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Find the number of real solutions to the system of equations $x=\frac{2z^2}{1+z^2},y=\frac{2x^2}{1+x^2},z=\frac{2y^2}{1+y^2}$ My approach is naive:
Given
$x=\frac{2z^2}{1+z^2},y=\frac{2x^2}{1+x^2},z=\frac{2y^2}{1+y^2}$,
$[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}]=\frac{1}{2}\cdot[\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}]+3$
What to do next?
Tried it using trigonometry by replacing $z^2$ by $tan^2\theta$ but could not get promising results.
Is there any trick to such genre of problems?
| From the system of equations, you can infer that $x,y,z \ge 0$ and consider the function $f(a) = \dfrac{2a^2}{1+a^2}=2 - \dfrac{2}{1+a^2}$. Thus if $a > b>0 \implies a^2>b^2\implies 1+a^2>1+b^2\implies \dfrac{2}{1+a^2} < \dfrac{2}{1+b^2}\implies -\dfrac{2}{1+a^2} > -\dfrac{2}{1+b^2}\implies 2-\dfrac{2}{1+a^2} > 2-\dfrac{2}{a+b^2}\implies f(a) > f(b)\implies f$ is strictly increasing. Thus if we have $x > y > z \implies f(x) > f(y) > f(z)\implies y > z > x $, contradication to $x > z$. Thus at least two variables must be equal to each other, say $x = y \implies f(x) = f(y) \implies y = z \implies x = y =z\implies x = \dfrac{2x^2}{1+x^2}\implies x^3-2x^2+x = 0 \implies x(x^2-2x+1) = 0\implies x(x-1)^2 = 0 \implies x = 0, 1 \implies x = y = z =1$ or $x = y = z = 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $1+\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{2018^3}<\frac{5}{4}$ Question: Prove that $1+\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{2018^3}<\frac{5}{4}$
Attempt: When I tried solving this problem, I paired up $1$ with $\frac{1}{2018^3}$, $\frac{1}{2^3}$ with $\frac{1}{2017^3}$ etc. but it wasn't useful.
| Another method that is generally applicable:
$\zeta(2n+1) = 1 + \frac{1}{2^{2n+1}} + \frac{1}{3^{2n+1}} + \frac{1}{4^{2n+1}} + \ldots$
$< 1 + \frac{1}{2^{2n+1}} + \frac{1}{3^{2n+1}} + \left(\frac{1}{4^{2n+1}} + \frac{1}{4^{2n+1}} + \frac{1}{4^{2n+1}} + \frac{1}{4^{2n+1}}\right) + \ldots $
$= 1 + \frac{1}{2^{2n+1}} + \frac{1}{3^{2n+1}} + \frac{1}{4^{2n}} + \frac{1}{8^{2n}} + \frac{1}{16^{2n}} + \ldots $
$ = 1 + \frac{1}{2^{2n+1}} + \frac{1}{3^{2n+1}} + \frac{1}{2^{2n}(2^{2n} -1)}$
For $n= 1$, we get $\zeta(3) < 1 + \frac{1}{8} + \frac{1}{27} + \frac{1}{12} = 1.24537 < 5/4$
| {
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"url": "https://math.stackexchange.com/questions/2723338",
"timestamp": "2023-03-29T00:00:00",
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I need the steps to solving this limit without using l´Hopital rule I've tried many ways of solving this limit without using l'Hopital and I just can't figure it out. I know the answer is $3/2 \sin (2a).$
$$\lim_{x\,\to\,0} \frac{\sin(a+x)\sin(a+2x) - \sin^2(a)} x$$
Thank you!
| \begin{align*}
&\dfrac{\sin(a+x)\sin(a+2x)-\sin^{2}a}{x}\\
&=2\cdot\dfrac{\sin(a+x)[\sin(a+2x)-\sin a]}{2x}+\dfrac{(\sin a)[\sin(a+x)-\sin a]}{x}\\
&\rightarrow 2\sin(a+0)(\sin x)'\bigg|_{x=a}+(\sin a)(\sin x)'\bigg|_{x=a}\\
&=2(\sin a)(\cos a)+(\sin a)(\cos a).
\end{align*}
The true without L'Hopital:
\begin{align*}
&\dfrac{\sin(a+x)\sin(a+2x)-\sin^{2}a}{x}\\
&=\dfrac{(\sin a\cos x+\cos a\sin x)(\sin a\cos 2x+\cos a\sin 2x)-\sin^{2}a}{x}\\
&=\dfrac{\sin^{2}a\cos x\cos 2x+\cos^{2}a\sin x\sin 2x+\sin a\cos a(\sin 2x\cos x+\cos x\sin 2x)-\sin^{2}a}{x}\\
&=\dfrac{\sin^{2}a(\cos x\cos 2x-1)}{x}+\cos^{2}a\cdot\dfrac{\sin x}{x}\cdot\sin 2x+\dfrac{\sin 2a}{2}\cdot 3\cdot\dfrac{\sin 3x}{3x},
\end{align*}
where
\begin{align*}
\dfrac{\cos x\cos 2x-1}{x}&=\dfrac{\cos x-1}{x}-2\cdot\dfrac{\sin x}{x}\cdot\cos x\\
&=\dfrac{-2\sin^{2}(x/2)}{x}-2\cdot\dfrac{\sin x}{x}\cdot\cos x\\
&\rightarrow 0
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve system of equations: $2x^2-4y^2-\frac{3}{2}x+y=0 \land 3x^2-6y^2-2x+2y=\frac{1}{2}$ $$
\begin{cases}
2x^2-4y^2-\frac{3}{2}x+y=0 \\
3x^2-6y^2-2x+2y=\frac{1}{2}
\end{cases}
$$
I multiplied the first with $-6$ and the second with $4$ and get two easier equations:
$9x-6y=0 \land -8x+8y=2 $ and out of them I get that $x=\frac{1}{2}$ and that $y=\frac{3}{4}$ but when I put it back into the original systems equation I dont get the right answer. Can somebody explain why?
| It looks like you took the two scaled equations $$-12x^2+24y^2+9x-6y = 0 \\ 12x^2-24y^2-8x+8y=2$$ and then simply lopped off the common quadratic parts to produce two linear equations that are entirely unrelated to the original system. Once you’ve scaled the two equations so that the coefficients of their squared terms match, you eliminate those terms by actually adding the two equations together, producing the single linear equation $x+2y=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2725579",
"timestamp": "2023-03-29T00:00:00",
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Evaluating $\prod^{100}_{k=1}\left[1+2\cos \frac{2\pi \cdot 3^k}{3^{100}+1}\right]$
Evaluate$$\prod^{100}_{k=1}\left[1+2\cos \frac{2\pi \cdot 3^k}{3^{100}+1}\right]$$
My attempt:
$$1+2\cos 2\theta= 1+2(1-2\sin^2\theta)=3-4\sin^2\theta$$
$$=\frac{3\sin \theta-4\sin^3\theta}{\sin \theta}=\frac{\sin 3\theta}{\sin \theta}$$
I did not understand how to solve after that. Help required.
| It's unclear if you are asking about $$\prod^{100}_{k=1}\left[1+2\cos \frac{2\pi \cdot 3^k}{3^{100}+1}\right]$$ or about $$\prod^{100}_{k=1}\left[1+2\cos \frac{2\pi k\cdot 3^k}{3^{100}+1}\right]$$
I will assume it is the former. At the moment, that is in the body of your question, while the latter is in the title. If it's the former, then using the trig identity you found, you have a telescoping product which then further simplifies nicely: $$
\begin{align}
\prod^{100}_{k=1}\frac{\sin\left(\pi\frac{3^{k+1}}{3^{100}+1}\right)}{\sin\left(\pi\frac{3^{k}}{3^{100}+1}\right)}&=\frac{\sin\left(\pi\frac{3^{2}}{3^{100}+1}\right)}{\sin\left(\pi\frac{3^{1}}{3^{100}+1}\right)}\frac{\sin\left(\pi\frac{3^{3}}{3^{100}+1}\right)}{\sin\left(\pi\frac{3^{2}}{3^{100}+1}\right)}\cdots\frac{\sin\left(\pi\frac{3^{101}}{3^{100}+1}\right)}{\sin\left(\pi\frac{3^{100}}{3^{100}+1}\right)}\\[1pc]
&=\frac{\sin\left(\pi\frac{3^{101}}{3^{100}+1}\right)}{\sin\left(\pi\frac{3^{1}}{3^{100}+1}\right)}\\[1pc]
&=\frac{\sin\left(\pi\frac{3\cdot3^{100}}{3^{100}+1}\right)}{\sin\left(\pi\frac{3}{3^{100}+1}\right)}\\[1pc]
&=\frac{\sin\left(\pi\frac{3\left(\cdot3^{100}+1\right)-3}{3^{100}+1}\right)}{\sin\left(\pi\frac{3}{3^{100}+1}\right)}\\[1pc]
&=\frac{\sin\left(3\pi-\pi\frac{3}{3^{100}+1}\right)}{\sin\left(\pi\frac{3}{3^{100}+1}\right)}\\[1pc]
&=\frac{\sin\left(\pi\frac{3}{3^{100}+1}\right)}{\sin\left(\pi\frac{3}{3^{100}+1}\right)}\\[1pc]
&=1
\end{align}$$
Note near the end that $\sin(3\pi-X)=\sin(X)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why can't I cancel $2x-3$ from $(2x-3)(x+5)=9(2x-3)$? Why are these simplifications wrong?
$$\begin{align}
(2x-3)(x+5)=9(2x-3) &\quad\to\quad \frac{(2x-3)(x+5)}{2x-3} = \frac{9(2x-3)}{2x-3} \quad\to\quad x+5 = 9\\[4pt]
x(x+2)=x(-x+3) &\quad\to\quad \frac{x(x+2)}{x} = \frac{x(-x+3)}{x} \quad\to\quad x+2=-x+3
\end{align}$$
| $$(2x-3)(x+5)=9(2x-3).$$ Since you are multiplying both sides by the same constant, namely $2x-3$, then they cancel out, thus bringing forward the following equation: $$x+5=9.\tag1$$
$$x(x+2)=x(-x+3).$$ In the same fashion as before, the $x$'s cancel out, i.e. $$x+2=-x+3.$$ Now, adding $x$ to both sides yeilds the following equation: $$2x + 2 = 3$$ $$\Downarrow$$ $$2x = 1.\tag2$$
Finally, solve for $x$ in Eq. $(1)$ and Eq. $(2)$. Also note that each of the common factors, $(2x-3)$ and $x$, can be equal to $0$, because for all values $n$, one always has that $0\cdot n = 0$. Thus, you have a bonus two other solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735413",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Find the maximum value of $y\cdot{x^2}+x$ if $y^2+x^2+x+y=100$. We know that $y^2+x^2+x+y=100$. Find the maximum value of $$y\cdot x^2+x$$
I tried to simplify it and use inequalities but I failed. Is there a way to solve it without calculus?
| You Can reduce your Problem in two variables in to one: from $$y^2+y-100+x^2+x=0$$ follows $$y=-\frac{1}{2}\pm\sqrt{\frac{1}{4}+100-x^2-x}$$
so you will get
$$h(x)=\left(-1/2\pm\sqrt{\frac{1}{4}+100-x^2-x}\right)x^2+x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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On the series $\sum \limits_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{2n-1} - \frac{1}{2n+1} \right )$ I managed to prove through complex analysis that
$$\sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{2n-1} - \frac{1}{2n+1} \right ) = 1 -2 \log 2$$
However, I'm having a difficult time proving this result with real analysis methods. Partial summation of the series gets nasty pretty quickly since it involves harmonic numbers. Another interesting fact to note about this series is that this part
$$\sum_{n=1}^{\infty} \left ( \frac{1}{2n-1} + \frac{1}{2n+1} \right )$$
diverges. This came as a complete surprise to me since I expected this to telescope. What remains now to prove the series I want is by using generating function.
\begin{align*}
\sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{2n-1} - \frac{1}{2n+1} \right )x^n &=\sum_{n=1}^{\infty} \frac{x^n}{n} - \sum_{n=1}^{\infty} \frac{x^n}{2n-1} - \sum_{n=1}^{\infty} \frac{x^n}{2n+1} \\
&= -\log \left ( 1-x \right ) - \sqrt{x} \;\mathrm{arctanh} \sqrt{x} - \frac{{\mathrm {arctanh} }\sqrt{x} - \sqrt{x}}{\sqrt{x}}
\end{align*}
I cannot , however, evaluate the limit as $x \rightarrow 1^-$ of the last expression. Can someone finish this up?
Of course alternatives are welcome.
| In the same spirit as achille hui,
$$S_p=\sum_{n=1}^p \left(\frac1n - \frac{1}{2n-1} - \frac{1}{2n+1}\right)$$
$$(2p+1)S_p=2 \gamma p+2 p-2 p \psi ^{(0)}\left(p+\frac{1}{2}\right)+2 p \psi ^{(0)}(p+1)+2 p
\psi ^{(0)}\left(\frac{1}{2}\right)-\psi ^{(0)}\left(p+\frac{1}{2}\right)+\psi
^{(0)}(p+1)+\gamma +\psi ^{(0)}\left(\frac{1}{2}\right)$$ where appears the digamma function.
Using generalized harmonic numbers, this reduces to
$$S_p=H_p-H_{p-\frac{1}{2}}+\frac{2 p}{2 p+1}-2\log (2)$$ Now, using the asymptotics
$$H_q=\gamma +\log \left({q}\right)+\frac{1}{2 q}-\frac{1}{12
q^2}+O\left(\frac{1}{q^4}\right)$$ and continuing with Taylor expansion
$$S_p=1-2\log (2)+\frac{1}{8 p^2}-\frac{1}{8
p^3}+O\left(\frac{1}{p^4}\right)$$ which allows quite accurate results even for small values of $p$.
For example $S_5=-\frac{5297}{13860}\approx -0.382179$ while the above expansion would give $\frac{251}{250}-2\log (2)\approx -0.382294$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding points on a line that are closest
Find the points that give the shortest distance between the lines$$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2-t\\-1+2t\\-1+t\end{pmatrix}\\\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}5+3s\\0\\2-s\end{pmatrix}$$
So I subtracted the second line from the first to get these two equations:
$\begin{pmatrix}-t-3s-3\\ -1+2t\\ t+s-3\end{pmatrix}\cdot \begin{pmatrix}-1\\ 2\\ 1\end{pmatrix}=0$
$\begin{pmatrix}-t-3s-3\\ -1+2t\\ t+s-3\end{pmatrix}\cdot \begin{pmatrix}3\\ 0\\ -1\end{pmatrix}=0$
I know I am supposed to rearrange them together two get a system of equations but I am not sure how.
Any help please?
| $$\begin{pmatrix}-t-3s-3\\ -1+2t\\ t+s-3\end{pmatrix}\cdot \begin{pmatrix}-1\\ 2\\ 1\end{pmatrix}=0\implies t+3s+3-2+4t+t+s-3 = 6t+4s=2$$
$$\begin{pmatrix}-t-3s-3\\ -1+2t\\ t+s-3\end{pmatrix}\cdot \begin{pmatrix}3\\ 0\\ -1\end{pmatrix}=0\implies -3t-9s+9-t-s+3=-4t-10s=-12$$
Sove above equations for $t$ and $s$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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When $2$ $6$-sided die are rolled, what is the probability that the first is odd and the difference is $0$? The probability the the first die is odd is $\frac{3}{6}=\frac{1}{2}$.
The probability that first is odd and diff is $0$ is $\frac{3}{6}\cdot \frac{1}{6} = \frac{3}{36} = \frac{1}{12}$.
Is $\frac{1}{12}$ correct?
| Set $X_1$: outcome of the first die. There are three successful outcomes: $X_1 = 1, X_1 = 3, X_1 = 5$; since the probabilities are uniform, $P(X_1= \{1,3,5\}) = \frac{1}{6}$. 'Difference equal to $0$' means the second outcome is the same; hence, we need
$P(X_2 \cap X_1) = P(X_2 = 1|X_1=1)P(X_1=1) + P(X_2 = 3|X_1=3)P(X_1=3) + P(X_2 = 5|X_1=5)P(X_1=5) = \frac{1}{6} \times \frac{1}{6} + \frac{1}{6} \times \frac{1}{6} + \frac{1}{6} \times \frac{1}{6} = \frac{3}{36}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2738700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding an invertible matrix P and some matrix C Find an invertible matrix $P$ and a matrix $C$ of the form
$C=\begin{pmatrix}a & -b\\b & a\end{pmatrix}$
such that the given matrix $A$ has the form $A = PCP^{-1}$
$A=\begin{pmatrix}5 & -2\\1 & 3\end{pmatrix}$
The first thing i tried to do was to find the eigenvectors of matrix $A$ and i got these vectors (which i glued together to get matrix $P$ and $P^{-1}$)
$P=\begin{pmatrix}1+ i& 1-i\\1 & 1\end{pmatrix}$
$P^{-1}=\begin{pmatrix}\frac{1}{2i} & \frac{-1+i}{2i}\\-\frac{1}{2i} & \frac{1+i}{2i}\end{pmatrix}$
Im not sure how to find the matrix $C$, i thought at first i could plug in the eigenvalues in the $C$ matrix, but i don't think that is what they problem i asking me to do.
Any help will be appreciated
| To find $P$ and $C$, note that
$$A = PCP^{-1}\iff AP=PC$$
since A and C are similar we have that
*
*$Tr(A)=Tr(C) \implies 2a=8 \implies a=4$
*$\det(A)=\det(C) \implies a^2+b^2=17 \implies b=\pm1$
then let $P=[v_1\, v_2]$ and we have
*
*$Av_1=av_1+bv_2$
*$Av_2=-bv_1+av_2$
and with $v_1=(x,y)\quad v_2=(z,w)$ we have for $b=1$
*
*$5x-2y=4x+z\implies x-2y-z=0$
*$x+3y=4y+w\implies x-y-w=0$
*$5z-2w=-x+4z\implies x+z-2w=0$
*$z+3w=-y+4w\implies y+z-w=0$
and we find $v_1=(2,1)\, v_2=(0,1)$ and finally
$$C=\begin{pmatrix}4 & -1\\1 & 4\end{pmatrix}\quad P=\begin{pmatrix}2 & 0\\1 & 1\end{pmatrix}\quad P^{-1}=\begin{pmatrix}\frac12 & 0\\-\frac12 & 1\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2740287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Inconsistency for solving $x' = x^{1/2}$ The proposed system $x' = x^{1/2}$ can be solved easily to obtain $x(t) = \frac{1}{4} (t^2 + t c + c^2)$, where $c$ is the integration constant.
However, differentiate the newly-found $x(t)$, one gets: $x(t)' = \frac{1}{2}t+\frac{1}{4}c$. This implies that $x^{1/2} = \frac{1}{2}t+\frac{1}{4}c$. However,
$$(x^{1/2})^2 = \left(\frac{1}{2}t+\frac{1}{4}c\right)^2 = \frac{1}{4}\left(t^2 + \frac{1}{2}tc + c^2\right) \neq \frac{1}{4} \left(t^2 + t c + c^2\right) = x(t)$$
Does anyone know why the inconsistency occurs? I understand there is another solution, but it is also inconsistent.
| You can do it as follows:
$\frac{dx}{dt}$ = $\sqrt{x}$. Rewrite this as follows:
$\frac{dx}{\sqrt{x}}$ = dt.
Let u = $\sqrt{x}$, then du = $\frac{dx}{2\sqrt{x}}$. After this, you'll get 2du = $\frac{dx}{\sqrt{x}}$ = dt, and an indefinite integration gives u = $\frac{1}{2}$(t + c) = $\sqrt{x}$, with c $\in$ $\mathbb{R}$. Finally, x = $\frac{1}{4}$(t+c)^2 = x(t). Take x', and you will see that x' = $\sqrt{x}$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\lim_{x\rightarrow \frac{\pi }{3}}\frac {\sin x-\sqrt {3}\cos x}{\sin 3x}$ has two different values? Given a limit like:
$$\lim_{x\rightarrow \frac{\pi }{3}}\frac {\sin x-\sqrt {3}\cos x}{\sin 3x}$$
How did I solve it:
$$\begin{align}\lim_{x\rightarrow \frac{\pi }{3}}\frac {-2 (-\frac {1}{2}\sin x+\frac {\sqrt {3}}{2}\cos x)}{\sin 3x} &= \lim_{x\rightarrow \frac{\pi }{3}}\frac {-2 (-\sin\frac {\pi}{6}\sin x+\cos \frac {\pi}{6}\cos x)}{\sin 3x}\\&= \lim_{x\rightarrow \frac{\pi }{3}}\frac {-2\cos (\frac{\pi }{6}+x)}{\sin 3x}\\&= \lim_{x\rightarrow \frac{\pi }{3}}\frac {-2\cos (\frac{\pi }{6}+x)}{\sin 3x}\\&= \lim_{x\rightarrow \frac{\pi }{3}}\frac {-2\cos \left[\frac{\pi }{2}-(\frac{\pi}{3}-x)\right]}{\sin 3x}\\&= \lim_{x\rightarrow \frac{\pi }{3}}\frac {-2\sin (\frac{\pi}{3}-x)}{\sin 3x}\\&= \lim_{t\rightarrow 0}\frac {-2\sin t}{\sin \left[3 (\frac{\pi}{3}-t)\right]}\\&= \lim_{t\rightarrow 0}\frac {-2\sin t}{-\sin 3t}\\&= \frac {2}{3}\end{align}$$
I don't know if this is correct but Wolfram Alpha points out it's $-\frac {2}{3}$ instead (L'hopital Rules).
Can anyone show me if there's any error above? Or the limit really has two answers?
Thanks in advance.
|
Can anyone show me if there's any error above?
Your mistake:
$$
\sin 3 (\frac{\pi}{3}-t)=\color{red}{-}\sin 3t
$$ instead of
$$
\sin\left[ 3 \left(\frac{\pi}{3}-t\right)\right]=\sin 3t.
$$ (one may recall that $\cos (3\pi)=-1$)
| {
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How to find the equation of the tangent to the parabola $y = x^2$ at the point (-2, 4)? This question is from George Simmons' Calc with Analytic Geometry. This is how I solved it, but I can't find the two points that satisfy this equation:
$$
\begin{align}
\text{At Point P(-2,4):} \hspace{30pt} y &= x^2 \\
\frac{dy}{dx} &= 2x^{2-1} \\
&= 2x = \text{Slope at P.}
\end{align}
$$
Now, the equation for any straight line is also satisfied for the tangent:
$$
\begin{align}
y - y_0 &= m(x - x_0) \\
\implies y - y_0 &= 2x (x - x_0) \\
\text{For point P, } x_0 &= -2 \text{ and } y_0 = 4 \\
\implies y - 4 &= 2x(x+2)\\
\implies y - 4 &= 2x^2 + 4x\\
\implies y &= 2x^2 + 4x +4\\
\end{align}
$$
This is where the problem occurs. If I were to try to solve for $y$ using:
$$
y = ax^2+bx+c \implies y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$
I'd get:
$$
\begin{align}
y &= 2x^2+4x+4 \text{ and, at x-intercept: }\\
x &= \frac{-4 \pm \sqrt{4^2 - (4\times2\times4)}}{2\times2} \\
x &= \frac{-4 \pm \sqrt{16 - 32}}{4} \\
x &= \frac{-4 \pm 4i}{4} \\
x &= -1 \pm i
\end{align}
$$
Is this the correct direction, or did I do something wrong?
| The derivative of the function $f(x)$ is the slope of the tangent line at that particular value of $x$. Meaning when you differentiate $f(x)=x^2$, the tangent line at that value is at $x=-2$ so$$f'(2)=-4$$Therefore you're tangent line is actually
$$y-4=-4(x+2)$$
| {
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how to compute Z^n? assuming $Z$ is a complex number i.e $a + bi$. we have
$Z^2 = (a + bi)^2 = (a^2-b^2)+(2ab)i$
$Z^3 = (a+bi)^3 = (a^3-3ab^2) - (b^3-3a^2b)i$
but what would be the general formula for computing $Z^n$ where n is any real number?
I wanted to use this method but i dont know how the computation goes for complex number $i$. also I want to do this computation on a computer so i have to keep real and imaginary parts separated from each other.
I know that i can only approximate numbers when $n$ is not an integer. but thats ok.
