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showing $1 + z + z^2 + \dots $ uniformly converges to $\frac{1}{1-z}$ for $|z| < 1$ What test can I use to show that $1 + z + z^2 + \dots $ uniformly converges to $\frac{1}{1-z}$ for $|z| < 1$.
I know $\displaystyle 1 + z + z^2 + \dots +z^n = \frac{1-z^{n+1}}{1-z}$ and as $n \to \infty$, $1 + z + z^2 + \dots = \frac{1}{1-z}$ for $|z| < 1$ but how to show uniform convergence using it's definition?
ADDED::
May be we could use something like this, not sure though ... please suggest correction.
$$\left |S_n(z) - \frac{1}{1-z} \right | = \left |\frac{1-z^{n+1}}{1-z} - \frac{1}{1-z} \right | =\left | \frac{z^{n+1}}{1-z} \right |$$
If $|1-z| = \delta : 0< \delta < 1 \implies 1-|z| \le \delta $ and we have $\displaystyle \left | \frac{z^{n+1}}{1-z} \right | \le \frac{(1-\delta)^n}{\delta} = \epsilon$
|
Let denote
$$S_n(z)=\sum_{k=0}^n z^k=\frac{1-z^{n+1}}{1-z}$$
the partial sum of the series so we have
$$\left|R_n(z)\right|=\left|S_n(z)-\sum_{k=0}^\infty z^k\right|=\frac{|z|^{n+1}}{|1-z|}$$
so it's clear that
$$\sup_{|z|<1}|R_n(z)|=+\infty$$
so $(S_n(z))$ does not converge uniformly to $\frac{1}{1-z}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Condition for fourth degree polynomial to have all real roots For what range of values of $a$ will the following fourth degree polynomial have all real roots:
$$x^4 - 2ax^2 + x + a^2 -a = 0$$
|
First transform this equation into the equation with unknown $a$:
$a^{2}-(2x^{2} + 1)a + x^{4} + x = 0$
Solving the equation with unknown obtain two second degree equations:
$x^{2} + x - a = 0 $ and $ x^{2} - x + 1 - a = 0$
For $a < - \frac{1}{4}$ equation has no real roots
For $a = - \frac{1}{4}$ equation has two equal real roots
For $- \frac{1}{4} < a < \frac{3}{4}$ equation has two different real roots
For $a=\frac{3}{4}$ equation has real four roots of which three are equal
For $ a> \frac{3}{4}$ equation has four different real roots
|
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|
Calculus Reduction Formula
For any integer $k > 0$, show the reduction formula
$$\int^{2}_{-2} x^{2k} \sqrt{4-x^2} \, dx = C_k \int^{2}_{-2} x^{2k-2} \sqrt{4-x^2} \, dx$$
for some constant $C_{k}$.
(original image)
I thought this would be fairly straightforward but im a little confused. Do I start out by doing a trig substitution?
|
Let $I_k = \displaystyle \int^{2}_{-2} x^{2k} \sqrt{4-x^2} \, dx$.
Let $u = x^{2k-1}$ and $dv = x\sqrt{4-x^2} \, dx$.
We then have $du = (2k-1)x^{2k-2} \, dx$ and
$\begin{align*}
v &= \int x\sqrt{4-x^2} \, dx\\
&=\displaystyle -\frac{1}{3} (4-x^2)^{3/2}\\
&=-\frac{1}{3}(4-x^2)\sqrt{4-x^2}\end{align*}$
Applying integration by parts $\left (\int u \, dv = uv - \int v \, du \right )$:
$$\begin{align*}
I_k&= \left [ x^{2k-1} \times -\frac{1}{3}\overbrace{(4-x^2)^{3/2}}^{\text{this term becomes 0}} \right ]^{2}_{-2} + \frac{1}{3}(2k-1)\int^{2}_{-2} x^{2k-2}(4-x^2)\sqrt{4-x^2} \, dx\\
&= 0 + \frac{1}{3}(2k-1)\int^{2}_{-2}4x^{2k-2}\sqrt{4-x^2} - x^{2k}\sqrt{4-x^2} \, dx\\
&= \frac{4}{3}(2k-1)\overbrace{\int^{2}_{-2}4x^{2k-2}\sqrt{4-x^2} \, dx}^{I_{k-1}} - \frac{1}{3}(2k-1) \overbrace{\int^{2}_{-2} - x^{2k}\sqrt{4-x^2} \, dx}^{I_k}\\[10pt]
\therefore 3I_k &= 4(2k-1)I_{k-1} - (2k-1)I_{k}\\\\
\end{align*}$$
Rearranging yields $I_k = \displaystyle \frac{4k-2}{k+1}I_{k-1}$, so
$$C_k = \displaystyle \frac{4k-2}{k+1}.$$
|
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|
Prove $\left(\frac{2}{5}\right)^{\frac{2}{5}}<\ln{2}$ Inadvertently, I find this interesting inequality. But this problem have nice solution?
prove that
$$\ln{2}>(\dfrac{2}{5})^{\frac{2}{5}}$$
This problem have nice solution? Thank you.
ago,I find this
$$\ln{2}<\left(\dfrac{1}{2}\right)^{\frac{1}{2}}=\dfrac{\sqrt{2}}{2}$$
following is my some nice methods,
use this inequality
$$\dfrac{x-y}{\ln{x}-\ln{y}}>\sqrt{xy},x>y$$
then we let
$x=2,y=1$
so $$\ln{2}<\dfrac{\sqrt{2}}{2}$$
solution 2:
since
$$\dfrac{1}{n+1}\le\dfrac{1}{2}\cdot\dfrac{3}{4}\cdots\dfrac{2n-1}{2n}$$
then
$$\ln{2}=\sum_{n=0}^{\infty}\dfrac{1}{(n+1)2^{n+1}}<\sum_{n=0}^{\infty}\dfrac{(2n)!}{(n!)^22^{3n+1}}=\dfrac{1}{\sqrt{2}}$$
solution 3
since
$$(1+\sqrt{2})^2(t+1)-(t+1+\sqrt{2})^2=t(1-t)>0$$
so
$$\ln{2}=\int_{0}^{1}\dfrac{1}{t+1}dt<\int_{0}^{1}\left(\dfrac{1+\sqrt{2}}{t+1+\sqrt{2}}\right)^2dt=\dfrac{\sqrt{2}}{2}$$
solution 4:
$$\ln{2}=\dfrac{3}{4}-\dfrac{1}{4}\sum_{n=1}^{\infty}\dfrac{1}{n(n+1)(2n+1)}<\dfrac{3}{4}-\dfrac{1}{4}\left(\dfrac{1}{1\times 2\times 3}-\dfrac{1}{2\times 3\times 5}\right)=\dfrac{7}{10}<\dfrac{\sqrt{2}}{2}$$
solution 5
$$\dfrac{1}{\sqrt{2}}-\ln{2}=\sum_{n=1}^{\infty}\dfrac{\sqrt{2}}{(4n^2-1)(17+2\sqrt{2})^n}>0$$
But $$\ln{2}>\left(\dfrac{2}{5}\right)^{\frac{2}{5}}$$ I can't have this nice solution
Thank you everyone can help.
|
Take $f (x) = \frac {5} {2} \log x + \log 5 - x$. Obviously, $f$ is defined on the interval $]0, \infty[$. Since $f ' (x) = \frac {5} {2x} - 1$, it is easy to see that $f$ increases on $]0, \frac {5} {2}]$, take its maximum at $x = \frac {5} {2}$ and decreases on $[\frac {5} {2}, \infty[$. Also, since $f_{\max} = f (x_{\max}) > 0$ and $f (0+) = f (\infty-) = - \infty$, there are exactly two non-negative real numbers $x_1 < \frac {5} {2}$ and $x_2 > \frac {5} {2}$ such that $f (x_1) = f (x_2) = 0$. Thus, we conclude that $f (x) \geqslant 0$ on $[x_1, x_2]$. Since $x_1 < (2/5)^{2/5} < x_2$, we should have $f \left ((2/5)^{2/5} \right) > 0$, that is, $$\log 2 - (2/5)^{2/5} >0,$$ as desired.
|
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|
How do you orthogonally diagonalize the matrix? How do you orthogonally diagonalize the matrix A?
Matrix A =
$$
\begin{bmatrix}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{bmatrix}
$$
|
Since the matrix $A$ is symmetric, we know that it can be orthogonally diagonalized. We first find its eigenvalues by solving the characteristic equation:
$$0=\det(A-\lambda I)=\begin{vmatrix}
1-\lambda & 1 & 1 \\
1 & 1-\lambda & 1 \\
1 & 1 & 1-\lambda
\end{vmatrix}=-(\lambda-3)\lambda^2 \implies \left\{\begin{array}{l l}
\color{red}{\lambda_1 = 0} \\
\color{green}{\lambda_2 = 0} \\
\color{blue}{\lambda_3 = 3}
\end{array}\right.$$
We now find the eigenvectors corresponding to $\lambda=0$:
$$\left(\begin{array}{ccc|c}
1 & 1 & 1 & 0 \\
1 & 1 & 1 & 0 \\
1 & 1 & 1 & 0
\end{array}\right) \implies \left(\begin{array}{ccc|c}
1 & 1 & 1 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{array}\right) \implies \mathbf{x}=\pmatrix{s\\t\\-s-t}=s\pmatrix{1\\0\\-1}+t\pmatrix{0\\1\\-1}$$
By orthonormalizing them, we obtain the basis
$$\left\{\color{red}{\frac{1}{\sqrt{2}}\pmatrix{1\\0\\-1}},\color{green}{\frac{1}{\sqrt{6}}\pmatrix{-1\\2\\-1}}\right\}$$
We finally find the eigenvector corresponding to $\lambda=3$:
$$\left(\begin{array}{ccc|c}
-2 & 1 & 1 & 0 \\
1 & -2 & 1 & 0 \\
1 & 1 & -2 & 0
\end{array}\right) \implies \left(\begin{array}{ccc|c}
0 & -3 & 3 & 0 \\
1 & -2 & 1 & 0 \\
0 & 3 & -3 & 0
\end{array}\right) \implies \left(\begin{array}{ccc|c}
0 & -1 & 1 & 0 \\
1 & -1 & 0 & 0 \\
0 & 0 & 0 & 0
\end{array}\right) \implies \mathbf{x}=\pmatrix{s\\s\\s}=s\pmatrix{1\\1\\1}$$
By normalizing it, we obtain the basis
$$\left\{\color{blue}{\frac{1}{\sqrt{3}}\pmatrix{1\\1\\1}}\right\}$$
Hence $A$ is orthogonally diagonalized by the orthogonal matrix
$$P=\pmatrix{\color{red}{1/\sqrt{2}}&\color{green}{-1/\sqrt{6}}&\color{blue}{1/\sqrt{3}}\\\color{red}{0}&\color{green}{2/\sqrt{6}}&\color{blue}{1/\sqrt{3}}\\\color{red}{-1/\sqrt{2}}&\color{green}{-1/\sqrt{6}}&\color{blue}{1/\sqrt{3}}}$$
Furthermore,
$$P^{T}AP=\pmatrix{\color{red}{0}&0&0\\0&\color{green}{0}&0\\0&0&\color{blue}{3}}$$
|
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|
How to find the number of real roots of the given equation?
The number of real roots of the equation $$2 \cos \left( \frac{x^2+x}{6} \right)=2^x+2^{-x}$$ is
(A) $0$, (B) $1$, (C) $2$, (D) infinitely many.
Trial: $$\begin{align} 2 \cos \left( \frac{x^2+x}{6} \right)&=2^x+2^{-x} \\ \implies \frac{x^2+x}{6}&=\cos ^{-1} \left( \frac{2^x+2^{-x}}{2} \right)\end{align}$$ Then I can't proceed.
|
If you consider $f(x)=2^x+2^{-x}$ and notice how it is symmetrical around $x=0$ and has its minimum there as for example $f(1)=2.5$ and $f(2)=4.25$, which the left hand side has a maximum value of 2, thus there is only one possible root though I'll leave that for you to see.
|
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|
Trigonometric Formula I am stuck with the simple expression
$$
\frac{\cos^2(\theta + \alpha)}{1 - \cos^2(\theta - \alpha)} = \text{const.}
$$
where $\theta$ is a variable and $\alpha$ is the number satisfying
$$
\alpha = \tan^{-1} (\frac{4}{3})\,.
$$
I cannot see it to be immediate, somehow I am missing a particular trigonometric identity. Or does this require a more detailed calculuation? Thanks for any hints, I'll fill in the details myself!
|
Note that $1 - \cos^2(\theta - \alpha) = \sin^2(\theta - \alpha)$
That gives you:
$$\frac{\cos^2(\theta + \alpha)}{1 - \cos^2(\theta - \alpha)} = \text{const.} = \frac{\cos^2(\theta + \alpha)}{\sin^2(\theta - \alpha)}$$
For the numerator: $\cos(\theta+\alpha) = \cos\theta\cos\alpha-\sin\theta\sin\alpha.\tag{1}$
For the denominator: $\sin(\theta - \alpha) = \sin \theta \cos \alpha - \cos \theta \sin \alpha \tag{2}$
Note that since $\tan \alpha = \dfrac 43$, we have a $3:4:5$ triangle, using the Pythagorean Theorem, and noting that the leg opposite $\alpha$ must be of length $4$, and the leg adjacent to $\alpha$ is length $3$. This gives us a hypotenuse of length $5$. Calculating $\cos \alpha = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac 35$. Likewise $\sin \alpha = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac 45.$
So $(1)$ becomes $\cos^2(\theta + \alpha) = \left(\dfrac 35 \cos\theta - \dfrac 45 \sin\theta\right)^2$.
And $(2)$ becomes $\sin^2(\theta - \alpha) = \left(\dfrac 35\sin\theta - \dfrac45 \cos\theta\right)^2$
|
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|
Find all function $f:\mathbb{R}\mapsto\mathbb{R}$ such that $f(x^2+y^2)=f(x+y)f(x-y)$. Find all function $f:\mathbb{R}\mapsto\mathbb{R}$ such that $f(x^2+y^2)=f(x+y)f(x-y)$.
Some solutions I found are $f\equiv0,f\equiv1$, $f(x)=0$ if $x\neq0$ and $f(x)=1$ if $x=0$.
|
Perform the substitution $x=\frac{a+b}{2}, y=\frac{a-b}{2}$ to get $f(\frac{a^2+b^2}{2})=f(a)f(b)$. We shall denote this statement with $P(a, b)$.
$P(a, a)$: $f(a^2)=f(a)^2$. In particular, when $a=0$, we have $f(0)=f(0)^2$ so $f(0)=0$ or $1$.
If $f(0)=0$, then $P(a, 0)$: $f(\frac{a^2}{2})=0$, so $f(x)=0 \, \forall x \geq 0$. Thus $f(x)^2=f(x^2)=0 \, \forall x \in \mathbb{R}$, so $f(x)=0 \, \forall x \in \mathbb{R}$.
Otherwise $f(0)=1$. Then $P(a, 0)$: $f(\frac{a^2}{2})=f(a)$. Thus $f(a^2)=f(a)^2=f(\frac{a^2}{2})^2$, so $f(2x)=f(x)^2 \, \forall x \geq 0$. Thus $f(x^2)=f(x)^2=f(2x)=f(\frac{(2x)^2}{2})=f(2x^2) \, \forall x \geq 0$, so $f(2x)=f(x) \, \forall x \geq 0$, so $f(x)=f(2x)=f(x)^2 \forall x \geq 0$. Thus $f(x)=0$ or $1 \, \forall x \geq 0$.
Note that $f(-a)=f(\frac{(-a)^2}{2})=f(\frac{a^2}{2})=f(a)$. If $f(c)=0$ for some $c>0$, then $P(a, c)$: $f(\frac{a^2+c^2}{2})=0$, so $f(x)=0$ for $x \geq \frac{c^2}{2}$. Consider an arbitrary $x>0$. Using $f(2x)=f(x)$, it is easy to prove by induction that $f(2^nx)=f(x) \, \forall n \in \mathbb{Z}^+$. Clearly there exists a positive integer $N_x$ s.t. $2^{N_x}x \geq \frac{c^2}{2}$, so $f(x)=f(2^{N_x}x)=0$. Therefore $f(x)=0 \, \forall x>0$, so $f(x)=0 \, \forall x \not =0, f(0)=1$.
Otherwise $f(x)=1 \, \forall x \geq 0$, so since $f$ is even, $f(x)=1 \, \forall x \in \mathbb{R}$
To conclude, we have 3 solutions: $f(x)=0 \, \forall x \in \mathbb{R}$, $f(x)=1 \, \forall x \in \mathbb{R}$, and $f(x)=0 \, \forall x \not =0, f(0)=1$. It is easily checked that these are all solutions.
|
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|
Showing that $\mathbb Q(\sqrt{17})$ has class number 1 Let $K=\mathbb Q(\sqrt{d})$ with $d=17$. The Minkowski-Bound is $\frac{1}{2}\sqrt{17}<\frac{1}{2}\frac{9}{2}=2.25<3$.
The ideal $(2)$ splits, since $d\equiv 1$ mod $8$. So we get $(2)=(2,\frac{1+\sqrt{d}}{2})(2,\frac{1-\sqrt{d}}{2})$ and $(2,\frac{1\pm\sqrt{d}}{2})$ are two ideals of norm $2$.
Now if we can show that $(2,\frac{1\pm\sqrt{d}}{2})$ are principal ideals, then we know that every ideal class contains a principal ideal, which shows that the class number is $1$.
But how can we show that $(2,\frac{1\pm\sqrt{d}}{2})$ are principal?
|
We show that $$\left\langle 2,\frac{1+\sqrt{17}}{2}\right\rangle = \left\langle \frac{5+\sqrt{17}}{2}\right\rangle.$$
Since, $$\frac{5+\sqrt{17}}{2}=2+\frac{1+\sqrt{17}}{2},$$ we have $$\frac{5+\sqrt{17}}{2}\in\left\langle 2,\frac{1+\sqrt{17}}{2}\right\rangle,$$ thus $$\left\langle 2,\frac{1+\sqrt{17}}{2}\right\rangle \subseteq \left\langle \frac{5+\sqrt{17}}{2}\right\rangle.$$
Now these two ideals have the same Norm--namely, $2$. Therefore, $$\left\langle 2,\frac{1+\sqrt{17}}{2}\right\rangle = \left\langle \frac{5+\sqrt{17}}{2}\right\rangle.$$
The proof of the principality of $\frac{5+\sqrt{17}}{2}$ is similar.
|
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|
Find all real numbers $x$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac76$ Find all real numbers $x$ for which $$\frac{8^x+27^x}{12^x+18^x}=\frac76$$
I have tried to fiddle with it as follows:
$$2^{3x} \cdot 6 +3^{3x} \cdot 6=12^x \cdot 7+18^x \cdot 7$$
$$ 3 \cdot 2^{3x+1}+ 2 \cdot 3^{3x+1}=7 \cdot 6^x(2^x+3^x)$$
Dividing both sides by $6$ gives us
$$2^{3x}+3^{3x}=7 \cdot 6^{x-1}(2^x+3^x)$$
Is this helpful? If so, how should I proceed form here? If not any hints would be greatly appreciated.
|
Let us put $2^x=a,3^x=b$ to remove the indices to improve clarity
So, $8^x=(2^3)^x=(2^x)^3=a^3$ and similarly, $27^x=b^3$
$12^x=(2^2\cdot3)^x=(2^x)^2\cdot3^x=a^2b$ and similarly, $18^x=ab^2$
So, the problem reduces to $$\frac{a^3+b^3}{ab(a+b)}=\frac76$$
$\displaystyle\implies 6(a^2-ab+b^2)=7ab$ as $a+b\ne0$
$\displaystyle\implies 6\left(\frac ab\right)^2-13\cdot\frac ab+6=0$
$\displaystyle\implies \frac ab=\frac32$ or $\dfrac23$
So, $\displaystyle\left(\frac23\right)^x=\frac32$ or $\dfrac23$
If $\displaystyle\left(\frac23\right)^x=\frac32\implies\left(\frac23\right)^x=\left(\frac23\right)^{-1}\iff\left(\frac23\right)^{x+1}=1$
Similarly, if $\displaystyle\left(\frac23\right)^x=\frac23, \left(\frac23\right)^{x-1}=1 $
Now if $\displaystyle u^m=1,$
either $\displaystyle m=0,u\ne0; $
or $\displaystyle u=1$
or $\displaystyle u=-1,m$ is even
|
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|
Why can I divide a fraction like this? Suppose I have a fraction: $$\frac{2^n}{2^{2n}+1}$$
I can simplify it to become: $$\frac{1}{2^{n}+\frac{1}{2^n}}$$
Now obviously, this is just dividing both the numerator and the denominator of the fraction by $2^n.$ My question is why I can do this. Can anyone explain the algebra behind this division to me?
EDIT: I tried thinking about the initial fraction as $2^n \cdot \frac{1}{2^{2n}+1}$ and the division operation as $\frac{2^n \cdot \frac{1}{2^{2n}+1}}{2^n}$, but I couldn't get anywhere with attempting to compute $\frac{\frac{1}{2^{2n}+1}}{2^n}$.
|
One way of looking at it, is that you are multiplying the numerator and the denominator by $\frac{1}{2^n}$.
In the numerator you get $\frac{2^n}{2^n}$ which is $1$.
In the denominator you get:
$$\frac{1}{2^n}\cdot (2^{2n} + 1) = \frac{2^{2n}}{2^n} + \frac{1}{2^n} = 2^{2n-n} + \frac{1}{2^n} = 2^n + \frac{1}{2^n}$$
and there you go:
$$\dfrac{1}{2^n + \dfrac{1}{2^n}}$$
|
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|
Finding $\lim_{x \to 0}\frac{\tan x-x}{x^3}$ Feeling like i did this wrong
$\displaystyle \lim_{x \to 0}\frac{\tan x-x}{x^3}$ $\to$ $\displaystyle \lim_{x \to 0}\frac{\sec^2x-1}{3x^2}$
$\displaystyle \lim_{x \to 0}\frac{2\tan x\sec^2x}{4x}$ $\to$ $\displaystyle \lim_{x \to 0}\frac{2\sec^2x\sec^2x+2\tan x(2\tan x\sec^2x)}{6}$
Simplified
$\displaystyle \lim_{x \to 0}\frac{\sec^2x(\sec^2x+4\tan^2x)}{3}$
Not really sure what to do at this point
|
The Pythagorean Identity is of help here: starting from your first application of l'Hopital's Rule, you could then write
$$\lim_{x \rightarrow 0} \ \frac{\sec^2 x \ - \ 1}{3 x^2} \ ^{*} \ = \ \lim_{x \rightarrow 0} \ \frac{(\tan^2 x \ + \ 1 ) \ - \ 1}{3 x^2} \ = \ \lim_{x \rightarrow 0} \ \frac{\tan^2 x }{3 x^2} $$
*
*there is an error in your differentiation of $x^3$
$$= \ \lim_{x \rightarrow 0} \ \frac{\sin^2 x}{3 x^2 \ \cdot \ \cos^2 x } \ = \ \lim_{x \rightarrow 0} \ \frac{1}{3} \ \cdot \ (\frac{\sin x}{x})^2 \ \cdot \ \frac{1}{\cos^2 x} \ , $$
with the limit of the middle factor having a familiar value.
|
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|
Prove $\int_0^{\infty } \frac{1}{\sqrt{6 x^3+6 x+9}} \, dx=\int_0^{\infty } \frac{1}{\sqrt{9 x^3+4 x+4}} \, dx$ Prove that:
$(1)$$$\int_0^{\infty } \frac{1}{\sqrt{6 x^3+6 x+9}} \ dx=\int_0^{\infty } \frac{1}{\sqrt{9 x^3+4 x+4}} \ dx$$
$(2)$$$\int_0^{\infty } \frac{1}{\sqrt{8 x^3+x+7}} \ dx>1$$
What I do for $(1)$ is (something trival):
$$\int_0^{\infty } \frac{1}{\sqrt{6 x^3+6 x+9}} \, dx=\int_0^{\infty } \frac{1}{\sqrt{x^3+36 x+324}} \, dx$$
$$\int_0^{\infty } \frac{1}{\sqrt{9 x^3+4 x+4}} \, dx=\frac{\sqrt3}3\int_0^{\infty } \frac{1}{\sqrt{x^3+4 x+12}} \, dx$$
so it remains to prove that
$$\int_0^{\infty } \frac{1}{\sqrt{x^3+36 x+324}} \, dx=\frac{\sqrt3}3\int_0^{\infty } \frac{1}{\sqrt{x^3+4 x+12}} \, dx$$
Thanks in advance!
|
We could use Carleman's Inequality which states that,
$$
\int_0^\infty f(x)dx \ge 1/e\int_0^\infty exp(1/x*\int_0^xln(f(x))dx)dx
$$
Then, by substituting our function as f(x), we can prove the statement by proving that
$$
1/e\int_0^\infty exp(1/x*\int_0^xln(f(x))dx)dx \gt 1
$$
This is done by noting that for all x >0
$$
9t^3 + 9 \gt 8t^3 + t + 7
$$
Simplifying the right hand integral might prove the case.
|
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|
Derivative of Trig Functions (Intuition Help?) Looking for some intuition help here.
I have the following exercise and these are the steps I take:
$$
y = \sin\left(\frac{1}{x}\right)
$$
$$
u=\frac{1}{x}
$$
$$
y = \sin u,\;\;\frac{dy}{du} = \cos u= \cos\left(\frac{1}{x}\right)
$$
$$
u=x^{-1};\;\frac{du}{dx} =-x^{-2}=-\frac{1}{x^2}
$$
$$
\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}=cos\frac{1}{x}\times-\frac{1}{x^2}
$$
This is incorrect but intuitively I want to multiply it this way.
$$
cos\frac{1}{x}\times-\frac{1}{x^2}=cos-\frac{1}{x^3}
$$
But the correct answer is:
$$
-\frac{cos-\frac{1}{x}}{x^2}
$$
Help me absorb the why so I can intuitively solve problems like these.
|
Here is where parentheses come in handy:
You found, correctly, $\dfrac{dy}{du}$ and $\dfrac{du}{dx}$.
