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Evaluating $\int_{0}^{\pi/2}\frac{x\sin x\cos x\;dx}{(a^{2}\cos^{2}x+b^{2}\sin^{2}x)^{2}}$ How to evaluate the following integral
$$\int_{0}^{\pi/2}\frac{x\sin x\cos x}{(a^{2}\cos^{2}x+b^{2}\sin^{2}x)^{2}}dx$$
For integrating I took $\cos^{2}x$ outside and applied integration by parts.
Given answer is $\dfrac{\pi}{4ab^{2}(a+b)}$.
But I am not getting the answer.
| You can use this property : $$\int_a^b f(x)\hspace{1mm}dx = \int_a^b f(a+b-x)\hspace{1mm}dx$$
To prove this property : Substitute $a+b-x = u$
Let us have $$I = \int_0^{\pi/2}\dfrac{x\sin x\cos x}{(a^2\cos^2 x+b^2\sin^2 x)^2}\hspace{1mm}dx\rightarrow (1)$$
After applying the property, you will get
$$I = \int_0^{\pi/2}\dfrac{(\pi/2-x)\sin x\cos x}{(a^2\cos^2 x+b^2\sin^2 x)^2}\hspace{1mm}dx\rightarrow (2)$$
After adding the two equation, you get
$$I =\dfrac{\pi}{4} \int_0^{\pi/2}\dfrac{\sin x\cos x}{(a^2\cos^2 x+b^2\sin^2 x)^2}\hspace{1mm}dx $$
Now you will substitute $a^2\cos^2x+b^2\sin^2x = u$
To get $$I = \dfrac{\pi}{8(b^2-a^2)}\int_{a^2}^{b^2} \dfrac{1}{u^2}\hspace{1mm}du$$
I hope you can Integrate from here
| {
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"url": "https://math.stackexchange.com/questions/1068649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Closed form of $\int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}$ Today I discussed the following integral in the chat room
$$\int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}$$
where $0\leq a, b\leq \pi$ and $k>0$.
Some users suggested me that I can use Frullani's theorem:
$$\int_0^\infty \frac{f(ax) - f(bx)}{x} = \big[f(0) - f(\infty)\big]\ln \left(\frac ab\right)$$
So I tried to work with that way.
\begin{align} I&=\int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}\\ &=\int_0^\infty \frac{\ln \left( x^2+2kx\cos b+k^2\right)-\ln \left( x^2+2kx\cos a+k^2\right)}{x}\mathrm dx\tag{1}\\ &=\int_0^\infty \frac{\ln \left( 1+\dfrac{2k\cos b}{x}+\dfrac{k^2}{x^2}\right)-\ln \left( 1+\dfrac{2k\cos a}{x}+\dfrac{k^2}{x^2}\right)}{x}\mathrm dx\tag{2}\\ \end{align}
The issue arose from $(1)$ because $f(\infty)$ diverges and the same issue arose from $(2)$ because $f(0)$ diverges. I then tried to use D.U.I.S. by differentiating w.r.t. $k$, but it seemed no hope because WolframAlpha gave me this horrible form. Any idea? Any help would be appreciated. Thanks in advance.
| Here is a complex-analytic method: Notice that
$$ \int_{0}^{\infty} \log\left( \frac{x^{2} + 2x\cos b + 1}{x^{2} + 2x\cos a + 1} \right) \frac{dx}{x}
= 2 \int_{0}^{\infty} \Re \frac{\log(1 + e^{ib}x) - \log(1 + e^{ia}x)}{x} \, dx. \tag{1} $$
Let $R$ be a positive large number. Then the function $z \mapsto \log(1+z)/z$ is analytic on $\Bbb{C} \setminus (-\infty, 1]$ with the standard branch cut and we get
\begin{align*}
\int_{0}^{R} \frac{\log(1 + e^{ib}x)}{x} \, dx
&=\int_{0}^{Re^{ib}} \frac{\log(1 + z)}{z} \, dz \qquad (z = e^{ib}x) \\
&= \int_{0}^{R} \frac{\log(1 + z)}{z} \, dz + \int_{R}^{Re^{ib}} \frac{\log(1 + z)}{z} \, dz \\
&= \int_{0}^{R} \frac{\log(1 + z)}{z} \, dz + i \int_{0}^{b} \log(1 + Re^{i\theta}) \, d\theta, \quad (z=Re^{i\theta})
\end{align*}
and likewise for the integral with $b$ replaced by $a$. This shows that
$$ \int_{0}^{R} \frac{\log(1 + e^{ib}x) - \log(1 + e^{ia}x)}{x} \, dx
= i \int_{a}^{b} \log(1 + Re^{i\theta}) \, d\theta. $$
Multiplying by 2 and taking real part, we get
$$ 2 \int_{0}^{R} \Re \frac{\log(1 + e^{ib}x) - \log(1 + e^{ia}x)}{x} \, dx
= - 2 \Im \int_{a}^{b} \log(R^{-1} + e^{i\theta}) \, d\theta. $$
(Here we exploited the fact that $\log R$ is real.) Taking $R \to \infty$, in view of the identity $\text{(1)}$ it follows that
$$ \int_{0}^{\infty} \log\left( \frac{x^{2} + 2x\cos b + 1}{x^{2} + 2x\cos a + 1} \right) \frac{dx}{x}
= - \int_{a}^{b} 2\theta \, d\theta
= a^{2} - b^{2}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1069376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
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Putnam definite integral evaluation $\int_0^{\pi/2}\frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}dx$
Evaluate $$\int_0^{\pi/2}\frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}dx$$ Source : Putnam
By the property $\displaystyle \int_0^af(x)\,dx=\int_0^af(a-x)\,dx$:
$$=\int_0^{\pi/2}\frac{(\pi/2-x)\sin x\cos x}{\sin^4 x+\cos^4 x}dx=\frac{\pi}{2}\int_0^{\pi/2}\frac{\sin x\cos x}{\sin^4 x+\cos^4 x}dx-\int_0^{\pi/2}\frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}dx$$
$$\Longleftrightarrow\int_0^{\pi/2}\frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}dx=\frac{\pi}{4}\int_0^{\pi/2}\frac{\sin x\cos x}{\sin^4x+\cos^4x}dx$$
Now I'm stuck. WolframAlpha says the indefinite integral of $\dfrac{\sin x\cos x}{\sin^4 x+\cos^4x}$ evaluates nicely to $-\frac12\arctan(\cos(2x))$.
I already factored $\sin^4 x+\cos^4 x$ into $1-\left(\frac{\sin(2x)}{\sqrt{2}}\right)^2$, but I don't know how to continue.. I suggest a substitution $u=\frac{\sin(2x)}{\sqrt{2}}$?
Could someone provide me a hint, or maybe an easier method I can refer to in the future?
| \begin{aligned}
I&= \int_{0}^{\frac{\pi}{2}} \frac{x \sin x \cos x}{1-\frac{\sin ^{2} 2 x}{2}} d x\\&=\int_{0}^{\frac{\pi}{2}} \frac{x \sin 2 x}{1+\cos ^{2} 2 x} d x
\\&= \int_{0}^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2}-x\right) \sin 2 x}{1+\cos ^{2} 2 x} d x \quad \text{ (By }x \mapsto \frac{\pi}{2}-x.)\\ &=\frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \frac{\sin 2 x}{1+\cos ^{2} 2 x} d x \\
&=-\frac{\pi}{8} \cdot\left[\tan ^{-1}(\cos 2 x)\right]_{0}^{\frac{\pi}{2}} \\
&=-\frac{\pi}{8}\left(-\frac{\pi}{4}-\frac{\pi}{4}\right) \\
&=\frac{\pi^{2}}{16} .
\end{aligned}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1072316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Proving inequality $(x^2+y^2)(y-1)+yx-y^2<0$ I have an inequality which came out of Lyapunov function for system of ODE's: $$(x^2+y^2)(y-1)+yx-y^2<0.$$ To prove stability of my solution, I have to prove that the inequalty is true in area $0<x^2 +y^2<1$. I know it must be simple, but I am still in trouble. Any help is appreciated.
| If $x = y$, then $$(x^2 + y^2)(y - 1) + yx - y^2 = 2x^2(x - 1) < 0$$ since $x < 1$. Now suppose $x \neq y$. If $x$ and $y$ are positive, then $0 < x^2 < x$ and $y^3 < y^2$, thus
\begin{align}(x^2 + y^2)(y - 1) + yx - y^2 &= x^2y + y^3 - x^2 - y^2 + yx - y^2\\
& < x^2y - x^2 - y^2 + yx \\
& < 2xy - (x^2 + y^2)\\
& < 0.\end{align}
If $x < 0 < y$ or $y < 0 < x$, then $yx - y^2 = y(x - y) < 0$; since $(x^2 + y^2)(y - 1) < 0$ (as $y < 1$) we have $(x^2 + y^2)(y - 1) + yx - y^2 < 0$. Finally, if $x$ and $y$ are both negative, then $y^3 < 0 < y^2$ and $x^2 > 0 > x$; like in the case when $x$ and $y$ are both positive, we have
$$(x^2 + y^2)(y - 1) + yx - y^2 < x^2y - x^2 - y^2 + yx < 2xy - (x^2 + y^2) < 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1072790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determinant of a 4x4 matrix with trigonometric functions I am stuck with my homework from math. I should calcutate the determinant of a matrix:
$$\begin{bmatrix}
\sin(x) & \sin(2x) & \cos(x) & \cos(2x)\\
\cos(x) & 2\cos(2x) & -\sin(x)& -2\sin(2x)\\
-\sin(x)& -4\sin(2x)& -\cos(x)& -4\cos(2x)\\
-\cos(x)& -8\cos(2x)& \sin(x) & 8\sin(2x)
\end{bmatrix}$$
I have no clue how to do it. I've tried to convert the matrix to the triangular matrix, but I failed to do so.
Then I tried the Laplacian expansion (I hope it is called so) and I was creating sub-matrices from the first and second column, but I got a really long row of sin and cos and I was unable to make it shorter.
Sorry if I wrote someting wrong, but I am not so good in English. Can you please help me? Is there any way how to solve it?
Thank you very much!!!
| Call the matrix $M$. Then by using @r9m's suggestion, we interchange the two middle columns (this switches the sign of the determinant) and apply two row replacements (this doesn't change the determinant) in order to obtain a zero lower left submatrix:
\begin{align*}
|M|
&= -\begin{vmatrix}
\sin x & \cos x & \sin 2x & \cos 2x \\
\cos x & -\sin x & 2\cos 2x & -2\sin 2x \\
-\sin x & -\cos x & -4\sin 2x& -4\cos 2x \\
-\cos x & \sin x & -8\cos 2x & 8\sin 2x
\end{vmatrix} \\
&= -\begin{vmatrix}
\sin x & \cos x & \sin 2x & \cos 2x \\
\cos x & -\sin x & 2\cos 2x & -2\sin 2x \\
0 & 0 & -3\sin 2x& -3\cos 2x \\
0 & 0 & -6\cos 2x & 6\sin 2x
\end{vmatrix} \\
&= -\begin{vmatrix}
\sin x & \cos x \\
\cos x & -\sin x
\end{vmatrix}
\begin{vmatrix}
-3\sin 2x & -3\cos 2x \\
-6\cos 2x & 6\sin 2x
\end{vmatrix} \\
&= -(-\sin^2 x - \cos^2 x)(-18\sin^2 2x - 18\cos^2 2x) \\
&= -18(\sin^2 x + \cos^2 x)(\sin^2 2x + \cos^2 2x) \\
&= -18
\end{align*}
This used the fact that if $A,B,D$ are square matrices and $0$ is a square matrix of all zeroes, then:
$$
\begin{vmatrix}
A & B \\
0 & D
\end{vmatrix}
= |AD - B0|
= |AD|
= |A||D|
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1073852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Exact value of $\frac{\arccos(1-2\tan^2\alpha)}{2\arcsin(\tan\alpha)}$ Let $\alpha\in\left(0,\dfrac\pi2\right)$. What is the exact value of
$$\dfrac{\arccos(1-2\tan^2\alpha)}{2\arcsin(\tan\alpha)}$$
Firstly, I tried to simplify $1-2\tan^2\alpha$ and got
$$\dfrac{3\cos^2\alpha-2}{\cos^2\alpha}$$
What is next step? Is there a formula to simplify $\arccos\dfrac{a}{b}$ for some $a,b\in\mathbb{R}$?
| Let $\theta = \arcsin (\tan \alpha) \to \sin \theta = \tan \alpha \to 1 - 2\tan^2\alpha = 1 - 2\sin^2\theta = \cos 2\theta \to \arccos \left(1-2\tan^2\alpha\right) = \arccos (\cos 2\theta) = 2\theta \to L = \dfrac{2\theta}{2\theta} = 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1074982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Evaluation of $\int \frac{\sqrt{1+x^4}}{1-x^4}dx$ Evaluation of $\displaystyle \int \frac{\sqrt{1+x^4}}{1-x^4}dx$
$\bf{My\; Try::}$ Given $\displaystyle \int\frac{\sqrt{1+x^4}}{1-x^4}dx\;,$ Then We can write the above Integral as $$\displaystyle \int\frac{\left(1+x^4\right)}{\left(1-x^4\right)\sqrt{1+x^4}}dx = \frac{1}{2}\int\left\{\frac{1}{1-x^2}+\frac{1}{1+x^2}\right\}\cdot \frac{(1+x^4)}{\sqrt{1+x^4}}dx$$
So Integral is $$\displaystyle = \frac{1}{2}\int\frac{1+x^4}{(1-x^2)\sqrt{1+x^4}}dx+\frac{1}{2}\int\frac{1+x^4}{(1+x^2)\sqrt{1+x^4}}dx$$
Now Let $$\displaystyle I = \int\frac{1+x^4}{(1-x^2)\sqrt{1+x^4}}dx$$ and $$\displaystyle J = \int\frac{1+x^4}{(1+x^2)\sqrt{1+x^4}}dx$$
Now How can I evaluate $I$ and $J\;,$ plz help me
Thanks
| Thanks Claude Leibovici and Lucian, Using Lucian Hint.
$\bf{Another \; Try::}$ Let $\displaystyle \mathcal{I} = \int \frac{\sqrt{1+x^4}}{1-x^4}dx\;,$ Now Let $x^2=\tan \phi\;,$
Then $\displaystyle 2xdx = \sec^2 \phi d\phi \Rightarrow dx = \frac{\sec^2 \phi}{2\sqrt{\tan \phi}}d\phi$.
So $$\displaystyle \mathcal{I} = \int\frac{\sec \phi}{1-\tan^2 \phi}\times \frac{\sec^2 \phi}{2\sqrt{\tan \phi}}d\phi = \frac{1}{2}\int \frac{\sec \phi}{\cos 2\phi}\times \frac{1}{\sqrt{\tan \phi}}d\phi = \frac{\sqrt{2}}{2}\int\frac{1}{\cos 2\phi\cdot \sqrt{\sin 2\phi}}d\phi$$
So $$\displaystyle \mathcal{I} = \frac{1}{2}\int\frac{\cos 2\phi}{\left(1-\sin^2 2\phi\right)\cdot \sqrt{\sin 2\phi}}d\phi$$
Now Let $\displaystyle \sin 2\phi = t^2\;,$ Then $\displaystyle \cos 2\phi d\phi = 2tdt$.
So $$\displaystyle \mathcal{I} = \frac{1}{2}\int\frac{1}{1-t^4}dt = \frac{1}{4}\int \left(\frac{1}{1-t^2}+\frac{1}{1+t^2}\right)dt = \frac{1}{4}\cdot \frac{1}{2}\ln \left|\frac{1+t}{1-t}\right|+\frac{1}{4}\tan^{-1}(t)+\mathcal{C}$$
So $$\displaystyle \mathcal{I} = \frac{1}{4}\cdot \frac{1}{2}\ln \left|\frac{1+\sqrt{\sin 2\phi}}{1-\sqrt{\sin 2\phi}}\right|+\frac{1}{4}\tan^{-1}(\sqrt{\sin 2\phi})+\mathcal{C}$$
Where $\phi = \tan^{-1}(x^2).$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Explanation solution partial-fraction of $\frac{x^2 + 2}{x^2 - 1}$ The partial fraction of $\dfrac{x^2+2}{x^2-1}$ is $1 + \dfrac{3}{2}\cdot(\dfrac{1}{x-1}-\dfrac{1}{x+1})$.
I understand how you get $\dfrac{3}{2}\cdot(\dfrac{1}{x-1}-\dfrac{1}{x+1})$ but from where does the $1 +$ come?
| Hint. you may write
$$
\frac{x^2+2}{x^2-1}=\frac{(x^2-1)+3}{x^2-1}=\frac{x^2-1}{x^2-1}+\frac{3}{x^2-1}=\color{red}{1+}\frac{3}{x^2-1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\int \ln(1 + e^x)\ \mathrm dx$ Evaluate the following indefinite integral.
$$\int\ln(1 + e^x) \mathrm dx$$
My attempt ::
Using integration by-parts,
\begin{align}
\int\ln(1 + e^x)\cdot 1\ \mathrm dx &= x\ln(1 + e^x) - \int \frac{x\cdot e^x\ \mathrm dx}{1 + e^x}\\
&= x\ln(1+e^x) - \frac{x^2\cdot e^x}{2(1+e^x)} + \int \frac{x^2\cdot e^x\ \mathrm dx}{2(1 + e^x)^2}\\
&= x\ln(1 + e^x) - \frac{x^2\cdot e^x}{2(1 + e^x)} + \frac{x^3\cdot e^x\ \mathrm dx}{6(1 + e^x)^2} - \int \frac{x^3 \cdot e^x(e^x-1)\ \mathrm dx}{6 (1+ e^x) ^3}\\
&= x\ln(1 + e^x) - \frac{x^2\cdot e^x}{2(1 + e^x)} + \frac{x^3\cdot e^x\ \mathrm dx}{6(1 + e^x)^2} - \frac{x^4\cdot e^x(e^x - 1)}{24(1 + e^x)^3}\\ &+ \int \frac{x^4 \cdot e^x(e^{2x} + e^x+1)\ \mathrm dx}{24 (1+ e^x) ^3}\\
&= \lim_{N \to \infty} \sum_{n=1}^N\left( \frac{x^n}{n!}\frac{\mathrm d^n}{\mathrm dx^n}\Big(\ln(1 + e^x)\Big)\right)
\end{align}
Obviously, this is a dead end (applicable for all n-differentiable functions) and I'm just not capable of finding the actual answer. Please educate me on how to properly do this integral.
| Let $$\mathcal{ I}=\int \ln(1 + e^x)\ \mathrm dx$$
By substituting $e^x=-t\iff e^x\,\mathrm dx=\dfrac{\mathrm dt}{t}$
$$\begin{align}
\mathcal{ I}
&=\int \frac{\ln(1 -t)}{t}\ \mathrm dt
=-\int \frac{1}{t}\sum_{n=1}^{\infty}\frac{t^n}{n}\ \mathrm dt
=-\int \sum_{n=1}^{\infty}\frac{t^{n-1}}{n}\ \mathrm dt\\
&=-\sum_{n=1}^{\infty}\int \frac{t^{n-1}}{n}\ \mathrm dt
=-\sum_{n=1}^{\infty}\frac{t^{n}}{n^2}
=-\text{Li}_2\left(t\right)
=-\text{Li}_2\left(-e^x\right)\\
\end{align}$$
$\text{Explanations}$
$$\ln(1-t)=-\sum_{n=1}^{\infty}\frac{t^n}{n}$$
$$
\operatorname{Li}_s(z) = \sum_{k=1}^\infty {z^k \over k^s} = z + {z^2 \over 2^s} + {z^3 \over 3^s} + \cdots \
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find Sum of $\sum_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)$. Prove that it converges. Question : For $$\sum_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)$$
a. Prove it converges
b. Find the sum
My Try
$
= \sum_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)\\
= \ln(1 - \frac{1}{4} ) + \ln(1 - \frac{1}{9} ) + \ln ( 1 - \frac{1}{16})\\
= \ln(\frac{3}{4}) + \ln (\frac{8}{9}) + \ln ( \frac{15}{16})\\
= -.287 + -.117 + -.064 = -.50
$
it converges to $-\frac{1}{2}$.
| Hint:
Rewrite it as:
$$\sum_{n=2}^{\infty} ln(\frac{n^{2} - 1}{n^{2}}) = \sum_{n=2}^{\infty} ln( \frac{(n-1)(n+1)}{n^{2}}) = \sum_{n=2}^{\infty} [ ln(n-1) + ln(n+1) - 2ln(n)]$$
So we have: $ln(1) + ln(3) - 2ln(2) + ln(2) + ln(4) - 2ln(3) + ...$
Think about this as a telescoping series.
| {
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"url": "https://math.stackexchange.com/questions/1079174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Calculate $\sin{55}-\sin{19}+\sin{53}-\sin{17}$ without calculator So here is a trigonometric series.
$$\sin{55^\mathrm{o}}-\sin{19^\mathrm{o}}+\sin{53^\mathrm{o}}-\sin{17^\mathrm{o}}$$
Strange isn't it, and I have to calculate the total result of the series (without calculator). I don't think Maclaurin series will help me any way. Further I tried almost all trigonometric identities (as per my knowledge) but so far I had no clue. Probably I am missing some kind of identity. Anyone can help me in this?
Note: It's not a homework question.
| You can use the following product-sum trig identity:
$$\sin (a+b) - \sin (a-b) = 2 \cos a \sin b$$
$$\sin (a+b) + \sin (a-b) = 2 \sin a \cos b$$
So, $$(\sin 55^\circ - \sin 17^\circ) + (\sin 53^\circ - \sin 19^\circ)$$
$$(2 \cos 36^\circ \sin 19^\circ) + (2 \cos 36^\circ \sin 17^\circ)$$
$$2 \cos 36^\circ (\sin 19^\circ + \sin 17^\circ)$$
$$4 \cos 36^\circ \sin 18^\circ \cos 1^\circ$$
We can find $\sin 18^\circ$ using the following method, and use $\cos 2x = 1 - 2 \sin^2 x$ to find $\cos 36^\circ$.
$\sin 72^\circ = 2 \sin 36^\circ \cos 36^\circ$
$\cos 18^\circ = 4 \sin 18^\circ \cos 18^\circ (1 - 2 \sin^2 18^\circ)$
$1 = 4x(1 - 2x^2)$ where $x = \sin 18^\circ$
$8x^3 - 4x + 1 = 0$
We can factor using a few methods, I think the easiest is to observe that $x = 1/2$ is a root and do polynomial long division of $(2x-1)$:
$(2x - 1)(4x^2 + 2x - 1) = 0$ We can't have $x = \frac{1}{2}$ because that's $\sin 30^\circ$, not $\sin 18^\circ$.
Quadratic formula: $x = \frac{-2 \pm \sqrt{20}}{8} = \frac{1}{4} (-1 \pm \sqrt{5})$
The root is positive, so it's $\frac{1}{4} (-1 + \sqrt{5})$. From this follows $\cos 36^\circ = \frac{1}{4}(1 + \sqrt{5})$
So, the above answer reduces to just $\cos 1^\circ$. This doesn't have a nice formula, but it is algebraic!
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the value of $\sum_{n=0}^{\infty}(-\frac{1}{8})^n\binom{2n}{n}$ What is the value of $$\sum_{n=0}^{\infty}\left(-\frac{1}{8}\right)^n\binom{2n}{n}\;?$$
EDIT
I bumped into this series when inserting $\overrightarrow{r_1}=\left(\begin{array} {c}0\\0\\1\end{array}\right)$ and $\overrightarrow{r}=\left(\begin{array} {c}1\\1\\0\end{array}\right)$ into $$\frac{1}{|\overrightarrow{r}-\overrightarrow{r_1}|}=\sum_{l=0}^\infty\frac{r_{<}^l}{r_{>}^{l+1}}P_{l}(cos\theta)\;.$$
See (Series representation of $1/|x-x'|$ using legendre polynomials).
So I knew the result, $\sqrt{\frac{2}{3}}$, but wished to find out what other approaches there would be to evaluate the series.
It transpires that I overlooked the relatively standard evaluation of $\sum_{n=0}^{\infty}\binom{2n}{n}x^n$ as being equal to $\sqrt{\frac{1}{1-4x}}$, following a calculation similar to that given by achille-hui below, which requires some complex function theory, in particular when proving that $\binom{2n}{n} = \frac{1}{2\pi}\int_0^{2\pi}\left(e^{i\theta}+e^{-i\theta}\right)^{2n}d\theta$, or that of alex.jordan below, requiring no more than Taylor expansion.
| As mentioned in Dr. Graubner's comment, the sum is $\sqrt{\frac23}$.
In fact, this is a special case of a sort of famous Taylor series expansion:;
$$\frac{1}{\sqrt{1-4z}} = \sum_{n=0}^\infty \binom{2n}{n} z^n$$
which appeared as limiting example of several theorems in complex analysis.