Is this even possible?
| If, when you say "I want to use this method", you mean that you really want to use the binomial expansion, rather than using polar form, then $(a+ ib)^n$ will be a sum of monomials of the form $\frac{a^(n-i)(ib)^i}{i!}$. "i" has "period 4". That is $i^1= i$, $i^2= -1$, $i^3= -1$, and $i^4= 1$ so that further powers are just repeats of that. You can separate that into one real part which will have powers that are multiples of 4 another subtracted with powers that are two more than multiples of 4, an imaginary part of two parts, the first with powers that are 1 more than multiples of 4, the second, subtracted with powers that are three more than powers of 4 : $\left(a^n+ \frac{a^{n-4}b^4}{4!}+ \frac{a^{n-8}b^8}{8!}+ \cdot\cdot\cdot+\frac{a^{n-4k}b^{4k}}{4k!}+ \cdot\cdot\cdot\right)- \left(a^{n-2}b^2+ \frac{a^{n- 6}b^6}{6!}+ \frac{a^{n-10}b^{10}}{10!}+ \cdot\cdot\cdot+\frac{a^{n-4k- 2}b^{4k+2}}{(4k+2)!}+ \cdot\cdot\cdot\right)+ i\left(a^{n-1}+ \frac{a^{n-5}b^5}{5!}+ \frac{a^{n-9}b^9}{9!}+ \cdot\cdot\cdot+\frac{a^{n-4k-1}b^{4k+1}}{(4k+1)!}+ \cdot\cdot\cdot\right)- i\left(a^n+ \frac{a^{n-4}b^4}{4!}+ \frac{a^{n-8}b^8}{8!}+ \cdot\cdot\cdot+\frac{a^{n-4k-3}b^{4k+3}}{(4k+3)!}+ \cdot\cdot\cdot\right)$.
| {
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if $d\mid n$ then $x^d-1\mid x^n-1$ proof How would you show that if $d\mid n$ then $x^d-1\mid x^n-1$ ?
My attempt :
$dq=n$ for some $q$. $$ 1+x+\cdots+x^{d-1}\mid 1+x+\cdots+x^{n-1} \tag 1$$ in fact, $$(1+x^d+x^{2d}+\cdots+x^{(q-1)d=n-d})\cdot(1+x+\cdots+x^{d-1}) = 1+x+x^2 + \cdots + x^{n-1}$$
By multiplying both sides of $(1)$ by $(x-1)$ we get that $1-x^d\mid 1-x^n$ which is the final result
Is this an ok proof?
| Let $\alpha$ be a solution to the equation $x^d = 1$. We then have that $\alpha^d$ = 1. Since $d|n$ we can write $n = d\cdot k$ for some integer $k$. Thus $$1 = 1^k = \left( \alpha^d \right)^k = \alpha^{d\cdot k} = \alpha^n$$
This shows that $\alpha$ is a solution to $x^n = 1$.
This shows that $x^d-1|x^n-1$.
| {
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A series of Wallis-like infinite products of fractions Would like to derive an analytic solution to the following two infinite products:
$$P_1=\frac{4}{5}\frac{8}{7}\frac{10}{11}\frac{14}{13}\frac{16}{17}...$$
$$=\prod_{n=1}^\infty \frac{3+6n+(-1)^n} {3+6n-(-1)^n}$$
and
$$P_2=\frac{8}{7}\frac{13}{14}\frac{20}{19}\frac{25}{26}\frac{32}{31}...$$
$$=\prod_{n=1}^\infty\frac{3+12n-(-1)^n} {3+12n+(-1)^n}$$
These are similar to the Willis product of
$$P_W=\frac{2}{1}\frac{2}{3}\frac{4}{3}\frac{4}{5}\frac{6}{5}...$$
$$=\prod_{n=0}^\infty\frac{3+2n+(-1)^n} {3+2n-(-1)^n}$$
featured on the latest episode of 3Blue1Brown which is known to converge to $\frac{\pi}{2}$.
Doing these numerical by hand, $P_1$ converges to around 0.87 and $P_2$ to around 1.09. Are the real convergent values irrational (thinking yes) and if so, are they multiples of powers of $\pi$?
Also, are the solutions extensible/describable via parametrization to a series?
| These products can be computed through standard Weierstrass products. For instance
$$ \prod_{n=1}^{2N}\frac{3+6n+(-1)^n}{3+6n-(-1)^n} = \prod_{n=1}^{N}\frac{4+12n}{2+12n}\cdot\frac{-4+12n}{-2+12n}=\frac{\sqrt{3}\,\Gamma\left(N+\tfrac{2}{3}\right)\,\Gamma\left(N+\tfrac{4}{3}\right)}{2\,\Gamma\left(N+\tfrac{5}{6}\right)\,\Gamma\left(N+\tfrac{7}{6}\right)} $$
converges to $\frac{\sqrt{3}}{2}$ as $N\to +\infty$, and
$$ \prod_{n=1}^{2N}\frac{3+12n+(-1)^n}{3+12n-(-1)^n} = \prod_{n=1}^{N}\frac{4+24n}{2+24n}\cdot\frac{-10+24n}{-8+24n}$$
converges to $\frac{\sqrt{\pi}\,\Gamma\left(\frac{2}{3}\right)}{2^{1/6}\,\Gamma\left(\frac{7}{12}\right)^2}$. TLDR: you only need the Weierstrass product for the $\Gamma$ function to manage those products.
| {
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Minimum length of the hypotenuse
A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Prove that the minimum length of the hypotenuse is $(a^{2/3}+b^{2/3})^{3/2}$.
My Attempt
$\frac{x}{y}=\frac{a}{CM}=\frac{AN}{b}$
$$
\frac{x}{y}=\frac{AN}{b}\implies y=\frac{xb}{\sqrt{x^2-a^2}}
$$
$$
h(x)=x+y=x+\frac{xb}{\sqrt{x^2-a^2}}
$$
$$
h'(x)=1+\frac{\sqrt{x^2-a^2}.b-xb.\frac{x}{\sqrt{x^2-a^2}}}{x^2-a^2}=1+\frac{x^2b-a^2b-x^2b}{(x^2-a^2)^{3/2}}\\
=1+\frac{-a^2b}{(x^2-a^2)^{3/2}}=\frac{(x^2-a^2)^{3/2}-a^2b}{(x^2-a^2)^{3/2}}
$$
$$
h'(x)=0\implies (x^2-a^2)^{3/2}=a^2b\implies (x^2-a^2)^{3}=a^4b^2\\
\implies x^6-3x^4a^2+3x^2a^4-a^6=a^4b^2\implies x^6-3x^4a^2+3x^2a^4-a^6-a^4b^2=0\\
$$
How do I proceed further and find $h_{min}$ without using trigonometry ? Or is there anything wrong with my calculation ?
| Following your solution from this point:
$$(x^2-a^2)^{3/2}=a^2b \Rightarrow x^2=(a^2b)^{2/3}+a^2 \Rightarrow x=a^{2/3}(a^{2/3}+b^{2/3})^{1/2}.$$
So:
$$y=\frac{xb}{\sqrt{x^2-a^2}}=\frac{a^{2/3}(a^{2/3}+b^{2/3})^{1/2}b}{\sqrt{((a^2b)^{2/3}+a^2)-a^2}}=(a^{2/3}+b^{2/3})^{1/2}b^{2/3}.$$
Hence:
$$x+y=a^{2/3}(a^{2/3}+b^{2/3})^{1/2}+(a^{2/3}+b^{2/3})^{1/2}b^{2/3}=\\
(a^{2/3}+b^{2/3})^{1/2}(a^{2/3}+b^{2/3})=(a^{2/3}+b^{2/3})^{3/2}$$
| {
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If $\frac{\cos^4 \alpha}{x}+\frac{\sin^4 \alpha}{y}=\frac{1}{x+y}$,prove that $\frac{dy}{dx}=\tan^2\alpha$ If $\frac{\cos^4 \alpha}{x}+\frac{\sin^4 \alpha}{y}=\frac{1}{x+y}$,prove that $\frac{dy}{dx}=\tan^2\alpha$
It is very long to direct differentiate it.Can someone help me?
| Denote $a=\sin \alpha$ and $b=\cos \alpha$. Then:
$$xy=(x+y)(xa^4+yb^4) \qquad (1) \Rightarrow \\
xy=x^2a^4+y^2b^4+xy(b^4+a^4) \Rightarrow \\
xy=x^2a^2(1-b^2)+y^2b^2(1-a^2)+xy(1-2a^2b^2) \overbrace{\Rightarrow}^{:a^2b^2} \\
(x+y)^2-\frac{x^2}{b^2}-\frac{y^2}{a^2}=0 \Rightarrow F(x,y)=0.$$
Take the derivative:
$$y'=-\frac{F_x}{F_y}=-\frac{2(x+y)-\frac{2x}{b^2}}{2(x+y)-\frac{2y}{a^2}}\overbrace{=}^{(1)}-\frac{\frac{xy}{xa^4+yb^4}-\frac{x}{b^2}}{\frac{xy}{xa^4+yb^4}-\frac{y}{a^2}}=\\
-\frac{xyb^2-x^2a^4-xyb^4}{xya^2-xya^4-y^2b^4}\cdot \frac{a^2}{b^2}=\frac{a^2}{b^2},$$
because:
$$-\frac{xyb^2-x^2a^4-xyb^4}{xya^2-xya^4-y^2b^4}=1 \iff \\
xy=(x+y)(xa^4+yb^4) \qquad (1)$$
| {
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Find the length and width of rectangle when you are given the area The area of a rectangle is $x^2 + 4x - 12$. What is the length and width of the rectangle?
The solution says the main idea is to factor $x^2 + 4x -12$.
So, since $-12 = -2 \times 6$ and $-2 + 6 = 4$, it can be written as $x^2 + 4x - 12 = (x - 2)(x + 6)$
since the length is usually the longer value, the length is $6$ and the width is $-2$.
I don't understand the logic to this solution at all. I understand $\text{length} \times \text{width} = \text{area}$, but outside of this information I don't understand how they got to this solution from the given information in the problem.
| The length is $(x+6)$ and the width is $(x-2)$.
The area is $(x+6)(x-2)=x^2+4x-12$
You may take this answer as the length is 6 units longer and the width is 2 units shorter than a give number $x$
For example, given $x=10$, you may get that the area is $10^2+4\times10-12=128$, or, the area is $(10+6)(10-2)=16\times8=128$
| {
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Show that the two straight lines $x^2(\tan^2 (\theta)+\cos^2 (\theta))-2xy\tan (\theta)+y^2.\sin^2 (\theta)=0$ Show that the two straight lines $x^2(\tan^2 (\theta)+\cos^2 (\theta))-2xy\tan (\theta)+y^2.\sin^2 (\theta)=0$ make with x axis such that the difference of their tangents is $2$.
My Attempt:
$$x^2(\tan^2 (\theta) +\cos^2 (\theta))-2xy\tan (\theta) + y^2 \sin^2 (\theta)=0$$
Let $y-m_1x=0$ and $y-m_2x=0$ be the two lines represented by the above equation. Their combined equation is:
$$(y-m_1x)(y-m_2x)=0$$
$$y^2-(m_1+m_2)xy+(m_1m_2)x^2=0$$
How do I proceed further?
| We have $$\left(\dfrac yx\right)^2\sin^2\theta-\dfrac yx(2\tan\theta)+\tan^2\theta+\cos^2\theta=0$$
If $m_1,m_2$ are the two roots
$m_1m_2=\dfrac{\tan^2\theta+\cos^2\theta}{\sin^2\theta}=\dfrac{s^2+c^4}{c^2s^2}$
$m_1+m_2=\dfrac{2s}{cs^2}=\dfrac2{cs}$
$$(m_1-m_2)^2=(m_1+m_2)^2-4m_1m_2=4\cdot\dfrac{1-(s^2+c^4)}{c^2s^2}=\dfrac{4c^2(1-c^2)}{c^2s^2}=4$$
| {
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Nonhomogenous variable coefficient How would you solve this ODE:
$$(1 + x^{2})y’’ + 4xy’ + 2y = 1/(1+x^{2})$$
I have the answer for homogenous part of this ODE but do not know how to do this, variation of parameter is a mess at integral step when applying to this
Any help would be appreciated. Thank you very much :)
| Using variation of parameters is actually not that messy.
Dividing by $(1+x^2)$, we get
$$ y'' + \frac{4x}{x^2+1}y' + \frac{2}{x^2+1} y = \frac{1}{(x^2+1)^2} $$
which has inhomogeneity of $g(x) = \frac{1}{(x^2+1)^2}$. The solutions to the homogeneous equation are $$y_1 = \frac{1}{x^2+1} \ \text{ and } \ y_2 = \frac{x}{x^2+1}$$which we can use to compute the Wronskian, $W = \frac{1}{(x^2+1)^2} \neq 0$. Then, the particular solution is $u_1y_1+u_2y_2$ where $u_1 = -\int \frac{y_2g}{W}dx$ and $u_2 = \int \frac{y_1 g}{W} dx$. The plugging in what we have,
$$ u_1 = -\int \frac{x}{x^2+1}dx = -\frac{1}{2}\log(x^2+1) $$
$$ u_2 = \int \frac{1}{x^2+1} = \tan^{-1}(x)$$
Thus, the solution is
$$ y(x) = \frac{c_1-\frac{1}{2}\log(x^2+1)+c_2x + \tan^{-1}(x)}{x^2+1} $$
| {
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Show that $x^3-a$ is irreducible in $\mathbb{Z}_7$ unless $a=0$ or $\pm1$ Show that $x^3-a$ is irreducible in $\mathbb{Z}_7$ unless $a=0$ or $\pm1$
My Idea:
Suppose $x^3-a=(Ax+b)(Bx^2+cx+d).$
Then $A=B=1$ or $A=B=-1$ WLOG $A=B=1.$
Then $x^3-a=x^3+x^2(b+c)+x(bc+d)+bd\\
\Rightarrow c+b=0,d+bc=0,bd=-a.$
I can't go further from here. Help...Thank You!
| It's a cubic, so you need not work so hard. Indeed, you can just check if it has a root.
By exhaustion: $x^3-a\cong 0\implies x^3\cong a$, so
$$
0^3=0=a\\
1^3=1=a\\
2^3=1=a\\
3^3=-1=a\\
(-3)^3=1=a\\
(-2)^3=-1=a\\
(-1)^3=-1=a
$$
where equality is in the the mod $7$ sense.
| {
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Prove $n$ is prime. (Fermat's little theorem probably)
Let $x$ and $n$ be positive integers such that $1+x+x^2\dots x^{n-1}$ is prime. Prove $n$ is prime
My attempt:
Say the above summation equal to $p$ $$1+x+x^2\dots x^{n-1}\equiv 0\text{(mod p)}\\
{x^n-1\over x-1}\equiv0\\
\implies x^n\equiv1\text{ (as $p$ can't divide $x-1$)}$$
How to proceed?
| This does not involve any Fermat's little theorem type tricks. Indeed, it follows from a very simple factorization trick.
In some sense, if $n = pq$, then we arrange the $n$ powers $x^0,...,x^{pq-1}$ in a $p \times q$ box fashion, and then collect terms.
So, write $n = pq$ for some $1 < p,q < n$ if $n$ is not prime. Then:
$$
1 + ... + x^{pq-1} = (1 + ... + x^{p-1}) + (x^{p} + ... + x^{2p-1}) + ... + (x^{p(q-1)} + ... + x^{pq - 1}) \\ = (1 + ... + x^{p-1}) (1 + x^p + x^{2p} + ... + x^{p(q-1)})
$$
for example, if $n = 6 = 2 \times 3$ then
$$1 + ... + x^5 = (1+x)(1 + x^2 + x^4)$$, and if $n = 9 = 3 \times 3$ then $$1 + ... + x^8 = (1+x+x^2)(1+x^3+x^6)$$
| {
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Area of polygon on complex plane formed by complex roots of a polynomial
Compute the area of the polygon whose vertices are the solutions in the complex plane to the polynomial $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$.
What I did was find the complex numbers by factoring and then using Surveyor's formula. However, that required me to use synthetic or polynomial division. I was wondering if there was a way to find the numbers without directly using division or whether there's a geometric approach to this?
| Simply note that the polynomial is $\frac{x^8-1}{x-1}$ has roots $\operatorname{cis}(2\pi k/8)$ for $k \in \{1, \ldots, 7\}$.
Thus the value is the area of the regular octagon minus the area of a triangle formed by two adjacent sides.
The area of an octagon (by splitting into triangles) with radius $1$ is $8 \cdot \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = 2\sqrt{2}$.
The is a unit square inside an octagon. Removing the square reveals $4$ little triangles. Thus the area of one little triangle is $\frac{2\sqrt{2} - 1}{4} = \frac{\sqrt{2}}{2} - \frac{1}{4}$.
$$2\sqrt{2} - \left(\frac{\sqrt{2}}{2} - \frac{1}{4}\right) = \frac{3\sqrt{2}}{2} + \frac{1}{4}$$
| {
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Range of $k$ for which equation has positive roots
Range of $k$ for which both the roots of the equation $(k-2)x^2+(2k-8)x+3k-17=0$ are positive.
Try: if $\alpha,\beta>0$ be the roots of the equation. Then $$\alpha+\beta=\frac{8-2k}{k-2}>0\Rightarrow k\in(2,4)$$
And $$\frac{3k-17}{k-2}>0\Rightarrow k\in(-\infty,2)\cup \bigg(\frac{17}{3},\infty\bigg)$$
I have got $k=\phi$. I did not understand where i am wrong , please explain, Thanks
| Use quadratic formula. $$(k-2)x^2+(2k-8)x+3k-17=0$$ gives $$x=\frac{-(2k-8)\pm\sqrt{(2k-8)^2-4(k-2)(3k-17)}}{2(k-2)}=-\frac{k-4}{k-2}\pm\frac{\sqrt{-8k^2+60k-72}}{2(k-2)}$$ Hence $$x=-1+\frac2{k-2}\pm\frac{\sqrt{(k-6)(3-2k)}}{k-2}$$ Now we need only find when the negative root is positive since the positive one would then be positive as well.
So $$\frac2{k-2}>1+\frac{\sqrt{(k-6)(3-2k)}}{k-2}\implies \cdots$$
| {
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Computing $\lim_{x\to{0+}}\frac{\tan(x)-x}{x^3}$ without L'Hôpital's rule.
Computing $\lim_{x\to{0+}}\frac{\tan(x)-x}{x^3}$ without L'Hopital
Say $\lim_{x\to{0+}}\frac{\tan(x)-x}{x^3} = L$
For $L$:
$$L=\lim_{x\to0}\frac{\tan x-x}{x^3}\\
L=\lim_{x\to0}\frac{\tan 2x-2x}{8x^3}\\
4L=\lim_{x\to0}\frac{\frac12\tan2x-x}{x^3}\\
3L=\lim_{x\to0}\frac{\frac12\tan{2x}-\tan x}{x^3}\\
=\lim_{x\to0}\frac{\tan x}x\frac{\frac1{1-\tan^2x}-1}{x^2}\\
=\lim_{x\to0}\frac{(\tan x)^3}{x^3}=1\\
\large L=\frac13$$
I found that in another Q, can someone tell me why
$$L=\lim_{x\to0}\frac{\tan x-x}{x^3}=\lim_{x\to0}\frac{\tan 2x-2x}{8x^3}$$
| Let $x=\arctan y.$ We have $y\to 0^+$ as $x\to 0^+.$
For $x\in (0,\pi /2)$ we have $\frac {\tan x-x}{x^3}=$ $\frac {y-\arctan y}{y^3}\frac {y^3}{x^3}=$ $\frac {(y-\arctan y)}{y^3}(\frac {\tan x}{x})^3.$
We have $\lim_{x\to 0^+}(\frac {\tan x }{x})^3=1.$
Consider $g(y)=y-\arctan y.$
We have $g'(y)-y^2=$ $(1-\frac {1}{1+y^2})-y^2=$ $\frac {-y^4}{1+y^2}\in [-y^4,0].$
Therefore $g(y)=g(y)-g(0)=$ $\int_0^y g'(t)dt=$ $y^3/3-\delta \cdot y^5/5$ for some $\delta \in [0,1].$
| {
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prove this nice inequality $\left|\prod_{i=1}^{n}(a_{i}-a_{i+1})\right|\le \frac{3\sqrt{3}}{16}$ Let $n$ be odd number, and $a_{i}\ge 0$, such that
$$2(a_{1}+a_{2}+\cdots+a_{n})=n$$
Show that
$$\left|\prod_{i=1}^{n}(a_{i}-a_{i+1})\right|\le \frac{3\sqrt{3}}{16}$$
where $a_{n+1}=a_{1}$
Seeing this inequality reminds me to use this conclusion to deal with it.
$$S_{AOB}=\frac{1}{2}|a_{1}b_{2}-a_{2}b_{1}|$$
where $A(a_{1},b_{1}),B(a_{2},b_{2}),O(0,0)$
It seems that this problem will involve a problem of maximum geometric area, perhaps it can be handled this way
| Here is a proof for $n=3$, perhaps that helps someone to solve the
general problem.
Let $a, b, c$ be non-negative real numbers with $a+b+c = \frac 32$.
Then
$$ f(a, b, c) = \vert (a-b)(b-c)(c-a) \vert \le \frac{3 \sqrt 3}{16} \, .
$$
Equality holds if and only if $(a, b, c)$ is a permutation of
$$
(0, \frac{3-\sqrt 3}{4}, \frac{3+\sqrt 3}{4} ) \, .
$$
Proof: Without loss of generality we can assume that $a \le b \le c$.
If $a > 0$ then we can replace $(a, b, c)$ with
$$
(a', b', c') = (0, b + \frac\delta 2, c + \frac\delta 2)
$$
where $\delta = \frac 32 - b - c$. Then $a'+b'+c' = \frac 32$ and
$$
f(a, b, c) = (b-a)(c-b)(c-a) < b'(c'-b')c' = f(a', b', c') \, .
$$
Therefore we can assume that $a=0$ and it remains to
maximize $f(0, b, c)$
for $0 \le b \le c, b+c = \frac 32$. It is convenient to set
$$
b = \frac 34 - x \, , \quad c = \frac 34 + x \, , \quad 0 \le x \le \frac 34 \, .
$$
Then
$$
f(0, b, c) = b (c-b) c = 2 x \left( \frac {9}{16} - x^2 \right) =: \phi(x) \, .
$$
An elementary analysis shows that $\phi$ has a unique maximum on $[0, \frac 34]$, it is attained at $x_0 = \frac{\sqrt 3}{4}$, and
$\phi(x_0) = \frac{3 \sqrt 3}{16}$.
This concludes the proof.
Remark: For arbitrary odd $n \ge 3$ the upper bound
$\frac{3 \sqrt 3}{16}$ is best possible. For $n=3$ that has been
demonstrated above. For odd $n \ge 5$ equality holds for
$$
(a_1, \ldots, a_n) = (0, \frac{3-\sqrt 3}{4}, \frac{3+\sqrt 3}{4}, 0, 1, \ldots, 0, 1) \, .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2761985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Show that if $2x + 4y = 1$ where x and y are real numbers. Show that if $2x + 4y = 1$ where $$x, y \in \mathbb R $$, then $$x^2+y^2\ge \frac{1}{20}$$
I did this exercise using the Cauchy inequality, I do not know if I did it correctly, so I decided to publish it to see my mistakes. Thank you!
If $x=2$ and $y=4$, then $$(2^2+4^2)(x^2+y^2)\ge (2x+4y)^2$$ iff $$\frac{x}{2}=\frac{y}{4}$$, then $$20\ge(2x+4y)^2$$
$$-4.47\le2x+4y\le4.47$$
| You have one big confusion here. It is always true:$$(2^2+4^2)(x^2+y^2)\ge (2x+4y)^2$$
with equality iff $\frac{x}{2}=\frac{y}{4}$. Now, since $2x+4y=1$ we get $$20(x^2+y^2)\ge 1$$
so $$x^2+y^2\ge \frac{1}{20}$$
and you are done.
Now if you are interested when eqaulity ocurres then just plug $y=2x$ in $2x+4y=1$ and you get $y=1/10$ and $x=1/5$.
| {
"language": "en",
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"source": "stackexchange",
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Tangent lines to the curve parameterized by $x = a \cos^4t$, $y = a \sin^4t$ The parametric equations of a curve are
$$x = a \cos^4t \qquad y = a \sin^4t$$
where $a$ is a positive constant.