But the scope of $\cos$ is restricted to its argument: $\left(\dfrac 1x\right)$ ONLY:
The FUNCTION $\dfrac{dy}{du} = \cos\left(\dfrac 1x\right)$ is multiplied by the function $\dfrac{du}{dx} = -\dfrac 1{x^2}$. That is not what you did. You multiplied argument of the $\cos$ function by the function $\dfrac{du}{dx} = -\dfrac{1}{x^2}$.
$$\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}=\left[cos\left(\frac{1}{x}\right)\right]\times\left(-\frac{1}{x^2}\right) $$ $$= -\frac 1{x^2} \cos\left(\frac 1x\right)
= -\dfrac{\cos\left(\frac 1x\right)}{x^2}$$
|
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|
Compute the remainder when $67!$ is divided by $71$. This is how far I've been able to get.
By using Wilson's Theorem:
$$\begin{align}
70! &\equiv -1 \pmod{71} \\
67!(68)(69)(70) &\equiv -1 \pmod{71} \\
67!(68)(69)(-1) &\equiv -1 \pmod{71} \\
67!(68)(69) &\equiv 1 \pmod{71} \\
\end{align}$$
EDIT: Here is how I proceeded using TMM's and Carl Mummert's hints.
$$\begin{align}
&68 \equiv -3 \pmod{71}\\
and\\
&69 \equiv -2 \pmod{71}\\
\end{align}$$
So:
$$\begin{align}
67!(-3)(-2) &\equiv 1 \pmod{71} \\
67! &\equiv 6^{-1} \pmod{71} \\
\end{align}$$
By using the Euclidean algorithm:
$$71 = 6 \cdot 11 + 5$$
$$6 = 5 \cdot 1 + 1$$
$$5 = 1 \cdot 5 + 0$$
Now, going backwards:
$$\begin{align}
1 &= 6 - 5 \\
&= 6 - (71 - 6 \cdot 11) \\
&= 6 + 6(11) - 71 \\
&= 6(1 + 11) - 71(1) \\
&= 6(12) - 71(1)
\end{align}$$
Therefore, $67! \equiv 6^{-1} \equiv 12 \pmod{71}$.
|
Recall that if
$$xy \equiv a \pmod{n}$$
then
$$x \equiv ay^{-1} \pmod{n}$$ if $y$ is invertible in $\pmod{n}$. In your case, $n$ is $71$ a prime, which guarantees that any $y \not \equiv 0 \pmod{71}$ is invertible in $\pmod{71}$. Hence,, we have
$$67! \times 68 \times 69 \equiv 1 \pmod{71} \implies 67! \equiv (69)^{-1} (68)^{-1} \pmod{71}$$
We have
\begin{align}
68 \times 47 - 71 \times 45 & = 1\\
69 \times 35 - 71 \times 34 & = 1
\end{align}
Hence,
$$(68)^{-1} \equiv 47 \pmod{71} \text{ and }(69)^{-1} \equiv 35 \pmod{71}$$
Hence,
$$67!\equiv 35 \times 47 \pmod{71} \equiv 12 \pmod{71}$$
|
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|
Solving recurrence relation, $a_n=6a_{n-1} - 5a_{n-2} + 1$ I'm trying to solve this recurrence relation:
$$
a_n = \begin{cases}
0 & \mbox{for } n = 0 \\
5 & \mbox{for } n = 1 \\
6a_{n-1} - 5a_{n-2} + 1 & \mbox{for } n > 1
\end{cases}
$$
I calculated generator function as:
$$
A = \frac{31x - 24x^2}{1 - 6x + 5x^2} + \frac{x^3}{(1-x)(1-6x+5x^2)} =
\frac{31x - 24x^2}{(x-1)(x-5)} + \frac{x^3}{(1-x)(x-1)(x-5)}
$$
(I'm not sure if that's right)
and its partial fractions decomposition looks like:
$$
A = \left(\frac{-7}{4} \cdot \frac{1}{x-1} - \frac{445}{4} \cdot \frac{1}{x-5}\right) +
\left( \frac{39}{16} \cdot \frac{1}{x-5} + \frac{3}{4} \cdot \frac{1}{(x-1)^2} - \frac{375}{16} \cdot \frac{1}{x-5} \right)
$$
(again - I'm not sure if it's ok)
I'm stuck here... From solutions I know that I should get:
$$
a_n = \frac{-21}{16} - \frac{1}{4}n + \frac{21}{16}5^n
$$
but I have no idea how it's solved... I hope somebody can help me (I spend more than 3h trying to solve this myself...)
|
I did not check your work, so I’ll outline what you need to do to finish. You have something of the form
$$A(x)=\frac{a}{1-x}+\frac{b}{5-x}+\frac{c}{(1-x)^2}=\frac{a}{1-x}+\frac{b/5}{1-\frac{x}5}+\frac{c}{(1-x)^2}\;.$$
Expand these three terms into power series:
$$\begin{align*}
A(x)&=a\sum_{n\ge 0}x^n+\frac{b}5\sum_{n\ge 0}\left(\frac{x}5\right)^n+c\sum_{n\ge 0}(n+1)x^n\\\\
&=\sum_{n\ge 0}\left(a+\frac{b}{5^{n+1}}+c(n+1)\right)x^n\;.
\end{align*}$$
Now you can read off the coefficient of $x^n$.
Added: I’ve now had a chance to check your work, and it appears to be a bit off. I’ll use my preferred approach, which begins by assuming that $a_n=0$ for all $n<0$. Then the recurrence can be written
$$a_n=6a_{n-1}-5a_{n-2}+1-[n=0]+4[n=1]\;,$$
for all $n\ge 0$, where the last two terms contain Iverson brackets and are added to make the recurrence give the correct values to $a_0$ and $a_1$. Now multiply through by $x^n$ and sum over $n\ge 0$:
$$\begin{align*}
\sum_{n\ge 0}a_nx^n&=\sum_{n\ge 0}\Big(6a_{n-1}-5a_{n-2}+1-[n=0]+4[n=1]\Big)x^n\\\\
&=6x\sum_{n\ge 0}a_{n-1}x^{n-1}-5x^2\sum_{n\ge 0}a_{n-2}x^{n-2}+\sum_{n\ge 0}x^n-1+4x\;.
\end{align*}$$
Since $A(x)=\sum_{n\ge 0}a_nx^n=\sum_{n\ge 0}a_{n-1}x^{n-1}=\sum_{n\ge 0}a_{n-2}x^{n-2}$ (remember the blanket assumption that $a_n=0$ for $n<0$), we have
$$A(x)=6xA(x)-5x^2A(x)+\frac1{1-x}-1+4x\;,$$
and hence
$$A(x)=\frac1{(1-x)^2(5-x)}+\frac{4x-1}{(1-x)(5-x)}\;.$$
I’ll leave the partial fraction decomposition to you, at least for now.
|
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|
A closed-form expression for the integral $\int_0^\infty\text{Ci}^{3}(x) \, \mathrm dx$ Is there a closed-form expression for this integral:
$$\int_0^\infty\text{Ci}^{3}(x) \, \mathrm dx,$$
where $\text{Ci}(x)=-\int_x^\infty\frac{\cos z}{z}\mathrm dz$ is the cosine integral?
$\text{Ci}(x)$ and $\text{Ci}^{2}(x)$ have primitives/antiderivatives that can be expressed in terms of the trigonometric integral functions.
So it's not too difficult to show that $$\int_0^\infty\text{Ci}(x) \, \mathrm dx =0$$ and $$\int_0^\infty\text{Ci}^{2}(x) \, \mathrm dx = \frac{\pi}{2}.$$
But $\text{Ci}^{3}(x)$ doesn't appear to have a primitive that can be expressed in terms of known functions.
|
Expanding on Start wearing purple's answer, the following is an evaluation of $$\int_{0}^{\infty} \frac{\sin (2x)}{x} \, \operatorname{Ci}(x) \, dx .$$
First notice that by making the substitution $ u = \frac{t}{x}$, we get
$$ \operatorname{Ci}(x) = - \int_{x}^{\infty} \frac{\cos (t)}{t} \, dt = - \int_{1}^{\infty} \frac{\cos (xu)}{u} \, du.$$
Therefore,
$$ \int_{0}^{\infty} \frac{\sin (2x)}{x} \, \operatorname{Ci}(x) \, dx = - \int_{0}^{\infty} \int_{1}^{\infty} \frac{\sin (2x)}{x} \frac{\cos (xu)}{u} \, du \, dx .$$
Since the iterated integral does not converge absolutely, changing the order of integration is not justified by Fubini's theorem.
But by integrating by parts, we get
$$ \begin{align} \int_{0}^{\infty} \frac{\sin (2x)}{x} \, \operatorname{Ci}(x) \, dx &= \int_{0}^{\infty} \frac{\sin (2x) \sin (x)}{x^{2}} \, dx - \int_{0}^{\infty} \int_{1}^{\infty} \frac{\sin (2x) \sin (xu)}{x^{2}u^{2}} \, du \, dx \\ &= \int_{0}^{\infty} \frac{\sin (2x) \sin (x)}{x^{2}} \, dx - \int_{1}^{\infty} \frac{1}{u^{2}}\int_{0}^{\infty} \frac{\sin (2x) \sin(ux)}{x^{2}} \, dx \, du \, . \end{align}$$
In general, for $a,b \ge 0$, we have $$ \int_{0}^{\infty} \frac{\sin (ax) \sin (bx)}{x^{2}} \ dx = \frac{\pi}{2} \min \{a,b \} .$$
Therefore,
$$ \begin{align} \int_{0}^{\infty} \frac{\sin 2x}{x} \, \operatorname{Ci}(x) \, dx &= \frac{\pi}{2} \, \text{min} \{2,1 \} - \frac{\pi}{2} \int_{1}^{\infty} \frac{1}{u^{2}} \, \text{min} \{2,u \} \, du \\ &= \frac{\pi}{2} - \frac{\pi}{2} \int_{1}^{2} \frac{u}{u^{2}} \, du - \frac{\pi}{2} \int_{2}^{\infty} \frac{2}{u^{2}} \, du \\ &= \frac{\pi}{2} - \frac{\pi}{2} \, \ln (2) - \frac{\pi}{2} \\ &= - \frac{\pi}{2} \, \ln (2) . \end{align}$$
UPDATE:
Integrating by parts wasn't necessary since $$- \int_{0}^{\infty} \int_{1}^{\infty} \frac{\sin (2x)}{x} \frac{\cos (xu)}{u} \, \mathrm du \, \mathrm dx = - \int_{1}^{\infty} \int_{0}^{\infty} \frac{\sin (2x)}{x} \frac{\cos (xu)}{u} \, \mathrm dx \, \mathrm du$$ is justified by Plancherel's theorem for the Fourier transform in the form $$\int_{\mathbb{R}^{2}} \hat{f}(x) g(x) \, \mathrm dx = \int_{\mathbb{R}^{2}} f(\omega) \hat{g}(\omega) \, \mathrm d \omega. $$ (Some textbooks refer to this as the multiplication formula.)
Therefore, we can also say that $$ \begin{align}\int_{0}^{\infty} \frac{\sin (2x)}{x} \, \operatorname{Ci}(x) \, \mathrm dx &= - \int_{0}^{\infty} \int_{1}^{\infty} \frac{\sin (2x)}{x} \frac{\cos (xu)}{u} \, \mathrm du \, \mathrm dx \\ &= - \int_{1}^{\infty} \int_{0}^{\infty} \frac{\sin (2x)}{x} \frac{\cos (xu)}{u} \, \mathrm dx \, \mathrm du \\ &= - \frac{1}{2} \int_{1}^{\infty} \frac{1}{u} \int_{0}^{\infty} \frac{\sin\left((2-u)x \right)+ \sin \left((2+u)x \right)}{x} \, \mathrm dx \, \mathrm du \\ &= -\frac{\pi}{4} \int_{1}^{\infty} \frac{1}{u} \left(\operatorname{sgn}(2-u) +1\right) \, \mathrm du \\ &= - \frac{\pi}{2} \int_{1}^{2} \frac{\mathrm du }{u} \\ &= - \frac{\pi}{2} \, \ln (2). \end{align}$$
In general, if $0 \le a \le 1$, then $$\int_{0}^{\infty} \frac{\sin (ax)}{x} \, \operatorname{Ci}(x) \, \mathrm dx = 0.$$
And if $a >1$, then $$\int_{0}^{\infty} \frac{\sin (ax)}{x} \, \operatorname{Ci}(x) \, \mathrm dx = - \frac{\pi}{2} \int_{1}^{a} \frac{\mathrm du}{u} = - \frac{\pi}{2} \, \ln (a). $$
|
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|
Integrating $\int \left( \frac{\cos\theta}{1+\sin^2\theta} \right)^n \, d\theta$, for $n=2$ and $n=3$ How do you integrate the following functions:
$$\int \left( \frac{\cos\theta}{1+\sin^2\theta} \right)^2 \, d\theta$$ and $$\int \left( \frac{\cos\theta}{1+\sin^2\theta} \right)^3 \, d\theta
$$
respectively?
Note: Initially, I tried integrating the function without the power and obtained the result below.
$$
\int \frac{\cos\theta}{1+\sin^2\theta} \, d\theta = \arctan(\sin\theta)+ C
$$
However, from here it is difficult to proceed. Integration by substitution doesn't seem to work.
How should I go on from here? Any pointers would be greatly appreciated.
|
I will be solving the first integral. If I can figure out the second integral I'll post my solution as another answer.
$$I=\int\bigg(\frac{\cos x}{1+\sin^2x}\bigg)^2\mathrm{d}x$$
$$I=\int\frac{\cos^2x}{(1+\sin^2x)^2}\mathrm{d}x$$
$$I=\int\frac{\sec^2 x\ \mathrm{d}x}{(2\tan^2x+1)^2}$$
$t=\tan x\Rightarrow \mathrm{d}t=\sec^2x\ \mathrm{d}x$:
$$I=\int\frac{\mathrm{d}t}{(2t^2+1)^2}$$
Apply the reduction formula
$$\int\frac{\mathrm{d}x}{(ax^2+b)^n}=\frac{x}{2b(n-1)(ax^2+b)^2}+\frac{2n-3}{2b(n-1)}\int\frac{\mathrm{d}x}{(ax^2+b)^{n-1}}$$
With $a=2,\ b=1,\ n=2$:
$$I=\frac{t}{2(2t^2+1)}+\frac12\int\frac{\mathrm{d}t}{2t^2+1}$$
Preform $u=\sqrt{2}\ t$:
$$I=\frac{t}{2(2t^2+1)}+\frac1{2\sqrt{2}}\int\frac{\mathrm{d}u}{u^2+1}$$
$$I=\frac{t}{2(2t^2+1)}+\frac1{2\sqrt{2}}\arctan u$$
$$I=\frac{t}{2(2t^2+1)}+\frac1{2\sqrt{2}}\arctan t\sqrt{2}$$
$$I=\frac{\tan x}{2(2\tan^2x+1)}+\frac1{2\sqrt{2}}\arctan(\sqrt{2}\tan x)\quad +C$$
|
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|
Solving two algebraic equations I'm trying to prove an identity from physics. I have the following two equations ($M$ is a constant):
$$e^2 = \left(1-\frac{2M}{r}\right)\left(1+\frac{l^2}{r^2}\right)$$ and $$r = \frac{l^2}{2M}\left[1 - \sqrt{1 - 12\left(\frac{M}{l}\right)^2}\right]$$ and I need to show that $$\frac{l}{e} = \sqrt{Mr}\left(1 - \frac{2M}{r}\right)^{-1}$$
I've tried rearranging the second eqation to get $l^2$ on its own and then dividing this by $e^2$ from equation 1 to get an expression for $l^2/e^2$ but this is quite messy and I can't see a clear way to simplify it. Can anyone see a straightforward way of doing this?
|
let $x=\dfrac{l^2}{M^2},a=\dfrac{2r}{M} \to a=x-\sqrt{x^2-12x} \to x^2-12x=(x-a)^2 \to x=\dfrac{a^2}{2a-12}$
$\sqrt{1-\dfrac{12}{x}}=\sqrt{\dfrac{a^2-12*2a+12*12}{a^2}}=\dfrac{|a-12|}{a}$
now you can simplify more easy.
ya,if 12 is wrong, you give a right number but the middle result is same.
But it seems the last one is not correct,there should a factor of $\sqrt{\left(1 - \dfrac{2M}{r}\right)^{-1}}$
edit: I am interesting to go further after seeing the source poster.
$l^2=\dfrac{(aM)^2}{2a-12}=\dfrac{4r^2}{\dfrac{4r}{M}-12}=\dfrac{r^2M}{r-3M}=\dfrac{rM}{1-\dfrac{3M}{r}}$
if $a>12$, we have :
$r=\dfrac{l^2}{2M}\left[1-\dfrac{a-12}{a} \right]=\dfrac{6l^2}{aM}=\dfrac{3l^2}{r} \to \dfrac{l^2}{r^2}=\dfrac{1}{3}\to e^2 = \left(1-\dfrac{2M}{r}\right)\left(1+\dfrac{1}{3}\right)=\left(1-\dfrac{2M}{r}\right)\dfrac{4}{3}$
$\dfrac{l^2}{e^2}=\dfrac{3l^2}{4}\left(1 - \dfrac{2M}{r}\right)^{-1}=\dfrac{3rM}{4}\left(1 - \dfrac{2M}{r}\right)^{-1}\left(1 - \dfrac{3M}{r}\right)^{-1}$
if $a<12$ ,we have :
$r=\dfrac{l^2}{2M}\left[1-\dfrac{12-a}{a} \right]=\dfrac{l^2}{2M}\dfrac{2a-12}{a}=\dfrac{a-6}{2r} \to \dfrac{l^2}{r^2}=\dfrac{2}{a-6}=\dfrac{M}{r-3M} \to$
$1+\dfrac{l^2}{r^2}=\dfrac{r-2M}{r-3M} \to e^2=\left(1 - \dfrac{2M}{r}\right)\dfrac{r-2M}{r-3M} \to $
$\dfrac{l^2}{e^2}=\dfrac{r-3M}{r-2M}*rM*\left(1 - \dfrac{2M}{r}\right)^{-1}\left(1 - \dfrac{3M}{r}\right)^{-1}=rM\left(1 - \dfrac{2M}{r}\right)^{-2}$
so that is the answer!
|
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|
What is the general equation of a cubic polynomial? I had this question:
"Find the cubic equation whose roots are the the squares of that of $x^3 + 2x + 1 = 0$"
and I kind of solved it. In that my answer was $x^3 - 4x^2 + 4x + 1$, but it was actually $x^3 + 4x^2 + 4x - 1 = 0$.
I took the general equation of a cubic equation, which was: $x^3 +bx^2/a + cx/a + d/a$.
Through simultaneous equations, I found what $b/a, c/a, d/a$ should equate to for my unknown cubic polynomial. Am I supposed to make $b/a, c/a, d/a$ all positive, then substitute it into the general formula?
Any help would be greatly appreciated, thanks.
|
Let $a$ be a root of $x^3+2x+1=0$ and $b=a^2$ be a root of the required equation
So, $a^3+2a+1=0\implies a\cdot b+2a+1=0\implies a=-\frac1{b+2}$
As $a$ be a root of $x^3+2x+1=0$, put this value of $a$ in $x^3+2x+1=0$
On simplification, I get $(b+2)^3-2(b+2)^2-1=0\iff b^3+4b^2+4b-1=0$
So, the required equation will be $y^3+4y^2+4y-1=0$
|
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|
Using the hypothesis $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$ to prove something else Assuming that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$$
Is it possible to use this fact to prove something like:
$$\frac{1}{a^{2013}}+\frac{1}{b^{2013}}+\frac{1}{c^{2013}}=\frac{1}{a^{2013}+b^{2013}+c^{2013}}$$
Just curious. Thanks for the help!
|
A simpler way to see that one of $a = -b$ or $b = -c$ or $c = -a$ is true.
wlog, we can assume $abc = 1$ (why?).
So we can assume $a,b,c$ are roots of $x^3 - px^2 + qx -1 = 0$.
Your relation gives us that $q = \frac{1}{p}$ (Vieta's formulas)
Thus we get $a,b,c$ are roots of
$$x^2(x - p) + \frac{1}{p}(x - p) = 0$$
Thus the roots are $$\pm\frac{1}{p}, p$$
|
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|
How can I find all the solutions of $\sin^5x+\cos^3x=1$
Find all the solutions of $$\sin^5x+\cos^3x=1$$
Trial:$x=0$ is a solution of this equation. How can I find other solutions (if any). Please help.
|
$\sin^5x\le \sin^2x, \cos^3x\le \cos^2x$, and $\sin^2x+\cos^2x=1$, so $\sin^5x+\cos^3x$
can be equal to 1 if and only if $$\sin^5x=\sin^2x$$ and $$\cos^3x=\cos^2x$$, i.e.
$$\sin^2x(1-\sin^3x)=0$$ and $$\cos^2x(1-\cos x)=0$$
The first equation has solution: $x=k\pi$ or $x=\pi/2+2m\pi, k,m $ integers.
The second equation has solution: $x=2n\pi$ or $x=\pi/2+p\pi, n,p$ integers.
Therefore the solution is either:
(i) $x=k\pi$ and $x=2n\pi$, or equivalently, $x=2n\pi$ for some integer $n$, or
(ii) $x=k\pi$ and $x=\pi/2+p\pi$ (impossible), or
(iii) $x=\pi/2+2m\pi$ and $x=2n\pi$ (impossible) or
(iv) $x=\pi/2+2m\pi$ and $x=\pi/2+p\pi$, or equivalently, $x=\pi/2+2m\pi$ for some integer $p$.
In conclusion, $x=2n\pi$ or $x=\pi/2+2n\pi$, $n$ integers, are the solutions.
|
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|
Finding the number of non-neg integer solutions? How would I find the number of non negative integer solutions to this problem?
$$x_1 + x_2 + x_3 + x_4 = 12$$ if $0 \leq x_1 \leq 2$.
|
$$x_1 + x_2 + x_3 + x_4 = 12, 0 \leq x_1 \leq 2$$
According to http://oeis.org/wiki/User:Adi_Dani_/Restricted_compositions_of_natural_numbers
there is derived formula
$${\binom{m}{k}}_{s}=\sum_{i=0}^{m}(-1)^{i}\binom{m}{i}\binom{m+k-si-1}{m-1}\,$$
for number of solutions of $$x_0+x_1 + x_2 +...+ x_{m-1} =k,0\leq x_i\leq s-1$$
in our case $k=12,s=3,m=4$
$${\binom{4}{12}}_{3}=\sum_{i=0}^{4}(-1)^{i}\binom{4}{i}\binom{15-3i}{3}=$$
$$=\binom{4}{0}\binom{15}{3}-\binom{4}{1}\binom{12}{3}+\binom{4}{2}\binom{9}{3}-\binom{4}{3}\binom{6}{3}+\binom{4}{4}\binom{3}{3}=$$
$$=\binom{15}{3}-4\binom{12}{3}+6\binom{9}{3}-4\binom{6}{3}+\binom{3}{3}$$
|
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|
Show that $8 \mid (a^2-b^2)$ for $a$ and $b$ both odd
If $a,b \in \mathbb{Z}$ and odd, show $8 \mid (a^2-b^2)$.
Let $a=2k+1$ and $b=2j+1$. I tried to get $8\mid (a^2-b^2)$ in to some equivalent form involving congruences and I started with
$$a^2\equiv b^2 \mod{8} \Rightarrow 4k^2+4k \equiv 4j^2+4j \mod{8}$$
$$\Rightarrow k^2+k-j^2-j=2m$$
for some $m \in \mathbb{Z}$ but I am not sure this is heading anywhere that I can tell.
Second attempt: Use Euler's Theorem and as $\gcd(a,8)=\gcd(b,8)=1$ and $\phi(8)=4$, $a^4 \equiv b^4 \equiv 1 \mod 8$ so $a^4-b^4\equiv 0 \mod{8}$.
I haven't gotten too much further are there any hints?
|
The given condition says that $a,b$ both are odd. So let, $a=2k+1$ and $b=2l+1$ where $k,l$ are some integers.
Now, $a^2-b^2=4k^2+4k+1-(4l^2+4l^2+1)=4(k^2-l^2+k-l)$
$$=4((k^2-k)(l^2-l))=4(k(k-1)+l(l-1)$$.
Notice that $k(k-1),l(l-1)$ are products of two consecutive integers and hence divisible by $2$. So, you can write them as $k(k-1)=2m, l(l-1)=2p$.
Therefore: $a^2-b^2=4(k(k-1)+l(l-1))=4(2m+2p)=8(m+p)$
|
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|
Show $60 \mid (a^4+59)$ if $\gcd(a,30)=1$
If $\gcd(a,30)=1$ then $60 \mid (a^4+59)$.
If $\gcd(a,30)=1$ then we would be trying to show $a^4\equiv 1 \mod{60}$ or $(a^2+1)(a+1)(a-1)\equiv 0 \mod{60}$. We know $a$ must be odd and so $(a+1)$ and $(a-1)$ are even so we at least have a factor of $4$ in $a^4-1$. Was thinking I could maybe try to show that there is also a factor of $3$ and $5$ necessarily giving that $a^4-1\equiv0 \mod{60}$.
Other things I was thinking was that as $Ord_n(a) \mid \phi(n)=\phi(60)=16$ that we just need to show that $Ord_n(a) \in \{1,2,4 \}$.
Any hints? I have the exam soon =/
|
Continue as you started: As $3 \nmid a$, we have $a \equiv \pm 1 \pmod 3$, giving
$$ a^4 - 1 \equiv 1 - 1 = 0 \pmod 3 $$
which means $3 \mid a^4 + 1$.
