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Prove that there are no integer solutions to: $x^4+6x^2+1=8y^4$
Prove that there are no integer solutions to:
$$x^4+6x^2+1=8y^4$$
where $x>1$.
My attempts:
Let $x^2=u, y^2=v$
$$u^2+6u+(1-8v^2)=0$$
$$\Delta=36-4(1-8v^2)=w_0^2$$
$$32v^2+32=(4w_1)^2$$
$$2v^2+2=w_1^2=(2w_2)^2$$
$$v^2+1=2w_2^2$$
$$v=w_2=1\implies x=y=1$$
I don't know how can I proceed.
| $x^4 + 6x^2 + 1 = (x^2 + 3)^2 + 1 - 9 = 8y^4 \implies (x^2 + 3)^2 = 8 + 8y^4 = (2y^2 + 2)^2 + (2y^2 - 2)^2$
So the three numbers $a = 2y^2 + 2, b = 2y^2 - 2, c = x^2 + 3$ must be a pythagorean triple and for this to be true, $a,b,c$ should be in the form $m^2 - n^2, 4mn, m^2 + n^2$ which is not the case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4563040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Calculate $\sum_{n=2}^{\infty}\left (n^2 \ln (1-\frac{1}{n^2})+1\right)$ I am interested in evaluating
$$\sum_{n=2}^{\infty}\left (n^2 \ln\left(1-\frac{1}{n^2}\right)+1\right)$$
I am given the solution for the question is $\,\ln (\pi)-\frac{3}{2}\,.$
$$\sum_{n=2}^{\infty}\left(n^2\ln\left(\!1\!-\!\frac{1}{n^2}\!\right)+1\right)=4\ln\left(\!\frac{3}{4}\!\right)+1+9\ln\left(\!\frac{8}{9}\!\right)+1+\ldots$$
Any tricks to solve it?
| Here's an elementary proof, assuming that we're allowed to use Stirling's formula:
$$\ln (n!) = \sum_{k=2}^n \ln(k) = n\ln n -n + \frac 1 2 \ln n + \frac 1 2 \ln 2\pi + \mathcal O(\frac 1 n)$$
Let us study the partial sums, denoting $u_k=k^2 \ln \left(1-\frac 1 {k^2} \right) + 1$:
$$\begin{align}
\sum_{k=2}^n u_k &=\sum_{k=2}^n \left(k^2 \ln \left(1-\frac 1 {k^2} \right) + 1 \right) \\
&=\sum_{k=2}^n \left(k^2\ln \left(\frac {(k+1)(k-1)} {k^2}\right) + 1 \right) \\
&=\sum_{k=2}^n \left(k^2 \ln(k+1) + k^2\ln(k-1) - 2k^2\ln k + 1 \right) \\
&=-\ln 2 + n^2\ln(n+1) - (n+1)^2\ln n +\sum_{k=2}^n \left((k-1)^2 \ln k + (k+1)^2\ln k - 2k^2\ln k + 1 \right) \\
&=-\ln 2 + n^2\ln(n+1) - (n+1)^2\ln n +\sum_{k=2}^n \left(2\ln k + 1 \right) \\
&=-\ln 2 + \left(n^2\ln n + n^2 \ln(1+\frac 1 n)\right) - \left(n^2\ln n + 2n\ln n + \ln n\right) \\ &\ \ \ \ +\left(2n\ln n - 2n + \ln n + \ln 2\pi + \mathcal O(\frac 1 n)\right) + (n-1) \\
&=-\ln 2 + n^2 \ln(1+\frac 1 n) - n - 1 + \ln 2\pi + \mathcal O(\frac 1 n) \\
&=n^2 \left(\frac 1 n - \frac 1 {2n^2} + O(\frac 1 {n^3})\right) - n - 1 + \ln\pi + \mathcal O(\frac 1 n) \\
&=- \frac 3 2 + \ln\pi + \mathcal O(\frac 1 n)
\end{align}$$
Therefore, as desired: $$\sum_{k=2}^{+\infty} u_k = \ln\pi - \frac 3 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4565763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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How to Evaluate the Integral? $\int_{0}^{1}\frac{\ln\left( \frac{x+1}{2x^2} \right)}{\sqrt{x^2+2x}}dx=\frac{\pi^2}{2}$ I am trying to find a closed form for
$$
\int_{0}^{1}\ln\left(\frac{x + 1}{2x^{2}}\right)
{{\rm d}x \over \,\sqrt{\,{x^{2} + 2x}\,}\,}.
$$
I have done trig substitution and it results in
$$
\int_{0}^{1}\ln\left(\frac{x + 1}{2x^{2}}\right)
{{\rm d}x \over \,\sqrt{\,{x^{2} + 2x}\,}\,} =
\int_{0}^{\pi/3}\sec\left(\theta\right)
\ln\left(\frac{\sec\left(\theta\right)}
{2\left[\sec\left(\theta\right) - 1\right]^{\,2}} \right){\rm d}\theta
$$
which doesn't help.
By part integration with
$\displaystyle u = \ln\left(\frac{x + 1}{2x^{2}} \right)$, $\displaystyle\,\,{\rm d}v=\frac{\displaystyle\,\,{\rm d}x}{\,\sqrt{\,{x^{2} + 2x}\,}\,}$ also makes it more complicated.
I appreciate any help on this problem.
| You can "simplify" the problem using a first integration by parts to get rid of the logarithm
$$u=\log \left(\frac{x+1}{2 x^2}\right)\quad \implies \quad du=-\frac{x+2}{x^2+x}$$
$$dv=\frac{1}{\sqrt{x^2+2 x}}\quad \implies \quad v=2 \tanh ^{-1}\left(\sqrt{\frac{x}{x+2}}\right)$$
Using the bounds $u\,v=0$ and we are left with
$$I=2\int_0^1\frac{(x+2) }{x^2+x}\tanh ^{-1}\left(\sqrt{\frac{x}{x+2}}\right)dx$$
Now
$$\sqrt{\frac{x}{x+2}}=t \implies x=\frac{2 t^2}{1-t^2}\implies dx=\frac{4 t}{\left(1-t^2\right)^2}$$
$$I=8\int_0^{\frac{1}{\sqrt{3}}}\frac{\tanh ^{-1}(t)}{t-t^5}\,dt$$ Now, using partial fraction decomposition
$$\frac{1}{t-t^5}=\frac{1}{t(1-t^2)(1+t^2)}=-\frac{t}{2 \left(t^2+1\right)}-\frac{1}{4 (t-1)}-\frac{1}{4
(t+1)}+\frac{1}{t}$$ and now would arrive a bunch of polylogarithm functions.
The simplest is
$$\int \frac{\tanh ^{-1}(t)}{t}\,dt=\frac{1}{2} (\text{Li}_2(t)-\text{Li}_2(-t))$$
Fortunately, between the given bounds everything simplify a lot.
I let you the pleasure of computing the pieces.
Edit
If we write
$$\frac{\tanh ^{-1}(t)}{t-t^5}=\frac{\tanh ^{-1}(t)}{t}+\sum_{n=1}^\infty t^{4n-1}\,\tanh ^{-1}(t)$$ we have
$$I=8\int_0^{\frac{1}{\sqrt{3}}}\frac{\tanh ^{-1}(t)}{t-t^5}\,dt=4
\left(\text{Li}_2\left(\frac{1}{\sqrt{3}}\right)-\text{Li}_2\left
(-\frac{1}{\sqrt{3}}\right)\right)+$$
$$\sum_{n=1}^\infty\frac{9^{-n} \log \left(2+\sqrt{3}\right)-B_{\frac{1}{3}}\left(2 n+\frac{1}{2},0\right)}{n}$$ that is to say
$$I=4
\left(\text{Li}_2\left(\frac{1}{\sqrt{3}}\right)-\text{Li}_2\left
(-\frac{1}{\sqrt{3}}\right)\right)+\log \left(\frac{9}{8}\right) \log \left(2+\sqrt{3}\right)-\sum_{n=1}^\infty \frac{B_{\frac{1}{3}}\left(2 n+\frac{1}{2},0\right)}{n}$$ Numerically, the sum of the first and second terms is $4.96991$ and the summation is only $0.03511$
Edit
There is something very strange : two different $CAS$ give as result
$$I =\frac{17 \pi ^2}{24}-\frac{1}{2} \left(8
\text{Li}_2\left(2-\sqrt{3}\right)-\text{Li}_2\left(-7+4
\sqrt{3}\right)+\log ^2\left(2-\sqrt{3}\right)\right)$$ without any further simplification while Wolfram Alpha gives $1.$ when it is written as
$$\int_0^1 \frac2{ \pi ^2} \frac{\log \left(\frac{x+1}{2 x^2}\right)}{\sqrt{x^2+2 x}}\,dx$$ Without the factor, it just return a decimal value.
Using RootApproximant[%] also behaves differently with or without the factor inside the integrand.
$$\frac{1}{2} \left(8
\text{Li}_2\left(2-\sqrt{3}\right)-\text{Li}_2\left(-7+4
\sqrt{3}\right)+\log ^2\left(2-\sqrt{3}\right)\right)=2.056167583560283045590519$$ Passed to the $ISC$, it is returned as $\frac{5 \pi ^2}{24}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 5,
"answer_id": 2
} |
Find the asymptotes of $xy^2 - y^2 - x^3 = 0$ I am lost with this question because there is no 3rd degree $y$ term and no 2nd degree $x$ term.
Can $y - x - 1/2 = 0$ and $y + x + 1/2 = 0$ be its asymptotes?
or $y - x - 1/3 = 0$ and $y + x + 1/2 = 0$
| Now, $y=\sqrt{\frac{x^3}{x-1}}$. Apart from the vertical asymptote $x=1$, we have two oblique asymptotes (OAs) in the form $y=ax+b$. To find these let us make the following limit zero:
$\begin{align}
\lim_{x\rightarrow \pm\infty}\sqrt{\frac{x^3}{x-1}}-(ax+b)&=\lim_{x\rightarrow \pm\infty}\frac{\frac{x^3}{x-1}-(ax+b)^2}{\sqrt{\frac{x^3}{x-1}}+(ax+b)}\\
&=\lim_{x\rightarrow \pm\infty}\frac{x^2+x+1+\frac{1}{x-1}-a^2x^2-2abx-b^2}{\sqrt{x^2+x+1+\frac{1}{x-1}}+(ax+b)}\\
&=\lim_{x\rightarrow \pm\infty}\frac{(1-a^2)x^2+(1-2ab)x+1-b^2+\frac{1}{x-1}}{\sqrt{(x+\frac{1}{2})^2+\frac{3}{4}+\frac{1}{x-1}}+ax+b}\\
&=\lim_{x\rightarrow \pm\infty}\frac{(1-a^2)x^2+(1-2ab)x+1-b^2+\frac{1}{x-1}}{|x+\frac{1}{2}|+ax+b}\\
\end{align}$
To make this limit zero at $\infty$: $a=1$ and $2b-1=0$. So, the OA at $\infty$ is $y=x+\frac{1}{2}.$
To make this limit zero at $-\infty$: $a=-1$ and $2b+1=0$. So, the OA at $-\infty$ is $y=-x-\frac{1}{2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4573468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Volume of the region determined by $x^2 + y^2 + z^2 \leq 10, z \geq 2$ This is question 13 on page 294 of Vector Calculus by Marsden and Tromba.
