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Let $A = \bigl( \begin{smallmatrix}2 & 2 \\ -1 & 5\end{smallmatrix}\bigr)$. Prove for $n \geq 1$ using induction that $A^n =$ ... Can someone check to see if my proof is correct? Feel free to nitpick, trying to get better at writing proofs. Here's the problem:
Let $A = \bigl( \begin{smallmatrix}2 & 2 \\ -1 & 5\end{smallmatrix}\bigr)$. Prove for $n \geq 1$ that $A^n = 4^n \bigl( \begin{smallmatrix}-1 & 2 \\ -1 & 2\end{smallmatrix}\bigr) + 3^n \bigl( \begin{smallmatrix}2 & -2 \\ 1 & -1\end{smallmatrix}\bigr)$ by induction.
Note: $A^n$ means $A$ matrix multiplied by itself $n$ times. So for example, $A^2 = \bigl( \begin{smallmatrix}2 & 2 \\ -1 & 5\end{smallmatrix}\bigr)\bigl( \begin{smallmatrix}2 & 2 \\ -1 & 5\end{smallmatrix}\bigr) = \bigl( \begin{smallmatrix}2 & 14 \\ -7 & 23\end{smallmatrix}\bigr)$
My attempt at the proof:
Base case, $n=1:$ $A^1 = \bigl( \begin{smallmatrix}2 & 2 \\ -1 & 5\end{smallmatrix}\bigr) = 4^1\bigl( \begin{smallmatrix}-1 & 2 \\ -1 & 2\end{smallmatrix}\bigr)+ 3^1\bigl( \begin{smallmatrix}2 & -2 \\ 1 & -1\end{smallmatrix}\bigr) \ \checkmark$
Assume the inductive hypothesis, $n=k: A^k = 4^k\bigl( \begin{smallmatrix}-1 & 2 \\ -1 & 2\end{smallmatrix}\bigr) + 3^k\bigl( \begin{smallmatrix}2 & -2 \\ 1 & -1\end{smallmatrix}\bigr)$
WTS that the proposition holds for $n=k+1: A^{k+1} = 4^{k+1}\bigl( \begin{smallmatrix}-1 & 2 \\ -1 & 2\end{smallmatrix}\bigr) + 3^{k+1}\bigl( \begin{smallmatrix}2 & -2 \\ 1 & -1\end{smallmatrix}\bigr)$
\begin{align*}
A^k A = \left(4^k\bigl( \begin{smallmatrix}-1 & 2 \\ -1 & 2\end{smallmatrix}\bigr) + 3^k\bigl( \begin{smallmatrix}2 & -2 \\ 1 & -1\end{smallmatrix}\bigr)
\right)\bigl( \begin{smallmatrix}2 & 2 \\ -1 & 5\end{smallmatrix}\bigr)
&= 4^k\bigl( \begin{smallmatrix}-1 & 2 \\ -1 & 2\end{smallmatrix}\bigr)\bigl( \begin{smallmatrix}2 & 2 \\ -1 & 5\end{smallmatrix}\bigr) + 3^k\bigl( \begin{smallmatrix}2 & -2 \\ 1 & -1\end{smallmatrix}\bigr)\bigl( \begin{smallmatrix}2 & 2 \\ -1 & 5\end{smallmatrix}\bigr) \\
&= 4^k\bigl( \begin{smallmatrix}-4 & 8 \\ -4 & 8\end{smallmatrix}\bigr)
+ 3^k\bigl( \begin{smallmatrix}6 & -6 \\ 3 & -3\end{smallmatrix}\bigr) \\
&= (4^k)(4)\bigl( \begin{smallmatrix}-1 & 2 \\ -1 & 2\end{smallmatrix}\bigr) +
(3^k)(3)\bigl( \begin{smallmatrix}2 & -2 \\ 1 & -1\end{smallmatrix}\bigr) \\
A^k A &= 4^{k+1}\bigl( \begin{smallmatrix}-1 & 2 \\ -1 & 2\end{smallmatrix}\bigr) + 3^{k+1}\bigl( \begin{smallmatrix}2 & -2 \\ 1 & -1\end{smallmatrix}\bigr) \\
A^k A &= A^{k+1}
\end{align*}
By induction the proposition holds for all $n \geq 1$
|
The proof looks perfect mathematically. Stylistically, it looks pretty good as well, although since you mentioned that you'd like some nitpicking, here goes:
Depending on the context in which you need to write this proof, the phrasing of the induction could be improved a little bit. (If this is for a class, and you've been taught to write induction arguments in a very particular manner, then of course you do not want to mess with it.) In particular, proofs are generally easier to read when written in sentence form, and so the flow of the argument might be improved by adding a few more words. What follows are the words I'd be likely to use, if writing the proof in a very detailed manner. This is of course not the only way to phrase the argument, and as you write more proofs you'll develop your own mathematical "tone." (In particular, you may read this as a bit overkill.)
We show the result by induction on $n$. For the base case of $n=1$, we have
$$\text{[computation]}$$.
Now, we show the inductive step. Assume that the proposition holds for some integer $n=k$, i.e. that [statement]. We wish to show that the proposition holds for $n=k+1$, i.e. that [statement]. Indeed,
$$\text{[computation]},$$
where we have used the inductive hypothesis in the first step. We have thus proven the proposition for all $n\geq 1$ by induction.
I'd also recommend moving the note that $A^kA=A^{k+1}$ to the beginning of the inductive-step computation, so that the computation reads
\begin{align*}
A^{k+1}&=A^kA\\
&=\ldots\\
&=4^{k+1}\bigl(\begin{smallmatrix}-1&2\\-1&2\end{smallmatrix}\bigr)+3^{k+1}\bigl(\begin{smallmatrix}2&-2\\1&-1\end{smallmatrix}\bigr)
\end{align*}
and the reader can see directly, just from glancing at the computation, that it is a sequence of steps verifying exactly what you want it to.
(Lastly, I'd also use \begin{pmatrix}\end{pmatrix} over \bigl\begin{smallmatrix}\end{smallmatrix}\bigr -- it's not like you have to pay for space and I think it looks less cluttered -- but this is purely a personal preference.)
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find all polynomials $P, Q$ such that $(x^2+ax+b)P(x^2+cx+d)=(x^2+cx+d)Q(x^2+ax+b)$
Find all polynomials $P(x), Q(x)$ such that
$(x^2+ax+b)P(x^2+cx+d)=(x^2+cx+d)Q(x^2+ax+b)$
where $a, b, c, d$ are all different real numbers.
I found this problem while solving functional equations in polynomials. Here's what I've found so far:
Suppose $u$ is a root of $x^2+ax+b$ which is not a root of $x^2+cx+d$. Then, substituting $x=u$ gives $Q(0)=0$.
Similarly, suppose $v$ is a root of $x^2+cx+d$ which is not a root of $x^2+ax+b$. Then, substituting $x=v$ gives $P(0)=0$.
Therefore, $0$ is a root of both $P$ and $Q$, so we can write $P(x)=xf(x)$ and $Q(x)=xg(x)$ for some polynomials $f, g$.
The original equation then becomes
$h(x)f(x^2+cx+d)=h(x)g(x^2+ax+b)$, where
$h(x)=(x^2+ax+b)(x^2+cx+d)$.
After canceling out $h(x)$, we get
$f(x^2+cx+d)=g(x^2+ax+b)$.
However, I can't find a way to make further progress. What should I do to finish solving this problem?
|
As Ng Chung Tak said in the comments above, the only solutions are $P = Q = \lambda X$, $\lambda\in \mathbb{R}$, if $a \neq c$ (though the answer seems more complicated for $a = c$, but this is not the object of this question).
Let $(P,Q) \in \mathbb{R}[X]^2$ be a solution of our polynomial equation, equation which we'll name $(*)$.
By taking the degrees in $(*)$, we get:
$$2 + 2\deg(P) = 2 + 2\deg(Q) \quad\Rightarrow\quad \deg(P) = \deg(Q) =: n$$
Similarly, by looking at the dominant coefficients of the polynomials on each side of $(*)$, we also have that $P$ and $Q$ have the same leading coefficient $\lambda$.
If $n = -\infty$, and thus $P = Q = 0$, we are done, and it is easily shown that $n$ can't be $0$ otherwise $a = c$ and $b = d$, so we can suppose $n \geq 1$.
Let $P =: \sum_{k = 0}^n \alpha_k X^k$ and $Q =: \sum_{k = 0}^n \beta_k X^k$ with $\alpha_n = \beta_n = \lambda \neq 0$. We then obtain:
$$\begin{align}(X^2 + aX + b)P(X^2 + cX + d) &= (X^2 + aX + b)\sum_{k=0}^n \alpha_k (X^2 + cX + d)^k =: S \\ (X^2 + cX + d)Q(X^2 + aX + b) &= (X^2 + cX + d)\sum_{k=0}^n \beta_k (X^2 + aX + b)^k =: T\end{align}$$
Let's look at the coefficient of $S$ that's in front of $X^{2n+1}$.
The highest power of $X$ appearing in $(X^2 + cX + d)^k$ for $k \leq n - 1$ is $X^{2n-2}$, which multiplied by $X^2$ only grants $X^{2n}$, hence only $(X^2 + cX + d)^n$ contributes, and the coefficient of $S$ in front of $X^{2n+1}$ is thus $\lambda (nc + a)$ (the only way to get $X^{2n+1}$ is to use $n$ times a factor in $X^2$ and exactly once a factor in $X$). In the same fashion, the corresponding coefficient for $T$ is $\lambda(na + c)$.
Therefore, since $(*)$ is just $S = T$:
$$\lambda(nc+a) = \lambda(na+c) \quad\Rightarrow\quad nc+a = na+c \quad\Rightarrow\quad n(c-a) = c - a \quad\Rightarrow\quad n = 1$$
As such, we have: $P = \lambda X + \alpha_0$ and $Q = \lambda X + \beta_0$, and so:
$$\begin{align}S &= (X^2 + aX + b) \big(\lambda(X^2 + cX + d) + \alpha_0\big) \\ T &= (X^2 + cX + d) \big(\lambda(X^2 + aX + b) + \beta_0\big)\end{align}$$
Which means that, by simplifying by $\lambda (X^2 + aX + b)(X^2 + cX + d)$:
$$\alpha_0(X^2 + aX + b) = \beta_0(X^2 + cX + d)$$
We can conclude that $\alpha_0 = \beta_0$ by looking at the coefficients in front of $X^2$, and then that $\alpha_0 a = \alpha_0 c$ by looking at $X$, which finally provides: $\alpha_0 = 0$, and so: $P = Q = \lambda X$.
Reciprocally, those are indeed solutions, and they are therefore exactly the solutions to $(*)$ when $a \neq c$ (not even any need to know anything about $b$ and $d$).
The case $a = c, b \neq d$ (which I'm aware is not really part of this question, but I still want to talk briefly about it) is actually way less trivial it seems:
Take $a = c = d = 0, b = -1$. You can prove that for any $\alpha \in \mathbb{R}$, the couple $(X^2 + \alpha X, X^2 + (\alpha + 1)X)$ is solution of $(*)$ (there's not even uniqueness modulo a multiplicative constant per degree in this case, $P$ is different from $Q$, and so on...). I wonder if solutions like this exist for all quadruplets such that $a = c$?
Moreover, the proof for $a \neq c$ can be generalised to all integral domains of characteristic zero, which $\mathbb{R}$ and $\mathbb{C}$ are, but not so much to rings of positive characteristic, though the common degree $n$ can at least be reduced to $1$ modulo said characteristic.
|
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"timestamp": "2023-03-29T00:00:00",
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A better lower bound for $f(t):=((t-a)^2+b)^2+d(t-c)^2$ on $[0,\infty[$ where $a,b,c,d>0$ and $c>a$ Let
$f(t):=((t-a)^2+b)^2+d(t-c)^2$ where $a,b,c,d>0$ and $c>a$, and $t\geq 0$. I am trying to find a better lower bound for $f$ than $b^2$, if it exists.
The parabola $t \mapsto (t-a)^2+b$ has its unique minimum $b$ at $t=a$. The same applies to its square $((t-a)^2+b)^2$. Also, the parabola $t\mapsto d(t-c)^2$ has its unique minimum value $0$ at $t=c$.
If we ignore the term $(t-a)^4$, and look instead at $g(t):=2b(t-a)^2+d(t-c)^2+b^2$, the minimum can be shown to be $b^2$ which is obtained at
$t=(2ab+cd)/(2b+d)$. I don't know if this helps get a sharper answer.
Is there a way around computing the critical point of $f$ ?
|
For a real-valued function, the maximum or minimum is taken where the first derivative is zero. To check whether there is a maximum or a minimum there is to test the sign of the second derivative.
For the set of four parameters as in the given function, the condition that there is an extremum is given by the following latex expression under the condition that $\{a,b,c,d\}$ are all reals.
$
\left\{\left\{t\to \text{ConditionalExpression}\left[\text{Root}\left[-2 a^3-2 a b^2-c d+\left(6 a^2+2 b^2+d \right) \text{ ss1}-6 a \text{ ss1}^2+2
\text{ ss1}^3\&,1\right],\left(9 a-9 c+\sqrt{6} \
\sqrt{-\frac{\left(2 b^2+d \right)^3}{d^2}}>0\&\&-2 b+\sqrt{2} \
\sqrt{-d}>0\&\&2 b+\sqrt{2} \sqrt{-d}>0\&\&d<0\&\&-9
a+9 c+\sqrt{6} \sqrt{-\frac{\left(2 b^2+d \right)^3}{d^2}}>0\right)\
\left\|\left(9 a-9 c+\sqrt{6} \sqrt{-\frac{\left(2 b^2+d \right)^3}{d^2}}<0\&\&-2
b+\sqrt{2} \sqrt{-d}>0\&\&2 b+\sqrt{2} \sqrt{-d}>0\&\&d<0\right)\right \|\left(-2 b+\sqrt{2} \sqrt{-d}>0\&\&2 b+\sqrt{2} \
\sqrt{-d}>0\&\&d<0\&\&-9
a+9 c+\sqrt{6} \sqrt{-\frac{\left(2 b^2+d \right)^3}{d^2}}<0\right)\
\|\left(-2 b+\sqrt{2} \sqrt{-d}<0\&\&d<0\right)\|\left(2 b+\sqrt{2} \
\sqrt{-d}<0\&\&d<0\right)\|d>0\right]\right\},\left\{t\to
\text{ConditionalExpression}\left[\text{Root}\left[-2 a^3-2 a b^2-c \
d+\left(6 a^2+2 b^2+d \right) \text{ ss1}-6 a \text{ ss1}^2+2 \text{ ss1}^3\&,2\right],2
b+\sqrt{2} \sqrt{-d}>0\&\&-2 b+\sqrt{2} \sqrt{-d}>0\&\&9 a-9 \
c+\sqrt{6} \sqrt{-\frac{\left(2 b^2+d \right)^3}{d^2}}>0\&\&-9 a+9 \
c+\sqrt{6} \sqrt{-\frac{\left(2
b^2+d \right)^3}{d^2}}>0\&\&d<0\right]\right\},\left\{t\to \text{ConditionalExpression}\left[\text{Root}\left[-2 a^3-2 a b^2-c \
d+\left(6 a^2+2 b^2+d \right)
\text{ ss1}-6 a \text{ ss1}^2+2 \text{ ss1}^3\&,3\right],2 \
b+\sqrt{2} \sqrt{-d}>0\&\&-2 b+\sqrt{2} \sqrt{-d}>0\&\&9 a-9 \
c+\sqrt{6} \sqrt{-\frac{\left(2
b^2+d \right)^3}{d^2}}>0\&\&-9 a+9 c+\sqrt{6} \sqrt{-\frac{\left(2 \
b^2+d \right)^3}{d^2}}>0\&\&d<0\right]\right\}\right\}
$
The immense effort stems from the fourth degree of the function is.
Simply pluck the t-values and calculate the extremum.
$ss1$ is a free parameter for the calculation.
In case this is understood and welcome there be more of this kind of extra-long and large expressions for the general solution. To my part, I think is already fairly hard. And harder as the problem is formulated just for staying reals.
|
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Show $a^3+b^3+c^3+d^3 \le 27$ Given $a, b, c, d$ are real numbers and $a^2+b^2+c^2+d^2 = 9$ show that
$$a^3+b^3+c^3+d^3\le27$$
I have tried to do the following:
$$(a^2+b^2+c^2+d^2)^{\frac{3}{2}} = 27 \\ \text{after simplifying} \\ (a^3+b^3+c^3+d^3) + \frac{3}{2}(a^2b^2+(a^2+b^2)(c^2+d^2)+c^2d^2)=27$$
By setting the following:
$$(a^3+b^3+c^3+d^3) + \frac{3}{2}(a^2b^2+(a^2+b^2)(c^2+d^2)+c^2d^2)=(a^3+b^3+c^3+d^3) \\ \frac{3}{2}(a^2b^2+(a^2+b^2)(c^2+d^2)+c^2d^2)=0 \\
(a^2+b^2)(c^2+d^2)=-(a^2b^2+c^2d^2) \\ (a^3+b^3+c^3+d^3)+\frac{3}{2}(a^2b^2+-(a^2b^2+c^2d^2)+c^2d^2)=27 \\ (a^3+b^3+c^3+d^3)=27$$
and so we have shown the inequality to hold - does this proof work?
|
Let $A=a^2, B=b^2, C=c^2, D=d^2$. You want to find the maximum of $\sum_{cyc}A^{3/2}$ on the set $\{A+B+C+D=9, A,B,C,D\geq 0\}$, i.e. the maximum of a convex function on a convex, closed and bounded set, namely a tetrahedron. By the properties of convex functions such maximum is attained at the vertices of the tetrahedron and it equals $9^{3/2}=27$.
|
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|
Find $\lim\limits_{(x,y)\to(0,0)}\frac{x^6+y^6}{x^3+y^3}$ using the $\epsilon-\delta$ definition? My textbook asks the question
$$f(x,y) = \frac{x^6+y^6}{x^3+y^3}$$
Does $f(x,y)$ have a limit as $(x,y) \rightarrow (0,0)$?
I used polar coordinates instead of solving explicitly in $\mathbb R^2 $, and it went as the following:
$$ x = r \cos \theta, \qquad y = r\sin\theta $$
Hence,
$$\lim_{(x,y) \to (0,0)} \frac{x^6+y^6}{x^3+y^3} = \lim_{r \to 0}\frac{{r^6\cos^6\theta + r^6\sin^6\theta}}{r^3\cos^3\theta + r^3\sin^3\theta}$$
This simplifies to,
$$ \lim_{r \to 0} \frac{r^3({\cos^6\theta + \sin^6\theta})}{\cos^3\theta + \sin^3\theta}$$
Now from the above, we find that as r→$0$ the limit is $0$.
So now i have got $0$ as a possible limit.I have tried to procceed further using $\epsilon-\delta$ definition but I cannot get anywhere.
Any thoughts on how to prove that $0$ is in fact the limit?
|
Top tip: don't trust wolfram! It gets the answer wrong sometimes!
Suppose we set $y = x$ and take $x \to 0$. Then
\begin{align*}
\lim_{(x,y(x)) \to (0,0)}\frac{x^6+y^6}{x^3+y^3} &= \lim_{x \to 0}\frac{2x^6}{2x^3}\\
&= 0
\end{align*}
Now let's suppose we take $y = -x$ and take $x \to 0$. Then
\begin{align*}
\lim_{(x,y(x)) \to (0,0)}\frac{x^6+y^6}{x^3+y^3} &= \lim_{x \to 0}\frac{2x^6}{x^3 - x^3}\\
&= \lim_{x \to 0}\frac{2x^6}{0}
\end{align*}
which is undefined. Since we have two approaches of $(x,y) \to (0,0)$ that give different results, the overall limit must not exist.
|
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|
The complex number $(1+i)$ is root of polynomial $x^3-x^2+2$. Find the other two roots. The complex number $(1+i)$ is root of polynomial $x^3-x^2+2$.
Find the other two roots.
$(1+i)^3 -(1+i)^2+2= (1-i-3+3i)-(1-1+2i) +2= (-2+2i)-(2i) +2= 0$.
The other two roots are found by division.
$$
\require{enclose}
\begin{array}{rll}
x^2 && \hbox{} \\[-3pt]
x-1-i \enclose{longdiv}{x^3 -x^2 + 2}\kern-.2ex \\[-3pt]
\underline{x^3-x^2- i.x^2} && \hbox{} \\[-3pt]
2 +i.x^2
\end{array}
$$
$x^3-x^2+2= (x-1-i)(x^2) +2+i.x^2$
How to pursue by this or some other approach?
|
As the coefficients are real thus the complex roots would appear in conjugate pair. Thus the second root would be $1-i$ and the third root (that is $-1$) can be calculated by using the fact that the sum of the roots of a cubic equation $ax^3+bx^2+cx+d$ is $-b/a$.
|
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Is difference equation $t_n = (a+b) t_{n-1} - ab \cdot t_{n-2}$ possible ($t_1 = a + b$ and $t_2 = a^2 + ab + b^2$)? I had problem solving difference equation $t_n = (a+b) t_{n-1} - ab \cdot t_{n-2}$ where $t_1 = a + b$ and $t_2 = a^2 + ab + b^2$.
Here is what I have done:
First, I've written down the equation as $$t^2 = (a+b) t - ab$$
which is $$t^2 - (a+b) t + ab = 0$$
Now solving this quadratic equation we get
$$t_{1,2} = \frac{2(a + b) \pm \sqrt{(a+b)^2 - 4ab}}{2} = \frac{2(a+b)\pm (a - b)}{2}$$
$$t_1 = \frac{3a+b}{2}$$
$$t_2 = \frac{3b+a}{2}$$
So, that leads us to
$$t_n = C_1\cdot t_1^n + C_2 \cdot t_2^n = C_1\cdot (\frac{3a+b}{2})^n + C_2 \cdot (\frac{3b+a}{2})^n$$
and we have to find $C_1$ and $C_2$ now.
Putting $n = 1$ and $n = 2$ in last equation, keeping in mind that $t_1 = a + b$ and $t_2 = a^2 + ab + b^2$ we'll get system of 2 equations with 2 variables
$$a + b = C_1\cdot \frac{3a+b}{2} + C_2 \cdot \frac{3b+a}{2}$$
$$a^2 + ab + b^2 = C_1\cdot (\frac{3a+b}{2})^2 + C_2 \cdot (\frac{3b+a}{2})^2$$
What was confusing to me was that, if we have $a = 1$ and $b = 1$ the system is not possible.
So, my question is what is wrong here and how to finish this?
|
Solving by characteristics method:
$t^2-(a+b) t+ab=0$
Solution $t=a, b$
Case 1 : $a\neq b$
Then $t_n=c_1a^n+c_2b^n$ for $c_1, c_2$ constants.
$t_1=a+b$ implies $c_1 a+c_2b=a+b \space \space ......(1) $
$t_2=a^2+ab+b^2$ implies $c_1 a^2+c_2b^2=a^2+ab+b^2..... (2) $
$(2) -a×(1) $ implies $c_2=\frac{b}{b-a}$ and using $c_2$ in $(1) $ , we get $c_1=\frac{q}{a-b}$
Hence solution : $t_n=\frac{a}{a-b}a^n+\frac{b}{b-a}b^n$
Case 2 : $a=b$
Then $t_n=(c_1+c_2n)a^n$
Solution : $t_n=(\frac{a^2+ab-b^2}{a^2}+\frac{b^2}{a^2}n)a^n$
|
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Systems of cubic equations I am looking for a general yet tailor-made methods to solve system of equations like that
$a^3 + 6ab^2 = 7$ and $3a^2b +2b^3 = 5$
that involve terms of $(a +b)^3$.
Does this have a name?
|
Let $\,u = a + b\sqrt{2}\,$ then $\,u^3 = a^3 + 6ab^2 + (3a^2b+2b^3) \sqrt{2}=7 + 5\sqrt{2}\,$, so the problem is equivalent to denesting $\,\sqrt[3]{5\sqrt{2} \pm 7} = b\sqrt{2} \pm a\,$.
The following steps are similar to my more general answer here.
*
*Let $\,v=\sqrt[3]{5 \sqrt{2} - 7}\,$, then $\,u^3-v^3=14\,$ and $\,uv = \sqrt[3]{5^2 \cdot 2 - 7^2}=1\,$.
*It follows that $\,(u-v)^3 = u^3-v^3 -3uv(u-v)=14 -3(u-v)\,$, so $\,t=u-v\,$ satisfies the cubic $\,t^3 + 3 t - 14 = 0\,$. The only real root is $\,t=2\,$, so $\,u-v=2\,$.
*Substituting $\,v=\dfrac{1}{u}\,$ in $\,u-v=2\,$ gives $\,u^2 - 2u - 1=0\,$ with roots $\,1 \pm \sqrt{2}\,$. Given that $\,u \gt 0\,$, it must be the positive root, so $\,u=\sqrt{2}+1\,$ and $\,v = \dfrac{1}{u} = \sqrt{2} - 1\,$.
It follows that $\,u,v=b\sqrt{2}\pm a=\sqrt[3]{5 \sqrt{2}\pm 7}=\sqrt{2}\pm 1\,$, so $\,a=b=1\,$.
[ EDIT ] $\;$ If the cubic at step $2$ did not have a "nice" root like $t=2$, the problem could still be solved the same way, but the following calculations would involve the cubic formula, thus be a lot uglier, and $\,a,b\,$ would not necessarily end up being "nice" rational values.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4466070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Determine the image of the unit circle $S^1$ by the action of the matrix $e^A$. We have:
$$e^{ \begin{pmatrix}
-5 & 9\\
-4 & 7
\end{pmatrix} }$$
I need to determine the image of the unit circle $S^1$ by the action of the matrix $e^A$.
I think that I know how to calculate $e^A$:
I get the Jordan decomposition:
$$A = \begin{pmatrix}
-5 & 9\\
-4 & 7
\end{pmatrix} =
\begin{pmatrix}
-6 & 1\\
-4 & 0
\end{pmatrix}
\cdot
\begin{pmatrix}
1 & 1\\
0 & 1
\end{pmatrix}
\cdot
\frac{1}{4}
\begin{pmatrix}
0 & -1\\
1 & -6
\end{pmatrix}
$$
With eigenvalues: $\lambda$ = 1, algebraic multiplicity = 2, eigenvecotrs: $\left\{ \begin{pmatrix}
1\\
0
\end{pmatrix}, \begin{pmatrix}
0\\
1
\end{pmatrix} \right\}$
$$ \displaystyle e^A = \sum^{\infty}_{i = 0} \frac{A^i}{i!}$$
$$e^A =
\begin{pmatrix}
-6 & 1\\
-4 & 0
\end{pmatrix}
\cdot
\left(
\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}
+
\displaystyle \sum^{\infty}_{i = 1} \frac{ \begin{pmatrix}
1 & 1\\
0 & 1
\end{pmatrix}}{i!}
\right)
\cdot
\frac{1}{4}
\begin{pmatrix}
0 & -1\\
1 & -6
\end{pmatrix}$$
$$e^A = \begin{pmatrix}
-5 & 9\\
-4 & 7
\end{pmatrix} =
\begin{pmatrix}
-6 & 1\\
-4 & 0
\end{pmatrix}
\cdot
\begin{pmatrix}
\displaystyle \sum^{\infty}_{i = 1} \frac{1}{i!}& \displaystyle \sum^{\infty}_{i = 1} \frac{2^{i-1}}{i!}\\
0 & \displaystyle \sum^{\infty}_{i = 1} \frac{1}{i!}
\end{pmatrix}
\cdot
\frac{1}{4}
\begin{pmatrix}
0 & -1\\
1 & -6
\end{pmatrix}$$
Where: $$\displaystyle \sum^{\infty}_{i = 1} \frac{2^{i-1}}{i!} = \frac{1}{2} \sum^{\infty}_{i = 1} \frac{2^{i}}{i!} = \frac{1}{2}(e^2 - 1)
$$
So:
$$e^A =
\begin{pmatrix}
-6 & 1\\
-4 & 0
\end{pmatrix}
\cdot
\begin{pmatrix}
e & \displaystyle \frac{e^2}{2} - \displaystyle \frac{1}{2}\\
0 & e
\end{pmatrix}
\cdot
\frac{1}{4}
\begin{pmatrix}
0 & -1\\
1 & -6
\end{pmatrix} =
\begin{pmatrix}
\displaystyle \frac{-3e^2 + e + 3}{4} & \displaystyle \frac{9e^2 - 9}{2}\\
\displaystyle \frac{-e^2 + 1}{2} & 3e^2 + e - 3
\end{pmatrix}
$$
Now, I don't know if I did it correctly up to this point and what I should do next - to operate on my unit circle.
Solution:
Because of @Oscar Lanzi we know that:
$$e^{\begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}}=e\begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}$$
Then because of that:
Equation of unit circle under linear transformation - can't understand role of inverse matrix
(answer by @Prototank)
We know that the image of unit circle in action of the matrix $A$ is given by: $$65x^{2}-166xy+106y^{2}=1$$
Now we need to scale by $e$ and we get the image of unit circle in action of the matrix $e^A$:
$$65x^{2}-166xy+106y^{2}=e^2$$
|
The matrix exponentiation is much simpler than it looks. When you find that $1$ is the only eigenvalue, render
$\begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}=\begin{pmatrix}1 & 0\\
0 & 1
\end{pmatrix}+\begin{pmatrix}-6 & 9\\-4 & 6\end{pmatrix}.$
The first matrix on the right just gives a factor of $e$ to the overall exponential. The second matrix is nilpotent and the series for its exponential is just
$\begin{pmatrix}1 & 0\\
0 & 1
\end{pmatrix}+\begin{pmatrix}-6 & 9\\-4 & 6\end{pmatrix}=\begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}.$
So
$e^{\begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}}=e\begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}.$
Continue from there.
|
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"url": "https://math.stackexchange.com/questions/4467835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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|
How to find $40! \mod 5^{10}$? I'm trying to solve this congruence. I found that in $40!$ there are $9$ powers of $5$, but I'll need one more for $40!=0 \mod 5^{10}$. I thought about Euler's theorem or Wilson's, but couldn't find any correlation.
|
Because $5^9$ divides $40!$, we can use $ka \bmod kb =k(a \bmod b)\,$ [mod distributive law] to get $40! \bmod 5^{10} = 5^9\cdot(\frac{40!}{5^9} \bmod 5)$. Hence we really only need to find the residue of $\frac {40!}{5^9} \bmod 5$. So separate $n \in 1 \cdots 40$ into two sets, those divisible by $5$ and those coprime to $5$:
The product of the $n$ coprime to $5$ is
$$ (1 \cdot 2 \cdot 3 \cdot 4)^8 \equiv 1 \pmod 5,$$
whereas the multiples of $5$ (with all the fives divided out) contribute
$$ 1 \cdot 2 \cdot 3 \cdot 4 \cdot 1 \cdot 1 \cdot 2 \cdot 3 \equiv (-1) \pmod 5.$$
Hence, we can write $$5^9 (\frac {40!}{5^9} \bmod 5) = 5^9 (1 \cdot -1 \bmod 5) = 5^9 \cdot 4$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4468130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Find the minimum value of a trigonometric function If the minimum value of $f\left(x\right)=\left(1+\frac{1}{\sin ^6\left(x\right)}\right)\left(1+\frac{1}{\cos ^6\left(x\right)}\right),\:x\:∈\:\left(0,\:\frac{\pi }{2}\right)$ is $m$, find $\sqrt m$.
How do I differentiate this function without making the problem unnecessarily complicated? If there are any other methods to finding the minimum value I am open to those too.
|
A more efficient answer than my previous approach:
Starting in the same way:
Since $\sin\left(\frac{\pi}{2}-x\right)=\cos(x)$, $f(x)$ is symmetric around $x=\frac{\pi}{4}$. This implies that $x=\frac{\pi}{4}$ is a critical point.
Writing $f(x)=(1+\sin(x)^{-6})(1+\cos(x)^{-6})$, we find that
$$
f'(x)=-6\sin(x)^{-7}\cos(x)(1+\cos(x)^{-6})+6\cos(x)^{-7}\sin(x)(1+\sin(x)^{-6}).
$$
This derivative can be factored as
$$
f'(x)=6\sin(x)^{-7}\cos(x)^{-7}(-\cos(x)^8-\cos(x)^2+\sin(x)^8+\sin(x)^2).
$$
Since $\cos(x)>\sin(x)$ for $0<x<\frac{\pi}{4}$ and $\cos(x)<\sin(x)$ for $\frac{\pi}{4}<x<\frac{\pi}{2}$, the cosine terms dominate for $0<x<\frac{\pi}{4}$ and the sine terms dominate for $\frac{\pi}{4}<x<\frac{\pi}{2}$.
Therefore, the derivative is negative for $0<x<\frac{\pi}{4}$ and positive for $\frac{\pi}{4}<x<\frac{\pi}{2}$. Hence, $x=\frac{\pi}{4}$ is the global minimum.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4474533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
How to solve $\sin(x) = \pm a$ for $a \not = 0$? I was solving the below equation: $\left|\sqrt{2\sin^2x + 18 \cos^2x} - \sqrt{2\cos^2x + 18 \sin^2x} \right| = 1$ for $x \in [0, 2\pi]$.
My attempt:
$$\begin{align}&\left|\sqrt{2\sin^2x + 18 \cos^2x} - \sqrt{2\cos^2x + 18 \sin^2x} \right| = 1\\\implies& \left|\sqrt{2\sin^2x + 2 \cos^2x + 16\cos^2x} - \sqrt{2\cos^2x + 2 \sin^2x + 16\sin^2x} \right| = 1
\\\implies& \left|\sqrt{2(\sin^2x + \cos^2x) + 16\cos^2x} - \sqrt{2(\cos^2x + \sin^2x) + 16\sin^2x} \right| = 1
\\\implies& \left|\sqrt{2+ 16\cos^2x} - \sqrt{2 + 16\sin^2x} \right| = 1\end{align}$$
Squaring both sides,
$$\begin{align}\implies& (2+ 16\cos^2x) + (2 + 16\sin^2x) - 2\sqrt{(2+ 16\cos^2x) (2 + 16\sin^2x)} = 1
\\\implies& 4 + 16(\sin^2x + \cos^2x) - 2\sqrt{2\cdot 2 \cdot (1+ 8\cos^2x) (1+ 8\sin^2x)} = 1
\\\implies &20 - 4\sqrt{1+8\cos^2x + 8\sin^2x + 8^2 \sin^2x \cos^2x } = 1
\\\implies &19 = 4\sqrt{1+8(\cos^2x + \sin^2x) + 16 \cdot (2\sin x \cos x)^2 }
\\\implies &\frac{19}{4} = \sqrt{9 + 16 \cdot (2\sin x \cos x)^2 }
\\\implies &\frac{19}{4} = \sqrt{9 + 16 \sin^2(2x) }\end{align}$$
Again squaring both sides,
$$\begin{align}\implies& \frac{361}{16} = 9 + 16 \sin^2(2x)
\\\implies& \frac{361}{16} - 9 = 16 \sin^2(2x)
\\\implies &\frac{361 - 144}{16} = 16 \sin^2(2x)
\\\implies& \frac{217}{256} = \sin^2(2x)
\\\implies& \pm\frac{\sqrt{217}}{16} = \sin(2x)\end{align}$$
Now I'm not getting any way to solve this equation. Although the question states to find only the solutions in the interval $[0, 2\pi]$, can we find the general form for all the solutions in $\mathbb{R}$ i.e. for all real numbers?