For $5$, we have either $a \equiv \pm 1 \pmod 5$, giving
$$ a^4 - 1 \equiv 1 - 1 \equiv 0 \pmod 5 $$
or $a \equiv \pm 2 \pmod 5$,
$$ a^4 - 1 \equiv 16 - 1 = 15 \equiv 0 \pmod 5$$
and we are done.
|
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|
Integration of volumes of revolution: bisector surface
$y = (16-x^2)^{0.5}$ is rotated around the $x$-axis to give a sphere of radius $4$ units. Find the equation of the straight line that passes through $(-4,0)$, such that when also rotated around the $x$-axis, will create a surface that bisects the volume of the sphere.
The equation will obviously take the form $y=kx+4k$. I tried to solve for the volume bounded by the revolutions of the semicircle and the straight line being equal to exactly half the volume of the sphere. The result was a very complicated series of substitutions and expansions which yielded no valid solution (it was a $6^{\text{th}}$ degree polynomial with only $1$ positive solution of $2.27$ for $k$, which is quite obviously wrong).
My question is whether there is a simpler, perhaps more ingenious way of finding $k$.
Again, thanks in advance and all contributions welcome!
|
The volume of your sphere is $\,\displaystyle{\frac43\pi 4^3=\frac{256\pi}{3}}\,$ . You need a line $\,y=kx+4k\,$ , which intersects $\,y=\pm\sqrt{16-x^2}\,$ at
$$16-x^2=k^2x^2+8k^2x+16k^2\implies(k^2+1)x^2+8k^2x+16(k^2-1)=0\implies$$
$$\Delta=64k^4-64(k^4-1)=8^2\implies$$
$$x_{1,2}=\frac{-8k^2\pm8}{2(k^2+1)}=\begin{cases}\;\;\;\;\;\;-4\\{}\\-\frac{4(k^2-1)}{k^2+1}\end{cases}$$
Thus, we want
$$\pi\int\limits_{-4}^{-\frac{4(k^2-1)}{k^2+1}}(kx+4k)^2x\,dx=\left.\frac\pi{3k}(kx+4k)^3\right|_{-4}^{-\frac{4(k^2-1)}{k^2+1}}=\frac\pi{3k}\frac{8^3k^3}{(k^2+1)^3}=\frac{512k^2\pi}{3(k^2+1)^3}$$
And now we want
$$\frac{2^9k^2\pi}{3(k^2+1)^3}=\frac{2^7\pi}{3}\iff4k^2=(k^2+1)^3$$
and according to WA the resulting sixtic equation doesn't even have any real roots...
|
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|
Determine the general solution for $2\cos 2x−5\cos x+2=0$ Determine the general solution for $2\cos 2x−5\cos x +2=0$
my answer I got was : $1.05+n\pi, 4.19+n\pi$
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First, notice that $\cos(2x) \equiv \cos^2x-\sin^2x$ and so
$$2\cos(2x)-5\cos x + 2 \equiv 2\cos^2x - 2\sin^2x - 5\cos x + 2$$
Then we know that $\sin^2x = 1-\cos^2x$, meaning that
$$2\cos(2x)-5\cos x + 2 \equiv 4\cos^2x - 5\cos x$$
We can factorise to give $(4\cos x - 5)\cos x$. The solutions are then just $\cos x = 0$ since $\cos x = \tfrac{5}{4}$ has no real solutions. Hence $x \in \{\tfrac{\pi}{2}+\pi n : n \in \mathbb{Z}\}.$
|
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|
Help finding eigenvectors? The given matrix is:
$$
\begin{pmatrix}3 & 1 & 6 \\ 2 & 1 & 0 \\ -1 & 0 & -3\end{pmatrix}\qquad
$$
I got the characteristic polynomial of $$x^3 - x^2 - 5x - 3 = 0$$
which factors down to $$(x+1)^2 * (x-3) = 0$$
I see that it has eigenvalues of -1 and 3.
I know I'm almost there, I plugged in the eigenvalues to $A-\lambda I$ but completely forgot how to find the eigenvectors after this.
When $\lambda$ = 3, I got:
$$\begin{pmatrix}0 & 1 & 6 \\ 2 & -2 & 0 \\ -1 & 0 & 0\end{pmatrix} \begin{pmatrix}x
_1 \\ x_2 \\ x_3\end{pmatrix}=0\qquad$$
and when $\lambda$ = -1, I got:
$$\begin{pmatrix}4 & 1 & 6 \\ 2 & 0 & 0 \\ -1 & 0 & 4\end{pmatrix}\begin{pmatrix}x
_1 \\ x_2 \\ x_3\end{pmatrix}=0\qquad$$
Where do I go from here? Row-reduce the $3x3$ matrices to solve?
|
it is enough you find:
$$\left(\begin{bmatrix}3 & 1 & 6 \\ 2 & 1 & 0 \\ -1 & 0 & -3\end{bmatrix}- 3I\right)\begin{bmatrix}x
_1 \\ x_2 \\ x_3\end{bmatrix}=0\qquad
\text{and} \qquad
\left(\begin{bmatrix}3 & 1 & 6 \\ 2 & 1 & 0 \\ -1 & 0 & -3\end{bmatrix}+ I\right)\begin{bmatrix}x
_1 \\ x_2 \\ x_3\end{bmatrix}=0 \ .$$
|
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|
Compute $\int_{0}^{1}\left[\frac{2}{x}\right]-2\left[\frac{1}{x}\right]dx$ The question is to find
$$\int_{0}^{1}\left(\left[\dfrac{2}{x}\right]-2\left[\dfrac{1}{x}\right]\right)dx,$$
where $[x]$ is the largest integer no greater than $x$, such as $[2.1]=2, \;[2.7]=2,\; [-0.1]=-1.$
Is there any nice method to solve this integral,Thank you everyone.
Some of my thoughts: I have if $a\in(0,1]$, then
$$\int_{0}^{1}\left[\dfrac{a}{x}\right]-a\left[\dfrac{1}{x}\right]dx=a\ln{a}.$$
What about $a>1$,
$$\int_{0}^{1}\left[\dfrac{a}{x}\right]-a\left[\dfrac{1}{x}\right]dx=?$$
and $$\int_{0}^{1}\left(\left[\dfrac{2}{x}\right]-2\left[\dfrac{1}{x}\right]\right)^{2}dx=?$$
becasue I have
$$\int_{0}^{1}\left(\dfrac{1}{x}-\left[\dfrac{1}{x}\right]\right)^2=-1-\gamma+\ln{(2\pi)}$$
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You just have to figure out where the integrand is nonzero. In this case, the integrand really takes the value
$$\left\lfloor \frac{2}{x} \right\rfloor - 2\left \lfloor \frac{1}{x} \right\rfloor = \begin{cases} \\0 & x \in \left [ \frac{2}{2 n+1},\frac{2}{2 n}\right )\\ 1 & x \in \left [ \frac{2}{2 n},\frac{2}{2 n-1}\right ) \end{cases}$$
for $n \ge 1$ and $x \in [0,1]$. Here is a plot generated in Mathematica:
Thus, the integral is just a sum of the interval lengths over which the integrand has the value of $1$:
$$\begin{align}\int_0^1 dx \, \left ( \left\lfloor \frac{2}{x} \right\rfloor - 2\left \lfloor \frac{1}{x} \right\rfloor \right ) &= \frac{2}{3} - \frac{2}{4} + \frac{2}{5} - \frac{2}{6} + \ldots \\ &= 2 \left [ \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} - \frac12 \right ]\\ &= 2 \log{2} - 1\end{align}$$
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|
Determine if vectors are linearly independent Determine if the following set of vectors is linearly independent:
$$\left[\begin{array}{r}2\\2\\0\end{array}\right],\left[\begin{array}{r}1\\-1\\1\end{array}\right],\left[\begin{array}{r}4\\2\\-2\end{array}\right]$$
I've done the following system of equations, and I think I did it right... It's been such a long time since I did this sort of thing...
Assume the following:
\begin{equation*}
a\left[\begin{array}{r}2\\2\\0\end{array}\right]+b\left[\begin{array}{r}1\\-1\\1\end{array}\right]+c\left[\begin{array}{r}4\\2\\-2\end{array}\right]=\left[\begin{array}{r}0\\0\\0\end{array}\right]
\end{equation*}
Determine if $a=b=c=0$:
\begin{align}
2a+b+4c&=0&&(1)\\
2a-b+2c&=0&&(2)\\
b-2c&=0&&(3)
\end{align}
Subtract $(2)$ from $(1)$:
\begin{align}
b+c&=0&&(4)\\
b-2c&=0&&(5)
\end{align}
Substitute $(5)$ into $(4)$, we get $c=0$.
So now what do I do with this fact? I'm tempted to say that only $c=0$, and $a$ and $b$ can be something else, but I don't trust that my intuition is right.
|
You just stopped too early:
Since you have 3 varibles with 3 equations, you can simply obtain $a,b,c$ by substituting $c = 0$ back into the two equations:
*
*From equation $(3)$, $c = 0 \implies b = 0$.
*With $b = 0, c = 0$ substituted into equation $(1)$ or $(2)$, $b = c = 0 \implies a = 0$.
So in the end, since
$$\begin{equation*}
a\left[\begin{array}{r}2\\2\\0\end{array}\right]+b\left[\begin{array}{r}1\\-1\\1\end{array}\right]+c\left[\begin{array}{r}4\\2\\-2\end{array}\right]=\left[\begin{array}{r}0\\0\\0\end{array}\right] \end{equation*}\implies a = b = c = 0, $$
the vectors are linearly independent, based on the definition(shown below).
The list of vectors is said to be linearly independent if the only $c_1,...,c_n$ solving the equation $0=c_1v_1+...+c_nv_n$ are $c_1=c_2=...=c_n=0.$
You could have, similarly, constructed a $3\times 3$ matrix $M$ with the three given vectors as its columns, and computed the determinant of $M$. Why would this help? Because we know that if $\det M \neq 0$, the given vectors are linearly independent. (However, this method applies only when the number of vectors is equal to the dimension of the Euclidean space.)
$$M = \begin{bmatrix} 2 & 1 & 4 \\ 2 & -1 & 2 \\ 0 & 1 & -2 \end{bmatrix}$$
$$\det M = 12 \neq 0 \implies\;\text{linear independence of the columns}.$$
|
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|
Solve the following in non-negative integers: $3^x-y^3=1$. Solve the following in non-negative integers: $$3^x-y^3=1$$
Of course $(x,y)=(0,0)$ is a trivial solution. After seeing that I proceeded like this:
$$3^x-y^3=1$$$$\implies3^x-1=y^3$$$$\implies2(3^{x-1}+3^{x-2}+ \cdots +3^1+1)=y^3$$$$\therefore2|y$$
So let $y=2k$ for non-negative $k$.
Then $$3^{x-1}+3^{x-2}+ \cdots +3^1+1=4y^3$$ So $4|LHS$ using a mod argument one can easily deduce that this implies that $x$ is even so let $x=2q$.
Then going back to our original expression we have $$9^q-1=8k^3$$$$\implies 9^{q-1}+9^{q-2}+ \cdots +9+1=k^3$$
Then again using mod arguments I deduced that $k \equiv 4 \pmod9$. And after that things started to get more and more yucky. This could be a completely bad approach so I also triad looking at $(y+1)(y^2-y+1)=3^x$ but that didn't lead me anywhere either.
Thanks in advance for any help.
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HINT:
If $d$ divides $y+1,y^2-y+1$ it will divide $y^2-y+1-y(y+1)=1-2y$
As $d$ divides $y+1,1-2y$ it will divide $2(y+1)+1-2y=3$
If $d=1,$ either $y+1=1$ or $y^2-y+1=1$ or both $=1$
If $d=3,$ either $y+1=3$ or $y^2-y+1=3$ or both $=3$
|
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|
Evaluate $\int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx$ and $\int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx$ Background: Evaluation of $\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^2}dx$
We can prove using the Beta-Function identity that
$$\int_0^\infty \frac{1}{(1+x^2)^\lambda}dx=\sqrt{\pi}\frac{\Gamma \left(\lambda-\frac{1}{2} \right)}{\Gamma(\lambda)} \quad \lambda>\frac{1}{2}$$
Differentiating the above equation with respect to $\lambda$, we obtain an expression involving the Digamma Function $\psi_0(z)$.
$$\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^\lambda}dx = \sqrt{\pi}\frac{\Gamma \left(\lambda-\frac{1}{2} \right)}{\Gamma(\lambda)} \left(\psi_0(\lambda)-\psi_0 \left( \lambda-\frac{1}{2}\right) \right)$$
Putting $\lambda=2$, we get
$$\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^2}dx = -\frac{\pi}{4}+\frac{\pi}{2}\log(2)$$
Question:
But, does anybody know how to evaluate $\displaystyle \int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx$ and $\displaystyle \int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx$?
Mathematica gives the values
*
*$\displaystyle \int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx = -\frac{G}{6}+\pi \left(-\frac{3}{8}+\frac{1}{8}\log(2)+\frac{1}{3}\log \left(2+\sqrt{3} \right) \right)$
*$\displaystyle \int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx = -\frac{\pi}{2}+\frac{\pi \log \left( 6+4\sqrt{2}\right)}{4}$
Here, $G$ denotes the Catalan's Constant.
Initially, my approach was to find closed forms for
$$\int_0^\infty \frac{1}{(1+x^2)^2(1+x^3)^\lambda}dx \ \ , \int_0^\infty \frac{1}{(1+x^2)^2(1+x^4)^\lambda}dx$$
and then differentiate them with respect to $\lambda$ but it didn't prove to be of any help.
Please help me prove these two results.
|
We can attack the other integral
$$I = \int_0^{\infty} dx \frac{\log{(1+x^3)}}{(1+x^2)^2}$$
in a manner similar to what @O.L. outlined in his answer for the other case, but with a different contour. To wit, consider
$$\oint_C dz \frac{\log{(1+z^3)} \log{z}}{(1+z^2)^2}$$
where $C$ is the following contour
This is a keyhole contour about the positive real axis, but with additional keyholes about the branch points at $z=e^{i \pi/3}$, $z=-1$, and $z=e^{i 5 \pi/3}$. There are poles of order $2$ at $z=\pm i$.
I will outline the procedure for evaluation. The integral about the circular arcs, large and small, go to zero as the radii go to $\infty$ and $0$, respectively. Each of the branch points introduces a jump of $i 2 \pi$ due to the logarithm in the integrand. By the residue theorem, we have
$$-i 2 \pi \int_0^{\infty} dx \frac{\log{(1+x^3)}}{(1+x^2)^2} - i 2 \pi \int_{e^{i \pi/3}}^{\infty e^{i \pi/3}} dt \frac{\log{t}}{(1+t^2)^2} \\ - i 2 \pi \int_{e^{i \pi}}^{\infty e^{i \pi}} dt \frac{\log{t}}{(1+t^2)^2} - i 2 \pi \int_{e^{i 5 \pi/3}}^{\infty e^{i 5 \pi/3}} dt \frac{\log{t}}{(1+t^2)^2} = \\ i 2 \pi \sum_{\pm} \frac{d}{dz} \left[\frac{\log{(1+z^3)} \log{z}}{(z\pm i)^2} \right]_{z=\pm i} $$
Without going into too much detail, I will illustrate how the integrals are done by evaluating one of them. Consider
$$\int_{e^{i \pi}}^{\infty e^{i \pi}} dt \frac{\log{t}}{(1+t^2)^2} = -\int_1^{\infty} dy \frac{\log{y}+i \pi}{(1+y^2)^2}$$
Now,
$$\int_1^{\infty} \frac{dy}{(1+y^2)^2} = \int_{\pi/4}^{\pi/2} d\theta \cos^2{\theta} = \frac{\pi}{8}-\frac14$$
$$\begin{align}\int_1^{\infty} dy\frac{\log{y}}{(1+y^2)^2} &= -\int_0^1 du \frac{u^2 \log{u}}{(1+u^2)^2}\\ &= -\sum_{k=0}^{\infty} (-1)^k (k+1) \int_0^1 u^{2 k+2} \log{u} \\ &= \sum_{k=0}^{\infty} (-1)^k \frac{k+1}{(2 k+3)^2} \\ &= \frac{G}{2} - \frac{\pi}{8}\end{align}$$
so that
$$\int_{e^{i \pi}}^{\infty e^{i \pi}} dt \frac{\log{t}}{(1+t^2)^2} = - \left ( \frac{G}{2} - \frac{\pi}{8} \right ) - i \pi \left ( \frac{\pi}{8}-\frac14\right ) $$
Along similar lines,
$$\int_{e^{i \pi/3}}^{\infty e^{i \pi/3}} dt \frac{\log{t}}{(1+t^2)^2} = \frac{G}{3}-\frac{\pi }{8}+\frac{1}{12} \pi \log \left(2+\sqrt{3}\right)+i
\left(\frac{1}{4} \log \left(2+\sqrt{3}\right)-\frac{\pi }{6}\right)$$
$$\int_{e^{i 5 \pi/3}}^{\infty e^{i 5 \pi/3}} dt \frac{\log{t}}{(1+t^2)^2} = \frac{G}{3}-\frac{\pi }{8}-\frac{5}{12} \pi \log \left(2+\sqrt{3}\right)+i
\left(-\frac{5 \pi }{6}+\frac{\pi ^2}{4}-\frac{1}{4} \log
\left(2+\sqrt{3}\right)\right)$$
Combining the integrals, I get
$$\frac{G}{6} -\frac{\pi}{8}-\frac{\pi}{3} \log{(2+\sqrt{3})} + i \left [-\frac{3 \pi}{4} + \frac{\pi^2}{8}\right ] $$
The sum of the residues on the RHS is relatively simple to evaluate; I get
$$\sum_{\pm} \frac{d}{dz} \left[\frac{\log{(1+z^3)} \log{z}}{(z\pm i)^2} \right]_{z=\pm i} = \frac{\pi}{2}-\frac{\pi}{8}\log (2)+i \left(\frac{3 \pi }{4}-\frac{\pi ^2}{8}\right)$$
The integral we seek is then the negative of the sum of the combined integrals and the sum of the residues, which gives us
$$\int_0^{\infty} dx \frac{\log{(1+x^3)}}{(1+x^2)^2} = -\frac{G}{6} - \frac{3\pi}{8} + \frac{\pi}{8} \log{2} + \frac{\pi}{3} \log{(2+\sqrt{3})} \approx 0.320555$$
which agrees with Mathematica. Note how the imaginary parts fortuitously canceled.
It should be understood that the above technique may be applied to the other integral. As O.L. has demonstrated, however, one may exploit symmetry and use a less computationally demanding technique for that particular case.
|
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|
Limit $\lim\limits_{n\to\infty}\left(1+\frac{1}{n}+a_n\right)^n=e$ if $\lim\limits_{n\to\infty}na_n=0$
Let $\{a_n\}$ be any sequence of real numbers such that $\lim_{n\rightarrow\infty}na_n=0$. Prove that $$\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}+a_n\right)^n=e$$
I thought about using binomial theorem. So $\left(1+\dfrac{1}{n}+a_n\right)^n = \left(1+\dfrac{1}{n}\right)^n + na_n\left(1+\dfrac{1}{n}\right)^{n-1} + \dbinom{n}{2}a_n^2\left(1+\dfrac{1}{n}\right)^{n-2}+\ldots$.
The first term has limit $e$, the second term has limit $0$ (because $na_n$ has limit $0$, and $\left(1+\dfrac{1}{n}\right)^{n-1} = \left(1+\dfrac{1}{n}\right)^{n}\left(1+\dfrac{1}{n}\right)^{-1}$ has limit $e$.) But for the other terms, it seems hard to find the limit.
|
Here is a technique
$$\left(1+\frac{1}{n}+a_n\right)^n= e^{n\ln\left(1+\frac{1}{n}+a_n\right)}= e^{n\left ( (\frac{1}{n}+a_n) -\frac{1}{2}(\frac{1}{n}+a_n)^2+\dots. \right) }=e^{\left( (1+na_n) -\frac{n}{2}(\frac{1}{n}+a_n)^2+\dots.\right)} $$
$$ = e^{\left( (1+na_n) -\frac{n}{2}\frac{(1+na_n)^2}{n^2}+\dots.\right)} .$$
Now, take the limit as $n\to \infty$ gives the limit $e$.
Note: We used the Taylor series of $\ln(1+x)$
$$ \ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\dots. $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to simplify $\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}$? $$\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}$$
I have been staring at it for ages and know that it simplifies to $x$, but have been unable to make any significant progress.
I have tried doing $(\frac{1-x}{1-2x})(\frac{1+2x}{1+2x})$ but that doesn't help as it leaves $\frac{1+x-2x^2}{1-4x^2}$ any ideas?
|
Notice that $$\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}} = \frac{\frac{-x}{1-2x}}{\frac{-1}{1-2x}}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\{x_1=1,\,x_{n+1}=\frac {x_n}2+\frac 1{x_n} \}$ converges when $n \to \infty$ I want to prove that the sequence defined by $\{x_1=1,\,x_{n+1}=\frac {x_n}2+\frac 1{x_n} \}$ has a limit.
By evaluating the sequence I notice that the sequence is strictly monotonically decreasing starting from $x_2=1.5$.
It seems to suggest itself to prove that the sequenced is bounded by $1\le\big( x_n \big)_{n\ge1} \le 1.5$ and to prove that it is strictly monotonically decreasing starting at $x_2$ which would imply convergence.
How would I proceed and could one prove the existence of the limit without first evaluating the values of the sequence to see how the sequence behaves?
|
$
\text {Assumption: } \big(x_n\big)_{n\ge2} \text { is strictly monotonically decreasing} \tag{1} \\
$
$
\text {Assumption: } \sqrt2 \le \big(x_n\big)_{n\ge2} \tag{2}
$
Strictly monotonic & bounded $\Longrightarrow$ convergent.
$$\begin{align*}
\text {ad (1):} && x_{n+1} &< x_n \\
&& \frac{x_n}2+\frac 1{x_n} &< x_n \\
&& \frac{x_n^2}2+1 &< x_n^2 \\
&& 1 &< \frac{x_n^2}2 \\
&& 2 &< x_n^2 \\
&& \sqrt2 &< x_n
\end{align*}
$$
If we can prove $\sqrt2<x_n$, then both $(1)$ and $(2)$ hold and the convergence is proven.
$$\begin{align*}
\text {Induction by } n \\
n=2\text : & \sqrt2 < 1.5 = x_2 \\
n\to n+1\text : & \text {Assume } \sqrt2 < x_n \text{ holds.} \\
& \sqrt2 < x_{n+1} \\
& \sqrt2 < \frac {x_n}2+\frac1{x_n}=\frac{x_n^2+2}{2x_n} \\
& \text{By assumption } x_n=\sqrt2+\delta \text{ for a positive } \delta. \\
& \sqrt2 < \frac{\left( \sqrt2+\delta \right)^2+2}{2\left( \sqrt2+\delta \right)} \\
& 4+2\sqrt2\delta < \left( \sqrt2+\delta \right)^2+2 \\
& 2+2\sqrt2\delta<2+2\sqrt2\delta+\delta^2 \\
& 0 < \delta^2 & \square \\
\end{align*}
$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Local maxima of Legendre polynomials When I plotted the (normalized) Legendre polynoials, I couldn't help noticing that all the local maxima lay on a really nice curve:
What is the equation of the curve (and how can we arrive to that equation)?
|
I claim that that curve is
$$y=\pm \frac{\sqrt{2/\pi}}{ \sqrt[4]{1-x^2}}.$$
This argument will not be rigorous, and will cite a source I haven't fully understood.
Take a look at Whittaker and Watson, A course in Modern Analysis, p. 316. They write:
$$P_n(\cos \theta) = \frac{4}{\pi} \frac{2 \cdot 4 \cdots (2n)}{3\cdot 5 \cdots (2n+1)} \left( \frac{\cos[(n+1/2) \theta - \pi/4]}{(2 \sin \theta)^{1/2}} + \frac{\cos[(n+3/2) \theta - 3\pi/4]}{2(2n+3) (2 \sin \theta)^{3/2}}+ \cdots \right)$$
... Shew also that the first few terms of this series give an approximate value of $P_n(\cos \theta)$ for all values of $\theta$ between $0$ and $\pi$ which are no nearly equal to either $0$ or $\pi$.
This approximation seems to be very good in practice. In the picture below, the blue curve is $P_5(x) \sqrt{\sin(\cos^{-1} x)}$ and the red curve is $512/(63 \sqrt{11} \pi)*\cos((11/2) \cos^{-1}(x) - \pi/4)$. (These are the same normalizations user79365 is using in his post.)
I assume that, in more modern language, the authors are saying that this is an asymptotic series for $P_n(\cos \theta)$, valid on $(0, \pi)$. I have taken the liberties of applying some algebraic rearrangements, replacing their parameter $\phi$ by its definition and stopping the sum after two terms rather than the four they give. Also, they are using the normalization where $\int_{-1}^1 P_n^2 = 2/(2n+1)$, so you'd want to multiply by $\sqrt{(2n+1)/2}$.
For fixed $\theta$, that second term is bounded by $c/n$. The later terms (not displayed) die off even faster as $n$ grows. So
$$\tilde{P}_n(\cos \theta)\approx \frac{4}{\pi} \sqrt{\frac{2n+1}{2}} \frac{2 \cdot 4 \cdots (2n)}{3\cdot 5 \cdots (2n+1)} \frac{\cos[(n+1/2) \theta - \pi/4]}{(2 \sin \theta)^{1/2}}$$
as $n \to \infty$. The $\sim$ over the $P$ indicates that I am now using your normalization. As long as $\theta/\pi$ is irrational, that $\cos$ term will swing between $1$ and $-1$, coming arbitrarily close to both ends. So
$$\lim \sup_{n \to \infty} \tilde{P}_n(\cos \theta) = \lim_{n \to \infty} \frac{4}{\pi} \sqrt{\frac{2n+1}{2}} \frac{2 \cdot 4 \cdots (2n)}{3\cdot 5 \cdots (2n+1)} \frac{1}{(2 \sin \theta)^{1/2}}$$
Taking $1/\sqrt{\sin \theta} =1/\sqrt[4]{1-x^2}$ out of everything, we need to compute
$$\lim_{n \to \infty} \frac{4}{\pi} \sqrt{\frac{2n+1}{2}} \frac{2 \cdot 4 \cdots (2n)}{3\cdot 5 \cdots (2n+1)} \frac{1}{\sqrt{2}}$$
This limit is similar to Wallis's product, and I think the second and third proofs on the Wikipedia page should be adaptable to evaluate it. I used the third proof, by Stirling's approximation:
$$ \sqrt{2n+1} \frac{2 \cdot 4 \cdots (2n)}{3\cdot 5 \cdots (2n+1)} = \sqrt{2n+1} \frac{2^{2n} (n!)^2}{(2n+1)!} \approx \sqrt{2n+1} \frac{2^{2n} (n/e)^{2n} (2 \pi n)}{((2n+1)/e)^{2n+1} \sqrt{2 \pi (2n+1)}}$$
$$=\frac{2 \pi n \sqrt{2n+1}}{(1+1/2n)^{2n} \cdot (2n+1)/e \cdot \sqrt{2 \pi (2n+1)}}=\frac{e \sqrt{2 \pi}}{(2+1/n) (1+1/2n)^{2n}} \approx \frac{\sqrt{2 \pi}}{2}.$$
Putting back in the other constants gave the result I state above.
