part of it here exactly as presented in the book</p>
<p>Theorem: If <span class="math-container">$f,g$</span> are nonnegative and <span class="math-container">$\int_Afd\mu=\int_Agd\mu$</span> for all <span class="math-container">$A$</span> in <span class="math-container">$\mathscr{F}$</span>, and <span class="math-container">$\mu$</span> is <span class="math-container">$\sigma$</span>-finite, then <span class="math-container">$f=g$</span> almost everywhere</p>
<p>Proof: Suppose that <span class="math-container">$f$</span> and <span class="math-container">$g$</span> are nonnegative and that <span class="math-container">$\int_Afd\mu\leq\int_Agd\mu$</span> for all <span class="math-container">$A$</span> in <span class="math-container">$\mathscr{F}$</span>. If <span class="math-container">$\mu$</span> is <span class="math-container">$\sigma$</span>-finite, there are <span class="math-container">$\mathscr{F}$</span>-sets <span class="math-container">$A_n$</span> such that <span class="math-container">$A_n\uparrow\Omega$</span> and <span class="math-container">$\mu(A_n)<\infty$</span>. If <span class="math-container">$B_n = [0\leq g<f, g\leq n]$</span>, then the hypothesized inequality applied to <span class="math-container">$A_n\cap B_n$</span> implies
<span class="math-container">$\int_{A_n\cap B_n}fd\mu\leq\int_{A_n\cap B_n}gd\mu< \infty$</span> (finite beacuse <span class="math-container">$A_n\cap B_n$</span> has finite measure and <span class="math-container">$g$</span> is bounded there) and hence <span class="math-container">$\int I_{A_n\cap B_n}(f-g)d\mu=0 \ldots$</span> (the proof continues)</p>
<p>I understand everything that follows except from one part when the author uses the fact that <span class="math-container">$B_n = [0\leq g<f, g\leq n]$</span> is a measurable set. Why is this a measurable set? Thanks in advance</p>
|
Sam
| 584,704 |
<p>The set <span class="math-container">$B_n$</span> can be written as <span class="math-container">$B_n = \{g \gt 0\}\cap\{g \lt f\} \cap\{g \le n\}$</span>. The first and third sets are measurable by definition. For the second one, notice that <span class="math-container">$g < f \iff \exists r \in \mathbb Q$</span> such that <span class="math-container">$g < r < f$</span>, so
<span class="math-container">$$\{g \lt f\} = \bigcup_{r\in\mathbb Q} \{g < r\}\cap\{f > r\}$$</span>
Which is a countable union of measurable sets.</p>
|
4,427,651 |
<p>In programming, we define an "array" (basically an ordered n-tuple) in the following way:</p>
<p><span class="math-container">$$a=[3,5].$$</span></p>
<p>Later on, if we want to refer to the first element of the predetermined array/pair/n-tuple, we write <span class="math-container">$a[0]$</span> (because in programming you start counting from <span class="math-container">$0$</span>, not <span class="math-container">$1$</span>), which will be equal to <span class="math-container">$3$</span>.</p>
<p>Is there a similar notation for doing this in mathematics? E.g something like <code>(3,5).firstElement</code>, which would equal <span class="math-container">$3$</span>? Or is such a reference only possible by using the set-theoretic definition of an ordered pair?</p>
