Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Show that the equation system $I:x^2-y^2=a;II: 2xy= b$ has always a solution $(x,y)\in \mathbb {R}^2$ The statement must be false because if $b=0$ then $x$ or $y$ must be $0$. If $x$ is $0$ we get from $I$ that $-y^2=a\iff y^2 = -a$. Wouldn't that be a contradiction to the statement? Because if $a>0$ then there is no solution for the equation System, vice versa if $y=0$ and $a<0$.
For the general case I have assumed that $b\neq 0$ and therefore from $II$ we get $y=\frac{b}{2x}$, substituting with $I \Longrightarrow x^2-(\frac{b}{2x})^2=a\iff 4x^4 - b^2 = 4x^2a \iff x^4 -ax^2 - \frac{b^2}{4}=0 (*)$
Completing the square
$(*)\iff x^4 - 2\frac{ax^2}{2}-\frac{b^2}{4} \iff x^4 - 2\frac{ax^2}{2} +\frac{a^2}{4}-\frac{b^2}{4}-\frac{a^2}{4}\iff (x^2 - \frac{a}{2})^2+\frac{-b^2-a^2}{4}=0$
$\Rightarrow x^2-\frac{a}{2}=\sqrt{\frac{b^2+a^2}{4}} \Rightarrow x^2=\sqrt{\frac{b^2+a^2}{4}}+\frac{a}{2} \Rightarrow x = \pm \sqrt{\sqrt{\frac{b^2+a^2}{4}}+\frac{a}{2}}$
$II\Longrightarrow y= \frac{b}{2(\pm \sqrt{\sqrt{\frac{b^2+a^2}{4}}+\frac{a}{2}})}$
Can somebody tell my whether this Solutions are Right or not because when I have tried to verify it I end up with that term:
$\frac{5a^2+a(5\sqrt\frac{b^2+a^2}{4})(4\sqrt{\frac{b^2+a^2}{4}}+\frac{a}{2})}{4}$ which should be equal to $a$
Thank you for your time.
| If $b=0$, then either $x$ or $y$ is zero.
Suppose $a\ge0$; then $x^2-y^2=a$ has the solution $x=\sqrt{a}$, $y=0$.
If $a<0$, then $x^2-y^2=a$ has the solution $x=0$, $y=\sqrt{-a}$.
In both cases, the constraint $xy=0$ is satisfied.
For the case $b\ne0$, you can substitute $y=b/(2x)$ and get the equation
$$
4x^4-4ax^2-b^2=0
$$
which is a biquadratic; the associated equation $4z^2-4az-b^2=0$ has a positive root, because $-b^2<0$, so also the biquadratic has a solution (actually two).
Your computations are wrong. The positive root of $4z^2-4az-b^2=0$ is
$$
\frac{4a+\sqrt{16a^2+16b^2}}{8}=\frac{a+\sqrt{a^2+b^2}}{2}
$$
Thus
$$
x=\pm\sqrt{\frac{\sqrt{a^2+b^2}+a}{2}}
$$
From $y=b/(2x)$ we get
\begin{align}
y^2=\frac{b^2}{4x^2}
&=\frac{b^2}{2(\sqrt{a^2+b^2}+a)} \\[6px]
&=\frac{b^2}{2(\sqrt{a^2+b^2}+a)}\frac{\sqrt{a^2+b^2}-a}{\sqrt{a^2+b^2}-a} \\[6px]
&=\frac{b^2(\sqrt{a^2+b^2}-a)}{2(a^2+b^2-a^2)}\\[6px]
&=\frac{\sqrt{a^2+b^2}-a}{2}
\end{align}
Therefore
$$
y=\pm\sqrt{\frac{\sqrt{a^2+b^2}-a}{2}}
$$
The signs are not arbitrarily chosen: if you take the $+$ for $x$, then you will need to take the $+$ or the $-$ for $y$, according whether $b>0$ or $b<0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3022475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
On the variance in the Bernoulli scheme I'm trying to solve the following problem.
There are $755$ cards, each of which has a number from $1$ to $755$ (each number occurs exactly once). Of these cards, $20$ pieces are randomly selected and a random value equal to the sum of the numbers written on all $20$ cards is considered. I want to find the variance of this random value.
I think in this problem there is a Bernoulli scheme in which the space of elementary events is a set of numbers from $1$ to $755$. But are there ways to calculate the parameters of such a distribution (e.g. variance)?
| With $n$ the number of cards and $k$ the number drawn we have the
OGF
$$[u^k] \prod_{q=1}^n (1+ux^q).$$
Differentiate once for the expectation
$$[u^k] \prod_{q=1}^n (1+ux^q)
\sum_{q=1}^n \frac{q u x^{q-1}}{1+ux^q}$$
and set $x=1$:
$$[u^k] \prod_{q=1}^n (1+u)
\sum_{q=1}^n \frac{q u}{1+u}
= [u^k] (1+u)^{n-1} u \frac{1}{2} n (n+1)
\\ = \frac{1}{2} n (n+1) {n-1\choose k-1}.$$
This gives for the expectation
$$\frac{1}{2} n (n+1) {n-1\choose k-1} {n\choose k}^{-1}
= \frac{1}{2} n (n+1) {n-1\choose k-1}
\frac{k}{n} {n-1\choose k-1}^{-1}$$
or
$$\bbox[5px,border:2px solid #00A000]{
\mathrm{E}[X] = \frac{1}{2} k (n+1).}$$
Differentiate again for the next factorial moment
$$[u^k] \prod_{q=1}^n (1+ux^q)
\left(\sum_{q=1}^n \frac{q u x^{q-1}}{1+ux^q}\right)^2
\\ + [u^k] \prod_{q=1}^n (1+ux^q)
\sum_{q=1}^n \frac{q (q-1) u x^{q-2}}{1+ux^q}
- [u^k] \prod_{q=1}^n (1+ux^q)
\sum_{q=1}^n \frac{q u x^{q-1}}{(1+ux^q)^2} q u x^{q-1}$$
and once more set $x=1$:
$$[u^k] (1+u)^n
\left(\sum_{q=1}^n \frac{q u}{1+u}\right)^2
\\ + [u^k] (1+u)^n
\sum_{q=1}^n \frac{q (q-1) u}{1+u}
- [u^k] (1+u)^n
\sum_{q=1}^n \frac{q u}{(1+u)^2} q u
\\ = {n-2\choose k-2} \frac{1}{4} n^2 (n+1)^2
\\ + {n-1\choose k-1} \frac{1}{3} (n-1) n (n+1)
- {n-2\choose k-2} \frac{1}{6} n (n+1) (2n+1).$$
This gives for the second factorial moment
$$ \frac{k(k-1)}{n(n-1)} \frac{1}{4} n^2 (n+1)^2
\\ + \frac{k}{n} \frac{1}{3} (n-1) n (n+1)
- \frac{k(k-1)}{n(n-1)} \frac{1}{6} n (n+1) (2n+1)
\\ = \frac{k(k-1)}{n-1} \frac{1}{4} n (n+1)^2
+ k \frac{1}{3} (n-1) (n+1)
- \frac{k(k-1)}{n-1} \frac{1}{6} (n+1) (2n+1).$$
or
$$\bbox[5px,border:2px solid #00A000]{
\mathrm{E}[X(X-1)] = \frac{1}{12} k (n+1)
(3 k n + 2 k + n - 6).}$$
Finally recall that
$$\mathrm{Var}[X] =
\mathrm{E}[X(X-1)] + \mathrm{E}[X] - \mathrm{E}[X]^2$$
so that
$$\bbox[5px,border:2px solid #00A000]{
\mathrm{Var}[X] = \frac{1}{12} k (n+1) (n-k).}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3022615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\sum_{n=0}^{\infty}\frac{\Gamma^2(n+1)}{\Gamma(2n+2)}=\frac{2\pi}{3^{3/2}}$ I am seeking alternate proofs for
$$\sum_{n\geq0}\frac{\Gamma^2(n+1)}{\Gamma(2n+2)}=\frac{2\pi}{3^{3/2}}$$
Here's mine:
Recall that, for $x\in(0,2)$,
$$\frac1x=\sum_{n\geq0}(1-x)^n$$
Hence we have that, for $\frac{1-\sqrt5}2<x<\frac{1+\sqrt5}2$,
$$\frac1{x^2-x+1}=\sum_{n\geq0}x^n(1-x)^n$$
Which gives
$$
\begin{align}
\int_0^1\frac{\mathrm dx}{x^2-x+1}=&\int_0^1\sum_{n\geq0}x^n(1-x)^n\mathrm dx\\
=&\sum_{n\geq0}\int_0^1x^n(1-x)^n\mathrm dx\\
=&\sum_{n\geq0}\frac{\Gamma^2(n+1)}{\Gamma(2n+2)}
\end{align}
$$
That last step was from the definition of the Beta function:
$$B(a,b)=\int_0^1t^{a-1}(1-t)^{b-1}\mathrm dt=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$
Setting $I=\int_0^1\frac{\mathrm dx}{x^2-x+1}$, we complete the square:
$$I=\int_0^1\frac{\mathrm dx}{(x-\frac12)^2+\frac34}$$
Preforming the substitution $x=\frac{1+3^{1/2}\tan u}2$, we have
$$I=\frac{3^{1/2}}2\int_{-\pi/6}^{\pi/6}\frac{\sec^2u\ \mathrm du}{\frac34+\frac34\tan^2u}$$
And with a little simplification, we arrive at
$$\sum_{n\geq0}\frac{\Gamma^2(n+1)}{\Gamma(2n+2)}=\frac{2\pi}{3^{3/2}}$$
Which brings me to my questions. How else can we prove this? What other series can be evaluated using similar tricks? Have fun!
| Still another way is through the Hypergeometric function.
Since the term of the sum is
$$
t_{\,n} = {{\Gamma \left( {n + 1} \right)^{\,2} } \over {\Gamma \left( {2n + 2} \right)}}
$$
and
$$
\eqalign{
& t_{\,0} = {{\Gamma \left( 1 \right)^{\,2} } \over {\Gamma \left( 2 \right)}} = 1 \cr
& {{t_{\,n + 1} } \over {t_{\,n} }} = {{\Gamma \left( {n + 2} \right)^{\,2} } \over {\Gamma \left( {2n + 4} \right)}}{{\Gamma \left( {2n + 2} \right)}
\over {\Gamma \left( {n + 1} \right)^{\,2} }} = {{\left( {n + 1} \right)^{\,2} } \over {\left( {2n + 3} \right)\left( {2n + 2} \right)}} \cr
& = {{\left( {n + 1} \right)\left( {n + 1} \right)} \over {\left( {n + 3/2} \right)}}{{1/4} \over {\left( {n + 1} \right)}} \cr}
$$
then
$$
\eqalign{
& \sum\limits_{n = 0}^\infty {{{\Gamma \left( {n + 1} \right)^{\,2} } \over {\Gamma \left( {2\left( {n + 1} \right)} \right)}}}
= {}_2F_{\,1} \left( {\left. {\matrix{ {1,\;1} \cr {3/2} \cr } \;} \right|\;1/4} \right) = \cr
& = {{\arcsin \left( {\sqrt {1/4} } \right)} \over {\sqrt {1 - 1/4} \sqrt {1/4} }} = {{4\pi } \over {6\sqrt 3 }} = {{2\pi } \over {3\sqrt 3 }} \cr}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3023224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Seeking Methods to solve $ I = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(\sin(x)\right)}{\sin(x)}\:dx$ I was wondering what methods people knew of to solve the following definite integral? I have found a method using Feynman's Trick (see below) but am curious as to whether there are other Feynman's Tricks and/or Methods that can be used to solve it:
$$ I = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(\sin(x)\right)}{\sin(x)}\:dx$$
My method:
Let
$$ I(t) = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(t\sin(x)\right)}{\sin(x)}\:dx$$
Thus,
\begin{align}
I'(t) &= \int_{0}^{\frac{\pi}{2}} \frac{\sin(x)}{\left(t^2\sin^2(x) + 1\right)\sin(x)}\:dx = \int_{0}^{\frac{\pi}{2}} \frac{1}{t^2\sin^2(x) + 1}\:dx \\
&= \left[\frac{1}{\sqrt{t^2 + 1}} \arctan\left(\sqrt{t^2 + 1}\tan(x) \right)\right]_{0}^{\frac{\pi}{2}} = \sqrt{t^2 + 1}\frac{\pi}{2}
\end{align}
Thus
$$I(t) = \frac{\pi}{2}\sinh^{-1}(t) + C$$
Now
$$I(0) = C = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(0\cdot\sin(x)\right)}{\sin(x)}\:dx = 0$$
Thus
$$I(t) = \frac{\pi}{2}\sinh^{-1}(t)$$
And finally,
$$I = I(1) = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(\sin(x)\right)}{\sin(x)}\:dx = \frac{\pi}{2}\sinh^{-1}(1) = \frac{\pi}{2}\ln\left|1 + \sqrt{2}\right|$$
| Using the following relation: $$\frac{\arctan x}{x}=\int_0^1 \frac{dy}{1+(xy)^2} \Rightarrow \color{red}{\frac{\arctan(\sin x)}{\sin x}=\int_0^1 \frac{dy}{1+(\sin^2 x )y^2}}$$ We can rewrite the original integral as:
$$I = \color{blue}{\int_{0}^{\frac{\pi}{2}}} \color{red}{\frac{\arctan\left(\sin x\right)}{\sin x}}\color{blue}{dx}=\color{blue}{\int_0^\frac{\pi}{2}}\color{red}{\int_0^1 \frac{dy}{1+(\sin^2 x )y^2}}\color{blue}{dx}=\color{red}{\int_0^1} \color{blue}{\int_0^\frac{\pi}{2}}\color{purple}{\frac{1}{1+(\sin^2 x )y^2}}\color{blue}{dx}\color{red}{dy}$$
$$=\int_0^1 \left(\frac{\arctan\left(\sqrt{1+y^2}\cdot\tan(x)\right) }{\sqrt{1+y^2}} \bigg|_0^\frac{\pi}{2}\right) dy=\frac{\pi}{2}\int_0^1 \frac{dy}{\sqrt{1+y^2}}=\frac{\pi}{2}\ln\left(1+\sqrt 2\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3024896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 2
} |
Projection matrices $\mathbf{A}^{+}\mathbf{A}$ and $\mathbf{A}\mathbf{A}^{+}$ We are learning about pseudoinverses using the Strang book and I am just confused as to how to interpret the pseudoinverse.
How come $\mathbf{A}^{+}\mathbf{A}$ projects into row space and $\mathbf{A}\mathbf{A}^{+}$ projects into column space? What does it mean when it says that $\mathbf{A}^{+}$ takes a matrix from the column space to the row space? Is it because if $\mathbf{A}x=b$, $\mathbf{A}^{+}$ produces $\mathbf{A}^{+}b = x$?
Would appreciate any help - thanks!
| The Singular Value Decomposition
The singular value decomposition of a matrix is
$$
\begin{align}
\mathbf{A} &=
\mathbf{U} \, \Sigma \, \mathbf{V}^{*} \\
%
&=
% U
\left[ \begin{array}{cc}
\color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
\end{array} \right]
% Sigma
\left[ \begin{array}{cccc|cc}
\sigma_{1} & 0 & \dots & & & \dots & 0 \\
0 & \sigma_{2} \\
\vdots && \ddots \\
& & & \sigma_{\rho} \\\hline
& & & & 0 & \\
\vdots &&&&&\ddots \\
0 & & & & & & 0 \\
\end{array} \right]
% V
\left[ \begin{array}{c}
\color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\
\color{red}{\mathbf{V}_{\mathcal{N}}}^{*}
\end{array} \right] \\
%
& =
% U
\left[ \begin{array}{cccccccc}
\color{blue}{u_{1}} & \dots & \color{blue}{u_{\rho}} & \color{red}{u_{\rho+1}} & \dots & \color{red}{u_{n}}
\end{array} \right]
% Sigma
\left[ \begin{array}{cc}
\mathbf{S}_{\rho\times \rho} & \mathbf{0} \\
\mathbf{0} & \mathbf{0}
\end{array} \right]
% V
\left[ \begin{array}{c}
\color{blue}{v_{1}^{*}} \\
\vdots \\
\color{blue}{v_{\rho}^{*}} \\
\color{red}{v_{\rho+1}^{*}} \\
\vdots \\
\color{red}{v_{n}^{*}}
\end{array} \right]
%
\end{align}
$$
The color blue denotes range space objects; the color red, null space.
The connection to the fundamental subspaces is direct:
$$
\begin{array}{ll}
%
column \ vectors & span \\\hline
%
\color{blue}{u_{1}} \dots \color{blue}{u_{\rho}} &
\color{blue}{\mathcal{R} \left( \mathbf{A} \right)} \\
%
\color{blue}{v_{1}} \dots \color{blue}{v_{\rho}} &
\color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} \\
%
\color{red}{u_{\rho+1}} \dots \color{red}{u_{m}} &
\color{red}{\mathcal{N} \left( \mathbf{A}^{*} \right)} \\
%
\color{red}{v_{\rho+1}} \dots \color{red}{v_{n}} &
\color{red}{\mathcal{N} \left( \mathbf{A} \right)} \\
%
\end{array}
$$
The vectors $\color{blue}{\{u_{k}\}}_{k=1}^{\rho}$, the column vectors of $\color{blue}{\mathbf{U}_{\mathcal{R}}}$, represent an orthonormal span of the row space. Similarly, the vectors $\color{blue}{\{v_{k}\}}_{k=1}^{\rho}$ span the column space.
The Moore-Penrose Pseudoinverse Matrix
The Moore-Penrose pseudoinverse matrix arises naturally (Singular value decomposition proof
) from using the SVD to solve the least square problem:
$$
\begin{align}
\mathbf{A}^{+} &= \mathbf{V} \, \Sigma^{+} \mathbf{U}^{*} \\
%
&=
% V
\left[ \begin{array}{cc}
\color{blue}{\mathbf{V}_{\mathcal{R}}} &
\color{red}{\mathbf{V}_{\mathcal{N}}}
\end{array} \right]
% Sigma
\left[ \begin{array}{cc}
\mathbf{S}^{-1} & \mathbf{0} \\
\mathbf{0} & \mathbf{0}
\end{array} \right]
%
% U
\left[ \begin{array}{c}
\color{blue}{\mathbf{U}_{\mathcal{R}}}^{*} \\
\color{red}{\mathbf{U}_{\mathcal{N}}}^{*}
\end{array} \right] \\
%
\end{align}
%
$$
The Fundamental Projectors
The four fundamental unitary projectors are
$$
\begin{align}
%
\mathbf{P}_\color{blue}{\mathcal{R}\left( \mathbf{A} \right)} &= \mathbf{A}\mathbf{A}^{\dagger} &
%
\mathbf{P}_\color{red}{\mathcal{N}\left( \mathbf{A}^{*} \right)} &=
\mathbf{I}_{m} - \mathbf{A}\mathbf{A}^{\dagger} \\
%
\mathbf{P}_\color{blue}{\mathcal{R}\left( \mathbf{A}^{*} \right)} &= \mathbf{A}^{\dagger}\mathbf{A}
&
%
\mathbf{P}_\color{red}{\mathcal{N}\left( \mathbf{A} \right)} &= \mathbf{I}_{n} - \mathbf{A}^{\dagger}\mathbf{A} \\
%
\end{align}
$$
Projection onto $\color{blue}{\mathcal{R}\left( \mathbf{A} \right)}$
Using the decomposition for the target matrix and the concomitant pseudoinverse produces
$$\mathbf{P}_\color{blue}{\mathcal{R}\left( \mathbf{A} \right)} = \mathbf{A}\mathbf{A}^{\dagger} = \left(
\left[ \begin{array}{cc}
\color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
\end{array} \right]
\left[ \begin{array}{cc}
\mathbf{S} & \mathbf{0} \\
\mathbf{0} & \mathbf{0}
\end{array} \right]
\left[ \begin{array}{c}
\color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\
\color{red}{\mathbf{V}_{\mathcal{N}}}^{*}
\end{array} \right]
\right)
\left(
\left[ \begin{array}{cc}
\color{blue}{\mathbf{V}_{\mathcal{R}}} &
\color{red}{\mathbf{V}_{\mathcal{N}}}
\end{array} \right)
\left[ \begin{array}{cc}
\mathbf{S}^{-1} & \mathbf{0} \\
\mathbf{0} & \mathbf{0}
\end{array} \right]
\left[ \begin{array}{c}
\color{blue}{\mathbf{U}_{\mathcal{R}}}^{*} \\
\color{red}{\mathbf{U}_{\mathcal{N}}}^{*}
\end{array} \right]
\right)
=
\color{blue}{\mathbf{U}_{\mathcal{R}}}
\color{blue}{\mathbf{U}_{\mathcal{R}}}^{*}
$$
The columns of the matrix $\color{blue}{\mathbf{U}}$ are an orthogonal span of the column space of $\mathbf{A}$.
Projection onto $\color{blue}{\mathcal{R}\left( \mathbf{A}^{*} \right)}$
Similar machinations will reveal
$$
\mathbf{P}_\color{blue}{\mathcal{R}\left( \mathbf{A}^{*} \right)} =
\color{blue}{\mathbf{V}_{\mathcal{R}}}
\color{blue}{\mathbf{V}_{\mathcal{R}}}^{*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3025109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding $\lim\limits_{n→∞}n^3(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt2)$ What is$$\lim_{n→∞}n^3(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt2)?$$So it is$$\lim_{n→∞}\frac{n^3(\sqrt{n^2+\sqrt{n^4+1}})^2-(n\sqrt{2})^2}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}=\lim_{n→∞}\frac{n^3(n^2+\sqrt{n^4+1}-2n^2)}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}.$$
I do not know what to do next, because my resuts is $∞$ but the answer from book is $\dfrac{1}{4\sqrt{2}}$.
| The expedite way:
$$\sqrt{1+\sqrt{1+n^{-4}}}=\sqrt{1+1+\dfrac12n^{-4}+o(n^{-4})}=\sqrt2\sqrt{1+\dfrac14n^{-4}+o(n^{-4})}=\sqrt2\left(1+\dfrac18n^{-4}+o(n^{-4})\right)$$
and the limit is
$$\frac{\sqrt2}8.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3025375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
What's wrong with this Penrose pattern? I programmed the Penrose tiling by projecting a portion of 5D lattice to 2D space, by the "cut and project" method described in
*
*Quasicrystals: projections of 5-D lattice into 2 and 3 dimensions, H. Au-Yang and J. Perk.
*Generalised 2D Penrose tilings, A. Pavlovitch and M. Kléman
The orthonormal basis is chosen as
$$
M=\sqrt{\frac{2}{5}} \begin{bmatrix}
\cos 0 & \cos \frac{2\pi}{5} & \cos \frac{4\pi}{5}& \cos \frac{6\pi}{5}& \cos \frac{8\pi}{5} \\
\sin 0 & \sin \frac{2\pi}{5} & \sin \frac{4\pi}{5}& \sin \frac{6\pi}{5}& \sin \frac{8\pi}{5} \\
\cos 0 & \cos \frac{4\pi}{5} & \cos \frac{8\pi}{5}& \cos \frac{12\pi}{5}& \cos \frac{16\pi}{5} \\
\sin 0 & \sin \frac{4\pi}{5} & \sin \frac{8\pi}{5}& \sin \frac{12\pi}{5}& \sin \frac{16\pi}{5} \\
\frac{1}{\sqrt {2}} & \frac{1}{\sqrt {2}} & \frac{1}{\sqrt {2}} & \frac{1}{\sqrt {2}} & \frac{1}{\sqrt {2}}\\
\end{bmatrix}
$$
Each row presents a basis vector, i.e.
$$
M_i\cdot M_j=0, \;\;\textrm{for } i<j.$$
and
$$||M_i||=1, \;\;\textrm{for } 1\leq i \leq 5.
$$
$M$ consists of the parallel operator (representing the physical space)
$$
A=\begin{bmatrix}
M_1\\
M_2 \\
\end{bmatrix}=
\begin{bmatrix}
\cos 0 & \cos \frac{2\pi}{5} & \cos \frac{4\pi}{5}& \cos \frac{6\pi}{5}& \cos \frac{8\pi}{5} \\
\sin 0 & \sin \frac{2\pi}{5} & \sin \frac{4\pi}{5}& \sin \frac{6\pi}{5}& \sin \frac{8\pi}{5} \\
\end{bmatrix}
$$
and the perpendicular operator
$$
B=\begin{bmatrix}
M_3\\
M_4 \\
M_5 \\
\end{bmatrix}=\begin{bmatrix}
\cos 0 & \cos \frac{4\pi}{5} & \cos \frac{8\pi}{5}& \cos \frac{12\pi}{5}& \cos \frac{16\pi}{5} \\
\sin 0 & \sin \frac{4\pi}{5} & \sin \frac{8\pi}{5}& \sin \frac{12\pi}{5}& \sin \frac{16\pi}{5} \\
\frac{1}{\sqrt {2}} & \frac{1}{\sqrt {2}} & \frac{1}{\sqrt {2}} & \frac{1}{\sqrt {2}} & \frac{1}{\sqrt {2}}\\
\end{bmatrix}
$$
The 5D lattice points are integer combinations of basis such as
$$
p=i \begin{bmatrix}
1\\
0\\
0\\
0\\
0\\
\end{bmatrix}
+ j\begin{bmatrix}
0\\
1\\
0\\
0\\
0\\
\end{bmatrix}
+\dots, \;\; i,j,\dots \in \mathbb{Z}
$$
A 5D cube (centered at origin) is projected into 3D as polytope
$$
v'= B v, \;\; v\in hypercube
$$
so that I can check whether a $p$ is inside this polytope (20 faces). This is called "cutting" the 5D lattice points.
The resultant 2d projection $Ap$ is
Everything works fine, however, my result differs from the "standard" one (e.g. in wiki page) as follows
Is this a mistake or an alternative view of the same tiling?
Finally, I find this image (from Vertex Frequencies in Generalized Penrose Patterns, by E. Zobetz and A. Preisinger)
where the center of standard tiling exhibits the "S" pattern, while the center of my version has the "ST" pattern. But what does it mean exactly?
| The Wikipedia article states "Therefore, a finite patch cannot differentiate between the uncountably many Penrose tilings, nor even determine which position within the tiling is being shown". Thus they could be the same tiling but differently centered.
In particular, aside from the two dimensional shifting of Penrose tilings, if a higher dimensional lattice is shifted before projecting to two dimensions, then a similar situation holds.
| {
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Show $\sqrt{x (1-y)} - \sqrt{y (1-x)} + y\sqrt{x} - x\sqrt{y} - \frac{x - y}{3} > 0$ Consider the inequality
$$
\sqrt{x (1-y)} - \sqrt{y (1-x)} + y\sqrt{x} - x\sqrt{y} - \frac{x - y}{3} > 0
$$
where $0<y<x<1$ and $x+y\leq 1$.
One can, for instance, consider optimizing the expression using the Karush-Kuhn-Tucker conditions. However, the computations involved are tedious and unwieldy.
Can the inequality be shown in a simpler way?
| For a traditional kind of proof I use beside $\,0<y<x<1\,$ and $\,x+y\leq 1\,$ also
first derivation ($:=0$) and second derivation ($\neq 0$) of a term to get minimum and maximum.
Equivalent to the given inequation is $\enspace\displaystyle \sqrt{x(1-y)}-\sqrt{y(1-x)} > x\sqrt{y}- y\sqrt{x} + \frac{x-y}{3}$
and we devide by $\,x-y\,$ to get $\enspace\displaystyle \frac{1}{\sqrt{x(1-y)}+\sqrt{y(1-x)}} > \frac{1}{\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{x}}} + \frac{1}{3} \,$ .
Using first and second derivation we get
$\displaystyle \sqrt{x(1-y)}+\sqrt{y(1-x)} \leq \max(\sqrt{x(1-y)}+\sqrt{y(1-x)}) = 1 \enspace$ and
$\displaystyle \frac{1}{\sqrt{y}}+\frac{1}{\sqrt{x}} \geq \frac{1}{\sqrt{1-x}}+\frac{1}{\sqrt{x}} \geq \min(\frac{1}{\sqrt{1-x}}+\frac{1}{\sqrt{x}}) = 2\sqrt{2}\enspace$ so that follows
$\displaystyle \frac{1}{\sqrt{x(1-y)}+\sqrt{y(1-x)}} \geq 1 > \frac{1}{2\sqrt{2}} + \frac{1}{3} \geq \frac{1}{\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{x}}} + \frac{1}{3} \enspace$ as wished.
| {
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Request for crazy integrals I'm a sucker for exotic integrals like the one evaluated in this post. I don't really know why, but I just can't get enough of the amazing closed forms that some are able to come up with.