Remark It's fun to note that
$$\int_{-1}^1 \left(\frac{\sqrt{2/\pi}}{\sqrt[4]{1-x^2}} \right)^2 dx = 2.$$
So $P_n^2$ is, on average, half the envelope above it.
Thanks to user79365 for providing the following image: The Legendre polynomials are in blue and the above curve is in red. It's also interesting to note the bunching up at values like $x=\pm 1/2$ and $x = 0$, when $\cos^{-1}(x)$ is a rational multiple of $\pi$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Determining a basis for solution set
For a given equation system (over field $R$), determine the basis for
the space $\operatorname{Sol}(U)$ (where $\operatorname{Sol}$ is a set of solutions):
\begin{cases} 2x + 7y + 3z+t=0 \\ 3x + 5y + 2z + 2t=0 \\ 9x + 4y + z+7t=0 \end{cases}
It's a new type of problem for me and I'm struggling a bit. I know what a basis is but am quite confused about the $\operatorname{Sol}(U)$ part. Does that mean I should just try to solve the system and then think of such a basis that it spans over the possible solutions?
|
Put the system into an augmented matrix:
$$\left[\begin{array}{rrrr|r}
2 & 7 & 3 & 1 & 0 \\
3 & 5 & 2 & 2 & 0 \\
9 & 4 & 1 & 7 & 0
\end{array}\right]$$
Use Gauss-Jordan elimination to get it into reduced row echelon form.
$$\left[\begin{array}{rrrr|r}
1 & 0 & -\frac{1}{11} & \frac{9}{11} & 0 \\
0 & 1 & \frac{5}{11} & -\frac{1}{11} & 0 \\
0 & 0 & 0 & 0 & 0
\end{array}\right]$$
Now you have a set of equations:
$$\begin{align}
x+0y+-\frac{1}{11}z+\frac{9}{11}t&=0 \\
0x+y+\frac{5}{11}z-\frac{1}{11}t &= 0
\end{align}$$
Finally, parameterize it with $u$ and $v$ as follows:
$$\begin{align}
x&=\space\space\space\frac{1}{11}u-\frac{9}{11}v \\
y&=-\frac{5}{11}u+\frac{1}{11}v \\
z&=u \\
t&=v \end{align}$$
The span of the solution set is therefore:
$$
u\begin{bmatrix}
\frac{1}{11} \\
-\frac{5}{11} \\
1 \\
0
\end{bmatrix}+
v\begin{bmatrix}
-\frac{9}{11} \\
\frac{1}{11} \\
0 \\
1
\end{bmatrix}
$$
|
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|
Find $x,y,z \in \mathbb Q$ such that $x + \frac 1y, y + \frac 1z, z+ \frac 1x \in \mathbb Z$
Find $x,y,z \in \mathbb Q$ such that: $$x + \frac 1y, y + \frac 1z, z+
\frac 1x \in \mathbb Z$$
Here is my thinking:
$$x + \frac 1y, y + \frac 1z, z+ \frac 1x \in \mathbb Z\\ \implies \left ( x + \frac 1y\right ) \left ( y + \frac 1z\right )\left ( z + \frac 1x\right )\in \mathbb Z \\ \iff \frac 1 {xyz} + xyz + \left( x+\frac 1y \right) + \left( y+\frac 1z \right) + \left( z+\frac 1x \right) \in \mathbb Z \\ \iff xyz + \frac 1 {xyz} \in \mathbb Z \\ \iff |xyz| = 1 \text{ (I proved it, easily)}$$
Case 1:
$xyz=1 \implies \exists a,b,c \in \mathbb Z$ such that: $x = \frac ab,y=\frac bc,z=\frac ca$
$$\implies \frac {a+c} b = x+\frac 1y \in \mathbb Z$$
and so on.
Now, I am stuck. Do you thing I am going a right way ?
|
It is most interesting to consider $x>0,y>0,z>0$, because there is a lots of solutions in other case.
A.
If $(x,y,z)$ is a solution, then one can construct other solutions:
$$\begin{array}{ccc}
(x,y,z) & (y,z,x) & (z,x,y) \\
(-x,-y,-z) & (-y,-z-,x) & (-z,-x,-y) \\
(\frac{1}{z},\frac{1}{y},\frac{1}{x}) & (\frac{1}{x},\frac{1}{z},\frac{1}{y}) & (\frac{1}{y},\frac{1}{x},\frac{1}{z}) \\
(-\frac{1}{z},-\frac{1}{y},-\frac{1}{x}) & (-\frac{1}{x},-\frac{1}{z},-\frac{1}{y}) & (-\frac{1}{y},-\frac{1}{x},-\frac{1}{z}). \\
\end{array}\tag{1}
$$
B.
For each pair $a,b\in \mathbb{N}$, such that $GCD(a,b)=1$, triple
$$
\Bigl(x = \frac{a}{b}, \quad y = -\frac{b}{a+b}, \quad z = -\frac{a+b}{a} \Bigr)
\tag{2}
$$
is a solution: $x+\frac{1}{y} = -1$, $y+\frac{1}{z} = -1$, $z+\frac{1}{x} = -1$.
Other derived triples (see A.) are solutions too.
C.
For each $n \in \mathbb{N}$ triple
$$
\Bigl(x = n, \quad y=-1, \quad z = -\frac{1}{n} \Bigr)
\tag{3}
$$
is a solution: $x+\frac{1}{y} = n-1$, $y+\frac{1}{z} = -n-1$, $z+\frac{1}{x} = 0$.
Other derived triples (see A.) are solutions too.
D. There are other (most interesting for me) solutions:
$\diamond \qquad (1,1,1)\quad$ (and derived modification $(-1,-1,-1)$);
$\diamond \qquad (1, \frac{1}{2}, 2)\quad$ (and modifications);
$\diamond \qquad (\frac{1}{2}, \frac{2}{3}, 3) \quad$ (and modfications).
They have $x,y,z$ all positive (or all negative).
Maybe this list has continuation (?).
|
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|
How can $\left({1\over1}-{1\over2}\right)+\left({1\over3}-{1\over4}\right)+\cdots+\left({1\over2n-1}-{1\over2n}\right)+\cdots$ equal $0$?
How can $\left({1\over1}-{1\over2}\right)+\left({1\over3}-{1\over4}\right)+\cdots+\left({1\over2n-1}-{1\over2n}\right)+\cdots$ equal $0$?
Let
$$\begin{align*}x &= \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} + \cdots\\
y &= \frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n-1} + \cdots\\
z &= \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots + \frac{1}{2n} + \cdots \end{align*}$$
so we have
$$x = y + z.$$
However, $x = 2\cdot z$, so $y$ = $z$ or
$$\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n-1} + \cdots = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots + \frac{1}{2n} + \cdots$$
This looks ok if I interpret it as
$$\frac{1}{1} = \left (\frac{1}{2} - \frac{1}{3} \right ) + \left (\frac{1}{4} - \frac{1}{5} \right ) + \left (\frac{1}{6} - \frac{1}{7} \right ) + \cdots + \left (\frac{1}{2n} - \frac{1}{2n+1} \right ) + \cdots$$
However, it's a bit weird if I write it as
$$\left (\frac{1}{1} - \frac{1}{2} \right ) + \left (\frac{1}{3} - \frac{1}{4} \right ) + \left (\frac{1}{5} - \frac{1}{6} \right ) + \cdots + \left (\frac{1}{2n-1} - \frac{1}{2n} \right ) + \cdots = 0.$$
How can a sum of positive numbers equal $0$?
|
Your proof is not right since $x=\infty$.
|
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|
Show that $\frac 1 2 <\frac{ab+bc+ca}{a^2+b^2+c^2} \le 1$
If $a,b,c$ are sides of a triangle, then show that $$\dfrac 1 2 <\dfrac{ab+bc+ca}{a^2+b^2+c^2} \le 1$$
Trial: $$(a-b)^2+(b-c)^2+(c-a)^2 \ge 0\\\implies a^2+b^2+c^2 \ge ab+bc+ca $$ But how I prove $\dfrac 1 2 <\dfrac{ab+bc+ca}{a^2+b^2+c^2}$ . Please help.
|
We know from triangle inequality
In a triangle the sum of any two sides is greater than the third side, from which it follows that
$(a+b-c)(a+c-b)+(b+c-a)(b+a-c)+(c+a-b)(c+b-a)>0$
Just expand this to get the inequality
$\Rightarrow a^2-b^2-c^2+2bc+b^2-c^2-a^2+2ac+c^2-a^2-b^2+2ab>0$
$\Rightarrow -a^2-b^2-c^2+2(ab+bc+ca)>0$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Given that $ x + \frac 1 x = r $ what is the value of: $ x^3 + \frac 1 {x^2}$ in terms of $r$? Given that $$ x + \cfrac 1 x = r $$
what is the value of: $$ x^3 + \cfrac 1 {x^2}$$ in terms of $r$?
NOTE: it is $\cfrac 1 {x^2}$ and not $ \cfrac 1 {x^3} $
Where I reached so far:
$$ \Big(x^3 + \cfrac 1 {x^2}\Big) + \cfrac 1 x \cdot\Big(x^3 + \cfrac 1 {x^2}\Big) = r^3 - r^2 -3r - 2 $$
Any hints??
|
If $g$ is a function such that $g(x)=f(x+\frac{1}{x})$ then $g(1/x)=g(x)$.
Your $g(x)=x^3+x^{-2}$ therefore cannot be written as $f(x+1/x)$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to derive compositions of trigonometric and inverse trigonometric functions? To prove:
$$\begin{align}
\sin({\arccos{x}})&=\sqrt{1-x^2}\\
\cos{\arcsin{x}}&=\sqrt{1-x^2}\\
\sin{\arctan{x}}&=\frac{x}{\sqrt{1+x^2}}\\
\cos{\arctan{x}}&=\frac{1}{\sqrt{1+x^2}}\\
\tan{\arcsin{x}}&=\frac{x}{\sqrt{1-x^2}}\\
\tan{\arccos{x}}&=\frac{\sqrt{1-x^2}}{x}\\
\cot{\arcsin{x}}&=\frac{\sqrt{1-x^2}}{x}\\
\cot{\arccos{x}}&=\frac{x}{\sqrt{1-x^2}}
\end{align}$$
|
Here is astart
$$ \sin({\arccos{x}})= \sqrt{1-\cos(\arccos(x))^2}=\sqrt{1-x^2}, $$
since $\cos(\arccos(x))=x.$ You need to use some trig. identities.
|
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|
CLT for random variables with varying distributions The following is not a homework problem. Here it is:
Suppose $\{X_k\}_{k=1}^{\infty}$ is a sequence of independent random variables such that
$$
P(X_k=1)=P(X_k=-1)=\frac{1}{2}-\frac{1}{2k^2}
$$
$$
P(X_k=k)=P(X_k=-k)=\frac{1}{2k^2}.
$$
Find the limiting distribution of $\frac{1}{\sqrt{n}} \sum\limits_{k=1}^n X_k$.
Now, I am guessing that the given probabilities are for $k\geq 2$, since $k=1$ would yield an inconsistent definition. Furthermore, I am assuming the case $k=1$ corresponds to only the second line.
The first sign of trouble comes from the fact that these random variables are not identically distributed (otherwise, we could just use CLT). One thing I noticed is that $\mathbb{E}(X_k)=0$ for each $k$. Also, it appears as though the limit of the sequence is a random variable $X$ with $P(X=1)=P(X=-1)=\frac{1}{2} \hspace{2mm}$ (not that this result is necessarily of much help).
I'd very much appreciate any suggestions on how to approach this problem. Thanks!
|
Suppose $\{X_k\}_{k=1}^{\infty}$ is a sequence of independent random
variables such that $$ P(X_k=1)=P(X_k=-1)=\frac{1}{2}-\frac{1}{2k^2}
$$
$$ P(X_k=k)=P(X_k=-k)=\frac{1}{2k^2}. $$ Find the limiting distribution of $\frac{1}{\sqrt{n}} \sum\limits_{k=1}^n X_k$.
The characteristic function of $X_k$ is
$$\phi_k(\xi)=(\frac{1}{2}-\frac{1}{2k^2})e^{i\xi}+(\frac{1}{2}-\frac{1}{2k^2})e^{-i\xi}+\frac{1}{2k^2}e^{ik\xi}+\frac{1}{2k^2}e^{-ik\xi}$$
$$=(1-\frac{1}{k^2})\cos \xi +\frac{1}{k^2}\cos k\xi = \cos \xi +\frac{1}{k^2}(\cos k\xi-\cos\xi)$$
Then the characteristic function of $X_k/\sqrt{n}$ is
$$Ee^{i\xi X_k/\sqrt{n}}=\cos \frac{\xi}{\sqrt{n}} +\frac{1}{k^2}(\cos \frac{k\xi}{\sqrt{n}}-\cos\frac{\xi}{\sqrt{n}})$$
and
$$Ee^{i\xi(\sum_{k=1}^n X_k)/\sqrt{n}}=\prod_{k=1}^n\left(\cos \frac{\xi}{\sqrt{n}} +\frac{1}{k^2}(\cos \frac{k\xi}{\sqrt{n}}-\cos\frac{\xi}{\sqrt{n}})\right)$$
$$=\exp\left[\log \sum_{k=1}^n\left(1-(1-\cos \frac{\xi}{\sqrt{n}} -\frac{1}{k^2}(\cos
\frac{k\xi}{\sqrt{n}}-\cos\frac{\xi}{\sqrt{n}}))\right)\right]$$
Then
$$\log \sum_{k=1}^n\left(1-(1-\cos \frac{\xi}{\sqrt{n}} -\frac{1}{k^2}(\cos
\frac{k\xi}{\sqrt{n}}-\cos\frac{\xi}{\sqrt{n}}))\right)$$
$$\sim -\sum_{k=1}^n(1-\cos \frac{\xi}{\sqrt{n}}-\frac{1}{k^2}(\cos \frac{k\xi}{\sqrt{n}}-\cos\frac{\xi}{\sqrt{n}}))$$
$$\sim -\sum_{k=1}^n(1-\cos \frac{\xi}{\sqrt{n}})+\sum_{k=1}^n\frac{1}{k^2}(\cos \frac{k\xi}{\sqrt{n}}-\cos\frac{\xi}{\sqrt{n}})=S_1+S_2$$
For $S_1$ we have
$$S_1=-n(1-\cos\frac{\xi}{\sqrt{n}})=-2n
\sin^2\frac{\xi}{\sqrt{n}}=-2n\frac{\sin^2\frac{\xi}{\sqrt{n}}}{\frac{\xi^2}{n}}\times\frac{\xi^2}{n}
\to -2\xi^2 , \ \ n \to \infty.$$
$$S_2=\sum_{k=1}^n\frac{1}{k^2}(\cos
\frac{k\xi}{\sqrt{n}}-\cos\frac{\xi}{\sqrt{n}})=-2\sum_{k=1}^n\frac{1}{k^2}\sin\frac{(k+1)\xi}{2\sqrt{n}}\sin\frac{(k-1)\xi}{2\sqrt{n}}$$
$$=-2\sum_{k=1}^n\frac{1}{k^2}\frac{(k^2-1)\xi^2}{n}\frac{\sin\frac{(k+1)\xi}{2\sqrt{n}}}{\frac{(k+1)\xi}{2\sqrt{n}}}\frac{\sin\frac{(k-1)\xi}{2\sqrt{n}}}{\frac{(k-1)\xi}{2\sqrt{n}} }$$
$$=-2\sum_{k=1}^n\frac{\xi^2}{n}\frac{\sin\frac{(k+1)\xi}{2\sqrt{n}}}{\frac{(k+1)\xi}{2\sqrt{n}}}\frac{\sin\frac{(k-1)\xi}{2\sqrt{n}}}{\frac{(k-1)\xi}{2\sqrt{n}} }+
2\sum_{k=1}^n\frac{1}{k^2}\frac{\xi^2}{n}\frac{\sin\frac{(k+1)\xi}{2\sqrt{n}}}{\frac{(k+1)\xi}{2\sqrt{n}}}\frac{\sin\frac{(k-1)\xi}{2\sqrt{n}}}{\frac{(k-1)\xi}{2\sqrt{n}} }$$
$$=S_{21}+S_{22}.$$
Now it is clear that $S_{21} \to -2\xi^2$ and $S_{22} \to 0$ as $n \to \infty$.
Therefore
$$Ee^{i\xi(\sum_{k=1}^n X_k)/\sqrt{n}} \to e^{-4\xi^2}, \ \ n \to \infty.$$
This is the characteristic function of r.v. $X=2Z \sim N(0,4)$.
|
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}
|
How does one [easily] calculate $\sum\limits_{n=1}^\infty\frac{\mathrm{pop}(n)}{n(n+1)}$? How does one [easily] calculate $\sum\limits_{n=1}^\infty\frac{\mathrm{pop}(n)}{n(n+1)}$, where $\mathrm{pop}(n)$ counts the number of bits '1' in the binary representation of $n$?
Is there any trick to calculate the sum? From what I already have, it definitely converges.
|
There is one more way to calculate this expression. Consider the following recurrence: $a_0 = 0, a_1 = 1, a_2 = 1$ and $$a_{2n} = a_n, \ a_{2n+1} = a_n + 1, \ n \ge 1.$$
It is easy to see that $a_n$ counts the number of ones in the binary expansion of $n$. We wish to calculate
\begin{align*}
\sum_{n = 1}^{\infty} \frac{a_n}{n(n+1)} &= \sum_{n=1}^{\infty} \frac{a_{2n}}{2n(2n+1)} + \sum_{n = 0}^{\infty} \frac{a_{2n+1}}{(2n+1)(2n+2)} \\
&= \sum_{n=1}^{\infty} \frac{a_{n}}{2n(2n+1)} + \sum_{n = 0}^{\infty} \frac{a_{n} + 1}{(2n+1)(2n+2)} \\
&= \frac{1}2 + \frac{1}2 \sum_{n = 1}^{\infty} \frac{a_n}{n(2n+1)} + \frac{1}2 \sum_{n=1}^{\infty} \frac{a_n}{(n+1)(2n+1)} + \frac{1}2 \sum_{n=1}^{\infty} \frac{1}{(n+1)(2n+1)} \\
&= \frac{1}2 + \frac{1}2 \sum_{n=1}^{\infty} a_n \left( \frac{1}{n(2n+1)} + \frac{1}{(n+1)(2n+1)} \right) + \frac{1}2 \sum_{n=1}^{\infty} \frac{1}{(n+1)(2n+1)} \\
&= \frac{1}2+ \frac{1}2 \sum_{n=1}^{\infty} \frac{a_n}{n(n+1)} + \frac{1}2\sum_{n=1}^{\infty} \frac{1}{(n+1)(2n+1)}.
\end{align*}
Thus,
$$\sum_{n=1}^{\infty} \frac{a_n}{n(n+1)} = 1 + \sum_{n=1}^{\infty} \frac{1}{(n+1)(2n+1)}.$$
Now
\begin{align*}
\sum_{n=1}^{\infty} \frac{1}{(n+1)(2n+1)} &= \sum_{n=1}^{\infty} \int_0^1 \frac{x^{2n}}{n+1} \ dx \\
&= \int_0^1 \sum_{n=1}^{\infty} \frac{x^{2n}}{n+1} \ dx \\
&= \int_0^1 -\frac{ \log(1-x^2)}{x^2} \ dx -1\\
&= \log(4) - 1.
\end{align*}
Therefore, our desired value is
$$1 + \log(4) - 1 = 2 \log(2).$$
|
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|
Non-homogenous recurrence relation I need to solve the recurrence relation $A(n)=2A(n-2)+ 2^{n-2}$. I tried writing out equations up to the $A(2)$ and multiplying by powers of two and adding all the equations together then all the terms cancelled but after that I couldn't find the sum of the powers of two. I forgot to mention that the base is $A(2)=2$ and n is even
|
It never hurts in such problems to start by gathering some numerical data:
$$\begin{array}{rcc}
n:&2&4&6&8&10&12\\
A(n):&2&8&32&128&512&2048\\
A(n):&2^1&2^3&2^5&2^7&2^9&2^{11}
\end{array}$$
There’s a very obvious pattern here, that leads to the conjecture that $A(n)=2^{\text{what function of }n?}$ if $n$ is even. Complete the conjecture correctly, and you should have little difficulty using mathematical induction to prove that it’s correct.
Alternatively, let $B(n)=A(2n)$; then $$B(n)=2B(n-1)+2^{2(n-1)}=2B(n-1)+4^{n-1}\;,$$
and ‘unwrap’ the recurrence:
$$\begin{align*}
B(n)&=2B(n-1)+4^{n-1}\\
&=2\Big(2B(n-2)+4^{n-2}\Big)+4^{n-1}\\
&=2^2B(n-2)+2\cdot 4^{n-2}+4^{n-1}\\
&=2^2\Big(B(n-3)+4^{n-3}\Big)+2\cdot 4^{n-2}+4^{n-1}\\
&=2^3B(n-3)+2^2\cdot4^{n-3}+2\cdot 4^{n-2}+4^{n-1}\\
&\;\vdots\\
&=2^kB(n-k)+2^{k-1}\cdot 4^{n-k}+2^{k-2}\cdot 4^{n-k+1}+\ldots+2\cdot 4^{n-2}+4^{n-1}\\
&\;\vdots\\
&=2^{n-1}B(1)+\sum_{i=1}^{n-1}2^{i-1}\cdot4^{n-i}\\
&=2^n+\sum_{i=1}^{n-1}2^{i-1}\cdot4^{n-i}\\
&=2^n+\sum_{i=1}^{n-1}2^{i-1}\cdot2^{2n-2i}\\
&=2^n+\sum_{i=1}^{n-1}2^{2n-i-1}\\
&=2^n+\sum_{i=n}^{2n-2}2^i\\
&=2^n+2^n\sum_{i=0}^{n-2}2^i\\
&=2^n\left(1+\sum_{i=0}^{n-2}2^i\right)\\
&=\;?
\end{align*}$$
|
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|
Trig substitution $\int x^3 \sqrt{1-x^2} dx$ $$\int x^3 \sqrt{1-x^2} dx$$
$x = \sin \theta $
$dx = \cos \theta d \theta$
$$\int \sin^3 \theta d \theta$$
$$\int (1 - \cos^2 \theta) \sin \theta d \theta$$
$u = \cos \theta$
$du = -\sin\theta d \theta$
$$-\int u^2 du$$
$$\frac{-u^3}{3} $$
$$\frac{\cos^3 \theta}{3}$$
With the triangle trick I get:
$$\frac{-\sqrt{1-x^2}^3}{3}$$
This is wrong but I am not sure where I went wrong.
|
You can avoid all the trig by making a much simpler substitution:$$u^2=1-x^2$$so: $$x^2=1-u^2$$ $$ u= \sqrt{1-x^2}$$ $$2u du=-2xdx$$
Rewriting the integral, factoring out one $x$:$$\int x^3 \sqrt{1-x^2} dx=\int x^2 \sqrt{1-x^2} xdx=-\int (1-u^2) u^2 du$$Multiply out the integrand, integrate with the power formula term by term and substitute back for $x$...
|
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|
Find the derivative of $y = f(x^2 - 2x + 7)$ where $f'(10) = 2$ Determine the derivative if $y = f(x^2 - 2x + 7)$ and $f'(10) = 2$
Ok so honestly, I dont know how to solve this, or even know where to start. All i know is that we are given a point $(10, 2)$. But what is $f(x^2 - 2x +7)$ supposed to mean?
|
Suppose $y=f(z)$ = well $f$ can be anything. Then we put $z=x^2-2x+7$ and get a new function $g(x)=f(x^2-2x+7)$. Some examples:
$$f(z)=z; f(x^2-2x+7)=x^2-2x+7=g(x)$$
$$f(z)=z^2+3; f(x^2-2x+7)=(x^2-2x+7)^2+3=g(x)$$
$$f(z)=\sin (z); f(x^2-2x+7)=\sin (x^2-2x+7)=g(x)$$
Now we want to see what we know about the derivative of $f(x^2-2x+7)$ when it could take any one of a myriad of forms. We use the chain rule - the derivative of $p(q(x))$ is $q'(x)p'(q(x))$ with $p=f$ and $q(x)=x^2-2x+7$ to obtain the derivative $$(2x-2)f'(x^2-2x+7)$$
Now we only know $f'(10)$, so we can only deal with $$x^2-2x+7=10$$ which reduces to $$(x-3)(x+1)=0$$
So $x=3$, or $x=-1$
With $x=3$ we get $4\times 2 =8$.
With $x=-1$ we get $-4 \times 2 = -8$.