(i) Express $\dfrac{dy}{dx}$ in terms of $t$. (3)
(ii) Show that the equation of the tangent to the curve at the point with parameter $t$ is
$$x \sin^2t + y \cos^2t = a \sin^2t \cos2t \qquad (3)$$
(iii) Hence show that if the tangent meets the $x$-axis at $P$ and the $y$-axis at $Q$, then
$$|OP| + |OQ| = a$$
where $O$ is the origin. (2)
My answer to the first part is $-\tan^3t \cot t$.
I can't figure out the second part.
| $$x = a \cos^4t \qquad y = a \sin^4t$$
This is what I got.
$$\dfrac{dy}{dx}
=\dfrac{4a \sin^3(t) \cos(t)}{-4a \cos^3(t) \sin(t)}
=-\dfrac{\sin^2(t)}{\cos^2(t)}\tag{i}$$
\begin{align}
y-a\sin^4(t) &= -\dfrac{\sin^2(t)}{\cos^2(t)}(x-a\cos^4(t)) \\
y\cos^2(t) - a\sin^4(t)\cos^2(t) &= -x\sin^2(t) + a\cos^4(t)\sin^2(t)\\
x\sin^2(t) + y\cos^2(t) &=a\sin^4(t)\cos^2(t)+a\cos^4(t)\sin^2(t)\\
x\sin^2(t) + y\cos^2(t) &=a\sin^2(t)\cos^2(t)[\sin^2(t)+\cos^2(t)]\\
x\sin^2(t) + y\cos^2(t) &=a\sin^2(t)\cos^2(t)\\
\dfrac{x}{a\cos^2(t)}+\dfrac{y}{a\sin^2(t)} &= 1\tag{ii}
\end{align}
Hence
the $x$-intercept is $P=(a\cos^2(t),0)$ and
the $y$-intercept is $Q=(0,a\sin^2(t))$. So
$$OP+OQ = |a|\cos^2(t)+|a|\sin^2(t)=|a|\tag{iii}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Sequence and series with specific nth term Let $a(1) = 2$ , $a(n+1) = a(n)^2-a(n) + 1$ for $n\geq 1$, Find $$\sum_{n=1}^{\infty} \frac{1}{a(n)}$$
| We first prove the following by induction:
$$
\frac{1}{a(n + 1)-1} + \sum_{k = 1}^n \frac{1}{a(k)} = 1.
$$
For $n = 1$, there stands
$$
\frac{1}{4 - 2 + 1 - 1} + \frac{1}{2} = 1,
$$
which holds. Suppose it holds for $n$. Then,
\begin{align*}
\frac{1}{a(n + 2) - 1} + \sum_{k = 1}^{n + 1} \frac{1}{a(k)} &= \frac{1}{a(n + 2) - 1} + \frac{1}{a(n + 1)} - \frac{1}{a(n + 1) - 1} + \frac{1}{a(n + 1) - 1} + \sum_{k = 1}^n \frac{1}{a(k)} \\
&= \frac{1}{a(n + 1)^2 - a(n + 1) + 1 - 1} + \frac{a(n + 1) - 1 - a(n + 1)}{a(n + 1)(a(n + 1) - 1)} + 1 \\
&= 1.
\end{align*}
Consequently,
$$
\sum_{k = 1}^\infty \frac{1}{a(k)} = \lim_{n \to \infty} \sum_{k = 1}^n \frac{1}{a(k)} = \lim_{n \to \infty} \left(1 - \frac{1}{a(n + 1) - 1}\right) = 1,
$$
since $a(n) \to \infty$ for $n \to \infty$.
EDIT: A little bit of explanation as to how I came up with the first step. I encountered this problem a while ago in a pdf on problem solving, but there was an extra part: it asked to show that $a(n)$ and $a(m)$ are coprime if $n \ne m$. To show that, it naturally came up to look at $a(n) - 1$. That was the first part of the problem, the second part was finding the series of the reciprocals. While trying to solve it, I calculated $\frac{1}{a(n)} + \frac{1}{a(n + 1)}$. When that did not give anything useful, I tried doing something with $\frac{1}{a(n + 1) - 1}$, as this kind of term was used in the first part.
Hopefully this gives a bit of an idea how I came up with my solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What are the local extreme values of $f(x,y)=x^3+x^2y-y^2-4y$? $f'_x=3x^2+2xy$,
$f'_y=x^2-2y-4$
so I have to solve the equation system of
$3x^2+2xy=0,x^2-2y-4=0$ as solutions I get that $((x=0),(y=0)),((x=-4),(y=6)),(x=1),((y=-\frac{3}{2}))$
After that:
$f''_x=6x+2y$,
$f''_y=-2$,
$f''_{xy}=2x=f''_{yx}$
so I get the determinant
$\begin{vmatrix}
6x+2y & 2x \\
2x & -2
\end{vmatrix}$
for $(x=4),(y=6)$
$\begin{vmatrix}
6 \cdot -4+2 \cdot6 & 2 \cdot -4 \\
2 \cdot -4 & -2
\end{vmatrix}$=$-40$ there is at this point no local extreme value
for $(x=1),(y=-\frac{3}{2})$
$\begin{vmatrix}
6 \cdot 1+2 \cdot-\frac{3}{2} & 2 \\
2 & -2
\end{vmatrix}$=$-10$ there is at this point no local extreme value
but what about $(x=0)(y=0)$? the value of the determinant will be $0$but it says nothing about the extreme value at this point.
Is my solution correct til this point?
| The stationary points are determined as the solutions for
$$
3x^2+2xy = 0\\
x^2-2y-4 = 0
$$
giving the set
$\{(-4,6),(0,-2),(1,-3/2)\}$
Now taking the Hessian
$$
H = \left(\begin{array}{cc}3x+y & x\\ x & -1\end{array}\right)
$$
To qualify the stationary points we should evaluate $H$ in such set, verifying it's eigenvalues. For both eigenvalues negative we have a local maximum. For both eigenvalues positive we have a local minimum and for eigenvalues with opposite sign we have a saddle point. In this case the point $(0,2)$ gives negative eigenvalues for $H$ so here we have a local maximum.
Attached a level contour plot showing the local maximum (red) and the two saddle points (blue)
| {
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"question_score": "1",
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} |
Calculating a determinant. $D_n$=\begin{vmatrix}
a & 0 & 0 & \cdots &0&0& n-1 \\
0 & a & 0 & \cdots &0&0& n-2\\
0 & 0 & a & \ddots &0&0& n-3 \\
\vdots & \vdots & \ddots & \ddots & \ddots&\vdots&\vdots \\
\vdots & \vdots & \vdots & \ddots & \ddots& 0&2 \\
0 & \cdots & \cdots & \cdots &0&a&1 \\
n-1 & n-2 & n-3 & \cdots & 2 & 1& a\\
\end{vmatrix}
I tried getting the eigenvalues for A =
\begin{vmatrix}
0 & 0 & 0 & \cdots &0&0& n-1 \\
0 & 0 & 0 & \cdots &0&0& n-2\\
0 & 0 & 0 & \ddots &0&0& n-3 \\
\vdots & \vdots & \ddots & \ddots & \ddots&\vdots&\vdots \\
\vdots & \vdots & \vdots & \ddots & \ddots& 0&2 \\
0 & \cdots & \cdots & \cdots &0&0&1 \\
n-1 & n-2 & n-3 & \cdots & 2 & 1& 0\\
\end{vmatrix}
For $a=0$ , the rank of the matrix is $2$ , hence $\dim(\ker(A)) = n-2 $
$m(0)>=n-2$
However, I was not able to determine the other eigenvalues.
Testing for different values of n :
for $n=2$ :
$D_2 = a^2-1$
for $n=3$ :
$D_3 = a^3 -5a$
$D_n$ seems to be equal to $a^n - a^{n-2}\sum_{i=1}^{n-1}i^2$ .
However I'm aware that testing for different values of $n$ is not enough to generalize the formula.
Thanks in advance.
| Develop with respect to the first column. Then
$$
\begin{aligned}
D_n &= aD_{n-1} -(n-1)\cdot (n-1)\cdot a^{n-2}
\\
&=aD_{n-1}-(n-1)^2a^{n-2}\ .
\end{aligned}
$$
This recursion, together with $D_1=a$ gives for $n\ge 2$ the solution
$$
D_n= (a^2-(1^2+2^2+\dots+(n-1)^2)a^{n-2}\ .
$$
(The sum in the first factor has a closed formula.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2765475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Question about irreducible polynomials over a finite field. If a polynomial $f(x)$ is irreducible over a finite field, does that mean the only factors are $\{1, f(x)\}$?
How would I go about proving a polynomial $f(x)$ is irreducible over a finite field? A bit of searching on StackExchange showed me this:
Irreducibility criterion: A polynomial $P\in\mathbf F_q[X]$ with degree $n$ is irreducible if and only if
*
*$P$ divides $X^{q^n}-X$;
*$P$ is coprime with all $X^{q^r}-X$, $\;r=\dfrac nd$, where $d$ is a prime divisor of $n$.
How would I apply this to $f(x) = x^8 + x^4 + x^3 + x + 1$ over $GF(2^8)$?
| You don't need a CAS for this low degree polynomial
$$f(x)=x^8+x^4+x^3+x+1.$$
A pencil & paper solution follows:
First we calculate the remainders $r_k(x)$ of the monomials $x^{2k}, 0\le k\le7$, modulo $f(x)$. These are:
$$
\begin{array}{c|c}
k& r_k\\
\hline
0&1\\
1&x^2\\
2&x^4\\
3&x^6\\
4&x^4+x^3+x+1\\
5&x^6+x^5+x^3+x^2\\
6&x^7+x^5+x^3+x+1\\
7&x^7+x^4+x^3+x
\end{array}
$$
Producing this table is easy. You get $r_{k+1}(x)$ as the remainder of $x^2r_k(x)$ modulo $f(x)$. You only need to do long division, when $x^2r_k(x)$ has degree $\ge8$.
With this table at hand for referrals we can then easily calculate the remainders $p_\ell(x)$ of $x^{2^\ell}$ modulo $f(x)$. This task is tailor-made for exponentiation by squaring. Clearly we get $p_{\ell+1}(x)$ as the remainder of $p_{\ell(x)}^2$ modulo $f(x)$. Remember that, by Freshman's dream in characteristic two, we can square a polynomial from $GF(2)[x]$ term-by-term. When calculating $p_{\ell+1}(x)$ a term $x^n$ in $p_\ell(x)$ produces a term $r_k\equiv x^{2k}$:
$$
\begin{aligned}
p_0(x)&\equiv&\equiv & x,\\
p_1(x)&\equiv p_0(x)^2\equiv r_1&\equiv &x^2,\\
p_2(x)&\equiv p_1(x)^2\equiv r_2&\equiv &x^4,\\
p_3(x)&\equiv p_2(x)^2\equiv r_4 &\equiv &x^4+x^3+x+1,\\
p_4(x)&\equiv p_3(x)^2\equiv r_4+r_3+r_1+r_0&\equiv&x^6+x^4+x^3+x^2+x,\\
p_5(x)&\equiv p_4(x)^2\equiv r_6+r_4+r_3+r_2+r_1&\equiv&x^7+x^6+x^5+x^2,\\
p_6(x)&\equiv p_5(x)^2\equiv r_7+r_6+r_5+r_2&\equiv&x^6+x^3+x^2+1,\\
p_7(x)&\equiv p_6(x)^2\equiv r_6+r_3+r_2+r_0&\equiv&x^7+x^6+x^5+x^4+x^3+x,\\
p_8(x)&\equiv p_7(x)^2\equiv r_7+r_6+r_5+r_4+r_3+r_1&\equiv&x.
\end{aligned}
$$
From this second table we see that
*
*The remainder of $x^{16}-x$ modulo $f(x)$ is $p_4(x)-x\neq0$ meaning that $f(x)$ is not a factor of $x^{16}-x$.
*The remainder of $x^{256}-x$ modulo $f(x)$ is $p_8(x)-x=0$ meaning that $f(x)$ is a factor of $x^{256}-x$.
Therefore we are done.
| {
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"source": "stackexchange",
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Find vector $x \not= 0$ that satisfies the equation $Ax = x$. Given a matrix A:
$$A = \begin{bmatrix}
0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \\
\frac{1}{3} & 0 & 1 & 0 & 0 \\
\frac{1}{3} & 0 & 0 & 0 & \frac{1}{2} \\
\frac{1}{3} & \frac{1}{2} & 0 & 0 & 0 \\
0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 \\
\end{bmatrix}$$
I have to find a vector $x \not= 0$ that satisfies the equation $Ax = x$.
I have no idea how to approach this. Any hints would be appreciated.
Thanks in advance.
| You want an $x\in\mathrm{Null}(A-I)$. Note that each column of the matrix $A-I$ adds up to $0$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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The shortest distance from a point to the graph of the function To compute the distance from the point (5,5) to the graph of xy=4. I choose an arbitrary point (u,v) on the graph of $xy=4$.
I get $d(u,v)=\sqrt{(u-5)^2+(v-5)^2}$ again $(u,v)$ satisfies equation of hyperbola so that $uv=4$. Now what shall i do next?
| You're after the minimum of $f(x)=(x-5)^2+\left(\frac4x-5\right)^2$, with $x\in\mathbb R$. Now,\begin{align}f'(x)&=-10-\frac{32}{x^3}+\frac{40}{x^2}+2x\\&=2\frac{x^4-5x^3+20x-16}{x^3}\\&=2\frac{(x^2-4)(x^2-5x+4)}{x^3}.\end{align}So, see what happens at the roots of $f'(x)$, which are $\pm2$, $1$ and $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2767570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving that this sequence is convergent Given $a_1 = 2$, and $ a_{n+1} = \frac{a_n+5}{4} $ for all $n > 1$ , is this sequence convergent? Give a formal proof in either case (converges or diverges).
Attempt: I do think this converges, but cannot say for sure.
$a_1 = 2$
$a_2 = \frac{7}{4}$
$a_3 = \frac{27}{16}$
$a_4 = \frac{107}{64}$
and so on. I can see that it is decreasing and seems to be bounded below by something. But I do not know how to present it formally.
Any help?
| Let $x \in \mathbb R$. If $x \gt 5/3$ then we always have
$\tag 1 \frac{5}{3} \lt \frac{x + 5}{4} \lt x$
Since our sequence $a_n$ is strictly decreasing and always greater than $5/3$, it has a limit which we denote by $\alpha$.
For the sequence to converge, $(a_{n+1} - a_n)$ must converge to zero, and
$\quad a_{n+1} - a_n = \frac{a_n + 5}{4} - a_n = \frac{-3 a_n + 5}{4} $
and applying the limit operation,
$\quad \lim_{n\to +\infty} \frac{-3 a_n + 5}{4} = \frac{-3 \alpha + 5}{4} = 0$
Solving we find the limit of the sequence is equal to $5/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2771447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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How is $2^{n-k-1} \cdot 3^k = (3/2)^k \cdot 2^{n-1}$? I'm currently reading through some litterature about difference equations, and i came across a relation i just can't seem to wrap my head around.
How come:
$2^{n-k-1}\cdot 3^k=(\frac{3}{2})^k\cdot 2^{n-1}$?
The original problem in the book is:
(1*) $y(n)=a^ny_0+\sum_{r=0}^{n-1}a^{n-r-1}g(r)$
Find a solution for the equation:
$x(n + 1) = 2x(n) + 3n$ , $x(1) = 0.5$
Solution From (1*), we have
(2) $x(n) = (\frac{1}{2})\cdot2^{n-1}+\sum_{r=1}^{n-1}2^{n-r-1}3^k$
(3) $x(n) = 2^{n-2}+2^{n-1}\sum_{r=1}^{n-1}(\frac{3}{2})^k$
(4) $x(n) = 2^{n-2}+2^{n-1}\frac{3}{2}(\frac{(\frac{3}{2})^{n-1}-1}{\frac{3}{2}-1})$
(5) $x(n) = 3^n -5\cdot2^{n-2}$
Although my question is more directed to the relation $2^{n-k-1}\cdot 3^k=(\frac{3}{2})^k\cdot 2^{n-1}$, a thorough explanation of the steps in the book would be appreciated as well.
I'm truly sorry if this question has been asked before or is in any way irrelevant to the tags, but I'm pretty green to this topic, so the relevance of the tags might be off.
| $$2^{n-k-1}\cdot 3^k=\frac{2^{n-1}}{2^k}\cdot 3^k=2^{n-1}\cdot \frac{3^k}{2^k} =\left(\frac{3}{2}\right)^k\cdot 2^{n-1}$$
| {
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Converting $\frac{(3+3i)^5 (-2+2i)^3}{(\sqrt 3 +i)^{10}}$ to trigonometric form
Convert complex to trig.
$$\frac{(3+3i)^5 (-2+2i)^3}{(\sqrt 3 +i)^{10}}$$
Let us consider
$$(3+3i)^5$$
Here $a = 3$ ,$b=3$
$$\sqrt{a^2+b^2}=\sqrt{3^2+3^2} = \sqrt{18}=3\sqrt{2}$$
$$\theta =\arctan\biggr(\frac{b}{a}\biggr) =\arctan\biggr(\frac{3}{3}\biggr) = \arctan (1)$$
$$\theta = \frac{\pi}{4}$$
Thus in trigonometric form we get
$$Z_1 = \biggr [3\sqrt2 \bigg(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}\bigg)\bigg]^5$$
$$Z_1 =(3\sqrt 2)^5 \biggr [\bigg(\cos \frac{5\pi}{4}+i\sin \frac{5\pi}{4}\bigg)\bigg]$$
Let us consider
$$(-2+2i)^3$$
$a = -2$, $b=2$
$$\sqrt{a^2+b^2}=\sqrt{(-2)^2+2^2} =2\sqrt{2}$$
$$\theta =\arctan\biggr(\frac{b}{a}\biggr) =\arctan\biggr(\frac{2}{-2}\biggr) = \arctan (1)$$
$$\theta = \frac{3\pi}{4}$$
$$Z_2 =(2\sqrt 2)^3 \biggr [\bigg(\cos \frac{9\pi}{4}+i\sin \frac{9\pi}{4}\bigg)\bigg]$$
Multiplying $Z_1 \cdot Z_2$ we get
$$Z_1 \cdot Z_2 = (3\sqrt 2)^5 \cdot (2\sqrt2)^3 \biggr [\bigg(\cos \frac{5\pi}{4}+\cos \frac{9\pi}{4}\bigg)+i\sin \bigg(\frac{5\pi}{4}+i\sin\frac{9\pi}{4}\bigg)\bigg]$$
$$Z_1 \cdot Z_2 = 31104\biggr [\bigg(\cos \frac{7\pi}{2}+i\sin \frac{7\pi}{2}\bigg)\bigg]$$
Let us consider
$$(\sqrt 3+i)^{10}$$
Here $a = \sqrt 3$, $b =1$
$$\sqrt{a^2+b^2}=\sqrt{(\sqrt3)^2+1^2} = \sqrt{4}=2$$
$$\theta = \arctan \biggr(\frac{1}{\sqrt 3}\biggr) = \frac{\pi}{6}$$
$$Z_3 = 2^{10} \biggr [ \cos \biggr(\frac{5\pi}{3}\biggr)+i\sin \biggr(\frac{5\pi}{3}\biggr)\biggr]$$
Now we have
$$\frac{Z_1 \cdot Z_2}{Z_3} = \frac{31104\biggr [\bigg(cos \biggr(\frac{7\pi}{2}\biggr) +i\sin \biggr(\frac{7\pi}{2}\biggr) \bigg]}{2^{10} \biggr [ \cos \biggr(\frac{5\pi}{3}\biggr)+i\sin \biggr(\frac{5\pi}{3}\biggr)\biggr]} = \boxed {30.375 \biggr [ \cos \biggr(\frac{11\pi}{6}\biggr)+i\sin \biggr(\frac{11\pi}{6}\biggr)\biggr]}$$
Is my assumption correct?
| I can't see any mistakes in your answers. Your calculations, however, are sometimes very wrong, not even leading the correct answers you pull out of them.
For $z^3=(-2+2i)^3$, $r=2\sqrt{2}$ as you stated. $\theta=\tan^{-1}{(-1)}=\frac{-\pi}{4}$. The reason the correct result is $\frac{3\pi}{4}$, is due to $z$'s position on the argand diagram, the angle of $\frac{-\pi}{4}$ would create $z=2-2i$, adding $\pi$ rotates the point to $-2+2i$.
When it comes to multiplying complex numbers, note this:
$$\cos A + \cos B \ne \cos (A+B)$$
Instead:
$$(\cos A+i\sin A)(\cos B+i\sin B)=\cos A \cos B-\sin A \sin B + i \sin A \cos B+ i\cos A \sin B)$$
$$=\cos (A+B)+i \sin (A+B)$$
Hence your overall result is correct, but your working is rather flawed.
| {
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"timestamp": "2023-03-29T00:00:00",
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Using de-Moivre's theorem to find the reciprocal of each number
Use de-Moivre's theorem to find the reciprocal of each number below.
$$\sqrt 3 - i$$
Given $\sqrt{3}-i$ , we need to find the reciprocal of it using de-Moivre's theorem.
$$\frac{1}{\sqrt 3-i} $$
$$= \frac{1(\cos0^c + i\sin 0^c)}{2\big(\frac{\sqrt3}{2}-\frac{i}{2}\big)}$$
$$= \frac{1}{2}\cdot \frac{\cos\big(\frac{\pi}{6}\big) - i\sin\big(\frac{\pi}{6})}{\big(\frac{\sqrt3}{2}-\frac{i}{2}\big)}$$
$$ = \frac{1}{2}\cdot \cos\biggr(\frac{\pi}{6}\biggr) - i\sin\biggr(\frac{\pi}{6}\biggr)$$
$$ = \frac{1}{2} \cdot \biggr(\frac{\sqrt3}{2} +i \frac{1}{2}\biggr )$$
$$\boxed { = \frac{\sqrt3}{4}+\frac{i}{4}}$$
Does my assumption seem correct?
| De Moivre's Theorem states that:
$$[r(\cos \theta + i \sin \theta)]^n=r^n(\cos (n\theta)+ i \sin (n\theta))$$
To find the reciprocal, take $n=-1$.
$$z=\sqrt{3}-i\to r=2, \theta=\tan^{-1}\bigg({\frac{-1}{\sqrt{3}}}\bigg)=\frac{-\pi}{6}$$
Hence $z^{-1}=2^{-1}(\cos{(-1)(\frac{-\pi}{6})}+ i\sin{(-1)(\frac{-\pi}{6})})=\frac{1}{2}(\cos \frac{\pi}{6}+i\sin\frac{\pi}{6})$
$$=\frac{\sqrt{3}}{4}+\frac{1}{4}i$$
as you achieved.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find Cauchy principal values of $\int_{-\infty}^{\infty}\frac{x}{(x^2+4)(x^2-2x+5)}\,dx$ I have been asked to find the Cauchy principal vlaues of the following problem using residues:
$\int_{-\infty}^{\infty}\frac{x}{(x^2+4)(x^2-2x+5)}\,dx$
So far I have taken $\oint_C\frac{z}{(z^2+4)(z^2-2z+5)}\,dz$
which gives me:
$2\pi i\sum \operatorname{Res}\biggl[\frac{z}{(z^2+4)(z^2-2z+5)}\biggl]$
Where do I go from here?
| factor the denominator.
$\frac {z}{(z+2i)(z-2i)(z-1+2i)(z-1-2i)}$
We only care about the poles in the upper half-plane.
You have poles at $z = 2i, z = 1+2i$
at $z=2i$
$Res f(z) = \lim_\limits{z\to 2i} (z-2i)f(z) = \frac {2i}{(4i)(-1+4i)(-1)} = \frac {1+4i}{2(17)} = \frac {1}{34} + \frac {2i}{17} $
and at $z = 1+2i$
$Res f(z) = \frac {1+2i}{(1+4i)(1)(4i)} = \frac {(1+2i)(1-4i)}{(17)(4i)} = \frac {9 - 2i}{(17)(4i)} = -\frac{9i}{68} - \frac {1}{34}$
add them together
$2\pi i \left(\frac {-i}{68}\right) = \frac {\pi}{34}$
| {
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show whether $\frac {xy}{x^2+y^2}$ is differentiable in $0$ or not? (multivariable) Q: $f(x,y)=\frac {xy}{x^2+y^2}$ if $(x,y)\not=(0.0)$, and $0$ if $(x,y)=(0,0)$.