So, what are your favorite exotic integral identities, and how do you prove them?
| Here are some of my favorites:
$$\int_0^\pi \sin^2\Big(x-\sqrt{\pi^2-x^2}\Big)dx=\frac{\pi}{2}$$
$$\int_0^\infty \frac{\ln(x)}{(1+x^{\sqrt 2})^\sqrt{2}}dx=0$$
$$\int_0^\infty \frac{dx}{(1+x^{1+\sqrt{2}})^{1+\sqrt{2}}}=\frac{1}{\sqrt{2}}$$
$$\int_{-\infty}^\infty \ln(2-2\cos(x^2))dx=-\sqrt{2\pi}\zeta(3/2)$$
$$\int_0^\infty \frac{\text{erf}^2(x)}{x^2}dx=\frac{4\ln(1+\sqrt{2})}{\sqrt{\pi}}$$
$$\int_0^\infty \frac{x^{3}\ln(e^x+\frac{x^3}{6}+\frac{x^2}{2}+x+1)-x^4}{\frac{x^3}{6}+\frac{x^2}{2}+x+1}=\frac{\pi^2}{2}$$
$$\int_0^{\pi/2} \ln(x^2+\ln^2(\cos(x)))dx=\pi\ln(\ln(2))$$
$$\int_0^\infty \frac{\arctan(2x)+\arctan(x/2)}{x^2+1}dx=\frac{\pi^2}{4}$$
$$\int_0^{\pi/2}\frac{\sin(x+100\tan(x))}{\sin(x)}dx=\frac{\pi}{2}$$
$$\int_0^1 \frac{x\ln(1+x+x^4+x^5)}{1+x^2}dx=\frac{\ln^2(2)}{2}$$
$$\int_0^{1/2}\sin(8x^4+x)\cos(8x^4-x)\cos(4x^2)xdx=\frac{\sin^2(1)}{16}$$
$$\int_0^{2\pi} \sqrt{2+\cos(x)+\sqrt{5+4\cos(x)}}dx=4\pi$$
And here are four extremely exotic scrumptious integrals:
$$\int_0^1 \frac{\sin(\pi x)}{x^x (1-x)^{1-x}}dx=\frac{\pi}{e}$$
$$\int_{-\infty}^\infty \frac{dx}{(e^x-x)^2+\pi^2}=\frac{1}{1+\Omega}$$
$$\int_0^\infty \frac{3\pi^2+4(z-\sinh(z))^2}{[3\pi^2+4(z-\sinh(z))^2]^2+16\pi^2(z-\sinh(z))^2}dz=\frac{1}{8+8\sqrt{1-w^2}}$$
$$\int_0^{\pi/2}\ln|\sin(mx)|\ln|\sin(nx)|dx=\frac{\pi^3}{24}\frac{\gcd^2(m,n)}{mn}+\frac{\pi \ln^2(2)}{2}$$
...where $\Omega$ is the Omega Constant, $w$ is the Dottie Number, and $m,n\in\mathbb N$.
| {
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How to calculate the indefinite integral of $(x-1)^{\frac 12}-(x-3)$? Why is the following integral wrong?
$$ \int[(x-1)^{\frac 12}-(x-3)]dx=\frac 23 (x-1)^{\frac32}-\frac12(x-3)^2+c $$
The answer given by my textbook is:
$$ \int[(x-1)^{\frac 12}-(x-3)]dx=\frac 23 (x-1)^{\frac32}-\frac12x^2+3x+c $$
I understand why the textbook's answer is right. But I don't understand what I did wrong in mine.
| Note that
\begin{align}
\frac{2}{3}(x-1)^{3/2}-\frac{1}{2}(x-3)^2+c &= \frac{2}{3}(x-1)^{3/2}-\frac{1}{2}(x^2-6x+9)+c\\ &= \frac{2}{3}(x-1)^{3/2}-\frac{1}{2}x^2+3x\underbrace{-\frac{9}{2}+c}_{\text{constant}},
\end{align}
so both answers are valid.
| {
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How to solve a system of ODE's using laplace transform? System of ODE's:
$$y_1^"=16y_2$$
$$y_2^"=16y_1$$
Initial conditions : $$y_1(0)=2,\space y_1^{'}(0)=12$$
$$y_2(0)=6, \space y_2^{'}(0)=4$$
The equations i get once i do the laplace transform are:
$$ s^2 \mathbf Y_1(s)-12s-2=16 \mathbf Y_2(s)$$
$$ s^2 \mathbf Y_2(s)-4s-6=16 \mathbf Y_1(s)$$
Then I put $\mathbf Y_2(s)$ in terms of $\mathbf Y_1(s)$
This gave me $$ \mathbf Y_2(s)=\frac{4s^3+6s^2+192s+32}{s^4-256}$$
I've tried to do partial fractions but keep getting the wrong answer, is there another way?
| If you do partial fractions this is what you should get
$$
{\bf Y}_2(x) = \frac{9}{2}\frac{1}{s - 4} + \frac{7}{2}\frac{1}{s + 4} - 2\frac{2 s - 1}{s^2 + 4^2}
$$
Use the transform for each term
$$
y_2(t) = \frac{9}{2}e^{4t} + \frac{7}{2}e^{-4t} - \frac{1}{2}\sin 4t + 4\cos 4t
$$
EDIT Note that $s^4 - 256 = (s^2 - 4^2)(s^2 + 4^2) = (s - 4)(s + 4)(s^2 + 16)$ so that
\begin{eqnarray}
\frac{4s^3 + 6s^2 + 192 s + 32}{s^4 - 256} &=& \frac{A}{s - 4} + \frac{B}{s + 4} + \frac{Cs + D}{s^2 + 16}\\
&=& \frac{A(s + 4)(s^2 + 16) + B(s - 4)(s^2 + 16) + (s - 4)(s + 4)(Cs + D)}{(s - 4)(s + 4)(s^2 + 16)} \\
&=& \frac{A(s^3 + 16s + 4s^2 + 64) + B(s^3 + 16s - 4s^2 - 64)}{s^4 - 256} \\
&& + \frac{Cs^3 + Ds^2 -16Cs - 16D}{s^4 - 256} \\
&=& \frac{s^3(A + B + C) + s^2(4A - 4B + D) + s(16A + 16B - 16C)}{s^4 - 256} \\
&& + \frac{64A - 64B - 16D}{s^4 - 256}
\end{eqnarray}
And from here you conclude
\begin{eqnarray}
A + B + C &=& 4 \\
4A - 4B + D &=& 6\\
16A + 16B - 16C &=& 192 \\
64A - 64B - 16D &=& 32
\end{eqnarray}
And from here is easy to get $A = 9/2$, $B = 7/2$, $C = -4$ and $D = -2$
| {
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How to factor $-x^3+x^2-2$ Is there an easy way to factor $$f(x)=-x^3+x^2-2\;?$$
I have checked step-by-step calculators that all use theorems I am not very familiar with.
It doesn't seem like a sum or difference of cubes, and grouping is not an option. I tried factoring out the $-x^2$ $$f(x)=-x^2(x-1)-2$$ but that didn't get me anywhere. I am trying to find the roots to graph this function. How do I approach this?
| If you can find a root, $r$ then it will factor to $(x-r)(ax^2 + bx + c)$ and then we can you complete the square or use quadratic formula to see if $ax^2 + bx + c$ has rational roots. If it has roots $q_1$ and $q_2$ this will factor to $a(x-r)(x-q_1)(x-q_2)$
So can we find any roots. By rational root theorem, any rational roots of $-x^3+x^2 - 2=0$ will be of the form $\pm 1, \pm 2$.
$x=-1$ is a root as $-(-1)^3 + (-1)^2 -2 = 0$ but $x=1$ is not. $x=2$ nor is $x = -2$.
Dividing by $(x- (-1))$ or $x+1$ we get $\frac {-x^3 +x^2 -2}{x + 1} = \frac {-x^3}{x+1} + \frac {x^2-2}{x+1}=$
$\frac {-x^3- x^2}{x+1} + \frac {x^2 + x^2-2}{x+1}=$
$-x^2 + \frac {2x^2 - 2}{x+1} = -x^2 +\frac {2x^2}{x+1} +\frac {-2}{x+1}=$
$-x^2 + \frac {2x^2 + 2x}{x+1} + \frac {-2x - 2}{x+1}=$
$-x^2 + 2x -2$
So $-x^3 +x^2 -2 = (x+1)(-x^2 + 2x -2)$.
Can we factor $-x^3 + x^2 -2$? We could use quadratic formula or complete the square but we do know from above that $x = -1$ is the possible rational solution. And it doesn't work. ($-(-1)^2 + 2(-1)-2 = -3 \ne 0$.)
So $-x^2 + 2x -2$ can not factor. (By quadratic equation the roots are $\frac {-2\pm \sqrt{4 -8}}{-2}$ and those aren't possible. But we didn't need to do that.)
So $-x^3 +x^2 -2 = (x+1)(-x^2 + 2x -2)$. We can, for style points, bring the negative term out to get $-x^3 + x^2 -2 = -(x+1)(x^2 - 2x +2)$. But we can't factor any further.
| {
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Find the marginal distribution of an point randomly chosen on an ellipse This exercise comes from rice 3.6 and states: A point is chosen randomly in the interior of an ellipse:
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
Find the marginal densities of the $x$ and $y$ coordinates of the point.
I was aware that I would need to use integration to find the marginal distribution from a joint distribution, but I was unsure how to start. I found a post here that outlines the solution as follows:
So this ought to be uniformly distributed, thus the density function for $( x , y )$ is $f_{x,y} =\frac{ 1}{ \pi a b} $ (where $\pi ab$ is the area of the ellipse)
the limits of integration are $− \frac{ b}{a} \sqrt{a^2 − x^2}$ and $\frac{ b}{a}
\sqrt{a^2−x^2}$
How are the limits of integration determined?
why is $f_{x,y} = \frac{ 1}{ \pi a b}$ ?
| The distribution is uniform so the density will be some constant $d$, making the probability of some area $A$ within the ellipse be $A\,d$. You want the probability of being within the whole ellipse to be $1$ which then requires $\pi a b\, d = 1$, and thus the density must be $d= \frac1{\pi a b}$
Within the ellipse you have $\frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1$, i.e. $y^2 \le \frac{b^2}{a^2}(a^2 - x^2)$, which implies $$− \frac{ b}{a} \sqrt{a^2 − x^2} \le y \le \frac{ b}{a} \sqrt{a^2 − x^2}$$ and these are then the limits of integration over $y$ for a particular $x$
| {
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Solution using multinomial theorem raised to a negative power The number of ways of selecting exactly $ 4 $ fruits out of $ 4 $ apples, $ 5 $ mangoes, $ 6 $ oranges is...
A) $ 10 $
B) $ 15 $
C) $ 20 $
D) $ 25 $
I did the solution writing all the possible ways, I am getting $ 15 $, which is correct. However, there is a way to solve this with a multinomial expansion of the negative power. Please explain.
The solution is: We need to find the $x^4$ in the expansion of $(1+x+x^2+x^3+x^4)^3$. I want to understand this in a detailed manner. My question is, why $x^4$? Why $(1+x+x^2+x^3+x^4)^3$ and not $$(1+x+x^2+x^3+x^4)(1+x+x^2+x^3+x^4+x^5)(1+x+x^2+x^3+x^4+x^5+x^6)?$$
| Your suggestion of
$$
(1+x+x^2+x^3+x^4)(1+x+x^2+x^3+x^4+x^5)(1+x+x^2+x^3+x^4+x^5+x^6)
$$
is, technically, the correct expression. There are 5 mangoes and six oranges available, and this is reflected in $x^5$ and $x^6$ appearing.
However, the fact that there are $5$ and $6$ mangoes and oranges isn't really relevant, because we are only picking four fruits in total. So the problem would have the same answer if you just removed one mango and two oranges from the setup, and just had four of each fruit available.
We see this in the algebra too: the $x^5$ and $x^6$ terms in the above expression cannot possibly contribute to the final coefficient of the $x^4$ term. So they may be removed to simplify the expression.
| {
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Show that $f(x) = \sqrt{x^3}+3$ for $\{x≥0 \ | \ x \in \mathbb{R}\}$ is differentiable and find its derivative. I'd like to find out if the function $f(x) = \sqrt{x^3}+3$ is differentiable for $\{x≥0 \ | \ x \in \mathbb{R}\}$ and if it is, then find it's derivative.
First I "simplified" the function $f(x)= (\sqrt {x^3}+3)^\frac{1}{2}$ and used the difference quotient to show that $f(x)$ is differentiable for $x>0$.
$$\frac{f(x) - f(x_0)}{x-x_0} = \frac{(\sqrt{x^3}+3)- \left(\sqrt{x_0^3}+3 \right)}{x-x_0}= \frac{\sqrt{x^3}- \sqrt{x_0^3}}{x-x_0}$$
We know that $x^3-x_0^3=\left(\sqrt{x^3} - \sqrt{x_0^3}\right)\left(\sqrt{x^3} + \sqrt{x_0^3}\right)$, therefore:
$$ \implies \frac{\left(\sqrt{x^3} - \sqrt{x_0^3} \right)}{\left({x^3} - \sqrt{x_0^3}\right)\left({x^3} + \sqrt{x_0^3}\right)} = \frac{1}{\left({x^3} + \sqrt{x_0^3}\right)}$$
Additionally, we show that $\lim_\limits{x\to x_0}\left({x^3} + \sqrt{x_0^3}\right) = 2\sqrt {x_0^3}$ and therefore:
$$\implies f'(x) = \frac{1}{2\sqrt {x_0^3}} $$
Therefore when $\{x>0 \ | \ x \in \mathbb{R}\}$, $f(x)$ is differentiable and now I'd have to show that $f(x)$ is differentiable for $x=0$. I do this by using the difference quotient.
$$\lim _\limits{x \to 0}\frac{f(x) - f(0)}{x-0}=\lim _\limits{x \to 0}\frac{\sqrt{x^3}+3 }{x}=\frac{\lim_\limits{x \to 0} \sqrt {x^3}+3}{\lim_\limits{x \to 0}x}$$
Since I get a situation where I would have to divide by $0$ I can't continue but in my exercise sheet it says that its $∞$ and I don't understand why. From the looks of it, I made a mistake somewhere or I have some understanding gap.
| You start off fine, somewhere your algebra goes a little wonky.
You are not quite explict with that algebra. So, I am having a had time putting my finger on the error.
This is how I would do it.
$f'(a) = \lim_\limits{x\to a} \frac {f(x) - f(a)}{x-a}\\
\lim_\limits{x\to a} \frac {(\sqrt{x^3}+3) - (\sqrt{a^3}+3)}{x-a}\\
\lim_\limits{x\to a} \frac {\sqrt{x^3} - \sqrt{a^3}}{x-a}\\
\lim_\limits{x\to a} \frac {(\sqrt{x^3} - \sqrt{a^3})(\sqrt{x^3} + \sqrt{a^3})}{(x-a)(\sqrt{x^3} + \sqrt{a^3})}\\
\lim_\limits{x\to a} \frac {x^3 - a^3}{(x-a)(\sqrt{x^3} + \sqrt{a^3})}\\
\lim_\limits{x\to a} \frac {(x-a)(x^2+ax+a^2)}{(x-a)(\sqrt{x^3} + \sqrt{a^3})}\\
f'(a) =\frac {3a^2}{2a\sqrt{a}}\\
f'(a) =\frac {3\sqrt a}{2}\\
f'(0) =0\\
$
| {
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prove by induction that $(a^n - b^n)$ is a multiple of $(a - b)$ for $n \geq 1$ Okay so I just finished a final in discrete mathematics and I could not figure out how to finish this proof:
"Prove by mathematical induction, that $(a^n - b^n)$ is a multiple of $(a - b)$ when $a$ and $b$ are integers and $n \geq 1$.
$\underline{Base case: n = 1}$
$(a^1 - b^1) = (a - b)x$
$x = 1$
$a - b = a - b$
$\underline{Inductive Hypothesis:}$
$n = k$ for some $k$
$(a^k - b^k) = (a - b)x$
$\underline{Inductive Step:}$
$n = k + 1$
$(a^{k + 1} - b^{k + 1}) = (a - b)x_1$
$(a \cdot a^{k}) - (b \cdot b^{k}) = (a - b) x_1$
$a \cdot [(a - b)x + b^k] + b \cdot [(a - b)x - a^k]= (a-b)x_1$
$(a -b)xa + ab^k + (a - b)xb - ba^k = (a - b)x_1$
$(a -b)xa + (a - b)xb - ab^k + ba^k = (a - b)x_1$
$(a -b)xa + (a - b)xb + ba^k - ab^k = (a - b)x_1$
.....
I don't know if I did it right but I couldn't get any further than this.
| 1)$n=1 : a-b ✓$
Hypothesis: $a^n-b^n$ is a multiple of $(a-b)$.
Step $n+1$:
$a^{n+1}-b^{n+1} =a a^n -b b^n= $
$aa^n-ba^n+ba^n -bb^n=$
$(a-b)a^n +b(a^n -b^n)=$
$(a-b)a^n +bk(a-b)=$
$ (a-b)(a^n +bk).$
(Since by hypothesis: $a^n-b^n =k(a-b)$).
| {
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range of $3x^2-2xy$ subjected to $x^2+y^2=1$ If $x^2+y^2=1$. then the range of expression $3x^2-2xy$ without trigonometric substitution method
what i have done try here is use arithmetic geometric inequality
$\displaystyle x^2+y^2\geq 2xy$
$\displaystyle -2xy\geq -(x^2+y^2)$
$\displaystyle 3x^2-2xy\geq 2x^2-y^2$
this will not help more
how do i solve it help me please
| Since $x^2+y^2=1$, you can substitute for $y=\pm\sqrt{1-x^2}$ in $f(x,y)=3x^2-2xy$ to get $g(x)=f(x,\pm\sqrt{1-x^2})=3x^2\pm2x\sqrt{1-x^2},-1\le x\le1$ and find the extrema using standard techniques.
| {
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Find eccentricity of ellipse $9x^2 +4xy+6y^2-22x-16y+9=0$ I tried to find the eccentricity by factorising $$9x^2 +4xy+6y^2-22x-16y+9=0$$ in the form of perpendicular distance from two lines but it is getting too lenghty as the $xy$ term is involved.
Is there a way to do it using partial derivatives?
| The first thing I would do is rotate the axes to eliminate the "4xy" term. Write $9x^2+ 4xy+ 6y^2= \begin{bmatrix}x & y \end{bmatrix}\begin{bmatrix}9 & 2 \\ 2 & 6 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}$. The characteristic equation for the matrix, $A= \begin{bmatrix}9 & 2 \\ 2 & 6 \end{bmatrix}$ is $\left|\begin{array}{cc}9- \lambda & 2 \\ 2 & 6- \lambda\end{array}\right|= (9- \lambda)(6- \lambda)- 4= \lambda^2- 15\lambda+ 50=$ $(\lambda- 5)(\lambda- 10)= 0$. The eigenvalues are 5 and 10.
To find the eigenvectors corresponding to eigenvalue 5, solve $\begin{bmatrix}9 & 2 \\ 2 & 6 \end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}$$= \begin{bmatrix} 5x \\ 5y \end{bmatrix}$ so 9x+ 2y= 5x and 2x+ 6y= 5y. Those both reduce to y= 2x. All eigenvectors corresponding to eigenvalue 5 are multiples of $\begin{bmatrix}1 \\ 2 \end{bmatrix}$. Similarly all eigenvectors corresponding to eigenvalue 10 are multiples of $\begin{bmatrix}-2 \\ 1 \end{bmatrix}$.
Form the matrix $P= \begin{bmatrix}1 & 2\\ -2 & 1\end{bmatrix}$ having those eigenvectors as columns. Then $P^{-1}= \begin{bmatrix}\frac{1}{5} & -\frac{2}{5} \\ \frac{2}{5} & \frac{1}{5}\end{bmatrix}$ and we have $P^{-1}AP= \begin{bmatrix}5 & 0 \\ 0 & 10\end{bmatrix}$, the diagonal matrix having the eigenvalues on the diagonal.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3044923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Evaluating $\int_{1/4}^1 \int_{\sqrt{x-x^2}}^{\sqrt x}\left(\frac{x^2-y^2}{x^2}\right)\,dy\,dx$ I am looking for an efficient way to evaluate
$$\int_{1/4}^1 \int_{\sqrt{x-x^2}}^{\sqrt x}\left(\frac{x^2-y^2}{x^2}\right)\,\mathrm{d}y\,\mathrm{d}x$$
I have
\begin{align}
I&=\int_{1/4}^1\left(\int_{\sqrt{x-x^2}}^{\sqrt x}\,dy-\frac{1}{x^2}\int_{\sqrt{x-x^2}}^{\sqrt x}y^2\,dy\right)\,dx
\\&=\int_{1/4}^1\left[\sqrt x-\sqrt{x-x^2}-\frac{(\sqrt{x})^3-(\sqrt{x-x^2})^3}{3x^2}\right]\,dx
\\&=\int_{1/4}^1\sqrt x\,dx-\int_{1/4}^1 \sqrt{x-x^2}\,dx-\frac{1}{3}\int_{1/4}^1 \frac{1}{\sqrt x}\,dx+\frac{1}{3}\int_{1/4}^1\frac{(x-x^2)^{3/2}}{x^2}\,dx
\\&=\frac{2}{3}\left(1-\frac{1}{4^{3/2}}\right)-\frac{1}{3}-\color{darkred}{\int_{1/4}^1 \sqrt{x-x^2}\,dx}+\frac{1}{3}\color{green}{\int_{1/4}^1\frac{(x-x^2)^{3/2}}{x^2}\,dx}
\end{align}
Using this answer,
$$\color{darkred}{\int_{1/4}^1 \sqrt{x-x^2}\,dx}=\frac{1}{4}\int_{-\pi/6}^{\pi/2}\cos^2\,dt=\frac{1}{8}\int_{-\pi/6}^{\pi/2}(1+\cos 2t)\,dt=\frac{1}{96}(3\sqrt 3+8\pi)$$
For $\color{green}{\int_{1/4}^1\frac{(x-x^2)^{3/2}}{x^2}\,dx}$ or even the indefinite integral, nothing comes to mind off the top of my head.
In a different approach, if I try to change the order of integration right at the start, it complicates matters for me to rewrite the region $$\sqrt{x-x^2}<y<\sqrt x\,,\,1/4<x<1$$
keeping a separate range of $y$ free of $x$ and bounding $x$ with $y$.
Any suggestion regarding specific substitution or change of variables would also be helpful.
| The substitution $u=\sqrt{x}$ gives
$$ \int_{1/4}^1 \frac{(x-x^2)^{3/2}}{x^2}dx = \int_{1/4}^1 \frac{(1-x)^{3/2}}{\sqrt{x}}dx = 2\int_{1/2}^1 (1-u^2)^{3/2} du $$
Then $u=\sin t$ should do the trick.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3046210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Computing $\sqrt[4]{28+16 \sqrt 3}$ I want to compute following radical
$$\sqrt[4]{28+16 \sqrt 3}$$
For that, I first tried to rewrite this in terms of exponential.
$$(28+16\cdot 3^{\frac{1}{2}})^{\frac{1}{4}}$$
We know that $ 28 = 2 \cdot 7^{\frac{1}{2}}$
$$(2 \cdot 7^{\frac{1}{2}}+16\cdot 3^{\frac{1}{2}})^{\frac{1}{4}}$$
However, I'm stuck at this step. Could you assist me?
Regards
| Hint.
*
*$28+16\sqrt3=12+16+2\cdot2\sqrt3\cdot4=(2\sqrt3)^2+4^2+2\cdot2\sqrt3\cdot4=(4+2\sqrt3)^2$
*$4+2\sqrt3=3+1+2\cdot1\cdot\sqrt3=(\sqrt3)^2+1^2+2\cdot1\cdot\sqrt3=(\sqrt3+1)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049263",
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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If $a$, $b$ are the roots of $x^2-2x+3$.Then the equation whose roots are $a^3-3a^2+5a-2$ and $b^3-b^2+b+5$ is:
If $a$, $b$ are the roots of $x^2-2x+3$.Then the equation whose roots are $a^3-3a^2+5a-2$ and $b^3-b^2+b+5$ is:
I have not been able to find a better method than to calculate $a$ and $b$ then substitute them into the roots for the new polynomial.
I believe this question can't be transformed in a similar manner as mentioned in this question as the new roots are asymmetrical.
Does a better method than the lackluster substitution, exist?
The answer is:
$x^2-3x+2$
| Since $x^2=2x-3$, we get that $x^3=2x^2-3x=x-6$. Then,
$$
\begin{align}
a^3-3a^2+5a-2
&=(a-6)-3(2a-3)+5a-2\\
&=1
\end{align}
$$
and
$$
\begin{align}
b^3-b^2+b+5
&=(b-6)-(2b-3)+b+5\\
&=2
\end{align}
$$
It is easy to find an equation which has roots of $1$ and $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3051402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How can I evaluate the left limit of this function? If $f(x) = {{\sqrt[4]{x^2\tan^2 4x}}\over{2x}}$
How can I evaluate
$$\lim_{x\to 0^-} f(x)$$
Here's what I tried:
\begin{align}
\lim_{x\to 0^-} {{\sqrt[4]{x^2\tan^2 4x}}\over{2x}}
& = {1\over 2}\lim_{x\to 0^-} {{\sqrt[4]{x^2\tan^2 4x}}\over{x}} \\
& = {1\over 2}\lim_{x\to 0^-} {{\sqrt[4]{x^2\tan^2 4x}}\over{-\sqrt[4]{x^4}}} \\
& = -{1\over 2}\lim_{x\to 0^-} \sqrt[4]{{x^2\tan^2 4x}\over{x^4}}
\end{align}
In the second step I wrote $x$ as $-\sqrt[4]{x^4}$ because we are working with the left limit which is on the left of $0$
\begin{align}
-{1\over 2}\lim_{x\to 0^-} \sqrt[4]{{x^2\tan^2 4x}\over{x^4}} & = -{1\over 2}\lim_{x\to 0^-} \sqrt[4]{{\tan^2 4x}\over{x^2}} \\
& = -{1\over 2}\lim_{x\to 0^-} \sqrt[4]{\left({\tan 4x}\over{x}\right)^2} \\
& = -{1\over 2} \sqrt[4]{\left( \lim_{x\to 0^-} {{\tan 4x}\over{x}}\right)^2} \\
& = -{1\over 2} \sqrt[4]{(4)^2} \\
& = -{1\over 2} \sqrt[4]{16} \\
& = -{1\over 2} × 2 = -1
\end{align}
Are all these steps correct?
I feel something wrong about the second step, because I found some other people who solved it $1$ and not $-1$ as my solution brought me.
| I would write $$\lim_{x\to 0^-}\frac{\sqrt{|x\tan(4x)|}}{2x}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Given $\left(x + \sqrt{1+y^2}\right)\left(y + \sqrt{1+x^2}\right) = 1$, prove $\left(x + \sqrt{1+x^2}\right)\left(y + \sqrt{1+y^2}\right) = 1$.
Let $x$ and $y$ be real numbers such that
$$\left(x + \sqrt{1+y^2}\right)\left(y + \sqrt{1+x^2}\right) = 1$$
Prove that
$$\left(x + \sqrt{1+x^2}\right)\left(y + \sqrt{1+y^2}\right) = 1$$
| We'll replace $y$ an $-y$.
Thus, the given it's $$\left(\sqrt{y^2+1}+x\right)\left(\sqrt{x^2+1}-y\right)=1$$ and we need to prove that:
$$\left(\sqrt{x^2+1}+x\right)\left(\sqrt{y^2+1}-y\right)=1.$$
Now, the condition gives
$$\left(\sqrt{y^2+1}+x\right)\left(\sqrt{x^2+1}-y\right)=\left(\sqrt{x^2+1}+x\right)\left(\sqrt{x^2+1}-x\right)$$ or
$$\left(\sqrt{y^2+1}+x\right)\left(\sqrt{x^2+1}-y\right)-\left(\sqrt{y^2+1}+x\right)\left(\sqrt{x^2+1}-x\right)+$$
$$+\left(\sqrt{y^2+1}+x\right)\left(\sqrt{x^2+1}-x\right)-\left(\sqrt{x^2+1}+x\right)\left(\sqrt{x^2+1}-x\right)=0.$$ or
$$(x-y)\left(\sqrt{y^2+1}+x\right)-\left(\sqrt{x^2+1}-x\right)\left(\sqrt{x^2+1}-\sqrt{y^2+1}\right)=0$$ or
$$(x-y)\left(\sqrt{y^2+1}+x-\frac{(x+y)\left(\sqrt{x^2+1}-x\right)}{\sqrt{x^2+1}+\sqrt{y^2+1}}\right)=0,$$ which gives $x=y$ because the second factor it's
$$\left(\sqrt{x^2+1}+x\right)\left(\sqrt{y^2+1}-y\right)+(x+y)^2+1>0.$$
Id est, $$\left(\sqrt{x^2+1}+x\right)\left(\sqrt{y^2+1}-y\right)=\left(\sqrt{x^2+1}+x\right)\left(\sqrt{x^2+1}-x\right)=1$$ and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Burnsides Theorem Application - Necklace Having a lot of trouble understanding this question: How many different necklaces can be made from eight black beads and four white beads?