|
Joe
| 623,665 |
<p>If <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> are sets, the function <span class="math-container">$f:X\times Y\to X$</span> given by <span class="math-container">$f(x,y)=x$</span> is called the <em>projection from <span class="math-container">$X\times Y$</span> onto <span class="math-container">$X$</span></em>; similarly, the function <span class="math-container">$g:X\times Y\to Y$</span> given by <span class="math-container">$g(x,y)=y$</span> is called the <em>projection from <span class="math-container">$X\times Y$</span> onto <span class="math-container">$Y$</span></em>. A variety of notations for projections are used, but, at least in the case <span class="math-container">$X=Y=\mathbb R$</span>, it seems common to write <span class="math-container">$\pi^1$</span> for <span class="math-container">$f$</span> and <span class="math-container">$\pi^2$</span> for <span class="math-container">$g$</span> (see for instance Spivak's <em>Calculus on Manifolds</em>). Using this notation, we can write <span class="math-container">$\pi^1(3,5)=3$</span> and <span class="math-container">$\pi^2(3,5)=5$</span>.</p>
<p>More generally, if <span class="math-container">$n$</span> is a positive integer, then the function <span class="math-container">$\pi^i:\mathbb R^n\to \mathbb R$</span> given by <span class="math-container">$\pi^i(x_1,\dots,x_n)=x_i$</span> is called the <em><span class="math-container">$i$</span>th projection function</em>. Although this is technically a different function for each value of <span class="math-container">$n$</span>, in practice we gloss over this point.</p>
|
512,591 |
<p>It is always confusing to prove with $\not\equiv$. Should I try contrapositive?</p>
|
Fly by Night
| 38,495 |
<p>If $a \not\equiv 0 \bmod 3$ then either $a \equiv 1 \bmod 3$ or $a \equiv 2 \bmod 3$.</p>
<ul>
<li>If $a \equiv 1 \bmod 3$ then $a^2 \equiv 1^2 = 1\equiv 1 \bmod 3$.</li>
<li>If $a \equiv 2 \bmod 3$ then $a^2 \equiv 2^2 = 4 \equiv 1 \bmod 3$.</li>
</ul>
<p>Hence, if $a \not\equiv 0 \bmod 3$ then $a^2 \equiv 1 \bmod 3$.</p>
|
1,638,051 |
<p>$$\int\frac{dx}{(x^{2}-36)^{3/2}}$$</p>
<p>My attempt:</p>
<p>the factor in the denominator implies</p>
<p>$$x^{2}-36=x^{2}-6^{2}$$</p>
<p>substituting $x=6\sec\theta$, noting that $dx=6\tan\theta \sec\theta$ </p>
<p>$$x^{2}-6^{2}=6^{2}\sec^{2}\theta-6^{2}=6^{2}\tan^{2}\theta$$</p>
<p>$$\int\frac{dx}{(x^{2}-36)^{3/2}}=\int\frac{6\tan\theta \sec\theta}{36\tan^{2}\theta}=\frac{1}{6}\int\frac{\sec\theta}{\tan\theta}$$</p>
<p>using trig identities:
$$\frac{1}{6}\int\frac{\sec\theta}{\tan\theta}=\frac{1}{6}\int \sin^{-1}\theta$$</p>
<p>now using integration by parts:
$$\frac{1}{6}\int \sin^{-1}\theta$$
$$u=\sin^{-1}\theta, du=\frac{1}{\sqrt{1-\theta^{2}}}, dv=1, v=\theta$$
using $uv-\int{vdu}$</p>
<p>$$\frac{1}{6}\bigg(\theta \sin^{-1}\theta-\int{\frac{\theta}{\sqrt{1-\theta^{2}}}}d\theta\bigg)$$</p>
<p>now using simple substitution:$$z=1-\theta^{2}, dz=-2\theta d\theta, -\frac{1}{2}du=\theta d\theta$$</p>
<p>it is apparent that</p>
<p>$$\frac{1}{6}\bigg(\theta \sin^{-1}\theta-\bigg(-\frac{1}{2}\int{\frac{dz}{\sqrt{z}}}\bigg)\bigg)$$</p>