Is $f$ differentiable at $(0,0)$?
Attempt: $$\lim_{(x,y) \to (0,0)} \frac {f(x,y)-f(0,0)}{||(x,y)-(0,0)||} = \lim_{(x,y) \to (0,0)} \frac {\frac{xy}{x^2+y^2}}{\frac {\sqrt {x^2+y^2}}{1}}$$
$$=\lim_{(x,y) \to (0,0)} \frac {xy}{(x^2+y^2)^{3/2}}.$$
Let $x=r\cos(\theta)$ and $y=r\sin(\theta)$. Then,
$$ \frac {xy}{(x^2+y^2)^{3/2}}= \frac {r^2\cos(\theta)\sin(\theta)}{r^{3/2}}=\sqrt r\cos(\theta)\sin(\theta).$$
What should be the next step? If I use polar coordinates, how can I transform $$\lim_{(x,y) \to (0,0)}\rightarrow\lim_{(r,\theta) \to (?,?)}$$??
| $$
f\left(\frac{1}{n},\frac{1}{n}\right)=\frac{1}{n^2}\frac{n^2}{2}=\frac{1}{2}
$$
And $\displaystyle \left(\frac{1}{n},\frac{1}{n}\right) \underset{n \rightarrow +\infty}{\rightarrow}(0,0)$, so it is not continuous at $(0,0)$. It cannot be differentiable.
| {
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"timestamp": "2023-03-29T00:00:00",
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USAMO 2018: Show that $2(ab+bc+ca) + 4 \min(a^2,b^2,c^2) \geq a^2 + b^2 + c^2$ Here is question 1 from USAMO 2018 Q1 (held in April):
Let $a,b,c$ be positive real numbers such that $a+b+c = 4 \sqrt[3]{abc}$.
Prove that: $$2(ab+bc+ca) + 4 \min(a^2,b^2,c^2) \geq a^2 + b^2 + c^2$$
This question is on symmetric polynomials. I recall as many facts as I can think of regarding inequalities. (This test was closed book).
*
*AM-GM inequality: $a + b + c \geq 3 \sqrt[3]{abc}$ so the equality there is strange.
*quadratic mean inequality suggests $3(a^2 + b^2 + c^2) > a + b + c $.
*The $\min(a^2,b^2,c^2)$ on the left side makes things difficult since we can't make it smaller.
*I am still looking for other inequalities that might work.
It's tempting to race through this problem with the first solution that comes to mind. I'm especially interested in some kind of organizing principle.
| Since our inequality and condition are symmetric and homogeneous,
we can assume that $abc=1$ and $a\geq b\geq c$.
Thus, $a+b+c=4$ and we need to prove that
$$2(ab+ac+bc)\geq a^2+b^2-3c^2$$ or
$$(a+b+c)^2\geq2(a^2+b^2-c^2)$$ or
$$8\geq a^2+b^2-c^2$$ or $$8\geq a^2+b^2-(4-a-b)^2$$ or
$$8\geq a^2+b^2-(16+a^2+b^2-8a-8b+2ab)$$ or
$$12-4(a+b)+ab\geq0$$ or $$12-4(4-c)+\frac{1}{c}\geq0$$ or
$$(2c-1)^2\geq0.$$
Done!
| {
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Maximizing $3\sin^2 x + 8\sin x\cos x + 9\cos^2 x$. What went wrong?
Let $f(x) = 3\sin^2 x + 8\sin x\cos x + 9\cos^2 x$. For some $x \in \left[0,\frac{\pi}{2}\right]$, $f$ attains its maximum value, $m$. Compute $m + 100 \cos^2 x$.
What I did was rewrite the equation as $f(x)=6\cos^2x+8\sin x\cos x+3$. Then I let $\mathbf{a}=\left<6\cos x,8\cos x\right>$ and $\mathbf{b}=\left<\cos x,\sin x\right>$.
Using Cauchy-Schwarz, I got that the maximum occurs when $\tan x=\frac{4}{3}$, and that the maximum value is $10\cos x$. However, that produces a maximum of $9$ for $f(x)$, instead of the actual answer of $11$.
What did I do wrong, and how do I go about finding the second part? Thanks!
| $f(x)=2\sin^2x+8\sin x\cos x+8\cos^2x+1=2(\sin x+2\cos x)^2+1$
Let $\displaystyle \alpha\in\left[0,\frac{\pi}{2}\right]$ and $\displaystyle\cos\alpha=\frac{2}{\sqrt{5}}$. Then $\displaystyle\sin\alpha=\frac{1}{\sqrt{5}}$ and
$$\sin x+2\cos x=\sqrt{5}(\sin x\sin\alpha+\cos x\cos\alpha)=\sqrt{5}\cos(x-\alpha).$$
attaining its maximum when $x=\alpha$.
So, $m=2(\sqrt{5})^2+1=11$ and $\displaystyle m+100\cos^2\alpha=11+100\left(\frac{4}{5}\right)=91$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inertia Tensor of an ellipsoid Given is the following inertia tensor of a certain mass distribution $\rho(\vec{r})$ :
$$ I_{ij} = \int dV \rho(\vec{r}) \left( \vec{r}^2 \cdot \delta_{ij} - r_ir_j \right) $$
I should compute the inertia tensor for the following ellipsoid:
$$x^2/a^2 + y^2/b^2 + z^2/c^2 \le 1$$
And I'm allowed to use $\rho(\vec{r}) = \rho_0$, and I should then give the result as a function of the mass $M$.
It's clear to me how I parameterize the Ellipsoid and how to compute the Jacobian, but I don't get how I should integrate that expression.
| You can use the parametrization
\begin{eqnarray}
x &=& a r \sin\theta\sin\phi \\
y &=& b r \sin\theta\cos\phi \\
z &=& c r \cos\theta
\end{eqnarray}
with $0<r<1$, $0<\theta<\pi$ and $0<\phi<2\pi$. The Jacobian for this choice of coordinates is $abc r^2\sin\theta$, so the integral becomes
\begin{eqnarray}
I_{ij} &=& \rho_0\int{\rm d}r~{\rm d}\theta~{\rm d}\phi ~ abc r^2\sin\theta [\underbrace{r^2\left(\sin^2\theta\left(a^2\cos^2\phi+b^2\sin^2\phi\right)+c^2
\cos^2\theta\right)}_{x^2+y^2+z^2}\delta_{ij} - x_i x_j] \\
&=& \frac{3M}{4\pi} \int{\rm d}r~{\rm d}\theta~{\rm d}\phi ~ r^2\sin\theta [r^2\left(\sin^2\theta\left(a^2\cos^2\phi+b^2\sin^2\phi\right)+c^2
\cos^2\theta\right)\delta_{ij} - x_i x_j]
\end{eqnarray}
where I used $\rho_0 = 3M/(4\pi a b c)$. Now it is a matter of integrating, for example
\begin{eqnarray}
I_{11} &=& \frac{3M}{4\pi} \int{\rm d}r~{\rm d}\theta~{\rm d}\phi ~ r^2\sin\theta [r^2\left(\sin^2\theta\left(a^2\cos^2\phi+b^2\sin^2\phi\right)+c^2
\cos^2\theta\right) - a br^2\sin^2\theta\sin\phi\cos\phi] \\
&\cdots& \\
&=& \frac{M}{5}(b^2 + c^2)
\end{eqnarray}
This is the result you should get
$$
{\bf I } = \frac{M}{5}\begin{pmatrix}
b^2+ c^2 & 0 & 0 \\
0 & a^2+ c^2 & 0 \\
0 & 0 & a^2+ b^2\\
\end{pmatrix}
$$
| {
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Finding the smallest n satisfying $S_n > 10$ Let $S_n = 1 + \frac 12 + \frac 13 + \cdots + \frac 1n$, where $n \in \{ 1,2,3,\cdots\}$ Find the smallest $n$ satisfying $S_n > 10$.
Sorry, it's my first time asking and I don't know how to format this thing. I still don't see anything even after staring at this for really long. Any clues?
| We have an asympotic expansion for $H_n$:
$$H_n = \ln n + \gamma + \frac{1}{2n} - \sum_{k=1}^{\infty} \frac{B_{2k}}{2 k n^{2k}} = \ln n + \gamma + \frac{1}{2n} -\frac{1}{12 n^2}+ \frac{1}{120 n^4} - \dots$$
Reference : Wikipedia on Harmonic number
So
$$\ln n + \gamma + \frac{1}{2n} -\frac{1}{12 n^2} < H_n < \ln n + \gamma + \frac{1}{2n}$$
Solving $\ln n + \gamma = 10$ for $n$, we find $n \approx 12367$. With this value of $n$, we compute
$$ H_n > \ln n + \gamma + \frac{1}{2n} -\frac{1}{12 n^2} = 10.000043$$
and $$H_{n-1} < \ln (n-1) + \gamma + \frac{1}{2(n-1)} = 9.999962144$$
so $H_{n-1} < 10$ and $H_n > 10$ for $n = 12367$.
| {
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Find all positive integers $x,y$ such that $\frac{x^2-1}{xy+1}$ is a non-negative integer. QUESTION: Find all positive integers $x,y$ such that $\frac{x^2-1}{xy+1}$ is a non-negative integer.
ATTEMPT (AND SOME SOLUTIONS): So, for a positive integer $x$, $x^2-1\ge0$. In the case $x^2-1=0$ i.e $x=1$ (since it's a positive integer), we get
$\frac{x^2-1}{xy+1}=0$ which is a non-negative integer, regardless of $y$. So, one solution is $(x,y)=(1,k)$ where $k$ is a positive integer.
Now, let's say $y=1$. Then $\frac{x^2-1}{xy+1}=\frac{x^2-1}{x+1}=x-1$ which is always a non-negative integer so $(x,y)=(k,1)$ is also a solution. However, I don't know how to find the other or prove that those are the only ones.
| Since $xy+1\mid xy+1$ we have $$xy+1\mid (x^2-1)+(xy+1)= x(x+y)$$
Now $\gcd(xy+1,x)=1$ so $$xy+1\mid x+y\implies xy+1\leq x+y$$
So $(x-1)(y-1)\leq 0$ and thus if $x>1$ and $y>1$ we have no solution.
Ergo $x=1$ or $y=1$...
| {
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Proving $a+a =0$ for Boolean ring
Suppose $R$ is a Boolean ring. Prove that $a+a=0$ for all $a\in R$.
Also prove that $R$ is commutative. Give an example (with explanation)
of a Boolean ring.
From what I know, a Boolean ring is a ring for which $a^2=a$ for all $a\in R$.
Under addition a ring is a commutative group.
$a + b = b + a$ (commutative)
$(a + b) + c = a + (b + c)$ (associative)
$a + (-a) = 0$ (inverse exists for every element)
$a + 0 = a$ (identity exists)
Where $a,b,c \in R$
But I'm not really sure how to proceed with the proof from here. Any idea?
| Take $x \in R$, with $R$ boolean ring, in particular is a ring, therefore $(x+x) \in R$ and $(x+x)^{2}=(x+x)$. Thus
\begin{align*}
(x+x)^{2} &= (x+x)(x+x) \\
&=(x+x)x + (x+x)x \\
&= x^2 + x^2 + x^2 + x^2
\end{align*}
since $x^2 = x$ and $(x+x)^2 = (x+x)$ because $R$ is boolean, then \begin{align*}
(x+x)^{2} &= x+x+x +x\\
&= (x+x) + (x+x)\\
&= (x+x)
\end{align*}
and finally we add both sides the right inverse of $-(x+x)$.
\begin{align*}
(x+x) + (x+x) -(x+x) &= (x+x) - (x+x)
\end{align*}
we conclude
\begin{align*}
(x+x) = 2x = 0
\end{align*}
| {
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Why is there a pattern to the last digits of square numbers? I was programming and I realized that the last digit of all the integer numbers squared end in $ 0, 1, 4, 5, 6,$ or $ 9 $.
And in addition, the numbers that end in $ 1, 4, 9, 6 $ are repeated twice as many times as the numbers that end in $ 0, 5$
I checked the numbers from $1$ to $1000$, and the results are:
$1.$ The numbers on the left are the last digit of each digit squared.
$2.$ The numbers on the right are the number of times that the last digit is repeated.
$$
\begin{array}{cc}
0: &100, \\
1: &200, \\
4: &200, \\
5: &100, \\
6: &200, \\
9: &200
\end{array}
$$
So, why does this happen? What is the property that all integers have?
| Note that any number can be written in the form $10a+(5 \pm b)$ where $0 \leq b \leq 5$. If n = $10a+(5 \pm b)$, then we can calculate $n^2$ as
$(10a)^2+2(10a)(5 \pm b) + (5 \pm b)^2=$$100a^2 + 100a \pm 20ab+(5 \pm b)^2$
$100a^2$, $100a$, and $\pm 20ab$ are all divisible by 10, so we can ignore them, and we're left with $(5 \pm b)^2 = 25 \pm 10b +b^2$. Again, we can ignore $\pm10b$, and we can reduce 25 to 5, leaving $5+b^2$. Note that the $\pm$ part has disappeared; $(10a+(5 + b))^2$ has the same last digit as $(10a+(5 - b))^2$. So, normally, each value of $b$ gives the same remainder twice, once for $+b$ and once for $-b$. But if $b=0$, then $+b$ and $-b$ are the same number, so it gives the remainder only once. And if $b=5$, then $5-b$ gives 0, and $5+b$ gives 10, which also corresponds to a last digit of 0.
So we have:
$b = 0$: last digit of $n$ is 5, last digit of $n^2$ is 5
$b = 1$: last digit of $n$ is 4 or 6, last digit of $n^2$ is 6
$b = 2$: last digit of $n$ is 3 or 7, last digit of $n^2$ is 9
$b = 3$: last digit of $n$ is 2 or 8, last digit of $n^2$ is 4
$b = 4$: last digit of $n$ is 1 or 9, last digit of $n^2$ is 1
$b = 5$: last digit of $n$ is 0, last digit of $n^2$ is 0
Hence, 0 and 5 show up once, while 1,4,6, and 9 show up twice.
| {
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If $\frac{\cos(\alpha -3\theta)}{\cos^3 \theta}=\frac{\sin(\alpha -3\theta)}{\sin^3 \theta}=m$ prove that $\cos\alpha=\frac{2-m^2}{m}$
If $$\frac{\cos(\alpha -3\theta)}{\cos^3 \theta}=\frac{\sin(\alpha -3\theta)}{\sin^3 \theta}=m$$
prove that
$$\cos\alpha=\frac{2-m^2}{m}$$
My approach:
$$\cos^2(\alpha-3\theta)+\sin^2(\alpha-3\theta)=m^2(\sin^6\theta+\cos^6\theta)$$
$$\Rightarrow \frac{1}{m^2}=\sin^6\theta+cos^6\theta=1-\frac{3}{4}\sin^22\theta$$
$$\Rightarrow \sin^22\theta=\frac{4}{3}᛫\frac{m^2-1}{m}$$
I can not proceed further, please help.
| $$\sin\alpha\sin3\theta+\cos\alpha\cos3\theta-m\cos^3\theta=0$$
$$\sin\alpha\cos3\theta-\cos\alpha\sin3\theta-m\sin^3\theta=0$$
$$-\dfrac{\sin\alpha}{m(\cos^3\theta\sin3\theta+\cos3\theta\sin^3\theta)}=-\dfrac{\cos\alpha}{m(\cos3\theta\cos^3\theta-\sin3\theta\sin^3\theta)}=-1$$
Using $\sin3\theta,\cos3\theta$ formula
$\dfrac{4\sin\alpha}m=(\cos3\theta+3\cos\theta)\sin3\theta+\cos3\theta(3\sin\theta-\sin3\theta)=3\sin(3\theta+\theta)$
Similarly, $\dfrac{4\cos\alpha}m=(\cos3\theta+3\cos\theta)\cos3\theta-\sin3\theta(3\sin\theta-\sin3\theta)=1-3\cos(3\theta+\theta)$
Use $\cos^24\theta+\sin^24\theta=1$
| {
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Divisibility of Sum of Equally Spaced Binomial Coefficients According to a numerical calculation I did for small values of $k$, it appears that the following is true.
$$4|\left[\sum_{j=1}^{n-1}\binom{3n}{3j}\right]$$ or $$\sum_{j=1}^{n-1}\binom{3n}{3j}=4p, p\in\mathbb{Z}$$
Ex.
If $n=2, \binom{6}{3}=20=4\cdot 5$
If $n=3, \binom{9}{3}+\binom{9}{6}=2\cdot 84=168=4\cdot 42$
If $n=4, \binom{12}{3}+\binom{12}{6}+\binom{12}{9}=2\cdot 220+924=1364=4\cdot341$
If $n=5, \binom{15}{3}+\binom{15}{6}+\binom{15}{9}+\binom{15}{12}=2\cdot 455+2\cdot 5005=10920=4\cdot 2730$
Is there a way to prove this? Using induction, as above I've shown the base case is true. Then if we assume that
$$S_m=\sum_{j=1}^{m-1}\binom{3m}{3j}=4q, q\in\mathbb{Z}$$
Then
$$S_{m+1}=\sum_{j=1}^{m}\binom{3m+3}{3j}=?$$
And I have no idea how to go forward. Perhaps its not true? Is there a counterexample?
| Starting with the binomial theorem,
$$(1+x)^{3m} = \sum_{k=0}^{3m} \binom{3m}{k} x^k,$$
we see that for the cube root of unity $\omega = e^{2\pi i/3}$,
$$(1+x)^{3m} + (1+\omega x)^{3m} + (1+\omega^2 x)^{3m} = 3 \sum_{\substack{k=0 \\ k \equiv 0 \pmod{3}}}^{3m} \binom{3m}{k} x^k.$$
The terms where $k \not\equiv 0 \pmod{3}$ are sifted away, since $1 + \omega + \omega^2 = 0$. Now evaluate at $x=1$:
$$2^{3m} + (1+\omega)^{3m} + (1+\omega^2)^{3m} = 3 \sum_{\substack{k=0 \\ k \equiv 0 \pmod{3}}}^{3m} \binom{3m}{k}.$$
Since $1+\omega = e^{\pi i/3}$ and $1+\omega^2 = e^{-\pi i/3}$ are 6-th roots of unity, it follows that $(1+\omega)^3 = (1+\omega^2)^3 = -1$.
The right-hand side is almost 3 times your sum $S_m$, but $S_m$ does not include the $k=0$ and $k=3m$ terms. Subtracting them away, we find
$$S_m = \frac{2^{3m} + 2((-1)^m-3)}{3}.$$
For $m \geqslant 1$, the numerator is always divisible by 4, which explains the observed divisibility.
| {
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"url": "https://math.stackexchange.com/questions/2795730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving first order differential equation with power series So, I was told solve the equation $y' - y = x^2$ using power series. Normal methods tell me that the solution is $y = c_{0}e^{x}-x^{2}-2x-2$, and this can be verified by plugging it back in. However, I am stuck on trying to solve this with power series.
We assume $y = \Sigma_{n=0}^{\infty}a_{n}x^{n}$. Thus, $y' = \Sigma_{n=0}^{\infty}a_{n+1}(n+1)x^{n}$. I plug these into the original equation.
$$\Sigma_{n=0}^{\infty}a_{n+1}(n+1)x^{n}-\Sigma_{n=0}^{\infty}a_{n}x^{n}=x^{2}
\\
a_{1}x^{0}+2a_{2}x^{1}+3a_{3}x^{2}+4a_{4}x^{3}+\dots-a_{0}x^{0}-a_{1}x^{1}-a_{2}x^{2}-a_{3}x^{3}-\dots = x^{2}$$
By equating powers of $x$, I find the following relations
$$a_{1}-a_{0}=0
\\
2a_{2}-a_{1}=0
\\
3a_{3}-a_{2}=1
\\
4a_{4}-a_{3}=0
\\
\vdots
\\
na_{n}-a_{n-1}=0
$$
So, I write the coefficients as
$$
a_{1}=a_{0}\\
a_{2}=\frac{a_{1}}{2}=\frac{a_{0}}{2}\\
a_{3}=\frac{1}{3}+\frac{a_{2}}{3}=\frac{a_{0}}{6}+\frac{1}{3}\\
a_{4}=\frac{a_{0}}{24}+\frac{1}{12}\\
\vdots\\
a_{n}=\frac{a_{n}}{n!}+\frac{2}{n!}
$$
Combining these to form $y$, I get
$$y=\Sigma_{n=0}^{\infty}\frac{a_{0}}{n!}x^{n}+\Sigma_{n=3}^{\infty}\frac{2}{n!}x^{n}
$$
The first bit gives me $a_{0}e^{x}$ as expected, but I don't see how to extract $-x-2x-2$ from the second half.
Is there something wrong in my approach that lead to an incorrect answer, or am I missing something in the manipulation of power series?
| Your approach is correct and you did no mistake
$$y=\sum_{n=0}^{\infty}\frac{a_{0}}{n!}x^{n}+\sum_{n=3}^{\infty}\frac{2}{n!}x^{n}$$
Transform the last part as an exponential and it will be aborbed by the first series excet for the three first terms
$$\sum_{n=3}^{\infty}\frac{2}{n!}x^{n}=\sum_{n=0}^{\infty}\frac{2}{n!}x^{n} -2-2x-x^2$$
Therefore
$$y=\sum_{n=0}^{\infty}\frac{a_{0}}{n!}x^{n}+\sum_{n=0}^{\infty}\frac{2}{n!}x^{n} -2-2x-x^2$$
$$y=\sum_{n=0}^{\infty}\frac{(a_{0}+2)}{n!}x^{n} -2-2x-x^2$$
$a_0$ is just a constant substitute $K=a_0+2$
$$y=\sum_{n=0}^{\infty}\frac{K}{n!}x^{n} -2-2x-x^2$$
$$y=K\sum_{n=0}^{\infty}\frac{x^{n}}{n!} -2-2x-x^2$$
Since $e^x=\sum_{n=0}^{\infty}\frac{x^{n}}{n!} $
$$y={K}e^x -2-2x-x^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2797579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Area of Triangle inside a Circle in terms of angle and radius A circle $O$ is circumscribed around a triangle $ABC$, and its radius is $r$. The angles of the triangle are $\angle CAB = a, \angle ABC = b$ and $\angle ACB = c$.
The area $\triangle ABC$ is expressed by $a, b, c$ and $r$ as:
$\Large r^2 \over\Large2$$\Bigg(\sin(x)+\sin(y)+\sin(z)\Bigg)$
find $x, y$ and $z$:
My approach:
Firstly, to make it clear, I set $\overline {AB} = A$, $\overline {BC} = B$ and $\overline {CA} = C$.
$\triangle ABC= \Large{Bh \over 2}$
where $h$ is the height
$\triangle ABC = \Large{BA\sin(c) \over2}$
then, using the law of sine:
$r= \Large{A\over 2 \sin(a)} = \Large{B\over 2 \sin(b)}$
$A = 2r\sin(a)$
$B = 2r\sin(b)$
replacing on the formula of area:
$\triangle ABC = 2r^2\sin(a)\sin(b)\sin(c)$
But that doesn't help to answer the question. Is my approach correct, or else, what am I missing?
| The area formula you derived is a good one to know, but if you want something in terms of a sum of sines, use the trigonometric sum-product relations. Along the way you will also note that the angles of the rriangle sum to 180°. And be careful with signs or this won't come out pretty.
Our starting point, taking $S$ as the area:
$S=2r^2 \sin a \sin b \sin c$
Plug in $\sin a \sin b =(1/2)(\cos (a-b) - \cos (a+b))$ (watch signs!):
$S=r^2 (\cos (a-b) - \cos (a+b)) \sin c$
$S=r^2(\cos (a-b) \sin c - \cos (a+b) \sin c)$
On each of these terms use $\cos u \sin v =(1/2)(\sin (u+v) - \sin (u-v))$ , then:
$S=(r^2/2)(\sin (a-b+c) - \sin (a-b-c) - \sin (a+b+c) + \sin (a+b-c))$
Now for the neat part where we use the angle sum being 180°. Then,
$\sin (a-b+c) = \sin(180°-2b) = \sin (2b)$
$\sin (a-b-c) = \sin(-180°+2a) = - \sin (2b)$ (watch signs!)