To compute the series ourselves, we can use following integral representation of
the binomial coefficients:
$$\binom{2n}{n} = \frac{1}{2\pi}\int_0^{2\pi} (e^{i\theta} + e^{-i\theta})^{2n} d\theta
= \frac{4^n}{2\pi}\int_0^{2\pi} \cos^{2n}\theta d\theta
$$
Substitute this into our sum, we find the sum is equal to
$$
\frac{1}{2\pi}\int_0^{2\pi} \sum_{n=0}^\infty \left(-\frac12\cos^2\theta\right)^n d\theta
= \frac{1}{2\pi}\int_0^{2\pi} \frac{1}{1+\frac12\cos^2\theta} d\theta
= \frac{2}{\pi}\int_0^{\pi/2} \frac{1}{1+\frac12\cos^2\theta} d\theta
$$
Introduce change of variable $t = \tan\theta$, this becomes
$$\frac{2}{\pi}\int_0^\infty \frac{1}{1 + \frac{1}{2(1+t^2)}}\frac{dt}{1+t^2}
= \frac{2}{\pi}\int_0^\infty \frac{dt}{t^2 + \frac32}
= \frac{2}{\pi}\sqrt{\frac23}\left[ \tan^{-1}\sqrt{\frac23} t \right]_0^\infty
= \sqrt{\frac23}$$
| {
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"url": "https://math.stackexchange.com/questions/1080635",
"timestamp": "2023-03-29T00:00:00",
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Integration by differentiating under the integral sign $I = \int_0^1 \frac{\arctan x}{x+1} dx$ $$I = \int_0^1 \frac{\arctan x}{x+1} dx$$
I spend a lot of my time trying to solve this integral by differentiating under the integral sign, but I couldn't get something useful. I already tried:
$$I(t) = \int_0^1 e^{-tx}\frac{\arctan x}{x+1} dx ; \int_0^1 \frac{(t(x+1)-x)\arctan x}{x+1} dx ; \int_0^1 \frac{\arctan tx}{x+1} dx ; \int_0^1 \frac{\ln(te^{\arctan x})}{x+1} dx $$
And something similar. A problem is that we need to get +constant at the very end, but to calculate that causes same integration problems.
To these integrals:
$$I(t) = \int_0^1 e^{-tx}\frac{\arctan x}{x+1} dx ; \int_0^1 \frac{\arctan tx}{x+1} dx$$
finding constant isn't problem, but to solve these integrals by itself using differentiating under the integral sign is still complexive.
Any ideas? I know how to solve this in other ways (at least one), but I particularly interesting in differentiating.
| $$I(a) = \int \frac{\arctan (ax)}{x+1}dx \Rightarrow I'(a) = \int \frac{x}{(x + 1)(a^2x^2+1)} dx$$
Now $$\frac{x}{(x + 1)(a^2x^2+1)} =\frac{A}{(x + 1)} + \frac{Bx + C}{(a^2x^2+1)}$$
then $A = \frac{-1}{a^2 + 1}$, $B = \frac{a^2}{a^2 + 1}$ and $C = \frac{1}{a^2+1}$. From this we have
$$\begin{align}I'(a) &= \int \frac{-1}{(a^2 + 1)(x+1)}dx + \int\frac{a^2x + 1}{(a^2 + 1)^2(a^2x^2+1)}dx \\&=\frac{-1}{(a^2+1)}\int \frac{1}{(x+1)}dx + \frac{1}{(a^2+1)^2}\Bigg[\int \frac{a^2x}{a^2x^2+1}dx +\int\frac{1}{a^2x^2+1}dx\Bigg]\end{align}$$
Where $\int \frac{a^2x}{a^2x^2+1}dx = x - \frac{\ln (a^2x^2 + 1)}{a}$ and $\int\frac{1}{a^2x^2+1}dx = \frac{\arctan (ax)}{a}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Find the value of : $\lim_{ x \to \infty} \left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)$ I need to calculate the limit of the function below:
$$\lim_{ x \to \infty} \left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)$$
I tried multiplying by the conjugate, substituting $x=\frac{1}{t^4}$, and both led to nothing.
| A useful result for a brute force solution to these sorts of computations is $\sqrt{1+ \theta} = 1+ {1 \over 2} \theta + r (\theta)$, where $\lim_{\theta \to 0} { r(\theta)\over \theta} = 0$
Note that $\sqrt{x+\sqrt{x+ \sqrt{x}}} = \sqrt{x} \sqrt{1+ \sqrt{{1 \over x}+\sqrt{1 \over x^3}
}}$, and so
$\sqrt{x+\sqrt{x+ \sqrt{x}}} - \sqrt{x} = \sqrt{x} (\sqrt{1+ \sqrt{{1 \over x}+\sqrt{1 \over x^3}
}} -1) $.
Hence
$d(x)=\sqrt{x} (\sqrt{1+ \sqrt{{1 \over x}+\sqrt{1 \over x^3}
}} -1) = \sqrt{x}({1 \over 2}\sqrt{{1 \over x}+\sqrt{1 \over x^3}}+r(\sqrt{{1 \over x}+\sqrt{1 \over x^3}}))$.
Simplifying gives
$d(x) = {1 \over 2} \sqrt{1+\sqrt{1 \over x}}+{1 \over 2} {r(\sqrt{{1 \over x}+\sqrt{1 \over x^3}}) \over \sqrt{1 \over x}}$.
Hence $\lim_{x \to \infty} d(x) = {1 \over 2}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why is $\cos(x/2)+2\sin(x/2)=\sqrt5 \sin(x/2+\tan^{-1}(1/2))$ true? According to Wolfram Alpha the following equality holds:
$$\cos\left(\frac{x}{2}\right)+2\sin\left(\frac{x}{2}\right)=\sqrt5 \sin\left(\frac{x}{2}+\tan^{-1}\left(\frac{1}{2}\right)\right)$$
I also checked it numerically. Why is it true?
| Let us consider the general case $$A=a \sin(x)+b\cos(x)$$ Without changing anything, we can rewrite $$A=\sqrt{a^2+b^2}\Big(\frac{a}{\sqrt{a^2+b^2}} \sin(x)+\frac{b}{\sqrt{a^2+b^2}} \cos(x)\Big)$$ Now, let us define$$\frac{a}{\sqrt{a^2+b^2}}=\cos(\phi)$$ so $$\sin^2(\phi)=1-\cos^2(\phi)=1-\frac{a^2}{a^2+b^2}=\frac{b^2}{a^2+b^2}$$ and so $$\sin(\phi)=\frac{b}{\sqrt{a^2+b^2}}$$ Replacing in the initial expression, we then have $$A=\sqrt{a^2+b^2}\Big(\cos(\phi) \sin(x)+\sin(\phi)\cos(x)\Big)=\sqrt{a^2+b^2}\sin(x+\phi)$$ and from the definitions, you can notice that $$\tan(\phi)=\frac{\sin(\phi)}{\cos(\phi)}=\frac ba$$ which makes $$\phi=\tan^{-1}\big(\frac ba\big)$$
Similar things could be done defining instead $$\frac{a}{\sqrt{a^2+b^2}}=\sin(\theta)$$ which make $$\cos(\theta)=\frac{b}{\sqrt{a^2+b^2}}$$ and then $$A=\sqrt{a^2+b^2}\Big(\sin(\theta) \sin(x)+\cos(\theta)\cos(x)\Big)=\sqrt{a^2+b^2}\cos(x-\theta)$$
So basically, using this approach allows to transform $A$ as a sine or a cosine depending on what you need for the remaining of the work.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $(1+a_1x+\ldots+a_rx^r)^k=1+x+x^{r+1}q(x)$
Fixed $k\ge 1$. Show that for each $r$, you can find $a_1,\cdot\cdot\cdot,a_r\in \mathbb{F}$ such that :$$(1+a_1x+\cdot\cdot\cdot+a_rx^r)^k=1+x+x^{r+1}q(x)$$
where $q(x)$ is a polynomial.
Any ideas?
I tried using induction as follows:
for the base case, $r=1:$
We wish that $$(1+a_1x)^k = 1+x+x^2q(x)$$
Setting $a_1 = 1$ we get:
$$1 + x + \text{other terms} = 1 + x + x^2q(x)$$
$$\text{other terms} = x^2q(x)$$
| If the field $\mathbb{F}$ has characteristic $p \neq 0$, and $p \mid k$, then the requested $a_1,\dots, a_r$ do not exist: writing $k=pk'$, we have
\begin{align*}
(1+a_1x+\cdots+a_rx^r)^k &= ((1+a_1x+\cdots+a_rx^r)^p)^{k'}\\
&= (1+a_1^px^p+\cdots+a_r^px^{rp})^{k'}
\end{align*}
which has a zero coefficient of $x$ and hence cannot be put into the form $1+x+x^2q(x)$.
So assume that $\mathbb{F}$ has characteristic 0 or that $p \nmid k$. To rephrase the problem, we are looking for a polynomial $f$ of degree $r$ with constant term 1 such that
$$f^k \equiv 1+x \pmod{x^{r+1}}$$
As suggested, we can find such a polynomial by induction on $r$. The base case $r=0$ is trivial (just take $f=1$). For the inductive step, assume we have a polynomial $f$ of degree $r$ satisfying the required conditions. Then
$$f^k \equiv 1+x+cx^{r+1} \pmod{x^{r+2}}$$
for some $c\in\mathbb{F}$. If we set $g=f+a_{r+1}x^{r+1}$ (where $a_{r+1}$ is yet to be determined), then by the binomial theorem, and using the fact that $f$ (and hence $f^{k-1}$) has constant term 1,
\begin{align*}
g^k &= (f+a_{r+1}x^{r+1})^k \\
&\equiv f^k + ka_{r+1}f^{k-1}x^{r+1} \\
&\equiv 1+x+(c+ka_{r+1})x^{r+1} \pmod{x^{r+2}}
\end{align*}
If we set $a_{r+1}=-c/k$, then $g^k$ satisfies the required conditions, completing the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
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Minimizing Sum of Reciprocals Find the minimum value, in terms of $k$ of $\frac{1}{x_1}+…+\frac{1}{x_n}$ if $x_1^2+x_2^2+…+x_n^2=n$ and $x_1+x_2+…+x_n=k$, where $\sqrt{n} < k \leq n$.
I tried the am-hm, but how to relate with the sum of squares?
| Checking the case $n = 2$
First condition:
$$
g(x,y) = x + y = k \iff y = - x + k
$$
Second condition:
$$
h(x,y) = x^2 + y^2 = 2 \iff y = \pm \sqrt{2 - x^2}
$$
This gives
\begin{align}
2 &= x^2 + (-x + k)^2=2 x^2 -2kx + k^2 \iff \\
1 &= x^2 - kx + k^2/2 = (x - k/2)^2 + k^2/4 \iff \\
x &= \frac{k\pm\sqrt{4-k^2}}{2} \quad \wedge \\
y &= \frac{-k\mp\sqrt{4-k^2}}{2} + k = \frac{k\mp\sqrt{4-k^2}}{2}
\end{align}
We got 2 points or 1 point ($k=2$) which satisfy the conditions, which agrees with an intersection of a line ($g=k$) and a circle ($h=2$). Their coordinates were expressed in terms of $k$. Inserting those points into $f$ we get:
\begin{align}
f(x,y) &= \frac{1}{x} + \frac{1}{y} \\
&= \frac{2}{k\pm\sqrt{4-k^2}} + \frac{2}{k\mp\sqrt{4-k^2}} \\
&= \frac{4k}{k^2 - (4 - k^2)} \\
&= \frac{2k}{k^2 - 2} \\
&= F(k)
\end{align}
This function is constant regarding $x$ and $y$, which includes the minimum.
(in progress)
The Lagrange function is
$$
L(x,\lambda, \mu) = f(x) +
\lambda \left( g(x) - k \right) +
\mu \left( h(x) - n \right)
$$
Gradient components are
\begin{align}
\frac{\partial L}{\partial x_j} &= -\frac{1}{x_j^2} + \lambda + 2 \mu x_j
\quad (j\in\{1,\ldots,n\}) \\
\frac{\partial L}{\partial\lambda} &= g(x) - k \\
\frac{\partial L}{\partial\mu} &= h(x) - n
\end{align}
At critical points $x^*$ the above components vanish. So
$$
f(x^*) = \sum_i \frac{1}{x^*_i}
= \sum_i \lambda x^*_i + 2 \mu (x^*_i)^2
= \lambda k + 2 \mu n
$$
The question is how to calculate $\lambda$ and $\mu$.
We have $n+2$ equations with $n+2$ unknowns, so there is a chance for a unique solution $(x, \lambda, \mu)$, but so far I found no way to extract it.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Integration by substitution. Integrating $$\int (x^2+2)(x-1)^7 dx$$
Let $ u = x^2 + 2$ then $ \frac{du}{dx} = 2x$
I can't see where to go after this?
I may have chosen the wrong substitution. Would it help if I let let $ u = \sqrt{u-2} $?
| First note that it would be possible to expand this polynomial and integrate term by term, though this would be fairly annoying. However, if we make the substitution $u=x-1$, we can get rid of some of the work:
$\begin{align} \int(x^2+2)(x-1)^7\,dx&=\int((u+1)^2+2)u^7\,du
\\&=\int(u^2+2u+3)u^7\,du
\\&=\int u^9+2u^8+3u^7\,du
\\&=\frac{1}{10}u^{10}+\frac{2}{9}u^9+\frac{3}{8}u^8+C
\\&=\frac{1}{10}(x-1)^{10}+\frac{2}{9}(x-1)^9+\frac{3}{8}(x-1)^8+C
\end{align}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit involving square roots, more than two "rooted" terms The limit is
$$\lim_{x\to\infty} \left(\sqrt{x^2+5x-2}-\sqrt{4x^2-3x+7}+\sqrt{x^2+7x+5}\right)$$
which has a value of $\dfrac{27}{4}$.
Normally, I would know how to approach a limit of the form
$$\lim_{x\to\infty}\left( \sqrt{a_1x^2+b_1x+c_1}\pm\sqrt{a_2x^2+b_2x+c_2}\right)$$
(provided it exists) by using the expression's conjugate, but this problem has me stumped.
I've considered using the conjugate
$$\sqrt{x^2+5x-2}-\sqrt{4x^2-3x+7}-\sqrt{x^2+7x+5}$$
and a term like this one,
$$\sqrt{x^2+5x-2}+\sqrt{4x^2-3x+7}-\sqrt{x^2+7x+5}$$
but that didn't seem to help simplify anything.
Edit: I stumbled across something at the last second that lets me use the conjugate approach. The expression can be rewritten as
$$\sqrt{x^2+5x-2}-\sqrt{4x^2-3x+7}+\sqrt{x^2+7x+5}\\
\sqrt{x^2+5x-2}-2\sqrt{x^2-\frac{3}{4}x+\frac{7}{4}}+\sqrt{x^2+7x+5}\\
\left(\sqrt{x^2+5x-2}-\sqrt{x^2-\frac{3}{4}x+\frac{7}{4}}\right)+\left(\sqrt{x^2+7x+5}-\sqrt{x^2-\frac{3}{4}x+\frac{7}{4}}\right)$$
which approaches
$$\frac{23}{8}+\frac{31}{8}=\frac{27}{4}$$
| Hint: $\sqrt{ax^2+bx+c} = \dfrac{\sqrt{a+\dfrac{b}{x}+\dfrac{c}{x^2}}}{\dfrac{1}{x}} = \dfrac{\sqrt{a+by+cy^2}}{y}$,and use L'hospitale rule. Note $y \to 0$. Repeat this process for each term of the sum and simplify to a common fraction and proceed to L'hospitale rule.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Divisibility of polynomial
Prove that:
$(x^2+x+1) \mid (x^{6n+2}+x^{3n+1}+1) $ and
$(x^2+x+1) \mid (x^{6n+4}+x^{3n+2}+1) $.
I saw proof in book with third roots of unity but i didn't understand it, so i want to see other solution but i dont have idea for that.
| $x^{6n+2}+x^{3n+1}+1 = (x^2+x+1) + x^2(x^{6n}-1) + x(x^{3n}-1)$
We know, $x^3-1\mid x^{3n}-1$ and $x^3-1 \mid x^{6n} - 1$ and $x^2+x+1 \mid x^3-1$ which proves the claim.
As for the second part note that $x^4+x^2+1 = (x^2+x+1)(x^2-x+1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\sin^3 a + \cos^3 a$, if $\sin a + \cos a$ is known
Given that $\sin \phi +\cos \phi =1.2$, find $\sin^3\phi + \cos^3\phi$.
My work so far:
(I am replacing $\phi$ with the variable a for this)
$\sin^3 a + 3\sin^2 a *\cos a + 3\sin a *\cos^2 a + \cos^3 a = 1.728$. (This comes from cubing the already given statement with 1.2 in it.)
$\sin^3 a + 3\sin a * \cos a (\sin a + \cos a) + \cos^3 a = 1.728$
$\sin^3 a + 3\sin a * \cos a (1.2) + \cos^3 a = 1.728$
$\sin^3 a + \cos^3 a = 1.728 - 3\sin a * \cos (a) *(1.2)$
Now I am stuck. What do I do next? Any hints?
| i see that you already have several hints pointing you towards an answer. i will try a different method, perhaps a lengthier one. we will use the addition formula $$\sin (a+b) = \sin a \cos b + \sin b \cos a, \cos (a+b) = \cos a \cos b - \sin a \sin b$$ for $\sin()$ and $\cos().$
from $$\cos(t -45^\circ) = \cos 45^\circ \cos t + \sin 45^\circ \sin 45 = {1.2 \over \sqrt 2} = \cos a$$
we can get two things: (i) $t = a + 45^\circ$ and (ii)
$\sin^2 a = 1 - \cos^2 a = 0.28$
now, we can evaluate $$\sqrt 2 \cos t = \sqrt 2 \cos(a+45^\circ) = \cos a - \sin a \text{ and } \sqrt 2 \sin t = \sqrt 2 \sin(a+45^\circ) = \cos a + \sin a$$
cubing and adding these you get $$ 2\sqrt 2(cos^3 t + \sin^3 t) = 2\cos a(\cos^2 a + 3 \sin^2 a) = 2*{1.2 \over \sqrt 2}(1 + 2*0.28)$$ so finally you get
$$ cos^3 t + \sin^3 t = 0.6*1.56 = 0.936$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 4
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When is $(x-1)(y-1)(z-1)$ a factor of $xyz-1$? Let $x$, $y$, $z$ be three natural numbers such that $1< x< y< z$. For how many sets of values of $(x,y,z)$, is $(x-1)(y-1)(z-1)$ a factor of $xyz-1$?
I noticed that $(x-1)(y-1)(z-1)=(xyz-1)-z(x+y-1)-xy+x+y$.
But i don't know how to proceed from here. Any clues?
| Well, at least there are solutions: $x = 2, y = 4, z = 8$ and $x = 3, y = 5, z = 15$
$(x-1)(y-1)(z-1) = x y z - 1 - p(x, y, z)$ with $p(x, y, z) = x y + y z + x z - x - y - z\,.$
If $(x-1)(y-1)(z-1)$ divides $x y z -1$, it also must divide $p(x, y, z)$.
We can prove easily that $(x-1)(y-1)(z-1) > \lvert p(x, y, z) \rvert$ for $x, y, z > 6$, so this is not possible and all solutions must have $x \leq 6$. Try to find similar bounds for $y$ in the case $x \leq 6$ and finally for $z$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove inequality $\;\frac{x+y+1}{\sqrt{x^2+x}+\sqrt{y^2+y}}\le2\;,\;\;\text{for}\;\;x,y\ge 1\;$ I have this great problem: to prove
$$\frac{x+y+1}{\sqrt{x^2+x}+\sqrt{y^2+y}}\le 2\;,\;\;\forall\;x,y\ge1\;?$$
Multiplicate by conjugated I get left side as
$$\frac{\sqrt{x^2+x}-\sqrt{y^2+y}}{x-y}\le 2$$
but I can't get any more, and I can't use derivatives or multiple limits since we didn't study these yet.
Thanks
| Since the denominator of the LHS is positive you are allowed to cross multiply. So it can be simplified as follows:-$$\begin{align}&(x+y+1)^2\le 4\left(\sqrt{x^2+x}+\sqrt{y^2+y}\right)^2\\&=>x^2+y^2+2xy+2x+2y+1\le 4x^2+4y^2+4x+4y+8\sqrt{(x^2+x)(y^2+y)}\\&=>2x^2+2y^2+(x-y)^2+2x+2y+8\sqrt{(x^2+x)(y^2+y)}-1\ge 0\end{align}$$ This is obviously true because $2(x+y)-1\ge 3$ as $x,y\ge 1$
| {
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"timestamp": "2023-03-29T00:00:00",
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Optimization—Finding the Area of the Largest Isoceles Triangle
I managed to solve $(a)$. Since the area of a triangle is determined by $\frac{1}{2}$ base $\times$ height, and we already know the height, we just have to solve for the base. Using Pythagorean theorem, we can deduce base as $\sqrt{36-h^2}$. Therefore, the area is $(6+h)(\sqrt{36-h^2})$. To find $h$, we will take the derivative of the area equation and set that equal to zero. It turns out $h = 3$. Plugging $h$ into the area equation gives us an area of $27\sqrt{3}$.
Can you guys help me do $(b)$? I believe for $(c)$, the largest type of triangle of maximum area is an equilateral.
Thanks.
| For solving $(b)$, we can find the total height $(h_T)$ of the isosceles triangle by adding the value for $h$ found in $(a)$ to the other height $6$:
$$h_T=6+3=9.$$
So, we can say that the total height of the triangle is $9$ units. Then we can find the base:
$$b=\sqrt{36-3^2}=\sqrt{36-9}=\sqrt{27}=3\sqrt{3}.$$
We can then divide the base by $2$ to get the length of the base for half the triangle (by the $AAS$ Similar Triangles Theorem) to get the base to be $\frac{3\sqrt{3}}{2}$. We can now focus on the right half of the triangle divided by the dotted line going through the vertex point of the circle. We can tell that the dotted line through the vertex makes a right angle. Thus, we can use Trigonometry to solve for $\alpha$. So,
$$\alpha=\tan\frac{\text{opposite}}{\text{adjacent}}=\tan\frac{\frac{3\sqrt{3}}{2}}{9}=\tan\frac{27\sqrt{3}}{2}.$$
Since we have found what $\alpha$ is, we can find the function of it with respect to $\alpha$:
$$\boxed{\alpha=\tan\left(\frac{(6+h)(\sqrt{36-h^2})}{2}\right)},$$
or as an incorporated function:
$$\frac12bh=\frac12(6\sin\alpha)(12\cos\alpha)=\boxed{36\sin\alpha\cos\alpha}.$$
And we are done with $(b)$!
For $(c)$, By the Isoperimetric Theorem, it states
Theorem: Among all triangles inscribed in a given circle, the equilateral one has the largest area.
Therefore, the equilateral triangle has the maximum area.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1094908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving the remainder when a polynomial is divided by an integer. How should I go around proving that $\forall x \in \mathbb{Z}$, the remainder when $x^2+2x$ is divided by $3$ is $0$ or $2$?
Do I use the division algorithm for this one?
| You have $x^2 + 2x = (x + 1)^2 - 1$. Now one of the following happens for any $x \in \mathbb{Z}$.
$$x \equiv 0 \mod 3 \Rightarrow x+ 1 \equiv 1 \mod 3 \Rightarrow (x+1)^2 - 1 \equiv 0 \mod 3$$
$$x \equiv 1 \mod 3 \equiv (x+1)^2 \equiv 4 \equiv 1 \mod 3 \Rightarrow (x+1)^2 - 1 \equiv 0 \mod 3$$
$$x \equiv 2 \mod 3 \Rightarrow (x+1)^2 \equiv 9 \equiv 0 \mod 3 \Rightarrow (x+1)^2 -1 \equiv -1 \equiv 2 \mod 3$$
And there you have it.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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A fair die is rolled $N > 6$ times. What is the probability the last 6 rolls were exactly $1,2,3,4,5,6$? You rolls a fair die $N > 6$ times and you want to rolls the sequence $1,2,3,4,5,6$ in this order.
What is the probability that the last 6 rolls were (in consecutive order) $1,2,3,4,5,6$, (So, on the Nth roll you get a 6, on the $(N-1)$th roll you get a 5, etc.), AND you did not roll this sequence any times before the last 6 rolls.
If I state another way, what is the probability that you roll the sequence 1,2,3,4,5,6 starting from the $(N-5)$th roll and you did not roll this sequence starting from any roll before the $(N-5)$th roll?