Find the volume of the region determined by $x^2 + y^2 + z^2 \leq 10, z \geq 2.$
I have attempted it as follows.
The region can be described by using spherical polar coordinates $(r,\theta,\phi)$ and we have $0 \leq r \leq \sqrt{10}$, $0 \leq \theta \leq 2 \pi$.
For the limits for $\phi$, I think that we can find it by saying that $z=r \cos \phi = \sqrt{10} \cos \phi \geq 2$ using that $r = \sqrt{10}$ on the surface. This gives $0 \leq \phi \leq \cos^{-1}(2/\sqrt{10}).$ Hence I get
\begin{align} \iiint_V dV &= \int_0^{2\pi} \int_0^{\cos^{-1}(2/\sqrt{10})}\int_0^{\sqrt{10}} r^2 \sin \phi drd\phi d\theta \\ &= 2\pi \int_0^{\cos^{-1}(2/\sqrt{10})} 10 \dfrac{\sqrt{3}}{3} \sin \phi d\phi \\&= 20\pi \dfrac{\sqrt{10}}{3} - 40\dfrac{\pi}{3}. \end{align}
However, the answer at the back of the book is
$20\pi \dfrac{\sqrt{10}}{3} - 52\dfrac{\pi}{3}$
but I have been unable to identify the mistake. Please, could someone help me? Thank you very much.
| From $z=r\cos\phi=2$ you get $r=\frac{2}{\cos\phi}$. The bounds of the first integral which is with respect to $r$ must have been $\frac{2}{\cos\phi}\leq r\leq\sqrt{10}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4574446",
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"source": "stackexchange",
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If $xyz = x+y+z$ for $x,y,z > 0$, then $\sqrt{1+x^2} + \sqrt{1+y^2} + \sqrt{1+z^2} \ge 6$
If $x,y,z \in \Bbb R_{>0}$ satisfy $xyz = x+y+z$, prove that $$\sqrt{1+x^2} + \sqrt{1+y^2} + \sqrt{1+z^2} \ge 6$$
We can express $1+x^2$ as
$$1+x^2 = (1-xy)(1-xz) = \frac{(y-xyz)(z-xyz)}{yz} = \frac{(x+z)(x+y)}{yz}$$
since $x + y + z = xyz$ implies $1+ x^2 + xy + xz = 1 + x^2yz$. Thus, our desired inequality boils down to
$$\sqrt{\frac{(x+z)(x+y)}{yz}} + \sqrt{\frac{(y+z)(y+x)}{xz}} + \sqrt{\frac{(z+x)(z+y)}{xy}} \ge 6$$
which looks more complicated. Perhaps I am not headed in the right direction? I'd appreciate any hints or solutions. Thank you!
| We have: $$xyz = x+y+z \ge 3\sqrt[3]{xyz} \Longrightarrow xyz \ge 3^{3/2}$$
and for $a\in \{x,y,z \}$:
$$\sqrt{1+a^2} = \sqrt{1+\frac{a^2}{3}+\frac{a^2}{3}+\frac{a^2}{3}}\ge\sqrt{4\sqrt[4]{\frac{a^6}{3^3}}}=\frac{2}{3^{3/8}}\cdot a^{3/4}$$
Then
$$\begin{align}
LHS &\ge \frac{2}{3^{3/8}}\cdot \left(x^{3/4}+y^{3/4}+z^{3/4} \right) \ge \frac{2}{3^{3/8}}\cdot3\cdot\sqrt[3]{(xyz)^{3/4}}= \frac{6}{3^{3/8}}(xyz)^{1/4}\\ & \ge \frac{6}{3^{3/8}} \cdot 3^{\frac{3}{2}\cdot \frac{1}{4}} = 6
\end{align}$$
The equality occurs when $ x=y=z= \sqrt{3}$.
Q.E.D
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integral Involving Harmonic Numbers: $\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx$ (Motivation) In an attempt to answer this question, I got stuck on evaluating a certain integral. I have made up the following conjecture:
$$\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx = \frac{\pi^{2}}{4}+\frac{3}{2}\ln\left(2\right)\ln\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right).$$
(Attempt) Let $H_n$ denote the n-th harmonic number $\displaystyle H_{n}=\sum_{k=1}^{n}\frac{1}{k}$. I will warn this process gets ugly, but it is the best I have so far. Expanding the integrand as a series, we get:
$$
\eqalign{
\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx &= \sum_{n=0}^{\infty}\int_{\sqrt{3}}^{\infty}\frac{4\ln\left(x\right)-H_{n}}{x^{4n+2}}dx+\sum_{n=0}^{\infty}\int_{\sqrt{3}}^{\infty}\frac{4\ln\left(x\right)-H_{n}}{x^{4n+4}}dx, \cr
}
$$
which simplifies down to
$$\sum_{n=0}^{\infty}\left(\frac{H_{n}\sqrt{3}^{-4n-1}}{-4n-1}-\frac{4\ln\left(\sqrt{3}\right)\sqrt{3}^{-4n-1}}{-4n-1}+\frac{4\sqrt{3}^{-4n-1}}{\left(-4n-1\right)^{2}}\right)+\sum_{n=0}^{\infty}\left(\frac{H_{n}\sqrt{3}^{-4n-3}}{-4n-3}-\frac{4\ln\left(\sqrt{3}\right)\sqrt{3}^{-4n-3}}{-4n-3}+\frac{4\sqrt{3}^{-4n-3}}{\left(-4n-3\right)^{2}}\right).$$
Since both series converge, we can split up the first series like
$$-\sum_{n=0}^{\infty}\frac{H_{n}\sqrt{3}^{-4n-1}}{4n+1}+4\ln\left(\sqrt{3}\right)\sum_{n=0}^{\infty}\frac{\sqrt{3}^{-4n-1}}{4n+1}+4\sum_{n=0}^{\infty}\frac{\sqrt{3}^{-4n-1}}{\left(4n+1\right)^{2}}$$
and the second series like
$$-\sum_{n=0}^{\infty}\frac{H_{n}\sqrt{3}^{-4n-3}}{4n+3}+4\ln\left(\sqrt{3}\right)\sum_{n=0}^{\infty}\frac{\sqrt{3}^{-4n-3}}{4n+3}+4\sum_{n=0}^{\infty}\frac{\sqrt{3}^{-4n-3}}{\left(4n+3\right)^{2}}.$$
Next, I found that
$$\sum_{n=0}^{\infty}\frac{\sqrt{3}^{-4n-1}}{4n+1}=\frac{\pi}{12}+\frac{1}{2}\operatorname{arctanh}\left(\frac{1}{\sqrt{3}}\right)$$
and
$$\sum_{n=0}^{\infty}\frac{\sqrt{3}^{-4n-3}}{4n+3}=\frac{1}{2}\operatorname{arctanh}\left(\frac{1}{\sqrt{3}}\right)-\frac{\pi}{12}.$$
However, after trying for a while, I am out of ideas for evaluating the other sums.
I realize I skipped a lot of steps, but that is because I don't want this question to be too long. So for your convenience, I put all of these into Desmos, so I believe my process is correct so far based on numerical approximations.
(Question) Does anyone have an idea of how to evaluate the integral in question, or how to evaluate the sums I am stuck on? Any hints and ideas are appreciated.
(Miscellaneous) Here are some other ideas I have:
$$
\eqalign{
\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx &= \int_{\operatorname{arcsec}\left(\sqrt{3}\right)}^{\frac{\pi}{2}}\frac{\ln\left(\left(\sec x\right)^{4}-1\right)}{\sec^{2}x-1}\sec\left(x\right)\tan\left(x\right)dx \cr
\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx &= \int_{0}^{\infty}\frac{\ln\left(\left(x+\sqrt{3}\right)^{4}-1\right)}{\left(x+\sqrt{3}\right)^{2}-1}dx.
}
$$
Maybe I could construct a keyhole contour for the last integral?
| This is not a complete solution but an extended comment.
I was particularly interested to find out what makes the lower integration border $\sqrt{3}$ so special as a prerequisite for an appreciable simplification of the final expression of the integral.
Hence I started with the same decomposition of our integral into the sum of three integrals as @dfnu, but now with a general lower integration border $a$, where $a>1$ to ensure convergence:
$$i_1(a) =\int_a^{\infty } \frac{\log (x-1)}{x^2-1} \, dx$$
$$i_2(a) =\int_a^{\infty } \frac{\log (x+1)}{x^2-1} \, dx$$
and looked particularly at the integral
$$i_3(a) = \int_{a}^{\infty}\frac{\log(x^2+1)}{x^2-1}$$
After a lengthy journey I found the following expression
$$\begin {align} i_3(a) = \frac{\pi ^2}{4}-2 \sum _{k=1}^{\infty } \frac{\left(\frac{\sqrt{a^2+1}}{\sqrt{2}}\right)^k \sin \left(\frac{\pi k}{4}\right) \sin \left(k \tan ^{-1}(a)\right)}{k^2}\\+\frac{1}{2} \log \left(\frac{a+1}{a-1}\right) \log \left(a^2+1\right)+\frac{1}{2} \pi \tan ^{-1}\left(\frac{1}{a}\right)\end{align}$$
Now to the special role of $\sqrt{3}$. The only value of $a=\sqrt{m}$ with positive integer $m$ making $\tan ^{-1}\left(\frac{1}{a}\right)$ a rational multiple of $\pi$ is $\sqrt{3}$, giving $\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi }{6}.$
The complete integral evaluates to
$$\begin{align} i(a) & = \int_{a}^{\infty}\frac {\log(x^4-1)}{x^2-1} \\
& = \frac{1}{2} \pi \tan ^{-1}\left(\frac{1}{a}\right)+\frac{\pi ^2}{3} \text{(*1*)}
\\
& -\frac{1}{4} \log ^2(a-1)+\frac{1}{2} \log ^2(a+1)
\\
& -\frac{1}{2} \log (2) \log (a-1)+\frac{\log ^2(2)}{4}
\\
& + \frac{1}{2} \log \left(\frac{a+1}{a-1}\right) \log \left(a^2+1\right) \text{(*2*)}
\\
& +\frac{1}{2} \text{Li}_2\left(\frac{1-a}{2}\right)+\frac{1}{2} \text{Li}_2\left(\frac{2}{a+1}\right)
\\
& +\frac{1}{2} \left(-\text{Li}_2\left(\left(\frac{1}{2}+\frac{i}{2}\right) (a-i)\right)
\\
+ \text{Li}_2\left(\left(-\frac{1}{2}+\frac{i}{2}\right) (a-i)\right)
\\
-\text{Li}_2\left(\left(\frac{1}{2}-\frac{i}{2}\right) (a+i)\right)
\\
+\text{Li}_2\left(\left(-\frac{1}{2}-\frac{i}{2}\right) (a+i)\right)\right)
\end{align}$$
As the two lines marked $(*1*)$ and $(*2*)$ are already similar to the final result for $a=\sqrt {3}$ there should be a lot of cancellation from the polylog functions.