Desmos shows that there are $8$ solutions over the interval $[0, 2\pi]$.
|
Use abbreviation $s = \sin x$ and $1-s^2 = \cos^2x$:
$$\begin{align}
1
&= \left|\sqrt{2\sin^2x + 18 \cos^2x} - \sqrt{2\cos^2x + 18 \sin^2x} \right| \\
&= \Big |\sqrt{2s^2+18(1-s^2)} - \sqrt{2(1-s^2) + 18 s^2} \Big| \\
&= \Big |\sqrt{18-16s^2} - \sqrt{2+16s^2} \Big| \\
&= \Big (\sqrt{18-16s^2} - \sqrt{2+16s^2} \Big)^2 \tag 1\\
&= 20 -2\sqrt{(18-16s^2)(2+16s^2)} \\
\end{align}$$
where $(1)$ squared the whole equation. Then isolate the square-root and square again to get rid of the root:
$$\begin{align}
4\cdot(18-16 s^2)\cdot(2+16s^2) &= 19^2 \tag 2\\
\end{align}$$
The absolute value on the left side can be dropped in $(2)$ assuming $|s^2|\leqslant1$, i.e. $x$ is real and the factors on the left side are positive. What remains is quadratic in $s^2$ which has 4 solutions:
$$s=\pm\frac14\sqrt{8\pm\frac{\sqrt{39}}2} \tag 3$$
Now it's clear that when $x$ is a solution, then
$-x$ and $\pi/2-x$ and $x\pm\pi$ are also solutions. This is because the original equation does not change under respective transformations.
So let $x_0$ be the smallest, non-negative solution, i.e.
$$x_0\approx 0.5849127361226382 \approx \arcsin(0.5521266272786072)$$
then all solutions are
$$\pm x_0+\frac\pi2 \Bbb Z\tag 4$$
*
*The 4 solutions that are in $[-\pi/2, \pi/2)$ are, in ascending order:
$-\pi/2+x_0, -x_0, x_0, \pi/2-x_0$.
*The 4 solutons that are in $[0, \pi)$: $x_0, \pi/2-x_0, \pi/2+x_0, \pi-x_0$.
*The 8 solutons that are in $[0, 2\pi)$: $x_0, \pi/2-x_0, \pi/2+x_0, \pi-x_0, \pi+x_0, 3\pi/2-x_0, 3\pi/2+x_0, 2\pi-x_0$.
As $(3)$ is constructible with straightedge and compass, so are the angles $x$ that are solutions.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
Find the limit $\lim _{x\to 0}\frac{\cos x-1+\frac{x}{2}\cdot \sin x}{\ln ^4(x+1)}$ I need to find $\displaystyle \lim _{x\to 0}\frac{\cos x-1+\frac{x}{2}\cdot \sin x}{\ln ^4\left(x+1\right)}$.
I tried using the following:
\begin{align*}
\ln(1+x)&\approx x,\\
\sin(x)&\approx x-\frac{x^3}{2},\\
\cos(x)&\approx 1-\frac{x^2}{2!}+\frac{x^4}{4!},
\end{align*}
I managed to get
$\displaystyle \lim _{x\to 0}\frac{\cos x-1+\dfrac{x}{2}\cdot \sin x}{\ln ^4\left(x+1\right)}=\lim _{x\to 0}\frac{\left(\dfrac{\left(-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}\right)}{x^{4}}+\dfrac{x^{2}-\frac{x^{4}}{3!}}{2x^{4}}\right)}{\left(\dfrac{\ln\left(1+x\right)-x}{x}+1\right)^{4}}$
but I can't see how to use that $\displaystyle \lim _{x\to 0}\frac{\ln\left(1+x\right)-x}{x}=0$ in here.
Am I missing something? I didn't learn the little-$\mathcal{O}$ notation yet.
Thanks
|
Write: $Q(x) =\dfrac{\cos x -1 +\dfrac{x}{2}\cdot\sin x}{\ln^4(1+x)} = \dfrac{-2\sin^2\left(\frac{x}{2}\right)+2\cdot\left(\dfrac{x}{2}\right)^2\cdot\dfrac{\sin x}{x}}{\ln^4(1+x)} = \dfrac{\dfrac{x^2}{2}}{\ln^2(1+x)}\cdot\dfrac{\left(\dfrac{\sin x}{x}-\dfrac{\sin^2\left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)^2}\right)}{\ln^2(1+x)}= f(x)g(x)$. Observe that when $x \to 0 \implies f(x) \to \dfrac{1}{2}$, and for $g(x)$, to make it simpler put $x=2y \implies g(x) = h(y) = \dfrac{\dfrac{\sin(2y)}{2y} - \dfrac{\sin^2y}{y^2}}{\ln^2(1+2y)}=\dfrac{\sin y}{y}\cdot \dfrac{\cos y - \dfrac{\sin y}{y}}{\ln^2(1+2y)}=\dfrac{\sin y}{y}\cdot \left(\dfrac{2y}{\ln(1+2y)}\right)^2\cdot \dfrac{1}{4}\cdot \dfrac{y\cos y - \sin y}{y^3}\to 1\cdot 1\cdot \dfrac{1}{4}\cdot \dfrac{-1}{3}= \dfrac{-1}{12}\implies f(x)g(x) \to \dfrac{-1}{24} \implies \displaystyle \lim_{x \to 0} Q(x) = \dfrac{-1}{24}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
help me to evaluate these integrals Find the value of $$\int\frac{x}{x^2-x+1} dx$$ and $$\int\frac{1}{x^2-x+1} dx$$ This was not the original question. Original question was tougher and I have simplified that to these two integrals. I am having a hard time evaluating these two integrals but what I know is that, in these two integrals there is arctan occurring in a term of both integrals. Maybe we have to do some operations in the numerators?
Any help will be greatly appreciated.
EDIT: The original question was: $$\int\frac{3x}{x^3+1}dx$$
EDIT: $$\begin{align}\int\frac{3x}{x^3+1}dx&=3\int\frac{x}{x^3+1}dx\\&=3\int\frac{x+1}{3x^2-3x+3}dx - 3\int\frac{1}{3x+3}dx\\&=\int\frac{x+1}{x^2-x+1}dx - \int\frac{1}{x+1}dx\end{align}$$
The value of last integral is $\ln|x+1|$ so we will consider it later. We will only consider the first integral now and we can write that as:
$$\int\frac{x}{x^2-x+1}dx + \int\frac{1}{x^2-x+1}dx$$ after this I am stuck.
|
The real problem is the second integral since the first one has the form
\begin{align}
\int{{x \over x^2-x+1}dx} & = \int{{x-{1 \over 2} \over x^2-x+1}dx}+{1 \over 2}\int{{1 \over x^2-x+1}dx} \\
& = {1 \over 2}\int{{1 \over x^2-x+1}d(x^2-x+1)}+{1 \over 2}\int{{1 \over x^2-x+1}dx} \\
& = {1 \over 2}\bigg(\log{(x^2-x+1)}+\int{{1 \over x^2-x+1}dx}\bigg)
\end{align}
To solve the second one, we notice $\int{{1 \over y^2+1}dy}=\int{{1 \over \tan^2{\theta}+1}d\tan{\theta}}=\int{{1 \over \sec^2{\theta}}\sec^2{\theta}d\theta}=\theta=\arctan{y}$. Therefore, it is easy with
\begin{align}
\int{{1 \over x^2-x+1}dx} & = \int{{1 \over (x-{1 \over 2})^2+{3 \over 4}}d(x-{1 \over 2})} \\
& = \sqrt{4 \over 3}\int{{1 \over \bigg(\sqrt{4 \over 3}(x-{1 \over 2})\bigg)^2+1}d\bigg(\sqrt{4 \over 3}(x-{1 \over 2})}\bigg) \\
& = \sqrt{4 \over 3}\arctan{\bigg(\sqrt{4 \over 3}(x-{1 \over 2})\bigg)}
\end{align}
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4478207",
"timestamp": "2023-03-29T00:00:00",
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|
Find the first three Laurent expansion terms of $\frac{1-z}{z^2} e^z$ I've this function $f(z)=\frac{1-z}{z^2} e^z$ and I've to find the first three Laurent expansion terms in $z=0$.
I've proceeded in this way:
First of all I've considered the expansion series of $e^z = \sum_{n=0}\frac{z^n}{n!}$ and I found the first three terms, so $$e^z=1+z+\frac{z^2}{2}+o(z^3)$$
At this point my function would be
$$f(z)=\frac{1}{z^2}(1-z)(1+z+\frac{z^2}{2}+o(z^3))$$
$$f(z)= \frac{1}{z^2}-\frac{1}{2}-\frac{z}{2}+o(z^3)$$
and it seems right, but using Wolfram Alpha, I've discovered that should be
$f(z)=\frac{1}{z^2}-\frac{1}{2}-\frac{z}{3}-\frac{z^2}{8}-\frac{z^3}{30}+o(z^4)$.
As you can see, the third term is different. What I've to do?
|
$$e^z=\sum_{n=0}^{\infty}\frac{z^n}{n!}$$
So:
$$\frac{e^z}{z^2}=\sum_{n=0}^{\infty}\frac{z^{n-2}}{n!}=\sum_{n=-2}^{\infty}\frac{z^n}{(n+2)!}=\frac{1}{z^2}+\sum_{n=-1}^{\infty}\frac{z^n}{(n+2)!}$$
and
$$\frac{e^z}{z}=\sum_{n=0}^{\infty}\frac{z^{n-1}}{n!}=\sum_{n=-1}^{\infty}\frac{z^n}{(n+1)!}$$
Thus, since both converges we can subtract them
$$e^z\frac{1-z}{z^2}=\frac{e^z}{z^2}-\frac{e^z}{z}=\frac{1}{z^2}+\sum_{n=-1}^{\infty}\frac{z^n}{(n+2)!}-\sum_{n=-1}^{\infty}\frac{z^n}{(n+1)!}=\\\frac{1}{z^2}-\sum_{n=-1}^{\infty}\frac{z^n(n+1)}{(n+2)!}=\frac{1}{z^2}-\frac{1}{2}-\frac{z}{3}-\frac{z^2}{8}+O(z^4)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Simplest proof of $|I_n+J_n|=n+1$ Answering another question, I realised that I "know" that the determinant of the sum of an identity matrix $I_n$ and an all-ones square matrix $J_n$ is $n+1$. I.e.
$$|I_n+J_n|=\left|\begin{pmatrix}
1 & 0 & 0 &\cdots & 0 \\
0 & 1 & 0 &\cdots & 0 \\
0 & 0 & 1 &\cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & 1
\end{pmatrix} +\begin{pmatrix}
1 & 1 & 1 & \cdots & 1 \\
1 & 1 & 1 & \cdots & 1 \\
1 & 1 & 1 & \cdots & 1 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 1 & 1 & \cdots & 1
\end{pmatrix}\right| =\left|\begin{matrix}
2 & 1 & 1 & \cdots & 1 \\
1 & 2 & 1 & \cdots & 1 \\
1 & 1 & 2 & \cdots & 1 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 1 & 1 & \cdots & 2
\end{matrix} \right|=n+1.$$
I can prove it by adding and subtracting rows and columns without changing the determinant:
*
*subtracting the first row from the others $
\left|\begin{matrix}
2 & 1 & 1 & \cdots & 1 \\
1 & 2 & 1 & \cdots & 1 \\
1 & 1 & 2 & \cdots & 1 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 1 & 1 & \cdots & 2
\end{matrix} \right|=
\left|\begin{matrix}
2 & 1 & 1 & \cdots & 1 \\
-1 & 1 & 0 & \cdots & 0 \\
-1 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
-1 & 0 & 0 & \cdots & 1
\end{matrix} \right|$
*subtracting the other rows from the first $\left|\begin{matrix}
2 & 1 & 1 & \cdots & 1 \\
-1 & 1 & 0 & \cdots & 0 \\
-1 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
-1 & 0 & 0 & \cdots & 1
\end{matrix} \right|=
\left|\begin{matrix}
n+1 & 0 & 0 & \cdots & 0 \\
-1 & 1 & 0 & \cdots & 0 \\
-1 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
-1 & 0 & 0 & \cdots & 1
\end{matrix} \right|$
*adding the other columns to the first $\left|\begin{matrix}
n+1 & 0 & 0 & \cdots & 0 \\
-1 & 1 & 0 & \cdots & 0 \\
-1 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
-1 & 0 & 0 & \cdots & 1
\end{matrix} \right|=
\left|\begin{matrix}
n+1 & 0 & 0 & \cdots & 0 \\
0 & 1 & 0 & \cdots & 0 \\
0 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & 1
\end{matrix} \right|$
*and clearly $\left|\begin{matrix}
n+1 & 0 & 0 & \cdots & 0 \\
0 & 1 & 0 & \cdots & 0 \\
0 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & 1
\end{matrix} \right| =n+1$
but I feel there should be something more intuitive.
What would be a more intuitive or insightful way to prove this?
|
Here's a geometric proof: whether it's simpler/more intuitive than the row manipulations proof is a matter of taste.
Let $e_i$ be the elementary vector with a $1$ in the $i$th entry and $0$ everywhere else, and let $u$ denote the all-ones vector. We seek to find the determinant of the matrix $M$ whose $i$th row is $v_i:=u+e_i$. Since $M$ scales volumes by $\det M$, this determinant is the ratio
$$\det M=\frac{\operatorname{Volume}(\text{simplex with vertices }\{0,v_1,\dots,v_n\})}{\operatorname{Volume}(\text{simplex with vertices }\{0,e_1,\dots,e_n\})},$$
since $Me_i=v_i$. For simplex volumes, we use the formula $\frac1n\text{base}\cdot\text{height}$, with bases of each determined by $\{e_1,\dots,e_n\}$ and $\{v_1,\dots,v_n\}$. Since the $(n-1)$-simplex with vertices $\{v_1,\dots,v_n\}$ is just a translation of the simplex with vertices $\{e_1,\dots,e_n\}$, their bases have the same volume, and so $\det M$ is the ratio of the distances from $0$ to the planes determined by $\{v_1,\dots,v_n\}$ and $\{e_1,\dots,e_n\}$.
These planes are preserved under any permutation of the axes, and so the projection of $0$ down to each of these planes must be preserved by such transformations as well, and must be some scalar multiple of the all-ones vector $u$. The multiple of $u$ on the plane spanned by $\{e_1,\dots,e_n\}$ is $\frac1nu$, while the multiple of $u$ on the plane spanned by $\{v_1,\dots,v_n\}$ is $(1+\frac 1n)u$. So,
$$\det M=\frac{\left(1+\frac 1n\right)\|u\|}{\frac 1n\|u\|}=n+1.$$
|
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|
trying to solve the nested sum $\sum_{n=1}^{\infty} \frac{1}{1^2+2^2+\dots+n^2} $ This sum grabbed my curiousity
$$
\sum_{n=1}^{\infty} \frac{1}{1^2+2^2+\dots+n^2}
$$
I want to solve it with Calc II methods (that's the math I have).
By getting a closed form for the expression in the denominator, it becomes this:
$$
6 \sum_{n=1}^{\infty} \frac{1}{n(n+1)(2n+1)}
$$
First thing I did was partial fractions
$$
6 \sum_{n=1}^{\infty} \left( \frac{1}{n} + \frac{1}{n+1}- \frac{4}{2n+1} \right)
$$
You can reindex the first two terms, but I couldn't reindex the $2n+1$ term to telescope with that combined term, my thought is it's not possible because it isn't in the form (n+integer).
So I tried expressing it as an integral, viewing each term as multiplied by an $x$ evaluated between 0 and 1. Then deriving with respect to $x$ to produce an integral.
$$
6 \sum_{n=1}^{\infty} \int_0^1 (x^{n-1} +x^n - 4x^{2n}) dx
$$
Switching the order of integration and summation, and doing some reindexing:
$$
6 \int_0^1 \left(3+ 2\sum_{n=0}^\infty x^n - 4 \sum_{n=0}^\infty (x^2)^n\right)dx
$$
I've been doing decimal approximations the whole time, checking that my steps work. They do up to and including that point, the answer is about $1.3$
Next I used the geometric series expansion, and this is where the approximation goes wrong, so I believe this is a problematic step:
$$
6 \int_0^1 \left(3+ \frac{2}{1-x} - \frac{4}{1-x^2} \right)dx
$$
This integral is easy to solve using partial fractions on the $x^2$ term. You get an answer of $6(3-2\ln2)$, which is about $9.6$
The correct answer, based on some website that doesn't show steps, is $6(3-4\ln2)$, which matches my approximation.
I'm really stumped on what I'm doing wrong and how to get that correct answer. I've checked my algebra over and over and learned some latex to write this post. Any help is appreciated!
|
We have
\begin{align}
\sum_{n=1}^\infty\left(\frac{1}{n}+\frac{1}{n+1}-\frac{4}{2n+1}\right) & =
\sum_{n=1}^\infty\left(\frac{1}{n}-\frac{2}{2n+1}+\frac{1}{n+1}-\frac{2}{2n+1}\right)\\
&= \sum_{n=1}^\infty\left(\frac{1}{n}-\frac{2}{2n+1}\right)+\sum_{n=1}^\infty\left(\frac{1}{n+1}-\frac{2}{2n+1}\right)\\
&= 2\sum_{n=1}^\infty\left(\frac{1}{2n}-\frac{1}{2n+1}\right)+2\sum_{n=1}^\infty\left(\frac{1}{2n+2}-\frac{1}{2n+1}\right)\\
&= 2\;(1-\log 2)+2\;(\tfrac{1}{2}-\log 2)\\
&=3-4\log 2
\end{align}
where we used $\displaystyle\log 2 = \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n} = \sum_{n=1}^\infty\left(\frac{1}{2n-1}-\frac{1}{2n}\right)$.
|
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|
Evaluating $\int_{0}^{1} \lim_{n \rightarrow \infty} \sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} x^{2k-1} dx$ for $n \in \mathbb{N}$ For $n \in \mathbb{N}$, evaluate
$$\int_{0}^{1} \lim_{n \rightarrow \infty} \sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} x^{2k-1} dx$$
I could not use wolframalpha, I do not know the reason.
For $n = 1$, the integrand $=x+x^3$
For $n = 2$, the integrand $=x+x^3-x^5-x^7+x^9+x^{11}$
For $n = 3$, the integrand $=x+x^3-x^5-x^7+x^9+x^{11}-x^{13}-x^{15}+x^{17}+x^{19}$
and so on.
For $n = 1000$, the integrand $=x+x^3-x^5-x^7+x^9+x^{11}-x^{13}-x^{15}+x^{17}+x^{19}-\dots+x^{7993}+x^{7995}$
Using MS-EXCEL with $n=1000$, I found that the value is approximately $0.56...$.
I do not know if $n \rightarrow \infty$ , will the required expression have a closed form or no.
Your help would be appreciated. THANKS!
|
Let
$$a = \int_{0}^{1} \lim_{n \rightarrow \infty} \sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} x^{2k-1} dx$$
Modified solution (now the main solution)
We have two choices: integral first then sum or vice versa.
Taking the integral first gives the following for the (finite) sum which we split immediately into even and odd terms
$$a=\sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} \frac{1}{2k}=\sum_{m=1}^{2n-1}(-1)^\frac{1+m}{2} \frac{1}{4m}+\sum_{m=1}^{2n-1}(-1)^\frac{1-m}{2} \frac{1}{4m-2}\tag{1}$$
Now we take the limit $n\to \infty$ to get the result $(4)$ of the original solution.
Claude Leibovici, in a comment, has taken the other altervative with the sum first.
Original solution
Exchanging integral and sum, and doing the limit in the sum gives
$$\sum_{k=1}^{\infty}(-1)^\frac{k^2+k+2}{2} \frac{1}{2k}$$
Now splitting even and odd summands
$k \to 2m$, $(-1)^{\frac{k^2+k+2}{2}}\to (-1)^{1+m+2m^2}\to (-1)^{1+m}$
$$s_{e} = \sum_{m=1}^{\infty} (-1)^{1+m} \frac{1}{4 m}=\frac{\log(2)}{4}\tag{2}$$
$k \to 2m-1$, $(-1)^{\frac{k^2+k+2}{2}}\to (-1)^{1-m+2m^2}\to (-1)^{1-m}$
$$s_{o} = \sum_{m=1}^{\infty} (-1)^{1+m} \frac{1}{4 m-2}=\frac{\pi}{8}\tag{3}$$
Hence we have
$$a = s_{e}+s_{o} = \frac{\log(2)}{4} +\frac{\pi}{8}\simeq 0.565986... \tag{4}$$
|
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|
If $a > 0$, $b > 0$ and $b^{3} - b^{2} = a^{3} - a^{2}$, where $a\neq b$, then prove that $a + b < 4/3$. If $a > 0$, $b > 0$ and $b^{3} - b^{2} = a^{3} - a^{2}$, where $a\neq b$, then prove that $a + b < 4/3$.
Now what I thought is to manipulate given result somehow to get something in the form of $a + b$:
\begin{align*}
b^{3} - b^{2} = a^{3} - a^{2} & \Longleftrightarrow b^{3} - a^{3} = b^{2} - a^{2}\\\\
& \Longleftrightarrow (b-a)(a^{2} + ab + b^{2}) = (b+a)(b-a)\\\\
& \Longleftrightarrow a^{2} + ab + b^{2} = b + a
\end{align*}
but what next?
|
Rearrange $a^3-a^2=b^3-b^2$ to get
$$
b^3-a^3=b^2-a^2.
$$
Since $a\neq b$ we can factor out $(b-a)$ from both sides to get
$$
a^2+ab+b^2=a+b
$$
which is equivalent to
$$
(a+b)^2-ab=a+b
$$
Let $Q\equiv a+b$ then by the AM-GM inequality $\sqrt ab\leq(a+b)/2=Q/2$, and since $a\neq b$ the inequality is strict ($\sqrt ab < (a+b)/2$). Squaring this gives $ab<Q^2/4$.
We now have
$$
Q^2-Q<Q^2/4
$$
and since $a>0$, $b>0$ we can factor out a $Q$ and rearrange to get
$$
3/4Q<1\equiv Q<4/3.
$$
|
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|
Question on calculating a Fibonacci Number using Matrix Exponentiation We know that $\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^k = \begin{pmatrix}F_{k + 1} & F_k \\ F_k & F_{k - 1}\end{pmatrix},$ of which there is a simple proof by induction. However, since matrix multiplication is associative, we should be able to split $k$ into $a + b$ so that,
$$ \begin{pmatrix}F_{a + 1} & F_a \\ F_a & F_{a - 1}\end{pmatrix} \begin{pmatrix}F_{b + 1} & F_b \\ F_b & F_{b - 1}\end{pmatrix} = \begin{pmatrix}F_{k + 1} & F_k \\ F_k & F_{k - 1}\end{pmatrix} .$$
The matrix multiplication gives,
$$ \begin{pmatrix}F_{a+1}F_{b+1} + F_aF_b & F_{a+1}F_b + F_aF_{b-1} \\ F_aF_{b+1} + F_{a-1}F_b & F_aF_b + F_{a-1}F_{b-1} \end{pmatrix}.$$
Is there a proof using more elementary Fibonacci identities that this is equal to $\begin{pmatrix}F_{k + 1} & F_k \\ F_k & F_{k - 1}\end{pmatrix}?$
|
I can't honestly call this proof "more elementary" than the proof you've given in the question because I find that proof to be quite elementary, but perhaps you'll find it more palatable.
Decomposing sequences defined by linear recurrences:
Let $A_k, B_k$ denote two sequences that satisfy the same recursive relation as $F_k$ but with the initial conditions $A_2 = B_1 = 1, A_1=B_2 = 0$.
Then for any $a$, the sequence $G_k = F_{a+1}A_k + F_aB_k$ satisfies
$G_{k+1} = G_k + G_{k-1}$, $G_2 = F_{a+1}$ and $G_1 = F_a$, so that $G_k = F_{k+a-1}$. That is, we've shown
$$F_{k+a-1} = F_{a+1}A_k+F_aB_k$$
Now we consider the identity of the $A,B$ sequences:
*
*$A$ satisfies the defining recurrence for the Fibonacci sequence. Also, $A_2 = 1=F_1$ and $A_3 = A_2+A_1 = 1+0 = 1=F_2$. That is, $A$ is just a re-indexed Fibonacci sequence, specifically $A_k = F_{k-1}$.
*$B$ satisfies the defining recurrence for the Fibonacci sequence. Also, $B_3 = B_2+B_1 = 0+1=F_1$ and $B_4 = B_3+B_2 = 1+0 = F_2$. That is, $B$ is another re-indexed Fibonacci sequence, specifically, $B_k = F_{k-2}$.
Put together, this shows
$$F_{k+a-1} = F_{a+1}F_{k-1}+F_aF_{k-2}$$
which is exactly the identity we were looking for. To make it look more alike, we can set $k=b+2$ to find
$$F_{a+b+1} = F_{a+1}F_{b+1}+F_aF_b$$
and changing $a$ to $a-1$ or $b$ to $b-1$ explicitly gives all the remaining identities.
|
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|
Irrational equation $\left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0$ I saw the problem from one math competition:
$$\left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0$$.
I tried to solve it this way:
\begin{align*}
& \left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0\\
\Leftrightarrow\,\, & \left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x\left(x^{3}-3x+1\right)+1=0\\
\Leftrightarrow\,\, & \left(x^{3}-3x+1\right)\left(\sqrt{x^{2}-1}+x\right)=-1
\end{align*}
Now I tried to use substitutions like $a=x^3-3x$ and $b=\sqrt{x^{2}-1}$ but I cannot get anything. I know that solutions are $x=1$ and $x=\pm\sqrt{2}$.
|
A complete solution, building on your last identity.
Multiply your last equation by $x-\sqrt{x^2-1}\neq0$ to obtain:
$$\begin{align}x^3-3x+1&=\sqrt{x^2-1}-x\\x^3-2x+1&=\sqrt{x^2-1}\\x(x^2-2)+1&=\sqrt{(x^2-2)+1}\end{align}$$
$1$ is a solution by inspection of the second equation, and $\pm\sqrt{2}$ are solutions by inspection of the third equation, which was hopefully well motivated from a natural rearrangement.
Now taking $x$ not equal to one of the already determined solutions, square both sides, potentially introducing extraneous solutions: $$\begin{align}(x^3-2x+1)^2&=x^2-1\\x^6-4x^4+2x^3+3x^2-4x+2&=0\end{align}$$
We know that we can divide by $(x-1)$ and also by $x^2-2$, so we reduce to: $$(x-1)(x^5+x^4-3x^3-x^2+2x-2)=0\\(x^2-2)(x^3+x^2-x+1)=0$$
Now we know any solution of our original problem is either a solution already found, or it is a (real) root of $x^3+x^2-x+1=0$.
For such a root, the quantity $x^3-2x+1$ is equal to $-(x^2+x)$. Suppose $x\gt0$: then $0\gt0+1-x,x\gt1$ but this implies $0\gt1+1+1-x$ and $x\gt3$, so on, we have a contradiction in the magnitude of $x$. Moreover $x\gt3$ is impossible since Cauchy’s root bound ensures $|x|\le2$. Lets check the quantity $x^2+x$ - it is negative iff. $x\lt-x^2$, which forces that $-1\lt x$ by a comparison of magnitude. But then $0\gt-1+x^2-x+1=x^2-x$ and $x\gt x^2$, a contradiction. Thus the quantity $x^2+x$ is positive for all roots of the equation $x^3+x^2-x+1$. Were this root a solution of our desired equation, then $-(x^2+x)=\sqrt{x^2-1}$, but this is an impossibility as the left hand side is negative and the right hand side is positive. Thus this other root is extraneous and we can ignore it, and $1,\pm\sqrt{2}$ are the only solutions.
|
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|
How to solve this function $f(x)$ for degree $3$ or $4$
If $f(x)=0$ is a polynomial whose coefficients are either $1$ or $-1$ and whose roots are all real, then the degree of $f(x)$ can be equal to$:$
$A$. $1$
$B$. $2$
$C$. $3$
$D$. $4$
My work$:$
For linear only four polynomials are possible which are $x+1$ and $-x+1$ and $x-1$ and $-x-1$. All of which have real roots. So the answer is $A$.
For quadratic we have $8$ polynomials which are $x^2+x+1$ and $x^2+x-1$ and $x^2-x+1$ and $x^2-x-1$ and $-x^2+x+1$ and $-x^2-x+1$ and $-x^2+x-1$ and $-x^2-x-1$ Here four of the polynomials have real roots which are $x^2+x-1$ and $-x^2-x+1$ and $-x^2+x+1$ and $x^2-x-1$. So we can say that $f(x)$ can be of degree two. So $B$ is also the answer.
For cubic and quartic I don't know how to proceed. Any hints or solutions are appreciated.
|
Let's analyze the cubic case.
Note that the general cubic equation
$$x^3+bx^2+cx+d$$ may be reduced to the so-called depressed cubic with the substitution $x = y-b/3$. After the algebra, we get the equivalent equation
$$y^3+py+q = y^3+(c-\frac{b^2}{3})y+(\frac{2}{27}b^3-\frac{bc}{3}+d) = 0.$$
The discriminant $\Delta$ of the cubic is given by $p^3/27+q^2/4$. This gives us
$$\Delta = \frac{b^3 d}{27} - \frac{b^2 c^2}{108} - \frac{b c d}{6} + \frac{c^3}{27} + \frac{d^2}{4}.$$
If $\Delta \leq 0$, the cubic has all three roots being real.
Notice that from the above equation, the smallest $\Delta$ can be is $0$ which occurs if $b^3d = -1,bcd = 1,c^3 = -1$. This requires $c = -1$ and $b,d$ to have opposite signs. This indeed is possible.
For example, $x^3-x^2-x+1 = 0$, has roots $\pm 1$ with $1$ being a double root.
|
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|
Parameterizing the intersection between $4x^2+y^2=z^2$ and $xy-1=z$ I would like to parameterize the intersection between the surfaces
\begin{align}
4x^2+y^2&=z^2\\
xy&=z+1
\end{align}
I started by noting that $z=xy-1$ and then substituting that into the first equation to get \begin{align*}
4x^2+y^2&=(xy-1)^2\\
&=x^2y^2-2xy+1
\end{align*}
Then, I introduced the substitution $u=xy$ and $v=2x+y$ to obtain
\begin{align*}
v^2-4u&=u^2-2u+1\\
\implies v^2=u^2+2u+1\implies v^2&=(u+1)^2\\
\therefore v=u+1
\end{align*}
Clearly, $z=u-1$.
Where do I go from here to obtain a parameterization of the curve of intersection?
|
You have that
$$
4x^2 + y^2 = \left( {xy - 1} \right)^2
$$
thus
$$
\left( {2x + y} \right)^2 = \left( {xy + 1} \right)^2
$$
This means
$$
2x + y = \pm \left( {xy + 1} \right)
$$
So you have two cases: the first
$$
\left\{ \begin{array}{l}
2x + y = xy + 1 \\
z = xy - 1 \\
\end{array} \right.
$$
and the second
$$
\left\{ \begin{array}{l}
2x + y = - xy - 1 \\
z = xy - 1 \\
\end{array} \right.
$$
Let us consider the first. You have
$$
y\left( {x - 1} \right) = 2x - 1
$$
From this you have
$$
\left\{ \begin{array}{l}
x = t \\
y = \frac{{2t - 1}}{{t - 1}} \\
z = t\frac{{2t - 1}}{{t - 1}} - 1 \\
\end{array} \right.\,\,\,\,\,\,t \ne 1
$$
The second case is similar.
|
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|
How to find $\int_0^1 x^4(1-x)^5dx$ quickly? This question came in the Rajshahi University admission exam 2018-19
Q) $\int_0^1 x^4(1-x)^5dx$=?
(a) $\frac{1}{1260}$
(b) $\frac{1}{280}$
(c)$\frac{1}{315}$
(d) None
This is a big integral (click on show steps):
$$\left[-\dfrac{\left(x-1\right)^6\left(126x^4+56x^3+21x^2+6x+1\right)}{1260}\right]_0^1=\frac{1}{1260}$$
It takes a lot of time to compute. How can I compute this quickly (30 seconds) using a shortcut?
|
Maybe it is not the fastest way to solve it, but it is faster than computing the original integral:
\begin{align*}
I = \int_{0}^{1}x^{4}(1 - x)^{5}\mathrm{d}x = \int_{0}^{1}(1 - x)^{4}x^{5}\mathrm{d}x & \Rightarrow 2I = \int_{0}^{1}x^{4}(1-x)^{4}(x + (1 - x))\mathrm{d}x\\\\
& \Rightarrow 2I = \int_{0}^{1}x^{4}(1-x)^{4}\mathrm{d}x\\\\
& \Rightarrow 2I = \int_{0}^{1}x^{4}(1 - 4x + 6x^{2} - 4x^{3} + x^{4})\mathrm{d}x\\\\
& \Rightarrow 2I = \int_{0}^{1}(x^{4} - 4x^{5} + 6x^{6} - 4x^{7} + x^{8})\mathrm{d}x\\\\
& \Rightarrow 2I = \frac{1}{5} - \frac{2}{3} + \frac{6}{7} - \frac{1}{2} + \frac{1}{9}\\\\
& \Rightarrow I = \frac{1}{1260}
\end{align*}
|
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|
Evaluating $\lim _{x \rightarrow 0} \frac{12-6 x^{2}-12 \cos x}{x^{4}}$ $$
\begin{align*}
&\text { Let } \mathrm{x}=2 \mathrm{y} \quad \because x \rightarrow 0 \quad \therefore \mathrm{y} \rightarrow 0\\
&\therefore \lim _{x \rightarrow 0} \frac{12-6 x^{2}-12 \cos x}{x^{4}}\\
&=\lim _{y \rightarrow 0} \frac{12-6(2 y)^{2}-12 \cos 2 y}{(2 y)^{4}}\\
&=\lim _{y \rightarrow 0} \frac{12-24 y^{3}-12 \cos 2 y}{16 y^{4}}\\
&=\lim _{y \rightarrow 0} \frac{3(1-\cos 2 y)-6 y^{2}}{4 y^{4}}\\
&=\lim _{y \rightarrow 0} \frac{3.2 \sin ^{2} y-6 y^{2}}{4 y^{4}}\\
&=\lim _{y \rightarrow 0} \frac{ 3\left\{y-\frac{y^{3}}{3 !}+\frac{y^{5}}{5 !}-\cdots \infty\right\}^{2}-3 y^{2}}{2 y^{4}}\\
&=\lim _{y \rightarrow 0} \frac{3\left[y^{2}-\frac{2 y^{4}}{3 !}+\left(\frac{1}{(3 !)^{2}}+\frac{2}{3 !}\right) y^{4}+\cdots \infty\right)^{2}-3 y^{2}}{2 y^{4}}\\
&=\lim _{y \rightarrow 0} \frac{3\left\{y^{2}-\frac{2 y^{4}}{3 !}+\left\{\frac{1}{(3 !)^{2}}+\frac{2}{5 !}+y^{4}+\cdots \infty\right)-3 y^{2}\right.}{2 y^{4}}\\
&=\lim _{y \rightarrow 0}\left[\frac{-\frac{6}{3 !}+3\left\{\frac{1}{(3 !)^{2}}+\frac{2}{5 !}\right\} y^{2}+\cdots \infty}{2}\right]\\
&=\lim _{y \rightarrow 0}\left[\frac{-1+3\left\{\frac{1}{(3 !)^{2}}+\frac{2}{5 !}\right\} y^{2}+\cdots \infty}{2}\right]\\
&=-\frac{1}{2} \text { (Ans.) }
\end{align*}
$$
Doubt
Can anyone please explain the 5,6,7 equation line? Thank you
|
$$\lim_{x\to0}\frac {12-6x^2-12\cos x}{x^4}=\lim_{x\to0}\frac{-12x+12\sin x}{4x^3}=\lim_{x\to0}\frac{-12+12\cos x}{12x^2}=\lim_{x\to0}\frac{-12\sin x}{24x}=\lim_{x\to0}-1/2\cos x=-1/2$$, by repeated application of L'hopital.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4496409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
}
|
If $\{v_1+v_2, v_2+v_3, v_1+v_3\}$ are linearly independent then $\{v_1, v_2, v_3\}$ are linearly independent Problem. Prove that
for $v_1, v_2, v_3 \in \mathbb{R}^3$, if $\{v_1+v_2, v_2+v_3, v_1+v_3\}$ are linearly independent then $\{v_1, v_2, v_3\}$ are linearly independent.