This is not a homework problems, just something I'm thinking about.
| For seven rolls the last six being $1,2,3,4,5,6$ rules out having $1,2,3,4,5,6$ before that, so the chance you get the first $1,2,3,4,5,6$ starting on roll $2$ is $1/6^6$. For any number of rolls below $12$ the odds of ending with the first $1,2,3,4,5,6$ are $1/6^6$. It is only with $12$ or more rolls that you can get $1,2,3,4,5,6$ before the end, so for $N\ge 12$ the probability is $P($no $1,2,3,4,5,6$ in the first $N-6$ rolls)$1/6^6$ Roughly speaking the chance of not getting $1,2,3,4,5,6$ in $M$ rolls is $(1-1/6^6)^{(M-5)}$ (this assumes independence, which is not quite true), so an approximate answer for the chance of getting $1,2,3,4,5,6$ for the first time at the end of $N$ rolls is $$\begin {cases} 0&N\lt 6\\1/6^6& 6\le N \lt 12 \\(1-1/6^6)^{N-11}/6^6&N \ge 12 \end {cases}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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System of recursive equation. Let's consider:
$$u_o = -1, v_0 = 3$$
$$\begin{cases} u_{n+1} = u_n + v_n \\ v_{n+1} = -u_n + 3v_n \end{cases}$$
I tried:
$$x^n = u_n , y^n = v_n$$
$$\begin{cases} x^{n+1} = x^n + y^n \\ y^{n+1} = -x^n + 3y^n \end{cases}$$
$$\begin{cases} x = 1 + \frac{y^n}{x^n} \\ y = -\frac{x^n}{y^n} + 3 \end{cases}$$
And I don't know how to continue. And it is the basic problem.
The second problem:
(*)The question which occured while attemption solving it, and basically it is not connected with main problem.
We have $y = 4-x$ and now $f =\frac{y^n}{x^n}= (\frac{4-x}{x})^n $
Lets observe that $n \to \infty$ so the $\lim_{n \to \infty} f = 0$
So we can ignore it in our system of equation? It is my doubt. I'm nearly convinced it is not correct but I can't convince of myself why.
Help me, please!
| I don’t necessarily recommend it, but in this case the original problem is easily solved by rather elementary ad hoc methods. Adding the two recurrences, we see that $u_{n+1}+v_{n+1}=4v_n$. From the first recurrence we know that $u_{n+1}+v_{n+1}=u_{n+2}$, so $u_{n+2}=4v_n$, and $u_n=4v_{n-2}$ for $n\ge 2$. Substituting this into the second recurrence, we find that
$$v_{n+1}=3v_n-4v_{n-2}\;.\tag{1}$$
The recurrence $(1)$ has the characteristic equation $x^3-3x^2+4=0$. A little work with the rational root test produces the root $2$, so we can divide out a linear factor $x-2$ and then factor the resulting quadratic:
$$x^3-3x^2+4=(x-2)(x^2-x-2)=(x+1)(x-2)^2\;.$$
Thus, we expect $(1)$ to have the general solution
$$v_n=(A+Bn)2^n+C(-1)^n\;.\tag{2}$$
We have the initial condition $v_0=3$, and we can compute $v_1=10$, $u_1=2$, and $v_2=28$; substituting $n=0$, $n=1$, and $n=2$ into $(2)$ gives us
$$\left\{\begin{align*}
&A+C=3\\
&2A+2B-C=10\\
&4A+8B+C=28\;,
\end{align*}\right.$$
which is easily solved: $A=3$, $B=2$, and $C=0$, so
$$v_n=(3+2n)2^n=3\cdot2^n+n2^{n+1}=(n+1)2^{n+1}+2^n\;.$$
(You can use whichever form seems most convenient.) Finally,
$$u_n=4v_{n-2}=4\left((n-1)2^{n-1}+2^{n-2}\right)=(n-1)2^{n+1}+2^n\;.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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The sum of square and cube roots is an algebraic function How can we prove that the complex valued function $\sqrt z + \sqrt[3] z$ is an algebraic function?
Context
According to my understanding, an algebraic function can be defined as a root of a polynomial equation, I just cannot find a polynomial equation which has this as a root.
| In general, the sum of two algebraic function is algebraic. To find an explicit algebraic equation with root $\sqrt z + \sqrt[3] z$ expand the product
$$\prod_{k \in \{0,1\}, l \in \{0,1,2\}}( u - ( (-1)^k \sqrt{z} + \omega^l \sqrt[3]{z} ))$$
$\omega = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$ is a third root of $1$. All the roots of $v^3 = z$ are $\omega^l \sqrt[3]{z}$, $l \in \{0,1,2\}$. You combine them all together in a product which will contain only integral powers of $z$. You get
$$u^6 - 3 z\, u^4 -2z\, u^3 + 3 z^2 u^2 - 6 z^2 u + (z^2 - z^3) =0$$
satisfied by $\sqrt z + \sqrt[3] z$ .
As an illustration, take a particular value $z = 2^6 = 64$. Then $$u^6 - 3 z\, u^4 -2z\, u^3 + 3 z^2 u^2 - 6 z^2 u + (z^2 - z^3) = \\ =u^6-192 u^4-128 u^3+12288 u^2-24576 u-258048$$ with roots
$$12 = \sqrt{64} + \sqrt[3]{64}\\
- 4 =-\sqrt{64}+ \sqrt[3]{64} \\
- 10 + 2 \sqrt{3} i = -\sqrt{64} + (-\frac{1}{2} + i \frac{\sqrt{3}}{2}) \sqrt[3]{64}\\
- 10 - 2 \sqrt{3} i = -\sqrt{64} + (-\frac{1}{2} + i \frac{\sqrt{3}}{2}) ^2\sqrt[3]{64}\\
6 + 2 \sqrt{3} i = \sqrt{64} + (-\frac{1}{2} + i \frac{\sqrt{3}}{2}) \sqrt[3]{64}\\
6 - 2 \sqrt{3} i = \sqrt{64} + (-\frac{1}{2} + i \frac{\sqrt{3}}{2}) ^2\sqrt[3]{64}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding big O of a function How do I find Big O of function which are polynomial fractions
$$f(x) = \frac {x^4 + x^2 + 1}{x^3 + 1}$$
The same question is posted here (Finding Big-O with Fractions) but i dont understand the explanation on how from the following we concluded that it is order x
$$f(x) = \frac {x^4 + x^2 + 1}{x^3 + 1}
= \frac{(x^4 + x^2 +1)/x^3}{(x^3 + 1)/x^3}
= \frac{x + \frac{1}{x} + \frac{1}{x^3}}{1 + \frac{1}{x^3}}$$
and you can prove directly from this that $f(x) = O(x)$.
Also, what are the witness variables for the above proof?
| For every $x\geqslant1$, $x\leqslant f(x)\leqslant x+1$. Can you conclude?
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $x+y-2=0$ and $y= x^3$ for $x$ and $y$ This is not a full question but needs to be solved to evaluate area using double integral. I need to solve
$$x+y=2$$
$$y=x^3$$
If I put $y= x^3$ in first equation I get messy $x+ x^3 -2=0$ to solve to find what $x$ is? How to do it using viable methods from calculus, or any other quick way.
| Clearly, $x=1$ satisfies $x^3+x-2=0$
Alternatively, $x^3+x-2=x^3-1+x-1=(x-1)(x^2+x+1)+(x-1)=\cdots$
the rest $\dfrac{x^3+x-2}{x-1}=0$ is a Quadratic equation, right?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find a basis for the subspace $\left\{\begin{bmatrix}x & y \\ z & t\end{bmatrix}, x-y-z = 0\right\}$ The exercise gives me the subspace
$$\left\{\begin{bmatrix}x & y \\ z & t\end{bmatrix}, x-y-z = 0\right\}$$
and ask me to show that these two sets are basis for this subspace:
$$B = \left\{\begin{bmatrix}1 & 1 \\ 0 & 0\end{bmatrix}, \begin{bmatrix}1 & 0 \\ 1 & 0\end{bmatrix}, \begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}\right\}$$
$$C = \left\{\begin{bmatrix}1 & 0 \\ 1 & 0\end{bmatrix}, \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}, \begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}\right\}$$
for the basis $B$ it's easy to see that if we solve for $x = y + z$ then a matrix
$$\begin{bmatrix}x & y \\ z & t\end{bmatrix}$$
can be written as
$$\begin{bmatrix}x & y \\ z & t\end{bmatrix} = y\begin{bmatrix}1 & 1 \\ 0 & 0\end{bmatrix} + z \begin{bmatrix}1 & 0 \\ 1 & 0\end{bmatrix} + t\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}$$
but what can I do to verify that basis $C$ generates the subspace? (I know I also must verify that these sets are linearly independent)
| This is similar to what you did with B:
$$
\begin{bmatrix}x & y \\ z & t\end{bmatrix} = x\begin{bmatrix}1 & 0 \\ 1 & 0\end{bmatrix} + (-y) \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix} + t\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit $\lim \limits_{x \to 2}{\frac{\sqrt{x^3 - 3x^2 + 4}-x +2}{x^2 - 4}}$ I try to calculate $\lim \limits_{x \to 2}{\frac{\sqrt{x^3 - 3x^2 + 4}-x +2}{x^2 - 4}}$. So, $\frac{\sqrt{x^3 - 3x^2 + 4}-x +2}{x^2 - 4} = \frac{(x-2)x}{(x+2)(x+\sqrt{(x-2)^2(x+1)}-2)}$ but I don't know what to do next.
| $$
\begin{aligned}
\lim _{x\to 2}\left(\frac{\sqrt{x^3\:-\:3x^2\:+\:4}-x\:+2}{x^2\:-\:4}\right)
& = \lim _{t\to 0}\left(\frac{\sqrt{\left(t+2\right)^3\:-\:3\left(t+2\right)^2\:+\:4}-\left(t+2\right)\:+2}{\left(t+2\right)^2\:-\:4}\right)
\\& = \lim _{t\to 0}\left(\frac{\left|t\right|\sqrt{\left(t+3\right)}-t}{t\left(t+4\right)}\right)
\\& = \lim _{t\to 0^+}\left(\frac{\left|t\right|\sqrt{\left(t+3\right)}-t}{t\left(t+4\right)}\right) = \color{red}{\frac{1}{4}\left(\sqrt{3}-1\right)}
\\& = \lim _{t\to 0^-}\left(\frac{\left|t\right|\sqrt{\left(t+3\right)}-t}{t\left(t+4\right)}\right) = \color{blue}{\frac{1}{4}\left(-\sqrt{3}-1\right)}
\end{aligned}
$$
Solved with substitution $\color{green}{t = x-2}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate to find the sum of an infinite series $∑_{n=1}^\infty$ $n\over2^{n-1}$
or
1 + $2\over2$ + $3\over4$ + $4\over8$ + $5\over16$ + $\ldots$
How to go about evaluating the above, showing that it sums to 4?
| $\sum_{n\geq 1} \frac{n}{2^{n-1}} = f'(1)$ where $f(x) = 2 \sum_{n\geq 0} \frac{x^n}{2^n} = 2 \frac{1}{1 - \frac{x}{2}}$ so that $f'(x) = 2 \frac{\frac{1}{2}}{\left(1-\frac{x}{2}\right)^2}$ and $f'(1) = 4$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Proving $\sin^2A \equiv \cos^2B + \cos^2C + 2 \cos A\cos B\cos C$ As the title,
By considering $\bigtriangleup$ABC, Prove
$$\sin^2A \equiv \cos^2B + \cos^2C + 2 \cos A\cos B\cos C$$
Thanks
| this is just the $\cos$ rule and the $\sin$ rule combined
$a^2 = b^2 + c^2 - 2bc \cos A$ and $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ where $R$ is the radius of the circumcircle of $ABC.$
the cosine rule becomes
$\begin{align}
\sin^2 A &= \sin^2 B + \sin^2 C-2\sin B \sin C \cos A \\
&=\cos^2 B + \cos^2 C - (\cos^2 B -\sin^2 B) - (\cos^2 C -\sin^2 C) - 2\sin B \sin C \cos A\\
&=\cos^2 B + \cos^2 C - (\cos 2B + \cos 2C)- 2\sin B \sin C \cos A\\
&=\cos^2 B + \cos^2 C - 2\cos(B +C)\cos(B-C)- 2\sin B \sin C \cos A\\
&=\cos^2 B + \cos^2 C + 2\cos A (\cos B\cos C + \sin B \sin C)- 2\sin B \sin C \cos A\\
&=\cos^2 B + \cos^2 C + 2\cos A \cos B\cos C\\
\end{align}$
use of $\cos(B+C) = \cos(180^\circ - A) = -\cos A$ was used in the one before the last.
| {
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"timestamp": "2023-03-29T00:00:00",
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Easiest way to calculate determinant 5x5 witx x I would like to calculate this determinant:
\begin{vmatrix}x&1&0&0&0\\4&x&2&0&0\\0&3&x&3&0\\0&0&2&x&4\\0&0&0&1&x\end{vmatrix}
| Hint By adding all the rows to first you get
$$\begin{vmatrix}x&1&0&0&0\\4&x&2&0&0\\0&3&x&3&0\\0&0&2&x&4\\0&0&0&1&x\end{vmatrix} = \begin{vmatrix}x+4&x+4&x+4&x+4&x+4\\4&x&2&0&0\\0&3&x&3&0\\0&0&2&x&4\\0&0&0&1&x\end{vmatrix} \\
=(x+4)\begin{vmatrix}1&1&1&1&1\\4&x&2&0&0\\0&3&x&3&0\\0&0&2&x&4\\0&0&0&1&x\end{vmatrix}=(x+4)\begin{vmatrix}1&1&1&1&1\\0&x-4&-2&-4&-4\\0&3&x&3&0\\0&0&2&x&4\\0&0&0&1&x\end{vmatrix} \\
=(x+4)\begin{vmatrix}x-4&-2&-4&-4\\3&x&3&0\\0&2&x&4\\0&0&1&x\end{vmatrix}$$
Now use expansion by a row or column.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\int_0^\infty\int_0^\pi\frac{k^2(e^{-it\sqrt{k^2+m^2}}-e^{it\sqrt{k^2+m^2}})\sin(\theta)}{e^{-ikx\cos{\theta}}\sqrt{k^2+m^2}}d\theta dk$ $$\int_0^\infty\int_0^\pi\frac{k^2\left(e^{-it\sqrt{k^2+m^2}}-e^{it\sqrt{k^2+m^2}}\right)\sin(\theta)}{e^{-ikx\cos{\theta}}\sqrt{k^2+m^2}}d\theta dk$$
I saw this Integral at Quora, and I have not idea how to evaluate it. Therefore, I thought of posting it here.
How do we Evaluate this?
| You can write:
$$\int_0^\infty\int_0^\pi\frac{k^2\left(e^{-it\sqrt{k^2+m^2}}-e^{it\sqrt{k^2+m^2}}\right)\sin(\theta)}{e^{-ikx\cos{\theta}}\sqrt{k^2+m^2}}d\theta dk$$
As:
$$\int_0^\infty\int_0^\pi\frac{k^2\left(-2 i \sin{(t \sqrt{k^2+m^2})}\right)\sin(\theta)}{e^{-ikx\cos{\theta}}\sqrt{k^2+m^2}}d\theta dk = \int_0^\infty \frac{k^2\left(-2 i \sin{(t\sqrt{k^2+m^2})}\right)}{\sqrt{k^2+m^2}} \int_0^\pi e^{ikx\cos{\theta}} \sin(\theta) d\theta dk = \int_0^\infty \frac{k^2\left(-4 i \sin{(t\sqrt{k^2+m^2})}\right)}{\sqrt{k^2+m^2}} \frac{\sin{k x}}{k x} dk=\int_0^\infty -4 ik^2 t sinc{(t\sqrt{k^2+m^2})} sinc(k x) dk= -4 t i \int_0^\infty k^2 sinc{(t\sqrt{k^2+m^2})} sinc(k x) dk $$
Now by using the relationship:
$$sinc{(t\sqrt{k^2+m^2})}= \prod_{n=1}^{+\infty}(1-\frac{(t\sqrt{k^2+m^2})^2}{n^2 \pi^2})$$
And:
$$sinc{(k x)}= \prod_{j=1}^{+\infty}(1-\frac{(k x)^2}{j^2 \pi^2})$$
We can write
$$-4 t i \int_0^\infty k^2 sinc{(t\sqrt{k^2+m^2})} sinc(k x) dk = -4 t i \lim_{\phi \rightarrow +\infty}\prod_{n=1}^{+\infty} \prod_{j=1}^{+\infty} \int_0^\phi k^2 (1-\frac{(t\sqrt{k^2+m^2})^2}{n^2 \pi^2}) (1-\frac{(k x)^2}{j^2 \pi^2}) = -4 t i \lim_{\phi \rightarrow +\infty}\prod_{n=1}^{+\infty} \prod_{j=1}^{+\infty} \left[ \frac{\phi^3}{3}(1-(\frac{m t}{n \pi})^2)+\frac{\phi^5}{5}((\frac{m t x}{j n \pi^2})^2-(\frac{t}{n \pi})^2-(\frac{x}{j \pi})^2)+\frac{\phi^7}{7}(\frac{t x }{j n \pi^2})^2\right]$$
Which seems to diverge
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Evaluation of a general trigonometric integral How can I evaluate the integral
$$\int\sin^k(x)\ dx$$
in which I don't know if $k$ is an even or an odd number?
| Let $f(x) = \sin^k x$.
We have $\sin x = \frac{e^{ix} - e^{-ix}}{2i}$. So
$$
\begin{align*}
f(x) &= \frac{1}{(2i)^k}(e^{ix} - e^{-ix})^k \\
&=\frac{1}{(2i)^k} \sum_{j = 0}^k \binom{k}{j} (e^{ix})^{k-j} (-e^{-ix})^j \\
&=\frac{1}{(2i)^k} \sum_{j = 0}^k (-1)^j \binom{k}{j} e^{(k-2j)ix} \\
&=\frac{1}{(2i)^k} \left(\sum_{0 \leq j < k/2} \left[(-1)^j\binom{k}{j}e^{(k-2j)ix} + (-1)^{k-j} \binom{k}{k-j}e^{(2j-k)ix} \right] + \varepsilon (-1)^{k/2}\binom{k}{k/2}\right),
\end{align*}
$$
where the term with $\varepsilon$ appears only if $k$ is even, and then $\varepsilon = 1$. On the last line, we've grouped together symmetric terms of the sum. From here on, we distinguish between even $k$ and odd $k$.
If $k$ is even, say $k = 2m$, then
$$
\begin{align*}
f(x) &= \frac{(-1)^m}{2^{2m}} \left[ \sum_{j = 0}^{m-1} (-1)^{j} \binom{2m}{j} \left(e^{2(m-j)ix} + e^{2(j-m)ix}\right) + (-1)^m\binom{2m}{m} \right]\\
&= \frac{1}{2^{2m}} \left[ \sum_{j=0}^{m-1} (-1)^{m-j} \binom{2m}{j} \cdot 2 \cos 2(m-j)x + \binom{2m}{m} \right] \\
&=\frac{1}{2^{2m-1}} \sum_{h = 1}^m (-1)^h \binom{2m}{m-h} \cos 2hx + \frac{1}{2^{2m}}\binom{2m}{m},
\end{align*}
$$
which is easy to integrate. (From $0$ to $\pi$, only the last term has a nonzero integral.) The last term is like half what the term for $h = 0$ would be. For example, for $k = 6$, we get
$$\sin^6 x = \frac{20}{64} - \frac{15}{32} \cos 2x + \frac{6}{32} \cos 4x - \frac{1}{32} \cos 6x $$
If $k$ is odd, say $k = 2m + 1$, then we have
$$
\begin{align*}
f(x) &= \frac{(-1)^{m}}{2^{2m}} \sum_{j = 0}^m (-1)^j \binom{2m+1}{j} \frac{e^{[2(m-j) + 1]x} - e^{-[2(m-j) + 1]x}}{2i} \\
&=\frac{1}{2^{2m}} \sum_{j = 0}^m (-1)^{m-j} \binom{2m+1}{j} \sin [2(m-j) + 1]x \\
&=\frac{1}{2^{2m}} \sum_{h = 0}^m (-1)^{h} \binom{2m+1}{m-h} \sin (2h + 1)x,
\end{align*}
$$
which is again easy to integrate. From $0$ to $\pi$, this last formula won't immediately give you the simplest form of the integral, which is twice the Wallis integral. Also, there are some circumstances where it might be preferable to integrate directly from one of the complex forms of $f(x)$, using the fact that the integral of $e^{inx}$ is $\displaystyle \frac{e^{inx}}{in}$.
For $k = 7$, we get
$$\sin^7 x = \frac{35}{64} \sin x - \frac{21}{64} \sin 3x + \frac{7}{64} \sin 5x - \frac{1}{64} \sin 7x.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1113610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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factor the following expression $25x^2 +5xy -6y^2$ How to factor
$$25x^2 +5xy -6y^2$$
I tried with $5x(5x+y)-6y^2$. I'm stuck here.
I can't continue.
| let me expand on my comment of using quadratic formula to factor an expression like this. of course, you need to have the quadratic formula that
$x = \dfrac{-b + \pm \sqrt{b^2 - 4ac}}{2a}$ are the solutions of
$ax^2 + bx + c = 0.$
we will look at first the equation $25x^2 +5xy - 6y^2 = 0$ later on we will see how we can use the solutions to factor the expression. matching $25x^2 +5xy - 6y^2 = 0$ with $ax^2 + bx + c = 0$ we can identify $a = 25, b = 5y, c = -6y^2.$ first the discriminant $b^2 - 4ac = (5y)^2 - 4*25*(-6y^2) = 625y^2 = (25y)^2$
using the formula $x = \dfrac{-5y \pm 25y}{50} = -\dfrac{3y}{5}, \dfrac{2y}{5}$
we can turn this into two equations into the factors
$k(x+\frac{3y}{5})(x - \frac{2y}{5}) = 25x^2 +5xy - 6y^2 = 0 $ matching the coeffcient of $x^2$ fixes $k = 25.$
what do we have now is $$25(x+\frac{3y}{5})(x - \frac{2y}{5})= (5x + 3y)(5x - 2y)= 25x^2 +5xy - 6y^2 = 0 $$
from the last equalities drop leftmost and the rightmost expressions and you have factored $25x^2 +5xy - 6y^2.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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For $c<1$, let $f_c(x)=x^2+c$, determine the period-2 points? Ok so I know I need to set $$f_c^2(x)=x$$ so:
$$(x_0^2+c)^2+c = x \iff x_0^4+2cx_0^2-x_0+c^2+c=0$$
But how do I then solve this? Ok I have solved for the four set of roots, now you can see there are two roots for $c<\frac{1}{4}$ and four for $c<-\frac{3}{4}$. How do I find the hyperbolicity and therefore the nature of these points?
| Notice there is no $x^3$ term.
Try something of the form $(x^2+Ax+B)(x^2-Ax+D)$
$$(x^2+Ax+B)(x^2-Ax+D)=x^4+(D+B-A^2)x^2+(AD-AB)x+BD$$
So we know
$$D+B-A^2=2c$$
$$AD-AB=-1$$
$$BD=c^2+c$$
Which has solution $A=1$, $B=c+1$, and $D=c$.
So $x^4+2cx^2-x+c^2+c=(x^2+x+c+1)(x^2-x+c)$
I assume you can take it from there.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is this equality true? Why? Why not? Let $$ \lim_{a\to 0} \frac{1}{2} \left( \left( \sum_{n=-\infty}^\infty \frac{1}{(n+a)^2} - \frac{1}{a^2} \right) \right) = \sum_{n=1}^\infty \frac{1}{n^2}$$
I already know that $\sum_{n=-\infty}^\infty \frac{1}{(n+a)^2} = \frac{\pi^2}{\sin^2(\pi a)}$. With that in mind, we know that the series converges in for every $a\in\mathbb{R}$ (Is that right?)