Because the form of $f$ is undefined, we can't say more than this - even that our function is differentiable in other places.
|
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|
Find all real numbers such that $\sqrt{x-\frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$ Find all real numbers such that
$$\sqrt{x-\frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$$
My attempt to the solution :
I tried to square both sides and tried to remove the root but the equation became of 6th degree.Is there an easier method to solve this?
|
Hints:
$$x=\sqrt{x-\frac1x}+\sqrt{1-\frac1x}\implies x\sqrt x=\sqrt{x^2-1}+\sqrt{x-1}\implies$$
$$ x^3=x^2+x-2+2\sqrt{x^3-x^2-x+1}\implies(x^3-x^2-x+2)^2=4(x^3-x^2-x+1)$$
Well, now you can put $\,w:=x^3-x^2-x+1\;$ and solve
$$(w+1)^2=4w\iff (w-1)^2=0\iff w=1\;\ldots\;\text{and etc.}\ldots$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Express $f(x) = x^2 \cos(2x) + \sin^2(x)$ as a power series
Express $f(x) = x^2 \cos(2x) + \sin^2(x)$ as a power series
What I know:
I know that $$x^2\cos(2x) = x^2 \cdot \sum_{n=0}^{\infty} {(-1)^n \cdot \frac{(2x)^{2n}}{(2n)!}} = \sum_{n=0}^{\infty} {(-1)^n \cdot \frac{2^{2n} \cdot x^{2n+2}}{(2n)!}}$$
But how do we find what $\sin^2(x)$ is in terms of a power series?
|
HINT
$$ \sin^2 {x} = 1 - \cos^2 {x} = \dfrac{1 - \cos(2x)}{2}$$
|
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|
How prove this $\frac{1}{\cos{A}}-\frac{\sin{\frac{A}{2}}}{\sin{\frac{B}{2}}\sin{\frac{C}{2}}}=4$ in $\Delta ABC $ not An equilateral triangle, let
$$\begin{vmatrix}
2\sin{A}\sin{C}-\cos{B}&2\sin{A}\sin{B}-\cos{C}\\
4\sin{\dfrac{A}{2}}\sin{\dfrac{B}{2}}\sin{\dfrac{C}{2}}+\cos{B}&4\sin{\dfrac{A}{2}}\sin{\dfrac{B}{2}}\sin{\dfrac{C}{2}}+\cos{C}
\end{vmatrix}=0$$
prove that
$$\dfrac{1}{\cos{A}}-\dfrac{\sin{\dfrac{A}{2}}}{\sin{\dfrac{B}{2}}\sin{\dfrac{C}{2}}}=4$$
my idea:
use this well known
$$4\sin{\dfrac{A}{2}}\sin{\dfrac{B}{2}}\sin{\dfrac{C}{2}}+1=\cos{A}+\cos{B}+\cos{C}$$
then we have
$$\Longrightarrow \begin{vmatrix}
2\sin{A}\sin{C}-\cos{B}&2\sin{A}\sin{B}-\cos{C}\\
\cos{A}+\cos{C}+2\cos{B}-1&\cos{A}+\cos{B}+2\cos{C}-1
\end{vmatrix}=0$$
then
$$(2\sin{A}\sin{C}-\cos{B})(\cos{A}+\cos{B}+2\cos{C}-1)-(2\sin{A}\sin{B}-\cos{C})(\cos{A}+\cos{C}+2\cos{B}-1)=0$$
folowing I can't work,Thank you everyone can help
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I'm getting the result without using the formula.
Note that, since $\Delta ABC$ is a triangle, $$2\sin A\sin C-\cos B=\cos (A-C)-\cos(A+C)-\cos C=\cos(A-C)$$
Similarly, $$2\sin A\sin B-\cos C=\cos (A-B)$$
Now, observe that if $B=C$, the determinant condition is automatically satisfied and in that case $$\large \cos A=-\cos 2B,\ \sin \frac{A}{2}=\cos B$$ So, if the answer holds then we will be able to find a solution for $B$ and hence for $C,A$! So the result is not a general one if $B=C$. So I'm assuming $B\ne C$.
Now, from the determinant, and with the assumption stated above along with the above equalities, we get,\begin{equation}
\begin{split}
\ &4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\left(\cos(A-C)-\cos(A-B)\right)\\
\ =&\cos B\cos(A-B)-\cos C \cos(A-C)\\
\ \Rightarrow & 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}2\sin\left(\frac{C-B}{2}\right)\sin\left(\frac{3A-\pi}{2}\right)\\
\ =&\frac{1}{2}\left(\cos A+\cos(A-2B)-\cos A-\cos (A-2C)\right)
\ =&\sin(C-B)\sin 2A\\
\ \Rightarrow & 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\cos\left(\frac{3A}{2}\right)\\
\ =&-\cos\left(\frac{B-C}{2}\right)\sin 2A\\
\ \Rightarrow & \sin \frac{B}{2}\sin \frac{C}{2}\cos\left(\frac{3A}{2}\right)\\
\ =& -\cos\left(\frac{B-C}{2}\right)\cos\left(\frac{A}{2}\right)\cos A\\
\end{split}
\end{equation}
So, now we get \begin{equation}
\begin{split}
\large
\frac{\cos \frac{3A}{2}}{\cos \frac{A}{2} \cos A}=& -\frac{\cos \frac{B-C}{2}}{\sin \frac{B}{2}\sin \frac{C}{2}}\\
\Rightarrow \frac{\cos \frac{3A}{2}-\cos \frac{A}{2} \cos A}{\cos \frac{A}{2} \cos A}=& \frac{-\cos \frac{B-C}{2}-\sin \frac{B}{2}\sin \frac{C}{2}}{\sin \frac{B}{2}\sin \frac{C}{2}}\\
\Rightarrow -\frac{\sin A \sin \frac{A}{2}}{\cos \frac{A}{2} \cos A}=&\frac{-\cos \frac{B}{2}\cos \frac{C}{2}-2\sin \frac{B}{2}\sin \frac{C}{2}}{\sin \frac{B}{2}\sin \frac{C}{2}}\ \\
\Rightarrow \frac{2\sin^2 \frac{A}{2}}{\cos A}= &\frac{\cos \frac{B}{2}\cos \frac{C}{2}}{\sin \frac{B}{2} \sin \frac{C}{2}}+2\\
\Rightarrow \frac{1}{\cos A}=3+\cot \frac{B}{2} \cot \frac{C}{2}
\end{split}
\end{equation}
Now, observe that $$\sin \frac{A}{2}=\cos \frac{B+C}{2}=\cos \frac{B}{2}\cos \frac{C}{2}-\sin \frac{B}{2}\sin \frac{C}{2}\\
\Rightarrow \frac{\sin \frac{A}{2}}{\sin \frac{B}{2}\sin \frac{C}{2}}=\cot \frac{B}{2} \cot \frac{C}{2}-1$$
Hence we get $$\Rightarrow \frac{1}{\cos A}=4+\frac{\sin \frac{A}{2}}{\sin \frac{B}{2}\sin \frac{C}{2}}$$
|
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|
How to find the sum of the sequence $\frac{1}{1+1^2+1^4} +\frac{2}{1+2^2+2^4} +\frac{3}{1+3^2+3^4}+.....$ Problem :
How to find the sum of the sequence $\frac{1}{1+1^2+1^4} +\frac{2}{1+2^2+2^4} +\frac{3}{1+3^2+3^4}+.....$
I am unable to find out how to proceed in this problem.. this is a problem of arithmetic progression... Please suggest how to proceed...Thanks..
|
HINT:
As $1+r^2+r^4=(1+r^2)^2-(r)^2=(1+r+r^2)(1-r+r^2)$
and $(1+r+r^2)-(1-r+r^2)=2r,$
$$\frac r{1+r^2+r^4}$$
$$=\frac12\cdot\frac{2r}{(1+r+r^2)(1-r+r^2)}$$
$$=\frac12\cdot\frac{(1+r+r^2)-(1-r+r^2)}{(1+r+r^2)(1-r+r^2)}$$
$$=\frac12\left(\frac1{1-r+r^2}-\frac1{1+r+r^2}\right)$$
Put the values of $r=1,2,\cdots.. n-1,n$ to find the partial sum
and recognize the Telescoping Sum/Series which is evident as $1-(r+1)+(r+1)^2=1+r+r^2$
Then set $n\to\infty$
|
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|
Simplifying compound fraction: $\frac{3}{\sqrt{5}/5}$ I'm trying to simplify the following:
$$\frac{3}{\ \frac{\sqrt{5}}{5} \ }.$$
I know it is a very simple question but I am stuck. I followed through some instructions on Wolfram which suggests that I multiply the numerator by the reciprocal of the denominator.
The problem is I interpreted that as:
$$\frac{3}{\ \frac{\sqrt{5}}{5} \ } \times \frac{5}{\sqrt{5}},$$
Which I believe is:
$$\frac{15}{\ \frac{5}{5} \ } = \frac{15}{1}.$$
What am I doing wrong?
|
In the expression $\frac{3}{\frac{\sqrt{5}}{5}}$, to find the reciprocal of that expression's denominator, $\frac{\sqrt{5}}{5}$, simply swap its numerator and denominator.
Thus the reciprocal of $\frac{\sqrt{5}}{5}$ is $\frac{5}{\sqrt{5}} $, and $$ \frac{3}{\frac{\sqrt{5}}{5}} = 3 \cdot \frac{5}{\sqrt{5}}$$
|
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|
How to evaluate the integral $\oint_{C}\frac{z + 2}{z^2(z^2-1)}dz$ I've been trying to solve the following complex-analysis problem:
Evaluate
$\oint_{C}\frac{z + 2}{z^2(z^2-1)}dz$
Where $C = \{z : |z + 1| = \frac{3}{2}\}$
What I've tried to do was the following: I detected that there are singularities at $z = -1$ and at $z = 0$, making the function non-analytic inside that domain. So I found the Laurent Series for $g(z) = z + 2$ around $-1$, which is $g(z) = 1 + (z - (-1))$, but from that point I can't find a way to use the coefficients $a_n$ to solve the integral with the Residue Method. I was not able to transform the function $g(z)$ into some $\frac{h(z)}{(z - (-1))} = \frac{z + 2}{z^2(z^2-1)}$ with $a_n$'s that are not functions of $z$.
|
You should expand $$\frac{z+2}{z^{2}(z^{2}-1)}$$ as $$\frac{1}{z(z^{2}-1)}+2(\frac{1}{z^{2}-1}-\frac{1}{z^{2}})=\frac{1}{2z}(\frac{1}{z-1}-\frac{1}{z+1})+(\frac{1}{z-1}-\frac{1}{z+1})-\frac{2}{z^{2}}$$where the first part can be further expanded as $$\frac{1}{2}(\frac{1}{z-1}-\frac{1}{z}-(\frac{1}{z}-\frac{1}{z+1}))=\frac{1}{2}(\frac{1}{z-1}+\frac{1}{z+1}-\frac{2}{z})$$So we are evaluating $$\frac{1.5}{z-1}-\frac{0.5}{z+1}-\frac{1}{z}-\frac{2}{z^{2}}$$
As others pointed out, the integral should be $-1.5*2\pi i=-3\pi i$.
|
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|
How many squares are there modulo a Mersenne prime? Mersenne primes are primes of the form $M_n = 2^n - 1$. I'm wondering how many distinct natural numbers result from squaring the naturals modulo $M_n$.
As an example, $M_3 = 7$. If we take the naturals less than seven, we get:
$$1^2 \equiv 1 \bmod 7$$
$$2^2 \equiv 4 \bmod 7$$
$$3^2 \equiv 2 \bmod 7$$
$$4^2 \equiv 2 \bmod 7$$
$$5^2 \equiv 4 \bmod 7$$
$$6^2 \equiv 1 \bmod 7$$
Thus, there are $3$ distinct results of squaring; namely $1$, $2$, and $4$. So I'm wondering, for a given Mersenne prime $M_n$, how many different squares can we get?
|
For any odd prime $p$, there are $\frac{p-1}{2}+1=\frac{p+1}{2}$ squares modulo $p$.
To show this, note first that $0$ is a square modulo $p$. It is, modulo $p$, $0^2$, or, if you prefer, it is congruent to $p^2$. (But it is not called a *quadratic residue of $p$.)
Now consider the squares of numbers in the interval $\left[1,\frac{p-1}{2}\right]$. These are all distinct modulo $p$. And since the numbers in the interval $\frac{p+1}{2}$ to $p-1$ are the negatives (modulo $p$) of numbers in the interval $\left[1,\frac{p-1}{2}\right]$, squaring them produces nothing new. You will observe this from the example you calculated. The squares of $1$, $2$, and $3$ were distinct modulo $7$, and after that you got nothing new.
To show that squares of numbers in the interval $\left[1,\frac{p-1}{2}\right]$ are all distinct modulo $p$, let $x$ and $y$ be numbers in the interval, with $x\gt y$. Suppose $x^2\equiv y^2\pmod{p}$. Then $(x-y)(x+y)$ is divisible by $p$. Thus one of them is. That's impossible, since each of $x-y$ and $x+y$ lies between $1$ and $p-1$.
|
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|
Deriving the generating function of a divide and conquer type recurrence relation I am working through Analysis of Algorithms by Sedgewick/Flajolet
On problem 3.44 I am given the recurrence, and I need to come up with a generating function. I have tried the various methods in the section (it isn't linear, I don't see how to do with a differential equation, nor does it appear to be a composition), but haven't had any success.
$f_{2n}=f_{2n-1}+f_n $ with $n>1$ and $f_0=0$
$f_{2n+1}=f_{2n}$ with $n>0$ and $f_1=1$
Thank you.
|
I think the first equation in the problem statement should be
$f_{2n}=f_{2n-1}+f_n $ with $n \ge 1$ and $f_0=0$.
Otherwise, as noted by RGB, there is no way to find $f_2$.
Let $F(x) = \sum_{n=0}^{\infty}f_{n} x^{n}$.
Note that we have
$\sum_{n=0}^{\infty}f_{2n} x^{2n}= \sum_{n=0}^{\infty} f_{2n-1} x^{2n} + \sum_{n=0}^{\infty} f_n x^{2n} \qquad$ so
$(1/2) [F(x)-F(-x)] = (1/2) \; x\; [F(x)-F(-x)] +F(x^2)$
$(1-x) F(x) + (1+x) F(-x) = 2 F(x^2) \qquad(*)$
Now starting from the second equation in the problem statement,
$\sum_{n=1}^{\infty} f_{2n+1} x^{2n+1} = \sum_{n=1}^{\infty} f_{2n} x^{2n+1}$ so
$\sum_{n=0}^{\infty} f_{2n+1} x^{2n+1} - x = \sum_{n=0}^{\infty} f_{2n} x^{2n+1}$
$(1/2) [F(x) - F(-x)] - x = (1/2) \; x \; [F(x) + F(-x)]$
$(1-x) F(x) -2x = (1+x) F(-x)$
Substituting into (*), we have
$2 (1-x) \; F(x) -2x = 2 F(x^2) \qquad$ so
$F(x) = \frac{F(x^2)}{1-x} +\frac{x}{1-x}$
Iterating this equation, we find
$F(x) = \frac{F(x^4)}{(1-x)(1-x^2)} +\frac{x}{1-x} + \frac{x^2}{(1-x)(1-x^2)}$
$= \frac{F(x^8)}{(1-x)(1-x^2)(1-x^4)} +\frac{x}{1-x} + \frac{x^2}{(1-x)(1-x^2)} + \frac{x^4}{(1-x)(1-x^2)(1-x^4)}$
which may allow us to guess
$F(x) = \frac{1}{(1-x)(1-x^2)(1-x^4)(1-x^8)\cdots}+\frac{x}{1-x} + \frac{x^2}{(1-x)(1-x^2)} + \frac{x^4}{(1-x)(1-x^2)(1-x^4)} + \dots$
which we can see does satisfy $(1-x) F(x) = F(x^2) + x$.
It would be nice to find a closed form in terms of a finite number of standard functions, but I don't see how to do that.
|
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Prove that $\sqrt 2 + \sqrt 3$ is irrational I have proved in earlier exercises of this book that $\sqrt 2$ and $\sqrt 3$ are irrational. Then, the sum of two irrational numbers is an irrational number. Thus, $\sqrt 2 + \sqrt 3$ is irrational. My first question is, is this reasoning correct?
Secondly, the book wants me to use the fact that if $n$ is an integer that is not a perfect square, then $\sqrt n$ is irrational. This means that $\sqrt 6$ is irrational. How are we to use this fact? Can we reason as follows:
$\sqrt 6$ is irrational
$\Rightarrow \sqrt{2 \cdot 3}$ is irrational.
$\Rightarrow \sqrt 2 \cdot \sqrt 3$ is irrational
$\Rightarrow \sqrt 2$ or $\sqrt 3$ or both are irrational.
$\Rightarrow \sqrt 2 + \sqrt 3$ is irrational.
Is this way of reasoning correct?
|
Note that $\sqrt{2}+\sqrt{3}$ is a solution to the equation: $$x^4-10x^2+1=0$$Does this polynomial have any rational roots?
Edit: To find this polynomial, note that if $x=\sqrt{2}+\sqrt{3}$, then: $$x^2=5+2\sqrt{6}$$and: $$x^4=49+20\sqrt{6}.$$You need $-10x^2$'s to get rid of the $20\sqrt{6}$ in $x^4$, and $x^4-10x^2=-1$, so you get: $$x^4-10x^2+1=0.$$
|
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|
Hypergeometric value Is their a closed form for the following
$${}_2F_1 \left(a,b;c;\frac{1}{2} \right)$$
I would use the following
$${}_2F_1 \left(a,b;c;x \right)= \frac{\Gamma(c)}{\Gamma(c-b)\Gamma(b)} \int^1_0 t^{b-1}(1-t)^{c-b-1} (1-xt)^{-a} \, dt $$
But it wasn't a success !
Edit: Corrected integral representation (swapped arguments in $\Gamma$ fraction)
|
There are closed forms for the 2F1 but their exact form depend on the values and the relations between the parameters.
*
*The trivial case is if $a$ or $b$ is a negative integer: the function becomes a finite sum of rational expressions in the parameters. if you choose $a=-4$ and if you name $d$ the value you have fixed at $1/2$, you got
$ 1 - \frac{4 b d}{c} +\frac{6 b (b+1) d^2}{c (c+1)} - \frac{4 b (b+1) (b+2) d^3}{c (c+1)
(c+2)} - \frac{(-b-3) b (b+1) (b+2) d^4}{c (c+1) (c+2) (c+3)}$
and setting $d = 1/2$:
$ 1 -\frac{2 b}{c} + \frac{3 (b+1) b}{2 c (c+1)} - \frac{(b+1) (b+2) b}{2 c (c+1)
(c+2)} - \frac{(-b-3) (b+1) (b+2) b}{16 c (c+1) (c+2) (c+3)} $
*
*if $a$ or $b$ is equal to $c$, they cancel each other, giving a simpler $_1F_0$.
$_2F_1( a, b ; b ; d) = _1F_0(a; d) = (1-d)^{-a} $
*for small rationals, there are more interesting closed forms involving elementary functions and elliptic functions and their inverses. Among classical cases
$_1F_0(1/2 ; d) = \frac{1}{\sqrt{1- d}} $
$_2F_1(1/2, 1/2 ; 3/2; d) = \frac{1}{\sqrt d} \sin^{-1} (\sqrt d )$
$_2F_1(1/2, 1 ; 3/2; d) = \frac{1}{\sqrt d} \tanh^{-1} (\sqrt d )$
|
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|
Partial Fractions I here would like to clear my doubt on the question below:
$$\frac{1}{x(x-1)(x-2)}\;,$$
that is, we want to bring it into the form:
$$\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x-2}\;,$$
in which the unknown parameters are $A,B$, and $C$. Multiplying these formulas by $x(x − 1)(x − 2)$ turns both into polynomials, which we equate:
$$A(x-1)(x-2) + Bx(x-2) + Cx(x-1) = 1\;,$$
or, after expansion and collecting terms with equal powers of $x$:
$$(A+B+C)x^2 - (3A+2B+C)x + 2A = 1\;.$$
At this point it is essential to realize that the polynomial $1$ is in fact equal to the polynomial $0x^2 + 0x + 1$, having zero coefficients for the positive powers of $x$. Equating the corresponding coefficients now results in this system of linear equations:
$$\left\{\begin{align*}
&A+B+C = 0\\
&3A+2B+C = 0\\
&2A = 1\;.
\end{align*}\right.$$
Solving it results in:
$$A = \frac{1}{2},\, B = -1,\, C = \frac{1}{2}\;.$$
So from my solving I had different values of $A,B$, and $C$ which gave me:
$$\left\{\begin{align*}
&A=\frac12\\
&B= 2\\
&C= -\frac52\;.
\end{align*}\right.$$
Can someone please tell me if these answers are correct because when I substitute these values into equation $A+B+C= 0$, it still gave me a zero.
But this time for the $2$nd equation, instead of $3A+2B+C= 0$, I used $-3A+2B+C= 0$, which then by substituting the values of $A, B$, and $C$ I had, also gave me a zero. Only $A= \frac12$ was the same as obtained from $2A= 1$.
Does this mean that the values that I have obtained for $A, B$, and $C$ are also correct? Kindly can someone please give a clear explanation to this?
Many thanks.
|
Examine the form $$A(x−1)(x−2)+Bx(x−2)+Cx(x−1)=1$$
This is supposed to be true for every value of $x$. If we try $x=0$ we get $2A=1$, $x=1$ gives $=B=1$, and $x=2$ gives $2C=1$.
This short-cut method (known as the "cover-up" rule) can be used to find the numerators for partial fractions where all the denominators are linear, and is very useful in other cases too.
|
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|
The values of $N$ for which $N(N-101)$ is a perfect square
For how many values of $N$ (integer), $N(N-101)$ is a perfect square number?
I started in this way.
Let $N(N-101)=a^2$
or $N^2-101N-a^2=0.$
Now if the discriminant of this equation is a perfect square then we can solve $N$.
But I can't progress further.
|
We want $N(N-101)$ to be a perfect square. Note that the gcd of $N$ and $N-101$ is $1$ or $101$.
Suppose it is $101$. Let $N=101M$. We get that $101M(101M-101)$ is a perfect square. This is the case iff $M(M-1)$ is a perfect square. That happens iff $4M^2-4M$ is a perfect square.
But $4M^2-4M$ is a perfect square iff $(2M-1)^2-1$ is a perfect square, say $a^2$, or equivalently iff $(2M-1)^2=a^2+1$. The only possibility is $2M-1=\pm 1$, giving $M=0$ or $M=1$, and therefore $N=0$ or $N=101$.
Next we deal with $N$ and $N-101$ relatively prime. Then each of $N$ and $N-101$ is a perfect square, say $N=a^2$ and $N-101=b^2$. We arrive at the equation $a^2-b^2=101$. There are $4$ possibilities: $a-b=1$, $a+b=101$; $a-b=101$, $a+b=1$; $a-b=-1$, $a+b=-101$; $a-b=-101$, $a+b=-1$.
For each of these, solve for $a$ and $b$. But note that $a-b=-1$, $a+b=-101$ yields the negatives of the solutions of $a-b=1$, $a+b=101$. Thus they give the same value of $N$, as do the other pair.
Remark: Your discriminant idea will work nicely, in basically the same way. (One could even argue that it will work more nicely). Setting $N(N-101)=a^2$, like you did, we find that the discriminant is $101^2+4a^2$. We want this to be a perfect square, say $b^2$. That gives $101^2=b^2-4a^2=(b-2a)(b+2a)$. So we want put $b-2a$ as one factor of $101^2$, and $b+2a$ as the complementary factor. We end up solving a few pairs of linear equations in two unknowns, just as in the proof given above.
|
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Initial value problem differential equation $y' = (x-1)(y-2)$ $$y' = (x-1)(y-2)$$
$y(2)= 4$
$$\frac{1}{y-2}dy = (x-1)dx$$
$$\int \frac{1}{y-2}dy =\int (x-1)dx$$
$$\ln(y-2) = \frac{x^2}{2} - x + c$$
$$y - 2 = e^{\frac{x^2}{2} - x + c} $$
$$y = e^{\frac{x^2}{2} - x + c} + 2$$
plug in the inital value
$$y = e^{\frac{2^2}{2} - 2 + c} + 2$$
$$y = e^c+ 2$$
I feel like this is wrong anyways, so where did I go wrong?
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For $x=2$ we have $y=4$ so $e^c=4-2=2$ and finally
$$y = 2e^{\frac{x^2}{2} - x } + 2$$
|
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|
Number of Solutions to Diophantine Equation $(a)$ Let $c < 2\pi$ be a positive real number. Show that
there are infinitely many integers $n$ such that the equation
$x^2 + y^2 + z^2 = n$
has at least $c\sqrt n$ integer solutions.
$(b)$ Find a constant $C > 0$ such that there are infinitely many $n$ for
which the equation
$x^5 + y^3 + z^2 = n$
has $\ge Cn^{1/30}$ nonnegative solutions
Note: this is not a homework problem
I am not really sure how to go about this problem. Any hints or a general strategy that can be applied to both problems would be very much appreciated.
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HINT:Consider $$F(t):=\sum_{n\le t} \#\{x^a+y^b+z^c=n\} $$ and find its asymptotic formula.