What I tried:
Let $m,n,p \in \mathbb{R}$ be such that
$$mv_1+nv_2+pv_3 = 0\;(\star)$$
From the hypothesis we know that if $a,b,c \in \mathbb{R}$ with $a(v_1+v_2)+b(v_2+v_3)+c(v_1+v_3) = 0$, then $a=b=c=0$.
First, every element $\begin{pmatrix}m\\n\\p \end{pmatrix} \in \mathbb{R}^3$ can be unique written in terms of $A = \biggl\{\begin{pmatrix}1\\0\\1 \end{pmatrix},\begin{pmatrix}1\\1\\0 \end{pmatrix},\begin{pmatrix}0\\1\\1 \end{pmatrix}\biggr\}$ because $A$ is a basis in $\mathbb{R}^3$, so we can let $\begin{cases} m=a+b \\ n=b+c \\ p=a+c \end{cases}$. So, from
$$
\begin{align}
(\star) \implies (a+b)v_1 + (b+c)v_2+(a+c)v_3=0 \\
\iff av_1+bv_1+bv_2+cv_2+av_3+cv_3=0 \\
\iff a(v_1+v_3)+b(v_1+v_2)+c(v_2+v_3)=0 \\
\implies a=b=c=0 \implies m=n=p=0
\end{align}$$
$\implies \{v_1, v_2, v_3\}$ are linearly independent
Please correct me if I am wrong or not. Thanks!
|
You are correct. But we can solve this in a general setting for any vector space. Let's rename the vectors in this way: $\alpha:=v_1+v_2, \beta:=v_2+v_3, \gamma:=v_3+v_1$, then we have $v_1=\frac{1}{2}(\alpha-\beta+\gamma), v_2=\frac{1}{2}(\beta-\gamma +\alpha), v_3=\frac{1}{2}(\gamma-\alpha+\beta)$. In other words, we have $$\frac{1}{2} \begin{pmatrix}
1 & -1 & 1\\
1 & 1 & -1\\
-1 & 1 & 1
\end{pmatrix} \begin{pmatrix}
\alpha\\
\beta\\
\gamma
\end{pmatrix}= \begin{pmatrix}
v_1 \\
v_2 \\
v_3
\end{pmatrix}.$$ Let's name the above 3 by 3 matrix $A$, then the matrix A is invertible. Now if we assume that $mv_1+nv_2+pv_3=0$, then we must have
$$ \begin{pmatrix}
m& n & p
\end{pmatrix}A \begin{pmatrix}
\alpha\\
\beta\\
\gamma
\end{pmatrix}=\begin{pmatrix}
m& n & p
\end{pmatrix} \begin{pmatrix}
v_1 \\
v_2 \\
v_3
\end{pmatrix}=0.$$
Therefore $\begin{pmatrix}
m& n & p
\end{pmatrix}A \begin{pmatrix}
\alpha\\
\beta\\
\gamma
\end{pmatrix}=0$. Since $\alpha, \beta, \gamma$ are linearly independent we must have $\begin{pmatrix}
m& n & p
\end{pmatrix}A=\begin{pmatrix}
0&0&0
\end{pmatrix}$ or $A^T\begin{pmatrix}
m\\
n\\
p
\end{pmatrix}=\begin{pmatrix}
0\\
0\\
0
\end{pmatrix}$. But $A^T$ is invertible so $\begin{pmatrix}
m\\
n\\
p
\end{pmatrix}=0$ which means $V_1, V_2 ,V_3$ are linearly independent.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4497624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
}
|
Calculating the center of mass of a lemniscate rotated around the x-axis? This is a problem I have been stuck on a while, it goes as follows:
A lemniscate has the equation $(x^2+y^2)^2 = 4(x^2-y^2)$. Let the part of the curve that lies in the first quadrant rotate around the $x$-axis to create an object. This object has an homogeneous massdistribution with the density $1$. Determine the coordinates of the center of mass $(x_T,y_T,z_T)$, if $x_T = \frac{1}{M}\int _K x dm$, where $M$ is the mass of the object.
I have so far managed to switch to polar coordinates and found the relation $r^2 = 4\cos{2\theta}$ and tried to calculate the volume (which equals the mass in this case due to the density being equal to 1) with the formula $\int _K \pi y^2 dx$. After the switch to polar coordinates, a point on the curve is $(x,y) = (r\cos{\theta},r\sin{\theta})$, which gives $\pi y^2 = \pi r^2\sin^2{\theta} = 4\pi\cos{2\theta}\sin^2{\theta}$.
I then reason that a small change $dx$ in polar coordinates, corresponds to a small change in $\theta$, hence I substitute $dx$ with $\frac{dx}{d\theta} 2\sqrt{\cos{2\theta}}\cos{\theta} \Leftrightarrow dx = -2(\frac{\sin{2\theta}\cos{\theta}}{\sqrt{\cos{2\theta}}}+ \sin{\theta}\sqrt{\cos{2\theta}}) d\theta$.
But when I try and calculate the integral
$\int^2_0 4\pi\cos{2\theta}\sin^2{\theta} \cdot (-2)(\frac{\sin{2\theta}\cos{\theta}}{\sqrt{\cos{2\theta}}}+ \sin{\theta}\sqrt{\cos{2\theta}}) d\theta $,
I get an incorrect volume/mass.
The correct solution is $(x_T,y_T,z_T) = (\frac{2}{3\sqrt{2}\ln{(\sqrt{2}+1)}-2},0,0)$
I would greatly appreciate any tips on how to proceed, thanks in advance!
|
Because you are rotating something around the $x$-axis, we will have $y=z=0$.
Now, for this case, I prefer to use the traditional coordinates. First we have ro solve for $y$ the equation:
$$(x^2+y^2)^2-4(x^2-y^2)=0\leftrightarrow \left[t=x^2\right] \leftrightarrow y=\pm\sqrt{\frac{-2x^2-4\pm\sqrt{(-2x^2-4)^2-4(x^4-4x^2)}}{2}}=\pm\sqrt{-x^2-2\pm\sqrt{1+2x^2}}$$
We are interested in the first quadrant, so:
$$y=\sqrt{-x^2-2+\sqrt{1+2x^2}}$$
Now, by definition of mass centre:
$$x_M=\frac{1}{M}\cdot\int_M xdm=\frac{1}{V}\cdot\int_V x dV$$
Where $dV$ is the volume of a disk with radius $y(x)$ and height $dx$.
We have:
$$V=\int_0^2 \pi \left(\sqrt{-x^2-2+\sqrt{1+2x^2}}\right)^2dx=\int_0^2 \pi\left(-x^2-2+\sqrt{1+2x^2}\right)dx=\pi\cdot\left[\ln\left(\frac{\sqrt{2x^2+1}+\sqrt2x}{\sqrt2}\right)+x\sqrt{2x^2+1}-\frac{x^3}{3}-2x+c\right]_0^2=\pi\cdot\left(\frac{\ln(3+2\sqrt2)}{\sqrt2}+6-\frac83-4\right)=\pi\cdot\left(\frac{\ln(3+2\sqrt2)}{\sqrt2}-\frac23 \right)$$
Also:
$$\int_0^2 \pi x\left(\sqrt{-x^2-2+\sqrt{1+2x^2}}\right)^2dx=\int_0^2 \pi x\left(-x^2-2+\sqrt{1+2x^2}\right)dx=\pi\left[\frac{\sqrt{(2x^2+1)^3}}{3}-\frac{x^4}{4}-x^2+c\right]_0^2=1-\frac13=\frac23\pi$$
Finally:
$$x_M=\frac{1}{M}\cdot\int_M xdm=\frac{1}{V}\cdot\int_V x dV=\frac{\frac23\pi}{\pi\cdot\left(\frac{\ln(3+2\sqrt2)}{\sqrt2}-\frac23 \right)}=\frac{2\sqrt2}{3\ln(3+2\sqrt2)-2\sqrt2}=\frac{2}{3\sqrt{2}\ln{(\sqrt{2}+1)}-2}$$
Where the last equality holds because you can pick up from numerator and denominator a $\sqrt2$ term and then notice that:
$$\sqrt{3+2\sqrt2}=\sqrt{\left( 1+\sqrt2 \right)^2}=1+\sqrt2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4498071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
$\int_0^1 \frac{\arcsin x\arccos x}{x}dx$ Someone on Youtube posted a video solving this integral.
I can't find on Math.stack.exchange this integral using search engine https://approach0.xyz
It is related to $\displaystyle \int_0^\infty \frac{\arctan x\ln x}{1+x^2}dx$
Following is a solution that is not requiring the use of series:
\begin{align}J&=\int_0^1 \frac{\arcsin x\arccos x}{x}dx\\
&\overset{\text{IBP}}=\underbrace{\Big[\arcsin x\arccos x\ln x\Big]_0^1}_{=0}-\underbrace{\int_0^1 \frac{\arccos x\ln x}{\sqrt{1-x^2}}dx}_{x=\cos t }+\underbrace{\int_0^1 \frac{\arcsin x\ln x}{\sqrt{1-x^2}}dx}_{x=\sin t}\\
&=\int_0^{\frac{\pi}{2}} t\ln(\tan t)dt\\
&\overset{u=\tan t}=\int_0^\infty \frac{\arctan u\ln u}{1+u^2}du\\
&\overset{\text{IBP}}=\underbrace{\left[\arctan u\left(\int_0^u \frac{\ln t}{1+t^2}dt\right)\right]_0^\infty}_{=0}-\int_0^\infty \frac{1}{1+u^2}\left(\underbrace{\int_0^u \frac{\ln t}{1+t^2}dt}_{y(t)=\frac{t}{u}}\right)du\\
&=-\int_0^\infty \left(\int_0^1 \frac{u\ln(uy)}{(1+u^2)(1+u^2y^2)}dy\right)du\\
&=-\int_0^\infty \left(\int_0^1 \frac{u\ln u}{(1+u^2)(1+u^2y^2)}dy\right)du-\int_0^1 \left(\int_0^\infty \frac{u\ln y}{(1+u^2)(1+u^2y^2)}du\right)dy\\
&=-\int_0^\infty \left[\frac{\arctan(uy)}{1+u^2}\right]_{y=0}^{y=1}\ln udu-\frac{1}{2}\int_0^1 \left[\frac{\ln\left(\frac{1+u^2}{1+u^2y^2}\right)}{1-y^2}\right]_{u=0}^{u=\infty}\ln ydy\\
&=-J+\int_0^1 \frac{\ln^2 y}{1-y^2}dy\\
&=\frac{1}{2}\int_0^1 \frac{\ln^2 y}{1-y}dy-\frac{1}{2}\underbrace{\int_0^1 \frac{y\ln^2 y}{1-y^2}dy}_{z=y^2}\\
&=\frac{7}{16}\int_0^1 \frac{\ln^2 y}{1-y}dy\\
&=\frac{7}{16}\times 2\zeta(3)=\boxed{\frac{7}{8}\zeta(3)}
\end{align}
NB:I assume $\displaystyle \int_0^1 \frac{\ln^2 y}{1-y}dy=2\zeta(3)$
Feel free to post your solution.
|
$$
\begin{align*}
\int_0^1 \frac{\arcsin(x)\arccos(x)}{x}\,dx& = \int_0^1 \frac{\arcsin(x)\left(\frac{\pi}{2}-\arcsin(x)\right)}{x}\,dx\\
&=\frac{\pi}{2}\int_0^1 \frac{\arcsin(x)}{x}\,dx-\int_0^1 \frac{\arcsin^2(x)}{x}\,dx\\
&=\frac{\pi}{2}\int_0^{\pi/2} \frac{x}{\sin(x)}\cos(x)\,dx-\int_0^{\pi/2} \frac{x^2}{\sin(x)}\cos(x)\,dx\\
&=\frac{\pi}{2}\int_0^{\pi/2} x\cot(x)\,dx-\int_0^{\pi/2} x^2\cot(x)\,dx\\
&=\frac{\pi}{2} J - K\\
&=\frac{\pi^2\ln(2)}{4}-\frac{\pi^2\ln(2)}{4}+\frac{7\zeta(3)}{8}\\
&=\frac{7\zeta(3)}{8} \qquad \blacksquare
\end{align*}
$$
$$
\begin{align*}
J&=\int_0^{\pi/2} x\cot(x)\,dx\\
&=x\ln(\sin(x))\Big|_0^{\pi/2}-\int_0^{\pi/2}\ln(\sin(x))\,dx\\
&=\ln(2)\int_0^{\pi/2} \,dx+\sum_{k=1}^\infty \frac{1}{k} \int_0^{\pi/2}\cos(2 k x)\,dx \\
&=\frac{\pi\ln(2) }{2}
\end{align*}
$$
$$
\begin{aligned}
K&=\int_0^{\pi/2} x^2\cot(x)\,dx\\
&=x^2\ln(\sin(x))\Big|_0^{\pi/2}-2\int_0^{\pi/2}x\ln(\sin(x))\,dx\\
&=-2\int_0^{\pi/2}x \ln\left(\sin(x)\right)\,dx\\
&=-2\left(-\ln(2)\int_0^{\pi/2}x \,dx-\sum_{k=1}^\infty \frac{1}{k} \int_0^{\pi/2}x \cos(2 k x)\,dx\right)\\
&=-2\left(-\frac{\pi^2\ln(2)}{8}-\sum_{k=1}^\infty \frac{1}{k} \left(\frac{x \sin(2 k x)}{2k}\Big|_0^{\pi/2} -\frac{1}{2k}\int_0^{\pi/2}\sin(2 k x)\,dx\right)\right)\\
&=-2\left(-\frac{\pi^2\ln(2)}{8}-\frac{1}{2}\sum_{k=1}^\infty \frac{1}{k^2} \left(\frac{\cos(2 k x)}{2k}\Big|_0^{\pi/2}\right)\right)\\
&=-2\left(-\frac{\pi^2\ln(2)}{8}-\frac{1}{4}\sum_{k=1}^\infty \frac{1}{k^3} \left((-1)^k-1\right)\right)\\
&=-2\left(-\frac{\pi^2\ln(2)}{8}+\frac{1}{4}\zeta(3)+\frac14\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^3}\right)\\
&=-2\left(-\frac{\pi^2\ln(2)}{8}+\frac{\zeta(3)}{4}+\frac{\eta(3)}{4}\right)\\
&=-2\left(-\frac{\pi^2\ln(2)}{8}+\frac{\zeta(3)}{4}+\frac{3\zeta(3)}{16}\right)\\
&=-2\left(-\frac{\pi^2\ln(2)}{8}+\frac{7\zeta(3)}{16}\right) \\
&=\frac{\pi^2\ln(2)}{4}-\frac{7\zeta(3)}{8}
\end{aligned}
$$
Where we used
$$ \arcsin(x)+\arccos(x)=\frac{\pi}{2}$$
$$\eta(s)=(1-2^{1-s})\zeta(s)$$
$$\ln\left(\sin(x)\right)=-\ln(2)-\sum_{k=1}^\infty \frac{\cos(2 k x)}{k}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4500073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 3
}
|
Limit of $n^2$ and a recurrence relation with ceiling function For all positive integer $n$ we define a finite sequence in the following way:
$n_0 = n$, then $n_1\geq n_0$ and has the property that $n_1$ is a multiple of $n_0-1$ such that the difference $n_1 - n_0$ is minimal among all multiple of $n_0 -1 $ that are bigger than $n_0$. More generally $n_k \geq n_{k-1}$ and has the property that is a multiple of $n_0-k$ such that the difference $n_k - n_{k-1}$ is minimal among all multiple of $n_0 -k $ that are bigger than $n_{k-1}$. We stop the procedure when $k=n-1$ so that we define $f$ to be $f(n_0) = n_{n-1} $.
Question: What is the value of $\lim_{n \to \infty} \frac{n^2}{f(n)} $ ?
I'm able to prove that $$ \frac{8}{3} \leq \lim_{n \to \infty} \frac{n^2}{f(n)} \leq 4 $$
but I can't do better. Someone has an idea? I think that the limit is $ \pi $ but don't know how to prove that.
My idea is to prove that
$$ f(n) = \frac{n^2}{\pi} + O(n) $$
the general term for $n_k$ is given by
$$ n_k = (n-k) \left \lceil \frac{n_{k-1}}{n-k} \right \rceil $$
So that
$$ n_k = (n-k) \left \lceil \frac{n-(k-1)}{n-k} \left \lceil \frac{n-(k-2)}{n-(k-1)} \left \lceil \ldots \left \lceil \frac{n-1}{n-2} \left \lceil \frac{n}{n-1} \right \rceil \right \rceil \right \rceil \right \rceil \right \rceil $$
|
Not sure but I think I did one direction in another way:
Claim: For $ n/2 < k < n$ and $n$ big enough we have that
$$ \left \lceil \frac{n_{k-1}}{n-k} \right \rceil \leq \frac{(k+1)^2}{\pi} \frac{ \Gamma(k+1/2)\Gamma(k+3/2)}{\Gamma(k+1)^2} $$
hence
$$ n_k = (n-k) \left \lceil \frac{n_{k-1}}{n-k} \right \rceil \leq (n-k)\frac{(k+1)^2}{\pi} \frac{ \Gamma(k+1/2)\Gamma(k+3/2)}{\Gamma(k+1)^2} $$
thus when $k=n-1$ we have
$$ f(n) \leq \frac{n^2}{\pi} \frac{ \Gamma(n+1/2)\Gamma(n-1/2)}{\Gamma(n)^2} = \frac{n^2}{\pi} + O(n) $$
Proof of the claim:
We have that the claim holds for $k=n/2+1$ in fact for $n > 6$ we have that
$ n_{n/2+1} = (n/2-1) (n/2+3)$. Hence $ \left \lceil \frac{n_{n/2}}{n/2-1} \right \rceil = (n/2+3)$ Then suppose that the claim holds for $ n/2 < k < n$, now we have that
$ \left \lceil x \right \rceil \leq x +1 $, then
$$ \frac{(k+1)^2}{\pi} \frac{ \Gamma(k+1/2)\Gamma(k+3/2)}{(n-k)\Gamma(k+1)^2} + 1 \leq \frac{(k+1)^2}{\pi} \frac{ \Gamma(k+1/2)\Gamma(k+3/2)}{\Gamma(k+1)^2} + 1 \leq \frac{(k+2)^2}{\pi} \frac{ \Gamma(k+1+1/2)\Gamma(k+1+3/2)}{\Gamma(k+2)^2} $$
where the last inequality cames from the fact that when $k$ is big enough the Laurent series of $ \frac{ \Gamma(k+1+1/2)\Gamma(k+1+3/2)}{\pi\Gamma(k+2)^2} $ and of $\frac{ \Gamma(k+1/2)\Gamma(k+3/2)}{\pi \Gamma(k+1)^2} $ is equal too
$$ \frac{1}{\pi} + O(1/k) $$
and we have that
$$ 1 \leq \left( (k+2)^2 - (k+1)^2 \right) \left( \frac{1}{\pi} + O(1/k) \right) $$
hence
$$ f(n) \leq \frac{n^2}{\pi} + O(n) $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4503801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
}
|
$\varepsilon-\delta$ proof of this multivariable limit We've to prove that
$$
\lim_{(x,y)\to(0,0)} \frac{x^3+y^4}{x^2+y^2} =0
$$
Kindly check if my proof below is correct.
Proof
We need to show there exists $\delta>0$ for an $\varepsilon>0$ such that
$$
\left| \frac{x^3+y^4}{x^2+y^2} \right| < \varepsilon \implies \sqrt{x^2+y^2}< \delta
$$
Start
$$
\left| \frac{x^3}{x^2+y^2} \right| <\left| \frac{x^3+y^4}{x^2+y^2} \right| < \varepsilon
$$
Note
$$
\left| \frac{x^3}{x^2+y^2} \right|= \frac{x^2|x|}{x^2+y^2}>\frac{(x^2-y^2)|x|}{x^2+y^2}
$$
Therefore
$$
\frac{x^2-y^2}{x^2+y^2}|x|<\varepsilon \tag{1}
$$
Note
$$
\frac{x^2-y^2}{x^2+y^2}|x|<|x|=\sqrt{x^2}<\sqrt{x^2+y^2}<\delta
$$
So
$$
\frac{x^2-y^2}{x^2+y^2}|x|<\delta \tag{2}
$$
From $(1)$ and $(2)$, we can say
$$
\delta=\varepsilon
$$
|
There is a simple strategy for dealing with such limits, which is just switching to polar coordinates. By setting $x=\rho\cos\theta, y=\rho\sin\theta$ we have $x^2+y^2=\rho^2$ and
$$ \left|\frac{x^3+y^4}{x^2+y^2}\right| = \rho\left|\cos^3\theta+\rho \sin^4\theta\right|. $$
The RHS is trivially bounded by $\rho(1+\rho)$, which is $\leq 2\rho$ if $\rho\leq 1$.
It follows that as long as $\rho\leq\frac{\varepsilon}{2}\leq 1$ we have $|f(x,y)|\leq \varepsilon$.
If you want a sharp bound you may consider that for any sufficiently small $\rho\in\mathbb{R}^+$ the maximum of $|\cos^3\theta+\rho\sin^4\theta|$ is indeed $1$, so $\rho\leq\varepsilon$ is sufficient to ensure $|f(x,y)|\leq\varepsilon$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4504768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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|
Finding $\sum_{n=0}^{\infty} (-1)^n\left(\frac{1}{(3n+2)^2}-\frac{1}{(3n+1)^2}\right)$ Recently, I stumbled upon a summation
$$S=\sum_{n=0}^{\infty} (-1)^n\left(\frac{1}{(3n+2)^2}-\frac{1}{(3n+1)^2}\right)$$
which can luckily be summed to a good number.
Use $\psi^1(z)=\sum_{n=0}^{\infty}\frac{1}{(n+z)^2}$ to write
$$ S=\frac{1}{36}\left(\left[\psi^1\left(\frac13\right)-\psi^{1}\left(\frac56\right)\right]-\left[\psi^1\left(\frac16\right)-\psi^1\left(\frac23\right)\right]\right)$$
Next use $\psi^1(z)+\psi^1(1-z)=\pi^2\csc^2(\pi z)$ to write https://en.wikipedia.org/wiki/Polygamma_function, we get
$$I=\frac{1}{36}\left[\frac{4\pi^2}{3}-4\pi^2\right]=-\frac{2\pi^2}{27}.$$
The question is what could be other ways of getting this result?
|
Combining consecutive terms, we have $$S = \sum_{m=0}^\infty \left(-\frac{1}{(6m+1)^2} + \frac{1}{(6m+2)^2} + \frac{1}{(6m+4)^2} - \frac{1}{(6m+5)^2}\right). \tag{1}$$
Now recall
$$\zeta(2) = \sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}, \tag{2}$$ hence
$$\sum_{m=1}^\infty \frac{1}{(3m)^2} = \frac{\zeta(2)}{9} = \frac{\pi^2}{54}. \tag{3}$$ We also have
$$\sum_{m=0}^\infty \frac{1}{(2m+1)^2} = \sum_{m=0}^\infty \frac{1}{(2m+1)^2} + \frac{1}{(2m+2)^2} - \sum_{k=1}^\infty \frac{1}{4k^2} = \frac{3}{4} \sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{8}. \tag{4}$$
It follows that
$$\begin{align}
\sum_{m=0}^\infty \frac{1}{(6m+2)^2} + \frac{1}{(6m+4)^2}
&= \frac{1}{4} \left(\!-\frac{\pi^2}{54} + \sum_{m=0}^\infty \frac{1}{(3m+1)^2} + \frac{1}{(3m+2)^2} + \frac{1}{(3m+3)^2}\! \right) \\
&= \frac{1}{4} \left( \sum_{k=1}^\infty \frac{1}{k^2} - \frac{\pi^2}{54} \right) \\
&= \frac{1}{4} \left(\frac{\pi^2}{6} - \frac{\pi^2}{54} \right) \\
&= \frac{\pi^2}{27}. \tag{5}
\end{align}$$
We apply the same trick again to $S$:
$$\begin{align}
S &= \sum_{m=0}^\infty \left(-\sum_{j=1}^6 \frac{1}{(6m+j)^2} + \frac{1}{(6m+3)^2} + \frac{1}{(6m+6)^2} \right) + 2 \frac{\pi^2}{27} \\
&= -\sum_{k=1}^\infty \frac{1}{k^2} + \sum_{m=0}^\infty \left(\frac{1}{9(2m+1)^2} + \frac{1}{36(m+1)^2} \right)+ \frac{2\pi^2}{27} \\
&= -\frac{\pi^2}{6} + \frac{\pi^2}{72} + \frac{\pi^2}{216} + \frac{2\pi^2}{27} \\
&= -\frac{2\pi^2}{27}.
\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4505167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Show that $\frac{\sin^3 \beta}{\sin \alpha} + \frac{\cos^3 \beta}{\cos \alpha} = 1$ with certain given $\alpha, \beta$ Let $$\frac{\sin (\alpha)}{\sin (\beta)} + \frac{\cos (\alpha)}{\cos (\beta)} = -1 \tag{$1$}$$ where $\alpha, \beta$ are not multiples of $\pi / 2$. Show that
$$\frac{\sin^3 (\beta)}{\sin (\alpha)} + \frac{\cos^3 (\beta)}{\cos (\alpha)} = 1\tag{$2$}$$
I've tried to rewrite $(1)$ and insert into $(2)$ to get
$$ - 1 - \frac{\sin^4 \beta \cos^2 \alpha + \cos^4 \beta \sin^2 \alpha}{\sin \alpha \sin \beta \cos \alpha \cos \beta} = 1 \\
\iff \frac{\sin^4 \beta \cos^2 \alpha + \cos^4 \beta \sin^2 \alpha}{\sin \alpha \sin \beta \cos \alpha \cos \beta} = -2$$
But I can't simplify any further, maybe there are some trig. identities I'm missing?
|
Eq.$(1)$ is equivalent to
$$\sin(a+b)=-\frac{1}2\sin(2b)\tag{3}$$
Now start from the LHS of Eq.$(2)$, and we will show it equals $1$
$$\begin{align}
\text{LHS}=\frac{\sin^3(b)\cos(a)+\cos^3(b)\sin(a)}{\sin(a)\cos(a)}\end{align}$$
Deal with the numerator:
$$\begin{align}
\text{Numerator}&=(1-\cos^2(b))\sin(b)\cos(a)+(1-\sin^2(b))\cos(b)\sin(a)\\
\\
&=\sin(a+b)-\cos^2(b)\cos(a)\sin(b)-\sin^2(b)\cos(b)\sin(a)\\
\\
&=\sin(a+b)-\frac{1}2\cos(a)\cos(b)\sin(2b)-\frac{1}2\sin(a)\sin(b)\sin(2b)~~~~~~~\text{use} ~~(3)\\
\\
&=\sin(a+b)+\sin(a+b)\cos(a-b)\\
\\
&=\sin(a+b)+\frac{1}2\sin(2a)+\frac{1}2\sin(2b)~~~~~~~\text{use} ~~(3)\\
\\
&=\frac{1}2\sin(2a)=\text{Denominator}
\end{align}$$
|
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|
Summation of reciprocal products When studying summation of reciprocal products I found some interesting patterns.
$$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)}=\frac{1}{1\cdot1!}-\frac{1}{1\cdot(N+1)}$$
$$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)(k+2)}=\frac{1}{2\cdot2!}-\frac{1}{2\cdot(N+1)(N+2)}$$
$$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)(k+2)(k+3)}=\frac{1}{3\cdot3!}-\frac{1}{3\cdot(N+1)(N+2)(N+3)}$$
The pattern is $$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)(k+2)(k+3)\cdot\cdot\cdot\cdot(k+i)}\space=\space\frac{1}{i\cdot i!}-\frac{1}{i\cdot(N+1)(N+2)(N+3)\cdot\cdot\cdot\cdot(N+i)}$$
which is easy to prove by induction.
As an easy consequence it follows that $$\lim_{N \to \infty} \sum_{k=1}^{N} \frac{1}{k\cdot(k+1)(k+2)(k+3)\cdot\cdot\cdot\cdot(k+i)}\space=\space\frac{1}{i\cdot i!}$$
I did not find this summation of reciprocal products in my mathbooks. Is this a known theorem?
Can anyone point me in the right direction for further study?
|
$\sum_{k=1}^{N} \frac{1}{k(k+1)\cdot\cdot\cdot(k+i)}=\frac{1}{i\cdot i!}-\frac{1}{i\cdot(N+1)\cdot\cdot\cdot(N+i)}$ is easy to prove by induction.
It is true for $N=1$ since
$\sum_{k=1}^{1} \frac{1}{k(k+1)\cdot\cdot\cdot(k+i)}=\frac{1}{(1+i)!}$ equals $\frac{1}{i\cdot i!}-\frac{1}{i\cdot 2\cdot 3\cdot\cdot\cdot(1+i)}=\frac{1}{i\cdot i!}-\frac{1}{i\cdot(1+i)!}= \frac{1}{(1+i)!}$
Assume the statement applies for a certain $N$.
Evaluating for $N+1$:
$\sum_{k=1}^{N+1}\frac{1}{k(k+1)\cdot\cdot\cdot(k+i)}=\frac{1}{i\cdot i!}-\frac{1}{i\cdot(N+1)\cdot\cdot\cdot(N+i)}+\frac{1}{(N+1)\cdot\cdot\cdot (N+1+i)}=$
$=\frac{1}{i\cdot i!}-\frac{N+1+i-i}{i\cdot (N+1)\cdot\cdot\cdot(N+i)}=\frac{1}{i\cdot i!}-\frac{1}{i\cdot (N+2)\cdot\cdot\cdot(N+1+i)}$
shows that it also applies for $N+1$ which ends the proof.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Rectangle of maximum area under the curve $y=x(x-1)^2$ A rectangle's bottom is $y=0$. While the top corners are on the curve $y=x(x-1)^2$ between $x=0$ and $x=1$. Find the maximum area of this rectangle.
My Progress
Defining $f(x) =x(x-1) ^2$
I first starting by assuming that there exist $a$ and $b$ such that $0<a, b<1$ and $f(a) =f(b)$.
The area of this rectangle is given by $|b-a|f(a) $.