What I know:
Burnsides Theorem is given by: $$\frac{1}{|G|} \sum_{g \in G} |\operatorname{fix}_{\Omega} (g)|$$
8+4=12 and this corresponds to the Dihedral group of order 12 (I use the notation $D_{2n})$ so I will note this as $D_{24}$, also known as the symmetric 12-gon.
So we know that $|G|=24$ and the total number of necklace configurations is given by $\binom{12}{4}=495$ so I'm guessing we have to $Fix(e)=495$? But I'm unsure how to apply the rest of Burnsides theorem from here and work out the configurations of rotations and reflections.
Any help would be great, thanks.
| Given that we have dihedral symmetry we suppose we are working with
bracelets (naming convention by OEIS). This requires the cycle index
$Z(D_{12})$ of the dihedral group $D_{12}.$ We have for the cyclic
group that
$$Z(C_{12}) = \sum_{d|12} \varphi(d) a_d^{12/d}
= \frac{1}{12}
(a_1^{12} + a_2^6 + 2 a_3^4 + 2 a_4^3 + 2 a_6^2 + 4 a_{12}).$$
We get twelve more permutations corresponding to flips about an axis
passing through opposite slots or opposite edges, for a result of
$$Z(D_{12}) = \frac{1}{24}
(a_1^{12} + a_2^6 + 2 a_3^4 + 2 a_4^3 + 2 a_6^2 + 4 a_{12})
+ \frac{1}{24} ( 6 a_1^2 a_2^5 + 6 a_2^6).$$
By the Polya Enumeration Theorem (PET) we are interested in the
quantity
$$[B^8 W^4] Z(D_{12}; B+W).$$
Working through the terms we find
$[B^8 W^4] (B+W)^{12} = {12\choose 8},$
$[B^8 W^4] (B^2+W^2)^{6} = [B^4 W^2] (B+W)^{6}
= {6\choose 4},$
$[B^8 W^4] 2 (B^3+W^3)^{4} = 0$
$[B^8 W^4] 2 (B^4+W^4)^{3} = [B^2 W] 2 (B+W)^{3}
= 2 {3\choose 2}$
$[B^8 W^4] 2 (B^6+W^6)^{2} = 0$
$[B^8 W^4] 4 (B^{12}+W^{12}) = 0$
We get from the reflections
$[B^8 W^4] 6 (B+W)^2 (B^2+W^2)^5
\\ = [B^6 W^4] 6 (B^2+W^2)^5 + [B^7 W^3] 12 (B^2+W^2)^5 +
[B^8 W^2] 6 (B^2+W^2)^5
\\ = [B^3 W^2] 6 (B+W)^5 + [B^4 W] 6 (B+W)^5
\\ = 6 {5\choose 3} + 6 {5\choose 4}$
$[B^8 W^4] 6 (B^2+W^2)^{6} = [B^4 W^2] 6 (B+W)^{6}
= 6 {6\choose 4}.$
Collecting everything we find for our result that it is
$$\frac{1}{24} \left({12\choose 8} + {6\choose 4} + 2 {3\choose 2}
+ 6{5\choose 3} + 6{5\choose 4} + 6 {6\choose 4}\right).$$
which yields
$$\bbox[5px,border:2px solid #00A000]{29.}$$
| {
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"url": "https://math.stackexchange.com/questions/3058262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Trying to simplify $\frac{\sqrt{8}}{1-\sqrt{3x}}$ to be $\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$ I am asked to simplify $\frac{\sqrt{8}}{1-\sqrt{3x}}$. The solution is provided as $\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$ and I am unable to arrive at this. I was able to arrive at $\frac{1+2\sqrt{2}\sqrt{3x}}{1-3x}$
Here is my working:
$\frac{\sqrt{8}}{1-\sqrt{3x}}$ = $\frac{\sqrt{8}}{1-\sqrt{3x}}$ * $\frac{1+\sqrt{3x}}{1+\sqrt{3x}}$ = $\frac{1+\sqrt{8}\sqrt{3x}}{1-3x}$ = $\frac{1+\sqrt{2}\sqrt{2}\sqrt{2}\sqrt{3x}}{1-3x}$ = $\frac{1+2\sqrt{2}\sqrt{3x}}{1-3x}$
Is $\frac{1+2\sqrt{2}\sqrt{3x}}{1-3x}$ correct and part of the way? How can I arrive at the provided solution $\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$?
| Well, in rationalizing the denominator, we arrive at the intermediate step $\frac{\sqrt{8}+\sqrt{24x}}{1-3x}$ which simplifies to the provided solution.
Your error comes in multiplying by the conjugate of the denominator. $\sqrt{8}*(1+\sqrt{3x})$ becomes $\sqrt{8}+\sqrt{24x}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
If $a^3 \equiv b^3\pmod{11}$ then $a \equiv b \pmod{11}$.
If $a^3 \equiv b^3\pmod{11}$ then $a \equiv b \pmod{11}$.
I verified the above statement for smaller values of $a$ and $b$ but I am unable to prove it in general. Please prove or disprove it. Can we replace $11$ by any other integer?
| Here is a general statement:
If $p$ is a prime and $m$ is coprime with $p-1$, then $a^m \equiv b^m \bmod p$ iff $a \equiv b\bmod p$.
This follows from Fermat's theorem by writing $(p-1)(-u)+mv=1$ for $u,v \in \mathbb N$.
In your example, $p=11$ and $m=3$. We can take $u=2$ and $v=7$ to get $10\cdot (-2) + 3\cdot(7)=1$. Write this as $1 + 10 \cdot 2 = 3\cdot 7 $. Then
$$
a \equiv a \cdot a^{10 \cdot 2} = a^{1 + 10 \cdot 2} = a^{3\cdot7}
\equiv
b^{3\cdot7} = b^{1 + 10 \cdot 2} = b \cdot b^{10 \cdot 2} \equiv b
\bmod 11
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3064839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solving for $x$ in $\sin^{-1}(2x) + \sin^{-1}(3x) = \frac \pi 4$
Given an equation: $$\sin^{-1}(2x) + \sin^{-1}(3x) = \frac \pi 4$$
How do I find $x$?
I tried solving by differentiating both sides, but I get $x=0$.
How do you solve it, purely using trigonometric techniques?
| Or this way using
*
*$\cos(a+b) = \cos a \cos b - \sin a \sin b$
*$\cos a = \sqrt{1-\sin^2 a}$
\begin{eqnarray*}
\sin^{-1}(2x) + \sin^{-1}(3x) & = & \frac \pi 4 \\
\sqrt{1-4x^2}\sqrt{1-9x^2} - 6x^2 & = & \frac{\sqrt{2}}{2} \\
(1-4x^2)(1-9x^2) &=& \left( \frac{\sqrt{2}}{2} +6x^2 \right)^2 \\
\frac{1}{2} & = & (6\sqrt{2}+13)x^2 \\
\end{eqnarray*}
The positive solution gives:
$$\boxed{x = \frac{1}{\sqrt{2(6\sqrt{2}+13)}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3065747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
Polynomial Long Division Confusion (simplifying $\frac{x^{5}}{x^{2}+1}$) I need to simplify \begin{equation}
\frac{x^{5}}{x^{2}+1}
\end{equation}
by long division in order to solve an integral.
However, I keep getting an infinite series:
\begin{equation}
x^{3}+x+\frac{1}{x}-\frac{1}{x^{3}}+...
\end{equation}
| Since $x^5 = \left(x^2 + 1\right)\left(x^3 - x\right) + x$, we have that
$$\cfrac{x^5}{x^2 + 1} = x^3 - x + \cfrac{x}{x^2 + 1} \tag{1}\label{eq1} $$
To do this in general, first note that $x^2$ divides into $x^5$ a total of $x^3$ times. However, this gives $x^3\left(x^2 + 1\right) = x^5 + x^3$, so it's too big by $x^3$. As such, you need to subtract an appropriate value, with it being $x$ here due to $x \times x^2 = x^3$. However, $x\left(x^2 + 1\right) = x^3 + x$, so subtracting this means that you are now $x$ too small, so you need to add that $x$ back. However, as the degree of $x$ is only $1$, which is less than the degree of $2$ in $x^2 + 1$, you leave that final $x$ as the remainder. This, in total, gives \eqref{eq1}.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3067188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Evaluate $\lim_{x\to \infty} (\frac {x+1}{x-1})^x$ Evaluate $$\lim_{x\to \infty} \left(\frac {x+1}{x-1}\right)^x$$
My method:
$$\lim_{x\to \infty} \left(\frac {x+1}{x-1}\right)^x=\lim_{x\to \infty} \left(\frac {1+1/x}{1-1/x}\right)^x=1$$
Is that right?
| Note that
*
*$\lim_{x\to +\infty}\left(1 + \frac{a}{x} \right)^x = e^a$
So you have
\begin{eqnarray*}\left(\frac{x+1}{x-1}\right)^x
& = & \left(\frac{x\left(1+\frac{1}{x} \right)}{x\left(1-\frac{1}{x} \right)}\right)^x \\
& = & \frac{\left(1+\frac{1}{x}\right)^x}{\left(1-\frac{1}{x}\right)^x} \\
& \stackrel{x \to +\infty}{\longrightarrow} & \frac{e}{e^{-1}} = e^2
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3067442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Calculating $\int \frac{dx}{(x^2+1)^3}$ integral Would anyone help me calculate the following integral? $\int \frac{dx}{(x^2+1)^3}$
During our lecutre we've done very similiar one, $\int \frac{dx}{(x^2+1)^2}$ like that:
$\int \frac{dx}{(x^2+1)^2} = \int \frac{x^2+1-x^2}{(x^2+1)^2}dx = \int \frac{1}{x^2+1}dx - \int \frac{x^2}{(x^2+1)^2}dx = $
$= \Biggr\rvert \begin{equation}
\begin{split}
& u = x \quad v' =\frac{x}{(x^2+1)^2} =\frac{1(x^2+1)'}{2(x^2+1)^2}\\
& u' = 1 \quad v = -\frac{1}{2} \frac{1}{x^2+1}
\end{split}
\end{equation} \Biggr\rvert$
$= \arctan x - (-x\frac{1}{2}\frac{1}{x^2+1} + \frac{1}{2} \int \frac{dx}{x^2+1})$
$= \arctan x + \frac{x}{2(x^2+1)} - \frac{1}{2}\arctan x + C = \frac{1}{2}\arctan x + \frac{x}{2(x^2+1)} + C$
Thank you.
| So why not use the same trick? Since you know from lecture what $$f_2(x) = \int \frac{dx}{\left(x^2+1\right)^2}$$ is, you now have
$$
\begin{split}
f_3(x)
&= \int \frac{dx}{\left(x^2+1\right)^3}\\
&= \int \frac{dx}{\left(x^2+1\right)^2} - \int x \cdot \frac{xdx}{\left(x^2+1\right)^3}\\
&= f_2(x) - \text{proceed by parts as before}
\end{split}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$n!=n(n-1)(n-2)\cdots 3\times 2\times 1 $ now let $n!=n(n-1)(n-2)\cdots 3\times 2\times 1 $
now let
$5!m!=n! , m\geq 4$ then $n+m=?$
My Try :
case$1$ Let $n=m$:
$$5!m!=m! \\ 5! \neq 1$$ so $n\neq m$
case$2$ Let $n>m$ :
$$5!=\dfrac{n!}{m!} \\ 5!=n(n-1)(n-1) . . . (n-m)$$
case$3$ Let $m>n$
$$5!m!=n! \\ m(m-1)(m-2)...(m-n)=\dfrac{1}{5!}=\dfrac{1}{120}$$
since $\dfrac{1}{120}$ is't natural number so $n>m$
Now what ?
| You can just write $5!=120=\frac{n!}{m!}=n(n-1)\cdots(m+1).$
Now just divide in cases: if $n-m=1$ then $n=120$, and by consequence $m=119.$
If $n-m=2$ we should write $120$ as $n(n-1),$ but this is impossible (cause $10\times11=110<120$ and $11\times12=132>120$).
If $n-m=3$ the solution is given by $120=6\times5\times4$ so that we get $n=6, m=3.$
If $n-m=4$ then $n(n-1)\cdots(n-3)=120$ which implies $n=5,$ and $m=1.$
Same as the last one, if $n-m=5$ then $n(n-1)\cdots(n-4)=120$ which implies $n=5,$ and $m=0.$
The difference can't be bigger because there would be too many factors to obtain $120.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3070344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Inverse of identity minus matrix exponential I am trying to analytically find the inverse of a matrix given by:
\begin{align}
W = \left( I - \alpha e^A \right)^{-1},
\end{align}
where $I$ is the identity matrix of appropriate size, $e^A$ denotes matrix exponential of $A$ and
\begin{align}
A = \begin{bmatrix}
\frac{ (1-1)^2 }{ \sigma^2 } & \frac{ (2-1)^2 }{ \sigma^2 } &
\frac{ (3-1)^2 }{ \sigma^2 } & \dots & \frac{ (n-1)^2 }{ \sigma^2 } \\
\frac{ (1-2)^2 }{ \sigma^2 } & \frac{ (2-2)^2 }{ \sigma^2 } &
\frac{ (3-2)^2 }{ \sigma^2 } & \dots & \frac{ (n-2)^2 }{ \sigma^2 } \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\frac{ (1-N)^2 }{ \sigma^2 } & \frac{ (2-N)^2 }{ \sigma^2 } &
\frac{ (3-N)^2 }{ \sigma^2 } & \dots & \frac{ (N-N)^2 }{ \sigma^2 } \\
\end{bmatrix}
\end{align}
Any help would be much appreciated!
Thank you very much,
Katie
| Define the variables
$$\eqalign{
\lambda &= \log(\alpha) &\implies \alpha = e^\lambda \cr
X &= -\tfrac{1}{2}(A+\lambda I) \cr
}$$
Then
$$\eqalign{
W^{-1} &= \Big(I-e^{-2X}\Big) &= 2e^{-X}\sinh(X) \cr
W &= \Big(I-e^{-2X}\Big)^{-1} &= \tfrac{1}{2}e^{X}{\,\rm csch}(X) \cr
}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3073155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the convergence of $ \sum_{n=1}^{\infty} (n+1)^\frac{1}{3} - n^\frac{1}{3}$ I want to find what the series $ \sum_{n=1}^{\infty} (n+1)^\frac{1}{3} - n^\frac{1}{3}$ converges to exactly or show that it diverges.
By taking the partial sum of the series $S_N$ = $ \sum_{n=1}^{N} (n+1)^\frac{1}{3} - n^\frac{1}{3}$ then $S_N = 2^\frac{1}{3} - 1 + 3^\frac{1}{3} - 2^\frac{1}{3} +4^\frac{1}{3}-3^\frac{1}{3} + ... + (N+1)^\frac{1}{3} - N^\frac{1}{3}$
And at the end I'm left with $S_N = -1 + (N+1)^\frac{1}{3}$ and $\lim_{N \to \infty} S_N = -1 + \infty= \infty $
So $ \sum_{n=1}^{\infty} (n+1)^\frac{1}{3} - n^\frac{1}{3} = \infty$
Is this correct is my first question and my second question is does there exist any other method of finding what series to converge exactly?
Thank you in prior.
| Your computation is right, up to this point:
$$S_N = 2^\frac{1}{3} - 1 + 3^\frac{1}{3} - 2^\frac{1}{3} +4^\frac{1}{3}-3^\frac{1}{3} + ... +N^{1/3}- (N-1)^{1/3}+ (N+1)^\frac{1}{3} - N^\frac{1}{3}$$
Note that once you cancel all terms you are left with
$$S_N=(N+1)^{\frac{1}{3}}-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3078607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why is $(x^{2n+1} - (2n+1)x^{n+1} + (2n+1)x^n - 1)/(x-1)^3$ irreducible? Consider the following polynomial
$$
f_n(x)=x^{2n+1} - (2n+1)x^{n+1} + (2n+1)x^n - 1
$$
Try the first $n$, I find that the $(x-1)^3$ is its factor:
$$
f_1(x)=(x-1)^3,\\
f_2(x)=(x-1)^3 (x^2+3x+1),\\
f_3(x)=(x-1)^3 (x^4+3x^3+6x^2+3x+1),\\
\cdots
$$
and that this is the complete factorization of $f_n(x)$ in $\mathbb{Z}[x]$.
But how can you prove that factoring out $(x-1)^3$ gives the complete factorization over $\mathbb{Z}$? That is, how can you prove that $f_n(x)/(x-1)^3$ is irreducible over $\mathbb{Z}$?
PS: I use Mathematica to test $n\leqslant 1000$, all is correct.
| $(x-1)^n$ is a factor of the polynomial $f(x)$ with real coefficients iff
$f(1)=f'(1)=f''(1)=\cdots=f^{(n-1)}(1)=0$.
Here
$$f(1)=1-(2n+1)+(2n+1)-1=0,$$
$$f'(1)=(2n+1)-(n+1)(2n+1)+n(2n+1)=0,$$
$$f''(1)=(2n+1)2n-(n+1)n(2n+1)+n(n-1)(2n+1)=0$$
and
$$f'''(1)=(2n+1)2n(2n-1)-(n+1)n(n-1)(2n+1)+n(n-1)(n-2)(2n+1)
=(2n+1)n(n+1)>0.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is it true that $\forall n \in \Bbb{N} : (\sum_{i=1}^{n} a_{i} ) (\sum_{i=1}^{n} \frac{1}{a_{i}} ) \ge n^2$ , if all $a_{i}$ are positive?
If $\forall i \in \Bbb{N}: a_{i} \in \Bbb{R}^+$ , is it true that $\forall n \in \Bbb{N} : \big(\sum_{i=1}^{n}a_{i}\big) \big(\sum_{i=1}^{n}
\frac{1}{a_{i}}\big) \ge n^2$ ?
I have been able to prove that this holds for $n=1$ , $n=2$, and $n=3$ using the following lemma:
Lemma 1: Let $a,b \in \Bbb{R}^+$. If $ab =1$ then $a+b \ge 2$
For example, the case for $n=3$ can be proven like this:
Let $a,b,c \in \Bbb{R}^+$. Then we have:
$(a+b+c)\big(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\big) = 1 + \frac{a}{b} + \frac{a}{c} + \frac{b}{a} + 1 + \frac{b}{c} + \frac{c}{a} + \frac{c}{b} + 1 $
$= 3 + \big(\frac{a}{b} + \frac{b}{a}\big) + \big(\frac{a}{c} + \frac{c}{a}\big) + \big(\frac{b}{c} + \frac{c}{b}\big) $
By lemma 1, $\big(\frac{a}{b} + \frac{b}{a}\big) \ge 2$, $ \big(\frac{a}{c} + \frac{c}{a}\big) \ge 2$ and $\big(\frac{b}{c} + \frac{c}{b}\big) \ge 2$ , therefore:
$3 + \big(\frac{a}{b} + \frac{b}{a}\big) + \big(\frac{a}{c} + \frac{c}{a}\big) + \big(\frac{b}{c} + \frac{c}{b}\big) \ge 3 + 2 + 2 +2 = 9 = 3^2 \ \blacksquare $
However I'm not sure the generalized version for all natural $n$ is true. I can't come up with a counterexample and when I try to prove it by induction I get stuck.
Here is my attempt:
Let $P(n)::\big(\sum_{i=1}^{n}a_{i}\big) \big(\sum_{i=1}^{n} \frac{1}{a_{i}}\big) \ge n^2$
Base case: $\big(\sum_{i=1}^{1}a_{i}\big) \big(\sum_{i=1}^{1} \frac{1}{a_{i}}\big) = a_{1} \frac{1}{a_{1}} = 1 = 1^2$ , so $P(1)$ is true.
Inductive hypothesis: I assume $P(n)$ is true.
Inductive step:
$$\left(\sum_{i=1}^{n+1}a_{i}\right) \left(\sum_{i=1}^{n+1} \frac{1}{a_{i}}\right) = \left[\left(\sum_{i=1}^{n}a_{i}\right) + a_{n+1}\right] \left[\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) + \frac{1}{a_{n+1}}\right]$$
$$=\left(\sum_{i=1}^{n}a_{i}\right) \left[\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) + \frac{1}{a_{n+1}}\right] + a_{n+1} \left[\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) + \frac{1}{a_{n+1}}\right]$$
$$=\left(\sum_{i=1}^{n}a_{i}\right)\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) +\left(\sum_{i=1}^{n}a_{i}\right) \frac{1}{a_{n+1}} + a_{n+1} \left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) +a_{n+1} \frac{1}{a_{n+1}}$$
$$=\left(\sum_{i=1}^{n}a_{i}\right)\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) +\left(\sum_{i=1}^{n}a_{i}\right) \frac{1}{a_{n+1}} + a_{n+1} \left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) +1 $$
$$\underbrace{\ge}_{IH} n^2 + \left(\sum_{i=1}^{n}a_{i}\right) \frac{1}{a_{n+1}} + a_{n+1} \left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) + 1$$
And here I don't know what to do with the $\big( \sum_{i=1}^{n}a_{i} \big) \frac{1}{a_{n+1}} + a_{n+1} \big(\sum_{i=1}^{n} \frac{1}{a_{i}}\big)$ term.
Is this inequality true? If it is, how can I prove it? If it isn't, can anyone show me a counterexample?
| Well you could just use AM-HM, Cauchy-Schwarz, Chebyshev or the rearrangement inequality or the like. However, I suppose the point of this exercise is to prove AM-HM so let's continue with your efforts. To use induction we need to show that $$A = \frac{1}{a_{n+1}} \sum_{i=1}^{n} a_i + a_{n+1}\sum_{i=1}^{n} \frac{1}{a_i} \ge 2n.$$ It's not hard to show that $x+y\ge2\sqrt{xy}$, hence $$A \ge 2\sqrt{(\sum_{i=1}^{n} a_i)(\sum_{i=1}^{n} \frac{1}{a_i})}.$$ So, by induction hypothesis, $A \ge 2n$. That's it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3081320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Find the smallest prime number $p$ such that $p\, | \,n^2-n-2023$ for some integer $n$. Find the smallest prime number $p$ such that $p\, | \,n^2-n-2023$ for some integer $n$.
since $n^2-n =n(n-1)$ is the product of two consecutive integers they must be even so the difference between an even and odd number is always odd so $n^2-n-2023$ is always odd which implies $p$ is not even and the only even prime is $2$ so $p\neq 2$ but after this I do not know what to do please help.
| Let $f(n)=n^2-n-2023$ and $f_p(n)=f(n)\pmod{p}$
$\begin{array}{c|cc}
n & 0 & 1\\ \hline
f_2(n) & 1 & 1\end{array}$ thus $f_2(n)\neq 0$ and $2$ is not a prime factor of $f(n)$
$\begin{array}{c|ccc}
n & 0 & 1 & 2\\ \hline
f_3(n) & 2 & 2 & 1\end{array}$ thus $f_3(n)\neq 0$ and $3$ is not a prime factor of $f(n)$
$\begin{array}{c|ccccc}
n & 0 & 1 & 2 & 3 & 4\\ \hline
f_5(n) & 2 & 2 & 4 & 3 & 4\end{array}$ thus $f_5(n)\neq 0$ and $5$ is not a prime factor of $f(n)$
And since $f(1)=-2023=-7\times 17^2$ then $p=7$ is the lowest prime dividing $f(n)$ for some $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3081953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
What's the answer to $\int \frac{\cos^2x \sin x}{\sin x - \cos x} dx$? I tried solving the integral $$\int \frac{\cos^2x \sin x}{\sin x - \cos x}\, dx$$ the following ways:
*
*Expressing each function in the form of $\tan \left(\frac{x}{2}\right)$, $\cos \left(\frac{x}{2}\right)\,$ and $\,\sin \left(\frac{x}{2}\right)\,$ independently, but that didn't go well for me.
*Multiplying and dividing by $\cos^2x$ or $\sin^2x$.
*Expressing $\cos^2x$ as $1-\sin^2x$ and splitting the integral, and I was stuck with $\int \left(\frac{\sin^3x}{\sin x - \cos x}\right)\, dx$ which I rewrote as $\int \frac{\csc^2x}{\csc^4x (1-\cot x) } dx,\,$ and tried a whole range of substitutions only to fail.
*I tried to substitute $\frac{1}{ \sin x - \cos x}$, $\frac{\sin x}{ \sin x - \cos x}$, $\frac{\cos x \sin x}{ \sin x - \cos x}$ and $\frac{\cos^2x \sin x}{\sin x - \cos x},$ independently, none of which seemed to work out.
*I expressed the denominator as $\sin\left(\frac{\pi}{4}-x\right)$ and tried multiplying and dividing by $\sin\left(\frac{\pi}{4}+x\right)$, and carried out some substitutions. Then, I repeated the same with $\cos\left(\frac{\pi}{4}+x\right)$. Neither of them worked.
| For integrands that are a rational function of sine and cosine, as a guide to what trigonometric substitution one may try the Bioche Rules can be used.
Here notice the differential form
$$w(x) = f(\sin x, \cos x) \, dx = \frac{\cos^2 x \sin x}{\sin x - \cos x} \, dx$$
is invariant under the substitution $x \mapsto \pi + x$, that is, $w(\pi + x) = w(x)$. This suggests a substitution of $t = \tan x$ can be used. As $dt = \sec^2 x \, dx$ we rewrite the integrand as the product between a rational function consisting of $\tan x$ terms and a $\sec^2 x$ terms. Doing so we have
\begin{align}
I &= \int \frac{\cos^2 x \sin x}{\sin x - \cos x} \, dx\\
&= \int \frac{\cos^2 x \sin x}{\sin x - \cos x} \cdot \frac{\sec^2 x}{\sec^2 x} \, dx\\
&= \int \frac{\tan x}{(\tan x - 1)(1 + \tan^2 x)^2} \cdot \sec^2 x \, dx.
\end{align}
Now let $t = \tan x$. Doing so yields
\begin{align}
I &= \int \frac{t}{(t - 1)(1 + t^2)^2} \, dt\\
&= \int \left [\frac{1}{4(t - 1)} - \frac{t + 1}{4(t^2 + 1)} + \frac{1 - t}{2(1 + t^2)^2} \right ] \, dt\\
&= \frac{1}{4} \ln |t - 1| - \frac{1}{8} \ln |1 + t^2| + \frac{t}{4(1 + t^2)} + C\\
&= \frac{1}{4}\ln |\tan x - 1| + \frac{1}{4} \ln |\cos x| + \frac{1}{4} (1 + \tan x) \cos^2 x + C.