<p>$$=\frac{1}{6}\bigg(\theta \sin^{-1}\theta-\bigg(-\frac{1}{2}(2\sqrt{z})\bigg)\bigg)=\frac{1}{6}\bigg(\theta \sin^{-1}\theta+\sqrt{1+\theta^{2}}\bigg)$$</p>
<p>$$=\frac{1}{6}\theta \sin^{-1}\theta+\frac{1}{6}\sqrt{1+\theta^{2}}+C$$</p>
<p>I have the following questions:</p>
<p>1.This integral seems tricky and drawn out to me, is there another method that reduces the steps/ methods of integration? I had to use trig substitution, integration by parts, and substitution in order to solve the integral, what can I do to find easier ways to complete integrals of this type?</p>
<p>2.Is this solution even correct? wolfram alpha says the solution to this integral is $-\frac{x}{36\sqrt{x^{2}-36}}+C$ how can i determine equivalence?</p>
|
Nikunj
| 287,774 |
<p>Your solution is incorrect as $$\frac{\sec\theta}{\tan\theta}$$ is not the same as $$\ sin^{-1}\theta$$ It is however, equal to $\csc\theta$, a standard integral that you had got after a few steps</p>
|
2,683,326 |
<p>I have a function $f(x)$ whose second order Taylor expansion is represented by $f_2(x)$. Is it true that $$f(x)>f_2(x)$$ for all $x$? Any help in this regard will be much appreciated. Thanks in advance.</p>
|
NewGuy
| 518,506 |
<p>similar to what @hardmath said </p>
<p>you can break the line into predetermined equal segments for example (10,20,....,40,...100 etc)</p>
<p>Example
lets break the line into 20 segments, the total number of ways 6 line segments can be selected is </p>
<p>total number of positive integers in the equation</p>
<p>$x_1+x_2+x_3+x_4+x_5+x_6 = 20$</p>
<p>= $\binom{19}{5}$</p>
<p>To get our event E: one segment is greater >= 10</p>
<p>let x_1 be that segment and possible cases for X_1 can be (10|11|12|13|14|15) </p>
<p>It can not be more than 15 because in that case value of other segment has to be 0 which is not possible.</p>
<p>when $x_1 = 10 : x_2+x_3+x_4+x_5+x_6 =10 : \binom{9}{4}$</p>
<p>$x_1 = 11 : x_2+x_3+x_4+x_5+x_6 =9 : \binom{8}{4}$</p>
<p>$x_1 = 12 : x_2+x_3+x_4+x_5+x_6 =8 : \binom{7}{4}$</p>
<p>$x_1 = 13 : x_2+x_3+x_4+x_5+x_6 =7 : \binom{6}{4}$</p>
<p>$x_1 = 14 : x_2+x_3+x_4+x_5+x_6 =6 : \binom{5}{4}$</p>
<p>$x_1 = 15 : x_2+x_3+x_4+x_5+x_6 =5 : \binom{4}{4}$</p>
<p>Total number of cases for event (E): 6*($\binom{4}{4}$+$\binom{5}{4}$+$\binom{6}{4}$+$\binom{7}{4}$+$\binom{8}{4}$+$\binom{9}{4}$) = 1512</p>
<p>Probability of event E: $\frac{1512}{11628}$ = .13</p>
<p>P(E) = .152 in case of 30 segments</p>
<p>P(E) = .167 in case of 50 segments</p>
<p>We can see that the probaility of E converges to some number A</p>
<p>our answer = 1 - A</p>
|
1,261,067 |
<p>$$\int \left(\frac15 x^3 - 2x + \frac3x + e^x \right ) \mathrm dx$$</p>
<p>I came up with
$$F=x^4-x^2+\frac{3x}{\frac12 x^2}+e^x$$
but that was wrong.</p>
|
MonkeyKing
| 225,981 |
<p>The first term and the third terms are wrong.</p>
<p>First term: $\int \frac{1}{5}x^3 dx= \frac{x^4}{20}$. Try differentiating and check $\int x^p dx = \frac{1}{p+1} x^{p+1}$ if $p \neq -1$.</p>
<p>Third term: $\frac{3}{x} = 3x^{-1}$, and $\int x^{-1} = \ln x$, so $\int 3x^{-1} = 3\ln x$.</p>