We know that $a(a-1) ^2=b(b-1) ^2$. Neglecting the $a=b$ case we get that $$a^2+b^2+ab-2(a+b) +1=0$$
Now it remains to maximise $$|b-a|a(a-1) ^2$$ under the constraint $$a^2+b^2+ab-2(a+b) +1=0$$
But now I'm pretty much stuck here.
|
There is really only one free variable, which is without loss of generality, the $x$ coordinate of the vertex of the rectangle that is closest to the origin. If we call this value $a$, then the other vertex coordinates are uniquely defined: in total, we have
$$(a,0), (a, a(1-a)^2), (b,a(1-a)^2), (b, 0)$$ where $$b = \frac{2-a - \sqrt{4a - 3a^2}}{2}$$ is the solution to $f(a) = f(b)$ for which $b > a$. The resulting area of the rectangle is
$$R(a) = a(1-a)^2 \cdot (b-a) = \frac{1}{2}\left( a(1-a)^2 \left(2-3a - \sqrt{4a-3a^2}\right)\right).$$ A tedious differentiation with respect to $a$ yields
$$R'(a) = 1 - 7a + 12a^2 - 6a^3 - (3 - 10a + 6a^2)(1-a) \sqrt{\frac{a}{4-3a}}.$$ So any critical points must satisfy $R'(a) = 0$, or $$(1-7a+12a^2-6a^3)^2(4-3a) = (3-10a+6a^2)^2(1-a)^2 a.$$ This eventually leads to the condition
$$0 = (1-a)^3 (1 - 14a + 58a^2 - 84a^3 + 36a^4).$$ Since $a < 1$, this latter factor yields the minimal polynomial $f(z) = 1 - 14z + 58z^2 - 84z^3 + 36z^4$ for $a$, which actually admits a solution by square roots. We find this by presuming that $f$ admits a factorization into quadratics of the form
$$f(z) = (1 + pz + 6z^2)(1 + qz + 6z^2)$$
for undetermined constants $p, q$. Then expanding and equating like coefficients yields the conditions $$p+q = -14, \quad pq + 12 = 58.$$ This gives $(p,q) = -7 \pm \sqrt{3}$ in any order, hence the roots of $f$ are
$$z \in \left\{ \frac{-p \pm \sqrt{p^2 - 24}}{12}, \quad \frac{-q \pm \sqrt{q^2 - 24}}{12} \right\}.$$
Numeric evaluation of the four roots shows that only one satisfies $0 < a < b < 1$, namely $$a = \frac{7 + \sqrt{3} - \sqrt{14(2+\sqrt{3})}}{12} = \frac{7 + \sqrt{3} - \sqrt{7} - \sqrt{21}}{12} \approx 0.12531031679570389137.$$
The corresponding maximum area is then $$R(a) = \frac{7 \sqrt{7} - 10}{108 \sqrt{3}} \approx 0.045547906756179342935.$$
By no means do I claim that this is the most elegant or simplest solution; it is simply what I was able to compute on a first try at this question.
|
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|
How many method are there to handle the integral $\int \frac{\sin x}{1-\sin x \cos x} d x$ $$
\begin{aligned}
\int \frac{\sin x}{1-\sin x \cos x} d x =& \frac{1}{2} \int \frac{(\sin x+\cos x)+(\sin x-\cos x)}{1-\sin x \cos x} d x \\
=& \int \frac{d(\sin x-\cos x)}{2-2 \sin x \cos x}-\int \frac{d(\sin x+\cos x)}{2-2 \sin x \cos x} \\
=& \int \frac{d(\sin x-\cos x)}{1+(\sin x-\cos x)^{2}}-\int \frac{d(\sin x+\cos x)}{3-(\sin x+\cos x)^{2}} \\
=& \tan ^{-1}(\sin x-\cos x)+\frac{1}{2 \sqrt{3}} \ln \left| \frac{\sin x+\cos x-\sqrt{3}}{\sin x+\cos x+\sqrt{3}} \right|+C
\end{aligned}
$$
Is there any other method?
|
Multiplying both numerator and denominator of the integrand by $1+\sin x \cos x$ transforms the integral into
$$
\begin{aligned}
I &=\int \frac{\sin x(1+\sin x \cos x)}{1-\sin ^{2} x \cos ^{2} x} d x \\
&=\int \frac{\sin x d x}{1-\sin ^{2} x \cos ^{2} x}+\int \frac{\sin ^{2} x \cos x}{1-\sin ^{2} x \cos ^{2} x} d x \\
&=- \underbrace{\int \frac{d c}{1-\left(1-c^{2}\right) c^{2}}}_{J}+ \underbrace{\int \frac{s^{2} d s}{1-s^{2}\left(1-s^{2}\right)}}_{K}
\end{aligned}
$$
where $c=\cos x$ and $s=\sin x.$
$$
\begin{aligned}
J&=-\int \frac{d c}{c^{4}-c^{2}+1}\\
&=-\int \frac{\frac{1}{c^{2}}}{c^{2}+\frac{1}{c^{2}}-1} d c
\end{aligned}
$$
Playing a little trick on the integrand gives
$$
\begin{aligned}
J &=-\frac{1}{2} \int \frac{\left(1+\frac{1}{c^{2}}\right)+\left(1-\frac{1}{c^{2}}\right)}{c^{2}+\frac{1}{c^{2}}-1} d c \\
&=-\frac{1}{2} \int \frac{d\left(c-\frac{1}{c}\right)}{\left(c-\frac{1}{c}\right)^{2}+1}+\frac{1}{2} \int \frac{d\left(c+\frac{1}{c}\right)}{\left(c+\frac{1}{c}\right)^{2}-3} \\
&=-\frac{1}{2} \tan ^{-1}\left(c-\frac{1}{c}\right)+\frac{1}{4 \sqrt{3}} \ln \left|\frac{c+\frac{1}{c}-\sqrt{3}}{c+\frac{1}{c}+\sqrt{3}}\right|+C_{1}
\end{aligned}
$$
Similarly,
$$
\begin{aligned}
K &=\int \frac{1}{s^{2}+\frac{1}{s^{2}}-1} d s \\
&=\frac{1}{2} \int \frac{\left(1+\frac{1}{s^{2}}\right)-\left(1-\frac{1}{s^{2}}\right)}{s^{2}+\frac{1}{s^{2}}-1} d s \\
&=\frac{1}{2} \tan ^{-1}\left(s-\frac{1}{s}\right)+\frac{1}{4 \sqrt{3}} \ln \left|\frac{s+\frac{1}{s}+\sqrt{3}}{s+\frac{1}{s}-\sqrt{3}}\right|+C_{2}
\end{aligned}
$$
Now we can conclude that
$$
\begin{aligned}
\therefore I=&-\frac{1}{2} \tan ^{-1}(\cos x-\sec x)+\frac{1}{2} \tan ^{-1}(\sin x-\csc x) \\
&+\frac{1}{4 \sqrt{3}} \ln \left|\frac{\cos ^{2} x-\sqrt{3} \cos x+1}{\cos ^{2} x+\sqrt{3} \cos x+1}\right|+\frac{1}{4 \sqrt{3}} \ln \left|\frac{\sin ^{2} x+\sqrt{3} \sin x+1}{\sin ^{2} x-\sqrt{3} \sin x+1}\right|+C
\end{aligned}
$$
|
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|
Given that $a>b>0$, can we find $\int_{0}^{2 \pi} \frac{\cos n x}{a+b \cos x} d x$ and $\int_{0}^{2 \pi} \frac{\sin n x}{a+b \cos x} d x?$ Let’s first consider
$$
\int_{0}^{2 \pi} \frac{\cos n x}{a+b \cos x} d x+i \int_{0}^{2 \pi} \frac{\sin n x}{a+b \cos x} d x = \int_{0}^{2 \pi} \frac{e^{n x i}}{a+b \cos x} dx=I \tag*{(*)}
$$
where $a>b>0$.
Next we are going to evaluate $I$ using contour integration by letting $z=e^{xi}$. Then
$$
I= \oint_{|z|=1} \frac{z^{n}}{a+\frac{b\left(z+z^{-1}\right)}{2}} \cdot \frac{d z}{i z} =\frac{2}{i} \oint_{|z|=1} \frac{z^{n}}{b(z-\alpha)(z-\beta)} d z,
$$
where $\alpha=\frac{-a+\sqrt{a^{2}-b^{2}}}{b}$, $\beta=\frac{-a-\sqrt{a^{2}-b^{2}}}{b}$ with $|\beta| =\frac{a+\sqrt{a^{2}-b^{2}}}{b} > \frac{a}{b}> 1 \Rightarrow |\alpha |<1$.
Therefore $\alpha $ is the unique simple pole in the unit circle $|z|=1.$ It’s residue is
$$\operatorname{Res}\left(\frac{z^{n}}{b(z-\alpha)(z-\beta)}, z=\alpha \right)=\lim _{z \rightarrow \alpha} \frac{z^{n}}{b(z-\beta)}=\frac{\alpha^{n}}{2 \sqrt{a^{2}-b^{2}}}$$
Hence
$$
\begin{aligned}
I &=\frac{2}{i} \cdot 2 \pi i \cdot \frac{\alpha^{n}}{2 \sqrt{a^{2}-b^{2}}}=\frac{2 \pi \alpha^{n}}{\sqrt{a^{2}-b^{2}}}\in R.
\end{aligned}
$$
Comparing the real and imaginary parts of both sides in (*) yields the result.
$$
\boxed{\int_{0}^{2 \pi} \frac{\cos n x}{a+b \cos x} d x=\frac{2 \pi}{\sqrt{a^{2}-b^{2}}}\left(\frac{-a+\sqrt{a^{2}-b^{2}}}{b}\right)^{n}
\textrm{ and }\int_{0}^{2 \pi} \frac{\sin n x}{a+b \cos x} d x=0}$$
Furthermore, if $a>-b>0$, then
$$\int_{0}^{2 \pi} \frac{\cos n x}{a+b \cos x} d x=\frac{2 \pi}{\sqrt{a^{2}-b^{2}}}\left(\frac{-a-\sqrt{a^{2}-b^{2}}}{b}\right)^{n}
\textrm{ and }\int_{0}^{2 \pi} \frac{\sin n x}{a+b \cos x} d x=0 $$
I really want to know whether there are other methods. Your comments and alternative solutions are highly appreciated!
|
Utilize the Fourier series
$$\eqalign{
\frac{1-r^2}{1+r^2+2r\cos x}=1+2\sum_{k>0}(-r)^k\cos k x
}
$$
and let $r=\frac {a-\sqrt{a^2-b^2}}b$ to integrate
\begin{align}
&\int_{0}^{2 \pi} \frac{e^{i n x}}{a+b \cos x} dx\\
=&\ \frac1{\sqrt{a^2-b^2}}\int_{0}^{2 \pi} \frac{(1-r^2) e^{i n x}}{1+r^2+2r\cos x} dx\\
=&\ \frac1{\sqrt{a^2-b^2}}\int_0^{2\pi}
e^{i n x}\bigg( 1+2\sum_{k>0}(-r)^k\cos k x\bigg)dx\\
= &\ \frac1{\sqrt{a^2-b^2}}\int_0^{2\pi}(-r)^n dx
=\frac{2 \pi \ (-r)^n}{\sqrt{a^{2}-b^{2}}}
\end{align}
whose real and imaginary parts are, respectively
$$\int_{0}^{2 \pi} \frac{\cos n x}{a+b \cos x} d x=\frac{2 \pi(-r)^n}{\sqrt{a^{2}-b^{2}}};\>\>\>\>\>
\int_{0}^{2 \pi} \frac{\sin n x}{a+b \cos x} d x=0 $$
|
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|
Solving recurrence relation $a_n = a_{n-1} - a_{n-2}$ I am given a sequence of determinants of matrices $M_n$, where the matrix elements $(M_n)_{ij}$ of $M_n$ are $0$ whenever $|i-j|>1$ and $1$ whenever $|i-j| ≤ 1$. Writing out the first five matrices, it becomes apparent that $\det(M_n) = \det(M_{n-1}) - \det(M_{n-2})$. I want a formula for the mapping $n ↦ \det(M_n)$, which I believe to be
$$a_n = \begin{cases} 0, & n ≡ 2 \mod 6 \,\, \vee n ≡ 5 \mod 6, \\ 1, & n ≡ 0 \mod 6 \,\, \vee n ≡ 1 \mod 6, \\ -1, & n ≡ 3 \mod 6 \,\, \vee n ≡ 4 \mod 6. \end{cases}
$$
This can quite readily be seen from the first 15 or so terms. Of course, this doesn't constitute a proof, which most likely will have to be performed by induction. I just fear that I am to embark on a six-piece proof by exhaustion, which I would like to avoid if there is a (much) quicker way to do it!
|
You are asking for the recurrence, I think, $a_0=a_1=1$, $a_n=a_{n-1}-a_{n-2}$. Let $T:\Bbb R^3\to\Bbb R^3$ be the linear operator with the following matrix (on the standard basis): $$\begin{pmatrix}1&-1&0\\1&0&0\\0&1&0\end{pmatrix}$$And let, $n\ge 0$, the vectors $v_n$ be given by: $$v_n=\begin{pmatrix}a_{n+2}\\a_{n+1}\\a_n\end{pmatrix}$$Then $Tv_n=v_{n+1}$ for all $n$, or equivalently, $v_n=T^nv_0$ for all $n$. If we can find a nice expression for $T^n$ then we are basically done (this is a standard trick!). The diagonal matrix for this operator is: $$\begin{pmatrix}0&0&0\\0&\zeta&0\\0&0&\overline{\zeta}\end{pmatrix}$$Where $\zeta=\frac{1}{2}(1-i\sqrt{3})=e^{-i\pi/3}$, so the diagonal matrix for $T^n$ is just the same with $\zeta^n,\overline{\zeta^n}$ instead, that is, $e^{\pm in\pi/3}$. The computation of the eigenvalues and the change-of-basis matrices is all doable by hand, in not too much time (less than $20$ minutes, much less if you're fast!), but I have skipped these steps as they are tedious.
We are interested only in $a_n$, the bottom row of $T^nv_0$. The computation looks as follows: $$\begin{align}v_n&=\begin{pmatrix}0&\zeta-1&\overline{\zeta}-1\\0&\zeta&\overline{\zeta}\\1&1&1\end{pmatrix}\begin{pmatrix}0&0&0\\0&\zeta^n&0\\0&0&\overline{\zeta^n}\end{pmatrix}\begin{pmatrix}1&-1&1\\-\frac{1}{2}+\frac{i}{6}\sqrt{3}&\frac{1}{2}+\frac{i}{6}\sqrt{3}&0\\-\frac{1}{2}-\frac{i}{6}\sqrt{3}&\frac{1}{2}-\frac{i}{6}\sqrt{3}&0\end{pmatrix}\begin{pmatrix}0\\1\\1\end{pmatrix}\\&=\begin{pmatrix}0&\zeta-1&\overline{\zeta}-1\\0&\zeta&\overline{\zeta}\\1&1&1\end{pmatrix}\begin{pmatrix}0&0&0\\0&\zeta^n&0\\0&0&\overline{\zeta^n}\end{pmatrix}\begin{pmatrix}0\\\frac{1}{2}+\frac{i}{6}\sqrt{3}\\\frac{1}{2}-\frac{i}{6}\sqrt{3}\end{pmatrix}\\&=\begin{pmatrix}0&\zeta-1&\overline{\zeta}-1\\0&\zeta&\overline{\zeta}\\1&1&1\end{pmatrix}\begin{pmatrix}0\\\zeta^n\left(\frac{1}{2}+\frac{i}{6}\sqrt{3}\right)\\\zeta^{-n}\left(\frac{1}{2}-\frac{i}{6}\sqrt{3}\right)\end{pmatrix}\end{align}$$And the bottom row is: $$a_n=\frac{1}{2}(\zeta^n+\zeta^{-n})+\frac{i}{\sqrt{3}}\frac{1}{2}(\zeta^n-\zeta^{-n})=\cos\frac{\pi n}{3}+\frac{1}{\sqrt{3}}\sin\frac{\pi n}{3}$$Since $\cos,\sin$ are periodic functions, you can read off the periodicity relations from here (modulo $6$, say).
Let's check our work: standard trigonometric identities give $a_0=1=a_1$ using the above formula. In general: $$\begin{align}a_{n+1}-a_n&=\left[\color{red}{\cos\frac{\pi n}{3}\cos\frac{\pi}{3}}-\sin\frac{\pi n}{3}\sin\frac{\pi}{3}\right]+\frac{1}{\sqrt{3}}\left[\sin\frac{\pi n}{3}\cos\frac{\pi}{3}+\cos\frac{\pi n}{3}\sin\frac{\pi}{3}\right]\\&-\left[\color{red}{\cos\frac{\pi n}{3}}+\frac{1}{\sqrt{3}}\sin\frac{\pi n}{3}\right]\\&=\color{red}{-\frac{1}{2}\cos\frac{\pi n}{3}}-\frac{\sqrt{3}}{2}\sin\frac{\pi n}{3}+\frac{1}{2\sqrt{3}}\sin\frac{\pi n}{3}+\frac{1}{2}\cos\frac{\pi n}{3}-\frac{1}{\sqrt{3}}\sin\frac{\pi n}{3}\\&=-\frac{2}{\sqrt{3}}\sin\frac{\pi n}{3}\end{align}$$And: $$\begin{align}a_{n+2}&=-\frac{1}{2}\cos\frac{\pi n}{3}-\frac{\sqrt{3}}{2}\sin\frac{\pi n}{3}+\frac{1}{\sqrt{3}}\left[-\frac{1}{2}\sin\frac{\pi n}{3}+\frac{\sqrt{3}}{2}\cos\frac{\pi n}{3}\right]\\&=0-\frac{4\sqrt{3}}{6}\sin\frac{\pi n}{3}\\&=-\frac{2}{\sqrt{3}}\sin\frac{\pi n}{3}\\&=a_{n+1}-a_n\end{align}$$As desired!
|
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|
Prove the following algebraic asymmetric inequality: $\sqrt{3abc(a + b + c)} + 2(a - c)^2 \geq a^2 + b^2 + c^2$
Consider $a \geq b \geq c \geq 0$ real numbers. Prove that:
$$\sqrt{3abc(a + b + c)} + 2(a - c)^2 \geq a^2 + b^2 + c^2$$
Source: RMM 2022
Comments & discussion:
A weaker form of the inequality may be obtained by using Newton's 3 variable inequality:
$$ab + bc + ca \geq \sqrt{3abc(a + b + c)}$$
This means that the new inequality would look like:
$$ab + bc - 3ac + a^2 + c^2 \leq b^2$$
Which will rewrite as:
$$(b - a)(b - c) \leq 0$$
Which is obviously true
However, I am unable to reduce the more tough form to a simillar expression.
|
Using $ab + bc + ca \ge \sqrt{3abc(a + b + c)}$, we have
$$\sqrt{3abc(a+b+c)}
\ge \frac{3abc(a+b+c)}{ab + bc + ca}.$$
Using $a \ge b \ge c$, we have
$$2(a-c)^2 = (a - c)^2 + (a - b + b - c)^2 \ge (a - c)^2 + (a - b)^2 + (b - c)^2.$$
It suffices to prove that
$$ \frac{3abc(a+b+c)}{ab + bc + ca} + (a - c)^2 + (a - b)^2 + (b - c)^2 \ge a^2 + b^2 + c^2$$
or
$$\frac{3abc(a+b+c)}{ab + bc + ca} + a^2 + b^2 + c^2 - 2(ab + bc + ca)\ge 0. $$
Let $p = a + b + c, q = ab + bc + ca, r = abc$. We need to prove that
$$\frac{3pr}{q} + p^2 - 4q \ge 0. \tag{1}$$
If $p^2 - 4q \ge 0$, (1) is true.
If $p^2 - 4q < 0$, using three degree Schur $r\ge \frac{4pq - p^3}{9}$, we have
$$\frac{3pr}{q} + p^2 - 4q
\ge \frac{3p}{q}\cdot \frac{4pq - p^3}{9} + p^2 - 4q
= \frac{(4q - p^2)(p^2 - 3q)}{3q} \ge 0$$
where we have used $p^2-3q = a^2 + b^2 + c^2 - ab - bc - ca \ge 0$.
We are done.
|
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|
Find the sum of radicals without squaring, Is that impossible?
Find the summation:
$$\sqrt {3-\sqrt 5}+\sqrt {3+\sqrt 5}$$
My attempts:
\begin{align*}
&A = \sqrt{3-\sqrt{5}}+ \sqrt{3+\sqrt{5}}\\
\implies &A^2 = 3-\sqrt{5} + 3 + \sqrt{5} + 2\sqrt{9-5}\\
\implies &A^2 = 6+4 = 10\\
\implies &A = \sqrt{10}
\end{align*}
So I was wondering about a way to find this sum without squaring? It seems impossible, but I still want to ask.
|
Write,
$$\sqrt {3\pm\sqrt 5}=\sqrt a\pm\sqrt b,\, a\ge b$$
and we obtain
$$A=\sqrt {3+\sqrt 5}+\sqrt {3-\sqrt 5}=2\sqrt a$$
Then, we want to find ratinal $a,b$ such that:
$$\sqrt {3\pm\sqrt 5}=\sqrt a\pm\sqrt b,\, a>b$$ holds.
We have
$$3\pm \sqrt 5=a+b+2\sqrt {ab}\\
\implies \begin{cases}a+b=3\\ ab=\frac 54\end{cases}$$
Using the Vieta's formulas, we have
$$t^2-3t+\frac 54=0\\
4t^2-12t+5=0\\
t_{1,2}=\frac {6\pm4}{4}\\
\implies a=\frac 52\\
\implies b=\frac 12.$$
This gives,
$$\sqrt {3\pm\sqrt 5}=\sqrt \frac 52\pm \sqrt \frac 12$$
and
$$A=2\sqrt a=\sqrt {10}.$$
|
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|
Ideals having the same norm as a prime ideal in ring of integers of a number field Let $K$ be a number field and $\mathcal{O}_K$ its ring of integers. Let $\mathfrak{p}$ be a prime ideal of $\mathcal{O}_K$, and $I$ be an ideal of $\mathcal{O}_K$ such that $N(I) = N(\mathfrak{p})$, where $N(\cdot)$ is the ideal norm (i.e., $N(I) = |\mathcal{O}_K/I|$). Is it true that:
(a) $I$ is a prime ideal;
(b) $[I]$ and $[\mathfrak{p}]$ have the same order in the ideal class group?
Any help appreciated.
|
The following is perhaps too long for a comment, so I'll put it in the answer. By looking at LMFDB I found that the cubic field $\mathbb{Q}(\sqrt[3]{11})$ has class number $2$. The Minkowski bound is $M = \dfrac{4}{\pi}\cdot \dfrac{6}{27} \sqrt{3267} = \dfrac{88}{\sqrt{3}\pi}<17$, so we need to consider the factorizations of $(p)$ for $p=2,3,5,7,11,13$ in $\mathcal{O}_{\mathbb{Q}(\sqrt[3]{11})}$. The LMFDB page shows that $7$ and $13$ remains inert in $\mathbb{Q}(\sqrt[3]{11})$. $3$ and $11$ factor into prime elements in $\mathcal{O}_{\mathbb{Q}(\sqrt[3]{11})}$: $3 = (-2+\sqrt[3]{11})^2(20+9\sqrt[3]{11}+4\sqrt[3]{121}), 11=(\sqrt[3]{11})^3$. The factorizations of $(2)$ and $(5)$ are
$$(2) = (2,5-\sqrt[3]{121})(2,25+11\sqrt[3]{11}+5\sqrt[3]{121})$$
$$(5) = (5,9-4\sqrt[3]{11})(5,81+36\sqrt[3]{11}+16\sqrt[3]{121})$$
The ideal $(2,5-\sqrt[3]{121})$ has norm $2$ and $(5,9-4\sqrt[3]{11})$ has norm $5$, so they are prime ideals. Since $p=2,5$ have Frobenius cycle type $2,1$, the other factors are also prime ideals. $(2,5-\sqrt[3]{121})$ is not principal: $(2,5-\sqrt[3]{121})(5,9-4\sqrt[3]{11}) = (-1+\sqrt[3]{11})$ is principal, if $(2,5-\sqrt[3]{121})$ is principal then $(2)$ and $(5)$ both factor into principal ideals, which would imply that $\mathbb{Q}(\sqrt[3]{11})$ has class number $1$, a contradiction.
As a result, if we pick $I = (5-\sqrt[3]{121})$ and $\mathfrak{p} = (2,25+11\sqrt[3]{11}+5\sqrt[3]{121})$, then $N(I)=N(\mathfrak{p}) = 4$, and $I$ is not a prime ideal; what's more, $I$ is principal while $\mathfrak{p}$ is not. (The same happens for $I = (9-4\sqrt[3]{11})$ and $\mathfrak{p} = (5,81+36\sqrt[3]{11}+16\sqrt[3]{121})$.)
|
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|
Solving floor function system of equations given $x \lfloor y \rfloor + y \lfloor x \rfloor =66$ and $x \lfloor x \rfloor + y \lfloor y \rfloor=144$
$x$ and $y$ are real numbers satisfying $x \lfloor y \rfloor + y \lfloor x \rfloor =66$ and $x \lfloor x \rfloor + y \lfloor y \rfloor=144.$ Find $x$ and $y.$
I first assumed $x>y.$ Then I approximated $x$ and $y$ respectively to be in ranges $11<x<12$ and $2<y<3.$ Then this means $\lfloor x \rfloor =11$ and $\lfloor y \rfloor =2.$ So I wrote the system of equations as $2x+11y=66$ and $11x+2y=144,$ but solving the system gave me solutions of $x$ and $y$ that didn't satisfy the $\lfloor x \rfloor =11$ and $\lfloor y \rfloor =2$ constraint. I'm not really sure how to approach this problem from here other than trying values. May I have some help?
|
By adding the given equations,
$$\begin{align*}
(x+y)(\lfloor x\rfloor + \lfloor y\rfloor) &= 210\\
(\lfloor x\rfloor+\lfloor y\rfloor + \{x\}+\{y\})(\lfloor x\rfloor +\lfloor y\rfloor) &= 210\\
\left(\lfloor x\rfloor+\lfloor y\rfloor+\frac{\{x\}+\{y\}}2\right)^2 - \left(\frac{\{x\}+\{y\}}2\right)^2 &= 210\\
\end{align*}$$
Since $0\le \{\ldots\} < 1$, then $0\le \frac{\{x\}+\{y\}}2 < 1$ and $0\le \left(\frac{\{x\}+\{y\}}2\right)^2 < 1$,
$$
210 \le \left(\lfloor x\rfloor+\lfloor y\rfloor+\frac{\{x\}+\{y\}}2\right)^2 < 211\\
\begin{align*}
14 < \lfloor x\rfloor+\lfloor y\rfloor+\frac{\{x\}+\{y\}}2 &< 15 &\text{or}&&-15 < \lfloor x\rfloor+\lfloor y\rfloor+\frac{\{x\}+\{y\}}2 &< -14\\
13 < \lfloor x\rfloor+\lfloor y\rfloor &< 15 &\text{or}&&-16 < \lfloor x\rfloor+\lfloor y\rfloor &< -14\\
\lfloor x\rfloor+\lfloor y\rfloor &=14 &\text{or}&& \lfloor x\rfloor+\lfloor y\rfloor &= -15\\
\end{align*}$$
Similarly, by subtracting the given equations,
$$\begin{align*}
(x-y)(\lfloor x\rfloor - \lfloor y\rfloor) &= 78\\
(\lfloor x\rfloor-\lfloor y\rfloor + \{x\}-\{y\})(\lfloor x\rfloor -\lfloor y\rfloor) &= 78\\
\left(\lfloor x\rfloor-\lfloor y\rfloor+\frac{\{x\}-\{y\}}2\right)^2 - \left(\frac{\{x\}-\{y\}}2\right)^2 &= 78\\
\end{align*}$$
Since $0\le \{\ldots\} < 1$, then $-\frac12 < \frac{\{x\}-\{y\}}2 < \frac12$ and $0\le \left(\frac{\{x\}-\{y\}}2\right)^2 < \frac14$,
$$
78 \le \left(\lfloor x\rfloor-\lfloor y\rfloor+\frac{\{x\}-\{y\}}2\right)^2 < 78.25\\
\begin{align*}
8.8 < \lfloor x\rfloor-\lfloor y\rfloor+\frac{\{x\}-\{y\}}2 &< 9 &\text{or}&&-9 < \lfloor x\rfloor-\lfloor y\rfloor+\frac{\{x\}-\{y\}}2 &< -8.8\\
8.3 < \lfloor x\rfloor-\lfloor y\rfloor &< 9.5 &\text{or}&&-9.5 < \lfloor x\rfloor-\lfloor y\rfloor &< -8.3\\
\lfloor x\rfloor-\lfloor y\rfloor &=9 &\text{or}&& \lfloor x\rfloor-\lfloor y\rfloor &= -9\\
\end{align*}$$
Considering the parity of $\lfloor x\rfloor$ and $\lfloor y \rfloor$, since their difference ($=9$) is odd, their sum is also odd:
$$\begin{align*}
\lfloor x\rfloor+\lfloor y\rfloor &= -15\\
\lfloor x\rfloor-\lfloor y\rfloor &= \pm 9\\
(\lfloor x\rfloor, \lfloor y \rfloor) &= (-3, -12) &\text{or}&& (\lfloor x\rfloor, \lfloor y \rfloor) &= (-12, -3)
\end{align*}$$
Substituting the first case $(\lfloor x\rfloor, \lfloor y \rfloor) = (-3, -12)$ into the original equations, to find and confirm the exact $x$ and $y$,
$$\begin{align*}
-12 x -3y &= 66 &\implies && 4x+y &= -22\\
-3x -12y &= 144 &\implies && x+4y &= -48
\end{align*}\\
\boxed{x = -\frac83} (\approx -2.67),\quad
\boxed{y = -\frac{34}3} (\approx -11.33)$$
Both $x$ and $y$ satisfy the original equations. Or similarly the other case by swapping $x$ and $y$.
|
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|
Finding the equation of a plane given three points Below is a problem I did from a Calculus text book. My answer matches the
back of the book and I believe my answer is right. However, the method
I used is something I made up. That is, it is not the method described
in the text book.
Is my method correct?
Problem:
Find the plane through the points $(1,1,-1)$, $(2,0,2)$ and $(0,-2,1)$.
Answer:
The general form of a plane is:
$$ Ax + By + Cz = D$$
Sometimes the following constrain is added:
$$ A^2 + B^2 + C^2 = 1$$
By inspection, we can see this plane is not parallel to the x-axis, the y-axis or the z-axis. Hence,
we can assume that the plane is of the form:
$$ Ax + By + Cz = 1 $$
Now we setup the following system of linear equations.
\begin{align*}
A + B - C &= 1 \\
2A + 2C &= 1 \\
-2B + C &= 1 \\
\end{align*}
To solve this system of equations, we get rid of $A$ and $B$ in the first equation.
\begin{align*}
2A &= 1 - 2C \\
A &= \frac{ 1 - 2C }{2} \\
-2B &= 1 - C \\
B &= \frac{ C - 1 }{2} \\
\left( \frac{ 1 - 2C }{2} \right) + \left( \frac{ C - 1 }{2} \right) - C &= 1 \\
1 - 2C + C - 1 - 2C &= 2 \\
- 2C + C - 2C &= 2 \\
-3C &= 2 \\
C &= -\frac{2}{3} \\
B &= \frac{ -\frac{2}{3} - 1 }{2} = -\frac{2}{6} - \frac{1}{2} \\
B &= -\frac{5}{6} \\
A &= \frac{ 1 - 2\left( -\frac{2}{3} \right) }{2} = \dfrac{1 + \dfrac{4}{3} }{2} \\
A &= \dfrac{7}{6}
\end{align*}
Hence the equation is:
$$ \left( \dfrac{7}{6} \right) A + \left( -\frac{5}{6} \right) B + \left( -\frac{2}{3} \right) C = 1 $$
Clearing the fraction, we get the final answer of:
$$ 7A - 5B - 4C = 6 $$
As pointed out by Paul, the correct answer is:
$$ 7x - 5y - 4z = 6 $$
|
You can also use standard solution which is the equation of a plane passing through a given point and a line. Suppose a plane passes through point M(x_0, y_0, z_0) and a line with following equation:
$\frac {x-x_1}{l}=\frac {y-y_1}{m}=\frac {z-z_1}{n}$
Then the equation of plane is:
$\begin {vmatrix}x-x_0&y-y_0& z-z_0\\x_1-x_0&y_1-y_0&z_1-z_0\\l&m&n\end {vmatrix}=0$
Now take for example points A an B; the components of line AB are:
$l=x_B-x_A=2-1=1$
$m=y_B-y_A=0-1=-1$
$n=z_B-z_A=2+1=3$
Take for example point $A (x_1=1, y_1=1, z_1=-1)$ and point $C(x_0=0, y_0=-2, z_0=1)$ and construct the determinant:
$\begin {vmatrix}x-0&y+2& z-1\\1-0&1+2&-1-1\\1&-1&3\end {vmatrix}=0$
Now solve this equation you finally get:
$$7x-5y-4z=6$$
|
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|
How to find this indefinite integral? $\int\frac{1+x^4}{(1-x^4)\cdot \sqrt{1+x^4}}dx$ I am thinking of a trig sub of $x^2 = \tan{t}$ but its not leading to a nice trigonmetric form, which i can integrate. Our teacher said that it can be computed using elementary methods, but I'm unable to think of the manipualtion.
|
\begin{aligned}\int{\frac{1+x^{4}}{\left(1-x^{4}\right)\sqrt{1+x^{4}}}\,\mathrm{d}x}&=\int{\frac{1+x^{4}}{\left(1-x^{2}\right)\left(1+x^{2}\right)\sqrt{1+x^{4}}}\,\mathrm{d}x}\\ &=\int{\frac{x^{2}+\frac{1}{x^{2}}}{\left(1-x^{2}\right)\left(x+\frac{1}{x}\right)\sqrt{x^{2}+\frac{1}{x^{2}}}}\,\mathrm{d}x}\\ &=\int{\frac{x^{2}+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}\left(x+\frac{1}{x}\right)\sqrt{x^{2}+\frac{1}{x^{2}}}}\left(\frac{1}{x^{2}}-1\right)\mathrm{d}x}\\ &=\int{\frac{\left(x+\frac{1}{x}\right)^{2}-2}{\left(\left(x+\frac{1}{x}\right)^{2}-4\right)\left(x+\frac{1}{x}\right)\sqrt{\left(x+\frac{1}{x}\right)^{2}-2}}\left(1-\frac{1}{x^{2}}\right)\mathrm{d}x}\\ &=\int{\frac{y^{2}-2}{y\left(y^{2}-4\right)\sqrt{y^{2}-2}}\,\mathrm{d}y}\end{aligned}
Using another substitution $ y = \sqrt{2}\cosh{t} $. We're left with the integral : $$ \frac{1}{2}\int{\frac{\tanh{t}}{\sinh^{2}{t}-1}\,\mathrm{d}t} $$
I'll let you take it from here.
Of caurse we made the assumption that we're working in a domain $ \subset \left]0,+\infty\right[ $, so we could have $ \sqrt{x^{2}}=x $. But the same approach will stand if we're working on a domain of $ \left]-\infty, 0\right[ $.
|
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|
Sum of squares of ratios of diagonal of a regular heptagon A problem from the 1998 Lower Michigan Math Competition...
A regular heptagon has diagonals of two different lengths. Let $a$ be the length of a side, $b$ the length of a shorter diagonal, and $c$ the length of a longer diagonal. Prove that
$$
\frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2} = 6 \quad \text{ and } \quad \frac{b^2}{a^2} + \frac{a^2}{c^2} + \frac{c^2}{b^2} = 5.
$$
What I have so far:
*
*Introduce coordinates so that the heptagon consists of the seventh roots of unity in the complex plane.
*Let $\omega = e^{2 \pi i/7}$ so that the vertices are $1, \omega, \omega^2, \dots, \omega^6$. Then $a = |1-\omega|$ and $b = |1-\omega^2| = |1-\omega||1+\omega|$ and $c = |1-\omega^3| = |1-\omega||1+\omega+\omega^2|$.