Take your problem (a) as an example.
$$\sum_{n\le t} \#\{x^2+y^2+z^2=n\}=\sum_{x,y,z\in\mathbb{Z}}\mathbb{1}_{\{x^2+y^2+z^2\le t\}}=Area\ of\{x^2+y^2+z^2\le t\}+o(Area\ of\{x^2+y^2+z^2\le t\})=\frac{4}{3}\pi t^\frac{3}{2}+o(t^\frac{3}{2})$$
Assume that $\exists c<2\pi$ with only finite many $n$ have has at least $c\sqrt{n}$ integer solutions. Then $$F(t)\le\sum_{n\le t}c\sqrt{n}+O(1)=\frac{2c}{3}t^\frac{3}{2}+o(t^\frac{3}{2})<\frac{4}{3}\pi t^\frac{3}{2}+o(t^\frac{3}{2})$$
A contradiction.
|
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|
Differential of normal distribution Let
$$f(x)=\frac{\exp\left(-\frac{x^2}{2\sigma^2}\right)}{\sigma\sqrt{2\pi}}$$
(Normal distribution curve) Where $\sigma$ is constant. Is my derivative correct and can it be simplified further?
$$\begin{align} f'(x)
&=\frac d{dx}\left(\frac{\exp\left(-\frac{x^2}{2\sigma^2}\right)}{\sigma\sqrt{2\pi}}\right)\\
&=\frac{1}{\sigma\sqrt{2\pi}}\frac d{dx}\left(\exp\left(-\frac{x^2}{2\sigma^2}\right)\right)\\
\end{align}$$
$$\begin{align}
\implies\frac d{dx}\exp\left(-\frac{x^2}{2\sigma^2}\right)&=\frac d{dx}\sum_{n=0}^\infty\frac{\left(-\frac{x^2}{2\sigma^2}\right)^n}{n!}
=\sum_{n=0}^\infty\frac d{dx}\frac{\left(-\frac{x^2}{2\sigma^2}\right)^n}{n!}\\
&=\sum_{n=0}^\infty \frac 1{n!}\frac d{dx}\left(-\frac{x^2}{2\sigma^2}\right)^n=\sum_{n=0}^\infty -\frac 1{n!}\frac{1}{2\sigma^{2n}}\frac d{dx}x^{2n}\\
&=\sum_{n=0}^\infty -\frac 1{n!}\frac{1}{2\sigma^{2n}}2nx^{2n-1}=\sum_{n=0}^\infty -\frac {1}{n!}\frac{2nx^{2n-1}}{2\sigma^{2n}}\\
\end{align}$$
$$\implies f'(x)=\frac{-\sum_{n=0}^\infty \left(\frac {1}{n!}\frac{2nx^{2n-1}}{2\sigma^{2n}}\right)}{\sigma\sqrt{2\pi}}$$
I attempted to simplify the infinite summation to the exponential function but I am failing. How would you simplify this further?
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If you want to use the method that you started, then you should note that we can only "pull out the negative sign" from $$\left(-\frac{x^2}{2\sigma^2}\right)^n$$ when $n$ is odd. It would be better to keep a $(-1)^n$ term around, because then, $$\begin{align}\frac d{dx}\exp\left(-\frac{x^2}{2\sigma^2}\right) &= \sum_{n=0}^\infty \frac 1{n!}\frac d{dx}\left(-\frac{x^2}{2\sigma^2}\right)^n\qquad[\text{as you showed}]\\ &= \sum_{n=0}^\infty \frac 1{n!}\frac{(-1)^n}{2^n\sigma^{2n}}\frac d{dx}\left[x^{2n}\right]\\ &= \sum_{n=0}^\infty \frac 1{n!}\frac{(-1)^n}{2^n\sigma^{2n}}\cdot 2nx^{2n-1}\\ &= \sum_{n=1}^\infty \frac 1{n!}\frac{(-1)^n}{2^n\sigma^{2n}}\cdot 2nx^{2n-1}\qquad[\text{the }n=0\text{ term vanishes}]\\ &= \sum_{n=1}^\infty \frac 1{(n-1)!}\frac{(-1)^n}{2^{n-1}\sigma^{2n}} x^{2n-1}\qquad[\text{cancellation}]\\ &= \sum_{n=0}^\infty \frac 1{n!}\frac{(-1)^{n+1}}{2^n\sigma^{2(n+1)}} x^{2(n+1)-1}\qquad[\text{reindex by }n\mapsto n+1]\\ &= \sum_{n=0}^\infty \frac 1{n!}\frac{(-1)^{n+1}}{2^n\sigma^{2n+2}} x^{2n+1}\\ &= -\frac{x}{\sigma^2}\sum_{n=0}^\infty \frac 1{n!}\frac{(-1)^n}{2^n\sigma^{2n}} x^{2n}\\ &= -\frac{x}{\sigma^2}\sum_{n=0}^\infty \frac 1{n!}\left(-\frac{x^2}{2\sigma^2}\right)^n\\ &= -\frac{x}{\sigma^2}\exp\left(-\frac{x^2}{2\sigma^2}\right).\end{align}$$
There are far too many places to go wrong, here, though. Better just to use the chain rule.
|
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|
A tricky integral (flux of a point charge through a disk) The integrals:
$$
\oint \frac{r\,dr\,d\phi}{\left(L^2+r^2+h^2+2Lr\cos\phi\right)^{3/2}}\\
\oint \frac{dx\,dy}{\left((L+x)^2+y^2+h^2\right)^{3/2}}
$$
If we have a point charge at the origin and we want to find the flux through a disk of radius $R$ which is located at $(x,0,z)$ and lies parallel to the $x-y$ plane we will have to do a tricky integral. We can write down the exact $E.dA$ or we can try and find the solid angle (turns out it is like finding the flux through an ellipse at $(0,0,z)$).
What I'm looking for is the answer to that integral or the solid angle. I thought the integral wasn't doable. But recently I came across a physical way of finding it using greens reciprocity theorem. If you can find it from that, then the integral should be doable as well!
To be clear: this is entirely a math problem. We have an inverse square vector field. We want its flux through a disk.
for example take the (x=0) situation.we can simply use gauss law and just find the solid angle to obtain the flux.ot write out the integral by using r and theta as variables.but even at this situation doing it with dxdy integral is tough.
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The solid angle can be expressed in terms of complete elliptic integrals of first and third kind.
I believe this is first derived by F. Paxton in 1959.
THE REVIEW OF SCIENTIFIC INSTRUMENTS. VOLUME 30, NUMBER 4 APRIL, 1959.
Solid Angle Calculation for a Circular Disk. F. PAXTON
A copy of the article can be found
here.
Below is my attempt to derive the same result in an alternate manner.
The integral is indeed trickier than I thought.
To avoid confusion with the usage of $x,y,z$ as coordinates, let us change the problem and assume the circular disk $C$ we are dealing with lies in the plane $z = b$, centered at $(a,0,b)$ with radius $s$. We will assume $a, b > 0$ and $a \ne s$.
In terms of ordinary spherical polar coordinates $(r,\theta,\phi)$, points on the plane $z = b$ can be parametrized as:
$$(x,y,z) = (b\tan\theta \cos\phi,\,b\tan\theta\sin\phi,\, b)$$
Let $\rho = \sqrt{x^2+y^2} = b\tan\theta$, the "element" for integrating the solid angle is given by:
$$d\Omega
= \sin\theta d\theta \wedge d\phi
=\frac{\tan\theta d\tan\theta}{\sqrt{1 + \tan\theta^2}^3} \wedge d\phi
= \frac{b \rho d\rho \wedge d\phi}{\sqrt{\rho^2+b^2}^3}
= \frac{b\;dx \wedge dy}{\sqrt{\rho^2+b^2}^3}
$$
Introduce complex coorindates $\eta = x + iy$ and $\bar{\eta} = x - iy$, the solid angle
extended by $C$ can be rewritten as:
$$\begin{align}
\Omega_{C}
= & \int_{C} d\Omega
= \frac{b}{2i}\int_{C} \frac{d\bar{\eta} \wedge d\eta}{\sqrt{\eta\bar{\eta}+b^2}^3}
=\color{blue}{^{[1]}} \frac{b}{i} \int_C d\left(\frac{1}{\eta}\left( \frac{1}{b} - \frac{1}{\sqrt{\eta\bar{\eta}+b^2}}\right)\right) \wedge d\eta\\
= & \frac{b}{i} \int_{\partial C} \left(\frac{1}{b} - \frac{1}{\sqrt{\eta\bar{\eta} + b^2}}\right) \frac{d\eta}{\eta}
= 2\pi \delta_{C} + ib \int_{\partial C} \frac{d\eta}{\eta \sqrt{\eta\bar{\eta} + b^2}}
\end{align}$$
where $\delta_{C} = 1 \text{ or } 0$ depends on whether $s > a$ or $< a$. In other words, whether $\partial C$ contains $0$ or not.
On $\partial C$, we can parametrize $\eta$ as $s e^{it} + a$ and $\bar{\eta}$ as $s e^{-it} + a$. Substitute this into above expression, we obtain:
$$\begin{align}
&\Omega_C - 2\pi \delta_{C}\\
= &ib \int_{-\pi}^{\pi} \frac{ dt }{\sqrt{ s^2 + a^2 + b^2 + 2sa\cos t}}\frac{ is e^{it}}{s e^{it} + a }\\
= & -b
\int_{0}^{\pi} \frac{ dt }{\sqrt{ s^2 + a^2 + b^2 + 2sa\cos t}}
\left( \frac{s e^{it}}{s e^{it} + a } + \frac{s e^{-it}}{s e^{-it} + a } \right)\\
= & -b
\int_{0}^{\pi} \frac{ dt }{\sqrt{ s^2 + a^2 + b^2 + 2sa\cos t}}
\frac{2s(s + a\cos t)}{s^2 + a^2 + 2sa\cos t}\\
= & -b
\int_{0}^{\pi} \frac{ dt }{\sqrt{ s^2 + a^2 + b^2 + 2sa\cos t}}
\left( 1 + \frac{s^2 - a^2}{s^2 + a^2 + 2sa\cos t} \right)\\
= & -2b
\int_{0}^{\pi/2} \frac{ dt }{\sqrt{ s^2 + a^2 + b^2 + 2sa\cos(2t)}}
\left( 1 + \frac{s^2 - a^2}{s^2 + a^2 + 2sa\cos(2t)} \right)\\
= & -2b
\int_{0}^{\pi/2} \frac{ dt }{\sqrt{ (s+a)^2 + b^2 - 4sa\sin^2(t)}}
\left( 1 + \frac{s^2 - a^2}{(s+a)^2 - 4sa\sin^2(t)} \right)
\end{align}$$
The last integral can be expressed in terms of the
complete elliptic integral
of first and third kind:
$$
K(k) = \int_{0}^{\frac{\pi}{2}} \frac{dt}{\sqrt{1-k^2\sin^{2}t}}
\quad\text{ and }\quad
\Pi(n,k) = \int_{0}^{\frac{\pi}{2}} \frac{dt}{(1-n\sin^{2}t)\sqrt{1-k^2\sin^{2}t}}
$$
The final results are:
$$\Omega_{C} = 2\pi\delta_C -\frac{2b}{\sqrt{(s+a)^2+b^2}}
\left(\;K(k) + \left(\frac{s-a}{s+a}\right) \Pi(n,k)\;\right)
$$
where $\displaystyle \;\;k = \sqrt{\frac{4sa}{(s+a)^2+b^2}}
\quad\text{ and }\quad
n = \frac{4sa}{(s+a)^2}
$.
Notes
$\color{blue}{[1]}$ When $C$ contains $(0,0,b)$, we need the $\frac{1}{b}$ term. It regularize the 1-form at $\eta = 0$ and make Stroke's theorem continue to work.
|
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|
If $n$ is an even perfect number $ n> 6$ show that the sum of its digits is $\equiv 1 (\bmod 9)$ If $n$ is an even perfect number $ n> 6$ show that the sum of its digits is $\equiv 1 \mod 9$.
I know perfect numbers are of the form $(2^{p-1})(2^{p}-1)$. I have a few trials that I have done and they check out but as far as a proof I have no idea where to start since I have to use the digits in the number.
|
Note that if $\mathfrak{d}(n)$ is digit sum of $n$, then $$\mathfrak{d}(n) \equiv n \pmod{9}.$$ Hence we wish to show that the perfect number $n$ is congruent to $1$ modulo $9$. Now note that $$2^0 \equiv 1 \pmod{9}$$$$2^1 \equiv 2 \pmod{9}$$ $$2^2 \equiv 4 \pmod{9}$$ $$2^3 \equiv 8 \pmod{9}$$ $$2^4 \equiv 7 \pmod{9}$$ $$2^{5} \equiv 5 \pmod{9}$$ $$2^{6} \equiv 1 \pmod{9}$$ $$\vdots$$ This exponentiation repeats modulo $6$. As you noted, perfect numbers are of form $2^{p - 1}(2^p - 1)$.
1) $p \equiv 0 \pmod{6}$, then $2^{p - 1}(2^p - 1) \equiv 5 \cdot (1 - 1) \equiv \color{red}0 \pmod{9}$
2) $p \equiv 1 \pmod{6}$, then $2^{p - 1}(2^p - 1) \equiv 1 \cdot (2 - 1) \equiv 1 \pmod{9}$
3) $p \equiv 2 \pmod{6}$, then $2^{p - 1}(2^p - 1) \equiv 2 \cdot (4 - 1) \equiv \color{red}6 \pmod{9}$
4) $p \equiv 3 \pmod{6}$, then $2^{p - 1}(2^p - 1) \equiv 4 \cdot (8 - 1) \equiv 1 \pmod{9}$
5) $p \equiv 4 \pmod{6}$, then $2^{p - 1}(2^p - 1) \equiv 8 \cdot (7 - 1) \equiv \color{red}3 \pmod{9}$
6) $p \equiv 5 \pmod{6}$, then $2^{p - 1}(2^p - 1) \equiv 7 \cdot (5 - 1) \equiv 1 \pmod{9}$
So only cases that could be troublesome are $p \equiv 0,2,4 \pmod{6}$. It is well-known since Euclid that if $2^{p - 1}(2^p - 1)$ is an even perfect number, then $p$ is prime. But then we have $$p = 6k + a \text{, where } a \in \{0,2,4\} \implies p \text{ is composite}$$ which is a contradiction.
|
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|
Need help with logarithmic differentitation I have the expression
$$y = \sqrt{x^2(x+1)(x+2)}.$$
I have tried looking at videos but I still cannot arrive at the correct answer and don't know how to get there.
By the way, the correct answer is
$$y' = \frac{4x^2+9x+4}{2\sqrt{(x+1)(x+2)}}.$$
Please, help.
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Alternatively, write $$y^2=x^2(x+1)(x+2)$$
Then $$2yy'=2x(x+1)(x+2)+x^2(x+1)+x^2(x+2)$$ whence
$$y'=\frac{2x(x+1)(x+2)+x^2(x+1)+x^2(x+2)}{2x\sqrt{(x+1)(x+2)}}$$
$$y'=\frac{2(x+1)(x+2)+x(x+1)+x(x+2)}{2\sqrt{(x+1)(x+2)}}\\=\frac{4x^2+9x+4}{2\sqrt{(x+1)(x+2)}}$$
assuming you wanted $x\sqrt{(x+1)(x+2)}$ since $\sqrt{x^2}=|x|$ and we get something not that nice with that.
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|
Simplify : $( \sqrt 5 + \sqrt6 + \sqrt7)(− \sqrt5 + \sqrt6 + \sqrt7)(\sqrt5 − \sqrt6 + \sqrt7)(\sqrt5 + \sqrt6 − \sqrt7) $ The question is to simplify $( \sqrt 5 + \sqrt6 + \sqrt7)(− \sqrt5 + \sqrt6 + \sqrt7)(\sqrt5 − \sqrt6 + \sqrt7)(\sqrt5 + \sqrt6 − \sqrt7)$ without using a calculator .
My friend has given me this challenge . I solved it by expanding $$(a+b+c)(-a+b+c)(a-b+c)(a+b-c) = -a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2-c^4$$ and then substituting $a,b,c=\sqrt 5 , \sqrt6 , \sqrt7$ respectively to get the answer $104$ .
But I suppose there is a more elegant and easy way to solve this problem .
Can anyone find it ?
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Here's a nice way to get the expansion the OP used:
The expression
$$P(a,b,c)=(a+b+c)(-a+b+c)(a-b+c)(a+b-c)$$
is clearly a homogeneous polynomial of degree $4$, symmetric in its three variables. It's also clear that the coefficient of $a^4$ (hence also $b^4$ and $c^4$) is $-1$. Moreover,
$$P(-a,b,c)=P(a,-b,c)=P(a,b,-c)=P(a,b,c)$$ which implies $P$ has no terms with any variable taken to an odd degree. Therefore $P$ must be of the form
$$P(a,b,c)=r(a^2b^2+b^2c^2+c^2a^2)-(a^4+b^4+c^4)$$
for some coefficient $r$. It's easy to determine $r$ by letting $a=b=c=1$, for which we have
$$3=3\cdot1\cdot1\cdot1=P(1,1,1)=r(1+1+1)-(1+1+1)=3r-3$$
so $r=2$. The rest of the answer follows what the OP did:
$$P(\sqrt5,\sqrt6,\sqrt7)=2(5\cdot6+6\cdot7+7\cdot5)-(5^2+6^2+7^2)=104$$
Added 8/12/13: Eric Jablow's invocation of Heron's formula inspires one more approach:
Consider the triangle formed by the origin and two vectors $x$ and $y$, with $|x|=a$, $|y|=b$, and $|x-y|=c$. There are two formulas for the area of the triangle: Heron's formula
$$\sqrt{s(s-a)(s-b)(s-c)}$$
where $s=(a+b+c)/2$, and the dot-product formula
$${1\over2}\sqrt{|x|^2|y|^2-(x\cdot y)^2}$$
Putting these together, we have
$$(a+b+c)(-a+b+c)(a-b+c)(a+b-c)=4((a^2b^2-(x\cdot y)^2)$$
What's nice is that this holds for vectors $x$ and $y$ in any dimension. So now let $x=(2,1,0,0)$ and $y=(0,2,1,1)$, so that $x-y=(2,-1,-1,-1)$. We have $a=\sqrt5$, $b=\sqrt6$, $c=\sqrt7$, and $x\cdot y=2$, and thus the OP's product of square roots simplifies to
$$4(5\cdot6-2^2) = 104$$
If you want to know why we went into the fourth dimension for the vectors $x$ and $y$, it's because $7$, like all positive integers, can be written as the sum of four squares, but not as the sum of three. Another possibility, which makes it clear one can handle arbitrary $a$, $b$, and $c$ (as long as the triangle inequality is satisfied, at least) is to let $x=(1,1,1,1,1,0,0,0,0)$ and $y=(0,0,0,1,1,1,1,1,1)$, so that $x-y=(1,1,1,0,0,-1,-1,-1,-1)$. (Edit: Actually, what I just said only works when the triangle inequality holds and $a+b+c$ is even.)
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|
Factorise: $2a^4 + a^2b^2 + ab^3 + b^4$ Factorize : $$2a^4 + a^2b^2 + ab^3 + b^4$$
Here is what I did:
$$a^4+b^4+2a^2b^2+a^4-a^2b^2+ab^3+b^4$$
$$(a^2+b^2)^2+a^2(a^2-b^2)+b^3(a+b)$$
$$(a^2+b^2)^2+a^2(a+b)(a-b)+b^3(a+b)$$
$$(a^2+b^2)^2+(a+b)((a^2(a-b)) +b^3)$$
$$(a^2+b^2)^2+(a+b)(a^3-a^2b+b^3)$$
At this point I don't know what to do and am feeling that my direction is wrong. Please help me.
( Wolfram alpha says that the answer is $(a^2-a b+b^2) (2 a^2+2 a b+b^2)$ but how? )
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After minar:
$2x^4+x^2+x+1= 2x^4+2x+x^2-x+1= 2x(x+1)(x^2-x+1)+x^2-x+1=(2x^2+2x+1)(x^2-x+1) $
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An irreducible polynomial of degree $4$ in $\mathbb{Z}_5[x]$ Q: Find an irreducible polynomial of degree four in $\mathbb{Z}_5[x]$.
My Answer: $x^{4} - 2$
Is my answer correct?
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This is correct, but you need to show it's irreducible. Here's one way to do it:
We immediately verify that $x^4-2$ has no roots in $\mathbb Z _5$. Thus, if it is reducible, it must split into two monic quadratic terms. Say $x^4-2=(x^2+ax+b)(x^2+cx+d)=x^4+x^3(a+c)+x^2(b+d+ca)+x(ad+bc)+bd$. Now, matching coefficients, we must have that $bd=-2=3, a+c=0,b+d+ca=0, ad+bc=0$. First, consider $bd=3$. This then holds true for $b=3,d=1$ or $b=4,d=2$. Thus, we must have $b+d+ac=0$ implying that $ac=1$, in the first case, or $ac=4$, in the second. The values for the pair $(a,c)$ are then $(1,4), (2,3), (0,0), (3,2), (4,1)$. Note that there are two pairs each where $ac=1$ or $4$. First, let $(b,d)=(3,1)$, so $(a,c)=(2,3)$ or $(3,2)$. Then $ad+bc=2+9=1$ or $ad+bc= 3+6=9$, neither of which are equal to $0$, so neither of these work. Now, let $(b,d)=(4,2)$, so $(a,c)=(1,4)$ or $(4,1)$. Again, we see $ad+bc = 4+8 =2$ or $16+2=3$, so neither of these work either. Thus, no two quadratic polynomials multiply to $x^4-2$, so it is irreducible.
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What is? $ \lim_{x \to \infty} \left(\frac{e}{\left( 1 + \frac {1}{x} \right)^x} \right)^x $
Find $$ \lim_{x \to \infty} \left( \dfrac{e}{\left( \left( 1 + \frac {1}{x} \right)^x \right)} \right)^x. $$
$ \lim_{x \to \infty} \left( \dfrac{e}{\left( \left( 1 + \frac {1}{x} \right)^x \right)} \right)^x = \lim_{x \to \infty} \left( \dfrac {e}{e} \right)^x = \lim_{x \to \infty} 1^x = \infty $
Why is this incorrect?
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Take the logarithm,
$$\begin{align}
\log \left(\frac{e}{\left(1+\frac1x\right)^x}\right)^x &= x\log \frac{e}{\left(1+\frac1x\right)^x}\\
&= x\left(1 - x \log \left(1+\frac1x\right)\right)\\
&= x\left(1 - x\left(\frac1x - \frac{1}{2x^2} + O(x^{-3})\right) \right)\\
&= x\left(1 - 1 + \frac{1}{2x} + O(x^{-2})\right)\\
&= \frac12 + O(x^{-1}).
\end{align}$$
And hence
$$\lim_{x\to\infty} \left(\frac{e}{\left(1+\frac1x\right)^x}\right)^x = e^{1/2}.$$
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|
Cyclic polynomial proof If$$ x+y+z = 0 $$
Then prove,
$$ (x^2+xy+y^2)^3+(y^2+yz+z^2)^3+(z^2+zx+x^2)^3$$
$$=3(x^2+xy+y^2)(y^2+yz+z^2)(z^2+zx+x^2)$$
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HINT:
Observe that if we put $x^2+xy+y^2=a$ etc., we need to prove $a^3+b^3+c^3=3abc$
From this, the above proposition will be true
either if $a+b+c=0$
or if $a=b=c$ for real $a,b,c$
Now, $x^2+xy+y^2-(y^2+yz+z^2)=x^2-z^2+xy-yz=(x-z)(x+z+y)=0 $
Alternatively eliminating $x,$
$x^2+xy+y^2=x(x+y)+z^2=\{-(y+z)\}(-z)+y^2$ as $x+y+z=0$
$\implies x^2+xy+y^2=y^2+yz+z^2$
Similarly we can prove, $ x^2+xy+y^2=z^2+zx+x^2 $
$$\implies x^2+xy+y^2=y^2+yz+z^2=z^2+zx+x^2$$
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Finding the transition matrix for the rational canonical form Let $A$ be the $3\times3$ matrix
$$\begin{bmatrix} 3 & 4 & 0 \\-1 & -3 & -2 \\ 1 & 2 & 1 \end{bmatrix}$$
The characterisitc and minimal polynomials are both $(x-1)^2(x+1)$
The eigenspace for $\lambda=1$ is
$$\left\{ \begin{bmatrix} 2 \\-1 \\ 1 \end{bmatrix} \right\}$$
The eigenspace for $\lambda=-1$ is:
$$\left\{ \begin{bmatrix} 2 \\-2 \\ 1 \end{bmatrix} \right\}$$
The rational canonical form $R$ is:
$$\begin{bmatrix} -1 & 0 & 0 \\0 & 0 & -1 \\ 0 & 1 & 2 \end{bmatrix}$$
I want to find the transition matrix $P$ such that $A=PRP^{-1}$
I thought we had to find $3$ independent vectors...one from the eigenspace of $1$, another from the eigenspace of $-1$, and then any other third vector such that the three would be linearly independent. So I chose P to be:
$$\begin{bmatrix} 2 & 2 & 1 \\-2 & -1 & 0 \\ 1 & 1 & 0 \end{bmatrix}$$
But when I multiplied $PRP^{-1}$, I did not get $A$...I'm not sure why.
I would appreciate it if anybody could tell me where I went wrong and how I can fix it.
Thanks in advance
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Hints:
We are given:
$$\tag 1 A = \begin{bmatrix} 3 & 4 & 0 \\-1 & -3 & -2 \\ 1 & 2 & 1 \end{bmatrix}$$
Jordan Form
*
*Correct on eigenvalues
*You need three independent eigenvectors, and one of those ends up being a generalized one.
*You would have, $A = P \cdot J\cdot P^{-1}$, where $J$ represents the Jordan Normal Form and is made up of the three eigenvalues plus the Jordan block and $P$ is made up of the three independent eigenvectors for the three eigenvalues.
We get:
$$A = P \cdot J \cdot P^{-1} = \begin{bmatrix} 2 & 2 & 1 \\ -2 & -1 & 0 \\ 1 & 1 & 0 \end{bmatrix} \cdot \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 0 & -1 & -1 \\ 0 & 1 & 2 \\ 1 & 0 & -2 \end{bmatrix}$$
Rational Canonical Form
The characteristic polynomial for $(1)$ is:
$$-1 + x + x^2 - x^3 = (-1 + x)^2 (1 + x)$$
If we do the invariant factor decomposition, we write:
$$xI - A = \begin{bmatrix} x-3 & -4 & 0 \\ 1 & x+3 & 2 \\ -1 & -2 & x-1 \end{bmatrix}$$
Using row and column operations (in $\mathbb{Q}[x]$), we arrive at:
$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 - x - x^2 + x^3 \end{bmatrix}$$
We see that since the minimal and characteristic polynomial have the same roots, the only possibilities for the minimal polynomial are $(-1 + x)^2 (1 + x)$. We can check that:
$$(A-I)^2(A+I) = \begin{bmatrix} 16 & 64 & 0 \\ -1 & -32 & -8 \\ 1 & 8 & 0 \end{bmatrix} \ne 0.$$
It follows that there are no other invariant factors for $A$ and the minimal polynomial is $(-1 + x)^2 (1 + x)$, which, of course, we see in the invariant factor decomposition.