\end{align}
or after playing around with a few trignometric identities
$$\int \frac{\cos^2 x \sin x}{\sin x - \cos x} \, dx = \frac{1}{4} \ln |\sin x - \cos x| + \frac{1}{8} (\cos 2x + \sin 2x) + C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3082128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Where did I go wrong in proving $\mathbb E[X^{2n}] = \prod_{1 \leq k \leq 2n, k \operatorname{odd}}k$ Let $X$ ~ $\mathcal{N}(0,1)$. Show that: $\displaystyle \mathbb E[X^{2n}] = \prod_{1 \leq k \leq 2n, k \operatorname{odd}}k$
Idea:
$$\mathbb E[X^{2n}]=\frac{1}{\sqrt{2\pi}}\int_{\mathbb R}x^{2n}e^{-\frac{x^2}{2}}dx=\frac{2}{\sqrt{2\pi}}\int_{0}^{\infty}x^{2n}e^{-\frac{x^2}{2}}dx~,$$ then set $t=x^2 / 2 \Rightarrow dt=xdx$ and
$$\frac{2}{\sqrt{2\pi}}\int_{0}^{\infty}x^{2n}e^{-\frac{x^2}{2}}dx=\frac{2^{n+1}}{\sqrt{\pi}}\int_{0}^{\infty}t^{n-\frac{1}{2}}e^{-t}dt=\frac{2^{n+1}}{\sqrt{\pi}}\Gamma(n+\frac{1}{2})$$
I know that in order to get the desired result, I need to show that:
$$\frac{2}{\sqrt{2\pi}}\int_{0}^{\infty}x^{2n}e^{-\frac{x^2}{2}}dx=\frac{2}{\sqrt{\pi}}\Gamma(n+\frac{1}{2})$$
But I am far off it, where did I go wrong?
| You were actually off only by a factor of 2. With your choice of change of variable,
\begin{align}
\frac2{\sqrt{2\pi}}\int_{0}^{\infty}x^{2n}e^{-\frac{x^2}2 }dx &= \frac2{\sqrt{2\pi}}\int_{0}^{\infty}x^{2(n - \frac12)} e^{-\frac{x^2}2 } \cdot xdx \\
&=\frac{ 2^{1/2} }{\sqrt{\pi}}\int_{0}^{\infty}(2t)^{n - \frac12} e^{-t^2 } \cdot dt \\
&= \frac{2^n}{\sqrt{\pi}}\int_{0}^{\infty}t^{n-\frac12}e^{-t}dt=\frac{2^n}{\sqrt{\pi}}\Gamma\Bigl( n + \frac12 \Bigr)
\end{align}
Invoking the definition of Gamma function that $\Gamma(x) = (x-1)\Gamma(x-1)$, for $n \in \mathbb{N}$ we have
\begin{align}
&\phantom{ {}={} }\frac{2^n}{\sqrt{\pi}}\Gamma\Bigl( n + \frac12 \Bigr) \\
&= \frac{ 2^{n-1} }{\sqrt{\pi}} \cdot 2\cdot \Bigl( n - \frac12 \Bigr)\cdot \Gamma\Bigl( n - \frac12 \Bigr) \\
&= \frac{ 2^{n-2} }{\sqrt{\pi}} \cdot (2n - 1) \cdot 2\Bigl( n - \frac32 \Bigr) \cdot \Gamma\Bigl( n - \frac32 \Bigr) \\
& \hspace{36pt}\vdots \\
&=\frac1{\sqrt{\pi}} \cdot (2n-1) \hspace{-10pt} \underbrace{(2n-3)}_{\text{from}~n-\frac12-(1)~\text{so 2nd}} \hspace{-24pt} \overbrace{(2n-5)}^{\text{from}~n-\frac12-(2)~\text{so 3rd}} \hspace{-12pt} \ldots 5\cdot 3\cdot \hspace{-24pt} \underbrace{\Bigl(2 \cdot \frac12 \Bigr)}_{\text{from}~n-\frac12-(n-1)~\text{so $n$ th}} \hspace{-24pt} \cdot\Gamma\Bigl( \frac12 \Bigr) \\
&= (2n-1)(2n-3)(2n-5)\cdots 5\cdot 3 \cdot 1
\end{align}
as desired, where it is well-known that $\Gamma(1/2) = \sqrt{\pi}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3085265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determinant of a matrix with trigonometric functions
Prove:
det
$\begin{pmatrix}
\cos(a-b) & \cos(b-c) & \cos(c-a) \\
\cos(a+b) & \cos(b+c) & \cos(c+a) \\
\sin(a+b) & \sin(b+c) & \sin(c+a)
\end{pmatrix}=-2\sin(a-b)\sin(b-c)\sin(c-a)
$.
I know that
$det\begin{pmatrix}
\cos(a-b) & \cos(b-c) & \cos(c-a) \\
\cos(a+b) & \cos(b+c) & \cos(c+a) \\
\sin(a+b) & \sin(b+c) & \sin(c+a)
\end{pmatrix}=2\cdot det\begin{pmatrix}
\sin a\sin b & \sin b\sin c & \sin c\sin a \\
\cos a\cos b & \cos b\cos c & \cos c\cos a \\
\sin(a+b) & \sin(b+c) & \sin(c+a)
\end{pmatrix}
$
Inspired by seeing the answers of Determinant of matrix with trigonometric functions, I try to tackle this problem likewise,but fail.
| $$det=\sum_{cyc}(\cos(a-b)\cos(b+c)\sin(a+c)-\cos(a-b)\sin(b+c)\cos(c+a))=$$
$$=\sum_{cyc}\cos(a-b)\sin(a+c-b-c)=\sum_{cyc}\cos(a-b)\sin(a-b).$$
In another hand,
$$-2\sin(c-a)\sin(b-c)\sin(a-b)=-(\cos(2c-a-b)-\cos(a-b))\sin(a-b)=$$
$$=\cos(a-b)\sin(a-b)-\frac{1}{2}(\sin(2c-2b)+\sin(2a-2c))=\sum_{cyc}\cos(a-b)\sin(a-b)$$
and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3089404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve $\frac{dx}{dy}+\frac{x}{\sqrt{x^2+y^2}}=y$ Solve the differential equation
Solve $$\frac{dx}{dy}+\frac{x}{\sqrt{x^2+y^2}}=y$$
My try:
I used $x=y \tan z$
$$\frac{dx}{dy}=\tan z+\sec^2 z\frac{dz}{dy}$$
So we get:
$$\tan z+\sec^2 z\frac{dz}{dy}+\sin z=y$$
Any clue from here?
| Hint:
Let $x=yu$ ,
Then $\dfrac{dx}{dy}=y\dfrac{du}{dy}+u$
$\therefore y\dfrac{du}{dy}+u+\dfrac{yu}{\sqrt{y^2u^2+y^2}}=y$
$y\dfrac{du}{dy}+u+\dfrac{u}{\sqrt{u^2+1}}=y$
$y\dfrac{du}{dy}=y-u-\dfrac{u}{\sqrt{u^2+1}}$
$\left(y-u-\dfrac{u}{\sqrt{u^2+1}}\right)\dfrac{dy}{du}=y$
This belongs to an Abel equation of the second kind.
Let $v=y-u-\dfrac{u}{\sqrt{u^2+1}}$ ,
Then $y=v+u+\dfrac{u}{\sqrt{u^2+1}}$
$\dfrac{dy}{du}=\dfrac{dv}{du}+1+\dfrac{1}{(u^2+1)^\frac{3}{2}}$
$\therefore v\left(\dfrac{dv}{du}+1+\dfrac{1}{(u^2+1)^\frac{3}{2}}\right)=v+u+\dfrac{u}{\sqrt{u^2+1}}$
$v\dfrac{dv}{du}+\left(1+\dfrac{1}{(u^2+1)^\frac{3}{2}}\right)v=v+u+\dfrac{u}{\sqrt{u^2+1}}$
$v\dfrac{dv}{du}=-\dfrac{v}{(u^2+1)^\frac{3}{2}}+u+\dfrac{u}{\sqrt{u^2+1}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3090832",
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"source": "stackexchange",
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Proving $2 \cdot 4 \cdot 6 \cdot \dots \cdot (p-1) \equiv (-1)^{(p-1)/4} (\frac{p-1}{2})! \text{ mod } p$ if $p \equiv 1 (\operatorname{mod} 4)$ My attempt:
$p \equiv 1 ( \operatorname{mod} 4) \iff (p-1)/2$ is even $\Rightarrow (-1)^{(p-1)/4} = ((-1)^{(p-1)/2}) ^{1/2} = 1.$
So we have to prove that $2 \cdot 4 \cdot 6 \cdot \dots \cdot (p-1) \equiv (\frac{p-1}{2})! \pmod p$ for $p \equiv 1 ( \operatorname{mod} 4)$.
RHS: $1 \cdot 2 \cdot 3 \cdot \dots \cdot \frac{p-1}2$; LHS: $2^{(p-1)/2}(1 \cdot 2 \cdot 3 \cdot \dots \cdot \frac{p-1}2)$. We just have to prove that $2^{(p-1)/2}\equiv 1$ mod $p$ ($p \equiv 1$ mod $4$). $p$ is prime, and thus we can say $ 2^{p-1} \equiv 1$ mod $p$ (Fermat's Little Theorem). So, $2^{(p-1)/2}\equiv \pm 1 \pmod p$.
What can I do now?
| We are given that
$$p \equiv 1 \pmod 4 \tag{1}\label{eq1}$$
The question doesn't state it explicitly, but it seems to assume that $p$ is a prime, e.g., for this to always work, plus as it also mentions using Fermat's Little Theorem. As the question basically already starts, we have that
$$2 \cdot 4 \cdot 6 \cdots \left(p - 1\right) = 2^{\left(p - 1\right)/2}\left(1 \cdot 2 \cdot 3 \cdots \cfrac{p - 1}{2}\right) = 2^{\left(p - 1\right)/2} \left(\cfrac{p - 1}{2}\right)! \tag{2}\label{eq2}$$
Thus, to prove
$$2 \cdot 4 \cdot 6 \cdots \left(p - 1\right) \equiv \left(-1\right)^{\left(p - 1\right)/4} \left(\cfrac{p - 1}{2}\right)! \pmod p \tag{3}\label{eq3}$$
we just need to prove that
$$\left(-1\right)^{\left(p - 1\right)/4} \equiv 2^{\left(p - 1\right)/2} \pmod p \tag{4}\label{eq4}$$
To do this, note that the second supplement to the law of quadratic reciprocity, such as stated in Legendre symbol and proven in Law of Quadratic Reciprocity, gives that
$$\left(\cfrac{2}{p}\right) = \left(-1\right)^{\frac{p^2 - 1}{8}} =
\begin{cases}
1 & p \equiv 1 \text{ or } 7 \pmod 8 \\
-1 & p \equiv 3 \text{ or } 5 \pmod 8
\end{cases} \tag{5}\label{eq5}$$
Also, Legendre's original definition was by means of an explicit formula, where for $2$ it is that
$$\left(\cfrac{2}{p}\right) \equiv 2^{\left(p - 1\right)/2} \pmod p \tag{6}\label{eq6}$$
As such, consider $p \equiv 1 \pmod 4$ means $p \equiv 1 \text{ or } 5 \pmod 8$. In the first case, i.e, $p \equiv 1 \pmod 8$, from \eqref{eq5} and \eqref{eq6}, we have
$$2^{\left(p - 1\right)/2} \equiv 1 \pmod p \tag{7}\label{eq7}$$
Also, $p = 8k + 1$ for some integer $k$, so $\left(p - 1\right) / 4 = 2k$. Thus,
$$\left(-1\right)^{\left(p - 1\right)/4} = \left(-1\right)^{2k} = \left(\left(-1\right)^{2}\right)^k = 1^k \equiv 1 \pmod p \tag{8}\label{eq8}$$
Thus, \eqref{eq4} holds when $p \equiv 1 \pmod 8$. For the second case of $p \equiv 5 \pmod 8$, from \eqref{eq5} and \eqref{eq6}, we have
$$2^{\left(p - 1\right)/2} \equiv -1 \pmod p \tag{9}\label{eq9}$$
Also, $p = 8k + 5$ for some integer $k$, so $\left(p - 1\right) / 4 = 2k + 1$. Thus,
$$\left(-1\right)^{\left(p - 1\right)/4} = \left(-1\right)^{2k + 1} = \left(-1\right)^{2k}\left(-1\right) \equiv -1 \pmod p \tag{10}\label{eq10}$$
Thus, \eqref{eq4} also holds when $p \equiv 5 \pmod 8$. As both possible cases have been checked, we can conclude that \eqref{eq3} is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3091324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find limit (type 0/0) I'm struggling to find the limit
$$I = \lim_{x\to 0}\frac{2\sqrt{1-x} - \sqrt[3]{8-x}}{x}$$
What I was trying:
$$ I = \lim_{x\to 0}\frac{1-x + 2\sqrt{1-x} + 1 - (1-x) - 1 - \sqrt[3]{8-x}}{x} $$
$$ = \lim_{x\to 0}\frac{(\sqrt{1-x}+1)^2 - (2-x)- \sqrt[3]{8-x}}{x} \qquad \quad $$
$$ = \lim_{x\to 0}\frac{(\sqrt{1-x}+1)^2 - (2-x+ \sqrt[3]{8-x})}{x} \qquad \qquad $$
$$ = \lim_{x\to 0}\frac{(\sqrt{1-x}+1)^2}{x} - \lim_{x\to 0}\frac{(2-x+ \sqrt[3]{8-x})}{x} \qquad $$
Thank all of you for your answers.
| In case Taylor series are allowed, then by Taylor expansion around $x=0$:
$$
2\sqrt{1-x} = 2\left(1 - {x\over 2} + O(x^2)\right) \sim 2 - x\\
\sqrt[3]{8-x} = 2 - {x\over 12} + O(x^2)\sim2-{x\over 12}
$$
Thus your limit becomes:
$$
\lim_{x\to0} \frac{2\sqrt{1-x} - \sqrt[3]{8-x}}{x} \sim \lim_{x\to0} \frac{2-x - (2-{x\over 12})}{x} = \lim_{x\to 0}\frac{-x + {x\over 12}}{x} = -{11\over 12}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3094942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\int_{0}^{\pi}\frac{x dx}{1+ e \sin x}=K\frac{\arccos e}{\sqrt{1-e^{2}}}$ $$\int_{0}^{\pi}\frac{x dx}{1+ e \sin x}=K\frac{\arccos e}{\sqrt{1-e^{2}}}, (e^{2}\lt1)$$
Find value of $K$ ?
I have solved till this step .
$$\int_{0}^{\pi}\frac{x dx}{1+ e \sin x}=\frac{1}{2}
\int_{0}^{\pi}\frac{ \pi dx}{1+ e \sin x}$$
| We have
$$\begin{split}
\int_{0}^{\pi}\frac{x dx}{1+ e \sin x} &= \int_{0}^{\frac \pi 2}\frac{x dx}{1+ e \sin x} + \int_{\frac \pi 2}^{\pi}\frac{x dx}{1+ e \sin x}\\
&= \int_{0}^{\frac \pi 2}\frac{x dx}{1+ e \sin x} + \int_{0}^{\frac \pi 2}\frac{(\pi - s) ds}{1+ e \sin s} \,\,\,\,\text{ (with } s=\pi-x\text{)}\\
&= \pi\int_{0}^{\frac \pi 2}\frac{dx}{1+ e \sin x}
\end{split}$$
Now, using $u=\tan \frac x 2$,
$$\begin{split}
\int_{0}^{\frac \pi 2}\frac{dx}{1+ e \sin x} &= 2\int_0^{1}\frac{du}{(1+e\frac{2u}{1+u^2})(1+u^2)}\\
&=2\int_0^{1}\frac{du}{(1+2eu + u^2)}\\
&= \frac 2 {1-e^2}\int_0^{1}\frac{du}{1 + \frac{(u+e)^2}{1-e^2}}\\
&= \frac 2 {\sqrt{1-e^2}}\int_{\frac{e}{\sqrt{1-e^2}}}^{\frac{1+e}{\sqrt{1-e^2}}} \frac{dv}{1+v^2}\,\,\,\text{(with }v=\frac{u+e}{\sqrt{1-e^2}}\text{)}\\
&=\frac 2 {\sqrt{1-e^2}} \left(\arctan\left(\frac{1+e}{\sqrt{1-e^2}}\right) - \arctan \left(\frac{e}{\sqrt{1-e^2}}\right)\right)\\
&=\frac 2 {\sqrt{1-e^2}} \arctan\left(\frac{\frac{1}{\sqrt{1-e^2}}}{1+\frac{1+e}{\sqrt{1-e^2}}\frac{e}{\sqrt{1-e^2}}}\right)\\
&=\frac 2 {\sqrt{1-e^2}} \arctan\left(\frac{\sqrt{1-e^2}}{1+e}\right)\\
&=\frac 1 {\sqrt{1-e^2}} \arccos(e)
\end{split}$$
Conclusion:
$$\int_{0}^{\pi}\frac{x dx}{1+ e \sin x}=\frac{\pi}{\sqrt{1-e^2}}\arccos(e)$$ and $K=\pi$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3096995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving $\lvert z\rvert^2+iz+i=0$ I want to solve the equation
$\lvert z\rvert^2+iz+i=0$
Let $z=x+iy$
$$\implies(x^2+y^2)+i(x+iy)+i=0 \\ \iff x^2+y^2-y+ix+i=0 \\ $$ Comparing left and right side:$$\iff (x^2+y^2-y)+i(x+1)=0+0i$$
$$\implies x+1=0 \implies \boxed{x=-1}\\\implies((-1)^2+y^2-y=0)\iff y^2-y+1=0 \\ \implies \boxed{y_{1,2}=\frac{1}{2} \pm \frac{\sqrt3}{2}i}$$
Substituting $x$ an $y$ back into $x+iy$, I get two complex solutions:
$$z_1=-1+i\left(\frac{1}{2} + \frac{\sqrt3}{2}i\right)=-1-\frac{\sqrt{3}}{2}+\frac{1}{2}i\\ z_2 =-1+i\left(\frac{1}{2} - \frac{\sqrt3}{2}i\right)=-1+\frac{\sqrt{3}}{2}+\frac{1}{2}i$$
However, these solutions don't seem to satisfy my equation:
$$\implies \left( \frac{-2-\sqrt{3}}{2}\right)^2+\left( \frac{1}{2}\right)^2+i\left( -1-\frac{\sqrt{3}}{2}+\frac{1}{2}i \right)+i\not=0 \\ $$
What am I doing wrong here?
| You forgot that $y$ is a real number. Therefore, the conclusion that you should have obtained is that the equation has no solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3097136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is $f(x,y)=\left(\frac{x^a(1-y)}{(1-x)}-\frac{y^{a}(1-x)}{(1-y)}\right)\frac{1}{x-y}$ convex? Is it possible to prove that,
$$f(x,y)=\left(\frac{x^a(1-y)}{(1-x)}-\frac{y^{a}(1-x)}{(1-y)}\right)\frac{1}{x-y}$$
is convex in the following range: $0<y<x<1$, where $a\ge2$ is an integer parameter.
Numerical testing shows that it is definitely convex, but can someone help to prove it?
If it is yet not convex I would be very thankful for a counter example.
The Hessian is:
$\frac{\partial^{2}f}{\partial x \partial y}=\frac{xy^{a}(a(x-y)+2y)+x^{a}y(ax-2x-ay)}{xy(x-y)^3}$
$\frac{\partial^{2}f}{\partial x \partial x}=\frac{(y-1) x^{a+2} \left(a^2 (y (y+4)+1)-3 a (y (y+4)+1)+2 \left(y^2+y+1\right)\right)-2 (a-2) a y \left(y^2-1\right) x^{a+1}-2 (a-3) (a-1) \left(y^2-1\right) x^{a+3}+(a-3) (a-2) (y-1) x^{a+4}+(a-1) a (y-1) y^2 x^a-2 x^5 y^a+6 x^4 y^a-6 x^3 y^a+2 x^2 y^a}{(x-1)^3 x^2 (x-y)^3}$
according to Mathematica.
Below you can find some developments but the problem is not solved yet.
| This is a partial answer.
Let us write
$$f(x,y) \:=\: \frac{x^a(1-y)^2-y^a(1-x)^2}{(1-x)(1-y)(x-y)}\tag{1}$$
and exploit that convex- or concavity is invariant under affine transformations.
Setting $\,u=1-y\,$ and $\,v=1-x\,$ leads to
$$\varphi(u,v,a) \:=\: \frac{u^2(1-v)^a-v^2(1-u)^a}{uv(u-v)}\tag{2}$$
with $\,0<v<u<1$.
As a side note: If $\,u=v>0\,$ then both the denominator and the numerator
vanish, hence these are removable discontinuities of this rational function. Furthermore, there's the symmetry $\,\varphi(v,u,a)=\varphi(u,v,a)\,$.
Expanding the $(1-\,?\,)^a$ terms in $(2)$, combining equal powers of $v$ and $u$, and simplifying then yields
$$\varphi(u,v,a) \:=\: \frac 1u+\frac 1v -a +\sum_{k=2}^{a-1}(-1)^k\binom a{k+1}
\underbrace{\frac{vu^k-uv^k}{u-v}}_{=\,\sum_{j=1}^{k-1}u^{k-j}v^j}\:.$$
Thus,
$$\varphi(u,v,a=2) \:=\:\frac 1u+\frac 1v -2$$
is convex because each summand is itself convex, and
$$\varphi(u,v,a=3) \:=\:\frac 1u+\frac 1v -3 +uv$$
is convex since its Hessian is
$$\begin{pmatrix}\frac 2{u^3} & 1\\ 1 & \frac 2{v^3}
\end{pmatrix} > 0$$
positive-definite.
Summary:
In the cases $a=2,3$ one gets that $f$ is a convex function.
The ansatz presented may pave the way for a complete answer
(which I currently do not see).
| {
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"url": "https://math.stackexchange.com/questions/3101723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Finding the value of $\prod^{6}_{r=0}\cos\bigg(\frac{\pi}{21}+\frac{r\pi}{7}\bigg)$
Find value of $$\prod^{6}_{r=0}\cos\left(\frac{\pi}{21}+\frac{r\pi}{7}\right)$$
What I try:
$$\cos \frac{\pi}{21}\cdot \cos \frac{4\pi}{21}\cdot \cos\frac{7\pi}{21}\cdot \cos \frac{10\pi}{21}\cdot \cos \frac{13\pi}{21}\cdot \cos \frac{16\pi}{21}\cdot \cos \frac{19\pi}{21}$$
$\displaystyle \cos \frac{19\pi}{21} = -\cos \frac{2\pi}{21}$ and $\displaystyle \cos \frac{13\pi}{21}=-\cos \frac{8\pi}{21}$
How do I solve it? Help me, please.
| $$\cos \frac{\pi}{21}\cdot \cos \frac{4\pi}{21}\cdot \cos\frac{7\pi}{21}\cdot \cos \frac{10\pi}{21}\cdot \cos \frac{13\pi}{21}\cdot \cos \frac{16\pi}{21}\cdot \cos \frac{19\pi}{21}=$$
$$=\cos \frac{\pi}{21}\cdot \cos \frac{4\pi}{21}\cdot\frac{1}{2}\cdot \cos \frac{10\pi}{21}\cdot \left(-\cos \frac{8\pi}{21}\right)\cdot \cos \frac{16\pi}{21}\cdot \left(-\cos \frac{2\pi}{21}\right)=$$
$$=\frac{32\sin\frac{\pi}{21} \cos\frac{\pi}{21}\cdot \cos \frac{2\pi}{21}\cdot \cos\frac{4\pi}{21}\cdot \cos \frac{8\pi}{21}\cdot \cos \frac{16\pi}{21}\cdot \cos \frac{10\pi}{21}}{64\sin\frac{\pi}{21}}=$$
$$=\frac{\sin\frac{32\pi}{21} \cdot \cos \frac{10\pi}{21}}{64\sin\frac{\pi}{21}}=-\frac{2\sin\frac{10\pi}{21} \cdot \cos \frac{10\pi}{21}}{128\sin\frac{\pi}{21}}=-\frac{\sin\frac{20\pi}{21} }{128\sin\frac{\pi}{21}}=-\frac{1}{128}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3102267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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In $\triangle ABC$, $D$ is an exterior point such that $AC = CD$ and $CE$ is parallel to $AF$. Find the area of $ABDF$.
In $\triangle ABC$, $CB$ is extended upto $D$ so that $AC$ = $CD$. An angle $\angle DCE$ is drawn at point $C$ so that is equal to $\angle CAB$ and $AB$ meets $CE$ at $I$.$E$ is such an external point that satisfies the term $\angle ACB$ = $\angle CDE$. The line parallel to $CE$ is drawn from $A$ meets extended $DE$ $F$. $AFEC$ is a parallelogram. $AB$ = $4IB$ and the area of $\triangle ABC$ is $\frac{9}{4}$$\sqrt{ 15}$. What is the area of $ABDF$?
What I try:
Here, $\triangle ABC$ $\cong$ $\triangle CDE$ (with the above condition). So the area of $\triangle CDE$ will be equal to $\frac{9}{4}$$\sqrt {15}$
Let denote the height of $\triangle ABC$ = $h$
As, $AI:IB$ = 4:1 and $h$ of both $\triangle ACI$ and $CIB$ are equal.
So, the area of $\triangle ACI$ = $\frac{4}{5}$×$\frac{9}{4}$$\sqrt {15} $= $\frac{9}{5}$$\sqrt { 15}$
Thus, the area of $BIED$ will also be equal to $\frac{9}{5}$$\sqrt {15}$, because $\triangle CIB$ is the common segment of both the triangle $\triangle ACB$ and $\triangle CDE$.
But, I got stuck whenever I try to find out the area of parallelogram of $AFEC$. I can't figure out even the length and height of paralleologram. I think that some parts of the diagram need to be showed as congruent. So, I can find the area of $ACEF$ somehow.
But I failed. How can I do this and how many ways it can be solved?
| Hint: You forgot to use the fact that $AFEC$ is a parallelogram. It doesn't directly follow from the rest of the statement, but is rather very informative.
Solution:
We have $\angle ACB = \angle CDE$. Since $AFEC$ is a parallelogram, $AC$ is parallel to $FD$, therefore, $\angle ACD + \angle CDE = 180^\text{o}$. From these two we can conclude that $\angle ACB$ is a right angle. Thus $CD$ is a height of $AFEC$
$S_{ABDF} = S_{ACFE} = AC \times CD = AC^2$ (both are $S_{ACDE} - S_{ABC}$).
So we just need to find $AC$. For the right triangle $ABC$, we have $\frac{AC}{AB} = \frac{AI}{AC}$ (proof), so $AC^2 = AB \times AI$. Similarly $BC^2 = AB \times BI$, so $\frac{AC^2}{BC^2} = \frac{AI}{BI} = 3$.
Recall $AC \times BC = \frac{9}{2}\sqrt{15}$. Since $AC = \sqrt{3}BC$, $AC \times BC = \sqrt{3}BC^2 = \frac{9}{2} \times \sqrt{15}$, so $BC^2 = \frac{9}{2} \sqrt{5}$. and so
$$S_{ABDF} = AC^2 = 3 BC^2 = \frac{27}{2}\sqrt{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3102395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Making 12 by adding 1, 3, and 5 It is given 3 numbers : 1, 3, and 5, you were told to write numbers by adding those 3 numbers, for example:
There are 8 ways of writing the number 6
6 = 1 + 5
6 = 5 + 1
6 = 3 + 3
6 = 1 + 1 + 1 + 3
6 = 1 + 1 + 3 + 1
6 = 1 + 3 + 1 + 1
6 = 3 + 1 + 1 + 1
6 = 1 + 1 + 1 + 1 + 1 + 1
How many ways are there to write the number 12.
The problem is, I need to finish it quick. I can do it manually, but it'll take some time. Can anybody help me? Just giving hints would help a lot! Thanks
| There are three distinct ways to write 12 as the sum of 1s, 3s and 5s:
1) Write 11 as the sum of 1s 3s and 5s and then add an extra 1
2) Write 9 as the sum of 1s, 3s and 5s and then add an extra 3
3) Write 7 as the sum of 1s, 3s and 5s and then add an extra 5
So if $f(n)$ is the number of ways of writing $n$ as the sum of 1s, 3s and 5s then we have
$f(12) = f(11) + f(9) + f(7)$
But by a similar argument we know that $f(11) = f(10) + f(8) + f(6)$. So
$f(12) = f(11) + f(9) + f(7) \\=f(10)+f(9)+f(8)+f(7)+f(6) \\=2f(9)+f(8)+2f(7)+f(6)+f(5) \\=3f(8)+2f(7)+3f(6)+f(5)+2f(4) \dots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3103813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Precalculus algebra exercise Hi I need to solve this problem and I don’t know how so I’d appreciate a hint.