<p>Finally, don't forget to <a href="http://spikedmath.com/434.html" rel="nofollow">PLUS $c$</a>!</p>
|
1,261,067 |
<p>$$\int \left(\frac15 x^3 - 2x + \frac3x + e^x \right ) \mathrm dx$$</p>
<p>I came up with
$$F=x^4-x^2+\frac{3x}{\frac12 x^2}+e^x$$
but that was wrong.</p>
|
Hasan Saad
| 62,977 |
<p>$\int cx^n=c\int x^n=c\frac{x^{n+1}}{n+1}$ when $n\neq -1$</p>
<p>Thus, $\int \frac{1}{5}x^3=\frac{1}{20}x^4$</p>
<p>Also, by the same rule, $\int -2x=-x^2$</p>
<p>$\int \frac{1}{x}=\ln(x)$</p>
<p>Thus, $\int \frac{3}{x}=3\ln(x)$</p>
<p>$\int e^x=e^x$</p>
<p>Thus, by above, the required integral is </p>
<p>$F(x)=\frac{1}{20}x^4-x^2+3\ln(x)+e^x+c$</p>
<p>Done.</p>
|
1,176,958 |
<p>I've been struggling with this for over an hour now and I still have no good results, the question is as follows:</p>
<blockquote>
<p>What's the probability of getting all the numbers from $1$ to $6$ by rolling $10$ dice simultaneously?</p>
</blockquote>
<p>Can you give any hints or solutions? This problem seems really simple but I feel like I'm blind to the solution.</p>
|
drhab
| 75,923 |
<p><strong>Hint</strong>:</p>
<p>If $X_{i}$ denotes the number of dice that show face $i$ then it
equals:</p>
<p>$1-P\left[X_{1}=0\vee X_{2}=0\vee X_{3}=0\vee X_{4}=0\vee X_{5}=0\vee X_{6}=0\right]$</p>
<p>Now apply <a href="http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle" rel="nofollow">inclusion/exclusion</a>.</p>
<hr>
<p>edit:</p>
<p>It leads to:$$1-\binom{6}{1}\left(\frac{5}{6}\right)^{10}+\binom{6}{2}\left(\frac{4}{6}\right)^{10}-\binom{6}{3}\left(\frac{3}{6}\right)^{10}+\binom{6}{4}\left(\frac{2}{6}\right)^{10}-\binom{6}{5}\left(\frac{1}{6}\right)^{10}=\frac{38045
}{139968}$$</p>
|
387,749 |
<p>This comes from Guillemin and Pollack's book Differential Topology. The book claims that one cannot parametrize a unit circle by a single map. I thought we could (by a single angle $\theta$). </p>
<p>I think one possible answer might be the fact that if we let $\theta$ in [0, 2$\pi$), when $\theta$ approaches 2$\pi$, $f(\theta)$ approaches $f(0)$. But is this a problem?</p>
<p>Also for $n$-spheres, why is there always a point that can't be covered by a single map (I know we can always cover it by two maps, one being the stereographic projection)?</p>
<p>Thanks a lot for your help!</p>
|
Henry T. Horton
| 24,934 |
<p>Let's look more closely at Guillemin and Pollack's definition of parametrization:</p>
<blockquote>
<p>Let $X$ be a subset of $\Bbb R^N$. A <strong>parametrization</strong> is a diffeomorphism $\phi: U \longrightarrow V$ from an open set $U \subset \Bbb R^k$ to an open subset $V \subset X$ (where we give $X$ the subspace topology inherited from $\Bbb R^N$).</p>
</blockquote>
<p>So we notice two things. First, a parametrization must be one-to-one (since it is a diffeomorphism) and it must be defined on an open subset of $\Bbb R^k$.</p>
<p>In your proposed parametrization, you have $U = [0, 2\pi)$, which is not open in $\Bbb R$. We at least need something like $U = (0, 2\pi + \varepsilon)$ for some $\varepsilon > 0$. But once we extend $U$ this far, the parametrization by angle is no longer injective, since the angles $\tfrac{\varepsilon}{2}$ and $2\pi + \tfrac{\varepsilon}{2}$, which are both in $(0, 2\pi + \varepsilon)$, correspond to the same point on the circle.</p>