(to be completed)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Show $a^2+b^2+c^2 \equiv 0 \pmod {3}$ if $a, b, c$ are *not* multiples of $3$. I was given the following problem:
Show $a^2+b^2+c^2 \equiv 0 \pmod {3}$ if $a, b, c$ are not multiples of $3$.
I would like a. verification of my proof (I self-study); b. alternative proofs, to enrich my appreciation of the problem. Here's what I did.
$\text{Lemma (demonstration skipped)} :$ $\forall j \in \mathbb{N}| a \equiv b \pmod{m} \implies a^j \equiv b^j \pmod{m}$.
$\text{Consideration :}$ I will use the $\binom{a}{b}$ notation to refer to the number of multisets of size $b$ that can be drawn out of $a$ elements; i.e., the number of collections with repetition but without order. This is due to the lack of a proper "\multiset" format in LaTex as provided in the site.
$\text{Solution}$. It is trivial to say
$$\begin{align}
a &\equiv r_1 \pmod{3}\\
b &\equiv r_2\pmod{3}\\
c &\equiv r_3\pmod{3}
\end{align}$$
where $r_1, r_2, r_3$ are the remainders of $a, b, c$ respectively in the division by $3$.
$I$. Let $S$ be the set of possible values of $r_i$. Because $3\nmid a, \space 3\nmid b, \space 3\nmid c$ we have $S=\{1, 2\}$.
$II.$ There are $2^3=8$ ways to draw $3$ elements of $S$. Since addition is conmutative, their ordering is irrelevant and therefore we have $\binom{2}{3}=\frac{4!}{3!}=4$ possible sums of the form $r_1+r_2+r_3$, with the same principle applying to $r_1^2+r_2^2+r_3^2$.
$III$. Let $S'$ be the set of the possible results of the sum $r_1^2+r_2^2+r_3^2$. These possible sums, as is easy to manually compute, are $S'= \{3, 6, 9, 12\}$, where $\forall s \in S' | s=3m, m\in\mathbb{Z}.$ It then follows
$$a^2+b^2+c^2 \equiv r_1^2+r_2^2+r_3^2 \equiv0 \pmod{3}$$
Thanks in advance.
| You're not wrong but simply observe that for any $a$ such that $a \not \equiv 0$ mod $3$, then we have that $a^2 \equiv 1$ mod $3$. This is easy enough to prove and follows from your lemma for instance.
Then $x^2, y^2, z^2 \equiv 1$ mod $3$ each, and so $x^2 + y^2 + z^2 \equiv 0 $ mod $3$.
| {
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How is $\ln (x-2) - \frac{1}{2} \ln (x-1) = \frac{1}{2} \ln \frac{(x-2)^2}{x-1}$
How is $\ln (x-2) - \frac{1}{2} \ln (x-1) = \frac{1}{2} \ln \frac{(x-2)^2}{x-1}$
Can someone enlighten me on how is these 2 actually equals and the steps taken? the left hand side is actually the answer for $\int \frac{x}{2 (x-2)(x-1)} dx$ but I need to combine the expression to the right hand side to continue with the steps in the question I am attempting
| $$\ln (x-2) - \frac{1}{2} \ln (x-1)$$
$$=\frac{2\ln (x-2) -\ln (x-1)}{2}$$
$$=\frac{\ln (x-2)^2 -\ln (x-1)}{2}$$
$$=\frac{\ln\frac{ (x-2)^2}{ x-1}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4578029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Find the volume between the regions $x^2 + y^2+ z^2 = 4$ and $x = 4-y^2$ I want the volume of the sphere $x^2 + y^2 + z^2 = 4$ from $x = 0$ to $x = 4-y^2$. The integral that gives this volume is
$$\int\limits_{-2}^2 \int\limits_0^{4-y^2} \int\limits_{-\sqrt{4-x^2-y^2}}^{\sqrt{4-x^2-y^2}}\ 1\ dz\ dx\ dy$$
I don't find $T$ such that $T\bar{u} = \bar{x}$ for every $\bar{x}\in D$ where $D$ is our desired volume.
I mean, I'm trying to find $D^*$ such that for every $\bar{u} \in D^*$, $T$ is a change of variables for our problem.
So, I'm trying the find the volume between the regions $x^2 + y^2 + z^2 = 4$ and $x = 4-y^2$ but I can neither find that scalar.
Any ideas please.
| Nobody commented on my scalar in the comments section so far.
The projection of the solid on the $xy$-plane is determined by curves, $x^2+y^2=4$, $x=4-y^2$ and $x=0$. The common solution of the circle and the parabola gives $(1,\pm \sqrt{3})$. Hence the quarter volume integral due to symmetries w.r.t $xy$-plane and $x$-axis, is
$$\frac{1}{4}V=\int_0^{\sqrt{3}}\int_0^{\sqrt{4-y^2}}\int_0^{\sqrt{4-x^2-y^2}}dzdxdy+\int_{\sqrt{3}}^2\int_0^{4-y^2}\int_0^{\sqrt{4-x^2-y^2}}dzdxdy$$
and
$$\frac{1}{4}V=\int_0^{\sqrt{3}}\int_0^{\sqrt{4-y^2}}\sqrt{4-x^2-y^2}dxdy+\int_{\sqrt{3}}^2\int_0^{4-y^2}\sqrt{4-x^2-y^2}dxdy$$
and
$$\frac{1}{4}V=\int_0^{\sqrt{3}}\frac{\pi}{4}(4-y^2)dy+\int_{\sqrt{3}}^2\left(\frac{x\sqrt{4-x^2-y^2}}{2}+\frac{(4-y^2)\arcsin(\frac{x}{\sqrt{4-y^2}})}{2}\right)\vert_0^{4-y^2}dy$$
and hence $V$ is
$$V=3\sqrt{3}\pi+2\int_{\sqrt{3}}^{2}\sqrt{y^2-3}\sqrt{4-y^2}+(4-y^2)\arcsin(\sqrt{4-y^2})dy$$
By WolframAlpha, $V\approx 16.324+0.482=16.806$. The second integral which gives the tiny volume is not nice.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4578958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is there any better way of finding the required value If $$z=\cos\theta+i\sin\theta$$ find the value of $$\frac{1+z}{1-z}$$
The solution that I have is this
$$z=\cos\theta+i\sin\theta \implies$$
$$\frac{1+z}{1-z}=\frac{1+(\cos\theta+i\sin\theta)}{1-(\cos\theta+i\sin\theta)}=\frac{(1+\cos\theta)+i\sin\theta}{(1-\cos\theta)+i\sin\theta}$$
$$=\frac{2\cos^2\frac{\theta}{2}+i\:\:2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}-i\:\:2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}=\frac{\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}}\cdot \frac{\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}}{\sin\frac{\theta}{2}-i\cos\frac{\theta}{2}}$$
$$=\cot\frac{\theta}{2}\cdot\frac{\color{red}{i}}{\color{red}{i}}×\frac{\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}}{\sin\frac{\theta}{2}-i\cos\frac{\theta}{2}}$$
$$=\color{red}{i}\cot\frac{\theta}{2}\cdot\frac{\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}}{\color{red}{i}\sin\frac{\theta}{2}-\color{red}{i}i\cos\frac{\theta}{2}}$$
$$=\color{red}{i}\cot\frac{\theta}{2}\cdot\frac{\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}+\color{red}{i}\sin\frac{\theta}{2}}$$
$$=\color{red}{i}\cot\frac{\theta}{2}$$
Now how on earth one will imagine the steps written in red. I am looking for an easy and a logical answer to this question. The solution that I have is impractical as you all can see. I tried it doing by $e^{i\theta}$ but no good happen.
Any help is greatly appreciated.
| I don't think the question is really accurate in the first place, the value of $\frac{1+z}{1-z}$ can just be determined by plugging in the expression for $z$. To bring it in the final form is a matter of taste, I believe it mostly comes from the experience and knowing which result you want to get. Multiplying by $1$ and adding $0$ are among the most common tricks, but I don't think there exists any universal method.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
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} |
Show that all solutions of $z^5 - z + 16 = 0$ satisfy $1 \lt |z| \lt 2$ I have to show that all solutions of $z^5 - z + 16 = 0$ satisfy $1 \lt |z| \lt 2$.
My attempt: By using Euler's formula I can rewrite the equation into $r^5e^{5i\phi} - re^{i\phi} = -16$ and then rewrite this into $$r^5 \cos(5\phi) - r \cos(\phi) + i(r^5 \sin(5\phi) -r \sin(\phi)) = -16 + i0 $$
so I get
*
*$r^5 \cos(5\phi) - r \cos(\phi) = -16$
*$r^5 \sin(5\phi) -r \sin(\phi) = 0$ which becomes $r^4 \sin(5\phi) = \sin(\phi)$
Here I don't know how to proceed. Any tips?
| We have $|z^5-z|=|-16|=16.$
If $|z|=2$ then $|z^5-z|\ge |z^5|-|z|=|z|^5-2=30>16.$
If $|z|>2$ then $|z-1|\ge |z|-|1|>1$ and $|z|^4>0,$ so $|z^5-z|=|z|^4\cdot |z-1|> |z|^4\cdot 1 > 2^4=16.$
If $|z|\le 1$ then $|z^5-z|\le |z^5|+|z|=|z|^5+|z|\le 1^5+1=2<16.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4582084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Why can't I use the form $\frac1{x^2+1}$ from the derivative of $\arctan(x)$ to convert the integral form in this situation? In the question that solves the integral $\displaystyle\int\frac1{6x^2 + 36x + 78} \,\mathrm{d}x$, I first tried to solve it by changing the denominator in a form of $\dfrac1{x^2 + 1}$ to apply $\;\arctan(x)$.
$\dfrac1{6\!\cdot\!\left(x^2 + 6x + 13\right)}=\dfrac16\!\cdot\!\dfrac1{(x + 3)^2 + 2^2}$
Now, in order to make it in a form of $\;\dfrac1{x^2 + 1}\;,\;$ I divide everything by $\,2^2$:
$=\dfrac16\!\cdot\!\dfrac1{\left(\frac{x+3}2\right)^2+1}\!\cdot\!\dfrac1{2^2}
=\dfrac1{24}\!\cdot\!\dfrac1{\left(\frac{x+3}2\right)^2+1}$
Then assume that $\;u=\dfrac{x+3}2\;,\;$ I thought I can apply arctan to get rid of the integral form:
$\dfrac1{24}\!\cdot\!\arctan\left(\dfrac{x+3}2\right)+c\;.$
But the correct answer is $\;\dfrac1{12}\!\cdot\!\arctan\left(\dfrac{x+3}2\right) + c\;,\;$ not $\;\dfrac1{24}\,.$
The answer also explained to use $\;\dfrac1{x^2 + k^2} = \left[\dfrac1k\!\cdot\!\arctan\left(\dfrac xk\right)\right]’$ (derivative), but I wonder why I cannot use the form of $\;\dfrac1{x^2 + 1}\;,\;$ which is the only equation I have known.
| A-ha! I think I found it.