*The first requested expression then becomes
$$
\frac{1}{|1+\omega|^2} + \frac{|1+\omega|^2}{|1+\omega+\omega^2|^2} + |1+\omega+\omega^2|^2.
$$
*All those modulus-squared expressions can be written as (thingy) times (conjugate of thingy), and the conjugate of a power of $\omega$ is another power of $\omega$. For example,
$$
|1+\omega|^2 = (1+ \omega)(1 + \overline{\omega}) = (1 + \omega)(1 + \omega^6).
$$
Using this, the first requested expression becomes
$$
\frac{1}{(1+\omega)(1+\omega^6)} + \frac{(1+\omega)(1+\omega^6)}{(1+\omega+\omega^2)(1+\omega^5+\omega^6)} + (1 + \omega + \omega^2)(1 + \omega^5 + \omega^6).
$$
Annnnd...that's about where I ran out of steam. Not sure how much of this is progress, but I really want the complex-roots-of-unity approach to pan out!
|
Use $\displaystyle e^{\tfrac{2\pi i}{7}}=\cos \frac{2\pi }{7}+i\sin \frac{2\pi }{7} $ to get $$|1+\omega|=\left|1+\cos \frac{2\pi }{7}+i\sin \frac{2\pi }{7}\right|$$$$=\left|2\cos^2 \frac{\pi }{7}+2i\sin \frac{\pi }{7}\cos \frac{\pi }{7}\right|=2\cos \frac{\pi }{7}\left|\cos \frac{\pi }{7}+i \sin\frac{\pi }{7} \right|$$$$=2\cos \frac{\pi }{7}.$$
Also, $\omega^2= e^{\tfrac{4\pi i}{7}} =\cos \dfrac{4\pi }{7}+i\sin \dfrac{4\pi }{7} $ and $$|1+\omega +\omega^2|=\left|\left(\color{red}{1}+\cos \frac{2\pi }{7}+\color{red}{\cos \frac{4\pi }{7}}\right)+i\left(\sin \frac{2\pi }{7}+\sin \frac{4\pi }{7}\right)\right|$$$$= \left|\left(\cos \frac{2\pi }{7}+2\cos^2\frac{2\pi}{7}\right)+i\left(\sin \frac{2\pi }{7}+2\sin \frac{2\pi }{7}\cos\frac{2\pi}{7}\right)\right|$$$$=1+2\cos\frac{2\pi}{7}.$$
|
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|
Show that for any two positive real numbers $a, b: \frac a{a + 2b} + \frac b{b + 2a} ≥ \frac12$ Question:
Show that for any two positive real numbers $a, b: \frac a{a + 2b} + \frac b{b + 2a} ≥ \frac12$.
So for this question, I began by expanding all terms and moving them all to one side. However, I do not know how to definitively say that the statement is proved.
This is my "work" so far:
$a, b: \frac a{a + 2b} + \frac b{b + 2a} ≥ \frac12$
$\frac {2a(b + 2a) + 2b(a + 2b) - 1(b + 2a)(a + 2b)}{2(a + 2b)(b+2a)} ≥ 0$
$\frac {4ab + 4a^2 + 4b^2 - 2b^2 - 5ab - 2a^2}{4b^2 + 10ab + 4a^2} ≥ 0$
$\frac {2a^2 - ab + 2b^2}{4b^2 + 10ab + 4a^2} ≥ 0$
|
One of $a$ or $b$ is non zero, or else the question is not well defined.
If $a=b$, then we get left hand side of inequality to be $2/3$, which is $\geq 1/2$ as desired.
So assume $a>b$. Then
$$ \frac{a}{a+2b} + \frac{b}{b + 2a}
\\ = \frac{a(b+2a)+b(a+2b)}{(a+2b)(b+2a)}
\\ = \frac{2ab+2a^2+2b^2}{3ab + (2ab+2a^2 + 2b^2)}
\\ = \frac{x}{3ab + x} \mbox{ where } x = 2ab+2a^2 + 2b^2 > 0
\\ = \frac{1}{\frac{3ab}{x} + 1}
$$
So we need to show that
$$\frac{3ab}{x} \leq 1 \iff 3ab \leq x \iff ab \leq 2a^2+2b^2 $$
But the furthest right inequality is obvious since
$$ ab < a^2 < 2a^2 < 2a^2 + 2b^2 $$
|
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|
Solving system of logarithmic equations. I have two equations
\begin{align}
\frac{1}{2} \ln(w+b-0.4)+\frac{1}{2}\ln(w-0.4) & =\ln(w)\\
\frac{1}{2} \ln(w+b-0.4)+\frac{1}{2}\ln(w-0.4) & =1.2
\end{align}
I want to find $b$ and $w$.
—-
I find $w=e^{1.2}$
But when I derive $b$, I obtain that $b= \frac{e^{2.4}}{e^{1.2}-0.4}-e^{1.2}+0.4$
But the value of $b$ Seems not good to me. Please share your answers for the value of $b$. Thank you.
|
From the first equation, we have
\begin{align}
\frac12\log(w+b-0.4)+\frac12\log(w-0.4)&=\log w\\
\implies\log\left(\sqrt{w+b-0.4}\sqrt{w-0.4}\right)&=\log w\\
\implies\sqrt{(w+b-0.4)(w-0.4)}&=w\\
\implies(w+b-0.4)(w-0.4)&=w^2\\
\implies w&=0.4\left(\frac{5b-2}{5b-4}\right)\\
\implies w&=0.4\left(1+\frac2{5b-4}\right)\tag1
\end{align}
We know that $\log w=1.2\implies w=e^{1.2}$. Using this to solve for $b$ from $(1)$, we have
$$e^{1.2}=0.4\left(1+\frac2{5b-4}\right)\implies b=0.2\left(\frac{2}{2.5e^{1.2}-1}+4\right)\approx0.854792,$$ and is the same result upon comparison to your answer.
|
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|
Methods of deriving $\sum^n_{x=0} xr^{x-1}$ To calculate this sum, we can differentiate $\sum^n_{x=0} r^{x}$:
$$ \sum^n_{x=0} xr^{x-1} = \frac{d}{dr} \sum^n_{x=0} r^{x} = \frac{d}{dr} \left(\frac{r^{n+1}-1}{r-1}\right) = \frac{(n+1)r^{n}}{r-1} - \frac{r^{n+1}-1}{(r-1)^2}
$$
However, this disagrees with the method of factoring out $r^{-1}$:
$$ \sum^n_{x=0} xr^{x-1} = \frac{1}{r}\sum^n_{x=0} xr^{x} = \frac 1r \left( \frac{ nr^{n+1}}{r-1} -\frac{r^{n+1}-r}{(r-1)^2} \right) = \frac{nr^n}{r-1} - \frac{r^n-1}{(r-1)^2}
$$
Which of the two methods should be used, or did I apply them incorrectly?
|
The two expressions are the same:
$$\begin{align} \frac{(n+1)r^{n}}{r-1} - \frac{r^{n+1}-1}{(r-1)^2} &= \frac{nr^{n}}{r-1} + \frac{r^n}{r-1}- \frac{r^{n+1}-1}{(r-1)^2}\\ &= \frac{nr^{n}}{r-1} + \frac{r^n(r-1)}{(r-1)^2}- \frac{r^{n+1}-1}{(r-1)^2}\\ &= \frac{nr^n}{r-1} - \frac{r^n-1}{(r-1)^2}. \end{align}$$
|
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|
If $g(x)=ax^2+bx+c$ and $f(x)= \begin{cases} {g(x)} & {x\ge k} \\ {g'(x)} & {x< k} \end{cases} $ , $\max (k)=?$ if $f$ is a differentiable function
If $g(x)=ax^2+bx+c$ and $f(x)= \begin{cases} {g(x)} & {x\ge k} \\
{g'(x)} & {x< k} \end{cases} $. If $f(x)$ is a differentiable
function, what is the maximum value of $k$, provided that $b+c=a$?
$1)\frac34\qquad\qquad2)1\qquad\qquad3)3\qquad\qquad4)4$
In order to $f(x)$ be differentiable function, we should have $g(k)=g'(k)=g''(k)$,
$$ak^2+bk+c=2ak+b=2a$$ $$(b+c)k^2+bk+c=(2b+2c)k+b=2b+2c$$
Here for each equation I tried to equate the coefficients of $k^2 , k^1 , k^0$ but I get $a=b=c=0$ which doesn't make sense at all. I don't know how to continue form here.
|
The original question looks wrong or at least badly worded. From continuity and differentiability at $k$, one indeed gets $g(k)=g'(k)=g''(k)$.
First assume that $a\neq 0$ and therefore $b+c\neq 0$.
Solving the differentiability condition gives $k=1-\frac{b}{2a}=\frac{b+2c}{2b+2c}$, using $b+c=a$.
The continuity condition gives the degree two equation in $k$, $(b+c)k^2 -(b+2c)k+c-b=0$, with discriminant $\Delta=5b^2+4bc$. The solutions are $k_{1,2}=\frac{b+2c\pm \sqrt{5b^2+4bc}}{2b+2c}$.
Comparing the two conditions, assuming $b+c\neq 0$, the only way a $k$ can exist is if $5b^2+4bc=0$, i.e., $b=0$ or $c=-\frac{5b}{4}$.
If $b=0$, then the unique solution is $k=1-\frac{0}{2a}=1$.
If $c=-\frac{5b}{4}$, then again there is a unique solution $k=\frac{b-\frac{5b}{2}}{2(b-\frac{5b}{4})}=3$.
Finally, if $a=b+c=0$, then $g(x)=bx-b$ and the differentiability and continuity condition boils down to $bk-b=b=0$, so $f$ is differentiable if and only if $g\equiv0$. And then clearly any $k$ works.
|
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How did people come up with the formula $a^3+b^3=(a+b)(a^2-ab+b^2)$? Every resource that I've read proves the formula
$$ a^3 + b^3 = (a+b)(a^2-ab+b^2) \tag1$$
by just multiplying $(a+b)$ and $(a^2 - ab + b^2)$.
But how did people come up with that formula? Did they think like, "Oh, let's just multiply these polynomials, I don't know why, let's just do it." I don't think that people just pointed a finger at the sky and came up with that formula.
So, how to prove $(1)$?
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Alternative approach:
Stealing the insight from the comment of Ethan Bolker.
Mathematicians discovered that
$$(1 + x + x^2 + \cdots + x^n) \times (1 - x) = 1 - x^{n+1}. \tag1 $$
For $0 \neq a,b$,
$a^3 + b^3$ can be rewritten as
$$a^3 \times \left[1 - \left(\frac{-b}{a}\right)^3\right]. \tag2 $$
Setting $~\displaystyle x = \left(\frac{-b}{a}\right)~$ (1) and (2) collectively imply that
$$a^3 + b^3 = a^3 \times \left[1 - \frac{-b}{a}\right] \times \left[ ~1 + \frac{-b}{a} + \left(\frac{-b}{a}\right)^2 ~\right]$$
$$ = a^3 \times \left[\frac{a + b}{a}\right] \times \left[\frac{a^2 - ab + b^2}{a^2}\right] = (a+b) \times (a^2 - ab + b^2).$$
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|
Calculate $\sin^5\alpha-\cos^5\alpha$ if $\sin\alpha-\cos\alpha=\frac12$ Calculate $$\sin^5\alpha-\cos^5\alpha$$ if $\sin\alpha-\cos\alpha=\dfrac12$.
The main idea in problems like this is to write the expression that we need to calculate in terms of the given one (in this case we know $\sin\alpha-\cos\alpha=\frac12$).
I don't see how to even start to work on the given expression as we cannot use $a^2-b^2=(a-b)(a+b)$ or $a^3-b^3=(a-b)(a^2+ab+b^2)$. So in other words, I can't figure out how to factor the expression (even a little).
The given answer is $\dfrac{79}{128}$.
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$$\dfrac1{2\sqrt2}=\sin\left(\alpha-\dfrac\pi4\right)=\sin x$$ where $\alpha-\dfrac\pi4=x$
$$\sin^5\alpha-\cos^5\alpha=\sin^5\left(x+\dfrac\pi4\right)-\cos^5\left(x+\dfrac\pi4\right)=\dfrac{(\cos x+\sin x)^5-(\cos x-\sin x)^5}{(\sqrt2)^5}$$
Now,
$$(\cos x+\sin x)^5-(\cos x-\sin x)^5=2\sin^5x+2\binom53\sin^3x\cos^2x+2\binom54\sin x\cos^4x$$
Use $\cos^2x=1-\sin^2x=1-\left(\dfrac1{2\sqrt2}\right)^2=?$
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Find the all possible values of $a$, such that $4x^2-2ax+a^2-5a+4>0$ holds $\forall x\in (0,2)$
Problem: Find the all possible values of $a$, such that
$$4x^2-2ax+a^2-5a+4>0$$
holds $\forall x\in (0,2)$.
My work:
First, I rewrote the given inequality as follows:
$$
\begin{aligned}f(x)&=\left(2x-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0\end{aligned}
$$
Then, we have
$$
\begin{aligned}
0<x<2\\
\implies -\frac a2<2x-\frac a2<4-\frac a2\end{aligned}
$$
Case $-1:\,\,\,a≥0 \wedge 4-\frac a2≤0$.
This leads,
$$
\begin{aligned}\frac {a^2}{4}>\left(2x-\frac a2\right)^2>\left(4-\frac a2\right)^2\\
\implies \frac {a^2}{4}+\frac {3a^2}{4}-5a+4>f(x)>\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4\end{aligned}
$$
For $f(x)>0$, it is enough to take $\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0$ with the restriction $a≥0\wedge 4-\frac a2≤0$.
Case $-2:\,\,\,a≤0 \wedge 4-\frac a2≥0$.
We have:
$$
\begin{aligned}\frac {a^2}{4}<\left(2x-\frac a2\right)^2<\left(4-\frac a2\right)^2\\
\implies \frac {a^2}{4}+\frac {3a^2}{4}-5a+4<f(x)<\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4\end{aligned}
$$
Similarly, for $f(x)>0$, it is enough to take $\frac{a^2}{4}+\frac {3a^2}{4}-5a+4>0$ with the restriction $a≤0\wedge 4-\frac a2≥0$.
Case $-3:\,\,\,a≥0 \wedge 4-\frac a2≥0$.
This case implies, $\left(2x-\frac a2\right)^2≥0$. This means, $f(x)≥\frac {3a^2}{4}-5a+4$
Thus, for $f(x)>0$, it is enough to take $\frac {3a^2}{4}-5a+4>0$ with the restriction $a≥0\wedge 4-\frac a2≥0$.
Finally, we have to combine all the solution sets we get.
I haven't done the calculation, because I want to make sure that the method I use is correct. Do you see any flaws in the method?
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Yes, (though strangely written) your method is correct. It can be reformulated noticing more explicitely that $f(x):=4x^2-2ax+a^2-5a+4$ is decreasing for $x\le a/4$ and increasing for $x\ge a/4$, since $f'(x)=8x-2a.$ We thus recover your 3 cases (and the same numerical results as Andreas, but more directly):
*
*Case 1: $a/4\ge2$. Then, $f>0$ on $(0,2)$ iff $f(2)\ge0$, i.e. $a^2-9a+20\ge0$, i.e. $a\le2$ or $a\ge4$. This or-condition is always satisfied since $a\ge8$ by hypothesis.
*Case 2: $0<a/4<2$. Then, $f>0$ on $(0,2)$ iff $f(a/4)>0$, i.e. $\frac34a^2-5a+4>0$, i.e. $a<a_1:=\frac{10-2\sqrt{13}}3\approx0.93$ or $a>a_2:=\frac{10+2\sqrt{13}}3\approx5.74.$ This or-condition is satisfied iff $0<a<a_1$ or $a_2<a<8.$
*Case 3: $a/4\le0$. Then, $f>0$ on $(0,2)$ iff $f(0)\ge0$, i.e. $a^2-5a+4\ge0,$ i.e. $a\le1$ or $a\ge4.$ This or-condition is always satisfied since $a\le0$ by hypothesis.
Putting your three cases together gives the whole set of solutions:
$a<a_1$ or $a>a_2.$
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|
In Tom Apostol Calculus: method of exhaustion for the area of a parabolic segment of $x^2$, proving $b^3/3$ is the only number between $s_n$ and $S_n$ In the method of exhaustion for the area of a parabolic segment of $x^2$, the part where the author proves that $b^3/3$ is the only number which satisfies $s_n <A<S_n$
The prove has come down to this
$\frac{b^3}{3}-\frac{b^3}{n}< A<\frac{b^3}{3}+\frac{b^3}{n} $ for every integer $n>=1$. ,.....(1)
Now there are three possibilities
$ A>\frac{b^3}{3}$ , $ A<\frac{b^3}{3}$ , $ A<\frac{b^3}{3}$
The author now shows that the first two inequalities lead to a contradiction, then we must have $ A=\frac{b^3}{3}$.
(what I find difficult to understand is the contradiction, I can't see which fact is being contradicted.)
The author says this:
Suppose the inequality $ A>\frac{b^3}{3}$ were true. Then from the second inequality in (1) we obtain
$ A-\frac{b^3}{3}<\frac{b^3}{n}$ for every integer $n \geq 1$.
Since $ A-\frac{b^3}{3}$ is positive, we may divide both side by $ A-\frac{b^3}{3}$ and then multiply by n to obtain the equivalent statement
$ n<\frac{b^3}{A-\frac{b^3}{3}}$ for every n.
But this inequality is obviously false when $ n \geq \frac{b^3}{A-\frac{b^3}{3}}$.
Hence the inequality $ A>\frac{b^3}{3}$ leads to contradiction.
I am confused about what contradicts what?
Is it saying that $ A>\frac{b^3}{3}$ is true only when $ n<\frac{b^3}{A-\frac{b^3}{3}}$. and it contradicts from the assumption that it should be true for all $n$?
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You have showed that, if we assume that $A > \frac{b^3}{3}$, then equation (1) is true only when $n < \frac{b^2}{A - \frac{b^3}{3}}$. However, the equation (1) must be true for all integers $n \geq 1$.
We have reached a contradiction since our assumption implies an upper bound for $n$ in equation (1), but the equation must be true for all integer $n \geq 1$.
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|
Is this proof of $\left\lfloor \frac{n}{m} \right\rfloor = \left\lceil \frac{n-m+1}{m} \right\rceil$ correct? I've been practicing proving things about floor and ceiling functions, so I thought I'd try to prove this well-known identity:
$$\left\lfloor \frac{n}{m} \right\rfloor = \left\lceil \frac{n-m+1}{m} \right\rceil$$
for all $n,m \in \mathbb{Z}$, $m>0$.
This is what I came up with. Is my proof correct?
Proof:
[see edit below]
Case 1: $m=1$ $$\left\lfloor \frac{n}{1} \right\rfloor = \lfloor n \rfloor = n$$ $$\left\lceil \frac{n-1+1}{1} \right\rceil = \lceil n \rceil = n$$
Case 2: $m>1$
If $\frac{n}{m}$ is an integer, then
$$\left\lceil \frac{n-m+1}{m} \right\rceil = \left\lceil \frac{n}{m} -1 + \frac{1}{m}\right\rceil = \frac{n}{m} -1 + \left\lceil \frac{1}{m} \right\rceil = \frac{n}{m} -1 + 1 = \frac{n}{m} = \left\lfloor \frac{n}{m} \right\rfloor$$
If $\frac{n}{m}$ is NOT an integer, then
$$\left\lfloor \frac{n}{m} \right\rfloor = \left\lceil \frac{n}{m} \right\rceil - 1 = \left\lceil \frac{n}{m} + \frac{1}{m} \right\rceil - 1 = \left\lceil \frac{n}{m} + \frac{1}{m} -1 \right\rceil = \left\lceil \frac{n-m+1}{m} \right\rceil$$
$\blacksquare$
If it's correct but you know a simpler/better way to prove it, please include that in your answer. Thank you.
EDIT: As pointed out by user peterwhy, "Case 1: $m=1$" is simply a special case of "$\frac{n}{m}$ is an integer" and therefore is not needed; hence I have grayed it out, and we don't need to separate the $m=1$ and $m>1$ cases anymore.
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Your proof looks okay (although the second equality in the $n/m$ not being integer might need some clarification). Notice that this identity can be proven essentially the same way as Prove that $\left\lceil \frac{n}{m} \right\rceil =\left \lfloor \frac{n+m-1}{m} \right\rfloor$ . Here is one variant.
By division with remainder, we can write $n=\lfloor \frac{n}{m} \rfloor m+(n \bmod m)$. Adding $-m+1$ to both sides and dividing by $m$ we get
$$
\frac{n-m+1}{m}=\Big\lfloor \frac{n}{m} \Big\rfloor +\frac{(n \bmod m)-m+1}{m}.\tag{*}
$$
Since $0 \leq n \bmod m \leq m-1$, we have $-m+1 \leq (n \bmod m)-m+1\leq 0$. So the rightmost fraction in $(*)$ (denote $x$) satisfies $-1<-1+\frac{1}{m}\leq x \leq 0$. Hence applying the ceiling function (and using that $\Big\lfloor \frac{n}{m} \Big\rfloor$ is an integer), we get
$$
\Big\lceil \frac{n-m+1}{m}\Big\rceil =\Big\lfloor \frac{n}{m} \Big\rfloor +\Big\lceil\frac{(n \bmod m)-m+1}{m}\Big\rceil=\Big\lfloor \frac{n}{m} \Big\rfloor.
$$
Note. Alternatively, if you already know the dual statement $\left\lceil \frac{n}{m} \right\rceil =\left \lfloor \frac{n+m-1}{m} \right\rfloor$, you can just use $\lfloor -x \rfloor=-\lceil x \rceil$ few times:
$$
\left\lfloor \frac{n}{m} \right\rfloor = -\Big\lceil \frac{-n}{m} \Big\rceil = - \left \lfloor \frac{-n+m-1}{m} \right\rfloor =-\Big(-\Big\lceil \frac{n-m+1}{m}\Big\rceil\Big)=\Big\lceil \frac{n-m+1}{m}\Big\rceil
$$
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Deriving $\gamma \approx H(n)-\ln(n+1)+\frac{1}{2(n+1)}+\frac{1}{12(n+1)^2}$ The Euler-Mascheroni constant can be represented geometrically by the infinite sum of the areas in blue in the following picture, which is the area between the curve $y=1/x$ and the harmonic numbers.
Thus, the total area can be approximated by taking a finite sum of the first $n$ areas, such that
$$\gamma \approx H(n)-\ln(n+1)$$
This approximation can be improved by noting that, as $n$ increases, these areas approach a triangle with base $\frac{1}{n}-\frac{1}{n+1}$ and height $1$. Therefore, the sum of the remaining areas is
$$A \approx\sum_{k=n+1}^{\infty}\frac{1}{2} \left(\frac{1}{n}-\frac{1}{n+1} \right)=\frac{1}{2(n+1)}$$
Thus,
$$\gamma \approx H(n)-\ln(n+1)+\frac{1}{2(n+1)}$$
Is there also a relatively simple way to derive the following even better approximation?
$$\gamma \approx H(n)-\ln(n+1)+\frac{1}{2(n+1)}+\frac{1}{12(n+1)^2}$$
I saw an almost identical formula in the harmonic numbers Wolfram MathWorld page but it seems that it comes from the rather complex Euler-Maclaurin formula.
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Consider one of the intervals from $n$ to $n+1$
After your $\frac{1}{2(n+1)}$ approximation, there is still some area between the $y=1/x$ curve and the hypotenuse of each triangle.
It's possible to approximate the function $y=1/x$ locally by a parabola with the Taylor series at $x=n+1/2$. Let this approximation be $f(x)$, then
$$f(x)=\frac{1}{n+1/2}-\frac{(x-n-1/2)}{(n+1/2)^2}+\frac{(x-n-1/2)^2}{(n+1/2)^3}$$
The hypotenuse of the triangle, represented by the segment AE in the image above, is given by
$$g(x)=\frac{1}{n}+(x-n)\left(\frac{1}{n+1}-\frac{1}{n}\right)$$
Therefore, the missing area is approximately
$$A\approx\int_{n}^{n+1}g(x)-f(x)\text{ }dx=\frac{8n^2+8n+3}{6n(n+1)(2n+1)^3}$$
Note that $(2n+1)^2=4n^2+4n+1$, and the numerator is $2(4n^2+4n+1.5)$. So a new approximation is
$$A\approx\frac{1}{3n(n+1)(2n+1)}=\frac{1}{6n(n+1)(n+1/2)}$$
So the total missing area would be the sum of $\frac{1}{6k(k+1)(k+1/2)}$ from $k=n+1$ to infinity. However, this does not have a "nice" solution due to the $1/2$ term, which is not an integer. Even so, we can get a lower and upper bound by considering the following twos series:
$$S_1=\sum_{k=n+1}^{\infty}\frac{1}{6k(k+1)(k+2)}=\frac{1}{12(n+1)(n+2)}$$
$$S_2=\sum_{k=n+1}^{\infty}\frac{1}{6k(k+1)(k-1)}=\frac{1}{12n(n+1)}$$
In which the $(n+1/2)$ term was changed to $(n+2)$ and $(n-1)$, respectively (-1 and 2 are the closest integers to $1/2$ that are not 0 or 1).
Now, it is interesting to consider the harmonic mean of both results. The reason is that the harmonic mean of $\frac{1}{6k(k+1)(k+2)}$ and $\frac{1}{6k(k+1)(k-1)}$ is exactly $\frac{1}{6k(k+1)(k+1/2)}$, which was the expression we were trying to use initially. So a better approximation for $A$ is to consider the harmonic mean of $S_1$ and $S_2$.
$$A\approx \frac{1}{12(n+1)^2}$$
Therefore,
$$\gamma \approx H(n)-\ln(n+1)+\frac{1}{2(n+1)}+\frac{1}{12(n+1)^2}$$
is a better approximation.
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Sum of Consecutive Powers I am trying to understand a cross between steps when finding the formula for the sum of consecutive powers.
I am following the steps from a webpage here http://mikestoolbox.com/powersum.html but will also provide them below:
Step 1: Sum = x^5 + x^4 + x^3 + x^2 + x + 1
Step 2: Sum · x + 1 = x^6 + x^5 + x^4 + x^3 + x^2 + x + 1
= x^6 + (x^5 + x^4 + x^3 + x^2 + x + 1)
= x^6 + Sum
Step 3: Rearranging the above equation we can then get:
Sum · (x − 1) = x^6 − 1
I am missing some very basic math knowledge that is blocking me from understanding how he is getting from step 2 to step 3. I.e how is he going from Sum * x + 1 = x^6 + Sum to the following step
Sum * (x-1) = x^6 - 1
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You do not need to add 1 after multiplying with $x$.
$$S=1+x+x^2+\ldots+x^n$$
$$xS=x+x^2+\ldots+x^n+x^{n+1}$$
Subtract both equations:
$$S(1-x)=1-x^{n+1}$$
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Evaluate $\sin^2x = \frac{2+ \sqrt{3}}{4}$
Evaluate $\sin^2x = \frac{2+ \sqrt{3}}{4}$
Find value of $2x$
I worked it out as such:
$\sin^2x = \frac{2+ \sqrt{3}}{4} \implies \frac{1- \cos 2x}{2} = \frac{2+ \sqrt{3}}{4}$
$2 - 2 \cos 2x = 2 + \sqrt{3}$
$- 2 \cos 2x = + \sqrt{3}$
$ \cos 2x = \frac{- \sqrt{3}}{2}$
And $ 2x = \frac{5\pi}{6}$
I was told that the angle $2x$ has another value: $2x = 2\pi - \frac{5\pi}{6} = \frac{7\pi}{6}$
Why is there another value and what is this value for or mean in this question context.
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It's because there are always two solutions for the equation $\cos x = a$ on $[0,2\pi)$ when $|a|<1$. If $x_1$ is one solution, then $x_2 = 2\pi - x_1$ is another solution. Indeed
$$\cos x_2 = \cos(2\pi - x_1) = \cos(-x_1) = \cos x_1 $$
where the second equality holds since $\cos$ is periodic with period $2\pi$ and the third because it's even function.
EDIT: To address the point Prem made in the comments. In your case you are solving $\cos 2x = a$. If it's expected that $2x\in[0,2\pi)$, then you have two solutions. However, if it's expected that $x\in[0,2\pi)$, and therefore $2x\in [0,4\pi)$, there are four solutions instead. The extra two can be obtained by adding $2\pi$ to the first two, i.e. just using periodicity.
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Representing $\frac{1}{x^2}$ in powers of $(x+2)$ I was asked to represent $f(x)=\frac{1}{x^2}$ in powers of $(x+2)$ using the fact that $\frac{1}{1 − x} = 1 + x + x^2 + x^3 + ...$. I am able to represent $\frac{1}{x^2}$ as a power series, but I am struggling withdoing it in powers of $(x+2)$. This is what I attempted.
$I.$ Firstly, I used the simple expansion
$$\ln(x+1) = \sum_{n=1}^\infty (-1)^n\frac{x^n}{n}$$
which comes from the fact that $\frac{1}{1 − x} = 1 + x + x^2 + x^3 + ...$. I then noticed that
$$\ln(x) = \ln(1+(x-1))=\sum_{n=1}^\infty (-1)^n\frac{(x-1)^n}{n}$$
$II$. $-f(x)=-\frac{1}{x^2}$ is the second derivative of $\ln(x)$, so that
$$\frac{d}{dx}\sum_{n=1}^\infty (-1)^n\frac{(x-1)^n}{n}= \sum_{n=1}^\infty(-1)^n \frac{(x-1)^{n-1}}{n^2}$$
and
$$\frac{d}{dx} \sum_{n=1}^\infty(-1)^n \frac{(x-1)^{n-1}}{n^2} = \sum_{n=1}^\infty(-1)^n \frac{(x-1)^{n-2}}{n^2(n-1)}=-\frac{1}{x^2}$$
from what plainly follows that $$\frac{1}{x^2} = -\sum_{n=1}^\infty(-1)^n \frac{(x-1)^{n-2}}{n^2(n-1)}$$
$III.$ Of course, this expansion is not in powers of $(x+2$). What I pressumed would be the logical thing to do is to repeat these steps but, instead of the using $\ln(1 + (x-1))$, using the equivalent
$$\ln(-2 + (x+2))=\ln((-2)(1+\frac{x+2}{(-2)})) = \ln(-2)+\ln(1+\frac{x+2}{(-2)})$$
I expected to be able to use this because the second term is of the form $\ln(1+ u)$ where $u$ is some function of $x$, and we know the expansion of such expression. However, $\ln(-2)$ is nonsense, since $\ln(x)$ is defined only for $\mathbb{R}^+$.
Is there an alternative, better way to do this or am I missing something? Thanks.
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The standard riff here is to first notice that $1/x^2$ is ($-1$ times) the derivative of $1/x$... and that differentiation preserves power series with a given center. Then some algebra gives
$$ 1/x \;=\; 1/(x+2-2) \;=\; {-1\over 2}{1\over 1-{x+2\over 2}}
\;=\; {-1\over 2} \sum_{n\ge 0} ({x+2\over 2})^n
$$
Differentiating (and multiplying through by $-1$) gives
$${1\over x^2} \;=\; {1\over 2}\sum_{n\ge 0} {n\over 2}\left({x+2\over 2}\right)^{n-1}
$$
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$\epsilon-N$ proof for the limit of $a_n=n\left(\sqrt{1+\frac{1}{n}}-1\right)$ We are told the limit is 1/2, and I know the definition for $\epsilon-N$ convergence, and generally what I am looking for, but just cannot find the right algebra as scratchwork to find how to choose $N$ at the actual top of the proof. So far, what seems to be most useful is this
\begin{align}
\left| n \left(\sqrt{1+\frac{1}{n}}-1\right)-\frac{1}{2}\right|&=\left| \frac{n \left(\sqrt{1+\frac{1}{n}}-1\right) \left(\sqrt{1+\frac{1}{n}}+1\right)}{\sqrt{1+\frac{1}{n}}+1}-\frac{1}{2}\right|\\
&=\left| \frac{1}{\sqrt{1+\frac{1}{n}}+1}-\frac{1}{2}\right|
\end{align}
I have been chasing myself in circles with the rest of the algebra here to get to something that can obviously translate to a choice of $N$.
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We have that $\frac{1}{\sqrt{1+\frac{1}{n}}+1}-\frac{1}{2} <0$ then we can consider, for $0<\varepsilon <\frac12$
$$\frac12-\frac{1}{\sqrt{1+\frac{1}{n}}+1} <\varepsilon \iff \frac{1}{\sqrt{1+\frac{1}{n}}+1} >\frac{1-2\varepsilon}{2}$$
$$\iff \sqrt{1+\frac{1}{n}}+1<\frac{2}{1-2\varepsilon} \iff \sqrt{1+\frac{1}{n}} <\frac{2}{1-2\varepsilon}-1$$
$$\iff 1+\frac{1}{n}<\left(\frac{1+2\varepsilon}{1-2\varepsilon}\right)^2 \iff \frac1n<\left(\frac{1+2\varepsilon}{1-2\varepsilon}\right)^2-1$$
$$\iff n>\frac1{\left(\frac{1+2\varepsilon}{1-2\varepsilon}\right)^2-1}$$
|
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|
Proving $\sum_{n=1}^{\infty}\frac{1}{1+n^2\pi^2} = \frac{1}{e^2-1}$ I hope I'm allowed to ask this question here, but I have to prove that $\sum_{n=1}^{\infty}\frac{1}{1+n^2\pi^2} = \frac{1}{e^2-1}$ using the following Fourier series:
$$
1-\frac{1}{e} + \sum_{n=1}^{\infty}\frac{2}{1+n^2\pi^2}\left[1-\frac{1}{e}(-1)^n\right]\cos(n{\pi}x) =
\begin{cases}
e^x & x\in[-1,0) \\
e^{-x} & x\in[0,1]
\end{cases}
$$
This is my progress so far:
Let $x=0$:
\begin{align*}
\therefore 1-\frac{1}{e} + \sum_{n=1}^{\infty}\frac{2}{1+n^2\pi^2}\left[1-\frac{1}{e}(-1)^n\right] &= 1 \\
\sum_{n=1}^{\infty}\frac{2}{1+n^2\pi^2}\left[1-\frac{1}{e}(-1)^n\right] &= \frac{1}{e} \\
\sum_{n=0}^{\infty}\frac{2}{1+(2n+1)^2\pi^2}\left(1+\frac{1}{e}\right) + \sum_{n=1}^{\infty}\frac{2}{1+(2n)^2\pi^2}\left(1-\frac{1}{e}\right) &= \frac{1}{e}
\end{align*}
Does anyone know what I should do next?
|
$\sum_{n=1}^{\infty}\frac{2}{1+n^2\pi^2}
=\frac{i}{\pi}\sum_{n=1}^{\infty}(\frac{1}{n+i/\pi}-\frac{1}{n-i/\pi})
=\frac{i(-\phi^{(0)}(\infty-i/\pi+1)+\phi^{(0)}(\infty+i/\pi+1)-\phi^{(0)}(1+i/\pi)+\phi^{(0)}(1-i/\pi))}{2\pi}=\frac{1}{2(e-1)}-\frac{1}{2(e+1)}=\frac{1}{e^2-1}$
where $\phi^{(0)}$ is the poly-gamma function of order $0$.
|
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"url": "https://math.stackexchange.com/questions/4539933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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|
Inverse of the function $f(x)=\sqrt{x-3}-\sqrt{4-x}+1$ I am trying to find the inverse of the function $$f(x)=\sqrt{x-3}-\sqrt{4-x}+1$$
First of all its domain is $[3,4]$
As far as my knowledge is concerned, since $f'(x)>0$, it is monotone increasing in $[3,4]$, so it is injective. Also the Range is $[0,2]$. So $$f:[3,4]\to [0,2],\:\:y=f(x)=\sqrt{x-3}-\sqrt{4-x}+1$$ is Invertible.