Also, all terms are positive so the series converges absolutely. Hence,
$$= \lim_{a\to 0} \frac{1}{2} \left( \left( \sum_{n=-\infty}^{-1} \frac{1}{(n+a)^2} \right) + \left( \sum_{n=-1}^{\infty} \frac{1}{(n+a)^2} \right) + \frac{1}{a^2} - \frac{1}{a^2} \right) = \lim_{a\to 0} \frac{1}{2} \left( \left( \sum_{n=-\infty}^{-1} \frac{1}{(n+a)^2} \right) + \left( \sum_{n=-1}^{\infty} \frac{1}{(n+a)^2} \right) \right) = \\ \frac{1}{2} \left( \left( \sum_{n=1}^{\infty} \frac{1}{(-n)^2} \right) + \left( \sum_{n=1}^{\infty} \frac{1}{n^2} \right) \right) = \\ \sum_{n=1}^\infty \frac{1}{n^2}$$
Now, how do I justify the continuity of the series which enabled me to do this algebraic trick?
| Since
$$\sum_{n\in\mathbb{Z}}\frac{1}{(n+a)^2}=\frac{\pi^2}{\sin^2(\pi a)}$$
as a function of $a$, is an even meromorphic function with a double pole in zero, as the square of a meromorphic function with a simple pole in zero:
$$ \frac{\pi}{\sin(\pi z)}=\frac{1}{z}+\zeta(2) z+\ldots $$
it happens that:
$$ \lim_{z\to 0}\left(-\frac{1}{z^2}+\frac{\pi^2}{\sin^2(\pi z)}\right)=\lim_{z\to 0}\left(-\frac{1}{z^2}+\frac{1}{z^2}+2\zeta(2)+\ldots\right)=2\zeta(2)$$
as expected.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How prove $\sin \left( \alpha+\frac{\pi }{n} \right) \cdots \sin \left( \alpha+\frac{n\pi }{n} \right) =-\frac{\sin n\alpha}{2^{n-1}}$? How prove
$$\prod_{k=1}^{n}\sin \left( \alpha+\frac{\pi k }{n}\right) =-\frac{\sin n\alpha}{2^{n-1}}$$
for $n \in N$?
| Using De Moivre's formula for odd $n=2m+1$,
and writing $\cos x=c,\sin x=s$
$$i\sin(2m+1)x=(i s)^{2m+1}+\binom{2m+1}2(i s)^{2m-1}c^2+\binom{2m+1}4(i s)^{2m-3}c^4+\cdots+\binom{2m+1}{2m}(is)c^{2m}$$
$$=i^{2m+1}[s^{2m+1}-\binom{2m+1}2s^{2m-1}(1-s^2)+\binom{2m+1}4s^{2m-3}(1-s^2)^2+\cdots+\binom{2m+1}{2m}(-1)^ms(1-s^2)^m]$$
$$\iff s^{2m+1}\left[1+\binom{2m+1}2+\binom{2m+1}4+\cdots+\binom{2m+1}{2m}\right]+\cdots= (-1)^m\sin(2m+1)x$$
$$\iff s^{2m+1}\left[1+1\right]^{2m+1-1}- (-1)^m\sin(2m+1)x=0$$
Now if $\sin(2m+1)x=\sin(2m+1)a,$
$(2m+1)x=2r\pi+(2m+1)a, x=a+\dfrac{2r\pi}{2m+1}$ where $-m\le r\le m$
$\implies2^{2m}\prod_{r=-m}^m\sin\left(a+\dfrac{2r\pi}{2m+1}\right)=-(-1)^m\sin(2m+1)a$
Now for $-m\le r<0, r=-t$(say) $\implies m\ge t>0$
$\sin\left(a+\dfrac{2r\pi}{2m+1}\right)=\sin\left(a-\dfrac{2t\pi}{2m+1}\right)$
$=-\sin\left(a-\dfrac{2t\pi}{2m+1}+\pi\right)=-\sin\left(a+\dfrac{(2m+1-2t)\pi}{2m+1}\right)$
As $m\ge t>0,-2m\le -2t<0\iff1\le2m+1-2t<2m+1$
$\implies2^{2m}\prod_{t=0}^{2m}\sin\left(a+\dfrac{t\pi}{2m+1}\right)(-1)^m=-(-1)^m\sin(2m+1)a$
Similarly for the even $n=2m$(say)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How do I simplify this expression about factorization? I am trying to simplify this
$$\frac{9x^2 - x^4} {x^2 - 6x +9}$$
The solution is
$$\frac{-x^2(x +3)}{x-3} = \frac{-x^3 - 3x^2}{x-3} $$
I have done $$\frac{x^2(9-x^2)}{(x-3)(x-3)} = \frac{x^2(3-x)(3+x)}{(x-3)(x-3)} $$
but I do not find a way to simplify
How can I simplify to get the answer?
I have to use
$${(a +b)(a -b)} = {a^2 - b^2} $$
$${(a +b)^2} = {a^2 + 2ab+b^2} $$
$${(a -b)^2} = {a^2 - 2ab+b^2} $$
| Observe that
$$
9x^2-x^4=-x^2(x^2-9)=-x^2(x-3)(x+3).
$$
Thus, we have that
$$
\frac{9x^2-x^4}{x^2-6x+9}=\frac{-x^2(x-3)(x+3)}{(x-3)^2}=-x^2\left(\frac{x+3}{x-3}\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1122635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that if $a$,$b$,$c$ are non-negative real numbers such that $a+b+c =3$, then $abc(a^2 + b^2 + c^2)\leq 3$ Prove that if $ a,b,c $ are non-negative real numbers such that $a+b+c = 3$, then $$ abc(a^2 + b^2 + c^2) \le 3 $$
My attempt :
I tried AM-GM inequality, tried to convert it to $a+b+c$, but I think I cannot get the manipulation of $abc$.
| Note that $(ab+bc+ac)^2 \ge 3abc(a+b+c) = 9abc$
Since, $(ab+bc+ac)^2 = \sum\limits_{cyc} a^2b^2 + 2\sum\limits_{cyc}ab^2c \ge 3abc(a+b+c)$
Hence, $$\begin{align}abc(a^2 + b^2 + c^2) &\le \frac{1}{9}(ab+bc+ac)^2(a^2 + b^2 + c^2) \tag{1} \\&\le \frac{1}{9}\left(\frac{a^2 + b^2 + c^2+2ab+2bc+2ac}{3}\right)^3\tag{2} \\ &= \frac{(a+b+c)^6}{3^5} = 3\tag{3}\end{align}$$
$(1)$ application of Am-Gm Inequality with three terms $$a^2+b^2+c^2,ab+bc+ca,ab+bc+ca$$
$(2)$ Using $a^2+b^2+c^2+2(ab+bc+ca) = (a+b+c)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the area of the triangle under certain preconditions
With vertices $(0, 0)$, $(b, a)$, $(x, y)$, prove the area of this triangle is $\frac{|by - ax|}{2}$.
We know area of a triangle = $\frac{rh}{2}$. ($r$ is the base.) Well, we have $r =$ the distance from the origin to $(x, y)$ and the height is the line perpendicular to $y = mx$, which is $y = -\frac{x}{m} + b'$, intersecting the points $(x_0, y_0)$ on the line $y = mx$ and $(b, a)$.
So then, the altitude from my calculations was $ h =\sqrt{1+\frac{1}{m^2}}|x_0 - b|$ The base is just the distance to the origin from $(x, y)$, which is $d = \sqrt{x^2+y^2}$. But I don't see how this equals what I'm trying to prove.
| If we denote the point $(x,y)$ by $(x_1,y_1)$ to avoid confusion, we can take the base of the triangle to be $\;\;\;B=\sqrt{x_1^2+y_1^2}$.
The line through $(0,0)$ and $(x_1,y_1)$ has equation $y=\frac{y_1}{x_1}x$ (assuming $x_1\ne0$), so the perpendicular line through $(b,a)$ has equation $y-a=-\frac{x_1}{y_1}(x-b)$ or $y=-\frac{x_1}{y_1}(x-b)+a$ (assuming $y_1\ne0$).
The two lines intersect where $\frac{y_1}{x_1}x=-\frac{x_1}{y_1}(x-b)+a$, so solving for $x$ gives
$\left(\frac{y_1}{x_1}+\frac{x_1}{y_1}\right)x=\frac{x_1 b}{y_1}+a$, so $x=\frac{x_1y_1}{x_1^2+y_1^2}\left(\frac{x_1 b}{y_1}+a\right)\implies x=\frac {x_1(bx_1+ay_1)}{x_1^2+y_1^2}\text{ and }y=\frac{y_1(bx_1+ay_1)}{x_1^2+y_1^2}.$
Then the height of the triangle is given by the distance from $(b,a)$ to this point, so
$\displaystyle h=\sqrt{\left(b-\frac {x_1(bx_1+ay_1)}{x_1^2+y_1^2}\right)^2+\left(a-\frac{y_1(bx_1+ay_1)}{x_1^2+y_1^2}\right)^2}$
$\displaystyle=\sqrt{\frac{y_1^{2}(b y_1-a x_1)^{2}+x_1^{2}(ax_1-by_1)^2}{(x_1^{2}+y_1^{2})^2}}=\sqrt{\frac{(by_1-ax_1)^2}{x_1^2+y_1^2}}=\frac{\lvert by_1-ax_1\rvert}{\sqrt{x_1^2+y_1^2}}$.
Then $\displaystyle A=\frac{1}{2}Bh=\frac{|by_1-ax_1|}{2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int_0^1 \frac {x^3}{\sqrt {4+x^2}}\,dx$ How do I evaluate the definite integral $$\int_0^1 \frac {x^3}{\sqrt {4+x^2}}\,dx ?$$ I used trig substitution, and then a u substitution for $\sec\theta$.
I tried doing it and got an answer of: $-\sqrt{125}+12\sqrt{5}-16$, but apparently its wrong.
Can someone help check my error?
| Hint:
$$\int_0^1\frac{x^3}{\sqrt{4+x^2}}dx=\frac{1}{2}\int_0^1\frac{x^22x}{\sqrt{4+x^2}}dx=\int_0^1\frac{(x^2+4-4)(4+x^2)'}{\sqrt{4+x^2}}dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How does $n(n-1)(n-2)\cdots(n-m+1) \cdot \frac{(n-m)(n-m-1)\cdots1}{(n-m)(n-m-1)\cdots1} = \frac{n(n-1)(n-2)\cdots1}{(n-m)(n-m-1)\cdots1} $ I'm having issues understanding how the previous line goes to the net line.
$$
\text{Assume } m \le n \\
n(n-1)(n-2)...(n-m+1) \cdot \frac{(n-m)(n-m-1)\cdots1}{(n-m)(n-m-1)\cdots1} \\
= \frac{n(n-1)(n-2)\cdots1}{(n-m)(n-m-1)\cdots1} $$
I just don't see how the numerator of the fraction in first line cancels out yet nothing in the denominator cancels out.
Also I'm not sure what tag this question should go under.
| Look at a concrete example, say $n=6$ and $m=3$. Then $n-m+1=6-3+1=4$, so we have
$$\begin{align*}
6\cdot5\cdot4&=6\cdot5\cdot4\cdot\color{brown}1\\\\
&=6\cdot5\cdot4\cdot\color{brown}{\frac{3\cdot2\cdot1}{3\cdot2\cdot1}}\\\\
&=\frac{6\cdot5\cdot4\cdot\color{brown}{3\cdot2\cdot1}}{\color{brown}{3\cdot2\cdot1}}\\\\
&=\frac{6!}{3!}\;.
\end{align*}$$
The general case is the same thing.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int^\infty_0 t^{a+b-1}(t+1)^{-b-1} U(a+2,a-b+2,ct)dt$
Evaluate
$$
\int^\infty_0 t^{a+b-1}\left(t+1\right)^{-b-1} U\left(a+2,a-b+2,ct\right)dt
$$
under the condition $a>0$, $b>0$ and $c>0$, where $U(\cdot,\cdot,\cdot)$ denotes the confluent hypergeometric function of the second kind. All of $a$, $b$ and $c$ are noninteger, but the stronger constraint on the range of $b$ is accepted.
Mathematica found the following form with the condition $b>\frac{1}{2}$, but I believe the simpler or shorter expression exists.
$$
\frac{2 c^{1-a} \, _2F_2(3,b+1;2-a,3-b;-c)}{a \left(a^2-1\right) (b-2) (b-1) b}-\frac{\sqrt{\pi } 4^{b-1} \csc (\pi b) \Gamma
\left(b-\frac{1}{2}\right) c^{-a+b-1} \Gamma (a-b+1) \, _2F_2(b+1,2 b-1;b-1,b-a;-c)}{b \Gamma (a+2) \Gamma (b-1)}+\frac{\pi \csc (\pi a)
\Gamma (-a+b-1) \Gamma (a+b) \, _2F_2(a+2,a+b;a,a-b+2;-c)}{\Gamma (a) \Gamma (b+1)^2}.
$$
| (Too big for comment)
It's probably a matter of taste whether this counts as simpler (and it certainly isn't shorter), but the second confluent hypergeometric function can be reduced to a linear combination of confluent hypergeometric functions of smaller rank by working through their respective integral representations:
$$\begin{align}
\small{{_2F_2}{\left(b+1,2b-1;b-1,b-a;-c\right)}}
&=\small{\frac{\Gamma{\left(b-a\right)}}{\Gamma{\left(2b-1\right)}\,\Gamma{\left(1-a-b\right)}}\int_{0}^{1}t^{2b-2}(1-t)^{-a-b}{_1F_1}{\left(b+1;b-1;-ct\right)}\,\mathrm{d}t}\\
&=\frac{c^2\,\Gamma{\left(b-a\right)}}{(b-1)b\,\Gamma{\left(2b-1\right)}\,\Gamma{\left(1-a-b\right)}}\int_{0}^{1}t^{2b}(1-t)^{-a-b}e^{-ct}\,\mathrm{d}t\\
&~~~~~ -\frac{2c\,\Gamma{\left(b-a\right)}}{(b-1)\Gamma{\left(2b-1\right)}\,\Gamma{\left(1-a-b\right)}}\int_{0}^{1}\frac{t^{2b-1}}{(1-t)^{a+b}}e^{-ct}\,\mathrm{d}t\\
&~~~~~ +\frac{\Gamma{\left(b-a\right)}}{\Gamma{\left(2b-1\right)}\,\Gamma{\left(1-a-b\right)}}\int_{0}^{1}t^{2b-2}(1-t)^{-a-b}e^{-ct}\,\mathrm{d}t\\
&=\frac{2(2b-1)c^2}{(a-b-1)(a-b)(b-1)}\,{_1F_1}{\left(2b+1;2-a+b;-c\right)}\\
&~~~~~ -\frac{2c(2b-1)}{(b-1)(b-a)}\,{_1F_1}{\left(2b;1-a+b;-c\right)}\\
&~~~~~ +{_1F_1}{\left(2b-1;b-a;-c\right)}.\\
\end{align}$$
The third hypergeometric term can be similarly "simplified" in the same manner. However, it seems the first hypergeometric term isn't as straightforward...
| {
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Let $f(x)=x^2+12x+30$. Solve $f(f(f(f(f(x)))))=0$ Here is my solve, is it correct?
I figured out that we can restate $f(f(x))$ as
$((x+r)(x+s)+r)((x+r)(x+s)+s)$
thus
$f(f(f(f(f(x)))))=0$
is
$(x+r)^2(x+s)^2(4s+3r)(4r+3s)$
from vieta's
$0=(x+r)(x+s)(36^2-r^2)$ and we get $x=-6$
Did I make a mistake somewhere?
I'd also like to see other solutions
EDIT: How can I solve it using vieta then?
| We have $f(x) = (x+6)^2 - 6$, therefore $f^2 (x) = (x+6)^4 - 6$, $f^3 (x) = (x+6)^8 - 6$, and so on.
So $f^5 (x) = (x+6)^{32} - 6 = 0$, giving the (real) solutions $x=-6 \pm \sqrt[32]{6}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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no. of real roots of the equation $ 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7} = 0$
The no. of real roots of the equation $\displaystyle 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7} = 0 $
$\bf{My\; Try::}$ First we will find nature of graph of function $\displaystyle 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7}$
So $$\displaystyle f(x) = 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7}.$$ Then Differentiate
both side w . r to $x\;,$ We get $$\displaystyle f'(x)=1+x+x^2+x^3+..........+x^6$$
Now for max. and Minimum Put $$f'(x) = 0\Rightarrow 1+x+x^2+x^3+x^4+x^5+x^6 = 0$$
We can write $f'(x)$ as $$\displaystyle \left(x^3+\frac{x^2}{2}\right)^2+\frac{3}{4}x^4+x^3+x^2+x+1$$
So $$\displaystyle f'(x) = \left(x^3+\frac{x^2}{2}\right)^2+\frac{1}{4}\left[3x^4+4x^3+4x^2+4x+4\right]$$
So $$\displaystyle f'(x) = = \left(x^3+\frac{x^2}{2}\right)^2+\frac{1}{4}\left[\left(\sqrt{2}x^2+\sqrt{2}x\right)^2+(x^2)^2+2(x+1)^2+2\right]>0\;\forall x\in \mathbb{R}$$
So $f'(x) = 0$ does not have any real roots. So Using $\bf{LMVT}$ $f(x) = 0$ has at most one root.
In fact $f(x) = 0$ has exactly one root bcz $f(x)$ is of odd degree polynomial and it
will Cross $\bf{X-}$ axis at least one time.
My question is can we solve it any other way, i. e without using Derivative test.
Help me , Thanks
| The simplest way of seeing that it can have only one real root is IMO to look at the derivative. To that end we use the formula for a geometric sum
$$
f'(x)=1+x+x^2+\cdots+x^6=\frac{x^7-1}{x-1}
$$
showing that $f'(x)>0$ whenever $x\neq1$, because the numerator and denominator both change signs only at $x=1$. Of course, $f'(1)=7>0$. So the function is everywhere increasing, and cannot have more than one zero. As you observed, being an odd degree polynomial it clearly has at least one zero.
| {
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$a^2 + b^2 + c^2 = 1 ,$ then $ab + bc + ca$ gives =? In a recent examination this question has been asked, which says:
$a^2+b^2+c^2 = 1$ , then $ab + bc + ca$ gives = ?
What should be the answer? I have tried the formula for $(a+b+c)^2$, but gets varying answer like $0$ or $0.25$, on assigning different values to variables.
How to approach such question?
| It depends on $a+b+c$:
$$
(a+b+c)^2-2(ab+bc+ca)=1\\
ab+bc+ca=\dfrac{(a+b+c)^2-1}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Proving $ab(a+b)+ac(a+c)+bc(b+c)$ is even
Prove that $\forall a,b,c\in \mathbb N: ab(a+b)+ac(a+c)+bc(b+c)$ is even
I tried to simplify the expression to something that would always yield an even number: $ (a+b+c)(ab+ac+bc)-3abc$ but that's just a sum of numbers that are divisible by $3$...
Is there a way to do this without checking every combination of the parity of $a,b,c$?
| We will use that for any $A$ the numbers $A$, $3A$ and $A^3$ all have the same parity. Then writing $S$ for our expression, we see
$$\begin{align*}
S &= ab(a+b)+bc(b+c)+ca(c+a) \\ &\equiv 3ab(a+b) + 3bc(b+c)+3ca(c+a) \\
&= (a+b)^3 - a^3 - b^3 + (b+c)^3 - b^3 - c^3 + (c+a)^3 - c^3 - a^3 \\
&\equiv (a+b)^3 + (b+c)^3 + (c+a)^3 \\
&\equiv (a+b) + (b+c) + (c+a) \equiv 0.
\end{align*}
$$
Here $\equiv$ denotes equality modulo $2$, i.e. $M \equiv N$ means that $M$ and $N$ have the same parity. So $S$ has the same parity as $0$, hence it is always even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to compute $\int_{0}^{\infty}dx\:\frac{\exp(-ax^2+bx)}{x+1}\:\text{ for }\: a>0, b\in \mathbb{C}$? As the title says I am trying to compute the integral $I=\displaystyle\int_{0}^{\infty}dx\:\frac{\exp(-ax^2+bx)}{x+1}$ where $a>0$ and $b$ is a complex number. For the special case of $b=-2a$, we have $I=-\displaystyle\frac{1}{2}\exp(a)\:\text{Ei}(-a)$ (where $\text{Ei}$ is the Exponential Integral function). However, for the general complex values of $b$ I am not sure how to compute this integral in terms of some known function.
| Hint:
$I(b)=\int_0^\infty\dfrac{e^{-ax^2+bx}}{x+1}dx$
$\dfrac{dI(b)}{db}=\int_0^\infty\dfrac{xe^{-ax^2+bx}}{x+1}dx$
$\therefore\dfrac{dI(b)}{db}+I(b)=\int_0^\infty e^{-ax^2+bx}~dx$
$=\int_0^\infty e^{-a\left(x^2+\frac{bx}{a}\right)}~dx$
$=\int_0^\infty e^{-a\left(x^2+\frac{bx}{a}+\frac{b^2}{4a^2}-\frac{b^2}{4a^2}\right)}~dx$
$=e^\frac{b^2}{4a}\int_0^\infty e^{-a\left(x+\frac{b}{2a}\right)^2}~dx$
$=e^\frac{b^2}{4a}\int_\frac{b}{2a}^\infty e^{-ax^2}~dx$
$=e^\frac{b^2}{4a}\left(\int_0^\infty e^{-ax^2}~dx-\int_0^\frac{b}{2a}e^{-ax^2}~dx\right)$
$=e^\frac{b^2}{4a}\left(\dfrac{1}{2}\sqrt{\dfrac{\pi}{a}}-\int_0^\frac{b}{2\sqrt a}e^{-x^2}~d\left(\dfrac{x}{\sqrt a}\right)\right)$
$=e^\frac{b^2}{4a}\left(\dfrac{1}{2}\sqrt{\dfrac{\pi}{a}}-\dfrac{1}{\sqrt a}\int_0^\frac{b}{2\sqrt a}e^{-x^2}~dx\right)$
$=\dfrac{1}{2}\sqrt{\dfrac{\pi}{a}}e^\frac{b^2}{4a}-\sum\limits_{n=0}^\infty\dfrac{b^{2n+1}n!}{2a^{n+1}(2n+1)!}$ (according to http://en.wikipedia.org/wiki/Dawson_function)
$e^bI(b)=\dfrac{1}{2}\sqrt{\dfrac{\pi}{a}}\int e^{\frac{b^2}{4a}+b}~db-\int\sum\limits_{n=0}^\infty\dfrac{b^{2n+1}e^bn!}{2a^{n+1}(2n+1)!}db$
$=\dfrac{1}{2}\sqrt{\dfrac{\pi}{a}}\int e^{\frac{1}{4a}(b^2+4ab)}~db-\int\sum\limits_{n=0}^\infty\dfrac{b^{2n+1}e^bn!}{2a^{n+1}(2n+1)!}db$
$=\dfrac{1}{2}\sqrt{\dfrac{\pi}{a}}\int e^{\frac{1}{4a}(b^2+4ab+4a^2-4a^2)}~db-\int\sum\limits_{n=0}^\infty\dfrac{b^{2n+1}e^bn!}{2a^{n+1}(2n+1)!}db$
$=\dfrac{1}{2}\sqrt{\dfrac{\pi}{a}}e^{-a}\int e^\frac{(b+2a)^2}{4a}~db-\int\sum\limits_{n=0}^\infty\dfrac{b^{2n+1}e^bn!}{2a^{n+1}(2n+1)!}db$
$=\int\sum\limits_{n=0}^\infty\dfrac{\sqrt\pi e^{-a}(b+2a)^{2n}}{2^{2n+1}a^{n+\frac{1}{2}}n!}db-\int\sum\limits_{n=0}^\infty\dfrac{b^{2n+1}e^bn!}{2a^{n+1}(2n+1)!}db$
$=\sum\limits_{n=0}^\infty\dfrac{\sqrt\pi e^{-a}(b+2a)^{2n+1}}{2^{2n+1}a^{n+\frac{1}{2}}n!(2n+1)}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{2n+1}\dfrac{(-1)^kb^ke^bn!}{2a^{n+1}k!}+C$ (can be obtained from http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions)
$I(b)=\sum\limits_{n=0}^\infty\dfrac{\sqrt\pi e^{-a-b}(b+2a)^{2n+1}}{2^{2n+1}a^{n+\frac{1}{2}}n!(2n+1)}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{2n+1}\dfrac{(-1)^kb^kn!}{2a^{n+1}k!}+Ce^{-b}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Taylor expansion of $\frac{(-1)^n}{\ln n(1+\frac{1}{n\ln n}+o(\frac{1}{n\ln n})}$ I can't get the right terms:
$$\frac{(-1)^n}{\ln n + \frac{(-1)^n}n + o(\frac1n)}=\frac{(-1)^n}{\ln n} - \frac1{n\ln^2 n}+o\left(\frac1{n\ln^2 n}\right)$$
My thoughts
$$\frac{(-1)^n}{\ln n(1+\frac{1}{n\ln n}+o(\frac{1}{n\ln n})}$$
note that
\begin{align*}
\dfrac{1}{1+x}&=1-x+x^2+o(x^2) \\
&=1-x+x^2+O(x^3)
\end{align*}
we work with o with $x=\frac{1}{n\ln n}+o(\frac{1}{n\ln n})$
then $(1+\frac{1}{n\ln n}+o(\frac{1}{n\ln n}))=1-\frac{1}{n\ln n}+o(\frac{1}{n\ln n})+(\frac{1}{n\ln n}+O(\frac{1}{n\ln n}))^2+o((\frac{1}{n\ln n}+O(\frac{1}{n\ln n})))^2$
finaly
$$\frac{(-1)^n}{\ln n(1+\frac{1}{n\ln n}+o(\frac{1}{n\ln n})}=\frac{ (-1)^n }{ ln(n)( 1-\frac{1}{n\ln n}+o(\frac{1}{n\ln n})+(\frac{1}{n\ln n}+o(\frac{1}{n\ln n}))^2+o((\frac{1}{n\ln n}+o(\frac{1}{n\ln n})))^2 ) ) }=$$
but i'm stuck
| We have $$\frac{(-1)^n}{\ln n + \frac{(-1)^n}{n} + o(\frac{1}{n})} = \frac{(-1)^n}{\ln n} \cdot \frac{1}{1 + \frac{(-1)^n}{n\ln n} + o(\frac{1}{n\ln n})} = \frac{(-1)^n}{\ln n}\left(1 - \frac{(-1)^n}{n\ln n} + o\left(\frac{1}{n\ln n}\right)\right),$$
which multiplies out to $$\frac{(-1)^n}{\ln n} - \frac{1}{n\ln^2 n} + o\left(\frac{1}{n\ln^2 n}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1130744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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I need the proving of $x\log(x)=(\frac{x-1}{x})+\frac{3}{2!}(\frac{x-1}{x})^2+\frac{11}{3!}(\frac{x-1}{x})^3+...\frac{S_{n}}{n!}(\frac{x-1}{x})^n$ I could find the role of Striling Numbers in the natural logarithm function as follows
$$x\log(x)=(\frac{x-1}{x})+\frac{3}{2!}(\frac{x-1}{x})^2+\frac{11}{3!}(\frac{x-1}{x})^3+...\frac{S_{n}}{n!}(\frac{x-1}{x})^n$$
Where $S_n$= absolute Striling Numbers of first kind (0,1,3,11,50,274....)
this series numerically checked without any problem, but I need the proving. Any help?
| Suppose we seek to show that
$$x\log x = \sum_{n=1}^\infty \frac{1}{n!} \left[n+1\atop 2\right]
\left(\frac{x-1}{x}\right)^n.$$
Recall the species equation for decompositions into disjoint cycles:
$$\mathfrak{P}(\mathcal{U}\mathfrak{C}(\mathcal{Z}))$$
which gives the generating function
$$G(z, u) = \exp\left(u\log\frac{1}{1-z}\right)$$
so that
$$\left[n+1\atop 2\right]
= \frac{1}{2} (n+1)! [z^{n+1}] \left(\log\frac{1}{1-z}\right)^2.$$
Substitute this into the sum to get
$$\frac{1}{2} \sum_{n=1}^\infty \frac{1}{n!}
(n+1)! \left(\frac{x-1}{x}\right)^n
[z^{n+1}] \left(\log\frac{1}{1-z}\right)^2
\\ = \frac{1}{2} \sum_{n=1}^\infty
(n+1) \left(\frac{x-1}{x}\right)^n
[z^{n+1}] \left(\log\frac{1}{1-z}\right)^2.$$
Now $\left(\log\frac{1}{1-z}\right)^2$ starts at $z^2$ so that we may
include the value for $n=0$ in the sum (zero contribution) to get
$$\frac{1}{2} \sum_{n=0}^\infty
(n+1) \left(\frac{x-1}{x}\right)^n
[z^{n+1}] \left(\log\frac{1}{1-z}\right)^2.$$
What we have here is an annihilated coefficient extractor that
evaluates to
$$\frac{1}{2}
\left. \frac{d}{dz} \left(\log\frac{1}{1-z}\right)^2
\right|_{z=(x-1)/x}.$$
This is
$$\left. \frac{1}{1-z} \log\frac{1}{1-z}
\right|_{z=(x-1)/x}$$
which finally yields
$$\frac{1}{1-(x-1)/x} \log\frac{1}{1-(x-1)/x}
= \frac{x}{x-(x-1)} \log\frac{x}{x-(x-1)}
\\ = x \log x.$$
There is another annihilated coefficient extractor at this
MSE link I and another one at this
MSE link II and at this MSE link III.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Does $\int_0^\infty \frac{\cos x}{1+x}$ absolutly converge?