You reduce your integral to
$$ \int \frac{1}{6x^2+36x+78} \ dx = \frac{1}{24} \int \frac{1}{\left( \frac{x+3}{2} \right)^2 + 1} \ dx $$
and wish to use $u = \frac{x+3}{2}$ so that $du = \frac{1}{2} \ dx.$ This makes $dx = 2 \ du$. There's your factor of two!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4585030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the value of $\int_0^1f(x)dx$
If $$f(x)=\binom{n}{1}(x-1)^2-\binom{n}{2}(x-2)^2+\cdots+(-1)^{n-1}\binom{n}{n}(x-n)^2$$
Find the value of $$\int_0^1f(x)dx$$
I rewrote this into a compact form.
$$\sum_{k=1}^n\binom{n}{k}(x-k)^2(-1)^{k-1}$$
Now,
$$\int_0^1\sum_{k=1}^n\binom{n}{k}(x-k)^2(-1)^{k-1}dx$$
$$=\sum_{k=1}^n\binom{n}{k}\frac{(1-k)^3}{3}(-1)^{k-1}-\sum_{k=1}^n\binom{n}{k}\frac{(-k)^3}{3}(-1)^{k-1}$$
$$=\sum_{k=1}^n\binom{n}{k}\frac{(1-k)^3}{3}(-1)^{k-1}+\sum_{k=1}^n\binom{n}{k}\frac{k^3}{3}(-1)^{k-1}$$
After this, I took $\dfrac13$ common and did some simplifications but nothing useful came out.
Any help is greatly appreciated.
| Solution 1. Here is another approach. Consider the shift operator $\Delta$ defined for functions on $\mathbb{R}$ by
$$ \Delta f(x) = f(x-1). $$
Then
\begin{align*}
\sum_{k=1}^{n} (-1)^{k-1} \binom{n}{k} (x-k)^2
&= \sum_{k=1}^{n} (-1)^{k-1} \binom{n}{k} \Delta^k x^2 \\
&= [\operatorname{id} - (\operatorname{id} - \Delta)^n] x^2.
\end{align*}
Here, $\operatorname{id}$ is the identity operator on functions, i.e., $\operatorname{id} f(x) = f(x)$. Now the crucial observation is as follows:
The backward difference operator $D = \operatorname{id} - \Delta$, when applied to a polynomial, results in another polynomial with degree decreased by at least one.
Intuitively, this is because $D$ behaves similar to the differential operator $\frac{\mathrm{d}}{\mathrm{d}x}$. In particular, when $n \geq 3$, it follows that $D^n x^2 = 0$. Hence it follows that
$$ \sum_{k=1}^{n} (-1)^{k-1} \binom{n}{k} (x-k)^2 = x^2 \qquad\text{for}\quad n \geq 3. $$
Now the rest computation is straightforward.
Solution 2. Here is yet another approach. Define the coefficient extraction operator $[x^n]$ by
$$ [x^n]\sum_{k=0}^{\infty} a_k x^k = a_k. $$
Note that $[x^n]$ is linear. Furthermore, we may rewrite the integral using this operator as:
\begin{align*}
\int_{0}^{1} f(x) \, \mathrm{d}x
&= \int_{0}^{1} \left( \sum_{k=1}^{n} (-1)^{k-1} \binom{n}{k} (x-k)^2 \right) \, \mathrm{d}x \\
&= \int_{0}^{1} \left( \sum_{k=1}^{n} (-1)^{k-1} \binom{n}{k} 2 [s^2] e^{(x-k)s} \right) \, \mathrm{d}x \\
&= 2 [s^2] \int_{0}^{1} \left( \sum_{k=1}^{n} (-1)^{k-1} \binom{n}{k} e^{(x-k)s} \right) \, \mathrm{d}x \\
&= 2 [s^2] \int_{0}^{1} e^{xs} \left( 1 - (1 - e^{-s})^n \right) \, \mathrm{d}x \\
&= 2 [s^2] \left( \frac{e^s - 1}{s} \left( 1 - (1 - e^{-s})^n \right) \right).
\end{align*}
Then, using the fact that
$$ \frac{e^s - 1}{s} = 1 + \frac{s}{2} + \frac{s^2}{6} + \mathcal{O}(s^3) $$
and
$$ 1-(1-e^{-s})^n = \begin{cases}
1 - s + \frac{s^2}{2} + \mathcal{O}(s^3), & n = 1, \\
1 - s^2 + \mathcal{O}(s^3), & n =2, \\
1 + \mathcal{O}(s^3), & n \geq 3,
\end{cases} $$
we can easily conclude that
$$ \int_{0}^{1} f(x) \, \mathrm{d}x = \begin{cases}
\frac{1}{3}, & n = 1 \text{ or } n \geq 3, \\
-\frac{5}{3}, & n = 2.
\end{cases} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4600131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
} |
Evaluation of $~\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta$ $$
I:=\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta
$$
My tries
$$\begin{align}
s&:=\sin\theta\\
c&:=\cos\theta\\
I&=\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta\\
&=\int_{0}^{2\pi}{c^2-s^2\over s^4+c^4}\mathrm d\theta\\
&=\int_{0}^{2\pi}{(1-s^2)-s^2\over s^4+(c^2)^2}\mathrm d\theta\\
&=\int_{0}^{2\pi}{1-2s^2\over s^4+(1-s^2)^2}\mathrm d\theta\\
&=\int_{0}^{2\pi}{1-2s^2\over s^4+(s^2-1)^2}\mathrm d\theta\\
&=\int_{0}^{2\pi}{1-2s^2\over s^4+s^4-2s^2+1}\mathrm d\theta\\
&=\int_{0}^{2\pi}\underbrace{\color{red}{\left({1-2s^2\over 2s^4-2s^2+1}\right)}}_{\text{I got stuck here}}\mathrm d\theta\\
\end{align}$$
I need your help.
| $$I=\int_0^{2\pi}\frac{\cos^2\theta-\sin^2\theta}{\cos^4\theta+\sin^4\theta}d\theta$$
Substitute $\theta→\pi-\theta$.
$$I=\int_{-\pi}^\pi\frac{\cos^2\theta-\sin^2\theta}{\cos^4\theta+\sin^4\theta}d\theta$$
Since the integrand is even, we can use even-odd properties.
$$I=2\int_{0}^\pi\frac{\cos^2\theta-\sin^2\theta}{\cos^4\theta+\sin^4\theta}d\theta$$
Substitute $\theta→\frac{\pi}{2}-\theta$
$$I=2\int_{-\pi/2}^{\pi/2}\frac{\sin^2\theta-\cos^2\theta}{\cos^4\theta+\sin^4\theta}$$
Knowing the integrand is even, we can use even odd properties again
$$I=4\int_{0}^{\pi/2}\frac{\sin^2\theta-\cos^2\theta}{\cos^4\theta+\sin^4\theta}d\theta$$
Substitute $\theta→\frac{\pi}{2}-\theta$ again
$$I=4\int_{0}^{\pi/2}\frac{\cos^2\theta-\sin^2\theta}{\cos^4\theta+\sin^4\theta}d\theta$$
Add two versions of I
$$I=4\int_{0}^{\pi/2}\frac{\sin^2\theta-\cos^2\theta}{\cos^4\theta+\sin^4\theta}d\theta+4\int_{0}^{\pi/2}\frac{\cos^2\theta-\sin^2\theta}{\cos^4\theta+\sin^4\theta}d\theta$$
We find the 2 integrals are negatives of each other, so the whole RHS cancels out, leaving zero. We achieve
$$I=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4603090",
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"source": "stackexchange",
"question_score": "7",
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} |
On generalizing $\frac{17-\sqrt{17}}8 =\sin^2(t)+\sin^2(2t)+\sin^2(4t)+\sin^2(8t)$? Given $\color{blue}{t = 2\pi/p}$ for the appropriate prime $p=4m+1$.
I. Sine
$$\begin{align}
\frac{5+\sqrt{5}}8 &=\sin^2(t)\\
\frac{13+\sqrt{13}}8 &=\sin^2(t)+\sin^2(3t)+\sin^2(4t)\\
\frac{17+\sqrt{17}}8 &=\sin^2(3t)+\sin^2(5t)+\sin^2(6t)+\sin^2(7t)\\
\frac{29+\sqrt{29}}8 &=\sum_{k=1}^7\sin^2(a_k\, t)
\end{align}$$
with the seven $a_k = 1,4,5,6,7,9,13.$ And so on for other prime $p=4m+1.$ For the opposite sign, one uses the remaining integers $b_k \leq \frac{p-1}2.$ For example,
$$\frac{17-\sqrt{17}}8 =\sin^2(t)+\sin^2(2t)+\sin^2(4t)+\sin^2(8t)$$
where the $a_k$ are simply $2^n$. (The next Fermat prime $p=257$ isn't so nice since it has $256/4 = 64$ sine terms.)
II. Cosine
This uses the same set of multipliers $a_k$.
$$\begin{align}
\frac{3-\sqrt{5}}8 &=\cos^2(t)\\
\frac{11-\sqrt{13}}8 &=\cos^2(t)+\cos^2(3t)+\cos^2(4t)\\
\frac{15-\sqrt{17}}8 &=\cos^2(3t)+\cos^2(5t)+\cos^2(6t)+\cos^2(7t)\\
\frac{27-\sqrt{29}}8 &=\sum_{k=1}^7\cos^2(a_k\, t)
\end{align}$$
with the same seven $a_k = 1,4,5,6,7,9,13.$ For the opposite sign,
$$\frac{15+\sqrt{17}}8 =\cos^2(t)+\cos^2(2t)+\cos^2(4t)+\cos^2(8t)$$
III. Conclusion
Given prime $p=4m+1$ and $t = 2\pi/p.$ The pattern clearly is,
$$\frac{p\pm\sqrt{p}}8 = \sum_{k=1}^m \sin^2(a_k\, t)$$
$$\frac{(p-2)\mp\sqrt{p}}8 = \sum_{k=1}^m \cos^2(a_k\, t)$$
Question: I used Mathematica's integer relations to find the above examples. But, for any prime $p=4m+1$, what is a clever and faster algorithm to derive the correct set of $a_k$?
| The ones that have all even or all odd quadratic residues relative to p, will do the trick.