Now to find $f^{-1}(x)$, we need to express $x$ in terms of $y$.
We have:
$$\begin{aligned}
&(y-1)^2=1-2 \sqrt{(x-3)(4-x)} \\
&\Rightarrow \quad 2\sqrt{(x-3)(4-x)}=1-(y-1)^2 \\
&\Rightarrow \quad 4(x-3)(4-x)=4 y^2+y^4-4 y^3 \\
&\Rightarrow \quad 4x^2-28 x+y^4-4 y^3+4 y^2+48=0
\end{aligned}$$
Which is a quadratic in $x$. But how to decide which root of $x$?
|
Complete the square from where you left off:
$$\begin{align}
4x^2-28x&=-y^4+4y^3-4y^2-48\\
4x^2-28x+49&=49-y^4+4y^3-4y^2-48\\
(2x-7)^2&=49-y^4+4y^3-4y^2-48\\
2x-7&=\pm\sqrt{1-y^4+4y^3-4y^2}\\
\end{align}$$
Since $x$ comes from $[3,4]$, then $2x-7$ comes from $[-1,1]$, sometimes positive, sometimes negative. So it is not as simple as choosing the $+$ or the $-$ in that expression. This is my answer to your question "But how to decide which root of $x$?"
At $x=3.5$ is the boundary between positive values of $(2x-7)$ and negative. This corresponds to $y=1$. You can conclude:
$$f^{-1}(y)=\begin{cases}
3.5&y=1\\
\frac{7-\sqrt{1-y^4+4y^3-4y^2}}{2}&y\lt1\\
\frac{7+\sqrt{1-y^4+4y^3-4y^2}}{2}&y\gt1
\end{cases}$$
It turns out the radicand factors, which you could find using the rational root theorem: $$1-y^4+4y^3-4y^2=(1-y)^2(1+2y-y^2)$$
So the the cases simplify:
$$f^{-1}(y)=\begin{cases}
3.5&y=1\\
\frac{7-\lvert1-y\rvert\sqrt{1+2y-y^2}}{2}&y\lt1\\
\frac{7+\lvert1-y\rvert\sqrt{1+2y-y^2}}{2}&y\gt1
\end{cases}$$
But then all these cases are equivalent to @Angelo's comment, because $\pm \lvert1-y\rvert$ is the same as $-(1-y)$ when conditioning around $y=1$:
$$f^{-1}(y)=\frac{7-(1-y)\sqrt{1+2y-y^2}}{2}$$
or if you prefer:
$$f^{-1}(y)=\frac{7+(y-1)\sqrt{1+2y-y^2}}{2}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
An estimate for $x/|x|$ I want to show that $$\left|\frac{x-y}{|x-y|}-\frac{x}{|x|}\right|\leq 4\frac{|y|}{|x|}$$ whenever $|x|>2|y|$ and $x,y\in\mathbb R^n$. This is at many places in the literature and related to singular integrals. Can anyone tell me how to prove this? I tried to prove this using mean value theorem but the derivative of the function $x/|x|$ does not seem to have the required bound.
|
You can notice that
$$
\left|\frac{x-y}{|x-y|} - \frac{x}{|x|}\right| = \left|\frac{|x| - |x-y|}{|x-y|}\,\frac{x}{|x|} - \frac{y}{|x-y|}\right|
$$
so that by the triangle inequality
$$
\left|\frac{x-y}{|x-y|} - \frac{x}{|x|}\right| \leq \frac{\left||x| - |x-y|\right|}{|x-y|} + \frac{|y|}{|x-y|} \leq \frac{2\left|y\right|}{|x-y|}.
$$
Now since $|x| - 2\left|y\right| \geq 0$, you get that $2\left|x-y\right| \geq 2\left|x\right|-2\left|y\right| \geq |x|$, and so $|x-y|^{-1}\leq 2\,\left|x\right|^{-1}$. Therefore,
$$
\left|\frac{x-y}{|x-y|} - \frac{x}{|x|}\right| \leq \frac{2\left|y\right|}{|x-y|} \leq 4 \,\frac{\left|y\right|}{|x|}.
$$
Edit: It seems one can actually get better. Write instead
$$
\left|\frac{x-y}{|x-y|} - \frac{x}{|x|}\right| = \left|\frac{|x|-|x-y|}{\left|x\right|}\,\frac{x-y}{|x-y|} - \frac{y}{|x|}\right|
$$
and so by the triangle inequality
$$
\left|\frac{x-y}{|x-y|} - \frac{x}{|x|}\right| \leq \frac{\left||x|-|x-y|\right|}{\left|x\right|} + \frac{|y|}{|x|} \leq 2\,\frac{|y|}{|x|}
$$
without needing the condition $|x| \geq 2\left|y\right|$. The constant $2$ is optimal (take $y=\lambda x$ with $\lambda > 1$ converging to $1$).
Edit 2: When $|x| \geq 2\left|y\right|$, the constant can be even further improved. Since $r+1/r\geq 2$ for any $r>0$, it follows that
$$\begin{align}
\left|\frac{x-y}{|x-y|} - \frac{x}{|x|}\right|^2 &= 2-2\,\frac{x\cdot (x-y)}{\left|x\right|\left|x-y\right|}
\\
&\leq \frac{|x|}{|x-y|} + \frac{|x-y|}{|x|} - 2\,\frac{x\cdot (x-y)}{\left|x\right|\left|x-y\right|} = \frac{|y|^2}{\left|x\right|\left|x-y\right|}
\end{align}$$
so that without condition on $x$ and $y$ (except of course $x\neq y$ and $x\neq 0$)
$$
\left|\frac{x-y}{|x-y|} - \frac{x}{|x|}\right| \leq \frac{|y|}{\sqrt{\left|x\right|\left|x-y\right|}}.
$$
In particular, when $|x| \geq 2\left|y\right|$, then $|x-y|^{-1}\leq 2\,\left|x\right|^{-1}$ and so
$$
\left|\frac{x-y}{|x-y|} - \frac{x}{|x|}\right| \leq \sqrt{2}\,\frac{|y|}{\left|x\right|}.
$$
Again, the constant $\sqrt{2}$ is optimal (take $x-y = z$ with $z \cdot x = 0$ and $|z|\to 0$).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4547891",
"timestamp": "2023-03-29T00:00:00",
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|
Is $\sum_i c_i (x^{n_i} + x^{-n_i} ) \ge 0$ for $\sum_i c_i =0$ and $\sum_i c_i n_i^2 > 0$? Let $ f(x) = \sum_i c_i (x^{n_i} + x^{-n_i} )$ for an index set $\{i\}$ with $n_i \in \cal{{N}}_0$; $\sum_i c_i = 0$, and $c_i >0 $ for the maximum $n_i$.
Further, it is required that $\sum_i c_i n_i^2 > 0$, which ensures that a local minimum is observed at $x=1$.
Then prove or disprove that $f(x) \ge 0$ for $x > 0$, where equality occurs only for $x=1$.
Some examples for explanation:
$f_1(x) = x^2 + x^{-2} -2$ satisfies the conditions with $c_1 = 1, n_1 = 2$ and $c_2 =-1, n_2 = 0$. Indeed, $f_1(x) = (x^1 - x^{-1})^2 \ge 0$ with equality only for $x = 1$.
$f_2(x) = x^4 + x^{-4} -3 x^2 -3 x^{-2} + 2 x + 2 x^{-1}$ satisfies the conditions with $c_1 = 1, n_1 = 4$ ; $c_2 = -3, n_2 = 2$ and $c_3 =2, n_3 = 1$. Indeed, $f_2(x) =(x - 1)^2 (x^6 + 2 x^5 + 2 x + 1)/x^4 \ge 0$ with equality only for $x = 1$.
A number of other trials showed no counterexamples.
Descartes rule of signs gives only a partial result since there can be numerous changes in the signs of the $\{c_i\}$.
|
A counterexample:
$$
\begin{align}
f(x) &= (x^3+x^{-3}) - 50 \cdot(x^2+x^{-2}) + 200 \cdot(x + x^{-1}) - 151\cdot 2 \\
&= \frac{(x-1)^2(x^4-48 x^3+103 x^2-48 x+1)}{x^3}
\end{align}
$$
satisfies the given conditions, but $f(2) = -51/8 < 0$.
Remark: This was inspired by the representation
$$\sum_ic_i\cosh tn_i=\frac{t^2}2\sum_ic_in_i^2+\frac{t^4}{24}\sum_ic_in_i^4+\cdots
$$
suggested by @TheSimpliFire in a comment to the question. While trying to prove or disprove that
$$
c_N > 0, \sum_{n=1}^N c_n = 0, \sum_{n=1}^N n^2 c_n > 0 \implies \sum_{n=1}^N n^4 c_n > 0
$$
I found the counterexample $(c_0, c_1, c_2, c_3) = (-151, 200, -50, 1)$ which satisfies
$$
c_0 + c_1 + c_2 + c_3 = 0 \, ,\\
c_1 + 4 c_2 + 9 c_3 = 9 > 0 \, , \\
c_1 + 16 c_2 + 81 c_3 = -519 < 0 \, . \\
$$
Luckily, that lead also to a counterexample for the original problem. More counterexamples can found in a similar way.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the value of $\int_{-\pi/4}^{\pi/4}\frac{(\pi-4\theta)\tan\theta}{1-\tan\theta}\,d\theta$. It is given that $$
I=\int_{-\pi/4}^{\pi/4}\frac{(\pi-4\theta)\tan\theta}{1-\tan\theta}\,d\theta=\pi\ln k-\frac{\pi^2}{w}
$$ and was asked to find the value $kw$. Here is my try on it:
Substituting $\theta = \dfrac{\pi}{4}+x$, we get
$$\begin{equation} \label{eq1}
\begin{split}
I &= \displaystyle\int_{-\frac{\pi}{2}}^0\dfrac{(-4x) \tan (\frac{\pi}{4}+x)}{1-\tan (\frac{\pi}{4}+x)} \, dx \\
&= -4\displaystyle\int_{-\frac{\pi}{2}}^0\dfrac{x.\dfrac{1+ \tan x}{1- \tan x}}{1- \dfrac{1+ \tan x}{1- \tan x}} \, dx \\
&= 2 \displaystyle\int_{-\frac{\pi}{2}}^0\dfrac{x(1+\tan x)}{\tan x} \, dx
\end{split}
\end{equation}$$
From here I am stuck on this, I don't know how to proceed from here to get answer in terms of $\pi \ln k-\frac{\pi^2}{w}$
Thank you for any help.
|
Note that
\begin{align}
\int_0^{\pi/2}\frac x{\tan x}dx
=&\int_0^{\pi/2}\int_0^1 \frac1{1+y^2\tan^2 x}dy \ dx
=\frac\pi2\int_0^1\frac1{1+y}dy=\frac\pi2\ln2
\end{align}
and
\begin{split}
I= &\ 2 \int_{-\frac{\pi}{2}}^0\dfrac{x(1+\tan x)}{\tan x} \, dx
= 2 \int^{\frac{\pi}{2}}_0\left(\frac x{\tan x}- x\right) dx
= \pi \ln2 -\frac{\pi^2}4
\end{split}
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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|
The value of $(a^2+b^2)(c^2+1)$. The question is:
Given real numbers $a,b,c$ that satisfy
$$ab(c^2-1)+c(a^2-b^2)=12$$
$$(a+b)c+(a-b)=7$$
Find the value of $(a^2+b^2)(c^2+1)$
From what I've done, I got $7(3ac+3a+3bc-b)-2ab(c+1)(c-1)=(a^2+b^2)(c^2+1)$. I think I'm inching further from the actual solution. Can anyone give me a hint? Thanks in advance :).
|
Since you are asking for the value of $(a^2+b^2)(c^2+1)$, we may assume that the value is equal for all $b$, so we can just set $b=0$, so that $(a-3)(a-4)=0$ and $c=12/a^2$. For $a=3$ we obtain $c=4/3$, so that $(a^2+b^2)(c^2+1)=16+9=25$. The other case, $a=4$ gives $c=3/4$ and again $(a^2+b^2)(c^2+1)=9+16=25$.
But even if we do not assume that there is only one value, we can generally conclude that one has either $a=3-bc$ or $a=4-bc$, so that again $(a^2+b^2)(c^2+1)=25$.
So the answer is $25$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve the equation $(x-1)^5+(x+3)^5=242(x+1)$ Solve the equation $$(x-1)^5+(x+3)^5=242(x+1)$$ My idea was to let $x+1=t$ and use the formula $$a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4),$$ but I have troubles to implement it. The equation becomes $$(t-2)^5+(t+2)^5=242t\\(t-2+t+2)\left[(t-2)^4-(t-2)^3(t+2)+(t-2)^2(t+2)^2-\\-(t-2)(t+2)^3+(t+2)^4\right]=242t$$ Let $A=(t-2)^4-(t-2)^3(t+2)+(t-2)^2(t+2)^2-(t-2)(t+2)^3+(t+2)^4.$
Then $$A=(t-2)^4-(t-2)^2(t^2-4+t^2+4t+4)-(t+2)^3(2-t+t+2)\\=(t-2)^4-2t(t+2)(t-2)^2-4(t+2)^3.$$
|
The “brute-force” way is to use the Binomial Theorem:
$$(x+a)^5 = x^5 + 5ax^4 + 10a^2x^3 + 10a^3x^2 + 5a^4x + a^5$$
Which gives us:
$$(x^5 - 5x^4 + 10x^3 - 10x^2 + 5x - 1) + (x^5 + 15x^4 + 90x^3 + 270x^2 + 405x + 243) = 242x + 242$$
$$2x^5 + 10x^4 + 100x^3 + 260x^2 + 410x + 242 = 242x + 242$$
$$2x^5 + 10x^4 + 100x^3 + 260x^2 + 168x = 0$$
$$2x(x^4 + 5x^3 + 50x^2 + 130x + 84) = 0$$
Clearly, $x=0$ is a solution. By the rational root theorem, roots of the quartic factor must be a factor of 84 (i.e., in $\pm\{ 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84\}$). Descartes' rule of signs rules out all positive solutions. Trial and error gives $x = -1$ and $x = -2$ as roots.
Dividing the quartic by the factors $(x+1)(x+2)$ gives the quadratic equation $x^2 + 2x + 42$, with the roots $-1 \pm i \sqrt{41}$.
Putting this all together, $x \in \{ 0, -1, -2, -1 + i\sqrt{41}, -1 - i\sqrt{41} \}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Describe the Jordan form of a linear operator $T:\mathbb{R}^7\to\mathbb{R}^7$ with characteristic polynomial $p_T(t)=(t-1)^2(t-2)^4(t-3)$
Describe the Jordan form of a linear operator $T:\mathbb{R}^7\to\mathbb{R}^7$ with characteristic polynomial $$p_T(t)=(t-1)^2(t-2)^4(t-3)$$ and such as $\dim(\ker(T-2I))=2, \dim(\ker(T-I))=1$ and $\ker(T-2I)^3\neq\ker(T-2I)^2$
My attempt:
As we have the characteristic polynomial and the dimension of the subspespaces associated to the eigenvalues, then I can say that the Jordan form to the operator is
\begin{bmatrix}
1 & & & & & & \\
1 & 1 & & & & & \\
& & 2 & & & & \\
& & 1 & 2 & & & \\
& & & & 2 & & \\
& & & & 1 & 2 & \\
& & & & & & 3 \\
\end{bmatrix}
Is this correct?
I don't understand the condition $\ker(T-2I)^3\neq\ker(T-2I)^2$, what does it means and what is the purpose of that? Thanks in advance!
|
The Jordan blocks associated with eigenvalues $1$ and $3$ are correct. To deduce the correct Jordan block associated with eigenvalue $2$, first note that $\dim(\ker(T - 2I)) = 2$ implies that there are two possibilities below:
\begin{align*}
& \text{(a).}\; J_a = \begin{bmatrix}
2 & 0 & 0 & 0 \\
1 & 2 & 0 & 0 \\
0 & 0 & 2 & 0 \\
0 & 0 & 1 & 2
\end{bmatrix}, \quad
\text{(b).}\; J_b = \begin{bmatrix}
2 & 0 & 0 & 0 \\
1 & 2 & 0 & 0 \\
0 & 1 & 2 & 0 \\
0 & 0 & 0 & 2
\end{bmatrix}.
\end{align*}
It is easy to verify that
\begin{align*}
(J_a - 2I_{(4)})^2 = (J_a - 2I_{(4)})^3 = 0,
\end{align*}
while
\begin{align*}
(J_b - 2I_{(4)})^2 = \begin{bmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix} \neq 0 = (J_b - 2I_{(4)})^3.
\end{align*}
Therefore, case (a) contradicts with the condition $\ker((T - 2I)^2) \neq \ker((T - 2I)^3)$, implying that case (b) is the unique compatible Jordan block.
|
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|
Find the line $l$ that goes through $P$ and intersects line $l_1$ and $l_2$. We have line $l_1:\begin{cases} x=1+t_1 \\ y=t_1 \\ z=-1+t_1\end{cases}$ and $l_2:\begin{cases} x=10+5t_2 \\ y=5+t_2 \\ z=2+2t_2\end{cases}$.
Find the line $l$ that goes through $P:(3, 2, −1)$ and intersects line $l_1$ and $l_2$.
I tried finding a point of intersection between line $l_1$ and $l_2$ but the equation I got is inconsistent.
$\begin{cases}
t_1-5t_2=9\\
t_1-t_2=5\\
t_1-2t_2=3
\end {cases}\iff \begin{pmatrix}1 & -5 & 9\\ 1 & -1 & 5\\ 1 & -2 & 3
\end{pmatrix} \iff \begin{pmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1
\end{pmatrix} $. So there is no point of intersection. How do I find the line that goes through $P$ and intersects $l_1$ and $l_2$?
|
First, find the plane that contains the point $Q=(3, 2, -1)$ and the line $\ell_1$. To do that, find a point on $\ell_1$, for example, by substituting $t_1 = 0 $, then $P_1 = (1, 0, -1)$ is on this plane as well as $Q=(3, 2, -1)$.
The direction vector of $\ell_1$ is $v_1 = (1, 1, 1)$. Now define the vector
$v_2 = P_1 - Q = (-2, -2, 0) $
So that the normal to the plane containing $Q$ and $\ell_1$ is
$ N = v_1 \times v_2 = (1, 1, 1) \times (-2, -2, 0) = (2 , -2 , 0 ) $
Hence, the equation of the plane is
$ (2, -2, 0) \cdot (p - Q) = 0 $
Next, find the intersection of $\ell_2$ with this plane, by substituting
$ p = (10, 5, 2) + t_2 (5, 1, 2) $
into the equation of the plane,
$ (2, -2, 0) \cdot ( (10, 5, 2) + t_2 (5, 1, 2) - (3, 2, -1) ) = 0$
From which
$ t_2 = \dfrac{ - 8 }{ 8 } = -1 $
Therefore, the point of intersection is $ P_2 = (10, 5, 2) - (5, 1, 2) = (5, 4, 0) $
Finally, the line connecting $Q$ and $P_2$ is the line we want, its equation is
$ \ell(t) = Q + t (P_2 - Q) = (3, 2, -1) + t ( 2, 2, 1 ) $
|
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|
Does $S_b$ become eventually greater than $S_{b+1}$? For a base $b \geq 2$, let $S_b$ be the sequence the $n$−th term of which is found by concatenating $n$ copies of the number $n$ written in base $b$, and then seeing the resulting string as a number in base $b$.
For example, given $b=2$, we get the sequence $1$, $1010_2$, $111111_2$, $100100100100_2$ etc, and for base ten we get $1$, $22$, $333$, $4444$, etc.
For fixed $b$, does the sequence $S_b$ become eventually greater than the sequence $S_{b+1}$? That is, does the $n$-th term of $S_b$ is greater than that of $S_{b+1}$ for all $n$ greater than some constant?
|
The proposition is false. Consider $S_2(n)$ and $S_3(n)$, where $n=3^k$, $ k\in\mathbb{N}.$
$ n = {1\underbrace{00\ldots 00}_{k \text{ zeros}} } (\text{ base } 3). $ This pattern ${1\underbrace{00\ldots 00}_{k \text{ zeros}} }$ repeated i.e. concatenated onto itself $\ 3^k$ times, $ \underbrace{100\ldots 00}_{1}\ \underbrace{100\ldots 00}_{2}\quad \ldots\quad \underbrace{100\ldots 00}_{3^k}\ $ in base $3$ is $\ > 3^{(k+1)3^k - 1}$.
On the other hand, $n=2^{k\log_2(3)\ }$ in base $2$ has at most $\ \left\lceil k \log_2(3) \right\rceil\ $ digits, and so $3^k$ in base $2$, repeated $3^k$ times - which has at most $\ 3^k\left\lceil k \log_2(3) \right\rceil\ $ digits - considered as a number in base 2, is:
$\ \leq 2^{\left\lceil k \log_2(3) \right\rceil\ \cdot3^k} \leq 2^{ (k \log_2(3) + 1) \cdot3^k } = 2^{ k3^k\left( 1 + \frac{1}{k\log_2(3)} \right) \log_2(3) } = 3^{ 3^k \left(k + \frac{1}{\log_2(3)}\right) },\ $ *which is $\leq 3^{3^k (k+1) - 1}\ $ for all $k\geq 1.\quad $
$*\quad$ The fact that $\ 3^k\left(k + \frac{1}{\log_2(3)}\right) \leq 3^k (k+1) - 1\ \forall\ k\geq 1\ $ is elementary: start by dividing both sides by $3^k$ and use the fact that $\ \frac{1}{\log_2(3)} \approx 0.63$.
|
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|
How to get the theta order of a non-homogeneous recurrence of Fibonacci sequence I'm recently learning recurrence and I'm stuck with how to find the theta order of a non-homogeneous Fibonacci sequence.
Given a Fibonacci sequence $F(N) = F(N-1)+F(N-2)+f(N)$. How can I determine the theta-order of this Fibonacci sequence if $f(N) = 1$? And what about $f(N) = N$?
I assume $F(0) = c_0$ and $F(1) = c_1$, and when $f(n) = 1$, I get $F(2) = c_0+c_1+1$, $F(3) = c_0+2 c_1+2$, $F(4) = 2 c_0+3 c_1+4$. And when $f(n) = n$, I get $F(2) = c_0+c_1+2$, $F(3) = c_0+2 c_1+3$,
$F(4) = 2 c_0+3 c_1+4$. From these I think maybe there are some relations between $f(N) = N$ and $f(N) = 1$?
|
Case $f(n) = n$
We have
$$F(n) = F(n-1)+F(n-2)+n$$
$$\Longleftrightarrow F(n)+ n + 3 = \left(F(n-1) + (n-1)+3\right)+\left(F(n-2) + (n-2)+3\right)$$
Denote $G(n) = F(n)+n+3$, then $G(n)$ is a Fibonacci sequence with $G(n) = G(n-1)+G(n-2)$. By consequence,
$$F(n) = G(n) - n-3 = A\varphi^n + B\varphi^{-n} - n-3$$
with the coefficients $A,B$ defined from the 2 initial values of $F(n)$.
Case $f(n) = 1$
$$F(n) = F(n-1)+F(n-2)+1$$
$$\Longleftrightarrow F(n)+ 1 = \left(F(n-1) +1\right)+\left(F(n-2) +1\right)$$
Denote $H(n) = F(n)+1$, and with same argument as in the case 1, we have
$$F(n) = H(n) -1 = A'\varphi^n + B'\varphi^{-n} - 1$$
with the coefficients $A',B'$ defined from the 2 initial values of $F(n)$.
|
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Determining the cartesian equation of an ellipse ( be it neither in standard position nor orientation) given center, one vertex and one semi -axis Desmos construction, with sliders waiting for being launched : https://www.desmos.com/calculator/vuou1gnese
Question : I obtained the final formula while thinking of center $C$
and vertex $V$ as points lying in the first quadrant, with $V$ located
to the right of $C$ Is the formula general in spite of this? - I
assumed that the " shifting terms" $h$ and $k$ operate , so to say, in
the rotated system, so that, intead of substituting $X_C$ for $h$ and
$Y_C$ for $k$ , I used the coordinates of center $C$ in the rotated
system. Is this assumption correct? Finally, can you think of other forms of this equation?
The givens are :
*
*$\text {Center} = C = (X_C, Y_C) $
*$ \text {Vertex} = V = ( X_V,Y_V) = (X_C+p, X_C+q) , \text {with }p, q$
$\in \mathbb R$
*$ m {\in\space \mathbb R^+} = \text {length of one semi axis ( perpendicular to the semi-axis [CV] }$
From what is given can be immediately deduced :
*
*$ M = \text {length of the semi axis [CV]}= \sqrt {p^2 + q^2}$
*$R = \text {inclination of the straight line (CV) on which lies the semi axis [CV] }= \arctan {\frac qp}$
*the equations of $X'$ and $Y'$, the rotated coordinate system in which lies the parabola , namely
$\space \space \space \space X' : y = \tan(R)x $
$\space \space \space \space Y' : y = (-1 / \tan(R))x $
Since the ellipse lies in a rotated coordinate system, and since its center is not necessarily at the origin its equation will be of the form :
$$\frac { (X(x,y) -h )^2 } { M^2 } + \frac { (Y(x,y)-k )^2 } { m^2 } =1$$
with
*
*$X(x,y) = x \cos(R) + y \sin (R) $
*$Y(x,y) = y \cos(R) - y \sin (R) $
*$ h$ and $k$ : the coordinates of center $C$ in the rotated coordinate system.
Determining the values of $h$ and $k$
*
*$h = X_C \cos (R) + Y_C \sin (R) $
*$k = Y_C \sin (R) - X_C \cos(R)$
Hence the formula for the ellipse of center $C= (X_C , Y_C) $, vertex $V$ and a semi-axis of length $m$ is :
$$\frac { (X(x,y) - (X_C \cos (R) + Y_C \sin (R)) )^2 } { M^2 } + \frac {( Y(x,y)-(Y_C \sin (R) - X_C \cos(R)) ) ^2 } { m^2 } =1$$
with, as said above $ M= \text {length of the semi- axis [CV]}$
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Rotate then Shift.
Starting from the algebraic equation of an ellipse in standard format
$ \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 $
If you define the position vector $\mathbf{p} = [x, y]^T $ , then the above equation can be written concisely as follows
$ \mathbf{p}^T \ D \ \mathbf{p} = 1\hspace{50pt}(*)$
where
$ D = \begin{bmatrix} \dfrac{1}{a^2} && 0 \\ 0 && \dfrac{1}{b^2} \end{bmatrix} $
Now in the end, you want to find the algebraic equation of the same ellipse but rotated by some angle $\theta$ (with respect to the original orientation) and shifted by a certain displacement vector $(Xc, Yc)$
So, first rotate the ellipse about the origin, then shift the resulting the ellipse by the vector $(Xc , Yc)$
The image of a point $\mathbf{p} = (x,y)$ on the original ellipse under a rotation by an angle $\theta$ is the point
$\mathbf{p'} = R \ \mathbf{p} $
where $R$ is the two-dimensional rotation matrix given by
$R = \begin{bmatrix} \cos \theta && - \sin \theta \\ \sin \theta && \cos \theta \end{bmatrix} $
And the image of $\mathbf{p'} $ after a shift by $C = [X_c, Y_c]^T $ is
$ \mathbf{p''} = \mathbf{p'} + \mathbf{C} = R \ \mathbf{p} + \mathbf{C} $
From this last equation, we deduce that $ \mathbf{p} = R^T (\mathbf{p''} - \mathbf{C} ) $
Substituting this into equation $(*)$, gives us
$ (\mathbf{p''} - \mathbf{C})^T \ R D R^T \ (\mathbf{p''} - \mathbf{C} ) = 1 \hspace{50pt} (**)$
And this is the algebraic equation of the rotated/shifted ellipse.
Explicitly, we have
$ R D R^T = \begin{bmatrix} \dfrac{1}{a^2} \cos^2 \theta + \dfrac{1}{b^2} \sin^2 \theta && \bigg( \dfrac{1}{a^2} - \dfrac{1}{b^2} \bigg) \sin \theta \cos \theta \\ \bigg( \dfrac{1}{a^2} - \dfrac{1}{b^2} \bigg) \sin \theta \cos \theta && \dfrac{1}{a^2} \sin^2 \theta + \dfrac{1}{b^2} \cos^2 \theta \end{bmatrix}$
And $\mathbf{p''} - \mathbf{C} = (x - X_c, y - Y_c) $
Therefore, the algebraic equation is
$\bigg(\dfrac{1}{a^2} \cos^2 \theta + \dfrac{1}{b^2} \sin^2 \theta \bigg) (x - X_c)^2 + 2 \bigg( \dfrac{1}{a^2} - \dfrac{1}{b^2} \bigg) \sin \theta \cos \theta
(x - X_c) (y - Y_c) + \bigg(\dfrac{1}{a^2} \sin^2 \theta + \dfrac{1}{b^2} \cos^2 \theta\bigg) (y - Y_c)^2 = 1 $
Using the double angle trigonometric formulas, this last equation can be expressed as follows
$(A + B \cos(2 \theta)) (x - X_c)^2 + 2 B \sin(2 \theta) (x - X_c) (y - Y_c) + (A - B \cos(2 \theta)) (y - Y_c)^2 = 1 $
where
$ A = \dfrac{1}{2} \bigg( \dfrac{1}{a^2} + \dfrac{1}{b^2} \bigg) $
$ B = \dfrac{1}{2} \bigg( \dfrac{1}{a^2} - \dfrac{1}{b^2} \bigg) $
|
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|
Find the area between line and curve Point $A$ lies on the curve $g(x) = x$ And lies on a value $x > 0$. From point $A$ a line is drawn parallel to the $y$-axis. The second drawn graph is $f(x)=x^3$.
Calculate the coordinates of point $A$ so that the area of $B$ and $C$ will be equal.
x must be less than 1, according to this picture, right? But I don't know how to get the point A.
Should I use this formula?
$$A_{C-B}=\int^{x}_{0} \pi x^{2}-\int^{x}_{0} \pi x^{6}$$, where $x<1$.
$$\int^{x}_{0} \pi x^{2}dx=\pi \int^{x}_{0} x^{2}dx=\pi \left[ \frac{x^{3}}{3} \right]^{x}_{0} =\pi \left[ \left( \frac{x^{3}}{3} -\frac{0^{3}}{3} \right) \right] =\frac{\pi x^{3}}{3}$$
$$\int^{x}_{0} \pi x^{6}dx=\pi \int^{x}_{0} x^{6}dx=\pi \left[ \frac{x^{7}}{7} \right]^{x}_{0} =\pi \left[ \left( \frac{x^{7}}{7} -\frac{0^{7}}{7} \right) \right] =\frac{\pi x^{7}}{7}$$
|
Let abscissa of $A$ be $A_x$.
We need to calculate coordinates of $A$ such that area of $B$ = area of $C$
$$\begin{align}\implies& \int_0^{A_x} x^3 \ dx = \int_0^{A_x} x - x^3\ dx\\\implies &\frac{(A_x)^4}{4 } = \frac{(A_x)^2}{2} - \frac{(A_x)^4}{4}\\\implies & \frac{(A_x)^4}{2} = \frac{(A_x)^2}{2}\\\implies & (A_x)^2(A_x^2 - 1) = 0\end{align}$$
Thus $A_x$ is $1$ indeed.
Therefore coordinates of $A$ are $(1, 1).$
I don't agree with the graph. But the area of regions $B$ and $C$ are equal when coordinates of $A$ are $(1, 1).$
|
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|
Calculate $\sum_{n=2}^{\infty}\left (n^2 \ln (1-\frac{1}{n^2})+1\right)$ I am interested in evaluating
$$\sum_{n=2}^{\infty}\left (n^2 \ln\left(1-\frac{1}{n^2}\right)+1\right)$$
I am given the solution for the question is $\,\ln (\pi)-\frac{3}{2}\,.$
$$\sum_{n=2}^{\infty}\left(n^2\ln\left(\!1\!-\!\frac{1}{n^2}\!\right)+1\right)=4\ln\left(\!\frac{3}{4}\!\right)+1+9\ln\left(\!\frac{8}{9}\!\right)+1+\ldots$$
Any tricks to solve it?
|
$$\begin{align}\sum^\infty_{n=2}n^2\ln\left(1-\frac{1}{n^2}\right)+1&=\sum^\infty_{n=2}n^2\ln\left(1-\frac{1}{n^2}\right)+\ln(e)=\\&=\sum^\infty_{n=2}\ln\left(1-\frac{1}{n^2}\right)^{n^2}+\ln(e)\;.\end{align}$$ $$\begin{align}\sum^\infty_{n=2}\ln\left(1-\frac{1}{n^2}\right)^{n^2}+\ln(e)&=\sum^\infty_{n=2}\ln\left(\left(1-\frac{1}{n^2}\right)^{n^2}e\right)=\\&=\ln\prod_{n=2}^\infty\left(\frac{n^2-1}{n^2}\right)^{n^2}\!\!e\,.\end{align}$$ Prove that the input of the logarithm is equal to $\pi/e^{3/2}$ and you would be good. Use this paper to get the answer.
|
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|
Find random variable $X$ with $\text{E}(X) = 0$, $\text{Var}(X) = 4$, and $\text{P}( \vert X \vert \geq 4 ) = 0.25$ As the title states, my task is to find a random variable $X$ with $\text{E}(X) = 0$, $\text{Var}(X) = 4$, and $\text{P}( \vert X \vert \geq 4 ) = 0.25$.
My first attempt was to consider $X \sim \text{Normal}(0,2^2)$, but with this we have
$$ \text{P}( \vert X \vert \geq 4 ) = 1 - \text{P}(-4 < X < 4) = 1 - \text{P}(-2 < Z < 2) \approx 0.0455 < 0.25 $$
and so this choice of $X$ is obviously incorrect.
My second thought was to try $X \sim \text{Uniform}(a,b)$ such that $\text{E}(X) = \frac{a+b}{2} = 0$, which is to say $a = -b$, and $\text{Var}(X) = \frac{(a-b)^2}{12} = \frac{b^2}{3} = 4$, which to say $b = \sqrt{12} \approx 3.4641$. However, using this we then have
$$ \text{P}( \vert X \vert \geq 4 ) = 1 - \text{P}(-4 < X < 4) = 1 - 1 = 0 $$
and so this choice of $X$ must also be incorrect.