$\sin (a+b-c) = \sin(180°-2c) = \sin (2c)$
$\sin (a+b+c) = \sin(180°) = 0$
So we get this elegant result:
$S=(r^2/2)(\sin (2a) + \sin (2b) + \sin (2c))$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2798370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate the limit with exponents using L'Hôpital's rule or series expansion Evaluate the limit$$\lim_{x\to 0}\dfrac{\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}
-\sqrt{ab}}{x}$$ It is known that $a>0,b>0$
My Attempt:
I could only fathom that $$\lim_{x\to 0}\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}=\sqrt{ab}$$
| We have that
*
*$a^x=e^{x\log a}=1+x\log a+\frac12x^2\log^2 a+o(x^2)$
*$b^x=e^{x\log b}=1+x\log b+\frac12x^2\log^2 b+o(x^2)$
then
$$\left( \frac{a^x+b^x}{2} \right)^{\frac{1}{x}}=\left(\frac{2+x\log ab+\frac12x^2(\log^2 a+\log^2 b)+o(x^2)}{2}\right)^{\frac{1}{x}}=\left(1+x\log \sqrt{ab}+\frac14x^2(\log^2 a+\log^2 b)+o(x^2)\right)^{\frac{1}{x}}=e^{\frac{\log\left(1+x\log \sqrt{ab}+\frac14x^2(\log^2 a+\log^2 b)+o(x^2)\right)}{x}}$$
and since
$$\log\left(1+x\log \sqrt{ab}+\frac14x^2(\log^2 a+\log^2 b)+o(x^2)\right)\\=x\log \sqrt{ab} +\frac14x^2(\log^2 a+\log^2 b)-\frac12x^2\log^2\sqrt{ab}+o(x^2)$$
we have that
$$e^{\frac{\log\left(1+x\log \sqrt{ab}+\frac14x^2(\log^2 a+\log^2 b)+o(x^2)\right)}{x}}=\sqrt {ab}(1+\frac14x(\log^2 a+\log^2 b)-\frac12x\log^2\sqrt{ab}+o(x))$$
therefore
$$\dfrac{\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}
-\sqrt{ab}}{x}= \sqrt {ab}\left(\frac14(\log^2 a+\log^2 b)-\frac12\log^2\sqrt{ab}+o(1)\right)\to \sqrt {ab}\left(\frac14(\log^2 a+\log^2 b)-\frac12\log^2\sqrt{ab}\right) =\sqrt {ab}\left(\frac14(\log^2 a+\log^2 b)-\frac18(\log a+\log b)^2\right)=\frac{\sqrt {ab}}8\left(\log a-\log b\right)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2799146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Determine the Taylor series for $f(x) = x^3 \cdot \ln{\sqrt{x}}$ around the point $a = 1$ and determine its radius of convergence. As the question states:
"Determine the Taylor series for $f(x) = x^3 \cdot \ln{\sqrt{x}}$ around the point $a = 1$ and determine its radius of convergence."
I have consulted this related question, and understand the steps to be:
*
*find the first few terms of the Taylor polynomial.
*Generalize the terms by making use of an infinite sum to represent the function as the Taylor series.
*use the infinite sum in the ratio test to find the radius of convergence.
Progress so far:
*
*The first 6 terms (n = 0 to n = 5) of the Taylor polynomial I have calculated to be:
$x^3 \cdot \ln{(\sqrt{x})} + \frac{1}{2}(x-a) + \frac{5}{4}(x-a)^2 + \frac{11}{12}(x-a)^3 + \frac{1}{8}(x-a)^4 - \frac{1}{40}(x-a)^5$
It is at this point however that I fall over. It is not intuitive to me how I can write the c-terms as a function without utilizing some sort of online maths engine for fitting the data to a curve.
Is there some sort of first-year-student-friendly technique for modelling these data points systematically? Alternatively, does someone have an intuition they would be willing to share for solving this problem?
| In fact a lot of the work needed is already done. Let's recapitulate. We have $f(x)=x^3\ln\left(\sqrt{ x}\right)$ and the Taylor expansion at $a=1$ is given as
\begin{align*}
f(x)=\sum_{n=0}^\infty \frac{1}{n!}f^{(n)}(a)(x-a)^n
\end{align*}
We obtain
\begin{align*}
f(x)&=x^3\ln\left(\sqrt{ x}\right)=\frac{1}{2}x^3\ln(x)\\
f^{\prime}(x)&=\frac{3}{2}x^2\ln(x)+\frac{1}{2}x^2\\
f^{\prime\prime}(x)&=3x\ln(x)+\frac{5}{2}x\\
f^{\prime\prime\prime}(x)&=3\ln(x)+\frac{11}{2}\\
f^{(4)}(x)&=\frac{3}{x}\\
f^{(5)}(x)&=-\frac{3}{x^2},\qquad f^{(6)}(x)=3\cdot\frac{2!}{x^3},\qquad f^{(7)}=-3\cdot \frac{3!}{x^4}\\
&\vdots\\
f^{(n)}(x)&=3(-1)^n\frac{(n-4)!}{x^{n-3}}\qquad\qquad n\geq 4\tag{1}
\end{align*}
From the fourth derivative $\frac{3}{x}$ we can relatively easy obtain higher derivatives and assume the general formula (1) which can be shown by induction.
Evaluated at $a=1$ we have
\begin{align*}
f(1)=0, f^{\prime}(1)=\frac{1}{2}, f^{\prime\prime}(1)=\frac{5}{2}, f^{\prime\prime\prime}(1)=\frac{11}{2}, f^{(n)}(1)=3(-1)^n(n-4)!\qquad n\geq 4
\end{align*}
We obtain from the derivatives above the Taylor series
\begin{align*}
\color{blue}{f(x)}&\color{blue}{=x^3\ln\sqrt{ x}}\\
&\color{blue}{=\frac{1}{2}(x-1)+\frac{5}{4}(x-1)^2+\frac{11}{12}(x-1)^3+3\sum_{n=4}^\infty\frac{(-1)^n}{n(n-1)(n-2)(n-3)}(x-1)^n}
\end{align*}
The radius $R$ of convergence is
\begin{align*}
R&=\lim_{n\to \infty}\left|\frac{a_{n}}{a_{n+1}}\right|=\lim_{n\to \infty}\left|\frac{3(-1)^n(n+1)n(n-1)(n-2)}{3(-1)^{n+1}n(n-1)(n-2)(n-3)}\right|\\
&=\lim_{n\to\infty}\left|\frac{n+1}{n-3}\right|\\
&=1
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2802024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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2014 AIME II: Cubic Question Real numbers $r$ and $s$ are roots of $p(x)=x^3+ax+b$, and $r+4$ and $s-3$ are roots of $q(x)=x^3+ax+b+240$. Find the sum of all possible values of $|b|$.
I assumed that the other root of p(x) would be t. From here, I used Vieta's formulas to get $-rst=b$ and $rs+st+rt=a$. Then, I assumed that the third root of q(x) would be y. Then, I used vieta's formulas to get that $-(r+4)(s-3)y=b+240$ and that $(r+4)(s-3)+y(s-3)+y(r+4)=a$. I am not sure how to proceed from here.
| Let $r$, $s$, and $-r-s$ be the roots of $p(x)$ (per Vieta's). Then $r^3 + ar + b = 0$ and similarly for $s$. Also, $q(r+4) = (r+4)^3 + a(r+4) + b + 240 = 12r^2 + 48r + 304 + 4a = 0$.
Set up a similar equation for $s$:
$q(s-3) = (s-3)^3 + a(s-3) + b + 240 = -9s^2 + 27s + 213 - 3a = 0.$
Simplifying and adding the equations gives $3r^2 - 3s^2 + 12r + 9s + 147 = 0$ and
$r^2 - s^2 + 4r + 3s + 49 = 0 (1)$
Plugging the roots $r$, $s$, and $-r-s$ into $p(x)$ yields a long polynomial, and plugging the roots $r+4$, $s-3$, and $-1-r-s$ into $q(x)$ yields another long polynomial. Equating the coefficients of x in both polynomials: $rs + (-r-s)(r+s) = (r+4)(s-3) + (-r-s-1)(r+s+1)$, which eventually simplifies to
$s = \frac{13 + 5r}{2}$
Substitution into (1) should give $r = -5$ and $r = 1$, corresponding to $s = -6$ and $s = 9$, and $|b| = 330, 90$, for an answer of $\boxed{420}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2802571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find a fraction $\frac{m}{n}$ which satisfies the given condition Find a fraction such that all of $\frac{m}{n}$, $\frac{m+1}{n+1}$, $\frac{m+2}{n+2}$, $\frac{m+3}{n+3}$, $\frac{m+4}{n+4}$, $\frac{m+5}{n+5}$ are reducible by cancellation. Condition: $m≠n$.
What I tried was... I wrote $$\frac{m}{n}=k$$ Then, I replaced $m$ in the fractions as $nk$. And after a bit of simple manipulation, I obtained $$k+\frac{1-k}{n+1},$$ $$k+\frac{2-k}{n+2},$$ $$k+\frac{3-k}{n+3},$$ $$k+\frac{4-k}{n+4},$$ $$k+\frac{5-k}{n+5}$$ Now I do not know how to proceed any further.
| Fixed $n$ a solution is given by
$$m=n+n(n+1)(n+2)(n+3)(n+4)(n+5)=n+\frac{(n+5)!}{(n-1)!}$$
indeed
$$\frac{m}{n}=\frac{n+n(n+1)(n+2)(n+3)(n+4)(n+5)}{n}\\=1+(n+1)(n+2)(n+3)(n+4)(n+5)$$
$$\frac{m+1}{n+1}=\frac{n+1+n(n+1)(n+2)(n+3)(n+4)(n+5)}{n+1}\\=1+n(n+2)(n+3)(n+4)(n+5)$$
$$\frac{m+2}{n+2}=\frac{n+2+n(n+1)(n+2)(n+3)(n+4)(n+5)}{n+2}\\=1+n(n+1)(n+3)(n+4)(n+5)$$
$$\frac{m+3}{n+3}=\frac{n+3+n(n+1)(n+3)(n+3)(n+4)(n+5)}{n+3}=\\1+n(n+1)(n+2)(n+4)(n+5)$$
$$\frac{m+4}{n+4}=\frac{n+4+n(n+1)(n+2)(n+3)(n+4)(n+5)}{n+4}\\=1+n(n+1)(n+2)(n+3)(n+5)$$
$$\frac{m+5}{n+5}=\frac{n+5+n(n+1)(n+2)(n+3)(n+4)(n+5)}{n+5}\\=1+n(n+1)(n+2)(n+3)(n+4)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2803295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How did root of the equation $x= \sqrt{(2-x)(3-x)} + \sqrt{(3-x)(5-x)} + \sqrt{(2-x)(5-x)}$ vanish while solving? The original question was to solve for $x \in \mathbb{R}$ in:
$$x= \sqrt{(2-x)(3-x)} + \sqrt{(3-x)(5-x)} + \sqrt{(2-x)(5-x)}$$
My Solution:
By domain requirements: $x \in (-\infty,2) \cup (5,\infty) $.
\begin{align}
x &= \sqrt{(2-x)(3-x)} + \sqrt{(3-x)(5-x)} + \sqrt{(2-x)(5-x)} \\
\Rightarrow (x - \sqrt{(2-x)(3-x)})^2 &= (\sqrt{(3-x)(5-x)} + \sqrt{(2-x)(5-x)})^2 \\
\Rightarrow 2x^2 - 5x + 6 - 2x\sqrt{(2-x)(3-x)} &= 2x^2 - 15x + 25 + 2(5-x)\sqrt{(2-x)(3-x)} \\
\Rightarrow x - 1.9 &= \sqrt{(3-x)(2-x)} \\
\Rightarrow x \geq 1.9 &\hspace{0.5 cm}\&\hspace{0.5 cm} (x - 1.9)^2 = (3-x)(2-x) \\
\Rightarrow x = \frac{239}{120}
\end{align}
Question: If I check for the answers of the original equation in WOLFRAM, I get two possible answers: $x = \frac{239}{120} $ and $x = \frac{1}{2}(5 + \sqrt{5} + \sqrt{8 + 3\sqrt{5}})$. My question is where (and why) in my solution did the second root vanish? I have only used squaring and simplification, and as far as I know squaring only creates an additional root, it does not destroy a root.
Link to Wolfram's solution: http://www.wolframalpha.com/input/?i=x%3D+%5Csqrt%7B(2-x)(3-x)%7D+%2B+%5Csqrt%7B(3-x)(5-x)%7D+%2B+%5Csqrt%7B(2-x)(5-x)%7D
| Your error occurs at the step where you removed the factor $(5-x)$.
It should be $|5-x|$, which leads to two cases.
If $x \le 5$, then $(5-x)$ is correct.
If $x > 5$, the correct factor is $(x-5)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Number of solutions to the congruence $x^2\equiv 121\pmod {1800}$ I'm trying to find the number of solutions to this congruence:
$$x^2\equiv 121\pmod {1800}$$
I thought about writing it as a system of congruences. As $1800=3^2 \cdot 5^2 \cdot 2^3$, we get:
$x^2\equiv 121\pmod {5^2} \;,\; x^2\equiv 121\pmod {3^2} \;,\; x^2\equiv 121\pmod {2^3}$
$\Downarrow$
$ x^2\equiv 21\pmod {5^2} \;,\; x^2\equiv 4\pmod {3^2} \;,\; x^2\equiv 1\pmod {2^3} $
Now I'm pretty stuck with how to solve each of the above quadratic congruences, is there a fast way to do it?
Thanks
| No, there's not a fast way. To solve $x^2\equiv a \pmod{p^n}$ we first solve $x^2\equiv a \pmod{p}$ and then use Hensel's Lemma to "lift" the solution to $\pmod{p^2}$ and then $\pmod{p^3}$, etc. Sometimes the solutions lift uniquely, sometimes not.
In the case of your problem, there are two solutions modulo $5$ which lift uniquely to two solutions modulo $25$. The exact same thing happens modulo $9$.
There is one solution modulo $2.$ It lifts to two solutions modulo $4.$ Each of these in turn lifts to two solutions modulo $8$, for a total of 4 solutions modulo $8.$
So in the end, there are $2\times 2\times 4 = 16$ solutions modulo $1800.$
| {
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"timestamp": "2023-03-29T00:00:00",
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The implication: $x+\frac{1}{x}=1 \implies x^7+\frac{1}{x^7}=1$ $$x+\frac{1}{x}=1 \implies x^7+\frac{1}{x^7}=1$$
The graphs/range: $\quad y_1(x)=x+\frac{1}{x}, \quad y_7(x)=x^7+\frac{1}{x^7} \quad$
and do not touch the line $\quad y=1\quad$. The relation $\quad x+\frac 1 x=1 \quad$ appears to only be true for certain complex values
Yet the reasoning for the implication itself is
\begin{align}
x+\frac {1}{x}&=1 \\
\color {green}{x^2}&=\color {green}{x-1 }\\
x^3&=\color {green}{x^2}-x \\
&=(\color {green}{x-1})-x \\
&=-1.
\end{align}
Hence $x^6=1$, which implies $x^7=x$.
I had just started reading this book when it was presented, asking you to prove the implication without finding the roots first.
So my question: is this implication really justified? Should the author have said to only prove it for an implicit (complex) specific value of $x$? If I didn't know any better, I'd say he presented it like it was true for some range of values...he couldn't of...
| The given equation implies $x^2-x+1=0$; multiplying by $x+1$, we get $x^3+1=0$. This implies $x^6=1$ and therefore
$$
x^7=x
$$
Your argument is sound as well.
The key is that a solution to $x+\frac{1}{x}=1$ is a 6th root of $1$. Indeed,
$$
x^6-1=(x^3-1)(x^3+1)=(x-1)(x+1)(x^2+x+1)(x^2-x+1)
$$
Thus any root of $x^2-x+1$ is also a root of $x^6-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2808056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Someone's Inequality ? $(a + b)^2 \le 2(a^2 + b^2) $ For real $a, b$ then $(a + b)^2 \le 2(a^2 + b^2) $
This fairly trivial inequality crops up a lot in my reading on (Lebesgue) integration, is it named after someone ? It extends rather obviously for positive reals to $a^2 + b^2 \le (a + b)^2 \le 2(a^2 + b^2) $.
Proof (if you need it):
$0 \le (a - b)^2 = a^2 + b^2 -2ab \implies 2ab \le a^2 + b^2$
$(a + b)^2 = a^2 + b^2 + 2ab$ which by previous $ \le 2(a^2 + b^2) $.
Application:
If $f, g$ are positive functions then $(f + g)^2$ is integrable $\iff$ $f^2, g^2$ are integrable since $f^2 + g^2 \le (f + g)^2 \le 2(f^2 + g^2) $ pointwise.
| That inequality probably does not have a name as it is so basic. In any case it can be viewed as a special case of Young's inequality.
| {
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Solution of the functional equation $f(x) + f(y) =f\left(x\sqrt{1-y^2 }+y\sqrt{1-x^2 }\right) $
If
$$f(x) + f(y) =f\left(x\sqrt{1-y^2 }+y\sqrt{1-x^2 }\right)\text,$$
prove that
$$f(4x^3 -3x) + 3f(x) =0\text.$$
I started by substituting $x = y$, in the expression and I get
$$2f(x)=f\left(2x\sqrt{1-x^2}\right)\text.$$
Then, I also substitute $y = \sqrt{(1-x^2)}$ and I get
$$f(x) +f\left(x\sqrt{1-x^2}\right) = 0\text.$$
How shall I proceed further and solve this problem?
| Easy way to prove that for every real number $ x $ with $ - 1 \le x \le 1 $, $ f ( x ) = 0 $ (using some steps already mentioned in the comments, but repeated here to be complete):
Let $ x = y = 0 $ in
$$ f ( x ) + f ( y ) = f \Big( x \sqrt { 1 - y ^ 2 } + y \sqrt { 1 - x ^ 2 } \Big) \tag { * } \label { * } $$
and you get $ f ( 0 ) = 0 $. Again, let $ x = y = 1 $ in \eqref{ * } and you get $ f ( 1 ) = 0 $. Thus, letting $ y = 1 $ in \eqref{ * } you'll have $ f ( x ) = f \big( \sqrt { 1 - x ^ 2 } \big) $. This also shows that $ f ( | x | ) = f ( x ) $. Now, substituting $ | x | $ for $ x $ and $ \sqrt { 1 - x ^ 2 } $ for $ y $ in \eqref{ * }, you get $ f ( x ) = 0 $.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Dodecahedron volume How can I derive the volume of a regular dodecahedron? A. P. Kiselev's Stereometry suggests that we cut it into a cube and congruent "tent-like" solids. I'm finding some difficulty to compute the volume of these "tent-like" solids.
| First, note that if $s$ is the side length of the dodecahedron, then the side length of the cube is $$c = \frac{s}{2} \csc \frac{\pi}{10} = s \left(\frac{1 + \sqrt{5}}{2}\right).$$
Now consider looking at the tent solid from above, it looks something like this:
where the dotted line $x$ denotes the slant height of the triangular face. Since $x, c/2, s$ form a right triangle, we can compute $x$ using the Pythagorean Theorem to be $$x = \sqrt{s^2 - \left(\frac{c}{2}\right)^2} = s\sqrt{1 - \left(\frac{1 + \sqrt{5}}{4}\right)^2} = s\sqrt{\frac{5 - \sqrt{5}}{8}}.$$
It is also easy to see that $$y = \frac{c - s}{2} = \frac{1}{2}s \left(\frac{1 + \sqrt{5}}{2} - 1 \right) = s \left(\frac{\sqrt{5} - 1}{4} \right)$$
Letting $h$ be the vertical height of the tent, note $y, h, x$ form a right triangle, so we have $$h = \sqrt{x^2 - y^2} = s\sqrt{\frac{5 - \sqrt{5}}{8} - \left(\frac{\sqrt{5} - 1}{4} \right)^2} = \frac{s}{2}.$$
We can now compute the volume as follows: cut the volume into three pieces along the dashed lines. The left and right pieces can be joined to form a pyramid whose base is a $2y$ by $c$ rectangle and whose height is $h$, and the piece in the middle is a prism whose base is a triangle with base $c$ and height $h$ and whose height is $s$.
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
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Exponential inequality solution I am trying to solve
$$ \displaystyle (2^{(3x-1)/(x-1)})^{1/3} < 8^{((x-3)/(3x-7))}$$
Here's how I proceeded.
$LHS = 2^{((3x-1)/(3(x-1))} , RHS = 2^{((3*(x-3))/(3x-7))}$
hence inequating the exponents ,
$$\frac {(3x-1)}{(3(x-1))} < 3\frac {(x-3)}{(3x-7)}$$
$$ \implies (3x-1)(3x-7) < 9(x-1)(x-3)$$
$$\implies 9x^2-24x+7 < 9x^2-36x+27$$
cancelling terms
$$12x <20$$
$$=> x < 5/3 $$
but the book states the solution as $(-\infty , 1)$ U $(5/3,7/3)$
Not sure how this has been arrived at , can someone help me understand ?
Thanks in advance.
Madavan.
| Be carefull when you multiply your inequality
$$\frac {(3x-1)}{(3(x-1))} < 3\frac {(x-3)}{(3x-7)}$$
Because $3x-7$ or $3(x-1)$ may be negative and change the order of the ineqaulity
You need to study the sign of $3(x-1)(3x-7)$...
A)$ \,3(x-1)(3x-7) $ is positive
$$\implies (3x-1)(3x-7) < 9(x-1)(x-3)$$
$$ \implies x<5/3$$
First case $3(x-1)>0$ and $3x-7>0 \implies x> \frac 73$
With the constraint $x>7/3$ there are no solution.
Second case $3(x-1)<0$ and $3x-7<0$
$$\implies x<1$$
with this constraint ($x<1$) you get a part of the result of the book $ \implies x \in [-\infty,1]$.
Now you have to study the case when $3(x-1)(3x-7)$ is negative to get the other part of the solution of the book...
B)$ \,3(x-1)(3x-7) $ is negative
$$ (3x-1)(3x-7) > 9(x-1)(x-3)$$
$$ \implies x>\frac 5 3$$
First case
$3(x-1)<0 \quad ; 3x-7>0 \implies x<1, x>7/3$ Impossible ..
Second case
$3(x-1)>0 \quad ; 3x-7<0 \implies x>1, x<7/3$ .
With the constraint you finally get $x \in (5/3,7/3)$
And the complete solution is $(-\infty,1)$ U $(5/3,7/3)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is $\lim_{n\to\infty} \left(\sum_{k=1}^n \frac1k\right) / \left(\sum_{k=0}^n \frac1{2k+1}\right)$? I have the following problem:
Evaluate
$$ \lim_{n\to\infty}{{1+\frac12+\frac13 +\frac14+\ldots+\frac1n}\over{1+\frac13 +\frac15+\frac17+\ldots+\frac1{2n+1}}} $$
I tried making it into two sums, and tried to make it somehow into an integral, but couldn't find an integral.
The sums I came up with,
$$ \lim_{n\to\infty} { \sum_{k=1}^n {\frac1k} \over {\sum_{k=0}^n {\frac{1}{2k+1}}}} $$
| One approach is as follows: it suffices to note that
$$
\sum_{k=2}^n\frac{1}{k} \leq \int_1^n \frac 1x \,dx \leq \sum_{k=1}^n\frac{1}{k}, \\
\sum_{k=2}^{n+1}\frac{1}{2k-1} \leq \int_1^{n+1} \frac 1{2x-1} \,dx \leq \sum_{k=1}^{n+1}\frac{1}{2k-1},
$$
and apply the squeeze theorem. In particular, we can use the above to get
$$
\frac{\ln(n)}{1 + \frac 12 \ln(2n + 1)}
\leq
\frac{1+\frac{1}{2}+\frac{1}{3}+…\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+…+\frac{1}{2n+1}}
\leq
\frac{1 + \ln(n)}{\frac 12 \ln(2n + 1)}.
$$
Another approach: note that adding a final $\frac 1{2n + 2}$ to $1/2$ times the numerator yields $\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n+2}$, and
$$
\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n+2} \leq \\
1+\frac{1}{3}+\frac{1}{5}+…+\frac{1}{2n+1} \leq \\
1 + \left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n+2}\right).