Does the following indefinite integral converge?
$\int_0^\infty \frac{\cos x}{1+x}$
converges, absolutely converges?
i can say that by the Dirichlet test it does converge.
i am trying to prove it diverges absolutely (seems close to $\frac{1}{x}$)
unsuccessfully so far
how do i prove /disprove it absolutely converge?
| Consider intervals $[2k\pi-\frac{\pi}{4},2k\pi+\frac{\pi}{4}]$ for $k \in \mathbb{N}$, then:
$$\int_{0}^{\infty}\frac{|\cos(x)|}{1+x}dx>\int_{\bigcup_{k \in \mathbb{N}}[2k\pi-\frac{\pi}{4},2k\pi+\frac{\pi}{4}]}\frac{|\cos(x)|}{1+x}dx=\sum_{k=1}^{\infty}\int_{2k\pi-\frac{\pi}{4}}^{2k\pi+\frac{\pi}{4}}\frac{|\cos(x)|}{1+x}dx$$
But on each interval $[2k\pi-\frac{\pi}{4},2k\pi+\frac{\pi}{4}]$ we have $\cos(x)>\frac{1}{2}$, so:
$$\sum_{k=1}^{\infty}\int_{2k\pi-\frac{\pi}{4}}^{2k\pi+\frac{\pi}{4}}\frac{|\cos(x)|}{1+x}dx>\sum_{k=1}^{\infty}\int_{2k\pi-\frac{\pi}{4}}^{2k\pi+\frac{\pi}{4}}\frac{1}{2(1+x)}dx$$
Now we have:
$$\int_{2k\pi-\frac{\pi}{4}}^{2k\pi+\frac{\pi}{4}}\frac{1}{2(1+x)}dx>\int_{2k\pi-\frac{\pi}{4}}^{2k\pi+\frac{\pi}{4}}\frac{1}{2(1+2k\pi+\frac{\pi}{4})}dx=\frac{\pi}{2}\frac{1}{2(1+2k\pi+\frac{\pi}{4})}$$
So:
$$\sum_{k=1}^{\infty}\int_{2k\pi-\frac{\pi}{4}}^{2k\pi+\frac{\pi}{4}}\frac{1}{2(1+x)}dx>\sum_{k=1}^{\infty}\frac{\pi}{2}\frac{1}{2(1+2k\pi+\frac{\pi}{4})}=\infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1132226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Find coefficient of x in a generating function The problem is as follows:
$\text{Determine the coef. of } x^{10} \text{ in } (x^3 + x^5 + x^6)(x^4 + x^5 + x^7)(1+x^5+x^{10}+x^{15}+...)$
I factored out some $x$'s, to get $x^3(1+x^2+x^3)x^4(1+x+x^3)(1+x^5+x^{10}+x^{15}+...)$and then combined the factored terms to get $x^7(1+x^2+x^4)(1+x+x^3)(1+x^5+x^{10}+x^{15}+...)$
Now I don't know what to do; usually it ends up factoring to $(1+x+x^2+...)$, but that doesn't appear to be the case here.
| Hint :
coefficient of $x^3$ in $(1+x^2+x^3)(1+x+x^4)(1+x^5+x^{10}+x^{15}+...)$ is $2$, because
$x^3=x^3\times 1 \times 1, x^3= x^2\times x \times 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1132304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For positive operators $A$ and $B$ with $A^6=B^6$ show that $A=B$ Since $A$ and $B$ are positive, I managed to show that $A^6$ and $B^6$ are positive.
Now, I can use the fact that there exists a unique square root of both of those and since they're equal, their roots must be equal, so $A^3=B^3$. But what now?
I'm guessing I need to show that $A^2=B^2$ somehow since then I can say
$A^3=B^3\Rightarrow AA^2=BB^2 \Rightarrow A=B$
How can I do that?
| Hint: Subtracting $A^2B$ on both sides leads to $A^2(A-B) = (B^2-A^2)B$. When subtracting $B^2A$ on both sides one obtains $(A^2-B^2)A = B^2(B-A)$. Since $A,B$ are arbitrary, $B-A$ vanishes if and only if $A^2-B^2 = 0$. Then apply the roots and you are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1135430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Having trouble solving part two of an equation: Part one consisted of proving that
$$\frac{x-1}{x-3} = 1+ \frac{2}{x-3}$$
I completed this and here is my working:
$$let \frac{x-1}{x-3}= LHS$$
$$RHS=\frac{x-3}{x-3} + \frac{2}{x-3}$$
$$\frac{2+(x-3)}{(x-3)}$$
$$\frac {x-3+2}{x-3}$$
$$\frac {x-1}{x-3} = LHS$$
Part two then asks me to hence solve:
$$\frac{x-1}{x-3}-\frac{x-3}{x-5} = \frac{x-5}{x-7}−\frac{x-7}{x-9}$$
I have attempted to sub in part one into part two but my answers always turn out to be incorrect.
Any tips in the matter would be much appreciated.
I'd like to thank anyone who comments and answers in advance, your help is much appreciated.
| The hence suggests:
Let $x=x-2$ the first equation becomes
$$
\frac{(x-2)-1}{(x-2)-3} = 1+\frac{2}{(x-2)-3} \\
\frac{x-3}{x-5}=1+\frac{2}{x-5}
$$
This process can be done with the other fractions in part 2. Notice that this will be of benefit because the $1$ will cancel because of the subtraction.
$$
1-\frac{2}{x-3}-\left(1-\frac{2}{x-5}\right)=1-\frac{2}{x-7}-\left(1-\frac{2}{x-9}\right)\\
-\frac{2}{x-3}+\frac{2}{x-5}=-\frac{2}{x-7}+\frac{2}{x-9}$$
Cancel the $2$ and simplify each side $$ \frac{-(x-5)+x-3}{(x-3)(x-5)} =\frac{-(x-9)+(x-7)}{(x-7)(x-9)}$$
Simplify and note the numerators are equal. The fractions can therefore only be equivalent if the denominators are equal. Note that $x\neq7$ for example. $$ (x-3)(x-5)=(x-7)(x-9) \\
x^2-8x+15=x^2-16x+63 \\
8x=48\\
x=6
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1137364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Having problems finding $x$ in terms of $a$ and $b$. I have attempted this question but have not found a solution. I am currently stuck. Hints on how I may go further would be helpful. Thank You in advance.
The Question:
$$\frac{a^2b}{x^2} + \left(1+\frac{b}{x}\right)a = 2b+ \frac{a^2}{x}$$
What I have done so far that I believe is correct:
$$\frac{a^2b}{x^2} + \frac{ab}{x} + a = 2b+ \frac{a^2}{x}$$
If you would like to see other work that I have attempted on this question I can also place it up.
| Bring $a\left(1+\frac{b}{x}\right)+\frac{a^2b}{x^2}$ together using the common denominator $x^2$. Bring $2b+\frac{a^2}{x}$ together using the common denominator $x$:
$$
\frac{a^2b+abx+ax^2}{x^2}=\frac{a^2+2bx}{x}.
$$
Now cross multiply:
$$
x(a^2b+abx+ax^2)=x^2(a^2+2bx).
$$
Then expand out the terms of the left- and right-hand sides:
$$
a^2bx+abx^2+ax^3=a^2x^2+2bx^3.
$$
Now subtract $a^2x^2+2bx^3$ from both sides to get
$$
a^2bx-a^2x^2+abx^2+ax^3-2bx^3=0.
$$
Collect everything in terms of $x$:
$$
a^2bx+x^2(ab-a^2)+x^3(a-2b)=0.
$$
Now notice that the left-hand side factors into a product with three terms:
$$
x(a-x)(ab-ax+2bx)=0.
$$
Split this into three equations:
$$
a-x=0\quad\text{or}\quad x=0\quad\text{or}\quad ab-ax+2bx=0.
$$
Simple manipulations to these three equations shows that
$$
x=a\quad\text{or}\quad x=0\quad\text{or}\quad x=-\frac{ab}{2b-a}.
$$
By the original expression, we know that $x\neq 0$; thus, the final answer is
$$
x=a\quad\text{or}\quad x=-\frac{ab}{2b-a}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to integrate $\frac{1}{x\sqrt{1+x^2}}$ using substitution? How you integrate
$$\frac{1}{x\sqrt{1+x^2}}$$
using following substitution? $u=\sqrt{1+x^2} \implies du=\dfrac{x}{\sqrt{1+x^2}}\, dx$
And now I don't know how to proceed using substitution rule.
| Maybe another substitution is interesting:
For $x>0, v=\frac{1}{x}$
$$
\int \frac 1 {x\sqrt{1+x^2}} \,dx = \int \frac {1} {x^2\sqrt{\frac{1}{x^2}+1}} \,dx = -\int\frac{dv}{\sqrt{v^2+1}} = - \ln(v+\sqrt{v^2+1})=$$
$$ =- \ln\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}+1}\right) + C.
$$
For $x<0, v=\frac{1}{x}$
$$
\int \frac 1 {x\sqrt{1+x^2}} \,dx = -\int \frac {1} {x^2\sqrt{\frac{1}{x^2}+1}} \,dx = \int\frac{dv}{\sqrt{v^2+1}} = \ln(v+\sqrt{v^2+1})+ C=$$
$$ = \ln\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}+1}\right) + C.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1137842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 4
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finding the max of $f(x)=\sqrt{(x^2-4)^2+(x-5)^2}-\sqrt{(x^2-2)^2+(x-1)^2}$ I need to find the max of $$f(x)=\sqrt{(x^2-4)^2+(x-5)^2}-\sqrt{(x^2-2)^2+(x-1)^2}$$
When $x$ is a real number.
What i did is to simplify: $$f(x)=\sqrt{x^4-7x^2-10x+41}-\sqrt{x^4-3x^2-2x+5}$$.
Then i compute: $$f'(x)=\frac{-5-7x+2x^3}{\sqrt{41-10 x-7 x^2+x^4}}+\frac{1+3x-2x^3}{\sqrt{5-2 x-3 x^2+x^4}}$$.
But failed to solve $f'(x)=0$ for finding $f(x)_{max}$.
I would be glad for your help.
Thanks.
| You have $\frac{-5-7x+2x^3}{\sqrt{41-10x-7x^3+x^4}}+\frac{1+3x-2x^3}{\sqrt{5-2x-3x^2+x^4}}=0$ so
$\frac{-5-7x+2x^3}{\sqrt{41-10x-7x^3+x^4}}=-\frac{1+3x-2x^3}{\sqrt{5-2x-3x^2+x^4}}$ then
$\frac{(-5-7x+2x^3)^2}{41-10x-7x^3+x^4}=\frac{(1+3x-2x^3)^2}{5-2x-3x^2+x^4}$ and finally $(5-2x-3x^2+x^4)(-5-7x+2x^3)^2=(41-10x-7x^3+x^4)(1+3x-2x^3)^2$
Its a large polynomial but you have to find the 10 roots (removing the strange roost that made $41-10x-7x^3+x^4<0$ or $5-2x-3x^2+x^4<0$) and check which of them is the max or if the max is in $-\infty$ or $\infty$. It will be a long way if you want to made it by your own...
Well you can remove $(x+1)^2$ by common factor in both sides so $x=-1$ are 2 roots and remain 8:
$(2x^2-2x-5)^2(x^4-3x^2-2x+5)=(2x^2-2x-1)^2(x^4-7x^3-10x+41)$
And by expanding you can get
$28x^7-84x^6+72x^5-108x^4+267x^3-155x^2-104x+84=0$ and dividing by $(x-1)$ (and therefore $x=1$ is another root)
$28x^6-56x^5+16x^4-92x^3+172x^2+20x-81=0$
only 6 roots remain and you can use methods to approach them to $x\approx-0.614946$ and $x\approx0.886814$
| {
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"url": "https://math.stackexchange.com/questions/1141449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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When is $2^x+3^y$ a perfect square?
If $x$ and $y$ are positive integers, then when is $2^x+3^y$ a perfect square?
I tried this question a lot but failed. I tried dividing cases into when $x,y$ are even/odd, but still have no idea what to do when they are both odd. I tried looking into when is $2^x+3^y$ is of the form $6k+1$ and so on .. but couldn't succeed. Can anyone help?
| If $x=1$, $2+3^y$ cannot be a square since it is $\equiv 2\pmod{3}$.
This gives $x\geq 2$, so $2^x+3^y\equiv (-1)^y\pmod{4}$ and $y$ must be even. Since in this case $3^y\equiv 1\pmod{8}$, we must have $x\geq 3$ and
$$ 2^x + 3^{2z} = A^2 $$
with $A$ odd, or:
$$ 2^x = (A-3^z)(A+3^z) \tag{1} $$
so both $A-3^x$ and $A+3^x$ must be powers of two, then:
$$ 2\cdot 3^z = 2^D-2^C\tag{2} $$
with $D>C$ and $C\geq 1$, since otherwise the RHS of $(2)$ is odd. But if $C>1$ then the RHS of $(2)$ is a multiple of four while the LHS is not, so $C=1$ and we are left with:
$$ 3^z = 2^E -1 \tag{3} $$
where $E$ must be even in order that the RHS is a multiple of three. But in such a case the only solution of:
$$ 3^z = (2^F-1)(2^F+1) \tag{4} $$
is given by $z=F=1$, since otherwise $2^F-1$ and $2^F+1$ cannot be both powers of three.
This gives that
$$ 2^4+3^2 = 5^2 $$
is the only solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof by induction and combinations I think I am stuck on this, I am not sure if I'm going down the correct path or not. I am trying to algebraically manipulate $p(k+1)$ so I can use $p(k)$ but I am unable to do so, so I am not sure if my math is bad or I am going about it the incorrect way.
15b.) Prove by induction that $(nCk) = \dfrac{(nC k - 1) \cdot (n - k +1)}{k}$
Starting step: prove $p(1)$
$$\begin{align}
p(1) & = {n \choose 1} = {n \choose 0} \cdot \frac{n + 0}{1} \\
& = \frac{n!}{1!(n-1)!} = \frac{n!}{0! (n!) \cdot (n)} \\
& = \left[\frac{n!}{1 \cdot (n-1)!}\right] = \left[\frac{n!}{1 \cdot (n)!}\right] \cdot (n) \\
& = \frac{n!}{(n-1)!} = \frac{n!}{n!} \cdot (n) \\
& = \frac{n!}{(n-1)!} = 1n \\
& = n = n
\end{align}$$
$p(k)$ is true: $$p(k) = \frac{n!}{[k! (n-k)!]} = \frac{n!}{[(k-1)!(n - k + 1)] \cdot [(n - k + 1) / k]}$$
Inductive step: show $p(k) \implies p(k+1)$
$$\begin{align}
p(k+1) & = {n \choose k + 1} = {n \choose k} \cdot \frac{n - k}{k+1} \\
& = \frac{n!}{(k+1)!(n - k + 1)!} = \frac{n!}{k! (n - k)!} \cdot \frac{n - k}{k+1} \\
& = \frac{n!}{(k+1)(k!)(n - k + 1)!} = \frac{n!}{k! (n - k)!} \cdot \frac{n - k}{k+1} \cdot k! (n - k)! \\
& = \frac{n! \cdot [k! (n - k)!]}{(k+1)(k!)(n - k + 1)!} = n! \cdot \frac{n - k}{k+1} \\
& = n! \cdot \frac{(n - k)!}{(k+1)(n - k + 1)!} = n! \cdot \frac{n - k}{k+1} \cdot (k+1) \\
& = n! \cdot (k+1) \cdot \frac{(n - k)!}{(k+1)(n - k + 1)!} = n! \cdot \frac{n - k}{k+1} \cdot (k+1) \\
& = n! \cdot \frac{(n - k)!}{(n - k + 1)!} = n! \cdot (n - k)
\end{align}$$
| I don't see how you can use induction here. Here are two direct proofs.
The first proof is just a calculation:
$$
\begin{align*}
\frac{n-k+1}{k} \binom{n}{k-1} &= \frac{n-k+1}{k} \frac{n!}{(k-1)!(n-k+1)!} \\ &=
\frac{n-k+1}{(n-k+1)!} \frac{1}{k(k-1)!} n! \\ &=
\frac{1}{(n-k)!} \frac{1}{k!} n! \\ &=
\frac{n!}{(n-k)!k!} \\ &= \binom{n}{k}.
\end{align*}
$$
The second proof is combinatorial. We will prove that
$$
(n-k+1) \binom{n}{k-1} = k \binom{n}{k}.
$$
On the left-hand side we have the number of possibilities of partitioning $\{1,\ldots,n\}$ into two sets $A,B$ of sizes $k-1,n-k+1$, and choosing an element $x \in B$. On the right-hand side we have the number of possibilities of partitioning $\{1,\ldots,n\}$ into two sets $C,D$ of sizes $k,n-k$, and choosing an element $y \in C$. The mapping
$$(A,B,x) \mapsto (A\cup\{x\},B\setminus\{x\},x)$$
maps a solution for the first problem to a solution for the second, bijectively.
| {
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Proof by induction; inequality $1\cdot3+2\cdot4+3\cdot5+\dots+n(n+2) \ge \frac{n^3+5n}3$ Ok so I'm kind of struggling with this:
The question is:
"Use mathematical induction to prove that
1*3 + 2*4 + 3*5 + ··· + n(n + 2) ≥ (1/3)(n^3 + 5n) for n≥1"
Okay, so P(1) is true as 1(1+2)=3 and (1/3)(1^3 + 5)=2
Assuming P(k) is true for k≥1 gives:
1*3 + 2*4 + 3*5 + ... + k(k+2) ≥ (1/3)(k^3 + 5k)
And we want to show that P(k+1) is true, that is;
1*3 + 2*4 + 3*5 + ... + (k+1)(k+3) ≥ (1/3)((k+1)^3 + 5(k+1)) (inequality 1)
This is where I'm not sure if I'm thinking along the right lines or not. Surely, by P(k) and the rules for inequalities we get:
(1*3 + 2*4 + 3*5 + ... + k(k+2)) + (k+1)(k+3) ≥ (1/3)(k^3 + 5k) + (k+1)(k+3) (inequality 2)
but the right hand side in this inequality does not equal (1/3)((k+1)^3 + 5(k+1)) which is what I am trying to show. There's obviously a flaw in my thinking somewhere so any help would be greatly appreciated.
Thanks
Edit: As far as I understand it the right hand side of inequality 1 should equal the right hand side of inequality 2 but this isn't the case when expanded
| Assume that statement is true for number unto k. We need to show that it is true for k+1.
$$
1*3+...+k(k+2)+(k+1)(k+3)\geq \frac{k^3+5k}{3}+(k+1)(k+3)=\frac{k^3+3k^2+17k+9}{3}\geq\frac{k^3+3k^2+8k+6}{3}=\frac{(k+1)^3+5(k+1)}{3}
$$
for last inequality expand both sides and see that
$$
\frac{(k+1)^3+5(k+1)}{3}=\frac{k^3+3k^2+8k+6}{3}
$$
and
$$
\frac{k^3+5k}{3}+(k+1)(k+3)=\frac{k^3+3k^2+17k+9}{3}
$$
put it in original inequation and get the result
| {
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Proof that the $\lim\limits_{x \to 2}\dfrac{1}{x} = \dfrac{1}{2}$ using the $\epsilon$-$\delta$ definition of limits (verification). Prove that the $\lim\limits_{x \to 2}\dfrac{1}{x} = \dfrac{1}{2}$ using the $\epsilon-\delta$ definition of limits.
$$
\\ \begin{align}
\\ &\textrm{Let } \forall \epsilon > 0
\\ &\textrm{Choose } \delta = \min{\{1, 2\epsilon \}}
\\ &\textrm{Assume } 0 < |x - 2| < \delta :
\\ \end{align}
$$
$$
\\ \begin{align}
\\ \left|\frac{1}{x} - \frac{1}{2}\right| &< \epsilon
\\ \frac{|2 - x|}{|2x|} &< \epsilon
\\ |-1(x - 2)| &< \epsilon|2x|
\\ |x - 2| &< \epsilon|2x|
\\ \end{align}
$$
$$
\\ \begin{align}
\\ |x - 2| &< 1
\\ -1 < x - 2 &< 1
\\ 1 < x &< 3
\\ \end{align}
$$
$$
\\ \begin{align}
\\ |x - 2| &< \epsilon|2(1)|
\\ |x - 2| &< 2\epsilon
\\ \end{align}
$$
$$
\\\therefore \delta \leq 2\epsilon
$$
Right, so I start by taking $|f(x)−L|<ϵ$. I then isolate $|x−2|$ to the left. I then limit $|x−2|$ to be less than one and then find a range of $x$ values which satisfy the inequality. Then I plug in the smallest $x$ value to minimise the value of $ϵ$. I make the conclusion that $δ≤2ϵ$. Am I excluding or misplacing steps? I'm fairly new to this whole thing.
| Here you have to show that for each $ \epsilon >0$, there exists $ \delta >0 $ such that for each $ x\in Domn(\frac{1}{x}) $ if $ 0<|x-2|<\delta $ then $ \left | \frac{1}{x}-\frac{1}{2}\right |<\epsilon $.
So begin with arbitrary $ \epsilon >0 $.
Notice that if $ 0<|x-2|<1 $ then $ 1<|x|<3 $ and hence $\frac{1}{3}<\frac{1}{|x|}<1$.
Now choose $ \delta =\min\{1,2\epsilon\} $. Then clearly $ \delta >0 $.
Now suppose $ 0<|x-2|<\delta $.
Then $ \left | \frac{1}{x}-\frac{1}{2}\right |=\frac{|x-2|}{2|x|}<\frac{|x-2|}{2}<\frac{2\epsilon}{2}=\epsilon $.