With the provided information, it must be true that $X$ can take both positive and negative values and I think it must be true that $X$ should be absolutely continuous (this may be incorrect, however). With these thoughts in mind and letting $f_X(x)$ represent the PDF of X for $x \in A \subseteq \mathbb{R}$, I know $X$ must satisfy the following:
$(i)$ $\text{ } \text{E}(X) = \int_{\mathbb{R}} x f_X(x) \text{d}x = 0$
$(ii)$ $\text{ } \text{Var}(X) = \text{E}(X^2) = \int_{\mathbb{R}} x^2 f_X(x) \text{d}x = 4$
$(iii)$ $\text{ } \int_{-4}^{4} f_X(x) \text{d}x = 0.25 $
I am unsure as to how to go about determining $f_X(x)$ under these conditions. I was hoping the distribution of $X$ would be something quite simple but it may be more complicated than I was thinking, perhaps $X$ is even a mix of absolutely continuous and discrete parts. Any direction would be greatly appreciated.
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Consider the discrete random variable $X$ with PMF given by
$$\text{P}(X = x) = \begin{cases}
\frac{3}{4} & \text{ for } x = 0 \\
0 & \text{ for } \vert x \vert = 1 \\
0 & \text{ for } \vert x \vert = 2 \\
0 & \text{ for } \vert x \vert = 3 \\
\frac{1}{8} & \text{ for } \vert x \vert = 4 \\
0 & \text{ for all other $x \in \mathbb{Z}$ }
\end{cases}$$
With this random variable, it is clear that
$$ \sum_{k = -4}^{4} \text{P}(X = k) = \frac{1}{8} + \frac{3}{4} + \frac{1}{8} = 1 $$
and the desired conditions are satisfied:
$$ \text{E}(X) = \sum_{k = -4}^{4} k \text{P}(X = k) = (-4)\frac{1}{8} + (4)\frac{1}{8} = 0 $$
$$ \text{Var}(X) = \sum_{k = -4}^{4} k^2 \text{P}(X = k) =(16)\frac{1}{8} + (16)\frac{1}{8} = 4 $$
$$ \text{P}(\vert X \vert \geq 4) = \text{P}(X = -4) + \text{P}(X = 4) = \frac{1}{4} $$
|
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|
Express the following permutation as a product of disjoint $2$ cycles :$(4,2,5,3)(1,5,2,4)$.
Express the following permutation as a product of disjoint $2$ cycles:
$$(4,2,5,3)(1,5,2,4).$$
I can find out the permutation as $(1,5,3)(2)(4)$. We can write $$(1,5,3)(2)(4)=(1,5)(1,3)(2)(4).$$
But how to express them as disjoint $2$ cycles? I am not quite getting it.
Is there a general way to do these problems? There may be posts in this site concerning the same topic probably, but I can't seem to find it . . .
|
Given are two permutations $\pi,\sigma$ with
\begin{align*}
\color{blue}{\pi}&\color{blue}{=(4,2,5,3)}\\
\color{blue}{\sigma}&\color{blue}{=(1,5,2,4)}\\
\end{align*}
Composition of $\pi\sigma$ (order right to left) gives using two notations
\begin{align*}
\color{blue}{\pi\sigma}&=(4,2,5,3)(1,5,2,4)=(1,3,4)(2)(5)\color{blue}{=(1,3,4)}\tag{*1*}\\
\pi\sigma
&=\begin{pmatrix}
1&2&3&4&5\\
1&5&4&2&3\\
\end{pmatrix}
\begin{pmatrix}
1&2&3&4&5\\
5&4&3&1&2\\
\end{pmatrix}
=
\begin{pmatrix}
1&2&3&4&5\\
3&2&4&1&5\\
\end{pmatrix}
\end{align*}
The right-most permutation in (*$1$*) is written without using cycles of length $1$. They can be omitted since they provide no additional information. A cycle of length $n>1$ can always be written as product of $n-1$ transpositions, i.e. cycles of length $2$. We obtain from (*$1$*)
\begin{align*}
\color{blue}{\pi\sigma}&=(1,3,4)\,\,\color{blue}{=(1,3)(3,4)}\tag{*2*}\\
\end{align*}
We do not need in (*$2$*) the usage of cycles of length $1$. We can write $\pi\sigma$ as product of two transpositions.
|
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|
Evaluate $\iint (\frac{(x-y)}{x+y})^4$ over the triangular region bounded by $x+y=1$, $x$-axis ,$y$-axis
Evaluate $\iint (\frac{(x-y)}{x+y})^4$ over the triangular region bounded by $x+y=1$, $x$-axis, $y$-axis.
My attempt: I tried using the change of variable concept:
Let $u=x-y$ and $v=x+y$ ,$|J|= \frac{1}{2}$
Then $x=\frac{u+v}{2}$ and $y=\frac{u-v}{2}$
Now we have to find the limit which is where I am stuck.
We were told in class that we could plug in the points and get the equations:
In this case the points are $(0,0),(0,1),(1,0)$.
So if we put the first point we get $u=0,v=0$ or $u=-v,u=v$
If we put in the second point we get
$u=-1,v=1$, $u=-v,u=v+2$
Similarly
$u=1,v=-1$,$u=v,u=v-2$
How do I proceed after this? This is making me even more confused on what values to take and what values not to take.can anyone help me out here with an easy explanation?
|
No change of variable seems necessary.I'll use Sage to explain.
Define the function :
$$f(x,y)=\left(\frac{x-y}{x+y}\right)^4$$
sage: f(x,y)=((x-y)/(x+y))^4
Find an antiderivative of this function with respect to $y$ :
sage: A1=integrate(f(x, y), y) ; A1
-8*x*log(x + y) + y - 8/3*(5*x^4 + 12*x^3*y + 9*x^2*y^2)/(x^3 + 3*x^2*y + 3*x*y^2 + y^3)
$$\int\left(\frac{x-y}{x+y}\right)^4\,dy\,=\,-8x\log(x + y) + y - \frac{8}{3}\frac{5x^4 + 12x^3y + 9x^2y^2}{x^3 + 3x^2y + 3xy2 + y^3}$$
This function is defined and continuous on the square $(0,\ 1],(0,\ 1]$ except in $(0,\ 0)$. Its limits give us the defined integral $\displaystyle\int_0^{1-x}f(x,y)\,dy$ :
sage: D1 = A1.limit(y=1-x)-A1.limit(y=0) ; D1
-16/3*x^4 + 16*x^3 - 24*x^2 + 8*x*log(x) + 37/3*x + 1
$$\int_0^{1-x}\left(\frac{x-y}{x+y}\right)^4\,dy\,=\,-\frac{16}{3}x^4 + 16x^3 - 24x^2 + 8x\log(x) + \frac{37}{3}x + 1$$
which is a function of $x$, defined and continuous on $\mathbb{R}^{*+}$, which has an antiderivative :
sage: A2 = integrate(D1, x) ; A2
-16/15*x^5 + 4*x^4 - 8*x^3 + 4*x^2*log(x) + 25/6*x^2 + x
$$\int\frac{16}{3}x^4 + 16x^3 - 24x^2 + 8x\log(x) + \frac{37}{3}x + 1\,dx\,=\,-\frac{16}{15}x^5 + 4x^4 - 8x^3 + 4x^2\log(x) + \frac{25}{6}x^2 + x$$
continuous on $\mathbb{R}$, whose limits give us the sought integral :
sage: D2 = A2.limit(x=1) - A2.limit(x=0) ; D2
1/10
BTW :
sage: integrate(integrate(f(x, y), y, 0, 1-x, algorithm="sympy"), x, 0, 1)
1/10
(The algorithm="sympy" parameter avoids Sage's default integrator (Maxima), whic is nosy about the value of $x$).
|
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|
$\lim_{n \to \infty} \frac{1}{\sqrt{n^k}} (1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}})^k$ by Stolz–Cesàro $$\lim_{n \to \infty} \frac{1}{\sqrt{n^k}} (1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}})^k, k \in \mathbb{N}$$
Putting $\sqrt{n^k}$ in denominator, we get:
$$\lim_{n \to \infty} \frac{(1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}})^k}{\sqrt{n^k}}$$
Apllying Stolz–Cesàro theorem:
$$\lim_{n \to \infty} \frac{(1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}})^k - (1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n-1}})^k }{\sqrt{n^k} - \sqrt{(n-1)^k}}$$
I don't know how to use properly multinomial formula, but the biggest number in numerator among denominators in first brackets with $k$ degree is $\sqrt{n^k}$, in second brackets $\sqrt{(n-1)^k}$, their coefficients are $1$
$$\lim_{n \to \infty} \frac{ \frac{1}{\sqrt{n^k}} - \frac{1}{\sqrt{(n-1)^k}} + \frac{1}{\sqrt{(n-1)^k}} - \cdots + 1 - 1 + \cdots}{\sqrt{n^k} - \sqrt{(n-1)^k}}$$
Multiplying numbers in first brackets by $\frac{1}{\sqrt{n}}$ gives us $\frac{1}{\sqrt{2^{k-1} \times n}} + \frac{1}{\sqrt{3^{k-1} \times n}} + \cdots + \frac{1}{\sqrt{(n-1)^{k-1} \times n}}$ etc., but I suspect we can't get same numbers in second brackets.
My guess, after evaluating sum (or difference) in numerator, we need to divide each number of our expression by $\sqrt{n^k}$.
In denominator we get 0, because $$\sqrt{\frac{n^k}{n^k}} - \sqrt{\frac{(n-1)^k}{n^k}} = 1 - \sqrt{(\frac{n-1}{n})^k} = 1 - 1 = 0, n \to \infty$$
Multiplication and division of denominator by conjugate ($\sqrt{n^k} + \sqrt{(n-1)^k}$) gives "slightly" different picture (unless I've done several big mistakes in a row):
$$\frac{ n^k - (n-1)^k }{ \sqrt{n^k} + \sqrt{(n-1)^k} } = \frac{ n^k - n^k + kn^{k-1} - \cdots }{ \sqrt{n^k} + \sqrt{(n-1)^k} } = \frac{kn^{k-1} - {k\choose 2}n^{k-2} +\cdots }{ \sqrt{n^k} + \sqrt{(n-1)^k} }, n \to \infty$$
Dividing by $n^{k-1}$:
$$\frac{k - 0 + 0}{ \sqrt{\frac{n^k}{n^{2(k-1)} } } + \sqrt{ \frac{(n-1)^k}{n^{2(k-1)}} }} = \frac{k - 0 + 0 - \cdots}{0 + 0}$$
So my guess was a bad guess.
Unfortunately, by this point I'm pretty much lost. Hence, I need some help... Thank you!
|
By Cesaro-Stolz theorem
$$\lim_{n \to \infty} \frac{\sum_{i = 1}^{n} \frac{1}{\sqrt{i}}}{\sqrt{n}} = \lim_{n \to \infty} \frac{\sum_{i = 1}^{n+1} \frac{1}{\sqrt{i}} - \sum_{i = 1}^{n} \frac{1}{\sqrt{i}}}{\sqrt{n+1} - \sqrt{n}} = \lim_{n \to \infty} \frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}} = \lim_{n \to \infty}\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}} = 2,$$
Thus
$$\lim_{n \to \infty} (\frac{\sum_{i = 1}^{n} \frac{1}{\sqrt{i}}}{\sqrt{n}})^k = 2^k.$$
|
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|
Proof verification: $\frac{a}{(bc-a^2)^2}+\frac{b}{(ca-b^2)^2}+\frac{c}{(ab-c^2)^2}=0$ The question states:
Let $a, b, c$ be real numbers such that $$\frac{1}{(bc-a^2)}+\frac{1}{(ca-b^2)}+\frac{1}{(ab-c^2)}=0$$ Prove that $$\frac{a}{(bc-a^2)^2}+\frac{b}{(ca-b^2)^2}+\frac{c}{(ab-c^2)^2}=0$$
Now this can be solved with algebraic manipulation, but I want to check if my solution is valid too:
By Cauchy-Schwarz,
$$\left (\frac{a^2}{(bc-a^2)^3}+\frac{b^2}{(ca-b^2)^3}+\frac{c^2}{(ab-c^2)^3}\right )\left (\frac{1}{(bc-a^2)}+\frac{1}{(ca-b^2)}+\frac{1}{(ab-c^2)}\right ) \ge \left (\frac{a}{(bc-a^2)^2}+\frac{b}{(ca-b^2)^2}+\frac{c}{(ab-c^2)^2}\right )^2 = P^2$$
Hence, $0 \ge P^2 \implies P = 0$.
|
C-S it's the following:
$$\sum_{i=1}^na_i^2\sum_{i=1}^nb_i^2\geq\left(\sum_{i=1}^na_ib_i\right)^2$$ for any reals $a_i$ and $b_i$,
which says that your solution is wrong because $bc-a^2$ may be negative and can not be square of real number. You can not use C-S so.
By the way, $$\sum_{cyc}\frac{1}{ab-c^2}=\frac{(ab+ac+bc)\sum\limits_{cyc}(a^2-bc)}{\prod\limits_{cyc}(a^2-bc)}$$ and
$$\sum_{cyc}\frac{a}{(bc-a^2)^2}=\frac{(a+b+c)(ab+ac+bc)\sum\limits_{cyc}(a^2-bc)\sum\limits_{cyc}(a^2b^2-a^2bc)}{\prod\limits_{cyc}(a^2-bc)^2},$$ which gives the solution.
|
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|
Find limit of trigonometric function with indeterminacy Find limit of the given function:
$$\lim_{x\rightarrow0} \frac{(4^{\arcsin(x^2)} - 1)(\sqrt[10]{1 - \arctan(3x^2)} - 1)}{(1-\cos\tan6x)\ln(1-\sqrt{\sin x^2})}
$$
I tried putting 0 instead of x
$$\lim_{x\rightarrow0} \frac{(4^{\arcsin(x^2)} - 1)(\sqrt[10]{1 - \arctan(3x^2)} - 1)}{(1-\cos\tan6x)\ln(1-\sqrt{\sin x^2})}
=
\frac{(4^{\arcsin(0)} - 1)(\sqrt[10]{1 - \arctan(0)} - 1)}{(1-\cos\tan0)\ln(1-\sqrt{\sin 0})}
=
\frac{(1 - 1)(\sqrt[10]{1 - 0} - 1)}{(1-1)\ln(1-\sqrt{0})}
=
\frac{0*0}{0*0}
=
\frac{0}{0}
$$
But as you can see at $0$ there is limit Indeterminacy ($0/0$). How to play around and solve it?
|
The limit of a product is the product of the limits (provided they exist). Multiply and divide by terms so that the resulting limmand is a product of known limits:
$$\frac{\left(4^{\arcsin(x^2)}-1\right)\left(\sqrt[10]{1-\arctan(3x^2)}-1\right)}{(1- \cos \tan 6x)\log(1-\sqrt{\sin x^2})}$$
$$= \frac{e^{\log 4 \arcsin x^2}-1}{\log 4\arcsin x^2}\cdot \log 4\cdot \frac{\arcsin x^2}{x^2}\cdot\frac{\left(\sqrt[10]{1-\arctan(3x^2)}\right)^{10}-1^{10}}{\arctan(3x^2)}\cdot\frac{1}{\left(\sqrt[10]{1-\arctan(3x^2)}\right)^{9}+\cdots+1^9}\cdot \frac{\arctan 3x^2}{3x^2} \cdot \frac{\tan^2 6x}{1-\cos \tan 6x}\cdot \frac{36x^2}{\tan^2 6x}\cdot \frac{\sqrt{\sin x^2}}{\log(1-\sqrt{\sin x^2})}\cdot\sqrt{\frac{x^2}{\sin x^2}}\cdot\frac{3}{36}\cdot|x|$$
$$
\longrightarrow 1 \cdot \log 4 \cdot 1 \cdot -1 \cdot \frac{1}{10} \cdot 1 \cdot 2 \cdot 1 \cdot 1 \cdot 1 \cdot \frac{1}{12} \cdot 0 = -\frac{\log 4}{60}\cdot 0$$
With an extra power of $|x|$ in the denominator, it converges to the values left over.
|
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|
Which one is bigger? $ \int_0^{\frac{\pi}{2}}\frac{\sin x}{1+x^2}dx$ or $ \int_0^{\frac{\pi}{2}}\frac{\cos x}{1+x^2}dx$ Which is bigger
$$ \int_0^{\frac{\pi}{2}}\frac{\sin x}{1+x^2}dx$$ or $$ \int_0^{\frac{\pi}{2}}\frac{\cos x}{1+x^2}dx~?$$
I let $x=\frac{\pi}{2}-t$ in the second integral, and I obtain this
$$\int_0^{\frac{\pi}{2}}\frac{\sin x}{1+(\frac{\pi}{2}-x)^2}dx$$
But it is still to decide which is the bigger.
|
Actually, this is rather crude:
$$\int_0^{\frac{\pi}{2}}\dfrac{\sin x}{1+x^2}dx<\int_0^{\frac{\pi}{2}}\dfrac{x}{1+x^2}dx = \dfrac{1}{2}\ln\dfrac{4+\pi^2}{4}\approx 0.62 < 0.72\approx \dfrac{3}{2}\arctan\frac{\pi}{2} - \frac{\pi}{4}=\int_0^{\frac{\pi}{2}}\dfrac{1-x^2/2}{1+x^2}dx <\int_0^{\frac{\pi}{2}}\dfrac{\cos x}{1+x^2}dx $$
|
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|
Cyclic inequality. $\sum_{cyc} \frac{x+2y}{\sqrt{z(x+2y+3z)}}\geq \frac{3\sqrt{6}}{2}$ Let $x,y,z>0$. Show that
$$\sum_{cyc} \frac{x+2y}{\sqrt{z(x+2y+3z)}}\geq \frac{3\sqrt{6}}{2}$$
Equality case is for $x=y=z$.
A hint I have is that I have to amplify with something convenient, and then apply some sort of mean inequality inequality. Also, how would one go about finding with what to amplify?
I've tried amplifying the given fraction with a lot of random stuff, but no succes in finding something where I can apply the AM $>$ GM inequality.
|
Apply AM-GM inquality now and Cauchy-Schwarz inequalities ( at later step ) we have: $\displaystyle \sum_{\text{cyclic}} \dfrac{x+2y}{\sqrt{6z(x+2y+3z)}}\ge \displaystyle \sum_{\text{cyclic}} \dfrac{2(x+2y)}{6z+(x+2y+3z)}= \displaystyle \sum_{\text{cyclic}} \dfrac{2(x+2y)}{(x+2y)+9z}$. Next, let $a = x+2y, b = y + 2z, c = z + 2x$, then solve for $x,y,z$ in terms of $a,b,c$ we have: and the above cyclic sum becomes: $\displaystyle \sum_{\text{cyclic}} \dfrac{2a}{4b-a+c} = \displaystyle \sum_{\text{cyclic}} \dfrac{2a^2}{4ab-a^2+ac} \ge \dfrac{2(a+b+c)^2}{5ab+5bc+5ca - a^2-b^2-c^2}\ge \dfrac{3}{2}\iff a^2+b^2+c^2 \ge ab+bc+ca \iff (a-b)^2+(b-c)^2+(c-a)^2 \ge 0$, which is clearly true. So the inequality is proven and $=$ occurs when $a = b = c $ or $x = y = z$.
Note: Observe that the denominators of the cyclic sum above are positive each because $4b - a + c = (x+2y) + 9z > 0$ since $x,y,z > 0$. And also: $x = \dfrac{a+4c-2b}{9}, y = \dfrac{b+4a-2c}{9}, z = \dfrac{c+4b-2a}{9}$.
|
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|
Taylor's Expansion Exercise I have a few troubles coming up with the solution of this limit.
$$ \lim_{x \to 0} \frac {8x^2(e^{6x}-1)}{2x-\sin(2x)}$$
I've tried using Taylor like this but I honestly have no idea if it's even remotely close
$$\lim_{x\to 0} \frac{8x^2(1+6x+o(x)-1)}{2x-2x+o(x)}$$
Thank you so much
|
Using
$$ e^{ax} - 1 = (a x) \, \left(1 + \frac{a x}{2} + \frac{a^2 x^2}{6} + \mathcal{O}(x^3) \right) $$
and
$$ b x - \sin(b x) = \frac{b^3 x^3}{3!} \ \left(1 - \frac{b^2 x^2}{20} + \mathcal{O}(x^4) \right) $$
then
\begin{align}
\frac{x \, \left(e^{a x} - 1\right)}{b x - \sin(b x)} &= \frac{ a x^3 \, \left(1 + \frac{a x}{2} + \frac{a^2 x^2}{6} + \mathcal{O}(x^3) \right) }{\frac{b^3 x^3}{6} \, \left(1 - \frac{b^2 x^2}{20} + \mathcal{O}(x^4) \right) } \\
&= \frac{6 a}{b^3} \, \left( 1 + \frac{a x}{2} + \left(\frac{b^2}{20} + \frac{a^2}{6}\right) \, x^2 + \mathcal{O}(x^3) \right).
\end{align}
The limit now takes the form
\begin{align}
\lim_{x \to 0} \, \frac{\alpha x \, \left(e^{a x} - 1\right)}{b x - \sin(b x)} &= \lim_{x \to 0} \, \frac{6 \alpha a}{b^3} \, \left( 1 + \frac{a x}{2} + \left(\frac{b^2}{20} + \frac{a^2}{6}\right) \, x^2 + \mathcal{O}(x^3) \right) \\
&= \frac{6 \alpha a}{b^3}.
\end{align}
The proposed limit has $\alpha = 8$, $a = 6$, and $b = 2$, which gives
$$ \lim_{x \to 0} \, \frac{8 x \, \left(e^{6 x} - 1\right)}{2 x - \sin(2 x)} = 36. $$
|
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|
How many ways to put $10$ distinguishable balls in $5$ identical boxes so that no box has more than $3$ balls? As the title says, I was asked to count the number of ways one can put $10$ different balls inside $5$ identical boxes so that no box has more than $3$ balls. I am really confused by the restriction and I can't even compute the number without the restriction (It confuses me that the boxes can't be distinguished but the balls are). Any explanation is more than welcome.
|
A comparison with distinct balls in distinct boxes$\;$ vs$\;$ identical boxes yields a simple transformation (although $5$ cases can't be avoided here)
For the pattern $3-3-3-1-0,\;$ for distinct balls to distinct boxes, we have the product of two factorials,
[Fill Pattern]$\times$[permute Pattern]$\;=
\large\frac{10!}{3!3!3!1!0!}\cdot\frac{5!}{3!1!1!}$
The simple transformation for identical boxes is to change the numerator of the second term to $1$,
thus $\quad\large\frac{10}{3!3!3!1!0!}\cdot\frac{1}{3!1!1!}$
So we simply apply this transform to each allowable pattern, viz.
$3-3-3-1-0:\;\;\large\frac{10!}{3!3!3!1!0!}\cdot\frac{1}{3!1!1!}$
$3-3-2-2-0:\;\frac{10!}{3!3~2!2!0!}\cdot\frac1{2!2!1!}$
$3-3-2-1-1:\;\frac{10!}{3!3!2!1!1!}\cdot\frac{1}{2!1!2!}$
$3-2-2-2-1:\;\frac{10!}{3!2!2!2!1!}\cdot\frac1{1!3!1!}$
$2-2-2-2-2:\;\frac{10}{2!2!2!2!2!}\cdot\frac1{5!}$
Computing and adding up, the answer is $\boxed{35245}$
|
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|
Find the $4 \times 4$ Vandermonde determinant I'm currently doing an exercise question from the textbook. The question is:
The goal of this problem is to find the $4 \times 4$ Vandermonde determinant.
$V_4 = \begin{bmatrix}1&a&a^2&a^3\\1&b&b^2&b^3\\1&c&c^2&c^3\\1&x&x^2&x^3\end{bmatrix}$.
(a). Explain why $V_4$ is a cubic polynomial in the variable $x$.
(b). Find three possible values $r_1, r_2, r_3$ for $x$ that make $V_4$ equal to $0$. These are the roots of $V_4$ as a polynomial in $x$.
(c). Explain why $V_4=A(x-r_1)(x-r_2)(x-r_3)$ for some value $A$, and show that the value of $A$ is the $3\times3$ Vandermonde determinant $(b-a)(c-a)(c-b)$.
The $3\times3$ Vandermonde matrix is $V_3 = \begin{bmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{bmatrix}$.
(d). Finally, write down a formula for $V_4$ in terms of $a, b, c, x$.
For question (a), I understand that $V_4$ is a cubic polynomial in the variable $x$ because the $x's$ are on the same row and they won't multiply each other. The highest degree is $3$. Therefore, the determinant is a cubic polynomial.
For question (b), three possible values $r_1, r_2, r_3$ are $a, b, c$. Because if $V_4 =0$, that means the matrix is non-invertible, which then means the rows are not linearly independent. If $x=a$ or $x=b$ or $x=c$, then the rows will cancel each other out and result in a zero row which then leads to the determinant being $0$.
However, I don't really understand $(c)$ and $(d)$. Can someone give me some guidance?
|
Answering my own question in a different way from
math.stackexchange.com/a/699339/42969
because it's easier to understand (for me and maybe for some other people).
(a). The determinant is a cubic polynomial because if we calculate the determinant by cofactor expansion across the last row, we have
$V_4= -1 \begin{vmatrix}a&a^2&a^3\\b&b^2&b^3\\c&c^2&c^3\end{vmatrix} + x\begin{vmatrix}1&a^2&a^3\\1&b^2&b^3\\1&c^2&c^3\end{vmatrix} - x^2\begin{vmatrix}1&a&a^3\\1&b&b^3\\1&c&c^3\end{vmatrix} + x^3\begin{vmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{vmatrix}$. This shows that the highest degree of this polynomial is $3$, and the coefficients of these variables are cofactors $C_41, C_42, C_43$ and $C_44$ which doesn't contain variable $x$.
(b). Three possible values $r_1, r_2, r_3$ for $x$ to make $V_4$ equal to $0$ are $a, b, c$. Because for $V_4$ to be $0$, the Vandermonde matrix has to be non-invertible, which means the rows have to be not independent. If $x=a$, $x=b$, or $x=c$, then there will be same rows which makes the matrix not independent.
(c). Since $V_4$ is some cubic polynomial, it can be factored as a product of three linear terms $V_4=A(x-r_1)(x-r_2)(x-r_3)$ for some value $A$ which is the coefficient of the cubic term. From my answer in (a), it shows that $A$ is the determinant of the $3\times3$ Vandermonde matrix.
(d). The formula for $V_4$ in terms of $a, b, c, x$ is $V_4=(b-a)(c-a)(c-b)(x-a)(x-b)(x-c)$.
|
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|
Runge-Kutta Method [ Butcher Tableau ] Show $A$-Stability. We consider the following Butcher Tableau:
\begin{array}{c|cc}
\frac{1}{3} & \frac{5}{12}& -\frac{1}{12} \\
1 & \frac{3}{4} & \frac{1}{4}\\
\hline
& \frac{3}{4} & \frac{1}{4} \\
\end{array}
What do I have to do?:
First I wrote down, everything I need:
For the $A$-stability I need:
*
*the stability function $g(z)$
*an explanation why $g$ has no pole in {$ z\in \mathbb{C}: \Re(z) \leq 0 $}
*$\lim_{|z| \rightarrow \infty } |g(z)| \leq 1$
*$ g(it) \leq 1 $ $ \forall t \in \mathbb{R} $
equation for the stability function $g(z)$ :
$ g(z) = 1 + zb^T(I_2 -zA)^{-1}e $
In general we shouldn't invert that matrix, but in this special case we have the formula for the inverse:
$\frac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$
So we get : $(I_2 -zA)^{-1} = \frac{1}{(1-\frac{5}{12}z)(1-\frac{1}{4}z)-(-\frac{1}{12}z)\frac{3}{4}z}\begin{bmatrix}1-\frac{1}{4}z&-\frac{1}{12}z\\\frac{3}{4}z&1-\frac{5}{12}z\end{bmatrix}$
simplify it a bit:
$ (I_2 -zA)^{-1} = \frac{6}{z^2-4z+6}\begin{bmatrix}1-\frac{1}{4}z&-\frac{1}{12}z\\\frac{3}{4}z&1-\frac{5}{12}z\end{bmatrix} $
Then:
$(I_2 -zA)^{-1}e = \begin{bmatrix}
\frac{2(3-z)}{z^2-4z+6}\\ \frac{2(3+z)}{z^2-4z+6}
\end{bmatrix} $
Then:
$ 1+ z (\frac{3}{4} , \frac{1}{4} )^T (I_2 -zA)^{-1}e = 1 + \frac{z(6-z)}{z^2-4z+6}$
Theres is a chance that I miscalculated somewhere ( if I find a mistake, I will edit this post ). But to be honest, the remaining steps for $ A $ Stability are my greater concern. I tried something for 3. and 4., but this wasnt effective.
|
Now examine the absolute value for $z=iy$
$$
|g(iy)|^2=\left|\frac{iy+6}{-y^2-4iy+6}\right|^2=\frac{y^2+36}{(6-y^2)+16y^2}=\frac{(y^2+1)+35}{(y^2+1)^2+35}\le 1.
$$
By the maximum modulus principle the maximum absolute value for the holomorphic domain $Re(z)\le 0$ has to be on the boundary (including here the infinite point), thus $|g(z)|\le 1$.
As the value at infinity is $0$, the method is, unlike the implicit trapezoidal method, A-stable and L-stable
|
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|
Solve the radical equation for all reals: $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1+x^2}$ Question:
Solve the radical equation for all reals: $$x\left(1+\sqrt{1-x^2}\right)=\sqrt{1+x^2}$$
My approach:
$$1+\sqrt {1-x^2}=\frac {\sqrt {1+x^2}}{x}\\1+2\sqrt {1-x^2}+1-x^2=\frac{1+x^2}{x^2}\\4(1-x^2)=\left(\frac{1+x^2}{x^2}+x^2-2\right)^2\\4x^4(1-x^2)=(x^4-x^2+1)^2$$
I don't know how can I proceed from here.
Is there a way so that not using the complicated expansion of polynomials?
I'm looking for methods that doesn't use $4$ or higher degree polynomial expansions.
|
We quickly observe that $x>0$.
Letting $1-x^2=u^2, \,u≥0$ we have:
$$x^2(1+2u+u^2)=1+x^2$$
This implies that,
$$\begin{align}&(1-u^2)(u^2+2u)=1\\
\implies &u(u+1)(u+2)(u-1)=-1\\
\implies &(u^2+u)(u^2+u-2)=-1\end{align}$$
Finally letting $t=u^2+u$, we have:
$$(t-1)^2=0\implies t=1$$
This leads to:
$$\begin{align}&u^2+u-1=0,\,u>0\\
\implies &u=1-u^2\\
\implies &x=\sqrt u=\sqrt {\frac{-1+\sqrt 5}{2}}~~~.\end{align}$$
|
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|
Show that $b^2+c^2-a^2\leq bc$. Let $a,b,c>0$ such that $b<\sqrt{ac}$, $c<\frac{2ab}{a+b}$. Show that $b^2+c^2-a^2\leq bc$.
I tried to construct a triangle with $a,b,c$ and to apply The cosine rule, but I am not sure that it's possible to construct it and also I have no idea how to prove that an angle it's greater than $60^{\circ}$.
|
Let $a<\sqrt{b^2-bc+c^2}.$
Thus, $$b^2<ac<c\sqrt{b^2-bc+c^2},$$ which gives
$$(b-c)(b^2(b+c)+c^3)<0$$ or
$$b<c.$$
Also, since $$a(2b-c)>bc,$$ we obtain: $$2b-c>0$$ and
$$\sqrt{b^2-bc+c^2}>a>\frac{bc}{2b-c},$$ which gives $$(b-c)(4b^3-4b^2c+4bc^2-c^3)>0$$ or
$$4b^3-4b^2c+4bc^2-c^3<0.$$
But $$4b^3-4b^2c+4bc^2-c^3=b(2b-c)^2+c^2(2b-c)+c^2b>0,$$
which is a contradiction.
|
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|
Proving $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}=2$ If $a,b,c,d \in \mathbb{N}$
Given pairwise distinct $a,b,c,d \in \mathbb{N}$, prove that $$E=2$$ if $E=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}$ is an integer.
My effort:
We have: $$\begin{aligned}
& E=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a} \\
& F=\frac{b}{a+b}+\frac{c}{b+c}+\frac{d}{c+d}+\frac{a}{a+d}
\end{aligned}$$
WLOG let $a>b>c>d$, so $$\begin{aligned}
& \frac{b}{a}<1 \\
& \Rightarrow \quad \frac{1}{1+\frac{b}{a}}>\frac{1}{2} \Rightarrow \frac{a}{a+b}>\frac{1}{2}
\end{aligned}$$
Like-wise:$$\frac{b}{b+c}>\frac{1}{2},\:\frac{c}{c+d}>\frac{1}{2} \Rightarrow E>1.5 \Rightarrow F<2.5 ----(1)$$
Which gives two possibilities $E=F=2$ or $E=3,F=1$.
How to rule out the second possibility?
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$$E>\frac a{a+b+c+d}+\frac b{a+b+c+d}+\frac c{a+b+c+d}+\frac d{a+b+c+d}=1
$$
Similarly, $F=\frac{b}{a+b}+\frac{c}{b+c}+\frac{d}{c+d}+\frac{a}{d+a}>1$.
Since $E+F=(\frac a{a+b}+\frac b{a+b})+(\frac b{b+c}+\frac c{b+c})+(\frac c{c+d}+\frac d{c+d})+(\frac d{d+a}+\frac a{d+a})=4$, we have $E=4-F<3$.
Since $E$ is an integer and $1<E<3$, we have $E=2$.
The proof above does not require any one of $a,b,c,d$ be integer. What is needed is $a,b,c,d>0$.
|
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|
Algebraic calculation with polynomial and complex root.
Let $f=X^{3}-7 X+7$ be in $\mathbb{Q}[X]$.
Let $\alpha \in \mathbb{C}$ be a root of $f$ and hence $1, \alpha, \alpha^{2}$ be a basis of the $\mathbb{Q}$ vector space $\mathbb{Q}(\alpha)$. Let $\beta=3 \alpha^{2}+4 \alpha-14$. Write $\beta^{2}$ and $\beta^{3}$ as linear combinations of $1, \alpha, \alpha^{2}$ over $\mathbb{Q}$ and conclude that $\beta$ is a zero of $f$. Is $\beta=\alpha$?
I found my mistake with your help.
The solution: We have
$$
\beta^2=(3\alpha^{2}+4 \alpha-14)^2=9 \alpha^4 + 24 \alpha^3 - 68 \alpha^2 - 112 \alpha + 196
$$
To remove the cubic and quartic powers, we use the fact that $\alpha^3-7\alpha+7=0$ holds and gradually eliminate partial polynomials.