<p>Note that if we could use a single parametrization $\phi: (a, b) \longrightarrow S^1$, then $S^1$ would be diffeomorphic to $\Bbb R$, which you likely know is not true.</p>
<p>Similarly, if we could use a single parametrization $\phi: U \longrightarrow S^n$, where $U \subset \Bbb R^n$ is open, then $S^n$ would be diffeomorphic to the open subset $U$ of $\Bbb R^n$, which is impossible since $S^n$ is compact.</p>
|
1,042,212 |
<p>If $\mathbb{Z}_p \leq K$ an algebraic extension, then $K$ has the identity $$\forall a \in K, \exists b \in K \text{ with } a=b^p$$</p>
<p>The proof is the following:</p>
<p>Let $a \in K$.</p>
<p>We take $\mathbb{Z}_p \leq \mathbb{Z}_p(a)$, $a$ algebraic over $\mathbb{Z}_p$.</p>
<p>So, $[\mathbb{Z}_p(a) : \mathbb{Z}_p ]=n<\infty$, so $dim_{\mathbb{Z}_p}\mathbb{Z}_p(a)=n<\infty$.</p>
<p>So, #$\mathbb{Z}_p(a)=p^n$, that means that it is a finite field of characteristic $p$.</p>
<p>From that it follows that $$\exists b \in \mathbb{Z}_p(a) \subseteq K \text{ with } a=b^p$$</p>
<p>Do we have that $[\mathbb{Z}_p(a) : \mathbb{Z}_p ]=n<\infty$ because of the fact that $a$ algebraic over $\mathbb{Z}_p$ ??</p>
<p>Also could you explain me how we conclude that #$\mathbb{Z}_p(a)=p^n$ ??</p>
|
Julián Aguirre
| 4,791 |
<p>Another example. Let $\Omega=(0,1)\subset\mathbb{R}$ and $f_n(x)=\sin(n\,\pi\,x)/n$. Then
$$
\|f_n\|_2\le\frac1n\to0\text{ as }n\to\infty.
$$
On the other hand
$$
\|f'_n\|_2^2=\pi^2\int_0^1\cos^2(n\,\pi\,x)\,dx=\frac{\pi^2}{2}.
$$</p>
|
1,042,212 |
<p>If $\mathbb{Z}_p \leq K$ an algebraic extension, then $K$ has the identity $$\forall a \in K, \exists b \in K \text{ with } a=b^p$$</p>
<p>The proof is the following:</p>
<p>Let $a \in K$.</p>
<p>We take $\mathbb{Z}_p \leq \mathbb{Z}_p(a)$, $a$ algebraic over $\mathbb{Z}_p$.</p>
<p>So, $[\mathbb{Z}_p(a) : \mathbb{Z}_p ]=n<\infty$, so $dim_{\mathbb{Z}_p}\mathbb{Z}_p(a)=n<\infty$.</p>
<p>So, #$\mathbb{Z}_p(a)=p^n$, that means that it is a finite field of characteristic $p$.</p>
<p>From that it follows that $$\exists b \in \mathbb{Z}_p(a) \subseteq K \text{ with } a=b^p$$</p>
<p>Do we have that $[\mathbb{Z}_p(a) : \mathbb{Z}_p ]=n<\infty$ because of the fact that $a$ algebraic over $\mathbb{Z}_p$ ??</p>
<p>Also could you explain me how we conclude that #$\mathbb{Z}_p(a)=p^n$ ??</p>
|
gerw
| 58,577 |
<p>You can take any orthonormal sequence $\{f_n\}$ in $H^1$. Then you have $f_n \rightharpoonup 0$, but $\|f_n\|_{H^1} = 1$. Using the compact embedding from $H^1$ to $L^2$, you have $f_n \to 0$ in $L^2$.</p>
<p>More generally, you can construct a counterexample by using any weakly but not strongly convergent sequence in $H^1$.</p>
|
2,154,608 |
<blockquote>
<p>Let $a$, $b$ and $c$ be non-negative numbers such that $a^3+b^3+c^3=3$. Prove that:
$$a^4b+b^4c+c^4a\leq3$$</p>
</blockquote>
<p>This inequality similar to the following.</p>
<blockquote>
<p>Let $a$, $b$ and $c$ be non-negative numbers such that $a^2+b^2+c^2=3$. Prove that:
$$a^3b+b^3c+c^3a\leq3,$$
which follows from the following identity.