The proof for this lies is polygon isomorphisms. The numbers are invariant if taken to an e power (ie an even quadratic), but not an odd power. So you end up with two numbers a, b whose sum would give p, and the difference is something in sqrt(p),
| {
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"url": "https://math.stackexchange.com/questions/4603836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Where did I go wrong with this integration? $$\int_0^{\frac {1}{\sqrt 3}} \frac {1}{\sqrt {2-3x^2}}$$
$$\int_0^{\frac {1}{\sqrt 3}} \frac {1}{\sqrt {2(1-\frac32x^2)}}$$
$$\frac 1{\sqrt 2}\int_0^{\frac {1}{\sqrt 3}} \frac {1}{\sqrt {1-\frac32x^2}}$$
$$\bigg(\frac 1{\sqrt 2}\sin^{-1}{\sqrt {\frac 32 x}}\bigg)\bigg|_0^\frac 1{\sqrt 3}$$
$$\frac 1{\sqrt 2} \times \frac {\pi}{4}$$
However, the answer is $\frac 1{\sqrt 3} \times \frac {\pi}{4}$. Where did I go wrong?
| In case you're not required to rely on trigonometric substitutions, we can make an alternative substitution of
$$t = \frac{\sqrt{2-3x^2}-\sqrt2}x \implies x=-\frac{2\sqrt2\,t}{3+t^2} \implies dx = -\frac{2\sqrt2(3-t^2)}{(3+t^2)^2} \, dt$$
Then
$$\begin{align*}
I &= \int_0^{\frac1{\sqrt3}} \frac{dx}{\sqrt{2-3x^2}} \\[1ex]
&= 2 \int_{\sqrt3-\sqrt6}^0 \frac{dt}{3+t^2}\\[1ex]
&= - \frac2{\sqrt3} \tan^{-1}\left(\frac{\sqrt3-\sqrt6}{\sqrt3}\right) \\[1ex]
&= \frac2{\sqrt3} \tan^{-1}\left(\sqrt2-1\right) = \boxed{\frac\pi{4\sqrt3}}
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4607606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Contest Math Question: simplifying logarithm expression further I am working on AoPS Vol. 2 exercises in Chapter 1 and attempting to solve the below problem:
Given that $\log_{4n} 40\sqrt{3} = \log_{3n} 45$, find $n^3$ (MA$\Theta$ 1991).
My approach is to isolate $n$ and then cube it. Observe:
\begin{align*}
\frac{\log 40\sqrt{3}}{\log 4n} = \frac{\log 45}{\log 3n} \\
\log 40\sqrt{3}\log 3n = \log 45\log 4n\\
\log 40\sqrt{3} \cdot (\log 3 + \log n) = \log 45 \cdot (\log 4 + \log n)\\
\log n \cdot (\log 40\sqrt{3} - \log 45) = \log 45\log 4 - \log 40\sqrt{3}\log 3
\end{align*}
Dividing through and putting the coefficients as powers, we have:
\begin{align*}
\log n &= \frac{\log 45^{\log 4} - \log \left[(40\sqrt{3})^{\log 3}\right]}{\log\left(\frac{40\sqrt{3}}{45}\right)}
=\frac{\log \left(\frac{45^{\log 4}}{(40\sqrt{3})^{\log 3}}\right) }{\log\left(\frac{40\sqrt{3}}{45}\right)} \\
&=\log \left(\frac{45^{\log 4}}{(40\sqrt{3})^{\log 3}}\right)^{\log\left(\frac{40\sqrt{3}}{45}\right)^{-1}}
\end{align*}
which shows that
\begin{align*}
n^3 = \left(\frac{45^{\log 4}}{(40\sqrt{3})^{\log 3}}\right)^{3\cdot\log\left(\frac{40\sqrt{3}}{45}\right)^{-1}}
\end{align*}
Somehow it feels like this answer may be simplified further. Are the steps shown so far correct and can the answer be expressed in a better way?
| $\displaystyle\frac{\ln(40\sqrt{3})}{\ln(4n)} = \frac{\ln(45)}{\ln(3n)}$
$\displaystyle \frac{\ln(3n)}{\ln(4n)}
= \frac{\ln(45)}{\ln(40\sqrt{3})} × \frac{k}{k}
= \frac{\ln(45^k)}{\ln[(40\sqrt{3})^k]}$
Find k, such that ratio inside ln/ln matched
$\displaystyle \frac{3n}{4n} = \left(\frac{45}{40\sqrt{3}}\right)^k = \left(\frac{3}{4}\right)^{3k/2}$
$→ k = \frac{2}{3}$
$3\,n = 45^{2/3}$
$27\,n^3 = 45^2$
$n^3 = 75$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4610313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
For $x,y,z∈ℝ^{+}$,without using Hölder's inequality prove that $\frac{x}{\sqrt{x^2+8yz}}+\frac{y}{\sqrt{y^2+8xz}}+\frac{z}{\sqrt{z^2+8xy}}\geq1$.
For $x,y,z∈ℝ^{+}$, prove that $\frac{x}{\sqrt{x^2+8yz}}+\frac{y}{\sqrt{y^2+8xz}}+\frac{z}{\sqrt{z^2+8xy}}\geq1$.
In this question solution used Hölder's inequality, but I am looking a solution without using Hölder's inequality.
I first tried using AM-GM inequality but not results clearly.
Then I used Cauchy-Schwarz inequality which gave,
$${(\sum_{cyc}x²)(\sum_{cyc}\frac{1}{x²+8yz})\geq(\sum_{cyc}\frac{x}{\sqrt{x²+8yz}})}^2$$
But I couldn't solve it. Can anyone help me out?
| Another way.
By AM-GM, C-S and AM-GM again we obtain:$$\sum_{cyc}\frac{x}{\sqrt{x^2+8yz}}=\sum_{cyc}\frac{2x(x+y+z)}{2\sqrt{(x+y+z)^2(x^2+8yz)}}\geq\sum_{cyc}\frac{2x(x+y+z)}{(x+y+z)^2+x^2+8yz}=$$
$$=\sum_{cyc}\frac{2x^2(x+y+z)}{x(x+y+z)^2+x^3+8xyz}\geq\frac{2(x+y+z)^3}{\sum\limits_{cyc}(x(x+y+z)^2+x^3+8xyz}=$$
$$=\frac{2(x+y+z)^3}{(x+y+z)^3+x^3+y^3+z^3+24xyz}\geq1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4611680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A $n \times n$ linear system Solve the linear system
$
\begin{array}{ c c c c c c c c c c c c c c }
& & x_{2} & + & x_{3} & + & \dotsc & + & x_{n-1} & + & x_{n} & = & 2 & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( 1)\\
x_{1} & & & + & x_{3} & + & \dotsc & + & x_{n-1} & + & x_{n} & = & 4 & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( 2)\\
x_{1} & + & x_{2} & & & + & \dotsc & + & x_{n-1} & + & x_{n} & = & 6 & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( 3)\\
\vdots & & \vdots & & \vdots & & \ddots & & \vdots & & \vdots & & \vdots & \ \ \ \ \ \ \ \ \ \ \ \\
x_{1} & + & x_{2} & + & x_{3} & + & \dotsc & + & x_{n-1} & & & = & 2n & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( n)
\end{array}
$
Attempt
I am not confident yet with my linear algebra, so I tried approaching this problem by adding all of the equations, which gives us
$
\begin{align*}
(n-1)(x_1+x_2+x_3+\ldots+x_{n-1}+x_n)=2+4+6+\ldots+2n \tag*{}
\end{align*}
$
since $x_i$ ($i \in \mathbb{N}, i \leq n$) always appear in every equation except the $i$-th equation. The right-hand side is the sum of first $n$ positive even number, which can be simplified further as
$
\begin{align*}
(n-1)(x_1+x_2+x_3+\ldots+x_{n-1}+x_n) &= n(n+1)
\\[.5em]
x_1+x_2+x_3+\ldots+x_{n-1}+x_n &= \dfrac{n(n+1)}{n-1} \tag*{}
\end{align*}
$
Subtracting respectively Eq. (1), Eq. (2), up to Eq. (n) from Eq. (A), gives us
$
\begin{align*}
x_1 &= \dfrac{n(n+1)}{n-1} - 2 = \dfrac{n(n+1) - 2(n-1)}{n-1} = \dfrac{n^2 - n + 2}{n-1} \tag*{}\\
x_2 &= \dfrac{n(n+1)}{n-1} - 4 = \dfrac{n(n+1) - 4(n-1)}{n-1} = \dfrac{n^2 - 3n + 4}{n-1} \\
\vdots \\
x_n &= \dfrac{n(n+1)}{n-1} - 2n = \dfrac{n(n+1) - 2n(n-1)}{n-1} = \dfrac{3n-n^2}{n-1}
\end{align*}
$
However, I'm rather concerned with the denominator ($n-1$) which causes division by zero when $n = 1$. Did I do a mistake, or maybe did I misunderstand something in the process? And what would be the correct approach to this problem? Thanks.
Note: I found the problem here.
| Let $A_n\in M_n(K)$ be the matrix with $0$ on the diagonal and entry $1$ everywhere else. Then $\det(A)\neq 0$ for $n\ge 2$. In fact, there is an easy formula for the determinant on this site, namely $\det(A)=(-1)^{n-1}\cdot (n-1)$, see here:
Determinant of a matrix with diagonal entries $a$ and off-diagonal entries $b$
The system then is given by $Ax=b$ with $b=(2,4,\ldots ,2n)^T$. The unique solution is given by
$$
x=A^{-1}b,
$$
because $Ax=b$ gives $x=A^{-1}Ax=A^{-1}b$. One can also find an explicit formula for the inverse of $A$, and thus obtain an explicit solution (which you can also derive directly from the equations).
| {
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"url": "https://math.stackexchange.com/questions/4615660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Prove that $ \frac{x^2}{1+x} + \frac{y^2}{1+y} + \frac{z^2}{1+z} \geq \frac{1}{2} $
Assume that positive numbers a, b, c, x, y, z satisfy $cy+bz =a;
az + cx = b$$bx + ay = c$.
Prove that $ \frac{x^2}{1+x} + \frac{y^2}{1+y} + \frac{z^2}{1+z} \geq \frac{1}{2} $
I've tried appling A.M.-G.M. inequality but it didn't work. Then applied jensen inequality assuming, $f(x)=\frac{x^2}{1+x}$
As the function is convex for $x>0$,
$$f\left(\frac{x+y+z}{3}\right)\leq \frac{f(x)+f(y)+f(z)}{3}$$
The thing we need to prove is $$f\left(\frac{x+y+z}{3}\right)=1/6$$ but it's not. How to prove it?
| Observations towards a solution
*
*As suggested by Siddharth, show that $ x = \frac{ -a^2 + b^2 + c^2 } { 2bc }$.
*As remarked by Abastro, $x, y, z$ are cosines of angles of the a-b-c triangle. In particular $x, y, z \leq 1$ (though we won't need this).