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve the given integral of the type: $\int_{0}^{\infty}\tfrac{1}{(a + b x)(c + d x) (1 + p x)}dx$ Can anyone give a hint how to solve integral of the following type:
$$\int_{0}^{\infty}\dfrac{1}{(a + b x)(c + d x) (1 + p x)}dx$$
The problem is if we proceed with partial fraction, we get
$$\dfrac{1}{(a + bx)(c + dx)(1 + px)} = \dfrac{b^2}{(bc - ad)(b - ap)(a + bx)} + \dfrac{d^2}{(bc - ad)(cp - d)(c + dx)}- \dfrac{p^2}{(b - ap)(cp-d)(1 + px)}$$
Now solving the integral for the first part,
$$\int_{0}^{\infty}\dfrac{b^2}{(bc - ad)(b - ap)(a + bx)} dx = \left.\dfrac{b \ln(a + bx)}{(bc - ad)(b - ap)}\right\vert_0^{\infty}$$
Hence a $\ln(\infty)$ term is encountered. Same is the case with all the three parts.
Any hints appreciated.
Edit: The final solution using Mathematica:
| For simplicity, write this as $$J = \dfrac{1}{bdp} \int_0^\infty \dfrac{dx}{(A+x)(B+x)(C+x)}$$
where $A = a/b$, and $B = c/d$, $C = 1/p$. We're assuming $a,b,c,d,p > 0$ and $A,B,C$ are all distinct. The partial fraction decomposition of the integrand is
$$ F(x) = \frac{1}{(A-B)(A-C)(A+x)} + \frac{1}{(B-A)(B-C)(B+x)} + \frac{1}{(C-A)(C-B)(C+x)}$$
so that
$$ J = \frac{1}{bdp} \lim_{R \to \infty} \left(\frac{\ln(A+R)-\ln(A)}{(A-B)(A-C)} + \frac{\ln(B+R)-\ln(B)}{(B-A)(B-C)} + \frac{\ln(C+R)-\ln(C)}{(C-A)(C-
B)}\right) $$
Now write $\ln(A+R) = \ln(R) + \ln(1+A/R)$ etc, and notice that (as Achille Hui remarked) the coefficients of $\ln(R)$ cancel (as they must, because the improper integral is known to converge), while $\ln(1+A/R), \ln(1+B/R), \ln(1+C/R) \to 0$
as $R \to \infty$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of digits when $99$ divides $n$ Let $n$ be a natural number such that $99\mid n$. Show that $S(n)\ge 18$.
It is clear that $9 \mid S(n)$, but I cannot apply the divisibility rule of $11$. Please help!
| If $99|n$ then $9|n$ and $11|n$.
Two well known rules.
1) If $9|n$ then $9|S(n)$.
So $S(n) = 9, 18,27,.... $ etc. So the the only way $S(n) \ge 18$ can be false is if $S(n) = 9$.
2) If $11|S(n)$ then the sum of the even positioned digits plus the sum of the odd positions digits are either equal, or off by a multiple of $11$.
So if $S_e$ the sum of the even positioned digits and $S_o$ is the sum of the odd positioned digits.
Then $S(n) = S_o + S_e = \begin{cases}2S_o &\text{if } S_o = S_e \\ 2S_o +11k &\text{if } S_e = S_o + 11k\text{ for some }k \ge 1 \\ 2S_e + 11k &\text{if } S_o = S_e + 11k\text{ for some }k \ge 1\end{cases}$
So either $S(n)$ is even. Or $S(n) \ge 11$ And $S(n) = 9,18,27,....$ so $S(n) \ge 18$.
====
Remains to prove those two basic rules.
You've seen proofs a million times, haven't you?
Well, here they are for the million + 1 time:
If $n = \sum\limits_{k=0}^m d_k 10^k$ where $d_k$ are the digits of $n$ then
1) $n = \sum\limits_{k=0}^m d_k (10^k - 1) + \sum\limits_{k=0}^m d_k$
$= \sum\limits_{k=0}^m d_k (10^k - 1) + S(n)$.
Now $10^k -1 = 9999....9$ and is divisible by $9$ so $\sum\limits_{k=0}^m d_k (10^k - 1)$ is divisible by $9$.
[Actually $(a^{k-1} + a^{k-2} + .... + a + 1)(a-1) = a^k - 1$ means $a-1$ always divides $a^k -1$ so $10-1=9$ always divides $10^k -1$; is a more sophisticated proof.]
So $9|n$ if and only if $9|S(n)$.
2) $n = \sum\limits_{k=0}^m d_k 10^k = \sum\limits_{k=0; k\text{ even}}^m d_k 10^k + \sum\limits_{k=0; k\text{ odd}}^m d_k 10^k$
$\sum\limits_{k=0; k\text{ even}}^m d_k (10^k -1) + \sum\limits_{k=0; k\text{ even}}^m d_k + \sum\limits_{k=0; k\text{ odd}}^m d_k (10^k + 1) -\sum\limits_{k=0; k\text{ odd}}^m d_k$
$= \sum\limits_{k=0; k\text{ even}}^m d_k (10^k -1)+ \sum\limits_{k=0; k\text{ odd}}^m d_k (10^k + 1)+S_e - S_o$.
Claim: $11|10^k - 1$ if $k$ is even and $11|10^k + 1$ if $k$ is odd.
If $k$ is even then $(a^{k-1} - a^{k-2} + a^{k-3} -....+a^3-a^2 +a -1)(a + 1) = a^k - 1$ so $a+1|a^k - 1$ if $k$ is even. So $10 + 1 = 11$ divides $10^k-1$ if $k$ is even.
If $k$ is odd then $(a^{k-1} - a^{k-2} + a^{k-3} -.... - a^3 +a^2 -a + 1)(a +1)=a^k + 1$ so $a+1|a^k + 1$ if $k$ is odd. So $10 + 1 = 11$ divides $10^k +1$ if $k$ is odd.
So $11|n$ if and only if $11$ divides $S_e - S_o$. Or in other words if $S_o$ and $S_e$ have a difference of a multiple of $11$.
| {
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$\frac{a^2} {1+a^2} + \frac{b^2} {1+b^2} + \frac{c^2} {1+c^2} = 2.$ Prove $\frac{a} {1+a^2} + \frac{b} {1+b^2} + \frac{c} {1+c^2} \leq \sqrt{2}.$ $a, b, c ∈ \mathbb{R}+.$
WLOG assume $a \leq b \leq c.$ I tried substitution: $x=\frac{1} {1+a^2}, y=\frac{1} {1+b^2}, z=\frac{1} {1+c^2},$ so $x \geq y \geq z$ and $(1-x)+(1-y)+(1-z)=2 \to x+y+z=1.$
We want to prove $ax+by+cz \leq \sqrt{2}.$ This somewhat looks like Cauchy-Schwarz so I tried that: $(a^2+b^2+c^2)(x^2+y^2+z^2) \geq (ax+by+cz)^2.$ The problem becomes $(a^2+b^2+c^2)(x^2+y^2+z^2) \geq 2,$ since $a,b,c,x,y,z>0.$
Expressing $a,b,c$ in terms of $x,y,z$: $(\frac {1}{x} + \frac {1}{y} + \frac {1}{z} - 3)(x^2+y^2+z^2)$
$= x+y+z+\frac{y^2}{x}+\frac{z^2}{x}+\frac{x^2}{y}+\frac{z^2}{y}+\frac{x^2}{z}+\frac{y^2}{z}-3(x^2+y^2+z^2) \geq 2.$
$\to \frac{y^2}{x}+\frac{z^2}{x}+\frac{x^2}{y}+\frac{z^2}{y}+\frac{x^2}{z}+\frac{y^2}{z}-3(x^2+y^2+z^2) \geq 1.$ Stuck here. Thinking about using AM-GM but not sure how. Help would be greatly appreciated.
| Cauchy-Schwartz ...
\begin{eqnarray*}
\left( \frac{a}{\sqrt{1+a^2}} \frac{1}{\sqrt{1+a^2}} + \frac{b}{\sqrt{1+b^2}} \frac{1}{\sqrt{1+b^2}} + \frac{c}{\sqrt{1+c^2}} \frac{1}{\sqrt{1+c^2}} \right)^2 \\\leq \left( \frac{a^2}{1+a^2} + \frac{b^2}{1+b^2} +\frac{c^2}{1+c^2} \right) \left( \frac{1}{1+a^2} + \frac{1}{1+b^2} +\frac{1}{1+c^2} \right)= 2.
\end{eqnarray*}
| {
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An interesting algebraic inequality involving a square root I came across the following inequality as a step in the proof of another: Let $x,\ y,\ z>0$, then this is true:
$$\sqrt{\frac{x^2 + 1}{y^2 + 1}} + \sqrt{\frac{y^2 + 1}{z^2 + 1}} + \sqrt{\frac{z^2 + 1}{x^2 + 1}}\le \frac{x}{y} + \frac{y}{z} + \frac{z}{x}.$$
I checked it numerically and it seems to be valid, but I don't see an obvious solution. For only two variables, it is easy to show the following:
$$\sqrt{\frac{x^2+\delta}{y^2+\delta}} + \sqrt{\frac{y^2+\delta}{x^2 + \delta}} \le \frac{x^2 + y^2 + 2\delta}{\sqrt{x^2y^2+2xy\delta + \delta^2}} = \frac{x^2 + y^2 + 2\delta}{xy + \delta}\le \frac{x^2+y^2}{xy} $$
since the left hand side of the last inequality is a decreasing function. However, I cannot extend this logic to the full inequality. Any help is appreciated.
| Let $z=\min\{x,y,z\}$.
Thus, we need to prove that
$$\frac{x}{y}+\frac{y}{x}-2+\frac{y}{z}-\frac{y}{x}+\frac{z}{x}-1\geq\sqrt{\tfrac{x^2+1}{y^2+1}}+\sqrt{\tfrac{x^2+1}{y^2+1}}-2+\sqrt{\tfrac{y^2+1}{z^2+1}}-\sqrt{\tfrac{y^2+1}{x^2+1}}+\sqrt{\tfrac{z^2+1}{x^2+1}}-1$$ or
$$\tfrac{(x-y)^2}{xy}+\tfrac{(z-x)(z-y)}{xz}\geq\tfrac{\left(\sqrt{x^2+1}-\sqrt{y^2+1}\right)^2}{\sqrt{(x^2+1)(y^2+1)}}+\tfrac{\left(\sqrt{z^2+1}-\sqrt{x^2+1}\right)\left(\sqrt{z^2+1}-\sqrt{y^2+1}\right)}{\sqrt{(x^2+1)(z^2+1)}},$$ which is true because
$$(x-y)^2\geq\left(\sqrt{x^2+1}-\sqrt{y^2+1}\right)^2,$$
$$x-z\geq\sqrt{x^2+1}-\sqrt{z^2+1},$$
$$y-z\geq\sqrt{y^2+1}-\sqrt{z^2+1},$$
$$\sqrt{(x^2+1)(y^2+1)}\geq xy$$ and
$$\sqrt{(x^2+1)(z^2+1)}\geq xz.$$
For example,
$$x-z\geq\sqrt{x^2+1}-\sqrt{z^2+1}$$ it's
$$x-z\geq\frac{x^2-z^2}{\sqrt{x^2+1}+\sqrt{z^2+1}}$$ or
$$\sqrt{x^2+1}+\sqrt{z^2+1}\geq x+z,$$ which is obvious.
| {
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"timestamp": "2023-03-29T00:00:00",
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Help setting up integral: $\iint_{\mathbb R^2} \frac{1}{(1+4x^2+9y^2)^2} dx dy$
Evaluate $\iint_{\mathbb R^2} \frac{1}{(1+4x^2+9y^2)^2} dx dy$
If I use polar coordinates I get
$$\int_0^{2\pi}\int_0^\infty \frac{r}{(1+4r^2\cos^2\theta+9r^2\sin^2\theta)^2} drd\theta.$$
However, I am not sure where to go from here.
| We substitute first $2x=s$, $3y=t$, so $6dx\wedge dy = ds\wedge dt$. The integral is after this easily computed.
$$
\begin{aligned}
\iint_{\mathbb R^2} \frac{1}{(1+4x^2+9y^2)^2} \; dx\; dy
&=
\frac 16\iint_{\mathbb R^2} \frac{1}{(1+s^2+t^2)^2} \;ds\; dt
\\
&=
\frac 16\int_0^\infty\int_0^{2\pi} \frac{1}{(1+r^2)^2} \; r\;dr\; d\theta
\\
&=
\frac 16\cdot2\pi\cdot \frac 12
\int_0^\infty \frac{1}{(1+r^2)^2} \;d(r^2)
\\
&=
\frac 16\cdot2\pi\cdot \frac 12
\left[\ -\frac{1}{1+r^2}\ \right]_0^\infty
\\
&=\frac \pi6\ .
\end{aligned}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluation of limits and finding the value of $\lambda$
If $f(x)= \lim_{n\to \infty}\dfrac{\left(1- \cos \left(1- \tan \left(\dfrac {\pi}{4}-x\right)\right)\right)(x+1)^n+ \lambda\sin((n - \sqrt{n^2 -8n})x)}{x^2(x+1)^n+x}$ ,$x\ne 0$ is continuous at $x=0$, then find the value of $(f(0)+ 2\lambda)$
Attempt:
I know that we are supposed to evaluate limit as $n \to \infty$ and $x \to 0$ but I am facing real trouble while doing that. It's $0/0$ form LHopital doesn't help here because differentiation of that function would be really long.
The term inside cos simplifies to $\dfrac{2\tan x}{1+\tan x}$
For $\dfrac {P(x)}{Q(x)}$'s limit when $x\to 0$ we see the ratio of the lowest powers of $x$ so according to that the ratio of lowest powers of $x$ is $\lambda(n- \sqrt{n^2-8n}) $ whose limit is $4\lambda$ but I am not sure we can use this method when we have double limits (limit on both $x$ and $n$).
Tl;dr:
Is the method of the last paragraph above allowed here? Why or why not?
What is the apt way to solve this problem?
| Rewrite the given function as:
$$f(x) = \lim_{n\to\infty}\frac{2\sin^2\left(\frac{\tan(x)}{1+\tan(x)}\right)(x+1)^n + \lambda \sin\left(\frac{8nx}{n+\sqrt{n^2-8n}}\right)}{x^2(x+1)^n+x}$$
We need left and right limits at $x=0$ to be equal as $n\to \infty$. First let us consider individual limits:
*
*For right hand limit, as $n\to \infty,$ and $x\to 0^+$, note that $\sin \left(\frac{8nx}{n+\sqrt{n^2-8n}}\right) \to 4x \to 0$. But $(x+1)^n \to \infty$. So the limit is of form $\infty / \infty$. Divide by $x^2(x+1)^n$ on numerator and denominator and the limit is
$$\lim_{x\to 0^+, n\to \infty} \frac{\frac{2\sin^2\left(\frac{\tan(x)}{1+\tan(x)}\right)(x+1)^n}{x^2 (x+1)^n}+\frac{\sin \left(\frac{8nx}{n+\sqrt{n^2-8n}}\right)}{x^2(1+x)^n}}{1+\frac{x}{x^2(1+x)^n }} \\
= \frac{\lim_{x\to 0^+, n\to \infty}\frac{2\sin^2\left(\frac{\tan(x)}{1+\tan(x)}\right)(x+1)^n}{x^2 (x+1)^n}+0}{1+0}
\\=\lim_{x\to 0^+} \frac{2\sin^2\left(\frac{\tan(x)}{1+\tan(x)}\right)}{x^2 } = 2$$
*For left limit, as $n\to \infty$ and $x\to 0^-$, $(x+1)^n \to 0$ , so limit is of form $0/0$:
$$\lim_{x\to 0^-,n\to \infty} \frac{2\sin^2\left(\frac{\tan(x)}{1+\tan(x)}\right)(x+1)^n + \lambda \sin\left(\frac{8nx}{n+\sqrt{n^2-8n}}\right)}{x^2(x+1)^n+x} \\
=\lim_{x\to 0^-, n\to \infty} \frac{2\sin^2\left(\frac{\tan(x)}{1+\tan(x)}\right)\frac{(x+1)^n}{x} + \frac{\lambda}{x} \sin\left(\frac{8nx}{n+\sqrt{n^2-8n}}\right)}{x(x+1)^n+1} \\
= \frac{0+4\lambda}{0+1}$$
For continuity we need $4\lambda = 2 = f(0)$ or $\lambda = 1/2$. So $f(0)+2\lambda = 2+1 = 3$.
| {
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Find for how many values of $n$. $I_n=\int_0^1 \frac {1}{(1+x^2)^n} \, dx = \frac 14 + \frac {\pi}8$
Find for how many values of $n$. $I_n=\int_0^1 \frac {1}{(1+x^2)^n} \, dx=\frac 14 + \frac {\pi}8$
My attempt (integration by parts):
\begin{align}
I_n & = \int_0^1 \frac 1{(1+x^2)^n}\,dx = \left. \frac {x}{(1+x^2)^n} \right|_0^1+2n\int_0^1 \frac {x^2+1-1}{(1+x^2)^{n+1}}\,dx \\[10pt]
& =\frac 1{2^n}+2n \times I_n-2n\times I_{n+1}\implies I_{n+1}= \frac 1{2^{n+1}n}+\frac {2n-1}{2n}I_n.
\end{align}
where $I_1=\frac {\pi}{4}.$
Finding out that $I_2=\frac {\pi}{8}+\frac 14.$ But how do I find out that this is the only solution?
| Hint. The sequence $I_n$ is strictly decreasing because for $x\in (0,1]$,
$$0<\frac {1}{1+x^2} <1.$$
.
| {
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Prove by induction that $n^4-4n^2$ is divisible by 3 for all integers $n\geq1$. For the induction case, we should show that $(n+1)^4-4(n+1)^2$ is also divisible by 3 assuming that 3 divides $n^4-4n^2$. So,
$$ \begin{align} (n+1)^4-4(n+1)&=(n^4+4n^3+6n^2+4n+1)-4(n^2+2n+1)
\\ &=n^4+4n^3+2n^2-4n-3
\\ &=n^4+2n^2+(-6n^2+6n^2)+4n^3-4n-3
\\ &=(n^4-4n^2) + (4n^3+6n^2-4n)-3
\end{align}$$
Now $(n^4-4n^2)$ is divisible by 3, and $-3$ is divisible by 3. Now I am stuck on what to do to the remaining expression.
So, how to show that $4n^3+6n^2-4n$ should be divisible by 3? Or is there a better way to prove the statement in the title? Thank you!
| $$n^4-4n^2=n^2(n^2-1-3)=n\underbrace{(n-1)n(n+1)}_{\text{product of three consecutive integers}}-3n^2$$
| {
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} |
Choose $y\in\mathbb{R}$ so that $\sin y=\frac{u}{\sqrt{u^2+v^2}}$ and $\cos y=\frac{v}{\sqrt{u^2+v^2}}$. I know the identity $\sin^2y+\cos^2y=1$. Also, I notice that
$$\left(\frac{u}{\sqrt{u^2+v^2}}\right)^2+\left(\frac{v}{\sqrt{u^2+v^2}}\right)^2=1$$
without $u,v$ vanishing simultaneously. But I am not sure whether $\exists y\in\mathbb{R}$ s.t. $\sin y=\frac{u}{\sqrt{u^2+v^2}}$ and $\cos y=\frac{v}{\sqrt{u^2+v^2}}$. Does anyone have an idea? Thanks a lot.
| It is clear that $\frac v{\sqrt{u^2+v^2}}\in[-1,1]$. Therefore, since $\cos0=1$ and $\cos\pi=-1$, the intermediate value theorem implies that there's a $y\in[0,\pi]$ such that $\cos y=\frac v{\sqrt{u^2+v^2}}$. Then\begin{align}\sin^2y&=1-\cos^2y\\&=1-\frac{v^2}{u^2+v^2}\\&=\left(\frac u{\sqrt{u^2+v^2}}\right)^2,\end{align}and therefore $\sin y=\pm\frac u{\sqrt{u^2+v^2}}$. If you got the $-$ sign, then replace $y$ by $-y$ and you're done.
| {
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Prove $ \frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\dots+\frac{1}{(n+1)\sqrt{n}}<2$
For any positive integer $n$ prove by induction that:
$$ \frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\dots+\frac{1}{(n+1)\sqrt{n}}<2.$$
The author says that it is sufficient to prove that
$$ \frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\dots+\frac{1}{(n+1)\sqrt{n}}<2-\frac{2}{\sqrt{n+1}}.$$
Why? Where this $\frac{2}{\sqrt{n+1}}$ term come from?
| The stronger inequality is easier to be proved by using induction than the original one. This is another example: in order to prove
$$\dfrac{1}{1^2}+ \dfrac{1}{2^2}+\cdots+\dfrac{1}{n^2} < 2$$ show that
$$\dfrac{1}{1^2}+ \dfrac{1}{2^2}+\cdots+\dfrac{1}{n^2} \le 2-\dfrac{1}{n}.$$
See Induction on inequalities: $\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\ldots+\frac1{n^2}<2$?
In our case, at the inductive step, it suffices to show that
$$2-\frac{2}{\sqrt{n+1}}+\frac{1}{(n+2)\sqrt{n+1}}<2-\frac{2}{\sqrt{n+2}}.$$
Can you take it from here?
| {
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How to show $ \sum\limits_{n=0}^{a_1} \dfrac{a_2 +n \choose a_2}{a_1+a_2 \choose a_2} $ identity? How do i show this identity:
$ \sum\limits_{n=0}^{a_1} \dfrac{a_2 +n \choose a_2}{a_1+a_2 \choose a_2} = \sum\limits_{n=0}^{a_1} \dfrac{a_1 \choose n}{a_1+a_2 \choose n} $
I don't think the partial sums match $ \dfrac{a_2 +n \choose a_2}{a_1+a_2 \choose a_2} ≠ \dfrac{a_1 \choose n}{a_1+a_2 \choose n}$
I was trying to understand why $\sum ^{a_{1}+a_{2}}_{n=1}\dfrac {\begin{pmatrix} a_{1} \\ n \end{pmatrix}}{\begin{pmatrix} a_{1}+a_{2} \\ n \end{pmatrix}}=\dfrac {a_{1}}{1+a_{2}}$,
Somebody kindly outlined the proof, using $\sum\limits_{m=0}^b {c+m \choose c} ={b+c+1 \choose c+1} = \frac{b+c+1}{c+1} {b+c \choose c}$ as $\sum\limits_{n=1}^{a_1+a_2} \dfrac{a_1 \choose n}{a_1+a_2 \choose n} = \sum\limits_{n=0}^{a_1} \dfrac{a_1 \choose n}{a_1+a_2 \choose n} - 1$ = $= \sum\limits_{n=0}^{a_1} \dfrac{a_2 +n \choose a_2}{a_1+a_2 \choose a_2} - 1$ = $= \dfrac{a_2+a_1+1}{a_2+1}-1$
I understand apart from that identity.
|
We start with the right-hand side and obtain
\begin{align*}
\color{blue}{\sum_{n=0}^{a_1}}&\color{blue}{\binom{a_1}{n}\binom{a_1+a_2}{n}^{-1}}\\
&=\sum_{n=0}^{a_1}\frac{a_1!}{n!(a_1-n)!}\cdot\frac{n!(a_1+a_2-n)!}{(a_1+a_2)!}\\
&=\frac{a_1!}{(a_1+a_2)!}\sum_{n=0}^{a_1}\frac{(a_1+a_2-n)!}{(a_1-n)!}\\
&=\frac{a_1!a_2!}{(a_1+a_2)!}\sum_{n=0}^{a_1}\frac{(a_2+n)!}{n!a_2!}\tag{1}\\
&\,\,\color{blue}{=\binom{a_1+a_2}{a_2}^{-1}\sum_{n=0}^{a_1}\binom{a_2+n}{a_2}}
\end{align*}
and the claim follows.
Comment:
*
*In (1) we change the order of summation $n\to a_1-n$ and expand with $a_2!$.