Therefore $$ \lim_{x\rightarrow 2}\frac{1}{x}=\frac{1}{2} .$$
| {
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"url": "https://math.stackexchange.com/questions/1147648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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inequalites of an acute triangle angles $ 180^{180}*a^b*b^c*c^a \le (a^2+b^2+c^2)^{180} $ If $a,b,c$ are an acute angle of triangle the prove that
$ 180^{180}*a^b*b^c*c^a \le (a^2+b^2+c^2)^{180} $
No idea
| To solve this you should know (this fact)[https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means]
Hint:
$180 = a + b + c$
Then we can use the fact that:
$\frac{a\cdot b + b\cdot c + c\cdot a}{a+b+c} \ge (a+b+c)(a^b\cdot b^c \cdot c^a)^{\frac{1}{a+b+c}} $
After you understood how to use all this facts you will have the inequality:
$\frac{a\cdot b + b\cdot c + a\cdot c}{a+b+c} \le a^2 + b^2 + c^2$
That much more easy to prove.
One more fact that might be helpful:
$\frac{a+b}{2} \le (\frac{a^2 + b^2}{2})^{\frac{1}{2}}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Two inequalities in a triangle I'm trying to prove that in a triangle with side lengths $a,b,c$, median lengths $m_a, m_b, m_c$ and circumdiameter $D$ the following inequality holds:
$$
\frac{a^2+b^2}{m_c}+\frac{b^2+c^2}{m_a}+\frac{c^2+a^2}{m_b}\le 6D
$$
In order to prove it, I tried to use $m_a\ge h_a$ etc. which was when I discovered that the following inequality should be true as well:
$$
\frac{a^2+b^2}{h_c}+\frac{b^2+c^2}{h_a}+\frac{c^2+a^2}{h_b}\ge 6D
$$
Where $h_a, h_b, h_c$ are the lengths of the hights of the triangle. My question is: how can we prove these inequalities?
| For the second one :
$ \frac{a^2+b^2}{h_c} = \frac{a^2+b^2}{\frac{2S}{c} } = \frac{c(a^2+b^2)}{2S} \geq \frac{2abc}{2S}= \frac{abc}{S} =4R$ because $a^2+b^2 \geq 2ab$ and $4R=\frac{abc}{S} $
analogously you prove for other 2 terms so we get :
$\sum_{cyc} \frac{a^2+b^2}{h_c} \geq 3 \cdot 4R = 12R = 6D $
For the other one:
Let $AD$ be median and it intersects circumcenter at $D'$.
Applying power of a point on point $D$ we get $AD \cdot DD' = BD \cdot DC \Rightarrow DD'= \frac{a^2}{4m_a}$ because $BD=DC=\frac{a}{2}$
Since $AD' \leq 2R \Rightarrow AD+DD' \leq 2R$ we get
$m_a+\frac{a^2}{4m_a} \leq 2R \Rightarrow \frac{4m_a^2+a^2}{4m_a} \leq 2R$ and putting $m_a^2=\frac{2b^2+2c^2-a^2}{4}$ we get $\frac{b^2+c^2}{m_a}\leq 4R $
And finally $\sum_{cyc} \frac{b^2+c^2}{m_a} \leq 3 \cdot 4R =12R = 6D $
| {
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"timestamp": "2023-03-29T00:00:00",
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Strong induction with Fibonacci numbers I have two equations that I have been trying to prove. The first of which is:F(n + 3) = 2F(n + 1) + F(n) for n ≥ 1.For this equation the answer is in the back of my book and the proof is as follows:1) n = 1: F(4) = 2F(2) + F(1) or 3 = 2(1) + 1, true.2) n = 2: F(5) = 2F(3) + F(2) or 5 = 2(2) + 1, true.3) Assume for all r, 1 ≤ r ≤ k: F(r + 3) = 2F(r + 1) + F(r)4) Then F(k + 4) = F(k + 2) + F(k + 3) = 5) 2F(k) + F(k - 1) + 2F(k + 1) + F(k) = 6) 2[F(k) + F(k + 1)] + [F(k - 1) + F(k)] = 7) 2F(k + 2) + F(k + 1)My first question here is how do I know how many values of n to test for? Here they chose two.My next question is how did they get from line 3 to line 4? I understand how the statement is correct but why is this chosen? I also understand that I need to prove it's true for all values of r because if I do that it implies that it is true for k + 1. Is it just to find a relation to F(r + 3) on line 3? If that was the case why not just have F(k + 3) = F(k + 2) + F(k + 1)?My final question about this is how did they get from line 4 to 5?The second equation I want to prove is:F(n + 6) = 4F(n + 3) + F(n) for n ≥ 1I'm able to prove n = 1 and n = 2 is true but I get stuck on going from what would be line 3 - 4 on this problem. As this is my problem for homework the answer is not in the back of the book.Now that I've gotten the help I just want to update this with the proof for my second equation (I haven't gotten the formatting down yet so bear with me):F(n + 6) = 4F(n + 3) + F(n)1) n = 1: F(7) = 4F(4) + F(1) or 13 = 12 + 1, true.2) n = 2: F(8) = 4F(5) + F(2) or 21 = 20 + 1, true.3) Assume for all r, 1 ≤ r ≤ k: F(r + 6) = 4F(r + 3) + F(r)4) Then F(k + 7) = 4F(k + 4) + F(k + 1) =5) F(k + 4) + F(k + 4) + F(k + 4) + F(k + 4) + F(k + 1) =6) F(k + 4) + F(k + 4) + F(k + 4) + F(k + 3) + F(k + 2) F(k + 1) =7) F(k + 4) + F(k + 4) + F(k + 4) +F(k + 3) + F(k + 3) =8) F(k + 5) + F(k + 5) + F(k + 4) =9) F(k + 6) + F(k + 5) =10) F(k + 7)
| They assume that $f(k+3) = 2f(k+1) + f(k)$, then consider $f((k+1)+3) = f(k+4)$.
Setting aside our hypothesis for a moment, we know by definition that
$$f((k+1) + 3) = f(k+4) = f(k+3) + f(k+2) = f((k+1)+2) + f((k+1)+1)$$
Then, substituting what we know about $f(k+3)$:
$$f((k+1) + 3) = f((k+1) + 1) + 2f(k+1) + f(k)$$
$$=f((k+1)+1) + [f(k+1) + f(k)] + f(k+1)$$
$$= 2f((k+1)+1) + f(k+1) = 2f(k+2) + f(k+1)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that f is unique, and express f(x) in closed form.
Let $f : [−1, 1] \to R$ be a continuous function such that
(i) $f(x) = \frac {2−x^2}{2}f(\frac {x^2}{2-x^2}) $ for every $x$ in [−1, 1],
(ii) $f(0) = 1$, and
(iii) $\lim_{x\to 1^{-}}\frac {f(x)}{\sqrt{1−x}}$ exists and is finite.
Prove that f is unique, and express f(x) in closed form.
Let $x=1$, we get $f(1)=0$. Let $\frac {x^2}{2-x^2}=1$, we have $x=\pm 1$, so let $x=-1$, we have $f(-1)=0$. It suffices to show that for any other function $g$ that satisfies the same conditions, $f(x)-g(x)=0$. We know that $f(x)-g(x)$ passes through the origin, thus we need to show that $f'(x)-g'(x)=0$. However, I don't know exactly what the third condition would imply. Anyone has a clue?
| The third condition is actually a hint.
It implies that $\sqrt{1-x}$ is a factor of $f(x)$,
and it is not difficult to figure out that
in order to fit the rest of conditions
the other factor should be $\sqrt{1+x}$.
Indeed, the closed form expression is:
\begin{align}
f(x)&=\sqrt{1-x^2}.
\end{align}
Check on conditions:
\begin{align}
\checkmark
(\mathrm{i})& f(x)=\frac{2-x^2}{2}
\sqrt{1-\left(\frac{x^2}{2-x^2}\right)^2}
=\frac{1}{2}\sqrt{(2-x^2)^2-x^4}
=\sqrt{1-x^2}.
\\
\checkmark
(\mathrm{ii})& f(0)=1.
\\
\checkmark
(\mathrm{iii})& \lim_{x\to 1^{-}}\frac {f(x)}{\sqrt{1-x}}=\sqrt{2}.
\end{align}
| {
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How to calculate sin(65) without a calculator. I know about the sum and difference formula but I can't think of two values which will be able to use for sin(65). Therefore, I come to the question: How to calculate sin(65) without a calculator.
| Let $i = \sqrt{-1}$. Then from this link, with the methodology explained here:
$$ \sin 65^\circ = -\left(-\frac12 + \frac i2 \sqrt{3}\right) \left( -\frac{1}{32} \sqrt{6} \left(1- \frac{\sqrt 3}{3}\right) + \frac{i}{32} \sqrt{-6 \left( 1 - \frac{\sqrt{3}}{3} \right)^2 + 16 }\right)^{1/3} -\left(-\frac12 - \frac i2 \sqrt{3}\right) \left( -\frac{1}{32} \sqrt{6} \left(1- \frac{\sqrt 3}{3}\right) + \frac{i}{32} \sqrt{-6 \left( 1 - \frac{\sqrt{3}}{3} \right)^2 + 16 }\right)^{1/3} $$
(Note that $65^\circ$ is not constructible, so we should not expect any expression involving a finite amount of additions, multiplications and square roots.)
| {
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Closer form for $\int_0^\infty\frac{(\arctan{x})^2\log^2({1+x^2})}{x^2}dx$ I Would like to know the value of this integral. $$\int_0^\infty\frac{(\arctan{x})^2\log^2({1+x^2})}{x^2}dx$$
I think
$$I=\frac{a}{b}(\pi^3\ln2)+\frac{c}{d}(\pi\ln^32)+\frac{e}{f}(\pi\ln^22)+\frac{g}{h}(\pi\ln2)+\frac{i}{j}(\pi^3)+\frac{k}{m}\zeta({3})??$$
Where a,b,c,d...are integers
Thanks.
| Let $\displaystyle\small\gamma=\lim_{R\to\infty}[-R,R]\cup Re^{i[0,\pi]}$. Observe that
$$\small\oint_\gamma\frac{\ln^4(1-iz)}{z^2}dz=\frac{1}{8}\int^\infty_0\frac{\ln^4(1+x^2)}{x^2}dx-3\int^\infty_0\frac{\ln^2(1+x^2)\arctan^2{x}}{x^2}dx+2\int^\infty_0\frac{\arctan^4{x}}{x^2}dx=0$$
since
*
*The integral over the arc vanishes as $\small\mathcal{O}\left(\dfrac{\ln^4{R}}{R}\right)$.
*The imaginary part is odd and vanishes over a symmetric interval.
*The singularity at $\small 0$ is removable.
Therefore,
\begin{align}
\small\int^\infty_0\frac{\ln^2(1+x^2)\arctan^2{x}}{x^2}dx
&\small=\frac{1}{24}\int^\infty_0\frac{\ln^4(1+x^2)}{x^2}dx+\frac{2}{3}\int^\infty_0\frac{\arctan^4{x}}{x^2}dx\\
&\small=\frac{1}{3}\int^\infty_0\frac{\ln^3(1+x^2)}{1+x^2}dx+\frac{2}{3}\int^\frac{\pi}{2}_0x^4\csc^2{x}\ dx\\
&\small=-\frac{8}{3}\int^\frac{\pi}{2}_0\ln^3(\cos{x})\ dx+\frac{8}{3}\int^\frac{\pi}{2}_0x^3\cot{x}\ dx\\
&\small=-\frac{1}{6}\frac{\partial^3\operatorname{B}}{\partial b^3}\left(\frac{1}{2},\frac{1}{2}\right)+\frac{16}{3}\sum^\infty_{n=1}\int^\frac{\pi}{2}_0x^3\sin(2nx)dx\\
&\small=2\pi\zeta(3)+\frac{\pi^3}{3}\ln{2}+\frac{4\pi}{3}\ln^3{2}+2\pi\sum^\infty_{n=1}\frac{(-1)^n}{n^3}+\frac{\pi^3}{3}\sum^\infty_{n=1}\frac{(-1)^{n-1}}{n}\\
&\small=2\pi\zeta(3)+\frac{\pi^3}{3}\ln{2}+\frac{4\pi}{3}\ln^3{2}+2\pi\left(-\frac{3}{4}\zeta(3)\right)+\frac{\pi^3}{3}\ln{2}\\
&\small=\frac{\pi}{2}\zeta(3)+\frac{2\pi^3}{3}\ln{2}+\frac{4\pi}{3}\ln^3{2}
\end{align}
| {
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How should you prove product rules by induction? For example:
$$\prod_{i=2}^n\left(1-\frac{1}{i^2}\right)=\frac{n+1}{2n}$$
For every $n$ greater than or equal to $2$
my approach for this was that I need to prove that:
$$ \left(1-\frac{1}{n^2}\right)\left(1-\frac{1}{(n+1)^2}\right)=\frac{n+1+1}{2(n+1)}$$
is this the right approach? Because when i try and work out the algebra i keep on hitting a wall.
\begin{align}
\left(1-\frac{1}{n^2}\right)\left(1-\frac{1}{(n+1)^2}\right)&=1-\frac 1{(1-n)^2}-\frac 1{n^2}-\frac 1{n^2(n+1)} \\
&=\frac{n^2}{(n+1)^2}-\frac 1{(n+1)^2} \\
&=\frac{n^2-1}{(n+1)^2}-\frac 1{n^2}+\frac 1{n^2(n+1)^2} \\
&=\frac{n^2(n^2-1)}{n^2(n+1)^2}-\frac{(n+1)^2}{n^2(n+1)^2} \\
&=\frac{n^2(n^2-1)-(n+1)^2}{n^2(n+1)^2}+\frac 1{n^2(n+1)^2} \\
&=\frac{n^2(n^2-1)-(n+1)^2+1}{n^2(n+1)^2}
\end{align}
| Hint $\ $ No ingenuity is required: by telescopy the proof reduces to this one-line calculation
$\qquad\ \ $ if $\rm\, \ f(k) = \dfrac{k\!+\!1}{2k\ }\ $ then $\rm\,\ \dfrac{f(k)}{f(k\!-\!1)} =\, \dfrac{k\!+\!1}{2k\ }\dfrac{2(k\!-\!1)}{k\, }\,=\,\dfrac{k^2\!-\!1}{k^2}\, =\, 1-\dfrac{1}{k^2}\,\ $ thus
Multiplicative Telescopy
$\ \ \rm\displaystyle f(a\!-\!1) \prod_{\large k\,=\,a}^{\large n} \dfrac{f(k)}{f(k-1)}\, =\ f(n) $
Proof $ $ Induct on $\rm\,n.\,$ Base is $\rm\, f(a\!-\!1)\frac{f(a)}{f(a-1)}=\,f(a)\,$ at $\rm\,n\!=\!a.\,$ Induction $\rm\,\color{#0a0}{P(n)}\Rightarrow\, P(n\!+\!1)\,$ is
$\quad\ \displaystyle\rm f(a\!-\!1)\prod_{\large k\,=\,a}^{\large n+1}\dfrac{f(k)}{f(k\!-\!1)}\, =\, \left[f(a\!-\!1)\prod_{\large k\,=\,a}^{\large n}\dfrac{f(k)}{f(k\!-\!1)}\right]\dfrac{f(n\!+\!1)}{f(n)}\,\overset{\rm\color{#0a0}{ P(n)}} =\, \color{brown}{f(n)}\dfrac{f(n\!+\!1)}{\color{brown}{f(n)}} \, =\, f(n\!+\!1) $
Remark $\ $ Unwinding the induction yields a vivid depiction of the telescopic cancellation
$\quad \rm\displaystyle f(a\!-\!1)\prod_{\large k\,=\,a}^{n} \frac{f(k)}{f(k\!-\!1)}\, = \ \frac{\color{#c00}{\rlap{---}f(a\!-\!1)}}{1}\frac{\color{green}{\rlap{--}f(a)}}{\color{#C00}{\rlap{---}f(a\!-\!1)}}\frac{\color{royalblue}{\rlap{---}f(a\!+\!1)}}{\color{green}{\rlap{--}f(a)}}\frac{\phantom{\rlap{--}f(3)}}{\color{royalblue}{\rlap{---}f(a\!+\!1)}}\, \cdots\, \frac{\color{brown}{\rlap{---}f(n\!-\!1)}}{\phantom{\rlap{--}f(n\!-\!1)}}\frac{f(n)}{\color{brown}{\rlap{---}f(n\!-\!1)}}\, =\ \frac{f(n)}{1} $
You can find many further examples of multiplicative telescopy in other posts here.
| {
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Un-Simplifying a fraction, i.e. computing partial fraction decomposition $\frac{3x^2+17x}{x^3+3x^2+-6x-8}$
I need to find the value of C in the form of
$\frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{x+4}$
which is based on the fraction give at the top.
I can get so far to do the following:
$A(x^2+2x-8) + B(x^2+5x+4) + C(x^2-x-2) = 3x^2+17x$
No clue on my next step or even if this is the right step.
| $$
\begin{align*}
\left(A+B+C\right) x^2 &= 3x^2 \\
\left(2A + 5B -C\right)x &= 17x \\
-8A + 4B - 2C &= 0
\end{align*}
$$
Dividing the first equation by $x^2$ and the second by $x$ will get you a linear system of equations.
| {
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How does one show that for $k \in \mathbb{Z_+},3\mid2^{2^k} +5$ and $7\mid2^{2^k} + 3, \forall \space k$ odd.
For $k \in \mathbb{Z_+},3\mid2^{2^k} +5$ and $7\mid2^{2^k} + 3, \forall \space k$ odd.
Firstly,
$k \geq 1$
I can see induction is the best idea:
Show for $k=1$:
$2^{2^1} + 5 = 9 , 2^{2^1} + 3 = 7$
Assume for $k = \mu$
so: $3\mid2^{2^\mu} + 5 , \space 7\mid2^{2^\mu} + 3$
Show for $\mu +2$
Now can anyone give me a hint to go from here? My problem is being able to show that $2^{2^{\mu+2}}$ is divisible by 3, I can't think of how to begin how to show this.
| We have $2^2\equiv 1\pmod 3$ now one notes that $$2^{2^k}=(2^2)^{2^{k-1}}\equiv 1\pmod 3$$ so $\forall k$ we have $$2^{2^k}+5\equiv 6\equiv 0\pmod 3$$
We also have $2^{2^2}\equiv 2\pmod 7$ this means $2^{2^k}\equiv 2^{2^{k-2}}\pmod 7$ So when $k$ is odd $2^{2^k}\equiv 2^2\pmod 7$ and therefore $2^{2^k}+3\equiv 0\pmod 7$
| {
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Expressing in terms of symmetric polynomials. How to express $$a^7+b^7+c^7$$ in terms of symmetric polynomials ${\sigma}_{1}=a+b+c$, ${\sigma}_{2}=ab+bc+ca$ and ${\sigma}_{3}=abc$ ?
| We can take a slightly different approach. We have that $a,b,c$ are the roots of the polynomial:
$$ p(x)=x^3-\sigma_1 x^2+\sigma_2 x-\sigma_3 $$
hence the eigenvalues of the companion matrix:
$$ M = \left(\begin{array}{ccc} 0 & 0 & \sigma_3 \\ 1 & 0 & -\sigma_2 \\ 0 & 1 & \sigma_1\end{array}\right) $$
so:
$$\begin{eqnarray*} a^7+b^7+c^7 &=& \operatorname{Tr}\left(M^7\right)\\&=&\sigma_1^7 - 7 \sigma_1^5 \sigma_2 + 14 \sigma_1^3 \sigma_2^2 - 7 \sigma_1 \sigma_2^3 + 7 \sigma_1^4 \sigma_3 - 21 \sigma_1^2 \sigma_2 \sigma_3 +
7 \sigma_2^2 \sigma_3 + 7 \sigma_1 \sigma_3^2.\end{eqnarray*} $$
| {
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Three terms from a sequence Find the number of ways of choosing three numbers from the set ${1,2,3,…,20}$, so that the sum of the three numbers is divisible by $3$. I wrote $a+b+c=3k$, where $k=1,2,...20$, till $k=6$, there were no problems, but then restrictions started($a<21$). Even if I come to know of the correct method from here, the summation of all the answers is a tedious job(the answers come in binomial coefficients). Therefore, s there a shorter way?
| We only need concern ourselves with congruence $\bmod 3$.
There are $7$ numbers that are $1\bmod 3$
There are $7$ numbers that are $2\bmod 3$
There are $6$ numbers that are $0\bmod 3$
We now count the ways to add to $0\bmod 3$:
Three zeros:
$0+0+0$ (There are $\binom{6}{3}$ of these)
Two zeros:
One zero:
$0+1+2$ (There are $7\cdot7\cdot 6$ of these )
No zeros:
$1+1+1$ (There are $\binom{7}{3}$ of these)
$2+2+2$ (There are $\binom{7}{3}$ of these)
Adding up the answer is $\binom{6}{3}+7\cdot7\cdot6+2\binom{7}{3}=384$
There are $\binom{20}{3}=1140$ ways to choose, $3$ numbers of $20$. A third of that would be $380$. So more than a third of the subsets of size three are $0\bmod 3$
| {
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Find the solutions of $x^2\equiv -1\pmod{5}$ and $x^2\equiv -1\pmod{13}$ Find the solutions of $x^2\equiv -1\pmod{5}$ and $x^2\equiv -1\pmod{13}$
I know that they are both soluble since $5\equiv 1\pmod{4}$ and $13\equiv 1\pmod{4}$
What is the method to solving this simultaneous equation.
Looking for a standard method to use with this type of problem.
| by Wilson's theorem, for prime $p$:
$$
(p-1)! \equiv_p -1
$$
if $p \equiv_4 1$ then $p-1 = 4n$ and
$$
(p-1)! = \prod_{k=1}^{2n} k \prod_{k=2n+1}^{4n} k =\prod_{k=1}^{2n} k\prod_{k=1}^{2n}(p- k) \equiv_p (-1)^{2n} \left( \prod_{k=1}^{2n} k \right)^2 = ((2n)!)^2
$$
i.e.
$$
\left( \left(\frac{p-1}2 \right)! \right)^2 \equiv_p -1
$$
so $(2!)^2 \equiv_5 -1$ and $(6!)^2 \equiv_{13} -1$, with $6! = 720 \equiv_{13} 720-650-65=5$
since, for any $m$ we have $j^2 \equiv_m (m-j)^2$ the square roots of $-1 \mod 5$ are $2,3$ and the square roots of $-1 \mod 13$ are $5,8$
| {
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Solving for $z^2 = x^2 -xy + y^2$ Recently, I came across the following solution to finding integer solutions for $z^2 = x^2 - xy + y^2$:
*
*$x = k(-n^2 -2mn)$
*$y = k(m^2 - n^2)$
*$z = k(mn + m^2 + n^2)$
I've been scratching my head trying to figure out how to derive this solution.
I initially thought that this was derived in the same way as the solution to the Pythagorean Triples based on:
$$(p^2 + q^2)^2 = (2pq)^2 + (p^2 - q^2)^2$$
So that, I get:
$(p^2 + q^2)^2 = z^2 + (2pq)(p^2 - q^2) = (2pq)^2 + (p^2 -q^2)^2$
But I don't see how to solve for $z$ based on the above.
Can anyone provide me the outline of how one figures out the general solution for $z^2 = x^2 -xy + y^2$?
Edit: Added clarification about solutions in integers
| Here is a way to find a parametrization.
First, notice that an integer solution means a rational solution $1=\left(\frac{x}{z}\right)^2-\frac{x}{z}\cdot\frac{y}{z}+\left(\frac{y}{z}\right)^2$ (when $z=0$ the only possible solution is $x=y=z=0$). So, we can reduce this problem to looking for rational points on the ellipse $1=a^2-ab+b^2$.
The point $(1,0)$ is on the ellipse, and every rational point $(a,b)$ determines a line through $(a,b)$ and $(1,0)$ with rational slope. If we solve the system $b-1=ma$ and $1=a^2-ab+b^2$ for $a$ in terms of $m$, we get $a=\frac{1-2m}{m^2-m+1}$ (with $m=\frac{1}{2}$ giving $a=0$), and from the equation of the line, $b=\frac{1-m^2}{m^2-m+1}$. So a rational slope also determines a rational $(a,b)$.
When $m=\frac{p}{q}$ with integers $p,q$ we have $a=\frac{q^2-2pq}{p^2-pq+q^2}$ and $b=\frac{q^2-p^2}{p^2-pq+q^2}$. With this, we can write $x=k(q^2-2pq)$, $y=k(q^2-p^2)$, and $z=k(p^2-pq+q^2)$ for arbitrary integers $k,p,q$ for solutions to $z^2=x^2-xy+y^2$. The $k$ comes from the fact we may scale solutions.