If
$a^2x^2 + b^2y^2 + c^2z^2 = 0$
$a^2x^3 + b^2y^3 + c^2z^3 = 0$
$\frac 1x - a^2 = \frac 1y - b^2 = \frac 1z - c^2$
Then $a^4x^3 + b^4y^3 + c^4z^3 = 0$
I think that $a^4x^3 + b^4y^3 + c^4z^3 = 0$ is a factor in an expression which can be found by manipulating the three given equations. I can see that
$\frac 1x - a^2 - \frac 1y + b^2 = 0$
So I tried to add, subtract, multiply given equations.
| $$a^2x^2 + b^2y^2 + c^2z^2 = 0\tag1$$
$$a^2x^3 + b^2y^3 + c^2z^3 = 0\tag2$$
$$\frac 1x - a^2 = \frac 1y - b^2 = \frac 1z - c^2=k\tag3$$
$(1)+(2)\implies$
$$a^2(x^2+x^3)+ b^2(y^2+y^3)+ c^2(z^2+z^3) = 0$$
$\iff $
$$a^2x^3\left(1+\frac 1x \right) + b^2y^3\left(1+\frac 1y\right) + c^2z^3\left(1+\frac 1z\right) = 0$$
Now using $(3) $, we can write, $$a^2x^3(1+k+a^2 ) + b^2y^3(1+k+b^2) + c^2z^3(1+k+c^2) = 0.$$
$$(a^2x^3 + b^2y^3 + c^2z^3)+k (a^2x^3 + b^2y^3 + c^2z^3)+a^4x^3 + b^4y^3 + c^4z^3 = 0.$$
Using $(2) $, we can conclude that
$$a^4x^3 + b^4y^3 + c^4z^3 = 0. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3107625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding $\int^{\frac{\pi}{8}}_{0}\ln(\tan x)\mathrm dx$
Finding $\displaystyle \int^{\frac{\pi}{4}}_{0}\ln(1+\cos x)\mathrm dx$
What I tried
Put $\displaystyle I =\int^{\frac{\pi}{4}}_{0}\ln(1+\cos x)\mathrm dx$
\begin{align*}
I&=\int^{\frac{\pi}{4}}_{0}\ln(1+\cos x)\mathrm dx\\
&=\int^{\frac{\pi}{4}}_{0}\ln\bigg(2\cos^2\frac{x}{2}\bigg)\mathrm dx\\
&=\frac{\pi}{4}\ln(2)+4\int^{\frac{\pi}{8}}_{0}\ln(\cos x)\mathrm dx\\
&=\frac{\pi}{4}\ln(2)+4\left[\left[x\ln(\cos x)\right]^{\frac{\pi}{8}}_{0}+\int^{\frac{\pi}{8}}_{0}\ln(\tan x)\mathrm dx\right]\\
&=\frac{\pi}{8}\ln(2)+\frac{\pi}{2}\ln\left(\cos \frac{\pi}{8}\right)+4\int^{\frac{\pi}{8}}_{0}\ln(\tan x)\mathrm dx
\end{align*}
How do I solve it? Help me, please.
| \begin{align}J=\int_0^{\frac{\pi}{8}}\ln\left(\tan x\right)\,dx\end{align}
Perform the change of variable $y=\left(\sqrt{2}+1\right)\tan x$,
\begin{align}J&=(\sqrt{2}-1)\int_0^{1}\frac{\ln\big((\sqrt{2}-1)x\big) }{1+(\sqrt{2}-1)^2x^2}\,dx\\
&=(\sqrt{2}-1)\int_0^{1}\frac{\ln\big(\sqrt{2}-1\big) }{1+(\sqrt{2}-1)^2x^2}\,dx+(\sqrt{2}-1)\int_0^{1}\frac{\ln x }{1+(\sqrt{2}-1)^2x^2}\,dx\\
&=\frac{\pi}{8}\ln\big(\sqrt{2}-1\big)+(\sqrt{2}-1)\int_0^{1}\frac{\ln x }{1+(\sqrt{2}-1)^2x^2}\,dx\\
&=\frac{\pi}{8}\ln\big(\sqrt{2}-1\big)+\frac{1}{2}\big(\sqrt{2}-1\big)\int_0^1\frac{\ln x}{1-i\big(1-\sqrt{2}\big)x}\,dx+\frac{1}{2}\big(\sqrt{2}-1\big)\int_0^1\frac{\ln x}{1-i\big(\sqrt{2}-1\big)x}\,dx\\
&=\boxed{\frac{\pi}{8}\ln\big(\sqrt{2}-1\big)+\frac{1}{2}i\text{Li}_2\left(i\big(\sqrt{2}-1\big)\right)-\frac{1}{2}i\text{Li}_2\left(i\big(1-\sqrt{2}\big)\right)}
\end{align}
NB:
For $0<|z|<1$ complex,
\begin{align}\int_0^1 \frac{\ln x}{1-zx}\,dx&=\int_0^1 \left(\sum_{n=0}^{\infty}(xz)^n\right)\ln x\,dx\\
&=\sum_{n=0}^{\infty} z^n\left(\int_0^1 x^n\ln x\,dx\right)\\
&=-\sum_{n=0}^{\infty} \frac{z^n}{(n+1)^2}\\
&=-\frac{1}{z}\sum_{n=0}^{\infty} \frac{z^{n+1}}{(n+1)^2}\\
&=-\frac{1}{z}\text{Li}_2(z)
\end{align}
PS:
if you prefer,
\begin{align}\boxed{J=\frac{\pi}{8}\ln\big(\sqrt{2}-1\big)-\Im{\left(\text{Li}_2\left(i\big(\sqrt{2}-1\big)\right)\right)}}\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3111547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Find all solutions in $\mathbb C$ for $z^4 = 1$ To start, I write the equation in polar form:
$$|z|^4(cos^4\theta + isin^4\theta) = 1(1 + 0i)$$
Next, I want to solve for $\theta$:
$$cos4\theta = 1 \textrm{ and } sin4\theta = 0$$
$$4\theta = cos^{-1}(1) = 0$$
$$4\theta = sin^{-1}(0) = 0$$
Here is where I get stuck.
I think that when solving for $\theta$, the $0$ also means $2\pi$ as well so I can get $4\theta = 2\pi + n2\pi$. This will turn into $\theta = \frac{\pi}{2} + \frac{n\pi}{2}$
My textbook says the values of $n$ yields values $\theta$ where $0 \leq \theta < 2\pi$ and it seems I feel like I am almost done with the problem, but I don't know how to deal with $n$ in the above equation.
There should $n$ solutions in $z^n$, so 4 solutions. I am not entirely sure why $n$ is added into the equation. I suspect it is link to the $i$ in the polar form. Am I missing a step between where I am with the $\theta$ value and the 4 solutions for $z^4$?
| I am not entirely sure why n is added into the equation.
===
Ah!
Important.
The thing is that $\theta \equiv \theta + 2n\pi$ for any integer value of $n$ in that $e^{i\theta} = \cos(\theta) + i \sin(\theta) = \cos(\theta + 2n\pi) + i\sin(\theta + 2n\pi) = e^{i\theta}$.
So if you have $c = r*e^{i\gamma}$ and you want
$z^k = c=r*e^{i\gamma}$ so you want to find the $k$th roots of $c$. (Note: the root $k$ that you want to find has nothing to do with the $n$ in the expression $\theta + 2n\pi$)
you want to figure $z = \sqrt[k]{r} e^{i\frac \gamma k}$.
But we have an issue. That isn't the only root.
$c = re^{i\gamma}= re^{i(\gamma + 2\pi)}$ as well.
So $\sqrt[k]{r}e^{i(\frac \gamma k + \frac {2\pi}k)}$ will also be a root.
So will $c = re^{i\gamma}= re^{i(\gamma + 2\pi)}=re^{i(\gamma + 4\pi)}$ and $\sqrt[k]{r}e^{i(\frac \gamma k + \frac {4\pi}k)}$ will also be a root.
In fact we are going to have $k$ roots of $\sqrt[k]{r} e^{i\frac \gamma k}, \sqrt[k]{r}e^{i(\frac \gamma k + \frac {2\pi}k)}, \sqrt[k]{r}e^{i(\frac \gamma k + \frac {4\pi}k)}, .... \sqrt[k]{r}e^{i(\frac \gamma k + \frac {2(k-1)\pi}k)}$.
We write this as the $k$ roots $\sqrt[k]{r} e^{i(\frac \gamma k) + \frac {2n\pi}k}$ for all the $n$ where $n = 0,....., k-1$.
...
So in your case you have $4\theta = 0, 2\pi, 4\pi, 6\pi$ or $2n \pi$ for $n\in \mathbb Z$ and you what
$\theta = \frac {2n\pi}4$ for $n\in \mathbb Z$. If we assume $0 \le \frac {2n\pi}4 < 2\pi$, we have $n = 0, 1,2,3$ and $\theta = 0, \frac \pi 2, \pi, \frac {3\pi} 2$.
===== old answer====
$\theta = \frac \pi 2 + \frac {n\pi}2$ for $n\in \mathbb Z$ so $n = ...... ,-4, -3, -2, -1, 0 , 1 ,2, 3, 4,....$
So $\theta =....., \frac \pi 2 - \frac {4\pi}2, \frac \pi 2 - \frac {3\pi}2, \frac \pi 2 - \frac {2\pi}2, \frac \pi 2 - \frac {1\pi}2, \frac \pi 2 - \frac {0\pi}2, \frac \pi 2 + \frac {1\pi}2, \frac \pi 2 + \frac {2\pi}2, \frac \pi 2 + \frac {3\pi}2, \frac \pi 2 + \frac {4\pi}2,....$
So $\theta = ...... -\frac {3\pi}2, -\pi, -\frac \pi 2, 0, \frac \pi 2, \pi, \frac {3pi}2,2\pi, \frac {5\pi}2,....$
but if we only consider the values of $\theta$ that are unique up to equivalent values between $0$ and up to $2\pi$ we have
$\theta = 0, \frac \pi 2, \pi, \frac {3\pi}2$ and so
The roots are $e^0 = 1; e^{i\frac \pi 2} = i; e^{i\pi} =-1; e^{i\frac {3\pi}2}=-i$
.... which hopefully were the four roots you were expecting.
Geting a value of $\frac {n\pi}2$ or whatever just means to go through the possible integer values of $n$ that give values within the range $[0,2\pi)$. That's all.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3112102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Evaluating an indefinite integral $\int \frac{dx}{\sqrt{x^3+2x+3}}$
Evaluate the following integral
\begin{equation}
J = \int \frac{dx}{\sqrt{x^3+2x+3}}
\end{equation}
I do not find suitable substitution to compute the above indefinite integral. Since $x^3+2x+3=(x+1)(x^2-x+3)$, substituting $z=\sqrt{x+1}$, we have $$J= 2\int \frac{dz}{\sqrt{z^4-3z^2+5}}.$$ I think this is not a good substitution. Any help is appreciated.
| Since the integral is just a cubic raised to the power $-\frac12$ and the cubic has no repeated roots, it is not elementary but may be expressed in terms of elliptic integrals.
Suppose we want to find the integral from the pole at $-1$ to some number $y$ greater than that – the indefinite integral from $0$ follows from this easily. The cubic's factorisation as given in the question is $x^3+2x+3=(x+1)(x^2-x+3)=(x+1)((x-1/2)^2+11/4)$, so Byrd and Friedman 239.00 gives for this
$$\int_{-1}^y\frac1{\sqrt{x^3+2x+3}}\,dx=5^{-1/4}F\left(\cos^{-1}\frac{\sqrt5-1-y}{\sqrt5+1+y},m=\frac{10+3\sqrt5}{20}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Area of a region delimited by chords and circular arcs.
Let $AB$ be the diameter of circle $O$, where $AB = 2$. Circle $P$ is internally tangent to circle $O$ at point $B$, and $PB$ = $\frac{2}{3}$.
Two different chords $AX$ and $AY$ are drawn tangent to circle
P. Let $R$ be the region bounded by $AX, AY$ , and arc $XBY$ . What is the area of the region inside R but outside circle $P$?
I was able to figure out that the sides of the other two sides were $\sqrt 3$ and $1$ using similar triangles. Then the segments were $\frac{π}{6}$-$\frac{\sqrt 3}{4}$. I added the area of the triangle then multiplied it by two and subtracted the smaller circle's area from the result. My answer is $\frac{\sqrt 3}{2}$-$\frac{π}{9}$. But the correct answer is $\frac{4\sqrt 3}{9}$-$\frac{4π}{27}$. Is there anything wrong with my solution?
| From the question AB = 2 and PB = 2/3. Then AP = 4/3 and PX = 2/3.
Then $AX = \sqrt{AP^2 - PX^2} = \frac{2\sqrt3}{3}$
Sum of area
$$AXPY = AX.XP = \frac{2\sqrt{3}}{3}.\frac{2}{3} = \frac{4\sqrt{3}}{9}$$
$$cos(angleXPA) = \frac{PX}{AP} = \frac{1}{2} \implies angle(XPY) = \frac{2\pi}{3}$$
Hence the area in question is
$$\frac{4\sqrt{3}}{9} - \frac{1}{2}(\frac{2}{3})^2\frac{2\pi}{3} = \frac{4\sqrt{3}}{9} - \frac{4\pi}{27}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find if terms are terms of the same arithmetic progression
Is it possible that numbers $\frac{1}{2}, \frac{1}{3}, \frac{1}{5}$ are (not necessarily adjacent) terms of the same arithmetic progression?
Hint: Yes. Try $\frac{1}{30}$ as a difference.
I was going back and forth how they found out that difference.
My idea was since we have an arithmetic sequence defined as $a, a+d, a+2d,...$ I thought I could solve for the difference $d=\frac{1}{30}$. Since:
$$\frac{1}{3} = \frac{1}{2}+nd$$
And
$$\frac{1}{5} = \frac{1}{3}+md$$
Then $nd = \frac{1}{3} - \frac{1}{2} = -\frac{1}{6}$ and $md = \frac{1}{5} - \frac{1}{3} = -\frac{2}{15}$
Since it is also part of the same sequence we can find:
$$nd + md = -\frac{1}{6} -\frac{2}{15} = -\frac{3}{10}$$
Now I'm stuck since I can't see how this brings me any closer to find $m, n, d$.
| Suggestion: Following the given hint, try writing each fraction with a common denominator of $30$:
$\frac12=\frac{15}{30}$
$\frac13=\frac{10}{30}$
$\frac15 = \frac{6}{30}$
Does that help?
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int_{y=-\infty}^1\int_{x=0}^\infty 4xy\sqrt{x^2+y^2}\mathrm dx\mathrm dy$ Evaluate $$\int_{y=-\infty}^1\int_{x=0}^\infty 4xy\sqrt{x^2+y^2}\mathrm dx\mathrm dy$$
I tried to solve this for hours without any success. I tried substitution method of $u = x^2+y^2$ but doesn't seem to work.
Here is my attempt:
$$\int_{-\infty}^1\int_0^\infty (16x^2y^2)^{\frac{1}{2}}(x^2 + y^2)^\frac{1}{2} \; \mathrm{d}x \mathrm{d}{y} = \int_{-\infty}^1 \int_0^\infty (16x^3y^2 + 16x^2 y^3)^{\frac{1}{2}} \; \mathrm{d}x\mathrm{d}{y}.$$
Let
\begin{align}
u &= 16x^3 y^2 + 16x^2 y^3 \\
\mathrm{d}u &= (48x^2y^2 + 32xy^3) \; \mathrm{d}{x} \\
\mathrm{d}u &= 16xy^2(3x + 2y) \; \mathrm{d}{x} \\
\mathrm{d}u &= 4xy(12xy + 8y^2) \; \mathrm{d}{x} \\
4xy \; \mathrm{d}{x} &= \frac{\mathrm{d}{u}}{12xy + 8y^2}.
\end{align}
$$\int_{-\infty}^1 \int_0^\infty 4xy \sqrt{x^2 + y^2} \; \mathrm{d}x \mathrm{d}y = \int_{-\infty}^1 \int_0^\infty \frac{\mathrm{d}u}{(12xy + 8y^2)}.$$
Attaching image for reference
EDITS
Original Post asked integrate: 4xy sqrt(x^2+y^2) where 0<x, y<1, and added this image. .
Later edits misread 0<x,y<1 as if 0<x<1and 0<y<1
| With constant $y$ the substitution $u=x^2+y^2$ implies $$\int_0^1\sqrt{x^2+y^2}xdx=\frac{1}{2}\int_{y^2}^{1+y^2}u^{1/2}du=\frac{1}{3}[u^{3/2}]_{y^2}^{1+y^2}=\frac{(1+y^2)^{3/2}-y^3}{3}.$$Now we apply the integration operator $4\int_0^1 dy\,y$, giving$$\frac{4}{3}\int_0^1[y(1+y^2)^{3/2}-y^4]dy=\frac{4}{15}[(1+y^2)^{5/2}-y^5]_0^1=\frac{4}{15}(2^{5/2}-2)\approx 0.975.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114531",
"timestamp": "2023-03-29T00:00:00",
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Second order differential equation with non constant coefficients and using a given function $z$ I need a step by step solution for this:
$$-(x^4+2x^2+1)y''+4x(x^2+1)y'+(x^4-4x^2+3)y=0$$
Given tip: introduce this function: $z(x)=\frac{y}{1+x^2}$
| Since the OP has solved the problem, I will post a full solution:
The tip you are given is indeed useful. Write $y=(1+x^2)z$. Then by the product rule, one obtains:
$$y'=2xz+(1+x^2)z'$$
$$y''=2z+4xz'+(1+x^2)z''$$
Substituting this into the ODE, we obtain:
$$\small -(1+x^2)^2[2z+4xz'+(1+x^2)z'']+4x(1+x^2)[2xz+(1+x^2)z']+(x^4-4x^2+3)(1+x^2)z=0$$
Dividing by $1+x^2\neq 0$, we obtain:
$$-(1+x^2)[2z+4xz'+(1+x^2)z'']+4x[2xz+(1+x^2)z']+(x^4-4x^2+3)z=0$$
A lot of terms will cancel after expansion. Grouping derivatives of $z$ gives:
$$(x^4+2x^2+1)z+(-1-2x^2-x^4)z''=0$$
Multiplying by $-1$ and dividing by $x^4+2x^2+1=(1+x^2)^2\neq 0$, we obtain the simple ODE:
$$z''-z=0 \tag{*}$$
This is an easy second order linear homogeneous ODE with constant coefficients, hence one can easily find that the solution to this is, where $c_1$ and $c_2$ are arbitrary constants:
$$z(x)=c_1 e^{-x}+c_2e^x$$
Thus, the general solution for the ODE is given by:
$$y(x)=c_1(1+x^2)e^{-x}+c_2(1+x^2)e^x$$
Addendum: In case you were wondering where the motivation for this substitution comes from, it comes from the fact that a linear differential equation of the form:
$$y''+P(x)y'+Q(x)y=0$$
Can be reduced to the form:
$$z''+R(x)z=0$$
By the substitution:
$$\ln(y)=\ln(z)-\frac{1}{2}\int P(x)~dx \tag{**}$$
One can show that $R(x)$ is then given by:
$$R(x)=Q(x)-\frac{1}{2}P'(x)-\frac{1}{4}P^2(x)$$
Here, $P(x)=-\frac{4x}{1+x^2}$, so by $(**)$, this indeed suggests the substitution $y=(1+x^2)z$.
Note that in general it is not guaranteed that $R(x)$ takes a simple form. In the case of the differential equation we are solving, we got lucky and got $R(x)=-1$.
| {
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"url": "https://math.stackexchange.com/questions/3115210",
"timestamp": "2023-03-29T00:00:00",
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Convergence of the series for $a \in \mathbb R$ $\sum_{n=1}^\infty\sin\left(\pi \sqrt{n^2+a^2} \right)$ Convergence of the series for $a \in \mathbb R$ $$\sum_{n=1}^\infty\sin\left(\pi \sqrt{n^2+a^2} \right)$$
I saw this problem in a calculus book and it gave a hint that says
HINT First show that $$\sin\left(\pi \sqrt{n^2+a^2} \right)=(-1)^n\sin\frac{\pi a^2}{\sqrt{n^2+a^2}+n}\sim(-1)^n\frac{\pi a^2}{2n}\qquad (n \to\infty)$$
I was able to show that $$\sin\left(\pi \sqrt{n^2+a^2} \right)=\sin\left(\pi \sqrt{n^2+a^2} -\pi n+\pi n\right)=\sin\left(\pi (\sqrt{n^2+a^2}-n)+\pi n \right)=(-1)^n\sin\left(\pi (\sqrt{n^2+a^2}-n) \right)=(-1)^n\sin\frac{\pi a^2}{\sqrt{n^2+a^2}+n}$$
But what I don't understand is how did they come up with that equivalence? First I thought that they used the limit comparison test but now I can see that you can't use that test because we're dealing with alternating series. Did they do a mistake or something?
Can somebody help me understand this hint and how to solve this problem?
| Hint:
$$
\begin{align}
\sum_{n=1}^\infty\sin\left(\pi\sqrt{n^2+a^2}\right)
&=\sum_{n=1}^\infty(-1)^n\sin\left(\pi\sqrt{n^2+a^2}-\pi n\right)\\
&=\sum_{n=1}^\infty(-1)^n\sin\left(\frac{\pi a^2}{\sqrt{n^2+a^2}+n}\right)
\end{align}
$$
where $\sin\left(\frac{\pi a^2}{\sqrt{n^2+a^2}+n}\right)$ decreases monotonically to $0$ for $n\ge a^2$. That is, for $n\ge a^2$, $\frac{\pi a^2}{\sqrt{n^2+a^2}+n}\le\frac{\pi a^2}{2n}\le\frac\pi2$ and $\sin(x)$ maps $[0,\pi/2]$ monotonically onto $[0,1]$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}+\frac{1}{1+d}=2$ if $\frac{1}{a+w}+\frac{1}{b+w}+\frac{1}{c+w}+\frac{1}{d+w}=\frac{2}{w}$
If $\frac{1}{a+\omega}+\frac{1}{b+\omega}+\frac{1}{c+\omega}+\frac{1}{d+\omega}=\frac{2}{\omega}$, where $a,b,c,d\in\mathcal{R}$ and $\omega$ is a non-real cube root of unity, then prove that $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}+\frac{1}{1+d}=2$
As it was asked as a multiple choice question(this expression was one option to pick) I really do not see an easy way to prove this.
$$
\frac{1}{a+\omega}+\frac{1}{b+\omega}+\frac{1}{c+\omega}+\frac{1}{d+\omega}=\frac{2}{\omega}\\
\frac{1}{a+\omega^2}+\frac{1}{b+\omega^2}+\frac{1}{c+\omega^2}+\frac{1}{d+\omega^2}=\frac{2}{\omega^2}\\
\frac{1}{a\omega^2+\omega}+\frac{1}{b\omega^2+\omega}+\frac{1}{c\omega^2+\omega}+\frac{1}{d\omega^2+\omega}=2\\
\frac{1}{a\omega+1}+\frac{1}{b\omega+1}+\frac{1}{c\omega+1}+\frac{1}{d\omega+1}=2\\
$$
Is it no coincidence that replacing $\omega$ with other two cube root of unity satisfy the equation ?
| Note that $(1+a_1)(1+a_2)...(1+a_n)= 1 + SP + DP + TP + ... nP$, where:
*
*SP stands for "single products": $a_1 + a_2 + ... + a_n$,
*DP for "double products" $a_1a_2 + a_1a_3 + ... $,
*TP for "triple products" $a_1a_2a_3 + ...$ and so on until nP which is $a_1a_2... a_n$
And, for the same reasons, $(\omega +a_1)(\omega+a_2)...(\omega+a_n)= \omega^n + \omega^{n-1}SP + \omega^{n-2}DP + \omega^{n-3}TP + ... nP$.
Then, you just want to sum everything:
$$0 = \dfrac{1}{a+\omega} + \dfrac{1}{b+\omega} +\dfrac{1}{c+\omega} +\dfrac{1}{d+\omega} - \dfrac{2}{w} = \dfrac{m\omega - n}{(a+\omega)(b+\omega)(c+\omega)(d+\omega)\omega},$$
where $m= 2 - abc - abd - acd - bcd$ and $n = 2abcd - a-b-c-d$ are real numbers. Please note that the numerator was reduced by using $\omega^3=1$ and $\omega^4=\omega$ when needed, and the double products cancel due to the "-2" term.
If $m\neq 0$, then $\omega = \dfrac{n}{m} \in \mathbb{R}$ which contradicts the hypotesis. Thus, $m=0$, and it follows that $n=0$ too.
In particular, $$0 = m-n = 2 - abc - abd - acd - bcd - (2abcd - a-b-c-d)$$ Then, $$ 0 = \dfrac{2 - abc - abd - acd - bcd - (2abcd - a-b-c-d)}{(1+a)(1+b)(1+c)(1+d)} $$ $$= \dfrac{1}{1+a} + \dfrac{1}{1+b} + \dfrac{1}{1+c} + \dfrac{1}{1+d} - 2, $$ so we finish.
PS. It would be nice to see proofs using other methods less rustic than this :)
| {
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Integral check. Is partial fractions the only way? I'm getting a different answer from wolfram and I have no idea where. I have to integrate:
$$\int_0^1 \frac{xdx}{(2x+1)^3}$$
Is partial fractions the only way?
So evaluating the fraction first:
$$\frac{x}{(2x+1)^3} = \frac{A}{2x+1} + \frac{B}{(2x+1)^2} + \frac{C}{(2x+1)^3}$$
$$x = A(2x+1)^2 + B(2x+1) + C$$
$$x = A(4x^2 + 4x + 1) + 2Bx + B + C$$
$$x = 4Ax^2 + 4Ax + A + 2Bx + B + C$$
$$x = x^2(4A) + x(4A+2B) + A + B + C$$
$4A = 0$ and $4A+2B = 1$ and $A + B + C = 0$ so $A = 0$ and $B=\frac{1}{2}$ and $C = \frac{-1}{2}$
Is the partial fraction part right?
So then I get:
$$\int_0^1 \frac{xdx}{(2x+1)^3} = \int_0^1 \frac{dx}{(2(2x+1)^2)} - \int_0^1 \frac{dx}{2(2x+1)^3}$$
for both, using $u = 2x+1$ and $\frac{du}{dx} = 2$ and $du = 2dx$ and $\frac{du}{2} = dx$,
$$\frac{1}{4} \int \frac{du}{u^2} - \frac{1}{4} \int \frac{du}{u^3}$$
$$ = [\frac{1}{4} - u^{-1} - \frac{1}{4} \cdot \frac{1}{-2} u^{-2} ]_1^3$$
I'm going the route of changing the limits to the new u and since $u = 2x+1$, when $x = 0, u = 1$ and when $x = 1, u = 3$. Is this the right path?
finally I get
$$[\frac{-1}{4}u^{-1} + \frac{1}{8}u^{-2} ]_1^3$$
I plug in numbers but I get a different answer than wolfram...
| You've computed the integral correctly, probably you made a mistake while substituting in the limits of integration. Everything else upto that point is correct.
Is partial fractions the only way?
No single way is the only way, you have tons of ways to do this integral. Have a look at the following approaches.
*
*Manipulation and U-Substitution
*
*Make the substitution $\begin{bmatrix}t \\ \mathrm dt \end{bmatrix}=\begin{bmatrix}2x+1 \\ 2\mathrm dx\end{bmatrix}$
*Use the General Power Rule for Integrals
$$I=\int_{0}^{1}\dfrac{x \mathrm dx}{(2x+1)^3}\implies 2I=\int_{0}^{1}\dfrac{2x+1-1}{(2x+1)^3}\mathrm dx$$ $$2I=\int_{0}^{1}\biggl[\dfrac{1}{(2x+1)^2}-\dfrac{1}{(2x+1)^3}\biggr] \mathrm dx $$
$$\implies I=\dfrac{1}{4}\biggl[-\dfrac{1}{t}+\dfrac{1}{2t^2}\biggr]_{1}^{3}=\dfrac{1}{18}$$
*Trigonometric Substitution
*
*Make the substitution $\begin{bmatrix}x \\ \mathrm dx\end{bmatrix}=\begin{bmatrix}1/2\cdot\tan^2 \theta\\\tan\theta\sec^2\theta\mathrm d\theta\end{bmatrix}$
*Use the Pythagorean Identity involving tangent and secant functions namely $\sec^2\theta=1+\tan^2\theta$ to get $2x+1=\sec^2\theta$.
$$I=\int_{0}^{1}\dfrac{x\mathrm dx}{(2x+1)^3}=\int_{0}^{\sqrt{2}}\dfrac{\tan^3\theta \sec^2\theta}{\sec^6\theta}\mathrm d\theta$$
Using the definitions of $\tan\theta=\sin\theta/\cos\theta$ and $\sec\theta=1/\cos\theta$. The integral simplifies to the following form: $$I=\int_{0}^{\sqrt{2}}\sin^3\theta\cos\theta\mathrm d\theta=\int_{0}^{\sqrt{2}}\sin^3\theta \cdot\mathrm d(\sin\theta)$$
Again use the General Power Formula for Integrals to get the following expression:
$$I=\dfrac{1}{8}\biggl[\sin^4\tan^{-1}(\sqrt{2x})\biggr]_{0}^{1}=\dfrac{1}{18}$$
To compute $\sin\arctan(\sqrt{2})$, make use of the following easy to prove identity: $$\sin\tan^{-1} x=\dfrac{x}{\sqrt{1+x^2}}$$
| {
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Relation between roots and coefficients - manipulation of identities
The polynomial $x^3+3x^2-2x+1$ has roots $\alpha, \beta, \gamma$ . Find $$\alpha^2(\beta + \gamma) + \beta^2(\alpha + \gamma) + \gamma^2(\alpha + \beta)$$
I tried finding the relation using $-b/a$, $c/a$ and $-d/a$. I couldn’t seem to find anything. I also tried solving for one root but it gave me back the polynomial but with the root as the variable. Also the polynomial can not be factorised.
| Since the polynomial has three roots and its highest degree is 3, we can write
$$
p(x) = (x-\alpha)(x-\beta)(x-\gamma) = x^3 +3x^2 - 2x + 1.
$$
It then follows from
$$
x^3 - (\alpha+\beta+\gamma)x^2 + (\alpha\beta + \beta \gamma + \gamma \alpha)x - \alpha\beta \gamma = x^3+3x^2-2x + 1
$$
that
$$
\alpha+\beta+\gamma = -3, \quad \alpha\beta + \beta \gamma + \gamma \alpha = -2, \quad
\alpha\beta \gamma = -1.
$$
Note that
$$
-2\alpha = \alpha(\alpha\beta + \beta \gamma + \gamma \alpha) = \alpha^2(\beta+\gamma) +\alpha\beta\gamma = \alpha^2(\beta+\gamma) - 1.
$$
Thus $\alpha^2(\beta + \gamma)=1-2\alpha$.