*In fact, the inequality holds for any triangle (with positive sides), without the condition that $ x, y, z \geq 0$. The proof still works, but you just have to keep track of signs (See point 6.)
*If we allow degenerate triangles, observe that $ (x,y, z) = (0, 0, 1)$ (and permutations), yield the equality condition. This implies that a simple AM-GM/Jensens, etc will likely not work, esp if it only yields the $ x = y = z = 1/2$ equality condition.
We likely need to use methods like Schur's inequality, Mixing Variable, Vasc's RCF theorem, etc, to get to the other equality case.
*The (naive) direct application of Titu's lemma yields
$$ \sum \frac{ x^2 } { 1 + x} \geq \frac{ ( x+y+z ) ^2 } { 3 + x + y + z}.$$
However, with $ (x, y, z ) \rightarrow (0,0,1)$, the RHS $\rightarrow \frac{1}{4}$, so we've over-optimized. We need to figure out how to weight these terms, which isn't that obvious.
*As suggested by Siddharth, using the substitution $ x = \frac{ -a^2 + b^2 + c^2 } { 2bc }$, we get $ \frac{ x^2 } { 1 + x } = \frac{ (-a^2 + b^2 + c^2 ) ^2 } { 2bc( -a^2 + b^2 + c^2 + 2bc)} $. Now, applying Titu's lemma gives us
$$ \sum \frac{ x^2 } { 1+x } \geq \frac{ ( a^2 +b^2 + c^2) ^2 } {\sum 2bc( -a^2 + b^2 + c^2 + 2bc) }. $$
Notes for this step:
*
*By clearing denominators, we've actually weighted the terms. As it turns out in the rest of the solution, we got lucky.
*Work through this if you want a solution to "all triangles" instead of just "acute triangles". While it seems like I'm using that $ x\geq 0 \Rightarrow b^2 + c^2 \geq a^2 $ in this part of the proof so that we don't flip the signs in the numerator, that isn't necessary. In the case of $ x < 0 \Rightarrow b^2 + c^2 < a^2$, I can still apply Titu as above because $1 + x > 0 $ and $ -x > x $.
*Now, observe that $ ( a^2 +b^2 + c^2) ^2 \geq \sum bc( -a^2 + b^2 + c^2 + 2bc)$ when expanded out is just the Schur's inequality $ \sum a^2 ( a-b)(a-c) \geq 0 $. (Verify this yourself, I'm too lazy to show the steps.)
Hence
$$ \sum \frac{ x^2 } { 1+x } \geq \sum \frac{ ( a^2 +b^2 + c^2) ^2 } {\sum 2bc( -a^2 + b^2 + c^2 + 2bc) } \geq \frac{1}{2}. $$
*The equality case is when
*
*All values are equal: $ a=b= c \Rightarrow x = y = z = \cos 60^\circ = 1/2$.
*Two of them are equal and the other is zero: $a=b, c =0$ is the degenerate isosceles triangle, which yields $(x,y,z) = (0, 0, 1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4616087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Sketching an arbitrary ellipse from its parametric equation I was solving a question, in which I was asked to solve the following system of differential equations:
$$
\dot x = 3x + 2y, \quad \dot y = -5x - 3y.
$$
I got the general solution in $\mathbb R^2$ to be
$$
\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} C & D \\ -\frac32 C + \frac12 D & -\frac12 C - \frac 32 D \end{pmatrix}\begin{pmatrix} \cos t \\ \sin t \end{pmatrix}, \quad C,D\in \mathbb R.
$$
Now, I want to sketch the phase portrait of this solution (apparently, as we change $C$ and $D$, only the scale of the ellipse changes, not its orientation or ratio between major and minor axes). My question is, how do I determine the orientation of axis lengths of this ellipse (in terms of $C$ and $D$)? Given an arbitrary ellipse of the form
$$
\begin{pmatrix} C & D \\ E & F \end{pmatrix}\begin{pmatrix} \cos t \\ \sin t \end{pmatrix},
$$
how would one sketch it?
| First write $\cos t$ and $\sin t$ in terms of $x$ and $y$:
$$
\begin{pmatrix} x \\ y\end{pmatrix}=\begin{pmatrix} C & D \\ E & F \end{pmatrix}\begin{pmatrix} \cos t \\ \sin t \end{pmatrix}\\
\begin{pmatrix} \cos t \\ \sin t \end{pmatrix}=\begin{pmatrix} C & D \\ E & F \end{pmatrix}^{-1}\begin{pmatrix} x \\ y\end{pmatrix}
$$
The inverse is a $2\times2$ matrix, say $\begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix}$.
You also know $\cos^2 t+\sin^2t=1$. So $$(\alpha x+\beta y)^2+(\gamma x+\delta y)^2=1$$
You can group the terms together, then use this answer to get the orientation, and something like this answer to find the length of the semiaxes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4617870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
Find $x$ such that $\sqrt{x+1} - \sqrt{1-x} = 1$ To solve this equation, I started by putting the condition $x\in [-1, 1]$, then squared a few times: $\sqrt{x+1} - \sqrt{1-x} = 1 \iff x + 1 +1-x-2\sqrt{1-x^2} =1 \iff 2\sqrt{1-x^2}=1 \iff 4(1-x^2)=1 \iff 4x^2=3 \iff x=\pm \frac{\sqrt{3}}{2}$
This, however, is not the right solution, as $-\frac{\sqrt{3}}{2}$ returns $-1$, not $1$. My question is where did I miss a condition that excludes the negative "solution"? I expect somewhere along the line I squared where I wasn't allowed to square without an additional condition, hoping that I don't have to check these solutions every time.
| Using a trigonometric substituition:
As $-1 \le x \le 1$, then $x = \cos \theta$:
$$\sqrt{1+x} = \sqrt{1-\cos \theta} = \sqrt{2\sin^2 \frac{\theta}{2}} = \sqrt{2}\left|\sin \frac{\theta}{2}\right|$$
$$\sqrt{1+x} = \sqrt{1+\cos \theta} = \sqrt{2\cos^2 \frac{\theta}{2}} = \sqrt{2}\left|\cos \frac{\theta}{2}\right|$$
Then the equation becomes
$$\left|\sin \frac{\theta}{2}\right| + \left|\cos \frac{\theta}{2}\right| = \frac{1}{\sqrt{2}}$$
As we care only about the value of $x$, we can restrict the value of $\theta$ to be at the interval $\left[0, \ \pi\right]$:
*
*If $\theta \in \left[0, \ \dfrac{\pi}{2}\right]$:
$$\sin \frac{\theta}{2} + \cos \dfrac{\theta}{2} = \frac{1}{\sqrt{2}}$$
$$\sqrt{2} \sin \underbrace{\left(\dfrac{\theta}{2} + \dfrac{\pi}{4}\right) }_{\alpha}= \dfrac{1}{\sqrt{2}}$$
$$\sin \alpha = \dfrac{1}{2} \Rightarrow \alpha = \dfrac{\pi}{3} \ \ \text{or} \ \ \alpha = \dfrac{2\pi}{3}$$
Then $\theta = 2\left(\alpha - \frac{\pi}{4}\right)$ can assume two values: $\theta = \frac{\pi}{6}$ or $\theta = \frac{5\pi}{6}$.
Only $\theta = \frac{\pi}{6}$ is on the interval $\left[0, \ \frac{\pi}{2}\right]$, then we exclude the second value. Therefore:
$$\boxed{x = \cos \frac{\pi}{6} = \dfrac{\sqrt{3}}{2}}$$
*
*If $\theta \in \left[\dfrac{\pi}{2}, \ \pi\right]$
$$\sin \frac{\theta}{2} - \cos \dfrac{\theta}{2} = \frac{1}{\sqrt{2}}$$
$$\sin \underbrace{\left(\dfrac{\theta}{2} - \dfrac{\pi}{4}\right) }_{\alpha}= \dfrac{1}{2}$$
$$\alpha = \frac{\pi}{3} \ \ \text{or} \ \ \alpha = \frac{2 \pi}{3}$$
$$\theta = \frac{7\pi}{6} \ \ \text{or} \ \ \theta = \frac{11 \pi}{6} $$
As both $\theta$ are outside the interval $\left[\dfrac{\pi}{2}, \ \pi\right]$, there's no solution for this interval.
Then we got only one solution:
$$\boxed{x = \dfrac{\sqrt{3}}{2}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 5
} |
Integral in two variables I have to prove that
$$\int_D xy\; dxdy=\frac{3}{8}$$
with $D=\{(x,y)\in \mathbb{R}^2: x^2+y^2 \geq 1,\ \frac{x^2}{4}+y^2 \leq 1,\ x \geq 0,\ y \geq 0 \}$.
So I have the intersection of ellipse $\frac{x^2}{4}+y^2 \leq 1$ and the ball $x^2+y^2 \leq 1$.
If I use the sets
$$\begin{eqnarray}
P&=&\{(x,y)\in\mathbb{R}^2: x^2+y^2 \leq 1,\ x \geq 0,\ y \geq 0 \}\\
E&=&\Big\{(x,y)\in\mathbb{R}^2: \frac{x^2}{4}+y^2 \leq 1,\ x \geq 0,\ y \geq 0 \Big\}
\end{eqnarray}$$
I obtain that $\displaystyle \int_D xy\; dxdy=\int_E xy\; dxdy - \int_P xy\; dxdy$.
The integral $\int_E xy\; dxdy$ can be calculated whith the elliptic coordinates $x=2\rho\cos\theta$, $y=\rho \sin \theta$ and the set E became
$$\Big\{\rho \leq 1,\ \theta \in \Big[0,\frac{\pi}{2}\Big]\Big\}.$$
So $\displaystyle\int_E xy\; dxdy
=\int_0^1 2\rho^3\; d\rho \int_0^{\frac{\pi}{2}}\cos\theta\sin\theta\;d\theta
=\left.2\frac{\rho^4}{4}\right|^1_0 \cdot
\left.\Big(-\frac{1}{2}\cos^2\theta\Big)\right|_0^{\pi/2}
=\frac{1}{4}\cdot$
In the same way we can use the coordinate $x=\rho \cos\theta$, $y=\rho \sin \theta$ and the set P became
$$\Big\{\rho\leq 1,\ \theta\in\Big[0,\frac{\pi}{2}\Big]\Big\}.$$
So $\displaystyle\int_P xy\; dxdy=\int_0^1 \rho^\; d\rho \int_0^{\frac{\pi}{2}}\cos\theta\sin\theta\; d\theta
=\left.\frac{\rho^4}{4}\right|^1_0 \cdot
\left.\Big(-\frac{1}{2}\cos^2\theta\Big)\right|_0^{\pi/2}
=\frac{1}{8}\cdot$
Then $\displaystyle\int_D xy\; dxdy=\frac{1}{4}-\frac{1}{8}=\frac{1}{8}\cdot$
Where am I doing wrong?
| It's better do the integral in Cartesian anyway
$$\int_0^1 \int_{\sqrt{1-y^2}}^{2\sqrt{1-y^2}}xy\:dxdy = \int_0^1\frac{3}{2}\left(y-y^3\right)\:dy = \frac{3}{4} - \frac{3}{8} = \frac{3}{8}$$
Don't always immediately go to polar coordinates whenever you see conic sections. The integrand made this nice since we had odd powers of $x$ and $y$ which, when integrated, will cancel out the square roots and give us nice polynomials.
| {
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"question_score": "4",
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How do we prove $x^6+3x^3+2x^2+x+1 \geq 0$ Question
How do we prove
$$x^6 + 3x^3+2x^2+x+1\geq0$$
My progress
$$x^6+3x^3+2x^2+x+1=(x+1)^2(x^4-2x^3+3x^2-x+1)$$
I appreciate your interest
| Case 1: $x$ is nonnegative.