$$(a^2+b^2+c^2)^2-3(a^3b+b^3c+c^3a)=\frac{1}{2}\sum_{cyc}(a^2-b^2-ab-ac+2bc)^2.$$</p>
</blockquote>
<p>I tried Rearrangement.</p>
<p>Let $\{a,b,c\}=\{x,y,z\}$, where $x\geq y\geq z$.</p>
<p>Hence, $$a^4b+b^4c+c^4a=a^3\cdot ab+b^3\cdot bc+c^3\cdot ca\leq x^3\cdot xy+y^3\cdot xz+z^3\cdot yz=$$
$$=y(x^4+y^2xz+z^4)$$
and I don't see what is the rest.</p>
<p>Thank you!</p>
|
Oscar Cunningham
| 1,149 |
<p>In any regular $4$-polytope, three or more regular polyhedra meet at each edge. The angles between faces on these polyhedra have to add to less than $2\pi$ (in the same way that angles meeting at a vertex in a polyhedron must add to less than $2\pi$). By calculating the angles between faces in the regular polyhedra, students can prove that the number of tetrahedra meeting at an edge could be $3$, $4$ or $5$, but the number of cubes, octohedra, or dodecahedra meeting at an edge can only be $3$. Icosahedra can't be made to fit around an edge even if there are only $3$ of them.</p>
<p>This shows that there are at most six regular $4$-polytopes, three made of tetrahedra, and one made of each of cubes, octohedra, and dodecohedra. The hardest part is showing that these exist. Clever students might be able to produce a list of $4$-dimensional coordinates and prove that these coordinates make the requisite shape, in the same way that the following coordinates give a dodecahedron:
$$(\pm 1,\pm 1,\pm 1)\\
(\pm\phi,\pm1/\phi,0)\\
(0,\pm\phi,\pm1/\phi)\\
(\pm1/\phi,0,\pm\phi)$$</p>
|
697,402 |
<p>I have this limit:</p>
<p>$$ \lim_{x\to\infty}\frac{x^3+\cos x+e^{-2x}}{x^2\sqrt{x^2+1}} $$ I tried to solve it by this:</p>
<p>$$ \lim_{x\to\infty}\frac{x^3+\cos x+e^{-2x}}{x^2\sqrt{x^2+1}} = \lim_{x\to\infty}\frac{\frac{x^3}{x^3}+\frac{\cos x}{x^3}+\frac{e^{-2x}}{x^3}}{\frac{x^2\sqrt{x^2+1}}{x^3}} = \frac{0+0+0}{\frac{\sqrt{\infty^2+1}}{\infty}}$$ I do not think that I got it right there... Wolfram also says that the answer is $1$, which this does not seems to be. How do I solve this?</p>
|
Aloizio Macedo
| 59,234 |
<p>For sufficiently large $x$</p>
<p>$\displaystyle \frac{x^3-1}{x^2\sqrt{x^2+1}} \leq \frac{x^3+\cos x+e^{-2x}}{x^2\sqrt{x^2+1}} \leq \frac{x^3+2}{x^2\sqrt{x^2+1}}$</p>
<p>Now, </p>
<p>$\displaystyle \frac{2}{x^2\sqrt{x^2+1}} \rightarrow 0$, and $\displaystyle \frac{-1}{x^2\sqrt{x^2+1}} \rightarrow 0$</p>
<p>Moreover,</p>
<p>$\displaystyle \frac{x^3}{x^2\sqrt{x^2+1}}=\frac{x^3}{x^2\sqrt{x^2(1+\frac{1}{x^2})}}=\frac{x^3}{x^3\sqrt{1+\frac{1}{x^2}}}=\frac{1}{1\sqrt{1+\frac{1}{x^2}}} \rightarrow 1$</p>
|
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