Similarly, $\beta^2(\alpha + \gamma) = 1-2\beta$ and $\gamma^2(\alpha + \beta) = 1-2\gamma$.
Therefore,
\begin{align}
\alpha^2(\beta + \gamma) + \beta^2(\alpha + \gamma) + \gamma^2(\alpha + \beta)
&= (1-2\alpha) + (1-2\beta) + (1-2\gamma) \\
&= 3 -2(\alpha+\beta+\gamma) = 3 +6 =9.
\end{align}
| {
"language": "en",
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Solution of the first order partial differential equation . The integral surface of the first order partial differential equation $$2y(z-3)\frac{\partial z}{\partial x}+(2x-z)\frac{\partial z}{\partial y} = y(2x-3)$$ passing through the curve $x^2+y^2=2x, z = 0$ is
*
*$x^2+y^2-z^2-2x+4z=0$
*$x^2+y^2-z^2-2x+8z=0$
*$x^2+y^2+z^2-2x+16z=0$
*$x^2+y^2+z^2-2x+8z=0$
My effort:
I find the mulipliers $x, 3y, -z$ and get the solution $x^2+3y^2-z^2=c_1$. How to proceed further? Please help.
| All surfaces passing through the curve $x^2+y^2=2x, z = 0$. Bat only
first surface $$x^2+y^2-z^2-2x+4z=0\tag{1}$$ is solution of pde. We check it
Let $z=z(x,y)$. We differentiate equation $(1)$ respect $x$ and $y$:
$$2x-2zz'_x-2+4z'_x=0,\\
2y-2zz'_y+4z'_y=0.
$$
Solving this system we get
$$z'_x=\frac{x-1}{z-2},\quad z'_y=\frac{y}{z-2}$$
Next we substitute this in to pde. We get
$$\frac{2 \left( x-1\right) y\, \left( z-3\right) }{z-2}+\frac{y\, \left( 2 x-z\right) }{z-2}=\left( 2 x-3\right) \, y,$$
$$(2x-3)y=(2x-3)y.$$
Then $(1)$ is solution.
| {
"language": "en",
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How can I solve this crazy limit? $\lim _{x\rightarrow0}\frac{1-\cos\left(\frac{1-\cos x \cos 2x}{x^2}-\frac {5}{2}\right)\cos2x}{x^2} $ Firstly, I think this can be done with equivalent infinitesimal, but it seems so much complicated. I'm not very brave to do L'Hospital's rule on this question. And I don't think trig formulas can simplify this..
$$\lim _{x\rightarrow0}\frac{1-\cos\left(\frac{1-\cos x \cos 2x}{x^2}-\frac {5}{2}\right)\cos2x}{x^2} $$
| Note that $\cos(x)\cos(2x)= \frac 1 2 (\cos(x) + \cos(3x))$. Hence
$$\frac{1-\cos(x)\cos(2x)}{x^2}-5/2= \frac{1-\cos(x)}{2x^2}+ \frac {1-\cos(3x)}{x^2}-5/2\\
=\frac {x^2/2+O(x^4)}{2x^2}+ \frac {9x^2/2+O(x^4)}{2x^2}-\frac 5 2=\frac 5 2 +O(x^2)-\frac 5 2=O(x^2). $$
Also observe that
$$ \cos(x^2)\cos(2x)= (1-\frac{x^4}{2}+O(x^6))(1-2x^2+O(x^3))=1-2x^2+O(x^3).$$
Therefore we know the nominator up to second order. But this is enough to calculate the limit:
$$\frac{1-\cos(\frac{1-\cos(x)\cos(2x)}{x^2}-5/2)\cos(2x)}{x^2} \\
=\frac{1-\cos(O(x^4))\cos(2x)}{x^2}\\ = \frac{1-[1-2x^2+O(x^3)]}{x^2}=2 $$
| {
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The least integer value of $x$, which satisfy $|x| + |\frac{x}{x - 1}| = \frac{x^2}{|x - 1|}$, is...
Question: The least integer value of $x$, which satisfy $|x| + |\frac{x}{x - 1}| = \frac{x^2}{|x - 1|}$, is...
MY ATTEMPT:
Case 1: Taking all the modulus positive and solving ...
Case 2: Taking all the modulus negative and solving...
Case 3: Taking different signs of different modulus at a time and solving...
But the Case 3 will arise to many different cases and this will take a lot of time to solve. So, definitely an alternate method is required. Please provide an efficient method to solve this question. Thank you!
P.S.: The answer to this question is x=0
This is a new edit made 8 hours ago. Anyone who posted their answers before that are not wrong and not flawed. Their answer will satisfy this question: The least integer value of $x$, which satisfy $|x| + |\frac{x}{x + 1}| = \frac{x^2}{|x - 1|}$, is.... Sorry for the confusion
| This is probably not the most efficient way of doing it, but in my opinion it is more fun
$$|x|+\frac{|x|}{|x+1|}=\frac{x^2}{|x-1|}$$
$$x^2+\frac{x^2}{(x+1)^2}+\frac{2x^2}{|x+1|}=\frac{x^4}{(x-1)^2}$$
So clearly $x=0$ is a solution. Taking out a factor of $x^2$, we get
$$1+\frac{1}{(x+1)^2}+\frac{2}{|x+1|}=\frac{x^2}{(x-1)^2}$$
$$\frac{x^2}{(x-1)^2}-\frac{1}{(x+1)^2}-1=\frac{2}{|x+1|}$$
$$\frac{x^4}{(x-1)^4}+\frac{1}{(x+1)^4}+1-\frac{x^2}{(x-1)^2(x+1)^2}-\frac{x^2}{(x-1)^2}+\frac{1}{(x+1)^2}=\frac{4}{(x+1)^2}$$
$$\frac{3}{(x+1)^2}=\frac{x^4(x+1)^4+(x-1)^4+(x+1)^4(x-1)^4-x^2(x+1)^2(x-1)^2-x^2(x-1)^2(x+1)^4}{(x-1)^4(x+1)^4}$$
So if we say $x\neq-1,1$, we can multiply through by $(x+1)^4(x-1)^4$:
$$3(x-1)^4(x+1)^2=x^4(x+1)^4+(x-1)^4+(x+1)^4(x-1)^4-x^2(x+1)^2(x-1)^2-x^2(x-1)^2(x+1)^4$$
So it will be the solutions of this polynomial except for $x=1,-1$
| {
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Proof of $ \sqrt{2}|z|\geq|Re(z)| + |Im(z)|$ I need to prove $\sqrt{2}|z|\geq|Re(z)| + |Im(z)|$
This is what I tried to do :
$Let$ $z=x+iy$ , this gives us $|Re(z)| = |x|$ and $|Im(z)|=|y|$
Also , $|z| = \sqrt{x^2+y^2}$
Then we have $\sqrt{2}\sqrt{x^2+y^2} \geq x +y $
$\Rightarrow$ $\sqrt{2x^2+2y^2} \geq x +y$
$\Rightarrow$ $2x^2+2y^2 \geq (x +y)^2$
$\Rightarrow$ $2x^2+2y^2 \geq x^2 +2xy +y^2$
$\Rightarrow$ $x^2+y^2 \geq 2xy $
Is this correct so far ? How do I proceed from here ?
| It should be $$\sqrt{2}\sqrt{x^2+y^2} \geq |x| +|y| $$
But if you write $a= |x|$ and $b= |y|$ then $$\sqrt{2}\sqrt{a^2+b^2} \geq a +b $$
and the proof goes as you did.
| {
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"timestamp": "2023-03-29T00:00:00",
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Adding sum of leftmost digits
Starting with a number of at least $9$ digits, every minute we add the sum of the leftmost $9$ digits of the current number to the number itself. Will we always, at some point, see $9$ consecutive numbers that are all not divisible by $9$?
The sum of the leftmost $9$ digits is at most $9\times 9=81$, so we never add more than $81$ each time. Consequently, when the number is large enough, the leftmost $9$ digits will stay the same for more than $9$ minutes. We can wait until the leftmost $9$ digits is divisible by $9$, to make sure that we have $9$ consecutive numbers that leave the same remainder when divided by $9$. If we can ensure that at some point this remainder is not $0$, we would be done.
| Let $a_n$ be the $n$th number. Clearly, the sequence $\{a_n\}_n$ is strictly increasing and hence unbounded. Let $M=10^m$ for some $m\ge 11$ and such that $M>a_1$. Then there is a minimal $n_0$ with $a_{n_0}\ge M$. As noted by the OP, $a_{n_0}\le a_{n_0-1}+ 81$, hence certainly $a_{n_0}$ has leading digits $100000000$. This means that the sequence will grow in steps of $1$ for a while, namely until we hit $10^m+10^{m-8}$ exactly. From here on, we will advance in steps of $2$ until we hit $10^m+2\cdot 10^{m-8}$ exactly. Now follow steps of $3$ until we hit $10^m+3\cdot 10^{m-8}+2$, then steps of size $4$ until we hit $10^m+4\cdot 10^{m-8}+2$, then steps of size $5$ until we hit $10^m+5\cdot 10^{m-8}+2$, then steps of size $6$ until $10^m+6\cdot 10^{m-8}+4$. Note that $10^m+5\cdot 10^{m-8}+2+k\cdot 6$ is not a multiple of $9$ (or even a multiple of $3$). Thus we have found $\approx \frac{10^{m-8}}6\gg 9$ consecutive terms that are not multiples of $9$
| {
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"timestamp": "2023-03-29T00:00:00",
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Constructing a homomorphism such that a given set is the kernel. Given the matrices with coefficients in $\mathbb Z_5$, I am asked to consider:
$$G= \left\{\begin{pmatrix}
a & 0 \\ 0 &d
\end{pmatrix} \mid ad\neq 0 \bmod 5 \right\}$$
and:
$$ H= \left\langle \begin{pmatrix}
1 & 0 \\ 0 &-1
\end{pmatrix}, \begin{pmatrix}
-1 & 0 \\ 0 &1
\end{pmatrix}\right \rangle$$
And then show that $G/H$ is a group. The standard approach would be to show that $H$ is the kernel of a homomorphism. The question would be, how would I construct such a group map/homomorphism?
I know $H$ consists of the elements:
$\begin{pmatrix}
1 & 0 \\ 0 &-1
\end{pmatrix} ^2 =\begin{pmatrix}
-1 & 0 \\ 0 &1
\end{pmatrix} ^2 = \begin{pmatrix}
1 & 0 \\ 0 &1
\end{pmatrix} $.
$ \begin{pmatrix}
1 & 0 \\ 0 &-1
\end{pmatrix} = \begin{pmatrix}
1 & 0 \\ 0 &4
\end{pmatrix}$
$ \begin{pmatrix}
-1 & 0 \\ 0 &1
\end{pmatrix}=\begin{pmatrix}
4 & 0 \\ 0 &1
\end{pmatrix} $
$\begin{pmatrix}
-1 & 0 \\ 0 &1
\end{pmatrix}\begin{pmatrix}
1 & 0 \\ 0 &-1
\end{pmatrix}= \begin{pmatrix}
-1 & 0 \\ 0 &-1
\end{pmatrix} = \begin{pmatrix}
4 & 0 \\ 0 &4
\end{pmatrix} $
I would like to map all these elements to the matrix
$\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}$
But I have trouble coming up with a good map of sorts.
| Consider the homomorphism $\phi:G\to H$ given by $\phi(g)=g^2\ \forall g\in G$. $\ker(\phi)=H$ since $H$'s elements have only $\pm1$ on the diagonal, which square to 1, yielding the identity matrix (in contrast, $(\pm2)^2=-1$ so the other elements of $G$ are not in the kernel). Thus $G/H$ is a group.
$G$ happens to be abelian, so all its subgroups are normal, including $H$.
| {
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Prove inequality between functions Show that $x^2<\tan{x}\arctan{x}$ if $x\in(0;\pi/2)$. I have thought about proving that $f(x) = \tan{x}\arctan{x} - x^2$ is monotonically increasing(its derivative is larger than 0), but its derivative ($f'(x) = \frac{\arctan{x}}{\cos^2x} + \frac{\tan x}{1 + x^2}-2x$) is hard to compare with 0. Please, can you give me a hint?
| Since $\tan x > x$ for all $x \in \left( 0,\dfrac{\pi}{2} \right)$, $\dfrac{\tan(x)}{x}> 1$. We claim that $g(x) = \dfrac{\tan(x)}{x}$ is strictly increasing on that interval.
$$g'(x) = \left(\frac{\tan x}x\right)'=\frac{\frac{x}{\cos^2 x}-\tan x}{x^2}=\frac{x-\sin x\cos x}{x^2\cos^2 x}=\frac{2x-\sin(2x)}{2x^2\cos^2x}>0,$$ Source: linked answer
Consider $g(\arctan x) < g(x)$, and you'll get the desired inequality.
$$\frac{x}{\arctan x}<\frac{\tan x}{x} \iff \tan x \arctan x > x^2$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the modulus and argument of complex numbers My problem:
Find the magnitude and argument of the following.
$$\frac{3+4i}{1-i} + \frac{2-i}{2+3i}$$
My solution so far:
\begin{align}
\frac{3+4i}{1-i} + \frac{2-i}{2+3i} &= \frac{3+4i}{1-i} \times \frac{1+i}{1+i} + \frac{2-i}{2+3i} \times \frac{2-3i}{2-3i}\\
&=\frac{3+3i+4i+4(-1)}{1+1} + \frac{4-6i-2i+3(-1)}{4+9(-1)}\\
&= \frac{3+7i-4}{2} + \frac{4-8i-3}{13} \\
&= \frac{-1+7i}{2} + \frac{1-8i}{13}\\
&= \frac{-13+91i+2-16i}{26}
\end{align}
Thus the complex number is given by
$$ z = \frac{-11+75i}{26}= \frac{-11}{26} + \frac {75}{26}i $$
Then, in order to find the modulus, I computed
$$|z| =\sqrt{x\cdot x + y\cdot y}$$
where $x = -11/26$ and $y = 75/26 $. Then,
$$|z| = \sqrt{\frac{-121}{676} + \frac {5625}{676}} = \sqrt{\frac{5746}{676}}$$
This is where I stopped.
How do I find the argument of $z$ from here?
| So as I touched on in the comments, we begin by "rationalizing the denominator" in a sense, in that we multiply each fraction's top and bottom by the conjugate of the denominator. Thus,
$$\frac{3+4i}{1-i} + \frac{2-i}{2+3i} = \frac{3+4i}{1-i} \left( \frac{1+i}{1+i} \right) + \frac{2-i}{2+3i} \left( \frac{2-3i}{2-3i} \right) \tag 1$$
The bottoms simplify nicely since $(a-b)(a+b)=a^2 - b^2$. Alternatively, since we have complex numbers, $z \bar z = |z|^2$; whichever you prefer to use.
Anyhow, $(1)$ becomes, with simplification,
$$\frac{3+4i}{1-i} \left( \frac{1+i}{1+i} \right) + \frac{2-i}{2+3i} \left( \frac{2-3i}{2-3i} \right) = \frac{-1+7i}{2} + \frac{1-8i}{13} \tag 2$$
We combine the real and imaginary parts to get
$$-\frac{11}{26} + \frac{75}{26}i$$
Up to this point, you've done everything correctly.
The number is a little messy but it can be worked with. Let $a = -11/26, b = 75/26$. Then the above number is $a+bi$. We know that for numbers in this form that
$$|a+bi| = \sqrt{a^2 + b^2} \;\;\;\;\;\; \arg(a+bi) = atan2(b,a)$$
Both of these are fairly evident if you imagine plotting $a+bi$ on the plane. Then its magnitude is the distance from the origin, which you can easily find from the Pythagorean theorem, and its argument is the angle it makes with the positive real axis (with counterclockwise rotation being positive). The $atan2$ function can calculate this directly if you like, but the more "intuitive" definition (angle from the positive real axis) may be easier to contend with for you, all depending - you don't want to memorize that complex formula after all. You will have to use a little trig, regardless.
A rough image for a generic $z$ in the complex plane, to help further what I mean by these intuitive notions:
In any event, substitute $a,b$ into the expressions above and do the arithmetic to find the magnitude and argument.
If it were me finding the argument, I'd note that, since $a<0,b>0$, then $z=a+bi$ is in the top-left quadrant of the complex plane. Then I would find the angle $z$ makes with the imaginary axis and add $\pi/2$ radians to it. This diagram makes it clear why:
| {
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A mistake on computing $\int \frac{dx}{\sqrt{x+1}+1}$ I have to find $\int \frac{dx}{\sqrt{x+1}+1}$. This was my attempt, which is wrong and I cannot find where exactly is the mistake.
First I write $\frac{1}{\sqrt{x+1}+1}=\frac{\sqrt{x+1}-1}{x}=\frac{\sqrt{x+1}}{x}-\frac{1}{x}$, therefore $\int \frac{dx}{\sqrt{x+1}+1}=\int \frac{\sqrt{x+1}}{x}dx-\log\left (x\right )$, so I only have to deal with $\int \frac{\sqrt{x+1}}{x}dx$.
Setting $u=\sqrt{x+1}$, we obtain $du=\frac{1}{2\sqrt{x+1}}dx$, therefore $dx=2udu$ and $x=u^2-1$, so
$\int\frac{\sqrt{x+1}}{x}dx=\int\frac{2u^2}{u^2-1}du=\int \left(2+\frac{2}{u^2-1}\right)du=2\sqrt{x+1}+\int \left(\frac{1}{u-1}-\frac{1}{u+1}\right)du=$
$=2\sqrt{x+1}+\log \left (\sqrt{x+1}-1\right )-\log \left (\sqrt{x+1}+1\right )=2\sqrt{x+1}+\log\left ( \frac{\sqrt{x+1}-1}{\sqrt{x+1}+1} \right )$
But then $\int \frac{dx}{\sqrt{x+1}+1}=2\sqrt{x+1}+\log\left ( \frac{\sqrt{x+1}-1}{\sqrt{x+1}+1} \right )-\log\left (x\right )=2\sqrt{x+1}+\log\left ( \frac{x+2-2\sqrt{x+1}}{x^2} \right )$, which is not what I am supposed to obtain. The actual answer is $2\sqrt{x+1}-2\log\left ( 1+\sqrt{x+1} \right )$.
Where is my mistake?
(I already know a correct way to solve it, I just want to know where I am committing a mistake).
| You're fine.
$$\log\left(\frac{\sqrt{x+1}-1}{\sqrt{x+1}+1}\right) -\log x $$
$$= \log\left(\frac{\sqrt{x+1}-1}{\sqrt{x+1}+1}\frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}\right) -\log x $$
$$=\log\left( \frac{x}{ (\sqrt{x+1}+1)^2}\right) -\log x$$
$$=\log x -2\log(\sqrt{x+1}+1)- \log x$$
$$=-2\log(\sqrt{x+1}+1).$$
| {
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Why aren't these two limits (calculated with CAS) equal? The first limit is
$$\lim_{(x,y) \to (0,0)} \frac{\partial^2}{\partial y \partial x} \frac{x^3 y -x y^3}{x^2 +y^2},$$
computed by
(%i17) limit(diff(diff((x^3*y-x*y^3)/(x^2+y^2),x,1),y,1),(y,x),(0,0));
(%o17) -1
The second limit is
$$\lim_{(y,x) \to (0,0)} \frac{\partial^2}{\partial x \partial y} \frac{x^3 y -x y^3}{x^2 +y^2},$$
computed by
(%i22) limit(diff(diff((x^3*y-x*y^3)/(x^2+y^2),y,1),x,1),(x,y),(0,0));
(%o22) 1
The CAS is Maxima.
Why do the two outputs differ? What does it result from?
| In addition to the judicious answer from Christian Blatter :
$$f(x,y)=\frac{\partial^2}{\partial x\partial y}\left(\frac{x^3y-xy^3}{ x^2+y^2}\right)=\frac{(x^2-y^2)(x^4+10 x^2 y^2+y^4)}{ (x^2+y^2)^3}\qquad
(x,y)\ne(0,0)$$
we observe that :
Case 1 :
$$f(x,0)=1$$
Thus if $y$ is set to $0$ and $x\to 0$ the limit is $1$.
Case 2 :
$$f(0,y)=-1$$
Thus if $x$ is set to $0$ and $y\to 0$ the limit is $-1$.
Even more, if $\quad y=cx\quad$ with $\quad c\ne 0\quad$ we get:
$$f(x,cx)=\frac{(1-c^2)(1+10c^2+c^4)}{(1+c^2)^3}$$
Thus if $x\to 0$ following the path $y=cx$, the limit is $\frac{(1-c^2)(1+10c^2+c^4)}{(1+c^2)^3}$
So, they are as many different limits as many manner to make $x\to 0$ and $y\to 0$ according to various paths relating $x$ and $y$.
In other words, there is no determined limit.
| {
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If nine coins are tossed, what is the probability that the number of heads is even? If nine coins are tossed, what is the probability that the number of heads is even?
So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.
We have $n = 9$ trials, find the probability of each $k$ for $k = 0, 2, 4, 6, 8$
$n = 9, k = 0$
$$\binom{9}{0}\bigg(\frac{1}{2}\bigg)^0\bigg(\frac{1}{2}\bigg)^{9}$$
$n = 9, k = 2$
$$\binom{9}{2}\bigg(\frac{1}{2}\bigg)^2\bigg(\frac{1}{2}\bigg)^{7}$$
$n = 9, k = 4$
$$\binom{9}{4}\bigg(\frac{1}{2}\bigg)^4\bigg(\frac{1}{2}\bigg)^{5}$$
$n = 9, k = 6$
$$\binom{9}{6}\bigg(\frac{1}{2}\bigg)^6\bigg(\frac{1}{2}\bigg)^{3}$$
$n = 9, k = 8$
$$\binom{9}{8}\bigg(\frac{1}{2}\bigg)^8\bigg(\frac{1}{2}\bigg)^{1}$$
Add all of these up:
$$=.64$$ so there's a 64% chance of probability?
| Nine coins, so that two events
$\mathscr{E}_1$ = #heads is even and
$\mathscr{E}_2$ = #tails is even
are mutually exclusive (the number of tails is 9 - number of heads, so former is even iff latter odd) and comprise all possibilities, thus $P(\mathscr{E}_1) + P(\mathscr{E}_2) =1$. But if the coins are fair, then the probabilities must be unchanged if we swap the roles of heads and tails. Hence $P(\mathscr{E}_1)= P(\mathscr{E}_2)$ and we immediately see both probabilities must be $\frac{1}{2}$.
Now you are wondering why your approach doesn't work, because it is basically sound. You've simply made a slip.
You're approach is: sum every second term in the 10 member (i.e. an even number of terms) sequence whose $n^{th}$ term is the probability of $n$ heads. So the sum is:
$$S_1=\sum_{k=0}^{N/2} \binom{N}{2\,k}\left(\frac{1}{2}\right)^N$$
with $N$ odd (here equal to 9).
But, by dint of $\binom{N}{2\,k} = \binom{N}{N-2\,k}$, this sum is equal to the sum of all the other terms
$$S_2 =\sum_{k=0}^{N/2} \binom{N}{N-2\,k}\left(\frac{1}{2}\right)^N$$
in the sequence that don't belong to the first sum. So $S_1=S_2$ and clearly $S_1+S_2=1$, because this sum is the sum of probabilities of all possible mutually exclusive outcomes, therefore 1, or, alternatively, call up the binomial theorem and see that $S_1+S_2=\left(\frac{1}{2} + \frac{1}{2}\right)^9=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3134991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "54",
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How can I factor $x^{12}+x^{11}+\cdots+x+1$ in $\mathbb F_3[x]$? I can prove that $\mathbb F_3[x]$ is a UFD, so $f(x)=x^{12}+x^{11}+\cdots+x+1$ can be factored. And because neither 0, 1, nor 2 is a root of $f(x)$, all factors are more than 1 degrees.
But I don’t know how to factor $f(x)=x^{12}+x^{11}+\cdots+x+1$, or as another simple example, $g(x)=x^4+x^3+x+2$.
How do I find out factorizations of $f(x)$ and $g(x)$ in $\mathbb F_3[x]$?
| $$x^4+x^3+x+2=x^4+x(x^2+1)-1=(x^2+1)(x^2+x-1).$$
$$x^{12}+x^{11}+x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1=$$
$$=x^{12}+x^{11}-2x^{10}-8x^9-5x^8+7x^7+16x^6+7x^5-5x^4-8x^3-2x^2+x+1=$$
$$=(x^3-x-1)(x^3+x^2-1)(x^3+x^2+x-1)(x^3-x^2-x-1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3140347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Use of generating function in combinatorial proof In Combinatorial proof of summation of $\sum\limits_{k = 0}^n {n \choose k}^2= {2n \choose n}$, the product
$\left(\sum_{a=0}^n\binom{n}{a}x^a\right)\left(\sum_{b=0}^n\binom{n}{b}x^{n-b}\right)=\sum_{c=0}^{2n}\left(\sum_{a+n-b=c}\binom{n}{a}\binom{n}{b}\right)x^c$
This approach uses a "generating function" and collects all the coefficients of $x^{c}$ for fixed $c$ into $\sum_{a+n-b=c}\binom{n}{a}\binom{n}{b}$, to get $\sum_{c=0}^{2n}\left(\sum_{a+n-b=c}\binom{n}{a}\binom{n}{b}\right)x^c$
I know that $\left(\sum_{a=0}^n\binom{n}{a}x^a\right)\left(\sum_{b=0}^n\binom{n}{b}x^{n-b}\right)= \sum_{a=0}^{n}\sum_{b=0}^{n}\binom{n}{a}\binom{n}{b}x^{a+n-b}$, but it remains unclear how one can obtain $\sum_{c=0}^{2n}\left(\sum_{a+n-b=c}\binom{n}{a}\binom{n}{b}\right)x^c$ from the initial product.
| Here we also have an application of the Cauchyproduct of series.
\begin{align*}
\left(\sum_{a=0}^n\binom{n}{a}x^a\right)\left(\sum_{b=0}^n\binom{n}{b}x^{n-b}\right)
=\sum_{c=0}^{\color{blue}{2n}}\left(\sum_{\color{blue}{{a+(n-b)=c}}\atop{a\geq 0,n-b\geq 0}}\binom{n}{a}\binom{n}{b}\right)x^c\tag{1}
\end{align*}
*
*We have the (blue marked) upper limit $2n$ which comes from $x^ax^{n-b}=x^{a+n-b}$ if $a=(n-b)=n$, i.e. $a=n$ and $b=0$.
*We have $x^c=x^ax^{n-b}$ with $0\leq c\leq 2n$. In order to get the coefficient $x^c=x^{a+(n-b)}$ we have the (blue marked) condition $a+(n-b)=c$ as lower limit in the inner sum.
| {
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"url": "https://math.stackexchange.com/questions/3141260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int \frac{x^2 + x + 1}{(x+1)^2(x+2)}dx$ via partial fractions $\int \frac{x^2 + x + 1}{(x+1)^2(x+2)}dx$
= $ \int \frac{Ax+B}{(x+1)} + \frac{Cx+B}{(x+1)^2} + \frac{Dx+E}{x+2}$
= $\int (Ax+B)(x+1)(x+2) + (Cx+B)(x+2) + (Dx+E)(x+1)^2$
= $ \int Ax^3 + 3Ax^2 + 2Ax + Bx^2 + 3Bx + 2B + Cx^2 + 2Cx + Bx + 2B + Dx^3 + 2Dx^2 + Dx + Ex^3 + 2Ex^2 + E$
= $ \int (A + D + E)x^3 + (3A + B + C + 2D + 2E)x^2 + (2A + 2C + 4B + D)x + (4B + E)$
Turn into matrix, find reduced row echelon form to solve system of equations:
$\begin{bmatrix}
0 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 1 & 1 & 0 \\
3 & 1 & 1 & 2 & 2 & 1 \\
2 & 4 & 2 & 1 & 0 & 1 \\
0 & 4 & 0 & 0 & 1 & 1 \\
\end{bmatrix}$
This is where things go wrong, apparently this doesn't reduce down properly :( and I have been relying on RREF to solve systems of equations everytime up until now.
I am also confused where the $x^2 + x + 1$ is supposed to go exactly. I know I put it into the system of equations later but until then, I feel like I kinda just ignored it and left it out of all my work up until then (is that okay?).