This is very similar to your parametrization. Perhaps if we selected a different point for the rational-sloped lines we would find yours.
| {
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To find the close form for the last value of a matrix product Suppose I have two matrices
$$A = \begin{bmatrix} 0 & 1 \\ a_2 & a_1\end{bmatrix}$$
$$B = \begin{bmatrix} 0 & 1 \end{bmatrix}$$
Then $AB^T = \begin{bmatrix} 1 \\ a_1\end{bmatrix}$, the last term is $a_1$
Now I expand matrix $A$ and $B$ a little bit to get
$$A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ a_3 & a_2 & a_1\end{bmatrix}$$
$$B = \begin{bmatrix} 0 & 0 & 1 \end{bmatrix}$$
Then I find $AB^T = \begin{bmatrix} 0 \\ 1 \\ a_1\end{bmatrix}$, the last term is $a_1$, furthermore
$A^2B^T = \begin{bmatrix} 1 \\ a_1 \\ a_1^2 + a_2 \end{bmatrix}$, the last term is $a_1^2 + a_2$, neato!
Finally let's define our matrix to be:
$$A = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ a_4 & a_3 & a_2 & a_1\end{bmatrix}$$
$$B = \begin{bmatrix} 0 & 0 & 0 & 1 \end{bmatrix}$$
As before, I find $AB^T = \begin{bmatrix} 0 \\ 0 \\ 1 \\ a_1\end{bmatrix}$, the last term is $a_1$, furthermore
$A^2B^T = \begin{bmatrix} 0 \\ 1 \\ a_1 \\ a_1^2 + a_2 \end{bmatrix}$, the last term is $a_1^2 + a_2$
Curiously, I can compute $A^3B^T$ and obtain $\begin{bmatrix} 1 \\ a_1 \\ a_1^2 + a_2 \\ a_3 + a_2a_1 + a_1(a_1^2+a_2) \end{bmatrix}$
Now suppose A and B are both N dimensional matrices.
Is there a closed form solution for the n by nth value of the product $A^{n-1}B^T$?
Note: $A$ and $B$ pair are what known as the canonical controllable form
| It's more helpful to write $\begin{bmatrix} 1 \\ a_1 \\ a_2 + a_1^2 \\ a_3 + 2 a_1a_2 + a_1^3 \end{bmatrix}$.
The next one is $\begin{bmatrix} 1 \\ a_1 \\ a_2 + a_1^2 \\ a_3 + 2 a_1a_2 + a_1^3\\a_4+2a_1a_3+3 a_1^2a_2 +a_2^2+a_1^4 \end{bmatrix}$.
The next one is $\begin{bmatrix} 1 \\ a_1 \\ a_2 + a_1^2 \\ a_3 + 2 a_1a_2 + a_1^3\\a_4+2a_1a_3+3 a_1^2a_2 +a_2^2+a_1^4\\a_5+2a_1a_4+2a_2a_3+3 a_1a_2^2+3a_3a_1^2 +4a_1^3a_2+a_1^5 \end{bmatrix}$.
Each term is linked to a partition of n.
So for example, one of the partitions of 5 is 2+2+1. Therefore one of the terms is the product of $a_2$ and $a_2$ and $a_1$, which is $a_1a_2^2$.
Each term has a coefficient that is equal to the number of $a_i$ that are multiplied together to make that term.
So for example the term above is the product of three values $a_2$ and $a_2$ and $a_1$, so the coefficient is 3.
The only exception to this coefficient rule is the last term, which is always $a_1^n$.
Partitions of 6 are:
6: $a_6$
1+5: $2a_1a_5$
2+4: $2a_2a_4$
3+3: $2a_3^2$
1+1+4: $3a_1^2a_4$
1+2+3: $3a_1a_2a_3$
2+2+2: $3a_2^3$
1+1+1+3: $4a_1^3a_3$
1+1+2+2: $4a_1^2a_2^2$
1+1+1+1+2: $5a_1^4a_2$
1+1+1+1+1+1: $a_1^6$
| {
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Closed-form solution for 3D rotation angles given pre- and post-image I'm working on some math involving a pinhole camera model, and I've run in to the following problem: given only $x$, $y$, $z$, $A$, $B$, and $C$, I need to solve for the angles $\theta$ and $\phi$ in the following system of equations:
$$
\begin{bmatrix} A \\ B \\ C \end{bmatrix} =
\begin{bmatrix} \cos\phi & 0 & -\sin\phi \\ 0 & 1 & 0 \\ \sin\phi & 0 & \cos\phi \end{bmatrix}
\begin{bmatrix} \cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}
\begin{bmatrix} x \\ y \\ z \end{bmatrix}
$$
under the assumption that $[x\ y\ z]^T$ and $[A\ B\ C]^T$ are both unit vectors. I know that $\phi$ must be in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$; the value of $\theta$ is unconstrained.
I am able to solve for $\theta$ and $\phi$ using Newton's method. Is there a closed-form solution for $\theta$ and $\phi$ in this case, and if so, what is it?
| There are two possible solutions for each angle, only one of which will work best
$$\begin{aligned}
\theta & = 2 \arctan\left( -\frac{\sqrt{x^2+y^2-B^2}+x}{y+B}\right) \\
\theta & = 2 \arctan\left( \frac{\sqrt{x^2+y^2-B^2}-x}{y+B}\right) \\
\end{aligned}$$
$$\begin{aligned}
\varphi & = 2 \arctan\left( -\frac{\sqrt{A^2+C^2-z^2}+A}{z+C}\right) \\
\varphi & = 2 \arctan\left( \frac{\sqrt{A^2+C^2-z^2}-A}{z+C}\right) \\
\end{aligned}$$
Why?
I used the Tan Half Angle substitution for $\theta = 2 \arctan(t)$ and $\varphi = 2 \arctan(s)$, $$\cos(\theta) = \frac{1-t^2}{t^2+1}$$ and $$\sin(\theta) = \frac{2 t}{t^2+1}$$ and similarly for $\cos(\varphi)$ and $\sin(\varphi)$.
Then I solved the two quadratic equations from the y and z components of the equation
$${\rm Rot}_Y(\varphi)[A,B,C] = {\rm Rot}_Z(-\theta) [x,y,z]$$
I used the standard convention for the 3×3 rotation matrices ${\rm Rot}_Y()$ and ${\rm Rot}_Z()$
The two equations I solved are
$$B = y \cos \theta-x \sin \theta \\ C \cos\varphi-A \sin\varphi=z$$
which get transformed to
$$ B = y \frac{1-t^2}{1+t^2}-x \frac{2 t}{1+t^2} \\
C \frac{1-s^2}{1+s^2} - A \frac{2 s}{1+s^2} = z $$
or
$$ B (1+t^2)=y (1-t^2)-2 t x \\ C (1-s^2) - 2 A s = z (1+s^2) $$
with solution
$$ t = \frac{\sqrt{x^2+y^2-B^2}-x}{y+B} \\ s = \frac{\sqrt{C^2+A^2-z^2}-A}{z+C} $$
| {
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Find point on x-axis that is a minimum distance from 2 points Find the point on the x-axis the sum of whose distances from (2, 0) and
(0, 3) is a minimum.
Let the point be denoted (x, 0) and the sum of the distances be D.
By Pythagoras' Theorem,
$$D = \sqrt{(x-2)^2+(0-0)^2} + \sqrt{(x-0)^2+(0-3)^2}$$
$$D = \sqrt{x^2-4x+4} + \sqrt{x^2+9}$$
The minimum distance is at the point where D'(x) = 0.
$$D'(x) = \frac{2x-4}{2\sqrt{x^2-4x+4}} + \frac{2x}{2\sqrt{x^2+9}}$$
$$D'(x) = \frac{x-2}{\sqrt{x^2-4x+4}} + \frac{x}{\sqrt{x^2+9}}$$
That is,
$$\frac{x-2}{\sqrt{x^2-4x+4}} = \frac{-x}{\sqrt{x^2+9}}$$
Square both sides:
$$\frac{(x-2)^2}{x^2-4x+4} = \frac{x^2}{x^2+9}$$
$$(x-2)^2(x^2+9) = (x^2)(x^2-4x+4)$$
$$(x^2-4x+4)(x^2+9) = (x^2)(x^2-4x+4)$$
$$x^4+9x^2-4x^3-36x+4x^2+36 = x^4-4x^3+4x^2$$
$$9x^2-36x = 0$$
$$x^2 = 4$$
$$x=2$$
But if
$$D = \sqrt{(x-2)^2+(0-0)^2} + \sqrt{(x-0)^2+(0-3)^2}$$
$$D = x-2 + \sqrt{(x-0)^2+(0-3)^2}$$
$$D'(x) = 1 + \frac{2x}{2\sqrt{x^2+9}}$$
$$1 = \frac{-x}{\sqrt{x^2+9}}$$
$$\sqrt{x^2+9} = -x$$
Square both sides:
$$x^2+9 = x^2$$
$$9 = 0$$
What's wrong here?
| This is a great algebra problem.
Even though Jason's remark is correct, it does NOT address the full issue here.
Why not use the TI for a change.
Put in for $y1$ the first D function with the square roots.
You can look at the graph and clearly see that the function makes an abrupt turn at its minimum $x=2$. In other words, even though $x=2$ represents a minimum, it is not differentiable at that value.
Now let's take the squareroot of $(x-2)^2$ and replace it by $x-2$ , put in for $y2$ and compare the graphs. For $x<2$ the graphs differ en the second graph no longer show $x=2$ as a minimum. Then replace $x-2$ by $2-x$ This time for $x>2$ the graphs are different with again the second graph showing no minimum at $x=2$.
In other words: The first method is correct, a simplification on the second method cannot be done. More important with optimization problems: Examine the graph with your graphing device to see if it makes sense!
| {
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solving cubic diophantine equation Can someone show me how to find all solutions in positive integers to the diophantine equation: $$x^3 + y^3 = 35$$ I know how to do it algebraically, but I want to know how you solve it in number theory.
My Algebraic Approach:
$x^3 + y^3 = (x + y)\cdot (x^2 - xy + y^2) = 35$.
The only integer factors of $35$ are $(1, 35)$ or $(5, 7)$. There are no integers $x$ and $y$ that add to $1$, so $x + y = 5$ or $7$.
Using $x + y = 5$ we get that $y = 5 - x$, so $y^3 = 125 - 75x + 15x^2 - x^3$ so
$x^3 + y^3 = 15x^2 - 75x + 125 = 35$
or
$$15x^2 - 75x + 90 = 0$$
$$x^2 - 5x + 6 = 0$$
$$(x - 3)(x - 2) = 0$$
So $x = 2$ or $x = 3$
Thus $y = 3$ or $y = 2$ respectively.
Using $x + y = 7$ we get that $y = 7 - x$, so $y^3 = 343 - 147x + 21x^2 - x^3$ so
$$x^3 + y^3 = 21x^2 - 147x + 343 = 35$$
$$21x^2 - 147x + 308 = 0$$
$$3x^2 - 21x + 44 = 0$$
or
$$x = \frac{21 \pm \sqrt{-87}}{6}$$
Since x is complex, this can't be a solution.
So $(x, y) = (2, 3)$ or $(x, y) = (3, 2)$
| COMMENT.-You have solved easily the problem algebraically and you want to know how solve it in number theory. Well your curve is an elliptic curve of rank $1$ whose generator is the point $(3,2)$ which is the only point of integer coordinates and it is determined using your algebraic way. For the rest it has an infinity of rational points whose set is dense in the curve and that can be calculated by the method of chords and tangents. That is what number theory in connection with elliptic curves tells us about this problem (algebraic number theory deals with the other non-rational algebraic points, i.e. non trascendental, of the curve).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1168613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Show that f is linear Let $f : \mathbb R \to \mathbb R$ be a solution of the additive Cauchy functional equation satisfying the condition
$$f(x) = x^2 f(1/x)\quad \forall x \in \mathbb R\setminus \{0\}.$$
Then show that $f(x) = cx,$ where $c$ is an arbitrary constant.
| Let $F(x)=f(x)-xf(1)$
For some $x\neq 0$, $F(\frac{1}{x})=f(\frac{1}{x})-\frac{1}{x}f(1)$.
Hence for $x\neq 0$ $$\begin{align}
x^2F(\frac{1}{x})&=x^2f(\frac{1}{x})-xf(1)\\&=f(x)-xf(1)\\&=F(x)\end{align}$$ and of course $F(1)=0$ and $F$ is additive.
Let us prove that $\forall x\in \mathbb R, F(x)=-F(-x)$
Indeed, $0=F(1)=F(x+1-x)=F(x)+F(1)+F(-x)=F(x)+F(-x)$
Also, for some $x\neq -1$,
$$\begin{align}
F(x)=F(x+1)
&=(x+1)^2F\left(\frac{1}{x+1}\right)\\
&=(x+1)^2F\left(1-\frac{x}{x+1}\right)\\
&=-(x+1)^2F\left(\frac{x}{x+1}\right)\\
&=-(x+1)^2\left(\frac{x}{x+1}\right)^2F\left(\frac{x+1}{x}\right)\\
&=-x^2F\left(1+\frac{1}{x}\right)\\
&=-x^2F\left(\frac{1}{x}\right)\\
&=-F(x)\\\end{align}$$
This also holds for $x=-1$ since $F(-1)=-F(1)$.
Hence $2F=0$.
Hence $F=0$ and we're done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1169058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
find all integer $x$ such that $7x\equiv 2x$ (mod 8) I am trying to find all integer $x$ such that $7x\equiv 2x$ (mod 8)
First, I have $$ 7x-2x=8k \hspace{0.1in} (\text{where} \hspace{0.1in}k\in\mathbb{z}) $$
$$5x=8k$$ $$x=\frac{8k}{5}$$
Does $x=\frac{8k}{5}$ right? if not, can someone give me a hit or a suggestion to solve it?
thanks
| I think you went one step too far, you already had the answer. When you got to $5x = 8k$, you should have realized that what you're looking for are the multiples of $8$. The only $x$ that will work are $x$ that are multiples of $8$.
And since $8$ is such a small number, it's no problem to check the eight possibilities one by one.
*
*If $x \equiv 1 \bmod 8$, then $7x \equiv 7 \bmod 8$ and $2x \equiv 2 \bmod 8$.
*If $x \equiv 2 \bmod 8$, then $7x \equiv 6 \bmod 8$ and $2x \equiv 4 \bmod 8$.
*If $x \equiv 3 \bmod 8$, then $7x \equiv 5 \bmod 8$ and $2x \equiv 6 \bmod 8$.
*If $x \equiv 4 \bmod 8$, then $7x \equiv 4 \bmod 8$ and $2x \equiv 0 \bmod 8$.
*If $x \equiv 5 \bmod 8$, then $7x \equiv 3 \bmod 8$ and $2x \equiv 2 \bmod 8$.
*If $x \equiv 6 \bmod 8$, then $7x \equiv 2 \bmod 8$ and $2x \equiv 4 \bmod 8$.
*If $x \equiv 7 \bmod 8$, then $7x \equiv 1 \bmod 8$ and $2x \equiv 6 \bmod 8$.
*If $x \equiv 0 \bmod 8$, then $7x \equiv 0 \bmod 8$ and $2x \equiv 0 \bmod 8$.
This confirms the answer.
| {
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"source": "stackexchange",
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Finding $\sin 2x, \cos 2x, \tan 2x$, & finding all solutions of $x$? I am a senior in high school at the moment taking a pre-calculus course. I was out a few days this week and last due to illness and I have a big test tomorrow. Please help me figure out how to do these two problems on my review sheet, and please show me the steps!
a) $8\sin x\cos x + 4 = 0$ , find all solutions for $x$ in the interval $(0, 4\pi]$
b) $\sin x = \frac{5}{7}$ , find $\sin 2x, \cos 2x$, and $\tan 2x$
cosine is negative since it's between $\pi$ and $\frac{\pi}{2}$.
| The double angle formulas for sine, cosine, and tangent are
\begin{align*}
\sin(2x) & = 2\sin x\cos x\\
\cos(2x) & = \cos^2x - \sin^2x\\
& = 2\cos^2x - 1\\
& = 1 - 2\cos^2x\\
\tan(2x) & = \frac{2\tan x}{1 - \tan^2x}
\end{align*}
Of course, if you know $\sin(2x)$ and $\cos(2x)$, you can compute $\tan(2x)$ by dividing $\sin(2x)$ by $\cos(2x)$.
For the first problem:
\begin{align*}
8\sin x\cos x + 4 & = 0\\
2\sin x\cos x + 1 & = 0\\
\sin(2x) + 1 & = 0 && \text{using the identity $\sin(2x) = 2\sin x\cos x$}\\
\sin(2x) & = -1\\
2x & = -\frac{\pi}{2} + 2n\pi, n \in \mathbb{Z}\\
x & = -\frac{\pi}{4} + n\pi, n \in \mathbb{Z}
\end{align*}
Since $x \in (0, 4\pi]$, $n = 1, 2, 3, 4$, which yields the solutions
$x = \dfrac{3\pi}{4}, \dfrac{7\pi}{4}, \dfrac{11\pi}{4}, \dfrac{15\pi}{4}$.
You can verify that these solutions are correct by direct substitution. For instance,
$$8\sin\left(\frac{3\pi}{4}\right)\cos\left(\frac{3\pi}{4}\right) + 4 = 8\left(\frac{\sqrt{2}}{2}\right)\left(-\frac{\sqrt{2}}{2}\right) + 4 = -4 + 4 = 0$$
For the second problem:
Since $\frac{\pi}{2} < x < \pi$, $x$ is a second-quadrant angle, so $\cos x < 0$. You know the value of $\sin x$. To use the double angle formulas for sine and cosine, you must determine $\cos x$. One option would be to draw a right triangle in the second quadrant with opposite side of length $5$ and hypotenuse of length $7$, then use the triangle to determine the cosine. Otherwise, you can use the Pythagorean Identity $\sin^2x + \cos^2x = 1$ to solve for $\cos x$, remembering to take the negative root. Once you find $\cos x$, substitute the values of $\sin x$ and $\cos x$ into the double angle formulas for $\sin(2x)$ and $\cos(2x)$, from which you can determine the values of the other trigonometric functions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$n^2(n^2-1)(n^2-4)$ is always divisible by 360 $(n>2,n\in \mathbb{N})$ How does one prove that $n^2(n^2-1)(n^2-4)$ is always divisible by 360? $(n>2,n\in \mathbb{N})$
I explain my own way:
You can factorize it and get $n^2(n-1)(n+1)(n-2)(n+2)$.
Then change the condition $(n>2,n\in \mathbb{N})$ into $(n>0,n\in \mathbb{N})$ that is actually equal to $(n\in \mathbb{N})$.
Now the statement changes into :
$$n(n+1)(n+2)^2(n+3)(n+4)$$
Then I factorized 360 and got $3^2 \cdot 2^3 \cdot 5$.
I don't know how to prove the expression is equal to $3^2 \cdot 2^3 \cdot5$.
Who can help me solve it?
| $n(n+1)(n+2)$, and $(n+2)(n+3)(n+4)$ are $2$ products of $3$ consecutive natural numbers hence each divisible by $3! = 6$, thus the product divisible by $6\cdot 6 = 36$, hence it is divisible by $9$, and it is divisible by $5! = 120$ since the product contains $5$ consecutive natural numbers $n(n+1)(n+2)(n+3)(n+4)$, thus it is divisible by $\text{lcm}(36,120) = 360$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Counting the number of integers less than $x$ that are relatively prime to a primorial $p\#$ Let $p \ge 5$ be a prime. Let $x \ge 20$ be an integer. Let $p\#$ be the primorial for $p$.
Let $|\{ i \le x \, \wedge \gcd(i,p\#)=1\}|$ be the count of integers less than or equal to $x$ that are relatively prime to $p\#$.
Is it true that:
$$|\{ i \le x \, \wedge \gcd(i,p\#)=1\}| \le \dfrac{x-2}{3}$$
If it is true, is there a more concise argument than my reasoning below?
Here's my reasoning:
*
*$\gcd(i,p\#)=1$ only if $\gcd(i,5\#)=1$
*$|\{ i \le x \, \wedge \gcd(i,p\#)=1\}| \le |\{ i \le x \, \wedge \gcd(i,5\#)=1\}|$
*$|\{ i \le 20 \, \wedge \gcd(i,5\#)=1\}| = 6 = \dfrac{20-2}{3}$
*Assume it is true up to $x \ge 20$
*If $\gcd(x+1,5\#) > 1$, then $|\{ i \le x+1 \, \wedge \gcd(i,5\#)=1\}| = |\{ i \le x \, \wedge \gcd(i,5\#)=1\}|$
*So, we can assume that $\gcd(x+1,5\#)=1$
*There exists $c,d$ such that $x+1 = 30d+c$ where $c \in \{1,7,11,13,17,19,23,29\}$
*We can assume $d \ge 1$ since:
$$|\{ i \le 23 \, \wedge \gcd(i,5\#)=1\}| = 7 = \dfrac{23-2}{3}$$
$$|\{ i \le 29 \, \wedge \gcd(i,5\#)=1\}| = 8 < \dfrac{29-2}{3}$$
*
*We can now show it is true for each value of $c$:
$$|\{ i \le 30d+1 \, \wedge \gcd(i,5\#)=1\}| = 8d+1 < \dfrac{30d-1}{3} = 10d - \dfrac{1}{3}$$
$$|\{ i \le 30d+7 \, \wedge \gcd(i,5\#)=1\}| = 8d+2 < \dfrac{30d+5}{3} = 10d + \dfrac{5}{3}$$
$$|\{ i \le 30d+11 \, \wedge \gcd(i,5\#)=1\}| = 8d+3 < \dfrac{30d+9}{3} = 10d + 3$$
$$|\{ i \le 30d+13 \, \wedge \gcd(i,5\#)=1\}| = 8d+4 < \dfrac{30d+11}{3} = 10d + \dfrac{11}{3}$$
$$|\{ i \le 30d+17 \, \wedge \gcd(i,5\#)=1\}| = 8d+5 < \dfrac{30d+15}{3} = 10d + 5$$
$$|\{ i \le 30d+19 \, \wedge \gcd(i,5\#)=1\}| = 8d+6 < \dfrac{30d+17}{3} = 10d + \dfrac{17}{3}$$
$$|\{ i \le 30d+23 \, \wedge \gcd(i,5\#)=1\}| = 8d+7 < \dfrac{30d+21}{3} = 10d + 7$$
$$|\{ i \le 30d+29 \, \wedge \gcd(i,5\#)=1\}| = 8d+8 < \dfrac{30d+27}{3} = 10d + 9$$
| Yes, it is true.
For $x = 6k \equiv 0 \bmod 6$, there are exactly $2k=\frac{x}{3}$ numbers $i\le x$ that are coprime to $6 = 3\#$ - every $i\equiv 1 \text{ or }5 \bmod 6$. Clearly for $x=3k$ it is also true that there are exactly $\frac{x}{3}$ numbers $i\le x$ that are coprime to $6$.
If we assume $x\ge 5$ and we also eliminate numbers that are multiples of $5$ (ie. only count numbers that are coprime to $5\#$), we have eliminated at least one more number ($5$) and the total drops to $\frac{x}{3}-1 < \frac{x-2}{3}$ for multiples of $3$. The only troublesome case in between multiples of $3$ occurs at $x\equiv 1 \bmod 6$, when a new coprime case is introduced before $\frac{x-2}{3}$ has reached the next integer.
Once we reach $x \ge 25$ we know that the maximum possible total at $x=3k$ is $\frac{x}{3}-2$ (because two multiples of $5$ that are coprime to $6$ have been included) and so the condition will always hold. The next smallest case of $x\equiv 1 \bmod 6 $ is $x=19$ so by choosing $x\ge 20$ we eliminate that last case.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How is $RN =\frac{2}{3}BC$ in this rectangle?
I tried this question first by letting $AB = 6p$ and $BC = 6q$. Then I let $R$ the midpoint of $GB$ and $N$ the midpoint of $FE$. I got stuck so I looked at the memo for a hint and it turns out I followed the memo exactly but the next step said that $RN = \frac{2}{3}BC$. How is this true? I couldn't get that into my head. Can someone please explain?
| We are free to assume that $ABCD$ is the unit square.
Then assuming $BE=x$ we have $GF=\frac{2}{3}x$ by Thales' theorem and we must have:
$$\frac{GF+BE}{MF+CE}=\frac{\frac{2}{3}x+x}{1-\frac{2}{3}x+1-x}=2,$$
from which it follows that $x=\frac{4}{5}$. That implies $IH=\frac{4}{15}$. Since $AI=\frac{1}{3}$,
$$\frac{[AIH]}{[ABC]}=\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{4}{15}=\frac{2}{45},$$
hence $(C)$ is the right answer. Since $RN=\frac{GF+BE}{2}$, it follows that $RN=\frac{5}{6}x=\frac{2}{3}$ as wanted.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $7$ divides $13^n- 6^n$ for any positive integer I need to prove $7|13^n-6^n$ for $n$ being any positive integer.
Using induction I have the following:
Base case:
$n=0$: $13^0-6^0 = 1-1 = 0, 7|0$
so, generally you could say:
$7|13^k-6^k , n = k \ge 1$
so, prove the $(k+1)$ situation:
$13^{(k+1)}-6^{(k+1)}$
$13 \cdot 13^k-6 \cdot 6^k$
And then I'm stuck....where do I go from here?
| We write
$n=1$ and get
$13 - 6 =7$( a multiple of $7$)
Moving to the induction hypothesis,
We assume that $n=k$, therefore
$13^k - 6^k = 7b$($b$ shows that $7$ is a multiple of $13^k -6^k$ by $b$ times)
Taking $k=1$ from above ($n=k=1$) then we get :
$13^1 - 6^1=13 - 6=7$.