Because the only invariant factor is $(-1 + x)^2 (1 + x)$, we can write the rational canonical form as:
$$R = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$$
There are several ways to find the matrix $P$, but I like using the row operations from the invariant factor decomposition. If we use those (work it to reduce to the matrix I show above), we get:
$$P = \begin{bmatrix} -2 & -6 & -18 \\ 0 & 0 & -8 \\ 1 & -1 & -7 \end{bmatrix}$$
From this we get:
$$\displaystyle P^{-1} = \begin{bmatrix} -\frac{1}{8} & \frac{3}{8} & \frac{3}{4} \\ -\frac{1}{8} & -\frac{1}{2} & -\frac{1}{4} \\ 0 & \frac{1}{8} & 0 \end{bmatrix}$$
We can verify:
$$P \cdot R \cdot P ^{-1} = \begin{bmatrix} 3 & 4 & 0 \\-1 & -3 & -2 \\ 1 & 2 & 1 \end{bmatrix} = A$$
From the comments, and from the book you mention (which I do not have), it should have worked, but I am not sure since I would like to read the details. There are several ways of approaching these and they are not trivial, nor are they unique.
Also, note the very important observation of the matrices $P$ for the Jordan versus the Rational Canonical forms, they are not the same.
Other references you might like to check out:
*
*Abstract Algebra, Dummit and Foote
*Linear Algebra, Edwards
*Algebra, Vivek Sahai, Bist Vikas
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a divisibility problem if $a$ is an integer , such that it is not divisible by 2 or 3, prove that $ 24 $ divides $ a^2+ 23$ .
I took cases ,case $1$ : when $a$ is divisible by only $2$ and not $3$ then we can write $a = 6k+2, 6k+4 $ and case 2: when $a= 6k+3$ and here $a$ is divisible by only $3$ and not $2$.
and then in each case,
I substituted values of $a$ in $a^2+23$ , but got nothing. I do think my case by- case analysis here is wrong, but surely I'm missing something .
please help by suggesting how can I improve my solution. No need to give me an elegant solution, only a few hints will be fine.
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Method $1:$
HINT:
Using Carmichael Function, $\lambda(24)=2$
If $(a,2)=(a,3)=1, (a,2^m\cdot3^n)=1$ for any integer $m\ge0,n\ge0$
So, $a^2\equiv1\pmod{24}$ if $(a,24)=1$
Now, $a^2+23\equiv a^2-1\pmod{24}$
Method $2:$
If $(a,3)=1, a\equiv\pm1\pmod 3$
$\implies a^2\equiv1\pmod 3\implies a^2+23\equiv24\pmod{3}\equiv0$
If $(a,2)=1, a=2b+1$(say),
$a^2=(2b+1)^2=8\frac{b(b+1)}2+1\equiv1\pmod 8$
$\implies a^2+23\equiv1+23\pmod 8\equiv 0$
So, $a^2+23$ is divisible by $3,8$ so will be divisible by lcm$(3,8)=24$
Method $3:$
HINT:
$$a^2+23=(a-1)(a+1)+24$$
Observe that $(a-1)a(a+1)$ being a product of $3$ three consecutive integers must be divisible by $3$
But as $(a,3)=1\implies 3$ divides $(a-1)(a+1)$
Again, as $(a,2)=1, a\pm1$ are even, one of them will be divisible by $2$, the other by $4$
$\implies 2\cdot4$ divides $(a-1)(a+1)$
So, $(a-1)(a+1)$ is divisible by lcm$(3,8)=24$
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How to show that $A^3+B^3+C^3 - 3ABC = (A+B+C)(A+B\omega+C\omega^2)(A+B\omega^2+C\omega)$ indirectly? I found this amazingly beautiful identity here. How to prove that $A^3+B^3+C^3 - 3ABC = (A+B+C)(A+B\omega+C\omega^2)(A+B\omega^2+C\omega)$ without directly multiplying the factors? (I've already verified it that way). Moreover, how could someone possibly find such a factorization using complex numbers? Is it possible to find such a factorization because $A^3+B^3+C^3 - 3ABC$ is a symmetric polynomial in $A,B,C$?
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HINT:
If $A+Bw+Cw^2=0$ where $w$ is one of the three cube roots of unity
$\implies -A=Bw+Cw^2$
Cubing we get, $(-A)^3=(Bw+Cw^2)^3$
$\implies -A^3=B^3w^3+C^3w^6+3\cdot Bw\cdot Cw^2(Bw+Cw^2)=B^3+C^3+3BC(-A)$
$\implies A+Bw+Cw^2$ is a factor of $A^3+B^3+C^3-3ABC$
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How to calculate the intersection of two planes? How to calculate the intersection of two planes ?
These are the planes and the result is gonna be a line in $\Bbb R^3$:
$x + 2y + z - 1 = 0$
$2x + 3y - 2z + 2 = 0$
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Let $x$ be in the intersection of the planes
\begin{equation*}
\langle m, x-b \rangle = 0,\\
\langle n, x-c \rangle = 0
\end{equation*}
and let $p = m \times n$.
Then for some $\lambda$,
\begin{equation*}
\begin{pmatrix}
m_0 & m_1 & m_2 \\
n_0 & n_2 & n_3 \\
p_0 & p_1 & p_2 \\
\end{pmatrix}
\begin{pmatrix}
x_0 \\ x_1 \\ x_2
\end{pmatrix}
=
\begin{pmatrix}
\langle m, b \rangle \\
\langle n, c \rangle \\
\lambda
\end{pmatrix}.
\end{equation*}
Using the vector identities
\begin{equation*}
\begin{split}
u \times (v \times w) &= \langle u, w \rangle v - \langle u, v \rangle w, \\
\Vert u \times v \Vert^2 &= \Vert u \Vert^2 \Vert v \Vert^2 - \langle u, v \rangle^2,
\end{split}
\end{equation*}
check that
\begin{equation*}
\begin{pmatrix}
m_0 & m_1 & m_2 \\
n_0 & n_1 & n_2 \\
p_0 & p_1 & p_2 \\
\end{pmatrix}
\begin{pmatrix}
q_0 & r_0 & p_0 \\
q_1 & r_1 & p_1 \\
q_2 & r_2 & p_2 \\
\end{pmatrix}
= \Vert p \Vert^2 I,
\end{equation*}
where
\begin{equation*}
\begin{split}
q &= -n \times (n \times m), \\
r &= -m \times (m \times n).
\end{split}
\end{equation*}
If $p \ne 0$ (that is, if $m$ and $n$ are linearly independent), it follows that
\begin{equation*}
\begin{pmatrix}
x_0 \\ x_1 \\ x_2
\end{pmatrix}
= \frac{1}{\Vert p \Vert^2}
\begin{pmatrix}
q_0 & r_0 & p_0 \\
q_1 & r_1 & p_1 \\
q_2 & r_2 & p_2 \\
\end{pmatrix}
\begin{pmatrix}
\langle m, b \rangle \\
\langle n, c \rangle \\
\lambda
\end{pmatrix},
\end{equation*}
or
\begin{equation*}
x = a + \lambda \frac{p}{\Vert p \Vert^2}
\end{equation*}
where
\begin{equation*}
a = (\langle m, b \rangle q + \langle n, c \rangle r) / {\Vert p \Vert^2}.
\end{equation*}
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How many zeros are there in $1 \times 2^2 \times 3^3 \times \cdots \times 100^{100}$? How many zeros are there at the end of $1 \times 2^2 \times 3^3 \times \cdots \times 100^{100}$ ?
I tried it by grouping all the $2$'s and $5$'s and $5$'s and $6$'s but cant get my answer...
|
To get a zero you need to multiply a factor $5$ by a factor $2$. There are clearly more factors of $2$ than of $5$ in the product, so you need to find a systematic way of counting the factors of 5.
Well these come from the terms $5^5\cdot 10^{10} \cdot 15^{15} \dots 100^{100}$
All of these terms provide a factor $5$ so we have $5^5\cdot 5^{10} \cdot 5^{15} \dots 5^{100}$, with exponent $5+10+15+ \dots +100$.
However the multiples of $25$ contribute twice, and we've only counted them once so far, so we need to add $25+50+75+100$ to the exponent.
We have no multiples of $5^3=125$ which would have contributed three factors.
The exponent counts the number of factors of $5$ and hence the number of zeros.
|
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|
On seventh powers $x_1^7+x_2^7+\dots+x_n^7 = 2$? We have,
$$(-6m^3 + 1)^3 + (6m^3 + 1)^3 + (-6m^2)^3 = 2$$
$$(-8m^5 + 1)^5 + (8m^5 + 1)^5 + (-8m^6 + 2m)^5 + (-8m^6 - 2m)^5 + 2(8m^6)^5=2$$
The first identity has been long known, while the second is by Ajai Choudhry. Anybody knows if there is anything similar for 7th powers?
|
The best I can do is in terms of the radical $\sqrt{3}$:
$$2=(9 m^7 + 1)^7 + (-9 m^7 + 1)^7 + (\sqrt{3} m - 9 m^8)^7 + (-\sqrt{3} m -
9 m^8)^7 + 2 (9 m^8)^7$$
|
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|
Direction and Magnitude of a Dog Running Problem: A dog in an open field runs 12.0m east and then 30.0m in a direction 54 degrees west of north.
Part A: In what direction must the dog then run to end up 12.0m south of her original starting point?
My Answer: Let the first path the dog runs A, the second path B, the third path C, and $R=A+B+C$.
$A_x=12$, $A_y=0$, $B_x=-30\cos(54)$, $B_y=30\sin(54)$, $R_x=0$, and $R_y=-12$
So $C_x=R_x-A_x-B_x=0+30\cos(54)-12$ and $C_y=R_y-A_y-B_y=-12-30\sin(54) + 0$.
Thus, $\arctan(30\cos(54)-12)/(-12-30\sin(54))$ which is equal to -8.8 east of south.
Is that correct?
Part B: How far must the dog then run to end up 12.0 south of her original starting point?
Using the components above, we find the magnitude of C.
The magnitude of C is $\sqrt{(30\cos(54)-12)^2+(-12-30\sin(54))^2}$ which is equal to 30.2m
Is that correct?
|
Almost. Note that the angle of path $B$ is $54^\circ$ West of North (not North of West). Hence, you mixed up your $\sin$ and $\cos$. You should obtain:
\begin{align*}
C_x &= 30\sin(54^\circ)-12 \\
C_y &= -30\cos(54^\circ)-12 \\
\theta &= \arctan(|C_x/C_y|) = \arctan \left(\frac{30\sin(54^\circ)-12}{30\cos(54^\circ)+12} \right) =22.49...^\circ \text{ East of South} \\
C &= \sqrt{C_x^2 + C_y^2} = \sqrt{(30\sin(54^\circ)-12)^2 + (-30\cos(54^\circ)-12)^2} = 32.07... \text{metres}
\end{align*}
|
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|
Compute $ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$ Compute $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$$
Of course we have $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \left( \frac{x}{1+x^2} \right) ^2 dx = \lim_{ A \to \infty } \int_{0}^{A} \left( \frac{x}{1+x^2} \right) ^2 dx $$
I think that firstly I should compute $ \int \left( \frac{x}{1+x^2} \right) ^2 dx $ but I don't have idea.
|
As the denominator contains $x^2+1$ we put $x=\tan y\implies dx=\sec^2ydy$
For $x^2-1$ with $x\ge1$ we need to put $x=\sec y$
and for $1-x^2$ with $x\le1$ we need to put $x=\sin y$
$$\implies \int\frac{x^2dx}{(x^2+1)^2}=\int\frac{\tan^2y\sec^2ydy}{\sec^4y}=\int\sin^2ydy=\frac12\int(1-\cos2y)dy=\frac{y}2-\frac{\sin2y}4+C$$
If it were indefinite integral, we had to replace back $y$ with $x$
$y=\arctan x$ and $\sin2y=\frac{2\tan y}{1+\tan^2y}=\frac{2x}{1+x^2}$
For definite integral, $x=0,y=\arctan 0=0$ and $x=\infty,y=\arctan \infty=\frac\pi2$
So, the required definite integral will be $$2\left[\frac{y}2-\frac{\sin2y}4+C\right]_0^{\frac\pi2}=2\cdot\frac\pi4=\frac\pi2$$
|
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|
Show that this relation is an implicit solution of the following differential equation Differential equation:
$$\frac{dy}{dx}=\frac{xy}{x^2+y^2}$$
Relation:
$$2y^2 \ln{y} - x^2 = 0$$
From this, I end up getting:
$$\frac{dy}{dx} = \frac{x}{2y\ln{y} + y} $$
The missing step would be to put in the form given at the top. How is this possible?
|
Multiply by $y/y$ first. Note from our relation $2y^2\log y-x^2=0$ that adding $x^2$ to both sides yields $2y^2\log y=x^2$. Substitute for $2y^2\log y$ and you are done.
$$\begin{align*}\frac{dy}{dx}&=\frac{x}{2y\log y+y}\\&=\frac{xy}{2y^2\log y+y^2}\\&=\frac{xy}{x^2+y^2}\end{align*}$$
|
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|
Integrating $\cos^3 (x) \, dx$ I am wondering whether I integrated the following correctly.
*
*$\int \cos^3 x \, dx$
I did
\begin{align}
\int \cos^3 x \, dx &= \int \cos(x)(1-\sin^{2}x) \, dx \\
&= \int \cos(x)-\sin^{2}x \cos x \, dx \\
&= \sin(x)-\frac{u^{3}}{3} + c, \quad(u=\sin(x)) \\
&= \sin(x)-\dfrac{\sin^{3}x}{3}+c
\end{align}
2.$\int \sin^{3}x \cos^{2}x\,dx$
\begin{align}
\int \sin^{3}x \cos^{2}x\,dx &= \int(1-\cos^2x)(\cos^2x)\sin(x)\,dx \\
&= \int \cos^2x\sin(x)-\cos^4x\sin(x)\,dx, \quad u=\cos(x) \\
&= \dfrac{u^3}{3}-\dfrac{u^5}{5}+c
\end{align}
And plug in my u.
|
Your first integral is correct.
The second has two sign errors: $$u = \cos x \implies du = -\sin x\,dx$$
So evaluating the second integral should yield
$$\begin{align} \int \Big(\cos^2x\sin x-\cos^4x\sin x\Big)\,dx & = -\int \cos^2x\Big(-\sin x\,dx\Big) - (-)\int \cos^4x\Big(-\sin x\,dx\Big) \\ \\& = -\int u^2 du + \int u^4\,du \\ \\ & = -\frac{u^3}{3} + \frac{u^5}{5}+c\quad\text{or}\quad \frac{u^5}{5} - \frac{u^3}{3}+c\end{align}$$
|
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|
How find this inequality $\sqrt{a^2+64}+\sqrt{b^2+1}$ let $a,b$ are positive numbers,and such $ab=8$ find this minum
$$\sqrt{a^2+64}+\sqrt{b^2+1}$$
My try:
and I find when $a=4,b=2$,then
$$\sqrt{a^2+64}+\sqrt{b^2+1}$$ is minum $5\sqrt{5}$
it maybe use Cauchy-Schwarz inequality
Thank you
|
by the Cauchy-Schwarz inequality,we have
$$(a^2+64)(1+4)\ge(a+16)^2$$
$$(b^2+1)(4+1)\ge (2b+1)^2$$
then
$$\sqrt{a^2+64}+\sqrt{b^2+1}\ge\dfrac{1}{\sqrt{5}}(a+2b+17)$$
and By AM-GM,we have
$$a+2b\ge 2\sqrt{2ab}=8$$
so
$$\sqrt{a^2+64}+\sqrt{b^2+1}\ge 5\sqrt{5}$$
|
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|
Inequalities question I've been having trouble solving these kind of inequalities ;
$\sqrt { -7x+1 } - \sqrt {x+10} \gt \ {7}$
Attempt at a Solution; We first find the official boundaries of the inequality; x is limited to the section: $\ -10 \le x \le \frac{1}{7} $
Then, squaring both sides we get:
${-7x+1} -2 \sqrt{(x+10)}\sqrt{(-7x+1)} +x+10 \gt 49 $
$-2 \sqrt{(x+10)}\sqrt{(-7x+1)} \gt 38+6x $
squaring again eventually yields:
$ -28x^2-276x+40\gt 1444+456x+36x^2 $
$-64x^2 -732x -1404 \gt 0 $
binomial roots are as specified below.
since the parabola is concave and in order this inequality is to be satisfied,
x must assume the values:$\ -9 \lt x \lt -2.4375 $ upon intersecting this with the range we get the same result and this is the (apparent) final solution.
We then resume to the usual manipulation of the inequality; squaring both sides, factoring etc. We get two roots; $-9$,$ -2.4375$ . The parabola we get after these manipulations is concave, so we get the solution: $\ -9 \lt x \lt -2.4375 $ . We then intersect this solution with the range for x, yielding the final solution $\ -9 \lt x \lt -2.4375 $. Upon testing it we find it does not fulfill the inequality we tried to solve, so we cancel it as a solution, leading to that there are no solutions to this inequality, as a final answer. Official answers state otherwise.
Any help?
|
From $\sqrt{-7x+1} - \sqrt{x+10} \gt 7$, your assessment of the interval $x \in [-10, \frac{1}{7}]$is clearly right.
Let us make sure we have positive quantities on both sides first, so rewrite as
$\sqrt{-7x+1} \gt 7 + \sqrt{x+10} $
Squaring, $- 7x + 1 > 49 + x + 10 + 14\sqrt{x+10} $
or $-4x -29 > 7\sqrt{x+10} $
Now it is easier to see that the LHS is positive only when $x < -\frac{29}{4}$, so the allowable values have shrunk to $x \in [-10, -\frac{29}{4}]$. In this interval, we have assurance of both sides being positive, so squaring again, we have
$16x^2 + 232x + 841 > 49(x+10) \implies 16x^2 + 183x + 351 > 0$ So we have
$$16\left(x+\dfrac{183}{32}\right)^2 > \dfrac{11025}{64} \implies x +\dfrac{183}{32} \not \in \left[- \frac{105}{32}, \frac{105}{32} \right] $$
Thus the allowable values are further shrunk to $x \in [-10, -9)$.
|
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|
Is this correct method to prove that $a^2 + b^2 + c^2 ≥ ab + bc + ac$, when $a,b,c \geq 0$? Can I prove it like this: Let's say that $a=b=c$ so we get "If $a \geq 0$ then $3a^2 ≥ 3a^2$" Now I take the negation of that statement and get "If $a \geq 0$ then $3a^2 < 3a^2$" The anti-thesis is obviously wrong which makes the original thesis right? Is this a correct way to do this, if not can you give some tips for what to do. I am not looking for complete solution as these are my homework and I really need to practice.
|
\begin{align}
a^2 + b^2 + c^2 - ab - bc - ca
&= (1/2)(2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca)\\
&= (1/2)(a^2 - 2ab + b^2 + b^2 - 2bc + c^2 + c^2 - 2ca + a^2)\\
&= (1/2)[(a-b)^2 + (b-c)^2 + (c-a)^2]
\end{align}
Now we know that, the square of a number is always greater than or equal to zero.
Hence, $(1/2)[(a-b)^2 + (b-c)^2 + (c-a)^2] ≥ 0 \implies$ it is always non-negative.
|
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|
How to get all solutions to equations with square roots I would like to find all solutions to
$$b-a\sqrt{1+a^2+b^2}=a^2(ab-\sqrt{1+a^2+b^2})$$
$$a-b\sqrt{1+a^2+b^2}=b^2(ab-\sqrt{1+a^2+b^2})$$
I found some solutions. For example, $a = 1, b = \pm i$ and $b = 1 , a = \pm i$. How can I find all solutions?
|
One technique with this kind of problem is to isolate the square root, and then square the equation (which brings in possible extra roots from the negative value of the square root).
[Now corrected for error pointed out by Anush in copying over the equations]
Beginning that process here we find that the first equation becomes: $$b(1-a^3)=(a-a^2)\sqrt{1+a^2+b^2}$$ and notice that the factor $a-1$ will cancel, provided that $a\neq 1$ so that$$b(1+a+a^2)=a\sqrt{1+a^2+b^2}$$
Similarly, from the second equation, if $b\neq 1$ we have $$a(1+b+b^2)=b\sqrt{1+a^2+b^2}$$
Now we can eliminate the square root by subrtacting $a\times$ the second equation from the $b \times$ first to obtain $$(b^2-a^2)+ab(b-a)=(b-a)(a+b+ab)=0$$
So that either $ab+a+b=0$ or $a=b$
This is substantial progress.
If $a=b$ we note the solution $a=b=0$, and assume that $a,b\neq 0, 1$. The two equations are the same and we need to solve $$(1+a+a^2)=\sqrt{1+2a^2}$$
If we square this we obtain $$1+2a+3a^2+2a^3+a^4=1+2a^2$$ which becomes $$2a+a^2+2a^3+a^4=0$$ or $$a(2+a)(a^2+1)=0$$
$a=0$ we know about. $a=-2, b=-2$ is another solution. $a=b=\pm i$ works in the original equation for $a=b=i$ but not for $a=b=-i$.
If now $a+b+ab=0$, the term $b(1+a+a^2)=b+ab+a^2b=-a + a^2b=a(ab-1)$ so we have
$a(ab-1)=a\sqrt{a^2+b^2+1}$ and if $a\neq 0$ we can cancel and square to obtain $$(ab-1)^2=a^2+b^2+1$$ Now we can write $ab=-(a+b)$ so that the equation becomes $$a^2+b^2+1+(2ab+2a+2b)=a^2+b^2+1$$ The term in brackets is zero, so the equation is an identity and gives no additional constraint.
Note that the condition $ab+a+b=0$ is equivalent to $(a+1)(b+1)=1$, and if $b\neq -1$ this is also $a=\cfrac {-b}{b+1}$
Some checking is required to identify which solutions get the right sign for the square root - assuming the solutions are real numbers (and complex solutions have been canvassed).
We know we have $a^2+b^2+1=(ab-1)^2$ so that $\sqrt {a^2+b^2+1}=|ab-1|$, so for a solution we need $ab\gt 1$. This in turn means $\frac{-b^2}{b+1} \gt 1$.
If $b+1\gt 0$ this reduces to $-b^2\gt b+1\gt 0$ which is impossible. So $b+1\lt 0$ (ignoring $b=-1$) and $$-b^2\lt b+1$$ or $b^2+b+1\gt 0$ which is equivalent to $(2b+1)^2+3\gt 0$, which is always true.
So we have the condition $b\lt -1$ and from the form $(a+1)(b+1)=1$ we see that this also implies $a\lt-1$
|
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Find $(a,b)$ if $x^2-bx+a = 0, x^2-ax+b = 0$ both have distinct positive integers roots
If $x^2-bx+a = 0$ and $x^2-ax+b = 0$ both have distinct positive integers roots, then what is $(a,b)$?
My Try:
$$\displaystyle x^2-ax+b = 0\Rightarrow x = \frac{a\pm \sqrt{a^2-4b}}{2}$$
So here $a^2-4b$ is a perfect square.
Similarly
$$\displaystyle x^2-bx+a = 0\Rightarrow x = \frac{b\pm \sqrt{b^2-4a}}{2}$$
So here $b^2-4a$ is a perfect square.
But I did not understand how can I solve after that.
|
Let the roots of $x^2 - ax +b$ be $r$ and $s$, and the roots of $x^2 - bx + a$ be $u$ and $v$. Then
$$\begin{align}
r+s &= a\\
rs &= b\\
u+v &= b\\
uv &= a.
\end{align}$$
Let, without loss of generality, $a \leqslant b$. Thus
$$\begin{align}
uv &\leqslant u+v\\
\iff uv - u - v + 1 & \leqslant 1\\
\iff (u-1)(v-1) & \leqslant 1.
\end{align}$$
So $u = 1$ or $v = 1$ or $u = v = 2$. But the roots are supposed to be distinct, hence $u = 1$ or $v = 1$. Without loss of generality, $u = 1$.
Thus $a = v = r+s$, $b = v+1 = rs$, so
$$rs = r+s+1 \iff (r-1)(s-1) = 2,$$
and that leaves $r = 2$, $s = 3$ (or vice versa), so $a = 5, b = 6$.
|
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|
Finding vertex and focus of parabola given an equation I am defeated to complete square on the following parabolic equation. Please help.
Find the vertex and focal width for the parabola: $$ x^2+6x+8y+1=0 $$
I am hoping to get an equation in this form
$$(x−h)^2=4p(y−k)$$
|
$$x^2 + 6x \color{blue}{+ 9} + 8y + 1 \color{blue}{- 9} = 0$$
$$(x + 3)^2 + 8(y - 1) = 0$$
$$(x + 3)^2 = -8(y - 1) = 4(-2)(y-1)$$
Now, you should be able to "read off" the vertex of the parabola. From there, see if you can find.