We get
$$
\beta^2=9 \alpha^4- 68 \alpha^2 + 24 \alpha^3 + 16\cdot (-7 \alpha) + 28\cdot 7
=9 \alpha^4- 68 \alpha^2 -8\cdot (-7 \alpha) \
+ 4\cdot 7\alpha(9 \alpha^3+10\cdot(-7) \alpha +8\cdot 7+2\alpha)+4\cdot 7=\alpha(-5\alpha -7) + 4\cdot 7=-5\alpha^2-7\alpha+28.
$$
Now we try $\beta^3$ with the same method
$$
\beta^3=\beta^2\beta=(3\alpha^{2}+4 \alpha-14)^2(3\alpha^{2}+4 \alpha-14)=(-5\alpha^2-7\alpha+28)(3\alpha^{2}+4 \alpha-14)\
=-15\alpha^4+ 126 \alpha^2 -(41 \alpha^3 +30\cdot(-7\alpha) + 56\cdot 7)=-15\alpha^4+ 126 \alpha^2 -( -11\cdot(-7\alpha) + 15\cdot 7)\
=-\alpha(15\alpha^3+18\cdot(-7\alpha) +7\cdot 11) - 15\cdot 7=-\alpha(3\cdot(-7\alpha) -4\cdot7) - 15\cdot 7=21\alpha^2 +28\alpha-105.
$$
And with this i get indeed
$$
f(\beta)=0.
$$
Thanks for your help! Maybe there are different reasonings for why $\beta\neq \alpha$?
|
Your mistake appears to be in the last line evaluating $\beta^3$. The expression
\begin{eqnarray*}
15\alpha^3+18⋅(−7\alpha)+77 &=& 15(7\alpha-7)-7*18\alpha+7*11\\
& =& 7(15-18)\alpha+7(-15+11) \\
& = & -21\alpha -28
\end{eqnarray*}
whereas you got $-21\alpha+62$. The expression for $\beta^3$ becomes $-21\alpha^2-28\alpha-105$. Plugging this new expression into the rest of your derivation shows that $\beta$ is a root of the polynomial. (btw. I cheated and used sympy to check your calculations...)
|
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|
Prove or disprove that the inequality is valid if $x,y,z$ are positive numbers and $xyz=1$. Prove or disprove that the inequality
$$\dfrac{1}{\sqrt{1+x}}+\dfrac{1}{\sqrt{1+y}}+\dfrac{1}{\sqrt{1+z}} \geq 1$$
is valid if $x,y,z$ are positive numbers and $$xyz=1.$$
My solution is:
Let $$x=\dfrac{a}{b}, y=\dfrac{b}{c}, z=\dfrac{c}{a}.$$
So we have
$$\dfrac{1}{\sqrt{1+\dfrac{a}{b}}}+\dfrac{1}{\sqrt{1+\dfrac{b}{c}}}+\dfrac{1}{\sqrt{1+\dfrac{c}{a}}}=\dfrac{1}{\sqrt{\dfrac{a+b}{b}}}+\dfrac{1}{\sqrt{\dfrac{b+c}{c}}}+\dfrac{1}{\sqrt{\dfrac{c+a}{a}}}=\sqrt{\dfrac{b}{a+b}}+\sqrt{\dfrac{c}{c+b}}+\sqrt{\dfrac{a}{a+c}}=\sqrt{\dfrac{bb}{b(a+b)}}+\sqrt{\dfrac{cc}{c(c+b)}}+\sqrt{\dfrac{aa}{a(a+c)}}.$$
Then we can use this:
$$\dfrac{1}{\sqrt{xy}}\geq\dfrac{2}{x+y}.$$
So we have
$$\sqrt{\dfrac{bb}{b(a+b)}}+\sqrt{\dfrac{cc}{c(c+b)}}+\sqrt{\dfrac{aa}{a(a+c)}} \geq \dfrac{2b}{2b+a}+\dfrac{2c}{2c+b}+\dfrac{2a}{2a+c}.$$
Then we can use this:
$$\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z} \geq 3 \sqrt[3]{\dfrac{abc}{xyz}}.$$
So we have
$$\dfrac{2b}{2b+a}+\dfrac{2c}{2c+b}+\dfrac{2a}{2a+c} \geq 3 \sqrt[3] {\dfrac{8abc}{(2b+a)(2c+b)(2a+c)}}.$$
The question is what should I do next? Is it already obvious that
$$3 \sqrt[3] {\dfrac{8abc}{(2b+a)(2c+b)(2a+c)}} \geq 1?$$
Any hint would help a lot! Thanks!
|
This is almost the same as the other answer and just slightly different: $\sqrt{\dfrac{b}{a+b}}+\sqrt{\dfrac{c}{c+b}}+\sqrt{\dfrac{a}{a+c}}\ge\sqrt{\dfrac{b}{a+b+c}}+\sqrt{\dfrac{c}{a+b+c}}+\sqrt{\dfrac{a}{a+b+c}}=\dfrac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{a+b+c}}> 1.$ Done.
|
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|
Show that $(x+1)^{p}x^{1-p}-(x+1)^{1-p}x^{p}$ is strictly increasing Let $1/2<p< 1$. I am asked to show that
$$f(x)=(x+1)^{p}x^{1-p}-(x+1)^{1-p}x^{p}$$
is strictly increasing for $x\geq 0$ and to compute $\lim_{x\to\infty} f(x)$.
I first computed the derivative, but I don't see why it must be positive:
$$\frac{d f(x)}{d x}=p(x+1)^{p-1}x^{1-p}+(1-p)(x+1)^px^{-p}-(1-p)(x+1)^{-p}x^p -p(x+1)^{1-p}x^{p-1}$$
Any ideas?
Thanks a lot for your help.
Case $p=3/4$. Then
$$f(x)=(x+1)^{1/4}x^{1/4}(\sqrt{x+1}-\sqrt{x})=\frac{(x+1)^{1/4}x^{1/4}}{\sqrt{x+1}+\sqrt{x}}$$
It suffices to show that $\ln f(x)$ is strictly increasing. We have
$$\ln f(x) =\frac{1}{4} \ln(x+1)+\frac{1}{4} \ln(x)-\ln(\sqrt{x+1}+\sqrt{x})$$
Taking derivative w.r.t. $x$ we get
$$\frac{d \ln f(x) }{d x}=\frac{1}{4}\frac{1}{x+1}+\frac{1}{4}\frac{1}{x}-\frac{1}{\sqrt{x+1}+\sqrt{x}}(\frac{1}{2}\frac{1}{\sqrt{x+1}}+\frac{1}{2}\frac{1}{\sqrt{x}})$$
Hence $\frac{d \ln f(x) }{d x}>0$ is equivalent to
$$(\frac{1}{4}\frac{1}{x+1}+\frac{1}{4}\frac{1}{x})(\sqrt{x+1}+\sqrt{x})>\frac{1}{2}\frac{1}{\sqrt{x+1}}+\frac{1}{2}\frac{1}{\sqrt{x}}$$
which is equivalent to
$$\frac{1}{4}\frac{\sqrt{x}}{x+1}+\frac{1}{4}\frac{\sqrt{x+1}}{x}>\frac{1}{4}\frac{1}{\sqrt{x+1}}+\frac{1}{4}\frac{1}{\sqrt{x}}$$
or
$$\frac{\sqrt{x+1}-\sqrt{x}}{x}>\frac{\sqrt{x+1}-\sqrt{x}}{x+1}$$
which holds. Hence $f(x)$ is strictly increasing.
|
Computing the limit is rather straight forward
\begin{align}
f(x)&=(x+1)^{p}x^{1-p}-(x+1)^{1-p}x^{p}=\frac{(x+1)^p}{x^p}x-(x+1)\frac{x^p}{(1+x)^p}\\
&=\frac{(1+x)^{2p}x-(x+1)x^{2p}}{x^p(1+x)^p}=\frac{x\Big(\big(1+\tfrac1x\big)^{2p} -1\Big)-1}{\big(1+\frac{1}{x}\big)^p}\\
&=\frac{\frac{\big(1+\tfrac1x\big)^{2p}-1}{\tfrac1x} -1}{\big(1+\tfrac1x\big)^p}\xrightarrow{x\rightarrow\infty}2p-1
\end{align}
As for monotonicity, it is easier to set $t=1/x$ to obtain that $f(x)=\phi(1/x)$ where
$$\phi(t)=\frac{(1+t)^{2p}-1-t}{t(1+t)^p}=\frac{(1+t)^{2p-1}-1}{t(1+t)^{p-1}}=\frac{(1+t)^p-(1+t)^{1-p}}{t}$$
for $t>0$. Notice that $\phi(t)>0$ for all $t>0$ and so, $\xi(t)=(1+t)^p-(1+t)^{1-p}>0$ for all $t>0$.
The function $\xi$ is concave (i.e. $-\xi$ is convex) since
\begin{align}
\xi''(t)&=p(p-1)(1+t)^{p-2}+p(1-p)(1+t)^{-p-1}\\
&=\frac{p(1-p)(1+t)^{-1}}{(1+t)^p(1+t)^{1-p}}\Big((1+t)^{1-p}-(1+t)^{-p}\Big)\\
&=-\frac{p(1-p)(1+t)^{-1}}{(1+t)^p(1+t)^{1-p}}\,\xi(t)<0
\end{align}
Consequently, the map $t\mapsto \frac{\xi(t)-\xi(0)}{t}=\phi(t)$ is monotone decreasing. Hence, $f(x)=\phi(1/x)$ is monotone increasing.
This is based on a well known fact that if $g$ is a convex function on an interval $I$ and $t_0\in I$, then $t\mapsto \frac{g(t)-g(t_0)}{t-t_0}$ is increasing on $(t_0,\infty)$.
|
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|
Solve $\int_0^1\frac{x(x+1)^b}{\alpha-x}dx$ using hypergeometric functions How can we solve this integral?
$$\int_0^1\frac{x(x+1)^b}{\alpha-x}dx$$
I used geometric series
$$=\int_0^1(x+1)^b\sum_{n=0}^\infty\left(\frac{x}{\alpha}\right)^{n+1}dx=\sum_{n=0}^\infty\int_0^1\left(\frac{x}{\alpha}\right)^{n+1}(x+1)^bdx$$
Using
$$\int_0^1t^b(1-t)^c(1-zt)^adt=\Gamma(b+1)\Gamma(c+1){_2}F_1(-a,b+1;c+b+2;z)$$
I got $$\sum_{n=0}^\infty a^{-n-1}\Gamma(n+2){_2}F_1(-b,n+2;n+3;-1)$$
The problem is I don't know how to simplify this any further. How can ${_2}F_1(-b,n+2;n+3;-1)$ be simplified?
|
$$\frac{x(x+1)^b}{\alpha-x}=\frac{(x-\alpha)(x+1)^b+\alpha(x+1)^b}{\alpha-x}=-(x+1)^b+\frac{\alpha(x+1)^b}{(\alpha+1)-(x+1)}$$
$$I=\int \frac{x(x+1)^b}{\alpha-x} \,dx=-\frac {(x+1)^{b+1}}{b+1}+\alpha J$$
$$J=\int \frac{(x+1)^b}{(\alpha+1)-(x+1)}\,dx$$
Let $$(x+1)=(\alpha+1) t\quad \implies \quad J=(1+\alpha)^b \int\frac {t^b}{1-t}\,dt $$
$$\int\frac {t^b}{1-t}\,dt=\frac{t^{b+1} }{b+1} \, \, \, _2F_1(1,b+1;b+2;t)=B_t(b+1,0)$$
Assuming $\alpha >1$, back to $x$
$$\color{red}{K=\int_0^1 \frac{x(x+1)^b}{\alpha-x} \,dx=\frac{1-2^{b+1}}{b+1}+}$$
$$\color{red}{\frac{\alpha }{(\alpha +1) (b+1)} \left(2^{b+1} \, _2F_1\left(1,b+1;b+2;\frac{2}{\alpha
+1}\right)-\, _2F_1\left(1,b+1;b+2;\frac{1}{\alpha
+1}\right)\right)}$$ or, simpler,
$$\color{blue}{K=\int_0^1 \frac{x(x+1)^b}{\alpha-x} \,dx=\frac{1-2^{b+1}}{b+1}+
\alpha (\alpha +1)^b \left(B_{\frac{2}{\alpha
+1}}(b+1,0)-B_{\frac{1}{\alpha +1}}(b+1,0)\right)}$$
|
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|
Is there another simpler method to evaluate the integral $\int_0^{2 \pi} \frac{1}{1+\cos \theta \cos x} d x , \textrm{ where } \theta \in (0, \pi)?$ Using ‘rationalization’, we can split the integral into two manageable integrals as:
$\displaystyle \begin{aligned}\int_0^{2 \pi} \frac{1}{1+\cos \theta \cos x} d x = & \int_0^{2 \pi} \frac{1-\cos \theta \cos x}{1-\cos ^2 \theta \cos ^2 x} d x \\= & \int_0^{2 \pi} \frac{d x}{1-\cos ^2 \theta \cos ^2 x}-\cos \theta \int_0^{2 \pi} \frac{\cos x}{1-\cos ^2 \theta \cos ^2 x} d x \\= & 4 \int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{\sec ^2 x-\cos ^2 \theta} d x+\int_0^{2 \pi} \frac{d(\cos \theta \sin x)}{\sin ^2 \theta+\cos ^2 \theta \sin ^2x} \\= & 4 \int_0^{\frac{\pi}{2}} \frac{d\left(\tan x\right)}{\sin ^2 \theta+\tan ^2 x}+\frac{1}{\sin \theta}\left[\tan ^{-1}\left(\frac{\cos ^2 \theta \sin x}{\sin \theta}\right)\right]_0^{2 \pi} \\= & \frac{4}{\sin \theta}\left[\tan ^{-1}\left(\frac{\tan x}{\sin \theta}\right)\right]_0^{\frac{\pi}{2}} \\= & \frac{4}{\sin \theta} \cdot \frac{\pi}{2} \\= & \frac{2 \pi}{\sin \theta}\end{aligned}\tag*{} $
Is there another simpler method to evaluate the integral?
Your comments and alternative methods are highly appreciated.
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For all intents and purposes, this amounts to the integral
$$\int_0^{2\pi} \frac{1}{1+ a \cos x} \, \mathrm{d}x$$
for $a := \cos \theta$. This is a rational function of sine and cosine, and hence the Weierstrass substitution may apply, if be a bit messy. Then
$$t = \tan \frac x 2 \implies \cos x = \frac{1-t^2}{1+t^2} , \, \mathrm{d}x = \frac{2 \, \mathrm{d}t}{1+t^2}$$
The integral becomes, after splitting up at $\pi$,
$$ \int_0^\infty \frac{1}{1 + a \frac{1-t^2}{1+t^2}} \frac{2 \, \mathrm{d}t}{1+t^2}
+ \int_{-\infty}^0 \frac{1}{1 + a \frac{1-t^2}{1+t^2}} \frac{2 \, \mathrm{d}t}{1+t^2}$$
Recombining and simplifying, we get
$$2 \int_{-\infty}^\infty \frac{\mathrm{d}t}{1+t^2 + a(1-t^2)} = 2 \int_{-\infty}^\infty \frac{1}{t^2(1-a) + (1+a)} \, \mathrm{d}t$$
This can be solved using that
$$\int \frac{1}{u^2 + \beta^2} \, \mathrm{d}u = \frac 1 \beta \arctan \frac u \beta + C$$
|
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|
Why does solving this integral using trigonometric substitution lead to the wrong answer? I was solving the integral $$\int \frac{\sqrt{x^2 - 16}}{x} \, dx,$$ and I admittedly attempted to solve it blindly using trigonometric substitution:
$$\begin{align}
&\int \frac{\sqrt{x^2 - 16}}{x} \, dx, \quad \text{let } x = 4\sec\theta \implies dx = 4\sec\theta\tan\theta \, d\theta \quad(\text{meaning} \ \ \theta=\sec^{-1}\frac{x}{4})\\[0.5em]
&\int \frac{\sqrt{16\sec^2\theta - 16}}{4\sec\theta} \cdot 4\sec\theta\tan\theta = \int 4\tan^2\theta \, d\theta = \int 4(\sec^2\theta - 1) \, d\theta = 4\tan\theta - 4\theta + c \\[0.5em]
&4\tan\theta - 4\theta + c \ \ \text{becomes} \ \ \sqrt{x^2-16} - 4\sec^{-1}\Bigl(\frac{x}{4}\Bigr) + c \ \ \text{so} \ \ \fbox{$\int \frac{\sqrt{x^2 - 16}}{x} \, dx = \sqrt{x^2-16} - 4\sec^{-1}\Bigl(\frac{x}{4}\Bigr) + c$}
\end{align}$$
So $\sqrt{x^2-16} - 4\sec^{-1}\Bigl(\frac{x}{4}\Bigr) + c$ is the family of antiderivates of $\frac{\sqrt{x^2 - 16}}{x}$, so if $$F(x) = \sqrt{x^2-16} - 4\sec^{-1}\Bigl(\frac{x}{4}\Bigr),$$ then $$F'(x) = \frac{\sqrt{x^2 - 16}}{x}$$, right? However, as the graph below shows, that's not the case.
For negative values, the two functions $F'$ and $f$ do not match at all. When setting
$$F(x) = \sqrt{x^2 - 16} - 4\tan^{-1}\biggl(\frac{\sqrt{x^2 - 16}}{4}\biggr),$$ then the two functions do seem to agree.
When I tried to ask an instructor, I was told that we always represent our solutions in terms of either inverse tangent or inverse sine, but not inverse secant. The reason I confidently chose to represent theta as inverse secant is that that's how we always did it, so when I compared my answers to the ones provided, I was extremely confused.
I fear that it might be an issue with the domain of the inverse secant, but I'm not exactly sure how.
|
$$
\begin{aligned}
\int \frac{\sqrt{x^2-16}}{x} d x=&\int \frac{x^2-16}{x \sqrt{x^2-16}} d x \\
= & \int \frac{x^2-16}{x^2} d\left(\sqrt{x^2-16}\right) \\
= & \int\left(1-\frac{16}{x^2}\right) d\left(\sqrt{x^2-16}\right) \\
= & \sqrt{x^2-16}-16 \int \frac{d\left(\sqrt{x^2-16}\right)}{\left(\sqrt{x^2-16}\right)^2+4^2} \\
= & \sqrt{x^2-16}-4 \tan ^{-1}\left(\frac{\sqrt{x^2-16}}{4}\right)+C \\
\end{aligned}
$$
|
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|
How To Prove $a^3b＋b^3c＋c^3a>a^2b^2＋b^2c^2＋a^2c^2 $ if $ a > b > c > 0\,$? How to prove this :
$$a^3b＋b^3c＋c^3a>a^2b^2＋b^2c^2＋a^2c^2 $$
if we know: $$ a > b > c > 0 $$
My attempt:
$$\frac {a^3b＋b^3a}{2}>a^2b^2
...(1)$$
$$\frac {b^3c＋c^3b}{2}>c^2b^2...(2)$$
$$\frac {a^3c＋c^3a}{2}>a^2c^2...(3)$$
(1) + (2)+ (3) :
$$\frac {a^3b＋b^3a＋a^3c＋c^3a＋b^3c＋c^3b}{2}>\;a^2b^2＋b^2c^2＋a^2c^2$$
$$a^3b＋b^3c＋c^3a>ab^3＋bc^3＋ca^3\\~\\ab(a^2－b^2)＋bc(b^2－c^2)>ca(a^2-c^2)\\~\\ab(a－b)(a＋b)＋bc(b－c)(b＋c)>ca(a－c)(a＋c)$$
I appreciate your help
|
We need to prove that:
$$\sum_{cyc}(a^3b-a^2b^2)>0$$ or
$$\sum_{cyc}(2a^3b-2a^2b^2)>0$$ or
$$\sum_{cyc}(a^3b+a^3c-2a^2b^2)+\sum_{cyc}(a^3b-a^3c)>0$$ or
$$\sum_{cyc}ab(a-b)^2+(a-b)(a-c)(b-c)(a+b+c)>0,$$ which is obvious.
I think, BW does not help for the following inequality.
Let $\frac{1}{\sqrt{a}},$ $\frac{1}{\sqrt{b}}$ and $\frac{1}{\sqrt{c}}$ be sides-lengths of a triangle. Prove that:
$$a^3b+b^3c+c^3a\geq a^2b^2+b^2c^2+c^2a^2.$$
|
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|
How to maximize the expression $xy(1-2x-3y)$ when $\{x>0, y>0, 2x+3y<1\}$ Find the greatest value of the following expression $xy(1-2x-3y)$ when $x>0$,$y>0$,$2x+3y<1$ and also determine in each case the value of the variables for which the greatest value is attained.
I have tried to solve it using AM-GM inequality considering the positive real numbers $xy$,$(1-2x-3y)$. What should I do after that? Or suggest any other way of solving the problem.
|
We have a one-line solution that can be obtained using the AM - GM inequality :
$$\begin{align}&\left(1-2x-3y\right)+2x+3y\\
&\geq 3\sqrt [3]{6xy(1-2x-3y)}\\
\implies &xy\left(1-2x-3y\right)\leq\frac {1}{162}\thinspace .\end{align}$$
The equality occurs iff, when
$$\begin{align}&2x=3y=1-(2x+3y)\\
\implies &2x=1-4x\\
\implies &\left(x,y\right)=\left (\frac 16, \frac 19\right)\thinspace .\end{align}$$
In this answer, we provide a alternative method that does not use some specific inequalities and known calculus tools.
Let $0<a=2x+3y<1$ and $y=\frac {a-2x}{3}$, then we have :
$$\begin{align}f(x,y):&=xy(1-2x-3y)\\
&=xy(1-a)\\
&=\frac {a-1}{24}\left(4x-a\right)^2+\frac {a^2-a^3}{24}\end{align}$$
Since $a-1<0$, the maximum occurs at $x=\frac a4,\thinspace y=\frac a6 \thinspace ,$ which makes $4x-a=0\thinspace.$ Thus, we need to find the $\max\{a^2-a^3\mid 0<a<1\}\thinspace.$
Using the method implemented here , we observe that :
$$\begin{align}\frac {4}{27}-\left(a^2-a^3\right)&=\frac {1}{27}(3a+1)(3a-2)^2\geq 0\end{align}$$
Therefore, we obtain :
$$\begin{align}xy(1-2x-3y)&\leq \frac {4}{27\cdot 24}\\
&=\frac {1}{162}\thinspace .\end{align}$$
where $\thinspace x,y>0$ and $2x+3y<1\thinspace .$
Equality holds iff, when $a=\frac 23\thinspace.$ This yields :
$$\begin{align}(x,y)=\left(\frac a4,\thinspace \frac a6\right)=\left (\frac 16, \frac 19\right)\thinspace .\end{align}$$
This completes the answer.
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"timestamp": "2023-03-29T00:00:00",
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|
Calculating double integral using variables substitution $\displaystyle
D = \left\lbrace \left. \rule{0pt}{12pt} (x,y) \; \right| \; 3 x^2 + 6 y^2 \leq 1 \right\rbrace$
Calculate $\displaystyle
\iint_D \frac{ x^2 }{ ( 3 x^2 + 6 y^2 )^{ 3/2 } } \; dx dy{}$.
Attempt:
$x=\frac{r}{\sqrt3}cost,y=\frac{r}{\sqrt6}sint \implies |J|=\sqrt{\frac{2}{3}}r$
$3 x^2 + 6 y^2 \leq 1 \implies 0\leq r \leq 1$
$\iint_D \frac{ x^2 }{ ( 3 x^2 + 6 y^2 )^{ 3/2 } } \; dx dy{} =\int _0^1\:\int _0^{2\pi }\:\frac{cos^2t\sqrt{2}}{3\sqrt{3}}dtdr = \frac{\sqrt{2}\pi }{3\sqrt{3}}$
My answer isn't corect , can't find out what is wrong.
Appreciate any help.
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With the change of variable $$(x,y)\to \left(\frac{1}{\sqrt{3}}r\cos t,\frac{1}{\sqrt{6}}r\sin t\right),$$ for $t\in [0,2\pi[$ and since $3x^2+6y^2\leqslant 1$ then $r\in [0,1]$.
The determinant of Jacobian is given by
$$\frac{\partial (x,y)}{\partial (r,t)}=\det\begin{pmatrix}\frac{\partial x}{\partial r}& \frac{\partial y}{\partial t}\\\frac{\partial y}{\partial r}&\frac{\partial y}{\partial t}\end{pmatrix}=\frac{r}{3\sqrt{2}}>0.$$
Therefore, the integral is $$\frac{1}{3^2\sqrt{2}}\int_{0}^{2\pi}\int_{0}^{1}\cos^{2}t\, drdt=\frac{\pi}{9\sqrt{2}}.$$
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"language": "en",
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|
Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$ Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$.
My effort: I tried using the fact $9^{\sqrt{2}}<9^{1.5}=27.$
Also We have $512 <729 \Rightarrow 2^9<27^2 \Rightarrow 2^{\frac{9}{2}}<27 \Rightarrow \sqrt{2}^9=2^{4.5}<27$. But both are below $27$.
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Remark: @achille hui posted a similar proof. But we got them independently.
The desired inequality is written as
$$3^{2\sqrt 2} < (2\sqrt 2)^3$$
or
$$2\sqrt 2\, \ln 3 < 3\ln (2\sqrt 2)$$
or
$$\frac{\ln 3}{3} < \frac{\ln(2\sqrt 2)}{2\sqrt 2}. \tag{1}$$
Let
$$f(x) := \frac{\ln x}{x}.$$
We have
$$f'(x) = \frac{1 - \ln x}{x^2}.$$
Thus, $f'(x) < 0$ on $(\mathrm{e}, \infty)$.
Since $\mathrm{e} < 2\sqrt 2 < 3$, (1) is true.
We are done.
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"timestamp": "2023-03-29T00:00:00",
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|
Question about limits at infinity: $f(x) < g(x)$ and $f(x) < \epsilon$ imply $g(x) < \epsilon$? I am going over the proof of $\lim\limits_{x \rightarrow \infty} \frac{x^2+2x}{2x^2+1} = \frac{1}{2}$. Let $f(x) = \frac{x^2+2x}{2x^2+1}$.
After setting $\epsilon >0$, we have $|f(x) - \frac{1}{2}| = |\frac{x^2+2x}{2x^2+1} - \frac{1}{2}| = |\frac{4x-1}{2(2x^2+1)}| < \epsilon$. We find that if $x > \frac{1}{4}$, then $\frac{4x-1}{2(2x^2+1)} < \epsilon$.
Now, the argument goes something like:
$$f(x) = \frac{4x-1}{2(2x^2+1)} < \frac{2x}{2x^2+1} < \frac{2x}{2x^2} = \frac{1}{x} = g(x)$$
And then it concludes that $f(x) < \frac{1}{x} = g(x) < \epsilon $ so $x > \frac{1}{\epsilon}$, and finally take $\max(\frac{1}{4}, \frac{1}{\epsilon})$ to prove the claim.
We have shown $f(x) < g(x)$ and we had $f(x) < \epsilon$. Why is it necessarily true that $g(x) < \epsilon$? Because in general if $a < b$ and $a < c$, we cannot conclude that $b < c$, since it is also possible that $c \leq b$. The inequality $g(x) < \epsilon$ does not look fully justified to me as is, especially because we have inequalities in the intermediate steps between $f(x)$ and $g(x)$ and not equalities.
|
The proof you are quoting seems to redefine $f(x)$. Worse, it is missing some connecting words to flesh out the argument. Some of the '<' you are seeing are aspirational: we want to assert inequality, but the inequality isn't true without additional conditions on $x$. The rest of the proof seeks to establish those conditions.
Here's how the proof should be read, with the connecting words filled in: First, we are given $\epsilon>0$. We seek a threshold $M$ so that if $x>M$, then $|f(x)-\frac12|<\epsilon$. Using algebra, we get
$$\left|f(x)-\frac12\right| = \left|\frac{4x-1}{2(2x^2+1)}\right|.$$ We want to arrive at the assertion $\left|\frac{4x-1}{2(2x^2+1)}\right|<\epsilon$. It isn't true for all $x$, so we seek conditions to make this assertion true. But the absolute value is getting in the way. We notice that if $x>\frac14$, then the numerator $4x-1$ is positive so we can drop the absolute value:
$$
\text{If $x>\frac14$, then}\qquad\left|f(x)-\frac12\right|=\frac{4x-1}{2(2x^2+1)}.\tag1$$
Next, we try to simplify the RHS of (1) to more easily see how to get it below $\epsilon$. Using algebra, we see
$$\frac{4x-1}{2(2x^2+1)}<\frac1x\tag2$$ for all $x$. So provided $\frac1x<\epsilon$, then the RHS of (2) will be less than $\epsilon$ and hence the LHS of (2) will be less than $\epsilon$ and hence the LHS of (1) will be less than $\epsilon$ (provided we also have $x>\frac14$). In summary: If $x>\frac14$ and $x>\frac1\epsilon$, then $|f(x)-\frac12|<\epsilon$. So we can take the threshold $M:=\max(\frac14, \frac1\epsilon$), proving the claim.
A key aspect of the proof is that we are seeking an upper bound on (1), so in (2) we must establish a chain of '<' inequalities, leading to the simple upper bound $\frac1x$. A chain of '>' inequalities, or a mixture of '>' and '<', won't get us anywhere. We can summarize the algebra in one line, with caveats:
$$\left|f(x)-\frac12\right|=\left|\frac{4x-1}{2(2x^2+1)}\right|\stackrel{(1)}=\frac{4x-1}{2(2x^2+1)}\stackrel{(2)}<\frac1x\stackrel{(3)}<\epsilon
$$ where the caveats are: (1) provided $x>\frac14$
(3) provided $x>\frac1\epsilon$. Note assertion (2) is true for all $x$.
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|
Is this logical deduction regarding some modular restrictions on odd perfect numbers valid? - Part II Let $p^k m^2$ be an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$. Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.
Here is the Abstract of Dris and San Diego's article titled "Some modular considerations regarding odd perfect numbers – Part II", published in NNTDM:
In this article, we consider the various possibilities for $p$ and $k$ modulo $16$, and show conditions under which the respective congruence classes for $\sigma(m^2)$ (modulo $8$) are attained, if $p^k m^2$ is an odd perfect number with special prime $p$. We prove that:
(1) $\sigma(m^2) \equiv 1 \pmod 8$ holds only if $p + k \equiv 2 \pmod {16}$.
(2) $\sigma(m^2) \equiv 3 \pmod 8$ holds only if $p - k \equiv 4 \pmod {16}$.
(3) $\sigma(m^2) \equiv 5 \pmod 8$ holds only if $p + k \equiv {10} \pmod {16}$.
(4) $\sigma(m^2) \equiv 7 \pmod 8$ holds only if $p - k \equiv 4 \pmod {16}$.
From this answer by mathlove and this earlier MSE question, we get the biconditional
$$\sigma(m^2) \equiv 3 \pmod 4 \iff p \equiv {k+4} \pmod {16}.$$
Also, we have the biconditional
$$\sigma(m^2) \equiv 1 \pmod 4 \iff p \equiv k \pmod 8,$$
per the characterization theorem of Chen and Luo.
Assume that the Descartes-Frenicle-Sorli Conjecture that $k=1$ holds.
If $p=5$, then this evidently implies that $\sigma(m^2) \equiv 3 \pmod 4$.
Here is our:
QUESTION: If $p=13$, then since $13 = p \equiv {k+4=5} \pmod 8$ holds, but the conjunction
$$(p = 13) \land (k=1)$$
obviously does not satisfy $p \equiv {k+4} \pmod {16}$ nor $p \equiv k \pmod 8$, does it follow that the implication $k = 1 \implies p \neq 13$ holds, since otherwise
$$\lnot\Bigg(p \equiv {k+4} \pmod {16}\Bigg) \iff \lnot\Bigg(\sigma(m^2) \equiv 3 \pmod 4\Bigg)$$
and
$$\lnot\Bigg(p \equiv k \pmod 8\Bigg) \iff \lnot\Bigg(\sigma(m^2) \equiv 1 \pmod 4\Bigg),$$
contradicting the fact that $\sigma(m^2)$ must be odd?
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Yes, it does.
Moreover, we can say that
$$k=1\implies p\not\equiv 13\pmod{16}$$
Proof :
Suppose that $p\equiv 13\pmod{16}$. Then, since we have
$$p \not\equiv {5} \pmod {16}\implies \sigma(m^2) \not\equiv 3 \pmod 4$$
and
$$p\not\equiv 1,9\pmod{16}\implies p\not\equiv 1\pmod 8\implies\sigma(m^2) \not\equiv 1 \pmod 4$$
we have $\sigma(m^2)\equiv 0,2\pmod 4$ which contradicts that $\sigma(m^2)$ is odd. $\ \blacksquare$
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|
Maximize $x^2 + y^2$ subject to $x^2 + y^2 = 1 - xy$ The equation $x^2 + y^2 = 1 - xy$ represents an ellipse. I am trying to show that its major axis is along $y=-x$ and find the vertex. To find the vertex I tried to find the vector in the ellipse with the greatest norm, which is equivalent to maximizing $x^2 + y^2$ subject to $x^2 + y^2 = 1 - xy$. How can I approach this?
(The maximum value of the objective function $x^2 + y^2$ seems to be 2. If I start out by assuming this I am able to show that the function cannot exceed this value. But this is kind of doing it in reverse as I start by evaluating the function at $y=-x=1.$ This also doesn't prove that $y=-x$ is the only optimal solution. How would I do it without "guessing" the solution first?)
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$\Delta \thinspace \thinspace \rm {method \thinspace\thinspace works\thinspace .}$
You have :
$$
\begin{cases}x^2+y^2=a,\thinspace a\in\mathbb R\\
x^2+y^2+xy=1\end{cases}
$$
This implies that,
$$
\begin{cases}y=\frac {1-a}{x}\\
x^2+\frac{(a-1)^2}{x^2}-a=0\end{cases}
$$
Then, letting $x^2=u$, we have :
$$
\begin{align}&u+\frac {(a-1)^2}{u}-a=0\\
\implies &u^2-ua+(a-1)^2=0\end{align}
$$
This leads to:
$$
\begin{align}&\Delta_u=a^2-4(a-1)^2\geq 0\\
\implies &\frac 23\leq a\leq 2\thinspace .\end{align}
$$
The equality $a=2$ holds, iff when :
$$\begin{align}&u=\frac a2=1\\
\implies &(x,y)=\left(\pm 1,\mp 1\right)\end{align}$$
Thus, we obtain a global maximum :
$$
\begin{align}\max\left\{x^2+y^2\mid x^2+y^2=1-xy\right\}&=2\\
\rm {at} \thinspace\thinspace\thinspace (x,y)=\left(\pm 1\thinspace,\mp 1\right)\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\end{align}
$$
Similarly, since $a\geq \frac 23$, then we obtain a global minimum :
$$
\begin{align}\min\left\{x^2+y^2\mid x^2+y^2=1-xy\right\}&=\frac 23\\
\rm {at} \thinspace\thinspace\thinspace (x,y)=\left(\pm \frac {\sqrt 3}{3},\thinspace \pm \frac {\sqrt 3}{3}\right)\end{align}
$$
This completes the answer .
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.