The last term is greater than or equal to $(x^2-x+1)^2,$ since it is $(x^2-x+1)^2 + x.$ By the Trivial Inequality, both factors are positive or zero, so the product is also positive or zero.
We can also see this directly from the question. It is clear that the $x^6+3x^3+2x^2+x+1 \geq 0,$ since all the summands are nonnegative.
Case 2: $x$ is negative.
We can write this as $(x^2 + 1)^2 + x^2 + (-2x^3-2x).$ Since the third term is positive (since $x$ is negative) and the first two terms are nonnegative, the second factor is positive. Since the first factor is nonnegative, we are done.
Alternatively, we can write the second term as a sum of positive numbers multiplied by squares. It is $(x^2-x)^2 + \frac{7}{4}x^2 + (1-\frac{x}{2})^2$.
| {
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What is the largest/least value of the function $f$ in $\mathbb{R}^2$? What is the largest/least value of the function $f=x^2ye^{-x^2-2y^2}$ in $\mathbb{R}^2$?
We have that $f'_x=y\left(2e^{-x^2-2y^2}x-2e^{-x^2-2y^2}x^3\right)$ and $f'_y=x^2\left(e^{-x^2-2y^2}-4e^{-x^2-2y^2}y^2\right)$. If $x=0$ then $f'_x=f'_y=0$ so $(0,y)$, where $y\in \mathbb{R}^2$ are stationary points.
Further $1-4y^2=0\iff y=+-1/\sqrt{2}$ and $2x-2x^3=0\iff2x(1-x^2)\iff x=0, x=+-1$ so $+-(1,1/\sqrt{2})$ are also stationary points.
The values for the stationary points are $f(0,y)=0$, $f(+-1,1/\sqrt{2})=\frac{e^{-2}}{\sqrt{2}}$ and $f(+-1,-1/\sqrt{2})=-\frac{e^{-2}}{\sqrt{2}}$.
Now how do I prove that $e^{-2}/\sqrt{2}$ is the largest value and $-e^{-2}/\sqrt{2}$ is the smallest? Would be nice with a blueprint that I can use for any function but I guess it doesn't exist.
Edit: Not looking for solutions that splits up the $f$ to $x^2e^{-x^2}ye^{-2y^2}$.
I wrote an answer and hopefully someone can check that it is right.
| The idea is to make a circle(ball) that contains the critical points. By the extreme value theorem there will be a greatest and least value within that circle. If we can show that the boundary points' f-value approaches 0 as the radius get larger then we have shown that for an arbitrary large circle that contains the critical points they will give greatest and least value. In our case the max and min value are $\frac {e^{-2}}{\sqrt{2}}$ and $\frac {e^{-2}}{\sqrt{2}}$ respectively.
Let $\begin{cases}x=r\cos\theta \\ y=r\sin\theta \end{cases}$ where $0 \leq\theta\leq 2\pi$. We need to show that $\lim _{r\rightarrow\infty}=r^3\cos^2\theta\sin\theta e^{-r^2\cos^2\theta-2r^2\sin^2\theta}=0$.
Since $-r^2\cos^2\theta -2r^2\sin^2\theta=-r^2(\cos^2\theta+2\sin^2\theta)=-r^2(1+sin^2\theta)\leq -r^2\iff e^{-r^2(1+\sin^2\theta)}\leq e^{-r^2}$ we have that $r^3\cos^2\theta\sin\theta e^{-r^2\cos^2\theta-2r^2\sin^2\theta}\leq r^3e^{-r^2}\rightarrow 0$ as $r\rightarrow \infty$. So we have shown that for large enough r, radius of the circle, the boundary points will have f-values close to 0 and as $r$ gets larger the boundary points of the f-values will get closer to 0. Since we can choose an arbitrarily large $r$ it means that $\frac {e^{-2}}{\sqrt{2}}$ and $\frac {e^{-2}}{\sqrt{2}}$ are the largest and least value respectively in $\mathbb{R}^2$.
| {
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Solve $\lim_{x\to\infty}(x-x^2 \ln{\frac{1+x}{x}})$ without L'Hopital I've seen the solution to $\lim_{x\to\infty}(x-x^2 \ln{\frac{1+x}{x}})$ using L'Hopital and I was wondering if there's a way to find out the result without it. My initial attempt was outright stupid of me because I tried to substitute the limit of $\frac{\ln{(1+\frac{1}{x})}}{\frac{1}{x}}$ as $x$ approaches $\infty$ with $1$, which results in the initial limit being $0$. That's obviously false as I ignored the fact that I cannot do such a substitution when the limit is in an indeterminate form. That being said, how could you solve this limit without L'Hopital?
| Use substitution $e^t = \frac{x}{1 + x}$;
$$x - x^2\ln\left(\frac{1+x}{x}\right) \quad \overset{e^t = \frac{x}{1 + x}}{\longrightarrow} \quad \frac{1}{e^{-t} - 1} + \frac{t}{\left( e^{-t} - 1 \right)^2} = \frac{t + e^{-t} - 1}{\left( e^{-t} - 1 \right)^2} \\ \quad \\ \quad \\ = \frac{te^{t} + 1 - e^{t}}{e^{t}\left( e^{-t} - 1 \right)^2} \\ \quad \\ \quad \\ = \frac{te^{t} + 1 - e^{t}}{\left( e^{\frac{-t}{2}} - e^{\frac{t}{2}} \right)^2} \\ \quad \\ \quad \\ = \frac{te^{t} + 1 - e^{t}}{\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} \right)^2 - 4} \\ \quad \\ \quad \\ = \frac{te^{t} - e^{\frac{t}{2}}\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} - 2 \right) - 2e^{\frac{t}{2}} + 2}{\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} - 2 \right)\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} + 2 \right)} \\ \quad \\ \quad \\ = \frac{te^{t} - 2e^{\frac{t}{2}} + 2}{\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} - 2 \right)\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} + 2 \right)} - \frac{e^{\frac{t}{2}}}{\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} + 2 \right)}$$
As $x \to \infty, t \to 0$. WLOG assume $te^{\frac{t}{2}} + 1 \approx e^{t}$ for small $t$. (The intention behind this was to be able to factor the numerator while keeping the integrity of limit equivalent. Since, $\color{red}{x}e^{x}$ would not be factored easily, the goal was to approximate $\color{red}{x}e^{x} \approx \color{blue}{k}e^{\frac{x}{\color{blue}{c}}} + \color{blue}{p}$, where $k = 1$ was preferred, for small $t$).
Hence, from where we left off,
$$\frac{te^{t} - 2e^{\frac{t}{2}} + 2}{\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} - 2 \right)\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} + 2 \right)} - \frac{e^{\frac{t}{2}}}{\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} + 2 \right)} \\ \quad \\ \quad \\ = \frac{e^{\frac{t}{2}} \left( te^{\frac{t}{2}} + 1 \right) - 3e^{\frac{t}{2}} + 2}{\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} - 2 \right)\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} + 2 \right)} - \frac{e^{\frac{t}{2}}}{\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} + 2 \right)} \\ \quad \\ \quad \\ \approx \frac{e^{\frac{t}{2}}e^{t} - 3e^{\frac{t}{2}} + 2}{\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} - 2 \right)\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} + 2 \right)} - \frac{e^{\frac{t}{2}}}{\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} + 2 \right)} $$
Now, let $u = e^{\frac{t}{2}}$;
$$\frac{e^{\frac{t}{2}}e^{t} - 3e^{\frac{t}{2}} + 2}{\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} - 2 \right)\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} + 2 \right)} - \frac{e^{\frac{t}{2}}}{\left( e^{\frac{-t}{2}} + e^{\frac{t}{2}} + 2 \right)} \\ \quad \\ \quad \\ \overset{u = e^{\frac{t}{2}}}{\longrightarrow} \frac{u^3 - 3u + 2}{\left( u^{-1} + u + 2 \right)\left( u^{-1} + u - 2 \right)} - \frac{u}{\left( u^{-1} + u + 2 \right)} \\ \quad \\ \quad \\ = \frac{u^2 (u - 1)^2 (u + 2)}{\left( 1 + u^2 + 2u \right)\left( 1 + u^2 - 2u \right)} - \frac{u}{\left( u^{-1} + u + 2 \right)} \\ \quad \\ \quad \\ = \frac{u^2 (u - 1)^2 (u + 2)}{(u + 1)^2(u - 1)^2} - \frac{u^2}{(u + 1)^2} \\ \quad \\ \quad \\ = \frac{u^2(u + 1)}{(u + 1)^2} = \frac{u^2}{u + 1}$$
As $x \to \infty \Rightarrow t \to 0 \Rightarrow u \to 1$
Hence and finally, $$\bbox[5px,border:2px solid black]{\lim\limits_{u \to 1}{\frac{u^2}{u + 1}} = \frac{1}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4622401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Can $\,9\!\cdot\!10^n+4\,$ be a perfect square? I think $\,9\!\cdot\!10^n+4\,$ can be a perfect square, since it is $0 \pmod 4$ (a quadratic residue modulo $4$), and $1 \pmod 3$ (also a quadratic residue modulo $3$).