Other note: I recognize that this is a proper rational fraction so no long division is nesecary and that this has irreducible factors that are repeated so that is why I split them up into the partial fractions up above in that manner. Did I miss any intermittent steps that made the RREF turn out wrong? I am not sure where I went wrong thus far either
| $\begin{align}\dfrac{x^2 + x + 1}{(x+1)^2(x+2)}=\dfrac{x^2 + 2x + 1-x}{(x+1)^2(x+2)}=&\dfrac{1}{(x+2)}-\dfrac{ (x+2-2)}{(x+1)^2(x+2)}\\=&\dfrac{1}{(x+2)}-\dfrac{ 1}{(x+1)^2}+\dfrac{2}{(x+1)^2(x+2)}\end{align}$
You may use partial fraction decomposition for $$\dfrac{2}{(x+1)^2(x+2)}=\dfrac{A}{x+1}+\dfrac{B}{(x+1)^2}+\dfrac{C}{x+2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3141438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simple Inequality Proof in Topology (Proving Open Rectangle Must Lie in Unit Disk) I am trying to complete the following exercise from "Topology Without Tears" by Sidney A. Morris (accessible at http://www.topologywithouttears.net/topbook.pdf). It is Exercise 2.2.1, and reads:
(i) Let $\langle a,b \rangle$ be any point in the disk $D = \{\langle a,b \rangle:x^2 + y^2 < 1\}$. Put $r = \sqrt{a^2 + b^2}$. Let $R_{\langle a,b \rangle}$ be the open rectangle with vertices at the points $\langle a \pm \frac{1-r}{8},b \pm \frac{1-r}{8} \rangle$. Verify that $R_{\langle a,b \rangle} \subset D$.
My Approach:
I am restricting myself to the case where $a>0$ and $b>0$. If I can prove that the top right vertex of the open rectangle lies in $D$, then the entire rectangle will lie in $D$. Thus, I must prove that $(a + \frac{1-r}{8})^2 + (b + \frac{1-r}{8})^2 < 1$. Expanding this, I acquire:
$$(a + \frac{1-r}{8})^2 + (b + \frac{1-r}{8})^2 = r^2 + (a+b)\frac{1-r}{4} + 2(\frac{1-r}{8})^2.$$
From here, I keep hitting dead-ends. Using $a+b<2$ gives the continuation:
$$(a + \frac{1-r}{8})^2 + (b + \frac{1-r}{8})^2 < r^2 + \frac{1}{2}(1-r) + 2(\frac{1-r}{8})^2.$$
I multiplied out this expression to acquire:
$$(a + \frac{1-r}{8})^2 + (b + \frac{1-r}{8})^2 < \frac{33}{32}r^2 - \frac{9}{16}r + \frac{17}{32}.$$
If I factor $r$ out from the first two terms on the RHS, we acquire:
$$(a + \frac{1-r}{8})^2 + (b + \frac{1-r}{8})^2 < r(\frac{33}{32}r - \frac{9}{16}) + \frac{17}{32}.$$
If $\frac{33}{32}r > \frac{9}{16}$, then $r < 1$ implies:
$$(a + \frac{1-r}{8})^2 + (b + \frac{1-r}{8})^2 < r(\frac{33}{32}r - \frac{9}{16}) + \frac{17}{32} < \frac{33}{32}r - \frac{9}{16} + \frac{17}{32}$$
$$<\frac{33}{32} - \frac{9}{16} + \frac{17}{32} = 1.$$
But if we don't have $\frac{33}{32}r > \frac{9}{16}$, we can't do this since the expression in the parentheses would be negative, and so removing $r$ would no longer make the expression larger. I'm thinking there must be a simpler way to prove this.
EDIT: Solution Possibly Found
With help from the answer I have marked as most helpful below, I believe I have arrived at a relatively complete solution. It is easy to see geometrically that we must have $a+b < \sqrt{2}r$, giving us a stronger bound than $a+b<2$. To see this more rigorously, I reference this other question on MathExchange: Prove that $2|ab| \leq a^2 + b^2$ and $|a|+|b| \leq \sqrt {2}(a^2+b^2)^{1/2}$.
Proceeding, we have:
$$(a + \frac{1-r}{8})^2 + (b + \frac{1-r}{8})^2 = r^2 + (a+b)\frac{1-r}{4} + 2(\frac{1-r}{8})^2$$
$$< r^2 + \sqrt{2}r\frac{1-r}{4} + 2(\frac{1-r}{8})^2 = (r+\sqrt{2}(\frac{1-r}{8}))^2 = ((1-\frac{\sqrt{2}}{8})r + \frac{\sqrt{2}}{8})^2 < 1^2 = 1$$
| Hint: Draw a picture. The line $x+y = \sqrt{2}$ is tangent to the unit circle, so $a + b < \sqrt{2}$. That should be enough stronger than $a+ b < 2$ to finish the proof. Maybe $a+b < \sqrt{2} < 3/2$ will do.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Question about partial fractions with irreducible quadratic factors Given this rational function: $$\frac{-4x^4-2x^3-26x^2-8x-44}{(x+1)(x^2 +3)^2}$$
The decomposition would look like this: $$\frac{A}{x+1} + \frac{Bx+C}{(x^2+3)} + \frac{Dx+E}{(x^2+3)^2}$$
And the final answer would be: $$\frac{-4}{x+1} - \frac{2}{(x^2+3)} - \frac{2}{(x^2+3)^2}$$
But, if you were to set it up like this: $$\frac{A}{x+1} + \frac{B}{(x^2+3)} + \frac{C}{(x^2+3)^2}$$
You end up with the same answer:
$$\frac{-4}{x+1} - \frac{2}{(x^2+3)} - \frac{2}{(x^2+3)^2}$$
I want to know why this is.
I thought that for irreducible quadratic factors, you needed a linear term in the numerator. Is this purely coincidence that it worked with constant terms in the numerator?
| In the first expansion B = D = 0, but with a different but similar question that would not necessarily happen.
I can provide an example easily by starting with values for A = B = C = D = E = 1, working backwards to a new question that will not have the coincidental property that your one has.
Here is the new question:
$$\frac{2x^4+2x^3+11x^2+8x+44}{(x+1)(x^2 +3)^2}$$
The decomposition will, as before, look like this:
$$\frac{A}{x+1} + \frac{Bx+C}{(x^2+3)} + \frac{Dx+E}{(x^2+3)^2}$$
and with the values substituted in,
$$\frac{1}{x+1} + \frac{x+1}{(x^2+3)} + \frac{x+1}{(x^2+3)^2}$$
Robert Z has added a neat answer that goes further and looks at the underlying theory.
Incidentally, the free online Wolfram Alpha is really handy for playing around with these sort of expressions and it'll let you check your answers from partial fraction questions.
Link https://www.wolframalpha.com/
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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} |
Integrate $\int \frac{dx}{\sqrt{x^2-9}}$ by trig substitution $$
\begin{align}
x = 3\sec\theta, dx &= 3\sec\theta\tan\theta d\theta\\\\
\int \frac{dx}{\sqrt{x^2-9}} &= \int \frac{3\sec\theta\tan\theta d\theta}{\sqrt{(3\sec\theta)^2 - 3^2}} \\\\
& = \int \frac{3\sec\theta\tan\theta d\theta}{\sqrt{3^2(\sec^2\theta -1)}} \\\\
&= \int \frac{3\sec\theta\tan\theta d\theta}{\sqrt{3^2\tan^2\theta}} = \int \sec\theta\\\\
&= \ln|\sec\theta + \tan\theta| + C = \ln| \frac{x}{3} + \frac{\sqrt{x^2-9}}{3}|
\end{align}
$$
However, wolphram alpha says the answer is $\ln |x+ \sqrt{x^2-9}$
I am wondering how did it get rid of the 3 in the denominator? This is pretty much how I got my answer:
$$
x = 3\sec\theta \\
\frac{x}{3} = \sec\theta \\
\frac{\sqrt{x^2-9}}{3} = \tan\theta
$$
| The indefinite integral of $f$, i.e. $\int f$ is not a function, but a set of functions:
$$\int f := \{g| g'=f \}$$
But we usually denote the set with one of it's element $+C$:
$$\int f = g + C \tag{1}$$
Or simply without the $+C$. Clearly, the $=$ sign in $(1)$ is not the same you'd write in $1=1$, for example. A really precise mathematician would use a different symbol, for example:
$$\int f \color{blue}{=} g$$
with $\color{blue}{=}$ means that $a\color{blue}{=}b$ iff $a'=b'$. So they would write:
$$\int x \mathrm{d}x \color{blue}{=} \frac{x^2}{2} \color{blue}{=} \frac{x^2}{2}+3$$
But we are lazy, and usually the meaning of $=$ is clear from the context.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3146542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $\sin(x) - \cos(x) = 1/3$ then determine $\sin(x)\cos(x)$
If
$$\sin(x) - \cos(x) = \frac{1}{3}$$
then determine
$$\sin(x)\cos(x)$$
I know that the expected solution is squaring both sides of equation and solving it this way:
\begin{gather}
\sin^2(x)+\cos^2(x)= 1 \\[4px]
(\sin(x) - \cos(x))^2 = \left(\frac{1}{3}\right)^2 \\[4px]
\sin^2(x) - 2\sin(x)\cos(x) + \cos^2(x) =\frac{1}{9} \\[4px]
-2\sin(x)\cos(x)=\frac{1}{9} -\sin^2(x)-\cos^2(x) \\[4px]
2\sin(x)\cos(x)=-\frac{1}{9} +\sin^2(x)+\cos^2(x) \\[4px]
2\sin(x)\cos(x)=-\frac{1}{9} +1\\[4px]
2\sin(x)\cos(x)=\frac{8}{9} \\[4px]
\sin(x)\cos(x)=\frac{4}{9}
\end{gather}
But assume I haven't noticed that I can solve it by squaring both sides in the first place. I can't figure it out how to solve it any other way.
| You can notice that
$$
\sin x-\cos x=\sqrt{2}\sin(x-\pi/4)
$$
Set $y=x-\pi/4$; then $\sin y=\frac{1}{3\sqrt{2}}$ and
\begin{align}
\sin x\cos x
&=\sin(y+\pi/4)\cos(y+\pi/4) \\[4px]
&=\left(\frac{1}{\sqrt{2}}\cos y+\frac{1}{\sqrt{2}}\sin y\right)
\left(\frac{1}{\sqrt{2}}\cos y-\frac{1}{\sqrt{2}}\sin y\right) \\[4px]
&=\frac{1}{2}(\cos^2y-\sin^2y) \\[4px]
&=\frac{1}{2}(1-2\sin^2y) \\[4px]
&=\frac{1}{2}\left(1-\frac{2}{(3\sqrt{2})^2}\right)
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Differentiate $11x^5 + x^4y + xy^5=18$ I am not sure how to differentiate $11x^5 + x^4y + xy^5=18$. I have a little bit of experience with implicit differentiation, but I'm not sure how to handle terms where both variables are multiplied together.
I have tried
$$\frac{d}{dx}(11x^5 + x^4y+xy^5) = \frac{d}{dx}(18)$$
$$\frac{d}{dx}(11x^5)+\frac{d}{dx}(x^4y) + \frac{d}{dx}(xy^5)=0$$
differentiating each term
$$\frac{d}{dx} (11x^5)=11\frac{d}{dy}(5x^4) = 55x^4$$
$$\frac{d}{dx} (x^4y) = [4x^3 \cdot y] + [1 \cdot x^4] = 4yx^3+x^4$$
$$\frac{d}{dx}(xy^5) = [1 \cdot y^5] + [x \cdot 5y^4] = y^5 + 5xy^4$$
finding $\frac{dy}{dx}$
$$\frac{dy}{dx}=\frac{-x}{y} = \frac{-[4yx^3+x^4] + [y^5+5xy^4]}{55x^4}$$
According to the website I'm using, "WebWork", this is wrong.
| you actually apply the derivative with respect to the variable $x$ in both members and develop using your knowledge of the product's derivative and chain rule. Note that to calculate implicit derivatives you must think of for example $y = y (x)$. This justifies the passage I marked with (*). For examples $\frac{d}{dx}(y^5)= 5y^4\frac{dy}{dx}$ exactly because $y = y (x)$. Then
$$\frac{d}{dx}(11x^5 + x^4y+xy^5) = \frac{d}{dx}(18)$$
$$\frac{d}{dx}(11x^5)+\frac{d}{dx}(x^4y) + \frac{d}{dx}(xy^5)=0$$
$$(55x^4)+ \left( 4x^3 y + x^4 \frac{dy}{dx} \right) +\left( y^5+5xy^4\frac{dy}{dx}\right) = 0 \,\,\,\,\,\,\,\,\,\,(*)$$
$$\left( x^4 \frac{dy}{dx} \right) +\left(5xy^4\frac{dy}{dx}\right) = -55x^4-4x^3 y- y^5$$
$$\left( x^4 +5xy^4\right)\frac{dy}{dx} = -55x^4-4x^3 y- y^5$$
$$\frac{dy}{dx} = \dfrac{-55x^4-4x^3 y- y^5}{\left( x^4 +5xy^4\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3147423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
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} |
$e^{\frac{1}{x}} < 1 + \frac{1}{x-1} $ I want to prove that $e^{1/x} < 1 + \frac{1}{x-1}$ for $x > 1$.
The first thing I tried is differentiating $f(x) = e^{1/x} - 1 - \frac{1}{x-1}$: this gives
$$ \frac{1}{x^2} \left( \left(1 + \frac{1}{x-1}\right)^2 - e^{\frac{1}{x}} \right) $$
If I could show that $\left(1 + \frac{1}{x-1} \right)^2 > e^\frac{1}{x} $ for $x>1$, then $f(x)$ would be increassing, and since $ \lim_{x \to \infty} f(x) = 0$ this would mean that $f(x) < 0$ for $x>1$. However, proving that inequality is very similar to the first one, and still involves bounding above $e^\frac{1}{x}$.
The other thing I tried is considering $f(x) = e^{x-1} - (x-1) - 1$ which is increasing for $x > 1$. Then $f(\frac{1}{x})$ is decreasing, so $e^{\frac{1}{x}-1} - \frac{1}{x}$ is decreasing; However this also does not seem to help too much.
Any ideas?
| HINT::
$$\begin{align}
e^{\frac{1}{x}} &< 1 + \frac{1}{x-1} \implies e < \left(\frac{x}{x-1}\right)^x\\
\end{align}$$
And
$$\lim_{x\to\pm\infty}\left(\frac{x}{x-1}\right)^x = \lim _{x\to \pm\infty }\left(e^{x\ln \left(\frac{x}{x-1}\right)}\right) = e$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
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Proving $\left|\begin{smallmatrix}1&1&1\\a&b&c\\a^3&b^3&c^3\end{smallmatrix}\right|=(b-a)(c-b)(c-a)(a+b+c)$
Prove that$$\begin{vmatrix}1&1&1\\a&b&c\\a^3&b^3&c^3\end{vmatrix}=(b-a)(c-b)(c-a)(a+b+c)$$
My attempt:
$$\begin{align}\begin{vmatrix}1&1&1\\a&b&c\\a^3&b^3&c^3\end{vmatrix}&=\begin{vmatrix}0&1&0\\a-b&b&c-b\\a^3-b^3&b^3&c^3-b^3\end{vmatrix}\\&=\begin{vmatrix}c-b&a-b\\c^3-b^3&a^3-b^3\end{vmatrix}\\&=(c-b)(a-b)\begin{vmatrix}1&1\\c^2+cb+b^2&a^2+ab+b^2\end{vmatrix}\\&=(c-b)(a-b)[(a^2+ab)-(c^2+cb)]\\\end{align}$$
Where did I go wrong?
| Note that $(a^2+ab)-(c^2+cb)=(a-c)(a+c)+b(a-c)=(a-c)(a+b+c)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3150198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Using Fermat's Little Theorem or Euler's Theorem to find the Multiplicative Inverse -- Need some help understanding the solutions here.
The answers to multiplicative inverses modulo a prime can be found without using the extended Euclidean algorithm.
a. $8^{-1}\bmod17=8^{17-2}\bmod17=8^{15}\bmod17=15\bmod17$
b. $5^{-1}\bmod23=5^{23-2}\bmod23=5^{21}\bmod23=14\bmod23$
c. $60^{-1}\bmod101=60^{101-2}\bmod101=60^{99}\bmod101=32\bmod101$
d. $22^{-1}\bmod211=22^{211-2}\bmod211=22^{209}\bmod211=48\bmod211$
The above is using Fermat's little theorem to find the multiplicative inverse of some modular functions. However, there is a final step just before arriving at the answer that I do not understand how to solve, except to solve it by factoring. Factoring takes a very long time.
Basically, I don't see how the answers move from the third step to the fourth step aside from arriving at the answer by factoring. There has to be a better way using Fermat's Theorem or Euler's Theorem.
| Bill's way seems great. Here's another approach (with the goal of finding easily reducible powers)
\begin{align}
8^{15}\pmod {17} &\equiv 2^{45}\\
&\equiv 2\cdot (2^{4})^{11}\\
&\equiv 2\cdot (-1)^{11} \tag{$16 \equiv -1$}\\
&\equiv 15 \tag{$15 \equiv -2$}
\end{align}
\begin{align}
5^{21}\pmod {23} &\equiv 5\cdot 5^{20}\\
&\equiv 5\cdot 25^{10}\\
&\equiv 5\cdot 2^{10} \tag{$25 \equiv 2$}\\
&\equiv 5\cdot 32^2\\
&\equiv 5\cdot 9^2 \tag{$32 \equiv 9$}\\
&\equiv 5\cdot 12 \tag{$81 \equiv 12$}\\
&\equiv 14 \tag{$60 \equiv 14$}
\end{align}
\begin{align}
60^{99}\pmod {101} &\equiv 10^{99}&\cdot 6^{99}\\
&\equiv 10\cdot 100^{49}&\cdot 6^4 \cdot (7776)^{19}\\
&\equiv -10&\cdot -6^4\\
&\equiv 12960\\
&\equiv 32
\end{align}
\begin{align}
22^{209}\pmod {211} &\equiv (2\cdot11)^{11\cdot19}\\
&\equiv ?\\
&\text{This is where the superiority of Bill's approach becomes obvious}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Range of function $f(x)=x \sqrt{1-x^2}$ How to find range of this function, it looks easy but somehow I am not able to do that . What is more important to me is , why is wolframalpha unable to find the range , is it not a simple enough function ?
| $1-x^2=(1-x)(1+x)$ needs to be positive because it is under the root. Hence,
$$(1-x)(1+x)\geq 0.$$
What can you conclude for possible values of $x$? The possible values of $x$ are limited on the interval $[-1,1]$. We have $f(-1)=0$ and $f(+1)=0$.
Then consider the derivative of the function
$$f'(x) = \sqrt{1-x^2}+x\dfrac{1}{2\sqrt{1-x^2}}\left(-2x \right)$$
$$=\dfrac{2(1-x^2)-2x^2}{2\sqrt{1-x^2}}=\dfrac{2(1-2x^2)}{2\sqrt{1-x^2}}$$
$$=\dfrac{1-2x^2}{\sqrt{1-x^2}}.$$
The derivative is $0$ for $1-2x^2=0\implies x = \pm \sqrt{1/2}$. The derivative is negative for $1\geq|x|\geq\sqrt{1/2}$ and positive for $|x|\leq \sqrt{1/2}$.
The extremal values are $f(\pm \sqrt{1/2})=\pm\sqrt{1/2}\sqrt{1-1/2}=\pm\sqrt{1/2}\sqrt{1/2}=\pm 0.5.$
As the continuous function vanishes on the bounds of $[-1,1]$ we know that the extremal values of the function will dictate the range of the function. The range is $[-1/2,1/2]$.
| {
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Prove that this map is a submersion (check my work please) There are some calculations but I would be very grateful if somebody could check whether my arguments are mathematically sound. I am new to differential geometry. You can also suggest easier solutions.
In the following,
$S_3$ denotes the 3-sphere in $\mathbb{C}^2$ : {$(x,y)\in \mathbb{C}^2 \space : \space |x|^2+|y|^2=1$}
and $\Omega=${$(x,y) \in S_3 \space | \space x^5+y^7 \neq0$}
Let $f:\Omega\rightarrow S_1$ be such that
$f(x,y)=\frac{x^5+y^7}{|x^5+y^7|}$
Prove that $f$ is a submersion.
Proof :
I will show that the differential of $f$ is a surjective map between tangent spaces.
$f$ is clearly not analytic but it should be smooth as a function of the real parts and imaginary parts of $x$ and $y$ (which are complex numbers here!).
Let $dx$ and $dy$ be «small» complex numbers (i.e. small real and imaginary parts).
$f(x+dx,y+dy)=\frac{(x+dx)^5+(y+dy)^7}{|(x+dx)^5+(y+dy)^7|}=\frac{x^5+y^7+5dx+7dy+o(dx,dy)}{|x^5+y^7|.|1+\frac{5dx+7dy}{x^5+y^7}|}$
Now, using $|z|=(z \bar z)^{1/2}$, I get that:
$$\color{red}{|1+\frac{5dx+7dy}{x^5+y^7}|^{-1}}=(1+\frac{5dx+7dy}{x^5+y^7})^{-1/2}(1+\overline{\frac{5dx+7dy}{x^5+y^7}})^{-1/2}=(1-\frac12 \frac{5dx+7dy}{x^5+y^7})(1-\frac12 \overline{\frac{5dx+7dy}{x^5+y^7}})+o(dx,dy)=1-\frac12 \frac{5dx+7dy}{x^5+y^7}-\frac12\overline{\frac{5dx+7dy}{x^5+y^7}}+o(dx,dy)=\color{red}{1-\Re(\frac{5dx+7dy}{x^5+y^7})+o(dx,dy)}$$
So,
$$\color{lime}{f(x+dx,y+dy)}=\frac{x^5+y^7+5dx+7dy+o(dx,dy)}{|x^5+y^7|}(1-\Re(\frac{5dx+7dy}{x^5+y^7})+o(dx,dy))=\color{lime}{\frac{x^5+y^7+5dx+7dy}{|x^5+y^7|}-\frac{x^5+y^7}{|x^5+y^7|}\Re(\frac{5dx+7dy}{x^5+y^7})+o(dx,dy)}$$
So identifying the linear parts, the differential of $f$ (viewed as a real function of the real and imaginary parts of $x$ and $y$, not complex!) is:
$$\color{blue}{d_{(x,y)}f(dx,dy)} = \frac{5dx+7dy}{|x^5+y^7|}-\frac{x^5+y^7}{|x^5+y^7|}\Re(\frac{5dx+7dy}{x^5+y^7})$$
$$=\frac12 \frac{5dx+7dy}{|x^5+y^7|}-\frac12 \frac{x^5+y^7}{|x^5+y^7|} \overline{\frac{5dx+7dy}{x^5+y^7}}$$
$$=\frac{1}{2|x^5+y^7|}(5dx+7dy-(x^5+y^7) \overline{\frac{5dx+7dy}{x^5+y^7}})$$
$$=\frac{x^5+y^7}{2|x^5+y^7|}(\frac{5dx+7dy}{x^5+y^7}-\overline{\frac{5dx+7dy}{x^5+y^7}})$$
$$=\frac{x^5+y^7}{2|x^5+y^7|}2i \Im(\frac{5dx+7dy}{x^5+y^7})$$
$$\color{blue}{=i\frac{x^5+y^7}{|x^5+y^7|}\Im(\frac{5dx+7dy}{x^5+y^7})}$$
$$\color{blue}{=if(x,y)\Im(\frac{5dx+7dy}{x^5+y^7})}$$
(so for this function I consider differentiability as in $\mathbb{R}^4 \rightarrow \mathbb{R}^2$ since in each of my "complex" variable I am in reality hiding two variables : the real and imaginary parts)
So now I just need to prove this constitutes a surjection from $T_{(x,y)}\Omega \rightarrow T_{f(x,y)}S_1$
We have that if $\theta$ is a regular real function then if $z(t)=e^{i\theta(t)}$, $z'(t)=i\theta'(t)e^{i\theta(t)}$ with $\theta'(t) \in \mathbb{R}$ so the tangent plane at $z$ of $S_1$ is just $T_zS_1=${${i \lambda z, \lambda \in \mathbb{R}}$}
With the same reasoning in higher dimension, $T_{(x,y)}\Omega =${$(i\lambda z, i\mu y), \lambda, \mu \in \mathbb{R}$}
When $dx, dy$ span $T_{(x,y)}\Omega$, $df$ clearly spans $T_{f(x,y)}S_1$ since we can just identify $\lambda =\Im(\frac{5dx+7dy}{x^5+y^7})$ which should span $\mathbb{R}$ I guess as $dx, dy$ vary. So the differential is a surjection between the tangent planes and therefore $f$ is a submersion. $\square$
Does my proof look sound to you?
| Going through that calculation... the differential of $x^5+y^7$ is $5x^4\,dx+7y^6\,dy$. You dropped the $x^4$ and $y^6$ factors there, in multiple places.
Now, the good news? The way you arranged your calculations, this is remarkably easy to fix. We can simply replace every instance of "$5\,dx+7\,dy$" with "$5x^4\,dx+7y^6\,dy$" and it will remain correct. The final differential is
$$df\cdot (dx,dy) = if(x,y)\operatorname{Im}\left(\frac{5x^4\,dx+7y^6\,dy}{x^5+y^7}\right)$$
On the sphere, $|f(x,y)| = 1$. In order for this differential to span the one-dimensional tangent space of $S^1$, it suffices to find some choice of $dx$ and $dy$ in the tangent space to $S^3$ so that the imaginary part there is nonzero. What is that tangent space?
$$|x|^2+|y|^2 = x\cdot \overline{x}+y\cdot \overline{y} = 1$$
$$\overline{x}\,dx+x\,\overline{dx}+\overline{y}\,dy+y\,\overline{dy} = 0$$
The tangent space is precisely the set of solutions to this linear equation, in what amounts to four variables. It's not just what you wrote down $\{(i\lambda z,i\mu y)\}$; that's the tangent space to the torus $S^1\times S^1$. Scaled copies of that $2$-manifold embed into the sphere, but there's another dimension in which $|x|$ and $|y|$ are allowed to vary.
All right, what does it take to find something in the tangent space for which the imaginary part is nonzero? Set $(dx,dy)=\left(\frac15ix,\frac17iy\right)$, so
$$\overline{x}\,dx+x\,\overline{dx}+\overline{y}\,dy+y\,\overline{dy} = \frac15i|x|^2 -\frac15i|x|^2+\frac17i|y|^2-\frac17i|y|^2=0$$
$$\frac{5x^4\,dx+7y^6\,dy}{x^5+y^7} =\frac{ix^5+iy^7}{x^5+y^7}=i$$
The first line verifies that this choice is indeed in the tangent space, and the second line verifies that the imaginary part we get out of it is nonzero. This is all valid as long as we're on the sphere and the denominator $x^5+y^7$ is nonzero - so all of $\Omega$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of arithmetic progressions There is this sum:
$$\sum_{i=0}^{n-1}\left(\sum_{j=i+1}^{n-1}(n-j-1)\right)=\frac{1}{6}(n-2)(n-1)n$$
I don't understand how the formula is derived. What I do currently understand is this: For each i, starting with 0 and ending with n-1 we have a sum like this:
$$(n-j_{i+1}-1) + (n-j_{i+2}-1)+...+(n-j_{n-1}-1)$$
All these small sums are of course added in a grand total. For a simple case like n = 5 we have these sums:
Sums computed for n = 5
I think the grand total could be expressed like this:
$$\frac{(n-2)(n-1)}{2} + \frac{(n-3)(n-2)}{2}+...+1$$
but I have no clue how to get from this expression to the formula of the sum. I need a step by step proof of how this formula is derived. I don't have a formal training in math.
| You can sum vertically instead of horizontally!
Refer to the table:
$$\begin{array}{c|c|c}
i/j&1&2&3&\cdots &n-3&n-2&n-1\\
\hline
0&n-2&n-3&n-4&\cdots&2&1&0\\
1&&n-3&n-4&\cdots&2&1&0\\
2&&&n-4&\cdots&2&1&0\\
\vdots&\cdots\\
n-3&&&&\cdots&\cdots&1&0\\
n-2&&&&\cdots&\cdots&\cdots&0\\
n-1&&&&\cdots&\cdots&\cdots&\cdots&
\end{array}\\
\sum_{i=0}^{n-1}\left(\sum_{j=i+1}^{n-1}(n-j-1)\right)=\sum_{i=0}^{n-3}(i+1)(n-i-2)=\\
\sum_{i=1}^{n-2}i(n-i-1)=(n-1)\sum_{i=1}^{n-2}i-\sum_{i=1}^{n-2}i^2=\\
\frac{(n-1)(1+n-2)}{2}\cdot (n-2)-\frac{(n-2)(n-1)(2(n-2)+1)}{6}=\\
\frac{(n-2)(n-1)[3(n-1)-(2n-3)]}{6}=\frac{1}{6}(n-2)(n-1)n\\$$
Note:
$$\sum_{i=1}^n i=\frac{1+n}{2}\cdot n; \ \ \sum_{i=1}^n i^2=\frac{n(n+1)(2n+1)}{6}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Fewest number of marbles in a bag such that drawing the probability of drawing 2 blue marbles is $\frac{1}{6}$.