The other identity
\begin{align*}
\sum_{n=1}^{a_1}\binom{a_1}{n}\binom{a_1+a_2}{n}^{-1}=\frac{a_1}{1+a_2}
\end{align*}
can be derived for instance using the Beta function with the identity
\begin{align*}
\binom{p}{q}^{-1}=(p+1)\int_0^1x^q(1-x)^{p-q}\,dx
\end{align*}
We obtain
\begin{align*}
\color{blue}{\sum_{n=1}^{a_1}}&\color{blue}{\binom{a_1}{n}\binom{a_1+a_2}{n}^{-1}}\\
&=(a_1+a_2+1)\int_0^1\sum_{n=1}^{a_1}\binom{a_1}{n}x^n(1-x)^{a_1+a_2-n}\,dx\\
&=(a_1+a_2+1)\int_0^1(1-x)^{a_1+a_2}\sum_{n=1}^{a_1}\binom{a_1}{n}\left(\frac{x}{1-x}\right)^n\,dx\\
&=(a_1+a_2+1)\int_0^1(1-x)^{a_1+a_2}\left[\left(1+\frac{x}{1-x}\right)^{a_1}-1\right]\,dx\\
&=(a_1+a_2+1)\int_0^1\left[(1-x)^{a_2}-(1-x)^{a_1+a_2}\right]\,dx\\
&=(a_1+a_2+1)\left[\frac{1}{a_2+1}-\frac{1}{a_1+a_2+1}\right]\\
&\,\,\color{blue}{=\frac{a_1}{a_2+1}}
\end{align*}
and the claim follows.
| {
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Probable a Binomial Probability Problem :) (not sure) I need to calculate the average profit in a month of a trading strategy.
My probability of a successful trade is 33%
My probability of a bad trade is 67%
The strategy has a Risk Return ratio of 1:3 meaning that I risk 1 to gain 3
So in a bad trade I loose one and in a good trade I gain 3.
I do 3 trades a day.
What is the expected return in a day and in a month, considering 22 days in a month?
My thought was:
$C\binom{3}{1} = \frac{3!}{1!(3-1)!}=\frac{6}{2}=3\\
P=C\binom{n}{k}\cdot p^{k}\cdot q^{n-k}\\
P=3\cdot \left( \frac{1}{3}\right )^1 \cdot \left( \frac{2}{3}\right )^{3-1}\\
P=3\cdot \left( \frac{1}{3}\right )^1 \cdot \left( \frac{2}{3}\right )^{2}\\
P=0.148$
However by doing this I do not take in consideration the risk return ratio so I thought that I should weight the equation with the risk return factor, and I am not sure if this is right.
When I introduce the Return R and Risk rsk I get
$C\binom{3}{1} = \frac{3!}{1!(3-1)!}=\frac{6}{2}=3\\
P=C\binom{n}{k}\cdot p^{k}\cdot q^{n-k}\\
P=3\cdot \left( \frac{1}{3}\right )^1 \cdot \left( \frac{2}{3}\right )^{3-1}\\
P=3\cdot R \cdot \left( \frac{1}{3}\right )^1 \cdot rsk \cdot \left( \frac{2}{3}\right )^{2}\\
P= 1.33$
Assuming that this is OK it means that after a day I will multiply the Value at Risk by 1.33. A superb return ratio of 33% a day and only feasible in math concept not in real world.
After the consideration of Laars Helenius I changed to
Probability of no Success trade:
$C\binom{n}{k} = \frac{n!}{k!(n-k)!}\\
C\binom{3}{0} = \frac{3!}{0!(3-0)!}=\frac{6}{6}=1\\
P=C\binom{n}{k}\cdot p^{k}\cdot q^{n-k}\\
P=1\cdot \left( \frac{1}{3}\right )^0 \cdot \left( \frac{2}{3}\right )^{3}\\
P=0.29\\$
Probability of one success trade:
$C\binom{n}{k} = \frac{n!}{k!(n-k)!}\\
C\binom{3}{1} = \frac{3!}{1!(3-1)!}=\frac{6}{2}=3\\
P=C\binom{n}{k}\cdot p^{k}\cdot q^{n-k}\\
P=3 \cdot \left( \frac{1}{3}\right )^1 \cdot \left( \frac{2}{3}\right )^{2}\\
P=0.44\\$
Probability of 2 success Trades:
$C\binom{n}{k} = \frac{n!}{k!(n-k)!}\\
C\binom{3}{2} = \frac{3!}{2!(3-2)!}=\frac{6}{2}=3\\
P=C\binom{n}{k}\cdot p^{k}\cdot q^{n-k}\\
P=3 \cdot \left( \frac{1}{3}\right )^2 \cdot \left( \frac{2}{3}\right )^{1}\\
P=0.22\\$
Probability of 3 success trades:
$C\binom{n}{k} = \frac{n!}{k!(n-k)!}\\
C\binom{3}{3} = \frac{3!}{3!(3-3)!}=\frac{6}{6}=1\\
P=C\binom{n}{k}\cdot p^{k}\cdot q^{n-k}\\
P=1 \cdot \left( \frac{1}{3}\right )^3 \cdot \left( \frac{2}{3}\right )^{0}\\
P=0.037\\$
How to consider the outcome of each possible situation and finally find out the average profit in a day to use it after in composite interest rate?
Second EDIT
I trade 5% of my capital. So in a \$2000 account I would trade \$100
Meaning that after each day my Value at Risk would be different.
| Expected payouts per day:
A = 3 losses = -3 units
B = 1 win / 2 losses = +1 unit
C = 2 wins / 1 loss = +5 units
D = 3 wins = +9 units
Probabilities:
$\Pr(A) = \binom{3}{0}(1/3)^0(2/3)^3=8/27$
$\Pr(B) = \binom{3}{1}(1/3)^1(2/3)^2=12/27$
$\Pr(C) = \binom{3}{2}(1/3)^2(2/3)^1=6/27$
$\Pr(D) = \binom{3}{3}(1/3)^3(2/3)^0=1/27$
Expected Value:
$A\cdot\Pr(A)+ B\cdot\Pr(B)+ C\cdot\Pr(C)+ D\cdot\Pr(D)=1\text{ unit}$
Then the linearity of expectation tells you to expect to make 22 units over 22 trading days.
| {
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Short way to evaluate $\int_{0}^{1}\ln(\frac{a-x^2}{a+x^2})\cdot\frac{dx}{x^2\sqrt{1-x^2}}$ I would like to evaluate this integral, $$I=\int_{0}^{1}\ln\left(\frac{a-x^2}{a+x^2}\right)\cdot\frac{\mathrm dx}{x^2\sqrt{1-x^2}}$$
An approach:
1. Using integration by parts
$u=\ln\left(\frac{a-x^2}{a+x^2}\right)$, $du=\frac{x^2+a}{x^2-a}\left[\frac{2x}{a+x^2}+\frac{2x(a-x^2)}{a+x^2}\right]dx$
$dv=\frac{dx}{x^2\sqrt{1-x^2}}$, $v=-\frac{\sqrt{1-x^2}}{x}$
$$I=2\int_{0}^{1}\frac{\sqrt{1-x^2}}{x^2+a}dx-2\int_{0}^{1}\frac{\sqrt{1-x^2}}{x^2-a}dx=2I_1-2I_2$$
Integral $I_1$: Make a substitution of: $x=\sin(v)$, $v=\arcsin(x)$, $dx=\cos(v)dv$
$$I_1=\int_{0}^{\pi/2}\frac{\cos^2(v)}{a+\sin^2(v)}dv$$
Using trig to rewrite: $$I_1=\int_{0}^{\pi/2}\sec^2(v)\frac{dv}{[1+\tan^2(v)][(a+1)\tan^2(v)+a]}$$
Make another substitiution of: $v=\tan(y)$, $dy=\frac{1}{\sec^2(v)}dv$ and using partial fraction decomp:
$$I_1=(a+1)\int_{0}^{\infty}\frac{dy}{a+(a+1)y^2}-\frac{\pi}{2}=(a+1)I_3-\frac{\pi}{2}$$
With this substitution: $t=\frac{\sqrt{a+1}}{\sqrt{a}}y$, $dy=\frac{\sqrt{a}}{\sqrt{a+1}}dt$
$$I_3=\frac{1}{\sqrt{a(a+1)}}\cdot \frac{\pi}{2}$$
$$I_1=\frac{\pi}{2}\left(\frac{\sqrt{a+1}}{\sqrt{a}}-1\right)$$
In the same manner, $$I_2=\frac{\pi}{2}\left(\frac{\sqrt{a-1}}{\sqrt{a}}-1\right)$$
Finally, $$I=\pi\left(\frac{\sqrt{a-1}-\sqrt{a+1}}{\sqrt{a}}\right)$$
I am looking for another short approach of evaluating integral $I$
| Under $x\to\sin x$, one has
$$I=\int_{0}^{1}\ln\left(\frac{a-x^2}{a+x^2}\right)\cdot\frac{\mathrm dx}{x^2\sqrt{1-x^2}}=\int_{0}^{\pi/2}\ln\left(\frac{a-\sin^2x}{a+\sin^2x}\right)\cdot\frac{\mathrm dx}{\sin^2x}.$$
Let
$$I(k)=\int_{0}^{\pi/2}\ln\left(\frac{a-k\sin^2x}{a+k\sin^2x}\right)\cdot\frac{\mathrm dx}{\sin^2x}$$
and then
\begin{eqnarray}
I'(k)&=&\int_{0}^{\pi/2}\left(-\frac1{a-k\sin^2x}-\frac1{a+k\sin^2x}\right)\mathrm dx\\
&=&-\frac{\pi}{2}\left(\frac1{\sqrt{a(a+k)}}+\frac1{\sqrt{a(a-k)}}\right).
\end{eqnarray}
So
$$ I=I(1)=-\frac{\pi}{2\sqrt a}\int_0^1\left(\frac1{\sqrt{a+k}}-\frac1{\sqrt{a-k}}\right)\mathrm dk=-\frac{\pi}{\sqrt a}(\sqrt{a+1}-\sqrt{a-1}).$$
| {
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Equation involving floor function and fractional part function How to solve $\frac{1}{\lfloor x \rfloor} + \frac{1}{\lfloor 2x \rfloor} = \{x\} + \frac{1}{3}$ , where $\lfloor \rfloor$ denotes floor function and {} denotes fractional part. I did couple of questions like this by solving for {x} and bounding it from 0 to 1. So here we will have, $$0\le\frac{1}{\lfloor x \rfloor} + \frac{1}{\lfloor 2x \rfloor} - \frac{1}{3}\lt1$$ adding throughout by $\frac{1}{3}$ $$\frac{1}{3}\le\frac{1}{\lfloor x \rfloor} + \frac{1}{\lfloor 2x \rfloor} \lt\frac{4}{3}$$ Now I am stuck.
| Let $x = n+z$ where $n$ is an integer and $0\leq z <1,$ and break into two cases.
Case 1: If $z<1/2$ the equation becomes
$$\frac{1}{n}+\frac{1}{2n} = z+\frac{1}{3},$$
which leads to
$$\frac{1}{3} \leq \frac{3}{2n} <\frac{1}{2} + \frac{1}{3} = \frac{5}{6}$$
which is equivalent to $1.8 < n \leq 4.5.$ So $n = 2, 3,$ or $4.$
Plug each of these into the original equation to get $z = 5/12, 1/6, 1/24,$ respectively, giving 3 solutions: $2 \frac{5}{12}, 3\frac{1}{6}, 4\frac{1}{24}$.
Case 2: If $z\geq 2$, the equation becomes
$$\frac{1}{n}+\frac{1}{2n} = z+\frac{1}{3},$$
and since $1/2 \leq z <1$ we have
$$\frac{5}{6} \leq \frac{1}{n}+\frac{1}{2n} < \frac{4}{3},$$
which has no integer solutions.
| {
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Position of Q in which the area of triangle CDQ is a third part of the area of square ABCD. Given
*
*$A(8,0)$
*$B(0,4)$
*$C(-4,-4)$
*$D(4,-8)$
*$P(2,3)$
Question
In the figure the triangle $CDQ$ is grayed out. There is a position of $Q$ in which the area of triangle $CDQ$ is a third part of the area of square $ABCD$. For this position, calculate exactly the coordinates of $Q$.
Answer:
I calculates that the height of triangle $CDQ$, with base $CD$, must be $2/3$ part of the side of the square.
I became step one of the solution, but I do not understand the following steps of the answer. Can someone help? I added a picture of the answer!
| The side of the square is $4\sqrt{5}$, so the area of the square is $80$.
The area of $\triangle_{QCD}=\frac{1}{2}\cdot4\sqrt{5}\cdot h=\frac{80}{3}$, so the height $h=\frac{8\sqrt{5}}{3}$, which is $\frac{2}{3}$ the side of the square. Now imagine drawing a line through the midpoint of $CD$, say $R$, through $Q$, to the midpoint of $AB$, say $S$, this bisects the square and has $RQ:QS=\frac{2}{3}:\frac{1}{3}$, or equivalently $RQ:QS=2:1$, and so $DQ:QP=2:1$ also.
Now find $\overrightarrow{DP}=\overrightarrow{DO}+\overrightarrow{OP}=-\overrightarrow{OD}+\overrightarrow{OP}=\binom{-4}{8}+\binom{2}{3}=\binom{-2}{11}$.
You can now find $\overrightarrow{OQ}=\overrightarrow{OD}+\frac{2}{3}\overrightarrow{DP}=\binom{4}{-8}+\frac{2}{3}\cdot\binom{-2}{11}=\binom{8/3}{-2/3}$, giving $Q=(\frac{8}{3},-\frac{2}{3})$ as required.
| {
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Equation involving inverse trigonometric function $$I=\cot^{-1}\frac{y}{\sqrt{1-x^2-y^2}} = 2\tan^{-1}\sqrt{\frac{3-4x^2}{4x^2}} - \tan^{-1}\sqrt{\frac{3-4x^2}{x^2}} $$
Expression of $I$ as a rational integer equation in $x$ and $y$ is
$$
(a) 27x^2=y^2(9-8y^2)
\\(b) 27y^2=x^2(9-8x^2)
\\(c) 27x^4 - 9x^2 + 8y^2=0
\\(d) 27y^4 - 9y^2 + 8x^2=0
$$
My attempt : I tried to assume $$x=\frac{\sqrt{3}}{2}sin\theta$$
That reduces the first arctan function to a simple expression $2\theta$, but the second one assumes the form $\tan^{-1}(2\tan\theta)$. How can I find the correct answer? Any help will be appreciated.
| Hint:
Let $\sqrt{\dfrac{3-4x^2}{x^2}}=2a$
$$2\tan^{-1}a-\tan^{-1}(2a)=\tan^{-1}a+\tan^{-1}a+\tan^{-1}(-2a)$$
$$\tan\left(\tan^{-1}a+\tan^{-1}a+\tan^{-1}(-2a)\right)=\dfrac{a+a-2a+4a^3}{1-a^2-a(-2a)-a(-2a)}$$
which is $$=\dfrac{\sqrt{1-x^2-y^2}}y$$
Now square both sides and replace the value of $a^2$
| {
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I need to find the real and imaginary part of this $Z_{n}=\left (\frac{ \sqrt{3} + i }{2}\right )^{n} + \left (\frac{ \sqrt{3} - i }{2}\right )^{n}$ I have a test tomorrow and i have some troubles understanding this kind of problems, would really appreciate some help with this
$$ Z_{n}=\left (\frac{ \sqrt{3} + i }{2}\right )^{n} + \left (\frac{ \sqrt{3} - i }{2}\right )^{n}
$$
$$
Z_{n}\epsilon \mathbb{C}
$$
| Note that $z=\left (\frac{ \sqrt{3} + i }{2}\right )^{n}=(a+bi)^n$ we have
$$|z|=\sqrt{(\sqrt{3}/2)^2+(1/2)^2}=1, \quad \theta_z=\arctan(\tfrac{b}{a})=\arctan(\tfrac{1}{\sqrt{3}})=\tfrac{\pi}{6}$$
and for $w=\left (\frac{ \sqrt{3} - i }{2}\right )^{n}=(a-bi)^n$ we have
$$|w|=\sqrt{(\sqrt{3}/2)^2+(1/2)^2}=1, \quad \theta_w=\arctan(-\tfrac{b}{a})=\arctan(-\tfrac{1}{\sqrt{3}})=-\tfrac{\pi}{6}$$
Then,
$$z=|z|^n(\cos\theta_z+i\sin\theta_z)^n=\left[\cos\left(\tfrac{\pi}{6}\right)+i\sin\left(\tfrac{\pi}{6}\right)\right]^n$$
$$w=|w|^n(\cos\theta_w+i\sin\theta_w)^n=\left[\cos\left(-\tfrac{\pi}{6}\right)+i\sin\left(-\tfrac{\pi}{6}\right)\right]^n$$
Using de Moivre's formula
$$z^n=|z|^n(\cos\theta+i\sin\theta)^n=|z|^n(\cos(n\theta)+i\sin(n\theta))^n$$
we get
$$z=\left[\cos\left(\tfrac{\pi}{6}\right)+i\sin\left(\tfrac{\pi}{6}\right)\right]^n=\left[\cos\left(\tfrac{n\pi}{6}\right)+i\sin\left(\tfrac{n\pi}{6}\right)\right]$$
$$w=\left[\cos\left(-\tfrac{\pi}{6}\right)+i\sin\left(-\tfrac{\pi}{6}\right)\right]^n=\left[\cos\left(-\tfrac{n\pi}{6}\right)+i\sin\left(-\tfrac{n\pi}{6}\right)\right]=\left[\cos\left(\tfrac{n\pi}{6}\right)-i\sin\left(\tfrac{n\pi}{6}\right)\right]$$
Finally,
$$Z_n=z+w=2\cos\left(\tfrac{n\pi}{6}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find the derivative of a function that contains a sum If $$f(x)=\sum_{n=1}^\infty{\frac{{(-1)}^{n+1}}{n\cdot3^n}{(x-3)}^n}$$ find $f''(-2)$.
I know that, by ratio test, the previous sum converges iff $x\in(0,6)$.
I did:
$$\begin{matrix}
f'(x)&=&\left(\displaystyle\sum_{n=1}^\infty{\frac{{(-1)}^{n+1}}{n\cdot3^n}{(x-3)}^n}\right)'&=&
\displaystyle\sum_{n=1}^\infty{\frac{{(-1)}^{n+1}}{n\cdot3^n}n{(x-3)}^{n-1}}&=&
\displaystyle\sum_{n=1}^\infty{\frac{{(-1)}^{n+1}}{3^n}{(x-3)}^{n-1}}
\\
f''(x)&=&\left(\displaystyle\sum_{n=1}^\infty{\frac{{(-1)}^{n+1}}{3^n}{(x-3)}^{n-1}}\right)'&=&
\displaystyle\sum_{n=1}^\infty{\frac{{(-1)}^{n+1}}{3^n}(n-1){(x-3)}^{n-2}},
\end{matrix}$$
so I did the radio test on the last sum:
$$\begin{matrix}
&&\displaystyle\lim_{n\to\infty}{\left|\dfrac{a_{n+1}}{a_n}\right|} \\
&=&\displaystyle\lim_{n\to\infty}{\left|\dfrac{{(-1)}^{n+2}n{(x-3)}^{n-1}}{3^{n+1}}\dfrac{3^n}{{(-1)}^{n+1}(n-1){(x-3)}^{n-2}}\right|}\\
&=&\dfrac{|x-3|}{3}\underbrace{\displaystyle\lim_{n\to\infty}{\left|\dfrac{n}{n-1}\right|}}_{=\:1}\\
&\Rightarrow&|x-3|<3\\
&\Rightarrow&x\in(0,6),&
\end{matrix}$$
and since $-2\not\in(0,6)$ we cannot find $f''(-2)$.
First question: is my reasoning right?
If yes, to my amazement the radius of convergence of both functions are the same. Out of curiosity, why is this happening? The convergence can be generalize to $n$-th derivative?
Thank you!
| In terms of finding $f''(-2)$ with
$$f(x)=\sum_{n=1}^\infty{\frac{{(-1)}^{n+1}}{n\cdot3^n}{(x-3)}^n}$$
then consider the following.
Method 1
Since
\begin{align}
f(x) &= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \, \left(1 - \frac{x}{3}\right)^{n} = \ln\left(\frac{x}{3}\right)
\end{align}
then
\begin{align}
f'(x) &= \frac{1}{x} \\
f''(x) &= - \frac{1}{x^2}
\end{align}
Method 2
\begin{align}
f(x) &= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \, \left(1 - \frac{x}{3}\right)^{n} \\
f'(x) &= \frac{1}{3} \, \sum_{n=1}^{\infty} \left( \frac{3 - x}{3} \right)^{n} = \frac{1}{3 \, \left(1 - \left( 1 - \frac{x}{3} \right) \right)} = \frac{1}{x} \\
f''(x) &= - \frac{1}{x^2}.
\end{align}
The value sought can now be obtained by setting $x = -2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Range of $(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$ is
If $a_{1},a_{2},a_{3},a_{4}\in \mathbb{R}$ and $a_{1}+a_{2}+a_{3}+a_{4}=0$ and $a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4}=1.$
Then Range of $$E =(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$$ is
Try:
From $$(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$$
$$=2(a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4})+2(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{4}+a_{4}a_{1})$$
$$=2-2(a_{1}+a_{3})(a_{2}+a_{4})=2+2(a_{1}+a_{3})^2\geq 2$$
and equality hold when $\displaystyle a_{1}=-a_{3}$ and $a_{2}=-a_{4}$
Could some help me how to find its upper bound, Thanks
| You can try also like this. Write $$a_1 =\cos \alpha \sin \beta$$
$$a_2 =\cos \alpha \cos \beta$$
$$a_3 =\sin \alpha \sin \beta$$
$$a_4 =\sin \alpha \cos \beta$$
for some $\alpha$ and $\beta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2842697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Is $ \lim_{x\to \infty}\frac{x^2}{x+1} $ equal to $\infty$ or to $1$? What is
$ \lim_{x\to \infty}\frac{x^2}{x+1}$?
When I look at the function's graph, it shows that it goes to $\infty$, but If I solve it by hand it shows that the $\lim \rightarrow 1$
$ \lim_{x\to \infty}\frac{x^2}{x+1} =$ $ \lim_{x\to \infty}\frac{\frac{x^2}{x^2}}{\frac{x}{x^2}+\frac{1}{x^2}} =$ $\lim_{x\to \infty}\frac{1}{\frac{1}{x}+\frac{1}{x^2}}=$$ \lim_{x\to \infty}\frac{1}{0+0} = \ 1$
what is wrong here?
| Writing the fraction as follows may clear it up:
$$
\frac{x^2}{x+1}
= \frac{x^2-1+1}{x+1}
= \frac{x^2-1}{x+1} + \frac{1}{x+1}
= x-1 + \frac{1}{x+1}
\to \infty
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2844097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Limit $\lim_{n\to\infty} n^2\left(\sqrt{1+\frac{1}{n}}+\sqrt{1-\frac{1}{n}}-2\right)$ Greetings I am trying to solve $$\lim_{n\to\infty} n^2\left(\sqrt{1+\frac{1}{n}}+\sqrt{1-\frac{1}{n}}-2\right)$$ Using binomial series is pretty easy: $$\lim_{n\to\infty}n^2\left(1+\frac{1}{2n}-\frac{1}{8n^2}+\mathcal{O}\left(\frac{1}{n^3}\right)+1-\frac{1}{2n}-\frac{1}{8n^2}+\mathcal{O}\left(\frac{1}{n^3}\right)-2\right)=\lim_{n\to\infty}n^2\left(-\frac{1}{8n^2}+\mathcal{O}\left(\frac{1}{n^3}\right)-\frac{1}{8n^2}+\mathcal{O}\left(\frac{1}{n^3}\right)\right)=-\frac{1}{4}$$ The problem is that I need to solve this using only highschool tools, but I cant seem too take it down. My other try was to use L'Hospital rule but I feel like it just complicate things. Maybe there is even an elegant way, could you give me some help with this?
| Hint: multiplying numerator and denominator by $\sqrt{1+1/n}+\sqrt{1-1/n}+2$ we get
$$2 n^2 \frac{\sqrt{1-1/n^2}-1}{\sqrt{1+1/n}+\sqrt{1-1/n}+2}$$ and then do the same with $$\sqrt{1-1/n^2}+1$$
you will get
$$\frac{n^2(2(\sqrt{1-1/n^2}-1))(\sqrt{1-1/n^2}+1)}{(\sqrt{1+1/n}+\sqrt{1-1/n}+2)(\sqrt{1-1/n^2}+1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2844601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the quotient and remainder Find the quotient and remainder when $x^6+x^3+1$ is divided by $x+1$
Let $f(x)=x^6+x^3+1$
Now $f(x)=(x+1).q(x) +R $ where r is remainder
Now putting $x=-1$ we get $R=f(-1)$
i.e $R=1-1+1=1$
Now $q(x)=(x^6+x^3)/(x+1)$
But what I want to know if there is another way to get the quotient except simple division.
| $$f(x)=x^6+x^3+1= x^6 +x^5 -x^5 +x^3+1\\ = x^5(x+1) -x^5-x^4+x^4+x^3+1 \\=x^5(x+1)-x^4(x+1)+x^3(x+1)+1\\= (x+1)(x^5-x^4+x^3)+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2847682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Prove $ \min \left(a+b+\frac1a+\frac1b \right) = 3\sqrt{2}\:$ given $a^2+b^2=1$ Prove that
$$ \min\left(a+b+\frac1a+\frac1b\right) = 3\sqrt{2}$$
Given $$a^2+b^2=1 \quad(a,b \in \mathbb R^+)$$
Without using calculus.