Take one side of the equation and use the equivalent to its multiple in each in the form of $13-7$ or $7+6$ : $\pm$ the divisor($7$).
Either $13^k(7+6) - 6^k(6)$ or $13^k(13) - 6^k(13-7)$.
Taking $ 13^k(13) - 6^k(13 -7)$ gives
$13^k(13) - 6^k(13) + 6^k(7)$.
Change the power $k$ on $6^k(13)$ to $13$ ;$13^k(6)$
$13^k(13) - 13^k(6) + 6^k(7)$
Factor out $13^k$ on the LHS
$13^k(13-6) + 6^k(7)$ giving
$13^k(7) + 6^k(7)$. This can be compressed to:
$7(13^k + 6^k)$
Therefore now $b= 13^k + 6^k$ making it $7|(13^n - 6^n)$ valid for all positive integers values of $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1181297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Stirling numbers: Combinatorial proof of an identity How to prove the following combinatorially ?
\begin{equation}
{n+1 \brace k+1}=\sum_{m=k}^{n}(k+1)^{n-m}{m \brace k}.
\end{equation}
My question is how are only ( n - m ) elements being considered to be placed on the (k + 1) boxes ? And that too without any combination like n C (n - m) ?
| By way of enrichment here is a proof using generating functions.
Suppose we seek to evaluate
$$\sum_{m=k}^n (k+1)^{n-m} {m\brace k}
= (k+1)^n \sum_{m=k}^n (k+1)^{-m} {m\brace k}.$$
This is
$$(k+1)^n \sum_{m=0}^{n-k} (k+1)^{-m-k} {m+k\brace k}$$
or
$$(k+1)^{n-k} \sum_{m=0}^{n-k} (k+1)^{-m} {m+k\brace k}.$$
What we have here is
$$(k+1)^{n-k} [z^{n-k}] \frac{1}{1-z}
\sum_{m\ge 0} {m+k\brace k} \frac{z^m}{(k+1)^m}.$$
Now recall the OGF of Stirling numbers with fixed $k$
which was evaluated e.g. at this
MSE link:
$$P(z) = \sum_{m\ge 0}{m\brace k} z^m
= \prod_{p=1}^k \frac{z}{1-pz}.$$
This is
$$\sum_{m\ge k}{m\brace k} z^m
= \sum_{m\ge 0}{m+k\brace k} z^{m+k}.$$
It follows that
$$\sum_{m\ge 0}{m+k\brace k} z^{m}
= \prod_{p=1}^k \frac{1}{1-pz}.$$
Substituting this into the sum yields
$$(k+1)^{n-k} [z^{n-k}] \frac{1}{1-z}
\prod_{p=1}^k \frac{1}{1-pz/(k+1)}
\\ = (k+1)^{n-k} [z^{n-k}]
\prod_{p=1}^{k+1} \frac{1}{1-pz/(k+1)}
\\ = (k+1)^{n-k} [z^{n-k}]
\prod_{p=1}^{k+1} \frac{k+1}{k+1-pz}
\\ = (k+1)^{n+1} [z^{n-k}]
\prod_{p=1}^{k+1} \frac{1}{k+1-pz}.$$
This is
$$\frac{(k+1)^{n+1}}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-k+1}}
\prod_{p=1}^{k+1} \frac{1}{k+1-pz} \; dz.$$
Put $z=(k+1) w$ so that $dz = (k+1)\; dw$ to get
$$\frac{(k+1)^{n+1}}{2\pi i}
\int_{|w|=\epsilon} \frac{k+1}{w^{n-k+1}(k+1)^{n-k+1}}
\prod_{p=1}^{k+1} \frac{1}{k+1-p(k+1)w} \; dw$$
which is
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{k+1}{w^{n-k+1}(k+1)^{-k}}
\prod_{p=1}^{k+1} \frac{1}{k+1-p(k+1)w} \; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^{n-k+1}(k+1)^{-(k+1)}}
\prod_{p=1}^{k+1} \frac{1}{k+1-p(k+1)w} \; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^{n-k+1}}
\prod_{p=1}^{k+1} \frac{1}{1-pw} \; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{w^{k+1}}{w^{n+2}}
\prod_{p=1}^{k+1} \frac{1}{1-pw} \; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^{n+2}}
\prod_{p=1}^{k+1} \frac{w}{1-pw} \; dw.$$
This last integral evaluates to
$${n+1\brace k+1}$$
by inspection.
| {
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"url": "https://math.stackexchange.com/questions/1181911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Lower bound on $|a+b \sqrt{2} + c \sqrt{3}|$ I have stumbled across this question:
Let $a,b,c$ be integers, not all $0$ such that $\max(|a|,|b|,|c|)<10^6$. Prove that $|a+b \sqrt{2} + c \sqrt{3}| > 10^{-21}$.
Could anybody help by solving this? Elementary solution is preferred.
| Let $$\begin{cases}f_{1}=a+b\sqrt{2}+c\sqrt{3}\\
f_{2}=a-b\sqrt{2}+c\sqrt{3}\\
f_{3}=a-b\sqrt{2}-c\sqrt{3}\\
f_{4}=a+b\sqrt{2}-c\sqrt{3}\end{cases}$$
It is clear $$f_{1}f_{2}f_{3}f_{4}\in Z,a,b,c\in Z$$
since $a,b,c$ are integer,and not all 0 ,so $f_{k}\neq 0,k=1,2,3,4$.and Note
$$\max\{|a|,|b|,|c|\}<10^6\Longrightarrow |f_{k}|<10^7,k=1,2,3,4$$ so we have
$$|f_{1}f_{2}f_{3}f_{4}|\ge 1\Longrightarrow |f_{1}|\ge\dfrac{1}{|f_{2}f_{3}f_{4}|}>10^{-21}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $∫_γ \frac{z^2+1}{z(z^2+4)} dz$ Where $γ(t)=re^{it}$ with $0≤t≤2π$ for all possible value of $r$, $0Evaluate
$∫_γ \frac{z^2+1}{z(z^2+4)} dz$
Where $γ(t)=re^{it}$ with $0≤t≤2π$ for all possible value of $r$, $0<r<2$ and $2<r<∞$
Theorem: Let $f: G \to \mathbb C$ be analytic, suppose $B(a,r) \subset G(r>0)$. If $\gamma(T)=a+re^{it}$, $0\leq t \leq 2\pi$ then
$$f(z)=\frac{1}{2\pi i}\int_\gamma\frac{f(w)}{w-z}dw$$
I guess I can use line integral to solve this problem. So first I need to find $\gamma '(t)=ire^{it}$. Plug into the formula, I have
$$\int_\gamma \frac{z^2+1}{z(z^2+4)} dz= \int _\gamma \frac{3z}{4(z^2+4)}+\frac {1}{4z} dz$$
$$=\int _\gamma \frac{3z}{4(z^2+4)} dz + \int _\gamma\frac {1}{4z} dz$$
$$=\int _0^{2\pi} \frac{3re^{it}}{4(r^2e^{2it}+4)} (ire^{it})dz + \int _0 ^{2\pi}\frac {1}{4re^{it}} (ire^{it})dt$$
$$=\int _0^{2\pi} \frac{3ir^2e^{2it}}{4(r^2e^{2it}+4)} dz + \int _0 ^{2\pi}\frac {i}{4} dt$$
$$=\frac{3i}{4}\int _0^{2\pi} \frac{r^2e^{2it}}{r^2e^{2it}+4} dz + \frac{\pi i}{2}$$
One more thing that concern me is the value of $r$, should I break this into $2$ integral?
| From you theorem, you can write
\begin{align}
g(z) &= \int_{\gamma}\frac{z^2+1}{z(z^2+4)}dz\\
& = 2i\pi\biggl[\int_{\gamma}\frac{(z^2+1)/(z^2+4)}{z}dz +\int_{\gamma}\frac{(z^2+1)/(z^2+2zi)}{z-2i}dz+\int_{\gamma}\frac{(z^2+1)/(z^2-2zi)}{z+2i}\biggr]\\
&= 2i\pi [f_1(0)+f_2(2i)+f_3(-2i)]\tag{1}
\end{align}
where $f_1(z) = \frac{z^2+1}{z^2+4}$, $f_2(z) = \frac{z^2+1}{z(z+2i)}$, and $f_3(z) = \frac{z^2+1}{z(z-2i)}$. In order to use equation $(1)$, we need to consider how many poles are in the contours. By Cauchy's theorem, a closed contour that encloses no poles, is equal to zero. Does this help you determine how to evaluate equation $(1)$ in light of $r$?
For $0<r<2$, only $z=0$ is in the contour. What can we conclude about the second and third integrals? For $r>2$, what poles would be enclosed?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1185642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculating arc length $y=x^2$ I picked this example for practice and got stuck with it. Someone moderate me if I am in the right path.
I need to calculate the length of arc s, on the section of the curve $y=x^2$ with $0≤x≤1$
My Workings:
The formula is $s=\int^b_a\sqrt{1+[f'(x)]^2}dx$
I work out everything under the $\sqrt{}$
my $f(x) =x^2$. I need to get $f'(x)$, which =$2x$
So our function becomes $s=\int^1_0\sqrt{1+[2x]^2}dx$
Then I work out $1+[f'(x)]^2 = 1+4x^2 $
$s=\int^1_0\sqrt{1+[f'(x)]^2}dx =\int^1_0\sqrt{(1+4x^2)}dx$
$s=\int^1_0\sqrt{(1+4x^2)}dx =\int^1_0(1+4x^2)^\frac{1}{2}dx$
I reckon I have to integrate $\int^1_0(1+4x^2)^\frac{1}{2}dx$
The integration part:
I used u-substitution:
let $ u = 1+4x^2$ then $du = 8x dx$
hence $dx = \frac{du}{8x}$
But, now there is trouble with the $8x$ because if I plug in $dx$, I get:
$\int^1_0(u)^\frac{1}{2}\frac{du}{8x} = \int^1_0\dfrac{(u)^\frac{1}{2}}{8x}du$
$= \dfrac{1}{8}\int^1_0\dfrac{1}{x}.{(u)^\frac{1}{2}}du$
Doesn't this give me $(\dfrac{1}{8}.ln|x|.\dfrac{2}{3}.u^\frac{3}{2})]^1_0$
...but... my answer doesn't look right......shouldnt' $x >0$?
How do I end this?
| Use the substitution $2x = \sinh u$. This gives you $\frac{dx}{du} = \frac{1}{2}\cosh u$.
So you get: $$ \int_0^1 (1+4x^2)^\frac{1}{2} dx= \int_0^{\sinh^{-1}2} (1+\sinh^{2} u)^\frac{1}{2} \frac{1}{2} \cosh udu = \frac{1}{2} \int_0^{\sinh^{-1}2}\cosh^{2} udu = \frac{1}{2} \int_0^{\sinh^{-1}2} \frac{1}{2} (\cosh2u +1)du = \frac{1}{4}[\frac{1}{2}\sinh 2u + u]_0^{\sinh^{-1}2}$$
This substitution is often very useful when you have an expression of the sort $(1+x^2)^{\frac{1}{2}}$ because of the hyperbolic identity $\cosh^{2} u = 1+\sinh^{2} u$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1187358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find centroid of region of two curves Find the coordinates (to three decimal places) of the centroid of $y=2^x$, and $y=x^2$.
EDIT: $(0\le x\le2)$
I understand this with a triangle, not with curves.
| The computation of the centroid in $R^2$, of a region bounded by two continuous functions, goes, by definition, as follows. (Note that, over $[0,2]$, $x^2 \le 2^x$.)
First, one has to calculate the area, $\mathscr a$, of the region $$A=\{x,y\ ;\ 0\le x\le 2, \ x^2\le y\le 2^x\}.$$
$ \color{white}{bbbbbbbbbbb}$
$$\text{The region with the centroid to be calculated below.}$$
$$\color{white}{nnn}$$
We have for the area:
$$\mathscr a= \iint_Adydx=\int_0^2\ \left[\int_{x^2}^{2^x}dy\right]\ dx=\int_0^2 2^xdx-\int_0^2x^2dx.$$
Then, for the coordinates of the centroid:
$$\overline x=\frac{1}{a}\int_0^2 x(2^x-x^2)\ dx=\frac{1}{\mathscr a}\left(\int_0^2x2^xdx-\int_0^2x^3dx\right),$$
$$\overline y=\frac{1}{2\mathscr a}\int_0^2 (2^x+x^2)(2^x-x^2)\ dx=\frac{1}{2\mathscr a}\left(\int_0^22^{2x}dx-\int_0^2x^4dx\right).$$
Now, we calculate the following integrals
$$\int_0^2 2^xdx=\frac{3}{\ln2},$$
$$\int_0^2x^2dx=\frac{8}{3},$$
$$\int_0^2 x2^x\ dx=\frac{8}{\ln2}-\frac{3}{(\ln2)^2},$$
$$\int_0^2x^3\ dx=4.$$
$$\int_0^22^{2x}\ dx=\frac{15}{2\ln2},$$
$$\int_0^2x^4\ dx=\frac{32}{5}.$$
Having done all this, one can easily put together the pieces:
$$a=\frac{3}{ln2}-\frac{8}{3}=1.6614,$$
$$\overline x=\frac{1}{\mathscr a}\left(\frac{8}{\ln2}-\frac{3}{(\ln2)^2}-4\right)=0.7809,$$
$$\overline y=\frac{1}{2\mathscr a}\left(\frac{15}{2\ln2}-\frac{32}{5}\right)=1.3303.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1187887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integrate $\int_{\frac{\sqrt{2}}3}^{\frac 23} \frac{dx}{x^5\sqrt{9x^2-1}}$
$$\int_{\frac{\sqrt{2}}3}^{\frac 23} \frac{dx}{x^5\sqrt{9x^2-1}}$$
What I did was trig substitution:
$$x=\frac 13 \sec \theta$$ $$dx=\frac 13 \sec \theta \tan \theta \, d\theta$$
Then my integral becomes $$\int_{\frac 13 \sec\frac{\sqrt{2}}{3}}^{\frac 13 \sec\frac 23} \frac{81}{\sec^4 \theta} \, d\theta$$
But I got stuck here. What can I do next?
| Use:
$$\frac{1}{\sec^4(x)} = \cos^4(x) = (\cos^2(x))^2 = (\frac{1 + \cos(2x)}{2})^2 = \frac14(1 + 2\cos(2x) + \cos^2(2x)) = \frac14(1 + 2\cos(2x) + \frac{1 + \cos(4x)}{2}) = etc. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1189974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to simplify $(a^2+ab+b^2)/(a+\sqrt{ab}+b)$ How can I simplify as much as possible:
$$\frac{a^2+ab+b^2}{a+\sqrt{ab}+b}$$
Also, first post here, looking forward to sticking around!
| $$\frac{a^2 + ab + b^2}{a + \sqrt{ab} + b}= \frac{(a^3-b^3)/(a-b)}{(a^{3/2} -b^{3/2})/(a^{1/2} - b^{1/2}) } =\frac{a^{3/2} + b^{3/2}}{a^{1/2} + b^{1/2}}=a-\sqrt{ab} + b$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1190659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
A sum expressed by a Kampe de Feriet function. Let $a_1$,$a_2$, $a_3$ and $b_1$,$b_2$, $b_3$ be real numbers subject to $1+b_1+b_2 - b_3 > 0 $. By generalizing the result from A sum involving a ratio of two binomial factors. we have shown that the following identity holds:
\begin{eqnarray}
&&S:= \sum\limits_{i=0}^{m-1} \frac{\binom{i+a_1}{b_1} \binom{i+a_2}{b_2}}{\binom{i+a_3}{b_3}} =\\
&& \left. \frac{\binom{m+a_1}{1+b_1} \binom{m+a_2}{b_2}}{\binom{m+a_3}{b_3}}
\sum\limits_{(q_2,q_3) \in {\mathbb N}_+^2}
\binom{q_2+q_3}{q_3} \frac{b_3^{(q_3)} (-b_2)^{(q_2)}}{(2+b_1)^{(q_2+q_3)}} \frac{(1+a_1+m)^{(q_2+q_3)} (1+a_2+m)^{(q_3)}}{(1+a_2-b_2+m)^{(q_2+q_3)}(1+a_3+m)^{(q_3)}} \right|_0^m =\\
&& \left. \frac{\binom{m+a_1}{1+b_1} \binom{m+a_2}{b_2}}{\binom{m+a_3}{b_3}}
\sum\limits_{q=0}^\infty \frac{(1+a_1+m)^{(q)}(1+a_2+m)^{(q)}b_3^{(q)}}{(1+a_3+m)^{(q)}(1+a_2-b_2+m)^{(q)}(2+b_1)^{(q)}}\\
F_{3,2}\left[
\begin{array}{rrr} -b_2 & -q & -a_3-m-q\\ 1-b_3-q & -a_2-m-q\end{array};1
\right]
\right|_0^m \\
\end{eqnarray}
The right hand side of the above equation can be formally expressed as a Kampe de Feriet function. It converges whenever $1+b_1+b_2 - b_3 > 0 $. Now, the question is how does the right hand side behave when $m\rightarrow \infty$?
| We will compute the sum in question in a slightly different way and only then analyze the large-$m$ behavior of the result. The idea is to take the second binomial factor in the numerator and rewrite it as a linear combination of factors that can be absorbed into the first binomial factor. We start with the following identity:
\begin{equation}
\tag{1} \binom{i+a_2}{b_2} = \frac{\sin(\pi (a_2+b_2))}{\sin(\pi a_2)} \cdot
\sum\limits_{l=0}^\infty \binom{i+a_1+l}{l} \binom{a_1-a_2+b_2}{b_2-l} (-1)^l
\end{equation}
The identity is a simple consequence of the Chu-Vandermonde identity and the identity $\binom{n}{k} = (-1)^k \binom{k-n-1}{k}$. If we multiply the said identity by $\binom{i+a_1}{b_1}$ we can absorb the terms involving $i$ into the binomial factor and we get:
\begin{equation}
\tag{2} \binom{i+a_1}{b_1} \binom{i+a_2}{b_2} =
\frac{\sin(\pi (a_2+b_2))}{\sin(\pi a_2)} \cdot
\sum\limits_{l=0}^\infty \binom{b_1+l}{l} \binom{a_1-a_2+b_2}{b_2-l} (-1)^l \binom{i+a_1+l}{b_1+l}
\end{equation}
Now we divide both sides of the above equation by $\binom{i+a_3}{b_3}$ and we sum over $i$ using A sum involving a ratio of two binomial factors. . We get:
\begin{eqnarray}
\tag{3}
&&S = \frac{\sin(\pi (a_2+b_2))}{\sin(\pi a_2)} \cdot
\sum\limits_{l=0}^\infty \binom{b_1+l}{l} \binom{a_1-a_2+b_2}{b_2-l} (-1)^l \cdot \\
&&\left\{\left.
\frac{\binom{a_1+l+m}{b_1+l+1}}{\binom{a_3+m}{b_3}}
F_{3,2}\left[
\begin{array}{rrr} 1 & b_3 & 1+a_1+l+m \\ 2+b_1+l & 1+a_3+m\end{array};1
\right]
\right|_0^m
\right\}
\end{eqnarray}
So far all the results above are exact. Now we will take the limit $m \rightarrow \infty$ and try to see if we get the same limit as in the first answer to this question. We have:
\begin{equation}
\tag{4}
F_{3,2}\left[\cdots ;1\right] \underset{m\rightarrow \infty}{=} F_{2,1} \left[\begin{array}{rr} 1 & b_3 \\ 2+b_2+l\end{array};1\right]+ O(\frac{1}{m}) =
\frac{1+b_1+l}{1+b_1-b_3+l} + O(\frac{1}{m})
\end{equation}
Now, inserting $(4)$ into $(3)$, absorbing the $1+b_1+l$ factor into the first binomial coefficient under the sum and then using identity $(2)$ along with the mean value theorem we easily get :
\begin{equation}
S \underset{m\rightarrow \infty}{=} \frac{\binom{m+a_1}{1+b_1} \binom{m+a_2}{b_2}}{\binom{m+a_3}{b_3}} \cdot \frac{1+b_1}{1+b_1+\theta b_2 - b_3}
\end{equation}
where $ \theta \in (0,1)$. So as we can see two different approaches lead to the same result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1192441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
I'm having trouble with induction. Prove $1 + 2^3 + 3^3 + ... + n^3 = \frac{((n^2)(n+1)^2)}4$ I started a new course and I'm expected to know this stuff, and I'm having trouble learning some on my own.
I'm stuck with this problem:
Prove $1 + 2^3 + 3^3 + ... + n^3 = \frac{((n^2)(n+1)^2)}4$ using induction (the $1+2^3...$ is written in sum notation, although I don't know how to enter that here, sorry).
I started substituting $n \rightarrow n+1$ but I don't know what to do next. Any help would be extremely appreciated.
| This question often causes a lot of confusion simply because of the cumbersome factoring.
Let $P(n)$ be the proposition that $$\sum_{i=1}^n i^3 = \frac{(n^2)(n+1)^2}{4}.$$
Base Case, $P(1)$: $\sum_{i=1}^1 i^3 = 1$ and $\frac{(1^2)(1+1)^2}{4} = \frac{4}{4} = 1$ so $P(1)$ is true.
Inductive Hypothesis: Assume that $P(k)$ is true: $1^3 + \cdots + k^3 = \frac{(k^2)(k+1)^2}{4}$
Now it remains to show that $P(k) \implies P(k+1)$. Start with the left side of the equation:
$$\sum_{i=1}^{k+1} i^3 = \left( \sum_{i=1}^k i^3 \right) + (k+1)^3 = \frac{(k^2)(k+1)^2}{4} + (k+1)^3$$
(by the IH above). Now we factor out a $(k+1)^2$: $$= (k+1)^2 \cdot (\frac{k^2}{4} + k + 1) = \frac{(k+1)^2(k^2+4k+4)}{4} = \frac{(k+1)^2((k+1)+1)^2}{4}$$
Which is equal to the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1196090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Inverse of this function I have given
$$ f(x) = \frac{1}{x}\left(\left(\frac{1}{x}\right)^{a} + 1\right)^{-\frac{1}{a}}$$
And am trying to invert it, but despairing.
Perhaps it helps (but I wouldn't know) that $f$ comes from
$$
g(m,n) = mn (m^a + n^a)^{-1/a}\\
f(x) = g(1/x, 1)\\
$$
| Hint:
$f(x) = \frac{1}{x} \bigg( \bigg( \frac{1}{x} \bigg)^a +1 \bigg)^{-\frac{1}{a}}$
$= \bigg( \bigg( \frac{1}{x} \bigg)^a \bigg)^{\frac{1}{a}} \bigg( \bigg( \frac{1}{x} \bigg)^a +1 \bigg)^{-\frac{1}{a}} $
$ = \bigg( \frac{ \big( \frac{1}{x} \big)^a}{\big( \frac{1}{x} \big)^a+1} \bigg)^{\frac{1}{a}}$
$=\bigg( 1- \frac{ 1}{\big( \frac{1}{x} \big)^a+1} \bigg)^{\frac{1}{a}}$
Now you only have one $x$ in the formula. Proceed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1197645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Continued Fraction for Root 5 How can I find the continued fraction expansion for the square root of 5. Do this without the use of a calculator and show all the steps.
| It is the same idea as $\sqrt{7}$ and $\sqrt{3}$, or any $a+b\sqrt{c}$.
First determine the integer part of $\sqrt{5}$. We know that $2<\sqrt{5}<3$.
*
*Then it is only about extracting the integer part of improper fractions,
*taking reciprocals of proper fractions
*and multiplying numerator and denominator by the conjugate number when there is an irrational number in a denominator.
The case of $\sqrt{5}$ is short:
We obtain $\sqrt{5}=2+(\sqrt{5}-2)=2+\frac{1}{\frac{1}{\sqrt{5}-2}}$.
With the fraction $\frac{1}{\sqrt{5}-2}=\frac{\sqrt{5}+2}{1}=4+(\sqrt{5}-2)=4+\frac{1}{\frac{1}{\sqrt{5}-2}}$.
Therefore $$\sqrt{5}=2+\frac{1}{\frac{1}{\sqrt{5}-2}}=2+\frac{1}{\frac{\sqrt{5}+2}{1}}=2+\frac{1}{4+(\sqrt{5}-2)}=2+\frac{1}{4+\frac{1}{\frac{1}{\sqrt{5}-2}}}.$$ Since the $\frac{1}{\sqrt{5}-2}$ has appeared before the continued fraction will continue to spit $4$'s. The result is that
$$\sqrt{5}=[2| 4,4,4,...]$$
For the case of quadratic irrational numbers (solutions of quadratic equations) there is always going to appear a periodicity (and conversely periodicity only appears in this cases [and rational numbers]). So, eventually you get a fraction that you have already treated before.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1198833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
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