With respect to completing the square: you have
$$(x + 3)^2 + 8 y + 1 = 9$$
Subtract $9$ from both sides of the equation. $$\begin{align} (x + 3)^2 + 8y + 1 - 9 = 0 & \iff (x+3)^2 + 8y - 8 = 0 \\ \\ & \iff (x+3)^2 + 8(y - 1) = 0 \end{align}$$
Then subtract $8(y - 1)$ from each side of the equation to get the form you need:
$$(x + 3)^2 = -8(y - 1) = 4(-2)(y - 1)$$
$$(x + 3)^2 = 4(-2)(y - 1)$$
So, $$p = -2, \;h = -3,\; k = 1$$
|
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|
Integral (square root function multiplied by exponential function) did i do it right? I'm trying to determine $\int x^3\sqrt{x^2 +1}\, dx$
I said that $u(x) = x^2 + 1$
and then that $dx = 2x\,dx$
so I rewrote the integral as
$$\int x^3\sqrt{x^2 +1}\,2x\,dx$$
which is also
$$\int2x^4\sqrt{x^2 +1} \,dx$$
and then it is easy to integrate, is that all legal to do?
|
\begin{align}
&\int x^{3}\,\sqrt{x^{2} + 1\,}\,{\rm d}x
=
\int x^{2}\,{\rm d}\left[{1 \over 3}\,\left(x^{2} + 1\right)^{3/2}\right]
\\[3mm]&=
x^{2}\,{1 \over 3}\left(x^{2} + 1\right)^{3/2}
-
\int{1 \over 3}\left(x^{2} + 1\right)^{3/2}
\,{\rm d}\left(x^{2} + 1\right)
\\[3mm]&=
{1 \over 3}\,x^{2}\left(x^{2} + 1\right)^{3/2}
-
{1 \over 3}\,{\left(x^{2} + 1\right)^{5/2} \over 5/2}
=
\left(x^{2} + 1\right)^{3/2}\left[%
{1 \over 3}\,x^{2}
-
{2 \over 15}\left(x^{2} + 1\right)
\right]
\\[3mm]&=
\left(x^{2} + 1\right)^{3/2}\,{3x^{2} - 2 \over 15}
\end{align}
$$
\begin{array}{|c|}\hline\\
\color{#ff0000}{\large\quad%
\int x^{3}\,\sqrt{x^{2} + 1\,}\,{\rm d}x
\color{#000000}{\ =\ }
{1 \over 15}\left(3x^{2} - 2\right)\left(x^{2} + 1\right)^{3/2}\
+\
\color{#000000}{\mbox{constant}}
\quad}
\\ \\ \hline
\end{array}
$$
|
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|
Can an integer of the form $4n+3$ written as a sum of two squares? Let $u$ be an integer of the form $4n+3$, where $n$ is a positive integer. Can we find integers $a$ and $b$ such that $u = a^2 + b^2$? If not, how to establish this for a fact?
|
Let's assume x^2+y^2 = 4n+3, then either x or y has to be even. Let's assume x = 2z and write
(2z)^2+y^2 = 4n+3. This can also be written as follows:
(2z+y)^2-4zy = 4n+3 by rearrangement we can write
(2z+y)^2-1^2 =4n+2+4zy
(2z+y)^2-1^2=2(2n+2zy+1) and further
(2z+y-1)(2z+y+1)=2(2n+2zy+1)
The left side is product of two even numbers the right side is the product of even and odd number. So the assumption is wrong, and no number in the form of 4n+3 can be a sum of two squares.
|
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|
Finding sequence that's defined by a recurrence relation This is the problem that I ran into when doing practice in my textbook.
How do I find the sequence (call it $a_n$) that's defined by recurrence relation whose generating function is $\frac 4 {-x^2-2x+3}$?
Help appreciated!
|
We use partial fractions. So we want to find integers $A$ and $B$ such that
$$\frac{4}{-x^2 -2x+3}=\frac{A}{3+x}+\frac{B}{1-x}.$$
There are general procedures, but we can see that $A=1$, $B=1$ work.
Now
$$\frac{1}{1-x}=1+x+x^2+x^3+\cdots=\sum_0^\infty x^n.$$
Expanding $\frac{1}{3-x}$ takes more work. Rewrite as
$$\frac{1}{3}\frac{1}{1-\left(-\frac{x}{3}\right)}.$$
When we expand we get
$$\frac{1}{3}\sum_0^\infty (-1)^n\frac{1}{3^n}x^n.$$
Put the pieces together to get the coefficient of $x^n$.
|
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The sum of an infinite series with integral $1+\dfrac{1}{9}+\dfrac{1}{45}+\dfrac{1}{189}+\dfrac{1}{729}+\dots=\sum\limits_{n=1}^\infty \dfrac{1}{(2n-1)\cdot 3^{n-1}}$
I got:
$\sum\limits_{n=1}^\infty \dfrac{1}{(2n-1)\cdot 3^{n-1}}=\sum\limits_{n=1}^\infty \dfrac{\int\limits_0^1 x^{2n-2}\,dx}{3^{n-1}}=\dots$
And no idea how to impove.
Thx!
|
After writing $\frac{1}{2n-1}$ as an integral, you have the series
$$\sum_{n=1}^\infty \int_0^1 \left(\frac{x^2}{3}\right)^{n-1}\,dx.$$
Since the geometric series
$$\sum_{k=0}^\infty \left(\frac{x^2}{3}\right)^k$$
converges uniformly on the interval $[0,1]$, we can interchange summation and integration, and obtain
$$\sum_{n=1}^\infty \frac{1}{(2n-1)3^{(n-1)}} = \int_0^1 \frac{3}{3-x^2}\,dx.$$
The integral can be evaluated in different ways. With partial fraction decomposition
$$\frac{3}{3-x^2} = \frac{\sqrt{3}}{2}\left(\frac{1}{\sqrt{3}-x} + \frac{1}{\sqrt{3}+x}\right)$$
we get
$$\int_0^1 \frac{3}{3-x^2}\,dx = \frac{\sqrt{3}}{2}\left[\log (\sqrt{3}+x) - \log (\sqrt{3}-x)\right]_0^1 = \frac{\sqrt{3}}{2}\log \frac{\sqrt{3}+1}{\sqrt{3}-1} = \frac{\sqrt{3}}{2}\log (2+\sqrt{3}).$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Proving $\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac19}-\sqrt[3]{\frac29}+\sqrt[3]{\frac49}$ I found the following two relational expressions in a book without any additional information:
$$\sqrt{\sqrt[3]{5}-\sqrt[3]{4}}=\frac13(\sqrt[3]{2}+\sqrt[3]{20}-\sqrt[3]{25})$$
$$\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac19}-\sqrt[3]{\frac29}+\sqrt[3]{\frac49}$$
Wolfram tells these are true, but I can't prove at all. Can anyone help?
|
For the identity in the title, use the identity $x^3+1=(x+1)(x^2-x+1)$.
Let $x=\sqrt[3]{2}$, then $(x^2-x+1)(x+1)=x^3+1=3$ hence the RHS is $\sqrt[3]{3}/(x+1)$ and the LHS divided by the RHS is the cube root of one third of $(x-1)(x+1)^3=(x^3-2)(x+2)+3$. Since $x^3=2$, this is $3$ and the result follows.
For the other identity:
Let $x=\sqrt[3]{5/4}$, then the identity to prove is $3\sqrt{x-1}=1+2x-2x^2$. Squaring both sides, this is equivalent to $1+4(x-x^2)+4(x-x^2)^2-9(x-1)=(5-4x^3)(2-x)$ being zero. Since $4x^3=5$, the result follows.
|
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Constant-Coefficient Systems. Find a real general solution of the following system. Find a real general solution of the following system. (Show the
details.)
$y'_1 = 9 y_1 + 13.5 y_2 \;\;\;\;\;\;\; y′_2 = 1.5 y_1 + 9 y_2$
$y' = \begin{pmatrix} 9& 13.5\\1.5& 9\end{pmatrix}$y
$\det$ (A - $\lambda$I$)= \begin{pmatrix} 9-\lambda& 13.5\\1.5& 9-\lambda\end{pmatrix}$ = $\lambda^2 - 18\lambda+60.75$
$\lambda \Rightarrow \lambda_1 = 13.5\;\;\;\lambda_2 = 4.5\;\;(eigen- values) $
Eigen vectors
$For \;\lambda = 13.5$
I get
$\begin{pmatrix} -4.5& 13.5\\1.5& -4.5\end{pmatrix}$$\begin{pmatrix} 0\\0 \end{pmatrix}$
Using Gaussian elimination I get:
$\begin{pmatrix} -1& 3\\0& 0\end{pmatrix}$$\begin{pmatrix} 0\\0 \end{pmatrix}$
Which implies that:
$-x_1 + 3x_2 =0 $
$Let\;\; x_2 = t$
$\therefore\;\;x_1 = 3t$
Hence
$\begin{pmatrix} x_1\\x_2 \end{pmatrix}$$\begin{pmatrix} 3\\1 \end{pmatrix}$t
Thus the eigen vectors are:
$x^{(1)}$=$\begin{pmatrix} 3\\1 \end{pmatrix}$
$For \;\lambda = 4.5$
$\begin{pmatrix} 4.5& 13.5\\1.5& 4.5\end{pmatrix}$$\begin{pmatrix} 0\\0 \end{pmatrix}$
Using Gaussian elimination I get:
$\begin{pmatrix} 1& 3\\0& 0\end{pmatrix}$$\begin{pmatrix} 0\\0 \end{pmatrix}$
Which implies that:
$x_1 + 3x_2 =0 $
$Let\;\; x_2 = t$
$\therefore\;\;x_1 = -3t$
Hence
$\begin{pmatrix} x_1\\x_2 \end{pmatrix}$$\begin{pmatrix} -3\\1 \end{pmatrix}$t
Thus the eigen vectors are:
$x^{(2)}$=$\begin{pmatrix} -3\\1 \end{pmatrix}$
My instructors solution for $\;for \;\lambda = 4.5$
$x^{(2)}$=$\begin{pmatrix} 3\\-1 \end{pmatrix}$
I just want to know if my working is correct or there is another simple and better
way. Just confused with instructors answers. Get confused when calculating eigen vectors because my solutions does not match instructors solution for other same type of questions.
|
We are given:
$$A = \begin{bmatrix}9 & 27/2\\3/2 & 9\end{bmatrix}$$
The characteristic equation and eigenvalues are found by solving $[A-\lambda I] = 0$, which gives:
$$1/4 (2 \lambda-27) (2 \lambda-9) = 0 \rightarrow \lambda_1 = \dfrac{27}{2}~,\lambda_2 = \dfrac{9}{2}$$
We now find the eigenvectors for each distinct eigenvalue by solving $[A-\lambda_i I]v_i = 0$.
For $\lambda_1 = 27/2$, we have the RREF of:
$$\begin{bmatrix} 1 & -3 \\ 0 & 0 \end{bmatrix}v_1 = 0$$
This leads to:
$a - 3b = 0 \rightarrow a = 3b, ~\mbox{so let}~ b = 1 \rightarrow a = 3$, so our eigenvector is:
$$v_1 = \begin{bmatrix}3\\ 1 \end{bmatrix}$$
Repeating this process for $\lambda_1 = 9/2$, we have the RREF of:
$$\begin{bmatrix} 1 & 3 \\ 0 & 0 \end{bmatrix}v_2 = 0$$
This leads to:
$a + 3b = 0 \rightarrow a = -3b, ~\mbox{so let}~ b = 1 \rightarrow a = -3$, so our eigenvector is:
$$v_2 = \begin{bmatrix} -3 \\ 1 \end{bmatrix}$$
Here is one approach to finding the exponential matrix using the eigenvalues / eigenvectors.
We can write the solution to this system as:
$$x(t) = \begin{bmatrix} x_1(t) \\ x_2(t) \end{bmatrix} = c_1 e^{27t/2 }\begin{bmatrix} 3 \\ 1 \end{bmatrix}+ c_2 e^{9t/2 }\begin{bmatrix} -3 \\ 1 \end{bmatrix}$$
This gives us a fundamental matrix of:
$$\phi(t) = \begin{bmatrix} 3e^{27t/2} & -3e^{9t/2} \\ e^{27t/2} & e^{9t/2} \end{bmatrix}$$
From this, we can find the matrix exponential using:
$$e^{A t} = \phi(t)(\phi(0))^{-1} = \begin{bmatrix}1/2(e^{9t/2} + e^{27t/2}) & 3/2 (-e^{9t/2} + e^{27t/2})\\ ~1/6(-e^{9t/2} + e^{27t/2}) & 1/2(e^{9t/2} + e^{27t/2}) \end{bmatrix}$$
|
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|
Convergence of $x_{n}=\frac{n^2}{\sqrt{n^{6}+1}}+\frac{n^2}{\sqrt{n^{6}+2}}+\cdots+\frac{n^2}{\sqrt{n^{6}+n}}$ Let $$x_{n}=\frac{n^2}{\sqrt{n^{6}+1}}+\frac{n^2}{\sqrt{n^{6}+2}}+\frac{n^2}{\sqrt{n^{6}+3}}+...+\frac{n^2}{\sqrt{n^{6}+n}}.$$ Then $(x_{n})$ converges to
(A)$1$
(B)$0$
(C)$\frac{1}{2}$
(D)$\frac{3}{2}$
My Try:
$$\lim_{n\to \infty}x_{n}=\frac{n^2}{n^{3}}(\frac{1}{\sqrt{1+\frac{1}{n^6}}}+\frac{1}{\sqrt{1+\frac{2}{n^{6}}}}+...+\frac{1}{\sqrt{1+\frac{1}{n^5}}})=0.$$
Am i right?
|
You can use the following inequality:
$$
\frac{n.n^2}{\sqrt{n^{6}+n}}\leq \frac{n^2}{\sqrt{n^{6}+1}}+\frac{n^2}{\sqrt{n^{6}+2}}+\frac{n^2}{\sqrt{n^{6}+3}}+...+\frac{n^2}{\sqrt{n^{6}+n}}\leq \frac{n.n^2}{\sqrt{n^{6}+1}}
$$
Then take the limit of both sides and you can see the limit is 1.
|
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|
Find all natural numbers n such that $7n^3 < 5^n$ Find all natural numbers $n$ such that $7n^3 < 5^n$.
I drew a graph which showed that $n \geq 4$
wolfram
How can I prove that? I guess I need to use induction with the base case $n=4$?
But I am stuck because the induction hypothesis uses a $\geq$ sign so I can not substitute...
|
Base Case: For $n=4$, we have $7(4)^3 = 448 < 625 = 5^4$, which works.
Induction Hypothesis: Assume that the claim holds true for all $n \in \{4,\ldots,k\}$, where $k>3$.
It remains to prove the inequality true for $n=k+1$. Indeed, observe that:
\begin{align*}
7(k+1)^3 &= 7(k^3 + 3k^2 + 3k + 1) \\
&< 7(k^3 + 3k^2 + 9k + 27) \\
&= 7(k^3 + (3)k^2 + (3)^2k + (3)^3) \\
&< 7(k^3 + (k)k^2 + (k)^2k + (k)^3) & \text{since }3 < k\\
&= 7(4k^3) \\
&= 4(7k^3) \\
&< 5(7k^3) \\
&< 5(5^k) & \text{by the induction hypothesis}\\
&= 5^{k+1}\\
\end{align*}
as desired. This completes the induction.
|
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Geometric/Visual Solution - Shortest Vector for which Dot Product = x + 2y = 5. (Strang P21 1.2.26) Much as I referenced this same exercise, I'm questing after an exclusively geometric solution.
Question: If $\color{#0070FF}{\vec{v} = (1,2)}$ draw all vectors $\vec{w} = (x,y)$ in the plane $x,y$ with $\color{#0070FF}{\vec{v}}\cdot\vec{w} = \color{#228B22}{x + 2y = 5}$. Which is the shortest $\vec{w}$?
Algebraic solution: It's given that $\mathbf{v \cdot w} = \color{#228B22}{x + 2y = 5}$. Thus, $\mathbf{v}$ forsooth lies on $x + 2y = 5$. We need an upper bound on $|\mathbf{w}|$, thus apply the Cauchy-Schwarz Inequality. $ \mathbf{v \cdot w} \leq |\mathbf{v}| \mid\mathbf{w}| \iff \cfrac{\mathbf{v \cdot w}}{|\mathbf{v}|} \leq |\mathbf{w}| \tag{CS$\neq$}$
Thus, $min \;\mathbf{w} \iff $ (CS$\neq$) becomes an equality $\iff$ $\mathbf{v}, \mathbf{w}$ are collinear $\iff k\mathbf{v} = \mathbf{w}$.
$\large{1.}$ How and why is $k = 1$?
$\large{2.}$ How can I solve this question with but the following picture?
Since $\mathbf{v \cdot w} := 5,$ thus $\mathbf{v \cdot w} > 0 \iff \cos(\text{angle between $\mathbf{v}$ & $\mathbf{w}$})>0$
$ \iff \text{angle between $\mathbf{v}$ & $\mathbf{w}$} \in (0, 90^{\circ}). $ So $\mathbf{w}$ must be right of/above $\color{magenta}{x + 2y = 0}$.
|
To answer your first question:
$$
\mathbf{v \cdot w} = 5 \Longrightarrow
\mathbf{v} \cdot (k\mathbf{v}) = 5 \Longrightarrow
k \|\mathbf{v}\|^2 = 5 \Longrightarrow k=1
$$
Regarding the second question ...
By definition, we have $\mathbf{v \cdot w} = \|\mathbf{v}\| \cdot \|\mathbf{w}\| \cos\theta$, where $\theta$ is the angle between $\mathbf{v}$ and $\mathbf{w}$.
But we know that $\mathbf{v \cdot w} = 5$, and $\|\mathbf{v}\| = \sqrt 5$, so
$$
\|\mathbf{w}\|\, \cos\theta = 5/{\sqrt 5} = \sqrt 5.
$$
Now $\|\mathbf{w}\| \cos\theta$ is the length of the projection of $\mathbf{w}$ onto $\mathbf{v}$. So, we are looking for the shortest possible vector $\mathbf{w}$ whose projected length along $\mathbf{v}$ is $\sqrt 5$. It's obvious (I think) that the shortest vector giving the desired projected length must be parallel to $\mathbf{v}$.
In other words, if the length of $\mathbf{w}$ is to be minimized, $\mathbf{w}$ must be a scalar multiple of $\mathbf{v}$, say $\mathbf{w} = k\mathbf{v}$. $\color{green}{\bigstar}$ But $\|\mathbf{v}\| = \sqrt 5$, so, to get a projected length of $\sqrt 5$, the scalar multiplier $k$ must be 1. So $\mathbf{w} = \mathbf{v}$.
Alternative methodology after $\color{green}{\bigstar}$: But $\mathbf{w} = k\mathbf{v} \Longrightarrow \|\mathbf{w}\|\ = k\|\mathbf{v}\|$,
where $\|\mathbf{w}\|\ \geq 0 \Longrightarrow LHS \geq 0 $ as well. So for the shortest $\mathbf{w}$ such that $\mathbf{v \cdot w} = 5$ , we must choose $k = 1$. Finally, because we already determined in the anterior paragraph $\mathbf{w} = k\mathbf{v},$ thus $ \mathbf{w} = \mathbf{v}.$
|
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|
How do I get the integral of $\frac{1}{(x^2 - x -2)}$ I'm working with this problem $$ \int \frac{1}{x^2 - x - 2}$$
I'm thinking I break up the bottom so that it looks like this $$\int \frac{1}{(x-2)(x+1)} $$
Then I do $$x^2 - x -2 = \frac{A}{x-2} + \frac{B}{x+1} $$
Multiple both sides by the common denominator and come out with $$ A(x+1) + B(x-2) = x^2 - x - 2 $$
Which equals $$Ax + A + Bx - 2B = x^2 - x - 2$$
Or $$ (A+B)x + (A-2B) = x^2 - x - 2$$
After that I tried to get values for my A and B but it doesn't seem right since I don't have anything for the $x^2$
Did I mess up somewhere?
|
Your error is in the step where you write:
$$x^2 - x -2 = \frac{A}{x-2} + \frac{B}{x+1}$$
What you have, actually, is $$\dfrac 1{x^2 - x - 2} = \dfrac A{x-2} + \frac B{x+1}$$
So $A(x+1) + B(x - 2) = 1$.
If $x = -1$, $$A(-1 + 1) + B(-1 - 2) = 1 \iff -3B = 1 \iff B = -\frac 13$$
If $x = 2$, $$A(2 + 1) + B(2 - 2) = 1 \iff 3A = 1 \iff A = \frac 13$$
$$\int \dfrac {dx}{x^2 - x - 2} = \int \left(\dfrac A{x-2} + \frac B{x+1}\right)\,dx = \int \left(\frac 1{3(x-2)} - \frac{1}{3(x + 1)}\right)\,dx$$
Can you take it from here?
|
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Is $X^2-Y^2/ \sqrt{X^2+Y^2}$ normal where $X,Y\sim N(0,1)$ Can I say that $(X^2-Y^2)/ \sqrt{X^2+Y^2}$ is normal since $X,Y\sim N(0,1)$ and $X,Y$ are normal and independent?
I am trying to do those problem using polar coordinates with $X=r \cos \theta$ and $Y=r \sin \theta$, but I got stuck when I was trying to simplify $(X^2-Y^2)/ \sqrt{X^2+Y^2}$. Can anybody please help me? Thanks!
|
Let $(R,\Theta)$ be the polar coordinates corresponding to the Cartesian coordinates $(X,Y)$, with, say, $0 \le \Theta < 2 \pi$. Thus $X = R \cos(\Theta)$, $Y = R \sin(\Theta)$,
$R = \sqrt{X^2 + Y^2}$. Then $R$ and $\Theta$ are independent, and $\Theta$ is
uniform on $[0,2\pi)$.
Now $(X^2 - Y^2)/\sqrt{X^2 + Y^2} = (R^2 \cos^2(\Theta) - R^2 \sin^2(\Theta))/R
= R \cos(2\Theta)$. But $\cos(2\Theta)$ has the same distribution as $\cos(\Theta)$ (i.e. $2 \Theta \mod 2 \pi$ is again
uniform on $[0,2\pi)$) and is independent of $R$. So $R \cos(2\Theta)$ has the
same distribution as $R \cos(\Theta) = X$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Limit $ \sqrt{2\sqrt{2\sqrt{2 \cdots}}}$
Find the limit of the sequence $$\left\{\sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, \dots\right\}$$
Another way to write this sequence is $$\left\{2^{\frac{1}{2}},\hspace{5 pt} 2^{\frac{1}{2}}2^{\frac{1}{4}},\hspace{5 pt} 2^{\frac{1}{2}}2^{\frac{1}{4}}2^{\frac{1}{8}},\hspace{5 pt} \dots,\hspace{5 pt}2^{\frac{1}{2^{n+1} - 2}}\right\}$$
So basically we have to find $\lim_{n\to\infty}2^{\frac{1}{2^{n+1} - 2}}$. This equates to $$\lim_{n\to\infty}2^{\frac{1}{2^{\infty+1} - 2}} \Longrightarrow 2^{\frac{1}{2^{\infty}}} \Longrightarrow 1$$
Is this correct? P.S. Finding that $S(n)$ was a pain!
|
Hint: take a logarithm of every element in your sequence.
|
{
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|
Find $\lim_{(x,y) \to (0,0)} \frac{\sin{(x^3 + y^5)}}{x^2 + y^4}$. Prove your result. Find $\lim_{(x,y) \to (0,0)} \frac{\sin{(x^3 + y^5)}}{x^2 + y^4}$. Prove your result.
I've attempted to apply the Squeeze Theorem as such:
$\frac{-1}{x^2 + y^4} \leq \frac{\sin{(x^3 + y^5)}}{x^2 + y^4} \leq \frac{1}{x^2 + y^4}$. Clearly, though, the leftmost and rightmost functions of this inequality tend to $\infty$ as $(x,y) \to 0$, so this result does not help.
|
Here's another hint/warm-up problem:
$$\lim_{(x,y)\to (0,0)}\frac{x^6+y^5}{x^4+y^4}.$$
|
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|
Probability of two consecutive head or tail or any one of them in a row? Fair coins are tossed and when either four consecutive heads and tails appear the process will be stopped. What is the probability of two consecutive head or tail or any one of them in a row?
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Because heads and tails play symmetric roles in the stopping criterion (and presumably have equal chances in each Bernoulli trial), it suffices to find the probability of getting two consecutive heads before the process stops.
If the process stops with four consecutive heads, obviously that means also we got two consecutive heads before the process stops. So we can focus on the probability $p$ that we get two consecutive heads before the process stops with four consecutive tails.
One way to compute this is by defining a Markov chain with two absorbing states, a) two consecutive heads and b) four consecutive tails. Finding $p$ amounts to finding the probability of reaching the first of these absorbing states (two consecutive heads).
The Wikipedia write-up of
absorbing Markov chains
may be overly concise, so here are some details. Ordering the
transient states before the absorbing states gives a
probability transition matrix having block structure:
$$ P = \begin{bmatrix} Q & R \\ 0 & I \end{bmatrix}$$
so that $Q$ gives the transition probabilities between the
transient states and $R$ the transition probabilities from
transient to absorbing states.
We define the fundamental matrix
$N = \sum_{k=0}^\infty Q^k = (I-Q)^{-1}$, and the product
$NR$ then gives the probabilities of eventually reaching an
absorbing state starting from a transient state.
In the case at hand it is convenient to use four transient
states (consecutive runs of one Head or of one to three
Tails, resp.) and two absorbing states (two consecutive
Heads or four consecutive Tails). Taking the states in
just this order gives:
$$ P = \begin{bmatrix} 0 & \frac{1}{2} & 0 & 0 & \frac{1}{2} & 0
\\ \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 & 0
\\ \frac{1}{2} & 0 & 0 & \frac{1}{2} & 0 & 0
\\ \frac{1}{2} & 0 & 0 & 0 & 0 & \frac{1}{2}
\\ 0 & 0 & 0 & 0 & 1 & 0
\\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} $$
$$ N = (I-Q)^{-1} = \begin{bmatrix}
\frac{16}{9} & \frac{8}{9} & \frac{4}{9} & \frac{2}{9}
\\ \frac{14}{9} & \frac{16}{9} & \frac{8}{9} & \frac{4}{9}
\\ \frac{4}{3} & \frac{2}{3} & \frac{4}{3} & \frac{2}{3}
\\ \frac{8}{9} & \frac{4}{9} & \frac{2}{9} & \frac{10}{9}
\end{bmatrix} $$
$$ NR = \begin{bmatrix} \frac{8}{9} & \frac{1}{9}
\\ \frac{7}{9} & \frac{2}{9}
\\ \frac{2}{3} & \frac{1}{3}
\\ \frac{4}{9} & \frac{5}{9} \end{bmatrix} $$
For economy I omitted an "empty" state, so we imagine our
Markov process to initialize in state one Head with probability
$1/2$ and likewise in state one Tail with probability $1/2$.
From the above computation it follows that our chance of
reaching two consecutive Heads before four consecutive Tails is:
$$ (1/2)\frac{8}{9} + (1/2)\frac{7}{9} = \frac{5}{6} $$
This is the same as our chance of reaching two consecutive
Tails before four consecutive Heads.
|
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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.