$\mathbf {My Attempt}$
I tried the AM-GM, but this gives $\min = 4 $.
I used Cauchy-Schwarz to get $\quad (a+b)^2 \le 2(a^2+b^2) = 2\quad \Rightarrow\quad a+b\le \sqrt{2}$
But using Titu's Lemma I get $\quad \frac1a+\frac1b \ge \frac{4}{a+b}\quad \Rightarrow\quad \frac1a+\frac1b \ge 2\sqrt{2}$
I'm stuck here, any hint?
| This is a summary of the arguments and comments above:
$a^2 + b^2 = 1$ and they symmetry argument that $a = b$ shows that $a = b = {1 \over \sqrt{2}}$, and thus $a + b + 1/a + 1/b = 3 \sqrt{2}$.
Calculus confirms it:
$$f(a) = a + 1/a + \sqrt{1 - a^2} + {1 \over \sqrt{1 - a^2}}$$
so
$${df(a) \over da} = -\frac{a}{\sqrt{1-a^2}}+\frac{a}{\left(1-a^2\right)^{3/2}}-\frac{1}{a^2}+1$$
Set this equal to zero and solve for $a$ to find $a = {1 \over \sqrt{2}}$ and the rest follows.
Here's a plot of $f(a)$:
Indeed, the minimum occurs at $a = {1 \over \sqrt{2}}$ and has value $f(a = 1/\sqrt{2}) = 3 \sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2850000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 3
} |
Find the Determinant when $p$ and $q$ are roots of $x^2-2x+5=0$ If $p$ and $q$ are roots of $x^2-2x+5=0$ Then find value of
$$\Delta=\begin{vmatrix}
1 & 1+p^2+q^2 & 1+p^3+q^3\\
1+p^2+q^2& 1+p^4+q^4 & 1+p^5+q^5\\
1+p^3+q^3 & 1+p^5+q^5 & 1+p^6+q^6
\end{vmatrix}$$
My try: I tried to express the determinant as product of two determinants but in vain.
Secondly i tried taking $$p=\sqrt{5}\frac{1+2i}{\sqrt{5}}=\sqrt{5}(\cos t+i\sin t)$$ and $q$ its conjugate, but still its lengthy
| HINT:
Note that by Vieta's formulas, $p+q=2$ and $pq=5$ so $$1+p+q=3$$ and $$1+p^2+q^2=1+2p-5+2q-5=2(p+q)-9=-5$$ giving $$1+p^3+q^3=(1+p+q)(1+p^2+q^2)-(p^2+q^2+p+q+pq(p+q))$$ and since $x^2=2x-5$, $$1+p^3+q^3=3(-5)-(2p-5+2q-5+2+5(2))$$ or $$1+p^3+q^3=-15-(2(p+q)+2)=-15-(6)=-21$$ and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2850392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\left(1+\frac1 n\right)^n > 2$ I'm trying to demonstrate that $\left( 1+\frac1 n \right)^n$ is bigger than $2$. I have tried to prove that $\left( 1+\frac1 n \right)^n$ is smaller than $\left( 1+\frac1{n+1} \right)^{n+1}$ by expanding
$\left( 1+\frac1n \right)^n = \sum\limits_{i=0}^n \left( \frac{n}{k} \right) \frac{1}{n^k}$ and $\left( 1+\frac1{n+1} \right)^{n+1} = \sum\limits_{i=0}^{n+1} \left( \frac{(n+1)}{k} \right) \frac{1}{(n+1)^k}$ but it doesn't seem to work.
What am I missing? Also, is there a method to demonstrate that without induction?
| \begin{align}
\left(1+\frac1n\right)^n &= \sum_{k=0}^n {n \choose k} \frac1{n^k} \\
&= \sum_{k=0}^n \frac{n(n-1)\cdots(n-k+1)}{k!n^k} \\
&= \sum_{k=0}^n \frac1{k!}\left(1-\frac1n\right)\left(1-\frac2n\right)\cdots \left(1-\frac{k-1}n\right)
\end{align}
so
\begin{align}
\left(1+\frac1{n+1}\right)^{n+1} &= \sum_{k=0}^{n+1} \frac1{k!}\left(1-\frac1{n+1}\right)\left(1-\frac2{n+1}\right)\cdots \left(1-\frac{k-1}{n+1}\right)\\
&> \sum_{k=0}^{n} \frac1{k!}\left(1-\frac1{n}\right)\left(1-\frac2{n}\right)\cdots \left(1-\frac{k-1}{n}\right)\\
&= \left(1+\frac1n\right)^n
\end{align}
On the other hand, for $n = 1$ we have
$$\left(1+\frac11\right)^1 = 2$$
so $\left(1+\frac1n\right)^n > 2, \forall n \ge 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2853989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Evaluating ${\Large\int} _{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sin(x)+\cos(x)}{\sin^4(x)-4}dx$ This integral is giving me hard times, could anyone "prompt" a strategy about? I tried, resultless, parameterization and some change of variables.
$${\Large\int} _{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sin(x)+\cos(x)}{\sin^4(x)-4}dx$$
| $$\int\dfrac{\sin x}{\sin^4 x-4}dx=-\int\dfrac{1}{(1-u^2)^2-4}du\\\int\dfrac{\cos x}{\sin^4 x-4}dx=\int\dfrac{du}{u^4-4}=0.25\int\dfrac{du}{u^2-2}-0.25\int\dfrac{du}{u^2+2}$$the first one is zero over a symmetric interval and for the second we have$$I=0.25\int_{-1}^{1}\dfrac{du}{u^2-2}-0.25\int_{-1}^{1}\dfrac{du}{u^2+2}=\dfrac{1}{4\sqrt 2}\ln\dfrac{\sqrt 2-1}{\sqrt 2+1}-{\sqrt{2}\over 4}\tan^{-1}\dfrac{1}{\sqrt 2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
a tough sum of binomial coefficients
Find the sum: $$\sum_{i=0}^{2}\sum_{j=0}^{2}\binom{2}{i}\binom{2}{j}\binom{2}{k-i-j}\binom{4}{k-l+i+j},\space\space 0\leq k,l\leq 6$$
I know to find $\sum_{i=0}^{2}\binom{2}{i}\binom{2}{2-i}$, I need to find the coefficient of $x^2$ of $(1+x)^4$ (which is $\binom{4}{2}$). But I failed to use that trick here. Any help appreciated!
| Computing the generating function:
$$
\begin{align}
&\sum_k\sum_l\sum_i\sum_j\binom{2}{i}\binom{2}{j}\binom{2}{k-i-j}\binom{4}{k-l+i+j}x^ly^k\\
&=\sum_k\color{#C00}{\sum_l}\sum_i\sum_j\binom{2}{i}\binom{2}{j}\binom{2}{k-i-j}\color{#C00}{\binom{4}{k-l+i+j}x^{l+4-k-i-j}}x^{k+i+j-4}y^k\\
&=\color{#C00}{(1+x)^4}\sum_k\sum_i\sum_j\binom{2}{i}\binom{2}{j}\binom{2}{k-i-j}x^{k+i+j-4}y^k\\
&=(1+x)^4\color{#090}{\sum_k}\sum_i\sum_j\binom{2}{i}\binom{2}{j}\color{#090}{\binom{2}{k-i-j}x^{k-i-j}y^{k-i-j}}x^{2i+2j-4}y^{i+j}\\
&=(1+x)^4\color{#090}{(1+xy)^2}\color{#00F}{\sum_i\sum_j\binom{2}{i}\binom{2}{j}x^{2i+2j-4}y^{i+j}}\\
&=\frac{(1+x)^4(1+xy)^2\color{#00F}{(1+x^2y)^4}}{\color{#00F}{x^4}}\\
&=\color{#CCC}{\frac1{x^4}+\frac{4+2y}{x^3}+\frac{6+12y+y^2}{x^2}+\frac{4+28y+12y^2}x}\\
&+\left(1+32y+44y^2+4y^3\right)+x\left(18y+76y^2+28y^3\right)+x^2\left(4y+69y^2+76y^3+6y^4\right)\\
&+x^3\left(32y^2+104y^3+32y^4\right)+x^4\left(6y^2+76y^3+69y^4+4y^5\right)\\
&+x^5\left(28y^3+76y^4+18y^5\right)+x^6\left(4y^3+44y^4+32y^5+y^6\right)\\
&\color{#CCC}{+x^7\left(12y^4+28y^5+4y^6\right)+x^8\left(y^4+12y^5+6y^6\right)+x^9\left(2y^5+4y^6\right)+x^{10}y^6}
\end{align}
$$
where the terms not in the requested range have been grayed out.
| {
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"url": "https://math.stackexchange.com/questions/2855695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $|z_{1}+z_{2}|^2\leq (1+c)|z_{1}|^2+\bigg(1+\frac{1}{c}\bigg)|z_{2}|^2$
If $z_{1},z_{2}$ are two complex numbers and $c>0.$ Then prove that
$\displaystyle |z_{1}+z_{2}|^2\leq (1+c)|z_{1}|^2+\bigg(1+\frac{1}{c}\bigg)|z_{2}|^2$
Try: put $z_{1}=x_{1}+iy_{1}$ and $z_{2}=x_{2}+iy_{2}.$ Then
from left side
$$(x_{1}+x_{2})^2+(y_{1}+y_{2})^2=x^2_{1}+x^2_{2}+2x_{1}x_{2}+y^2_{1}+y^2_{2}+2y_{1}y_{2}$$
Could some help me how to solve it further, Thanks in Advance.
| You are on the right track. Following your approach the inequality becomes
$$x^2_{1}+x^2_{2}+2x_{1}x_{2}+y^2_{1}+y^2_{2}+2y_{1}y_{2}\leq (1+c)(x_1^2+y_1^2)+\bigg(1+\frac{1}{c}\bigg)(x_2^2+y_2^2)$$
that is
$$2x_{1}x_{2}+2y_{1}y_{2}\leq c(x_1^2+y_1^2)+\frac{1}{c}(x_2^2+y_2^2).$$
Is it true that $2uv\leq cu^2+\frac{v^2}{c}$? (Hint. use AM-GM inequality).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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find all the points on the line $y = 1 - x$ which are $2$ units from $(1, -1)$ I am really struggling with this one. I'm teaching my self pre-calc out of a book and it isn't showing me how to do this. I've been all over the internet and could only find a few examples. I only know how to solve quadratic equations by converting them to vertex form and would like to stick with this method until it really sinks in. What am I doing wrong?
1.) Distance formula $\sqrt{(x-1)^2 + (-1 -1 + x)^2}=2$
2.) remove sqrt, $(x - 1)(x - 1) + (x - 2)(x - 2) = 4$
3.) multiply, $x^2 - 2x +1 + x^2 -4x +4 = 4$
4.) combine, $2x^2 -6x +5 = 4$
5.) general form, $2x^2 -6x +1$
6.) convert to vertex form (find the square), $2(x^2 - 3x + 1.5^2)-2(1.5)^2+1$
7.) Vertex form, $2(x-1.5)^2 -3.5$
8.) Solve for x, $x-1.5 = \pm\sqrt{1.75}$
9.) $x = 1.5 - 1.32$ and $x = 1.5 + 1.32$
10.) $x = 0.18$ and $2.82$
When I plug these two $x$ values back into the vertex form of the quadratic equation, I'm getting $y = 0.02$ for both $x$ values. These points are not on the line. Can someone tell me what I'm doing wrong please?
| Since you have the general form $2x^2-6x+1=0$
Now solve for x,
$$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$
$$x=\dfrac32\pm\dfrac12\sqrt{7},$$
Now try to plug in these values of $x$.
Edit:
$$2(x-1.5)^2-3.5=0$$
$$2(x^2+2.25-3x)=3.5$$
$$x^2-3x+2.25=1.75$$
$$x^2-3x+0.5=0$$
$$2x^2-6x+1=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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Find $\log _{24}48$ if $\log_{12}36=k$ Find $\log _{24}48$ if $\log_{12}36=k$
My method:
We have $$\frac{\log 36}{\log 12}=k$$ $\implies$
$$\frac{\log 12+\log 3}{\log 12}=k$$ $\implies$
$$\frac{\log3}{2\log 2+\log 3}=k-1$$ So
$$\log 3=(k-1)t \tag{1}$$
$$2\log 2+\log 3=t$$ $\implies$
$$\log 2=\frac{(2-k)t}{2} \tag{2}$$
Now $$\log _{24}48=\frac{\log 48}{\log 24}=\frac{4\log 2+\log 3}{3\log 2+\log 3}=\frac{2(2-k)+k-1}{3\left(\frac{2-k}{2}\right)+k-1}=\frac{6-2k}{4-k}$$
is there any other approach?
| We have
$$\log _{24}48=\frac{\log _{12}48}{\log _{12}24}=\frac{\log _{12}36+\log _{12}\frac43}{\log _{12}36+\log _{12}\frac23}=\frac{k+\log _{12}\frac43}{k+\log _{12}\frac23}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Prove: $\sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}=\frac{\sqrt2}{2}$
Prove: $$\sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}=\frac{\sqrt2}{2}$$
ok, what I saw instantly is that:
$$\sin\frac{\pi}{20}+\sin\frac{3\pi}{20}=2\sin\frac{2\pi}{20}\cos\frac{\pi}{20}$$
and that,
$$\cos\frac{\pi}{20}-\cos\frac{3\pi}{20}=-2\sin\frac{2\pi}{20}\sin\frac{\pi}{20}$$
So,
$$2\sin\frac{2\pi}{20}(\cos\frac{\pi}{20}-\sin\frac{\pi}{20})=\frac{\sqrt2}{2}=\sin\frac{5\pi}{20}$$
Unfortunately, I can't find a way to continue this, any ideas or different ways of proof?
*Taken out of the TAU entry exams (no solutions are offered)
| Hint: multiply both sides with $\frac{\sqrt 2}{2} $: $$\underbrace{\sin\frac{\pi}{20}+\cos\frac{\pi}{20}}+\underbrace{\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}}=\frac{\sqrt 2}{2}\;\;\;/\cdot \frac{\sqrt 2}{2} $$
so we have to prove:
$$\sin\frac{3\pi}{10}-\sin\frac{\pi}{10}=\frac{1}{2} $$
| {
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"url": "https://math.stackexchange.com/questions/2858498",
"timestamp": "2023-03-29T00:00:00",
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Is there a simpler way to determine m, n, p, such that the following holds for all reals? I am given the following equation and I am asked to find $m$, $n$ and $p$, so that the equation holds for all reals:
$$ \sin^4x + \cos^4x + m(\sin^6x + \cos^6x) + n(\sin^8x + \cos^8x) + p(\sin^{10}x + \cos^{10}x) = 1, \space \forall x \in \mathbb R $$
I have managed to solve it by, in advance, calculating the following power reduction formulas, and then applying them to the equation:
$$
\sin^4x + \cos^4x = 1 - \frac12\sin^2{2x} \\
\sin^6x + \cos^6x = 1 - \frac34\sin^2{2x} \\
\sin^8x + \cos^8x = 1 - \sin^2{2x} + \frac18\sin^4{2x} \\
\sin^{10}x + \cos^{10}x = 1 - \frac54\sin^2{2x} + \frac5{16}\sin^4{2x}
$$
As a result, I managed to simplify it to the following, which is only in terms of powers of $\sin{2x}$:
$$ \left( 1+m+n+p \right) - \left( \frac12 + \frac{3m}4 + n + \frac{5p}4 \right) \sin^2{2x} + \left( \frac{n}8 + \frac{5p}{16} \right) \sin^4{2x} = 1, \space \forall x \in \mathbb R $$
I then came to the conclusion that the only possible way for which this can be true is if:
$1+m+n+p=1$, $ \frac12 + \frac{3m}4 + n + \frac{5p}4 = 0 $ and $ \frac{n}8 + \frac{5p}{16} = 0 $.
By solving the system of equations below I arrive at the solutions $m=6$, $n=-10$, $p=4$.
$$
\left\{
\begin{aligned}
1 + m + n + p = 1 \\
\frac12 + \frac{3m}4 + n + \frac{5p}4 = 0 \\
\frac{n}8 + \frac{5p}{16} = 0
\end{aligned}
\right.
$$
My question is if my reasoning is correct and if there exists any simpler way to solve the problem.
| Your reasoning is correct. You may save a little algebraa by formulating it as follows. Define $\, f(t) := t^2 + m t^3 + n t^4 + p t^5. \,$ We want the equation
$\, f(t) + f(1-t) - 1 = 0\,$ to be true for all $\, t = \sin^2(x). \,$ If you expand the left side, it is a fourth degreee polynomial in $\,t.\,$ All of its coefficients must be zero. This gives five linear equations in the four unknowns and the solution is as you found. The advantage of this approach is that it avoids the need to express $\, \sin^{2n}(x) + \cos^{2n}(x) \,$ in terms of even powers of $\, \sin(2x). \,$ All that is needed is to know the identity $\, \sin^2 x + \cos^2 x = 1. $
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\min\{x \in\mathbb{N} : xn+1 \text{ is a square}\} \geq \frac{n}{4}$ then n is prime.
Let $n$ be a odd natural number greater than 3 and $k=\min\{x \in\mathbb{N} : xn+1 \text{ is a square}\}$ and $t=\min\{x \in\mathbb{N} : xn \text{ is a square}\}.$ If $k>\frac{n}{4}$ and $t>\frac{n}{4}$ then n is prime.
I proved second direction easily, so I don't cite it. If $n$ is composite and isn't squarefree the hypothese about $t$ is false and it's very easy from prime decomposition. I have problems wiith a part about composite squarefree $n$. I tried prime decomposition, I tried find explicitly k to contradict. I spent a few hours. I would be very grateful for any hint.
| If $n\in \mathbb{N}$ is odd and not a prime power, then there is always an $a$ such that $1 < a \leqslant \frac{n-1}{2}$ and $a^2 \equiv 1 \pmod{n}$. Then it follows that
$$k \leqslant \frac{a^2-1}{n} < \frac{n^2/4}{n} = \frac{n}{4}\,.$$
To see that such an $a$ exists, let $p$ be a prime dividing $n$, and write $n = p^r\cdot m$ with $p \nmid m$. Since $n$ is not a prime power, we have $m > 1$. By the Chinese remainder theorem, there is a $0 < b < n$ such that $b \equiv 1 \pmod{p^r}$ and $b \equiv -1 \pmod{m}$. Since $b \equiv 1 \pmod{p^r}$ we have $b \neq n-1$, and since $b \equiv -1 \pmod{m}$ we have $b \neq 1$, thus $1 < b < n-1$. If $b \leqslant \frac{n-1}{2}$, put $a = b$, else put $a = n-b$. Then $1 < a \leqslant \frac{n-1}{2}$, and $a^2 \equiv (\pm 1)^2 \equiv 1 \pmod{p^r}$ and $a^2 \equiv (\mp 1)^2 \equiv 1 \pmod{m}$, so we indeed have $a^2 \equiv 1 \pmod{n}$.
If $n$ is an odd prime power and not a prime, then $n = p^r$ for some odd prime $p$ and $r > 1$. If $r$ is even, then $n$ is already a square and so $t = 1 < \frac{n}{4}$. If $r$ is odd, then $pn$ is the smallest square multiple of $n$, and $t = p \leqslant \frac{n}{p^2} \leqslant \frac{n}{9} < \frac{n}{4}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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The value of $a+b=ab=a^2-b^2$ If $a+b=ab=a^2-b^2$ and $a$ and $b$ are real numbers then what do all of the expressions evaluate to?
*
*$(1+\sqrt5)/2$
*$(3+\sqrt5)/2$
*2
*$\sqrt5$
Not sure at all how to find the answer. I tried $a=ab/b$ and $a=(a^2-b^2)/b$ but after that I am lost.
| Any time you see $a^2-b^2$ you should think $(a+b)(a-b)$. Here if $a+b \neq 0$ you can divide it out to get $1=a-b$ Then $a=b+1, a+b=2b+1=(b+1)b$ and feed it to the quadratic formula.
If $a+b=0$ we must have $a=b=0$ and the sum is again $0$, which should have been one of the choices.
| {
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"timestamp": "2023-03-29T00:00:00",
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How many fair dice of this kind exist? I am not talking about the shape of the dice here, I am talking about another type. You will see what I mean soon.
For example, when there are 1 dice, a normal dice is a fair dice, because the probability of getting each number is the same by $ \frac{1}{6} $
When 2 normal dice are thrown, you can get numbers from 2 to 12, but the probability of getting each of them are different. Think of Monopoly, the probability of getting 7 is $ \frac{6}{36} $ while the probability of getting 2 is $ \frac{1}{36} $ the goal is to make the probability of getting each number the same.
I got an example:
Dice 1: $ [1,2,3,4,5,6] $
Dice 2: $ [0,6,12,18,24,30] $
When these two dice are thrown and add the two number, the possible outcome can be every number from 1 to 36 and the probability of each is
So the question is, How many different pairs of dice of this kind are possible so that the possible outcome can be every number from 1 to 36 and the probability of each is $ \frac{1}{36} $
The number on the dice can be negative too.
Example:
Dice 1: $ [-1,1,11,13,23,25] $
Dice 2: $ [2,3,6,7,10,11] $
Same number, different order is considered the same.
$$ [-1,1,11,13,23,25] [2,3,6,7,10,11] $$
$$ [1,11,13,23,25,-1] [2,3,6,7,10,11] $$
$$ [2,3,6,7,10,11] [-1,1,11,13,23,25] $$
They are all the same
So how many are there?
Is there a way to find it without trying one by one?
| Add $1$ to each number in one of the dice:
$A: (0,1,2,3,4,5),(0,6,12,18,24,30)$
$B: (0,1,2,6,7,8),(0,3,12,15,24,27)$
$C: (0,1;2,9,10,11),(0,3,6,18,21,24)$
$D: (0,1,2,18,19,20),(0,3,6,9,12,15)$
$E: (0,1,4,5,8,9),(0,2,12,14,24,26)$
$F: (0,1,6,7,12,13),(0,2,4,18,20,22)$
$G: (0,1,12,13,24,25),(0,2,4,6,8,10)$
36 is the product of four prime numbers: 2, 2, 3, 3. Arrange them in any of six ways, for example $a=2, b=3,c =3, d=2$.
$a=2$ so I let $A=\{0,1\}$.
( If $a$ were 3 I would let $A=\{0,1,2\}$. )
$b=3$ so $B=\{0,a,2a\}=\{0,2,4\}$.
$c=3$ so $C=\{0,ab,2ab\}=\{0,6,12\}$.
$d=2$ so $D=\{0,abc\}=\{0,18\}$
Form one die from $A+B$ and the other from $C+D$; or else $A+C$ and $B+D$.
Here, $A+C=\{0,1,6,7,12,13\}$ and $B+D=\{0,2,4,18,20,22\}$.
Finally add 1 to either die so they go from 1 to 36 instead of 0 to 35.
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"timestamp": "2023-03-29T00:00:00",
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