Two marbles are randomly selected without replacement from a bag
containing blue and green marbles. Probability of drawing both blue is
$\frac{1}{6}$. If three marbles are drawn, then the probability all
three are blue is $\frac{1}{21}$. What is the fewest number of marbles
that must have been in the bag before any were drawn?
I use the criteria to make two equations:
$$\frac{x}{y}\cdot\frac{x-1}{y-1}=\frac{1}{6}$$
for drawing 2 marbles$$\frac{x}{y}\cdot\frac{x-1}{y-1}\cdot\frac{x-2}{y-2}=\frac{1}{21}$$
Where $x$ is the number of blue marbles and $y$ is the total number of marbles.
Simplifying these equations gives:
$$6x^2-6x=y^2-y$$ for the first equation and $$21x^3-63x^2+42x=y^3-3y^2+2y$$
I can't seem to solve this set of equations.
Is my approach correct? If so, how should I continue? If not, how would one solve this problem?
Thanks! Your help is appreciated!
Max0815
| As Sean Lee and InterstellarProbe note, dividing the second equation by the first leaves
$${x-2\over y-2}={6\over21}={2\over7}$$
It follows that $x-2=2k$ and $y-2=7k$ for some $k\not=0$. Plugging $x=2k+2$ and $y=7k+2$ into $6x^2-6x=y^2-y$, we have
$$6(2k+2)^2-6(2k+2)=(7k+2)^2-(7k+2)$$
which expands and simplifies initially to
$$24k^2+36k+12=49k^2+21k+2$$
and then to
$$25k^2-15k-10=5(5k+2)(k-1)=0$$
This has two solutions, $k=1$ and $k=-2/5$. The first gives a meaningful solution in integers, $x=4$, $y=9$. The second gives a meaningless non-solution, $x=1/5$, $y=-4/5$. Thus $9$ marbles (with $4$ of them blue) is not only the fewest number of marbles you can have to start with, it's the only number.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\lim\limits_{x\to 0}\left (\frac{1^x+2^x+3^x+\dots+n^x}{n} \right)^{\frac1x}$ Consider the following expression.
$$\lim\limits_{x\to 0} \left (\frac{1^x+2^x+3^x+\dots+n^x}{n} \right )^{\frac 1 x}$$
How to solve this?
Let $y= \left (\frac {1^x+2^x+\cdots +n^x} {n} \right)^{1/x}$
I tried taking $\ln$ on both sides. We get that $$\ln(y)=\frac{1}{x}\ln \left (\frac {1^x+2^x+\cdots +n^x} {n} \right ).$$
Taking $\lim$ on both sides we get $$\ln(y)=\lim_{x\to 0}\frac{1}{x}\ln \left (\frac {1^x+2^x+\cdots +n^x} {n} \right ).$$
Now applying the LH rule, we get $$\ln(y)=\lim_{x\to 0}\frac{n}{1^x+2^x+\cdots +n^x}({1^x\ln(1)+\cdots +n^x\ln(n)})$$
Is this a right way to go?
| Actually, $\ln(y)=\frac{1}{x}\ln{\frac{(1^x+2^x+...+n^x)}{n}}$. Whenever after putting $x=0$ you get the form $1^{\infty}$, you need to take $\ln$ and find the limit.
Here, after putting $x=0$, we have $\big(\frac{1^0+2^0+\cdots +n^0}{n}\big)^{1/0}=1^{\infty}$. So taking $\ln$ in both side and finding limit is correct approach.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving $4\sin^2{x}\cos^2{x}-\cos^2{x}=0$
Does anyone know why these answers are wrong?
$$4\sin^2{x}\cos^2{x}-\cos^2{x}=0$$
$(4\sin^2x)(\cos^2x)=\cos^2x$
$4\sin^2x=1$
$\sin x=1/2$
$x=30° , 150°$
Thank you
| You did three mistakes in your proposed solution.
First, you are allowed to divide both sides of an equation for an expression if the expression is not null. In the second step of your solution you divide both sides by $\cos^2 x$, thus $4\sin^2 x\cdot \cos^2 x=\cos^2 x$ is equivalent to $4\sin^2 x = 1$ only if $\cos^2 x\neq 0$. But note that the values of $x$ such that $\cos^2 x = 0$ are solution of your equation and hev to be included in the answer.
Second, $4\sin^2x = 1$ is equivalent to $2\sin x=1\vee 2\sin x=-1$.
Third, both the equations $4\sin^2 x=1$ and $\cos^2 x=0$ have infinitely many solutions in $\mathbb{R}$. For instance, consider $\cos x=0$. This equation has $x=\frac \pi 2 \vee x=\frac 32 \pi$ as solutions in $[0,2\pi)$, that is an interval of periodicity for $f(x)=\cos x$. As a consequence, by periodicity, $\left\{\frac \pi 2+2k\pi, \frac 32 \pi+2k\pi \,:\, k\in \mathbb{Z}\right\}$ are all the solutions of $\cos x =0$.
| {
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Inequality in a triangle: A line segment dividing opposite side with ratio 3:1
In triangle $\triangle ABC$, $D$ is a point on side $BC$ such that $|DC|=3 |BD|$. Also $|AB|=24$ and $|AC|=18$. Find the interval of all possible values of length of line segment $AD$.
I wrote $|DC|=3k$ and $|BD|=k$. Triangle inequality gives
$$ 24-18<4k<24+18\ \ \ \Longrightarrow 1.5<k<10.5$$
Then I used Stewart theorem to express $x$ in terms of $k$. The relation was $$x^2=513-3k^2$$Which gave me $13.5<x<22.5$.
Can we find a solution without Stewart theorem (and without trigonometry). I feel that there must be an easier (or more basic) solution around.
| We can use the Heron's formula:
$$S_{\Delta ABC}=\frac{1}{4}\sqrt{\sum_{cyc}(2a^2b^2-a^4)}.$$
Now, let $BD=y$, $DC=3y$ and since $S_{\Delta ADC}=3S_{\Delta ABD},$ we obtain:
$$2(18^2x^2+54^2y^2+9x^2y^2)-18^4-x^4-81y^4=9(2(24^2x^2+24^2y^2+x^2y^2)-24^4-x^4-y^4)$$ or
$$(x^2+3y^2-513)(x^2-3y^2-702)=0$$ and since $x^2=3y^2+702$ is impossible, we obtain:
$$x^2+3y^2=513.$$
Can you end it now?
| {
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$\int \cos^4{x}dx$ unsolvable with $t = \tan{x}$? I have been told that "universal substitution always works", so I wanted to give it a try on this specific integral.
$\int \cos^4{x}dx$
For some reason it does not work. Please note that I solved this integral in a normal way, just don't understand why the universal substitution does not work.
$\Bigg(t=\tan{x}, \cos^2{x} = \frac{1}{1+t^2},dx=\frac{dt}{1+t^2} \Bigg)$
$\int \cos^4xdx = \int (\cos^2x)^2dx = \int \Big(\frac{1}{1+t^2} \Big)^2 \frac{dt}{1+t^2} = \int \frac{dt}{(1+t^2)(1+t^2)(1+t^2)}$
$\frac{1}{(1+t^2)^3} = \frac{At +B}{1+t^2} + \frac{Ct+D}{(1+t^2)^2} + \frac{Et+F}{(1+t^2)^3}$
$1 = (At+B)(t^4+2t^2+1)+(Ct+D)(1+t^2)+Et+F$
$1 = At^5 + 2At^3 + At + Bt^4 + 2Bt^2 + B + Ct^3 + Ct + Dt^2 + D + Et + F$
Now this boils down to six linear equations:
$0 = A$
$0 = B$
$0 = 2A + C$
$0 = 2B + D$
$0 = A + C + E$
$1 = B + D + F$
Which results in: $A = 0, B = 0, C = 0, D = 0, E = 0, F = 0$ making it unsolvable for me.
Any ideas what went wrong?
| Your attempt at partial fraction decomposition is the problem; you can't remedy repeated factors like that. (Well, not usefully; see my comment.)
Instead you could cube $\frac{1}{1+t^2}=\frac{1}{2i}\left(\frac{1}{t-i}-\frac{1}{t+i}\right)$ and simplify further. I don't recommend it, though. The simplest solution for the original problem is $$\int\cos^4 xdx=\frac{1}{4}\int(1+\cos 2x)^2dx=\frac{1}{8}\int(3+4\cos 2x+\cos 4x)dx=\frac{3x+2\sin 2x+\frac{1}{4}\sin 4x}{8}+C.$$
| {
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$x^2 + y^2+xy = 1$ , then find the minimum of $x^3 y + xy^3 +4$ x and y belongs to real numbers. $ x^2 + y^2+xy = 1 $. then find the minimum value of $x^3 y + xy^3 +4$.
I assume $ x = r \sin (w)$ and $ y = r\cos(w) $. $ x^3 y + xy^3 +4 = L $ which give me $ \frac{2}{3} \le r^2 \le 2 $ I am stuck after that.Its my Humble request to help me after that.
| Your method of solution is good. You just need to continue. By setting $x=r\cos\theta$ and $y=r\sin\theta$, we get$$\cos\theta\sin\theta=\frac1{r^2}$$ and so $\tfrac23\leqslant r^2\leqslant2$, as you found. We have $$x^3y+xy^3+4=r^2-r^4+4=\tfrac{17}4-(r^2-\tfrac12)^2,$$which is minimum when $|r^2-\frac12|$ is maximum, namely when $r^2=2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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inequality $(a+b+c+d)(a^3+b^3+c^3+d^3) > (a^2+b^2+c^2+d^2)^2$ I am trying to prove the inequality:
$$(a+b+c+d)(a^3+b^3+c^3+d^3) > (a^2+b^2+c^2+d^2)^2$$ given $a,b,c,d$ are positive and unequal.
starting from LHS
since AM of mth power > mth power of AM
$$(a^3+b^3+c^3+d^3)/4 > ((a+b+c+d)/4)^3$$
multiplying both sides by $(a+b+c+d)$
$$(a+b+c+d)(a^3+b^3+c^3+d^3)/4 > (a+b+c+d)^4/(4^3)$$
$\implies$
$$(a+b+c+d)(a^3+b^3+c^3+d^3) > (a+b+c+d)^4/16$$
now taking expression on the RHS and using AM of mth power > mth power of AM
$$(a^2+b^2+c^2+d^2)/4 > ((a+b+c+d)/4)^2$$
squaring both sides
$$(a^2+b^2+c^2+d^2)^2/16 > (a+b+c+d)^4/(4^4)$$
$$\implies (a^2+b^2+c^2+d^2)^2 > ((a+b+c+d)^4)/16$$
Now I have proved that both LHS and RHS are greater than $$((a+b+c+d)^4)/16$$
still unable to prove LHS > RHS. Please help with this.
| Hint: Use Cauchy inequality: $$(x_1+x_2+...+x_n)(y_1+y_2+...+y_n)\geq (\sqrt{x_1y_1}+...+\sqrt{x_ny_n})^2$$
| {
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Double Factorial expression in terms of regular factorials I have the following double factorial expression:
$$\frac{(x-2)!!(x-2k-1)!!}{(x-1)!!(x-2k)!!}$$
if x is an odd number
I am trying to rewrite this without double factorials by using the following expressions for double factorials:
$$n!!=2^{n/2}(n/2)!$$
for even $n$ and $$\frac{(n+1)!}{2^{(n+1)/2}(\frac{n+1}{2})!}$$
for odd $n$.
What is the equivalent expression using these definitions? Is there a better way to write the original expression without double factorials?
| First of all:
$$
\frac{(x-2)!!(x-2k-1)!!}{(x-1)!!(x-2k)!!}
=
\frac{(x-2)!!}{(x-1)!!} \times \frac{(x-2k-1)!!}{(x-2k)!!}
$$
For the first factor:
$$
\begin{align}
\frac{(x-2)!!}{\left(x-1\right)!!}& =
\frac
{\frac{(x-1)!}{2^{(x-1)/2}(\frac{x-1}{2})!}}
{2^{(x-1)/2}\left(\frac{x-1}2\right)!}\\
& =
\frac{(x-1)!}{\left(2^{(x-1)/2}(\frac{x-1}{2})!\right)^2}\\
& =
\frac{(x-1)!}{2^{x-1}\left(\left(\frac{x-1}{2}\right)!\right)^2}
\end{align}
$$
A similar sequence with the second factor gives:
$$
\begin{align}
\frac{(x-2k-1)!!}{(x-2k)!!}& =
\frac
{2^{(x-2k-1)/2}\left(\frac{x-2k-1}2\right)!}
{\frac{(x-2k+1)!}{2^{(x-2k+1)/2}\left(\frac{x-2k+1}{2}\right)!}}\\
& =
\frac
{2^{(x-2k-1)/2}\left(\frac{x-2k-1}2\right)! \;
{2^{(x-2k+1)/2}\left(\frac{x-2k+1}{2}\right)!}}
{(x-2k+1)!}\\
& =
\frac
{2^{x-2k}\left(\frac{x-2k-1}2\right)! \; {\left(\frac{x-2k+1}{2}\right)!}}
{(x-2k+1)!}
\end{align}
$$
Multiply them together and get:
$$
\frac
{2^{x-2k}\left(\frac{x-2k-1}2\right)! \; {\left(\frac{x-2k+1}{2}\right)!}}
{(x-2k+1)!}
\frac{(x-1)!}{2^{x-1}\left(\left(\frac{x-1}{2}\right)!\right)^2} =
\frac{2^{x-2k}}{2^{x-1}}
\frac
{\left(\frac{x-2k-1}2\right)!}
{\left(\frac{x-1}{2}\right)!}
\frac{\left(\frac{x-2k+1}{2}\right)!}{\left(\frac{x-1}{2}\right)!}
\frac{(x-1)!}{(x-2k+1)!}
=
2^{1-2k}
\frac
{\left(\frac{x-2k-1}2\right)!}
{\left(\frac{x-1}{2}\right)!}
\frac{\left(\frac{x-2k+1}{2}\right)!}{\left(\frac{x-1}{2}\right)!}
\frac{(x-1)!}{(x-2k+1)!}
$$
That is the best I will do for now. The ratio of two factorials (with occurs three times in this expression) often has meaning as a product of consecutive integers, not beginning with one.
Hope this helps!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3173930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is $\lim_{x \rightarrow \infty}(3^x+7^x)^{\frac{1}{x}} ?$ $\lim_{x \rightarrow \infty}(3^x+7^x)^{\frac{1}{x}}$
I did in this way:
$\lim_{x \rightarrow \infty}(3^x+7^x)^{\frac{1}{x}}$
$=\lim_{x \rightarrow \infty}[(1+\frac{1}{3^x}+1-\frac{1}{7^x})(3^x\cdot7^x)]^{\frac{1}{x}}$
$=21\lim_{x \rightarrow \infty}(1+\frac{1}{3^x}+1-\frac{1}{7^x})^{\frac{1}{x}}$
$=21\cdot1=21$
But the limit given in the solution is $7.$
I'm not getting where is wrong!!
| Hint
$$b^x<a^x+b^x\le2b^x$$ for $b\ge a>0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3174673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proof $\sum_{k=1}^n (-1)^{k+1} \binom{n}{k}\frac{1}{k} = H_n$ by induction I found interesting task:
Calculate
$$\sum_{k=1}^n (-1)^{k+1} \binom{n}{k}\frac{1}{k}$$
I have calculated some first values and I see that it is $H_n$. I found
there tip that it can be solved by induction or by "integral" trick by considering $\sum_{k=1}^n(-1)^k{n\choose k}x^{k-1}$
I don't know what is that trick so I decided to solve it by induction.
Let $S_n = \sum_{k=1}^n (-1)^{k+1} \binom{n}{k}\frac{1}{k} $
$$ S_1 = 1 = H_1 \text{ ok.} $$
$$S_{n+1} = \sum_{k=1}^{n+1} (-1)^{k+1} \binom{n+1}{k}\frac{1}{k} = \\
-\sum_{k=0}^{n} (-1)^{k+1} \binom{n+1}{k+1}\frac{1}{k+1}$$
but I have problem with use induction assumption.
$$-\sum_{k=0}^{n} (-1)^{k+1} \binom{n}{k}\frac{n+1}{(k+1)^2} = \\
-(n+1)\sum_{k=0}^{n} (-1)^{k+1} \binom{n}{k}\frac{1}{(k+1)^2}$$
but know I have $\frac{1}{(k+1)^2} $ instead of something like $\frac{1}{k}$
|
We show by induction the following is valid for $n\geq 1$:
\begin{align*}
\sum_{k=1}^n(-1)^{k+1}\binom{n}{k}\frac{1}{k}=H_n
\end{align*}
Base step: $n=1$
\begin{align*}
\sum_{k=1}^1(-1)^{k+1}\binom{1}{k}\frac{1}{k}=1=H_1
\end{align*}
Induction hypothesis: $n=N$
We assume the validity of
\begin{align*}
\sum_{k=1}^N(-1)^{k+1}\binom{N}{k}\frac{1}{k}=H_N\tag{1}
\end{align*}
Induction step: $n=N+1$
We have to show
\begin{align*}
\sum_{k=1}^{N+1}(-1)^{k+1}\binom{N+1}{k}\frac{1}{k}=H_{N+1}\
\end{align*}
We obtain for $N\geq 1$:
\begin{align*}
\color{blue}{f_{N+1}}&\color{blue}{=\sum_{k=1}^{N+1}(-1)^{k+1}\binom{N+1}{k}\frac{1}{k}}\\
&=\sum_{k=1}^{N+1}(-1)^{k+1}\left[\binom{N}{k}+\binom{N}{k-1}\right]\frac{1}{k}\tag{2}\\
&=f_{N}+\sum_{k=1}^{N+1}(-1)^{k+1}\binom{N}{k-1}\frac{1}{k}\tag{3}\\
&=f_{N}-\frac{1}{N+1}\sum_{k=1}^{N+1}(-1)^k\binom{N+1}{k}\tag{4}\\
&=f_{N}-\frac{1}{N+1}\left[(1-1)^{N+1}-1\right]\\
&=f_{N}+\frac{1}{N+1}\\
&\,\,\color{blue}{=H_{N+1}}
\end{align*}
and the claim follows.
Comment:
*
*In (2) we use the binomial identity $\binom{p+1}{q}=\binom{p}{q}+\binom{p}{q-1}$.
*In (3) we apply the induction hypothesis (1).
*In (4) we use the binomial identity $\frac{p+1}{q+1}\binom{p}{q}=\binom{p+1}{q+1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3176104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
2D Matrix Transformation Invariance The transformation $\textbf{T}$ maps points $(x,y)$ of the plane into image points $(x', y')$ such that
$$\begin{align*} x' &= 4x + 2y + 14 \\ y' &= 2x + 7y + 42 \end{align*}$$
Find the coordinates of the invariant point of $\textbf{T}$. Hence express $\textbf{T}$ in the form $$\begin{pmatrix} x' \\ y' + k \end{pmatrix} = \textbf{A} \begin{pmatrix} x \\ y + k \end{pmatrix}$$ where $k$ is a positive integer and $\textbf{A}$ is a $2 \times 2$ matrix.
I found that the invariant point is $(0, -7)$ by setting $x = x'$ and $y = y'$. But I am a bit confused about what to do now. I tried some rearrangement to obtain $$\begin{pmatrix} x' \\ y' \end{pmatrix} + \begin{pmatrix} 0 \\ k \end{pmatrix} = \textbf{A} \begin{pmatrix} x \\ y \end{pmatrix} + \textbf{A}\begin{pmatrix} 0 \\ k \end{pmatrix}$$ Let's assume that $\textbf{A}$ is the matrix which represents the transformation $\textbf{T}$, then $$\begin{pmatrix} 0 \\ -7 \end{pmatrix} = \textbf{A} \begin{pmatrix} 0 \\ -7 \end{pmatrix} \implies \begin{pmatrix} 0 \\ 7 \end{pmatrix} = \textbf{A} \begin{pmatrix} 0 \\ 7 \end{pmatrix}$$ So from the earlier equation, $$\begin{pmatrix} x' \\ y' \end{pmatrix} = \textbf{A} \begin{pmatrix} x \\ y \end{pmatrix}$$ But the issue is that I cannot find any $\textbf{A}$ which represents the transformation. Can anyone please help me find $\textbf{A}$?
| A transformation of the form $A\begin{bmatrix} x\\ y\end{bmatrix} $ is linear: it maps the origin to the origin; that's not what your $T$ does. For an affine transformation, you need to add the constant term. So your $T$ is of the form
$$
T\begin{bmatrix} x\\ y\end{bmatrix}=\begin{bmatrix} 4&2\\ 2&7\end{bmatrix} \begin{bmatrix} x\\ y\end{bmatrix}+ \begin{bmatrix} 14\\ 42\end{bmatrix}.
$$
So you have $$Tv=Av+r.$$A fixed point will satisfy $$Av+r=v,$$ which we may rewrite as $$(A-I)v=-r.$$ So that's the linear system you are looking for:
$$
\begin{bmatrix} 4-1&2\\ 2&7-1\end{bmatrix} \begin{bmatrix} x\\ y\end{bmatrix}=-\begin{bmatrix} 14\\ 42\end{bmatrix}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3180934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inequality related with $abcd=(1-a)(1-b)(1-c)(1-d)$ Given that $0<a,b,c,d<1$ satisfying $abcd=(1-a)(1-b)(1-c)(1-d)$. Prove that $$(a+b+c+d)-(a+c)(b+d)\geq 1.$$
First, I have already done a quite similar exercise as below:
"Given that $a^2+b^2+c^2+d^2=1$. Prove that $(1-a)(1-b)(1-c)(1-d)\geq abcd$". The solution is to use the remark of $(a+b-1)^2\geq 0$, which leads to $2(1-a)(1-b) \geq 1-a^2-b^2 \geq 2cd$.
Then, I use the same method for this problem and I have showed that $a^2+b^2+c^2+d^2 \geq 1$. I don't know what to do next with $(a+b+c+d)-(a+c)(b+d)$.
Many thanks!
| Note that the inequality $(a+b+c+d)-(a+c)(b+d) \geq 1$ is equivalent to $(a+c-1)(b+d-1) \leq 0$. To notice this, we let $a+c = x$, and $b+d = y$, and therefore $x+y-xy \geq 1 \implies xy-x-y+1 \leq 0 \implies (x-1)(y-1) \leq 0$.
Now, we have two cases - if $a+c > 1$, and if $a+c < 1$ (if $a+c = 1$, then the inequality is obviously true.)
If $a+c > 1$, we have that $a+c-ac-1 > -ac \implies -(a-1)(c-1) > -ac \implies ac > (a-1)(c-1)$, and therefore $(b-1)(d-1) > bd \implies b+d < 1$, and therefore $(a+c-1)(b+d-1) < 0$.
We continue similarly if $a+c < 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3181762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Finding two tangential intersection points of an order four polynomial and a circle I have two equations:
(1) $y=0$ and
(2) $x^2 + y^2 = 2^2$
I also have another equation: (3) $x^4 + (y-a)^4 = b^4$
$a$ and $b$ are chosen such that (3) intersects with (1) at $x=y=0$ and intersects with (2) at exactly two points and at these points (2) and (3) are tangential . I do not know how to approach finding these two points. Can someone please give me a clue how to get started?
Image of my problem
| If (3) intersects (1) at $x=y=0$, that says $a^4 = b^4$. Since this is "geometry", I'm assuming $a$ and $b$ are real, so either $a=b$ or $a=-b$. Since (3) is unchanged by replacing $b$ by $-b$, I'll assume $a=b$.
The resultant of $x^2 + y^2 - 2^2$ and $x^4 + (y-b)^4 - b^4$ with respect to $y$ is
$$(16 -4 b^3 y + (6 b^2-8) y^2 - 4 b y^3 + 2 y^4 )^2$$
Thus an intersection of (2) and (3) has $y$ satisfying $g(y) = 16 -4 b^3 y + (6 b^2-8) y^2 - 4 b y^3 + 2 y^4 = 0$. Each $y$ satisfying this will correspond to two points $(x,y)$ and $(-x,y)$. In order for the two curves to be tangential, the discriminant of $g(y)$ must be $0$. That will give you a polynomial in $b$ of degree $12$ (but even, so of degree $6$ in $b^2$), which has two real roots: approximately $b = \pm 0.9720586597$. This sextic has Galois group $S_6$, so no solutions in radicals.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3184242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to do Given's rotation for $3×2$ matrix? (QR decomposition) I need to solve $Ax=b$ in lots of ways using QR decomposition.$$A=\begin{bmatrix}1&1\\-1&1\\1&2\end{bmatrix},\ b=\begin{bmatrix}1\\0\\1\end{bmatrix}.$$I know how to do Given's rotation for $3×3$ matrices as here: Which matrices to use in Given's rotation in QR decomposition But how do I do it for this matrix?
| You may find three $2\times2$ rotation matrices $G_1,G_2$ and $G_3$ such that
\begin{align}
\left[\begin{array}{rr}1\\ &G_1\end{array}\right]A
&=\left[\begin{array}{rr}1\\ &G_1\end{array}\right]
\left[\begin{array}{rr}1&1\\ \hline-1&1\\ 1&2\end{array}\right]
=\left[\begin{array}{rr}1&1\\ \hline \ast&\ast\\ 0&\ast\end{array}\right]
=B,\\
\left[\begin{array}{rr}G_2\\ &1\end{array}\right]B
&=\left[\begin{array}{rr}G_2\\ &1\end{array}\right]
\left[\begin{array}{rr}1&1\\ \ast&\ast\\ \hline 0&\ast\end{array}\right]
=\left[\begin{array}{rr}\ast&\ast\\ 0&\ast\\ \hline 0&\ast\end{array}\right]
=C,\\
\left[\begin{array}{rr}1\\ &G_3\end{array}\right]C
&=\left[\begin{array}{rr}1\\ &G_3\end{array}\right]
\left[\begin{array}{rr}\ast&\ast\\ \hline 0&\ast\\ 0&\ast\end{array}\right]
=\left[\begin{array}{rr}\ast&\ast\\ \hline 0&\ast\\ 0&0\end{array}\right]
=R.
\end{align}
Then
$$
A=\underbrace{
\left[\begin{array}{rr}1\\ &G_1^T\end{array}\right]
\left[\begin{array}{rr}G_2^T\\ &1\end{array}\right]
\left[\begin{array}{rr}1\\ &G_3^T\end{array}\right]}_Q\ R=QR.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3185146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Absolute Value within an Epsilon Delta Problem I'm having trouble with an epsilon-delta proof. I'm asked to prove the limit as $x$ goes to $-2$ of $f(x)=|x^2-9|/(x^2+3x+1)$ is $-5$, so we have $|(|x^2-9|/(x^2+3x+1))-5|<ϵ$. I've seen that $|x^2-9|$ is $|x-3||x+3|$ but I'm not really sure how to proceed to
"obtain" the desired $|x-a|=|x+2|<δ$. Am I supposed to place a limitation on either $|x-3|$ or $|x+3|$? Or is there some other way to handle this problem? Thanks in advance.
| Using the advice provided by @Robert Shore, we can get this
$$
\vert x^2-9\vert = \vert 9 - x^2 \vert = 9-x^2 \iff x^2\leq9 \implies -3\leq x\leq 3
$$
since we are checking the limit as $x\rightarrow2$, restrict $\delta < \frac{1}{2}$. Now
$$
\vert\dfrac{\vert 9 - x^2 \vert}{x^2+3x+1} + 5\vert < \varepsilon
$$
we can modify this by bringing everything to a common denominator
$$
\vert\dfrac{ 9 - x^2 + 5(x^2+3x+1)}{x^2+3x+1}\vert = \vert\dfrac{4x^2+15x+14}{x^2+3x+1}\vert = \vert\dfrac{(4x+7)(x+2)}{x^2+3x+1}\vert
$$
now if $\delta<\frac{1}{2}$, then we can know that $|x+2|<\frac{1}{2} \implies -2.5<x<-1.5$. It turns out that the bottom polynomial has a root at $x\approx -2.6$ which is why we had to restrict $\delta<\frac{1}{2}$ instead of $1$
Now that we have a range of $x$, lets maximize the Left Hand Side by maximizing the numerator and minimizing the denominator.
$|4x+7|$ has a maximum on the interval at $x=-2.5$ which is equal to $3$
$|x^2+3x+1|$ has a minima at $x=-2.5$ equal to $\frac{1}{4}$ so we can say
$$
\vert\dfrac{(4x+7)(x+2)}{x^2+3x+1}\vert < |\dfrac{3(x+2)}{\frac{1}{4}}| = |12(x+2)| < \varepsilon \implies|x+2|<\dfrac{\varepsilon}{12}
$$
Now
$$
\forall \varepsilon > 0, \exists \delta = \min\left(\frac{1}{2}, \frac{\varepsilon}{12}\right) \vert |x+2|<\delta \implies |f(x)+5|<\varepsilon
$$
Q.E.D.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3185691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
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.