But when I tried to find if $\;9\!\cdot\!10^n+4\,$ is a perfect square, I didn’t succeed. Can someone help me see if $\;9\!\cdot\!10^n+4\,$ can be a perfect square ?
| Comment:
Lets try construction such a number. Suppose we have:
$k^2-9\times 10^n=4$
We use following known Pell's equation:
$x^2-Dy^1=1$
For $D=10$ we have $x=19$ and $y=6$ such that:
$19^2-10\times 6^2=1$
multiplying both sides by $2^2$ we get"
$38^2-10\times 12^2=4$
we rewrite this as:
$38^2-9\times(4^2\times 10)=4$
multiplying both sides by $25^2$ we get:
$(25\times 38)^2-9\times 10^5=4\times 25^2=2500=2496+4$
Or:
$[(25\times 38)^2-2496]-9\times 10^5=4$
Or generally:
$A=[(25\times 38)\times 10^m-(25\times 10^{2m}+96)]$
here in equation $k^2-9\times 10^n=4$, $n=2m+1$
$A$ must be perfect square.May be by brute force we can find such a number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4622956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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In $pqr$ method, if $q+r=4$, show that $p^2-2q+3r\ge2p$
Positive numbers $x$, $y$ and $z$ satisfy $xy+yz+zx+xyz=4$, show that
\[x^2+y^2+z^2+3xyz\ge2(x+y+z).\]
I thought of $pqr$ method, let $\begin{cases}p=x+y+z,\\q=xy+yz+zx,\\r=xyz.\end{cases}$ so we have $q+r=4$, and needed to prove
\[p^2-2q+3r\ge2p.\]
Transform into $(p-1)^2-2q+3r\ge1$. Since $q$, $r$ have a condition, I wanted to get rid of $p$. Clearly $p>1$ (actually $p\ge3$), by $p^2\ge3q$, it suffices to prove
\[\left(\sqrt{3q}-1\right)^2-2q+3(4-q)\ge1.\]This resulted in $q+\sqrt{3q}\le6$, which is false because we can prove $q\ge3$. If we change $p$ into something else it will become only weaker so it cannot work. Perhaps we couldn't use $pqr$ method here.
Any method is okay, except for Lagrange Multipliers, of course.
| pqr method:
It suffices to prove that
$$p^2 - 2q + 3r - 2p + (4 - q - r) \ge 0$$
or
$$p^2 - 2p + 4 + 2r - 3q\ge 0.$$
Degree three Schur's inequality
yields $p^3 - 4pq + 9r \ge 0$
which results in
$q \le \frac{p^3 + 9r}{4p}$.
It suffices to prove that
$$p^2 - 2p + 4 + 2r - 3\cdot \frac{p^3 + 9r}{4p}\ge 0$$
or
$$p(p-4)^2 - (27-8p)r \ge 0. \tag{1}$$
If $27 - 8p \le 0$, clearly (1) is true.
If $27 - 8p > 0$, using $p^3 \ge 27r$,
we have
$$p(p-4)^2 - (27-8p)r
\ge p(p-4)^2 - (27-8p)\cdot \frac{p^3}{27} = \frac{8}{27}p(p+6)(p-3)^2 \ge 0.$$
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4623974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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How far from home, can my robot roam? It has constant step size, and turns by increasing amounts. A robot's step size is always $1$. Between steps it turns right, by increasing amounts: $\frac{1\pi}{2},\frac{2\pi}{3},\frac{3\pi}{4},\frac{4\pi}{5},...$
What is the robot's maximum distance from the origin?
Here are the first 15 steps, starting in the lower-left corner (switching color every three steps, for clarity).
Superimpose cartesian coordinates, with the first step from $(0,0)$ to $(0,1)$. After the $n$th step, the robot's coordinates are:
$$x=\sum_{k=1}^n \sin{\left(\sum_{i=1}^k \pi\left(1-\frac{1}{i}\right)\right)}$$
$$y=\sum_{k=1}^n \cos{\left(\sum_{i=1}^k \pi\left(1-\frac{1}{i}\right)\right)}$$
Its distance from the origin is
$$d(n)=\sqrt{x^2+y^2}$$
Here is the graph of $d(n)$ against $n$.
It looks like the maximum value of $d(n)$ is $d(2)=\sqrt{2}$, which corresponds to the red point in the first diagram. But we cannot check the entire graph, because it goes forever. How can we know the maximum value of $d(n)$?
EDIT
It would be enough to prove the following:
Lemma: The circle through three consecutive vertices of the robot's path, encloses all subsequent vertices.
The circle through the first three vertices $(0,0),(0,1),(1,1)$ is $(x-\frac12)^2+(y-\frac12)^2=\frac12$. All points on this circle are within $\sqrt{2}$ from the origin. So we would know that $\sqrt{2}$ is the maximum distance form the origin. But I don't know how to prove the lemma.
EDIT2
@Intelligentipauca's comment pointed out that the lemma is false. For example, the circle through the points for $n=2,3,4$ does not enclose the point for $n=6$.
| Identify Euclidean plane $\mathbb{R}^2$ with complex plane $\mathbb{C}$.
For each $k \in \mathbb{Z}_{+}$, let
$u_k = i(-1)^k e^{i \pi H_k}$ where $H_k = \sum\limits_{\ell=1}^k \frac{1}{\ell}$ are the Harmonic numbers.
In terms of $u_k$, the position after $n^{th}$ move, $(x_n,y_n)$, is given by the formula:
$$a_n \stackrel{def}{=} x_{n} + iy_{n}= \sum_{k=1}^n u_k$$
Let $b_n$ and $c_n$ be the averages and averages of averages of successive locations of $a_n$. ie.
$$b_n = \frac12(a_{n+1} + a_n)\quad\text{ and }\quad c_n = \frac12(b_{n+1} + b_n)$$
Let $K = \frac{\pi\sqrt{\pi^2+4}}{4}$, notice
$$\begin{align}
|a_n - b_n|
&= \frac12|u_{n+1}| = \frac12\\
|b_n - c_n |
&= \frac14|u_{n+2} + u_{n+1}| = \frac14\left| e^{i\frac{\pi}{n+2}} - 1\right|\\
&= \frac12\sin\frac{\pi}{2(n+2)} < \frac{\pi}{4(n+2)}\\
|c_{n+1} - c_n|
&= \frac14|u_{n+3} + 2u_{n+2} + u_{n+1}|
= \frac14\left|e^{i\frac{\pi}{n+3}} - 2 + e^{-i\frac{\pi}{n+1}}\right|\\
&\stackrel{\color{blue}{[1]}}{\le}
\frac14\sqrt{\left(\frac{\pi^2}{(n+1)(n+3)}\right)^2 + \left(\frac{\pi}{n+1} - \frac{\pi}{n+3}\right)^2}\\
&< \frac{K}{(n+1)(n+2)}
\end{align}
$$
So for all $m \ge n$, we have
$$\begin{align}|a_m - c_n|
&\le |a_m - b_m| + |b_m - c_m| + \sum_{k=n}^{m-1}|c_{k+1}-c_k|\\
&\le \frac12 + \frac{\pi}{4(m+2)} + \sum_{k=n}^{m-1}\frac{K}{(k+1)(k+2)}\\
&< \frac12 + \frac{K}{n+1}
\end{align}
$$
which in turn implies
$$|a_m| < |c_n| + \frac12 + \frac{K}{n+1}$$
For $n = 55$, the expression on RHS $\sim
1.399210680593634 < \sqrt{2} = |a_{2}|$. This means for all $m \ge 55$, $a_m$ is closer to origin than $a_2$.
By brute force, one can verify $|a_1|, |a_3|,\ldots |a_{54}|$ are all smaller than $\sqrt{2}$. This means $\sqrt{2}$ is indeed the maximum distance from origin.
Notes
*
*$\color{blue}{[1]}$ - For any $\frac{\pi}{2} \ge u > v> 0$, let $u = p + q$ and $v = p-q$, we have
$$\begin{align}|e^{iv} - 2 + e^{-iu}|
&= |e^{ip} - 2e^{iq} + e^{-ip}|\\
&= 2\sqrt{(\cos(p) - \cos(q))^2 + \sin(q)^2}\\
&= \sqrt{\left(4\sin\frac{u}{2}\sin\frac{v}{2}\right)^2 + (2\sin(q))^2}\\
&\le \sqrt{(uv)^2 + (2q)^2}\\
&= \sqrt{(uv)^2 + (u-v)^2}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4624148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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} |
How to solve a system of equations over a finite field? I need to solve a system of equations over $\mathbb{Z}_{11}$.
My system is:
$$ \left\{
\begin{array}{l}
2x + 5y + z = 8 \\
7x + 6y + 8z = 10 \\
10x + 3y + 4z = 6
\end{array}
\right.
$$
In matrix form:
$$
\begin{bmatrix}
2 & 5 & 1\\
7 & 6 & 8\\
10 & 3 & 4\\
\end{bmatrix}
\begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix} =
\begin{bmatrix}
8\\
10\\
6\\
\end{bmatrix}
$$
after $R_1*6$ :
$$
\begin{bmatrix}
1 & 8 & 6\\
7 & 6 & 8\\
10 & 3 & 4\\
\end{bmatrix}=
\begin{bmatrix}
4\\
10\\
6\\
\end{bmatrix}
$$
$R_2-7*R_1$ and $R_3-10*R_1$:
$$
\begin{bmatrix}
1 & 8 & 6\\
0 & 5 & 10\\
0 & 0 & 1\\
\end{bmatrix}=
\begin{bmatrix}
4\\
4\\
10\\
\end{bmatrix}
$$
$R_2*9$ :
$$
\begin{bmatrix}
1 & 8 & 6\\
0 & 1 & 2\\
0 & 0 & 1\\
\end{bmatrix}=
\begin{bmatrix}
4\\
3\\
10\\
\end{bmatrix}
$$
$$\left\{
\begin{array}{l}
z = 10 \\
y = 3 - 2z = 3 - 20 = -17=5 \\
x = 4 - 8y - 6z = 4 - 8*5 - 6*10 = -96 = 3
\end{array}
\right.$$
But after substitution found values in the initial system I have an error:
$$ \left\{
\begin{array}{l}
2x + 5y + z = 2*3 + 5*5+10= 41 = 8 \\
7x + 6y + 8z = 7*3 + 6*5 + 8*10 = 131 = 10 \\
10x + 3y + 4z = 10*3 + 3*5 + 4*10= 85 = 8 \neq 6
\end{array}
\right.
$$
Do I have an arithmetic error or the whole way is wrong? Is there an easier and faster way to solve it?
| You made a calculation error when calculating $R_3-10* R_1$. The last row in the matrix should be $[0,0,10|10]$ rather than $[0,0,1|10]$. With this fix, you get that the solution to the system is $x=y=z=1$, which you can easily check actually solves the system.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4624324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve $7^a+b!=13^c$ over $\mathbb{Z}^{+}$
Solve the following Diophantine equation $7^a+b!=13^c$ over positive
integers.
Clearly $7^1+3!=13^1$ and $7^2+5!=13^2$. Are there any more solutions? Here are some thoughts.
$b!$ must be even, so $b>1$.
Case 1: $c$ is even: let $c=2k$. Then, $b!\equiv13^c\equiv13^{2k}\equiv1\pmod 7\implies b=5$. So we're looking for solutions to $7^a+120=13^{2k}$. In modulo 4, this reduces to $(-1)^a\equiv1$ which means that $a$ is even: $a=2n$ and then we have $7^{2n}+120 = 13^{2k} \implies(13^k-7^n)(13^k+7^n)=120$. The possible even factors (they must both be even) are:
*
*$2\cdot60 \implies 13^k=31, 7^n=29$;
*$4\cdot30 \implies 13^k=17, 7^n=13$;
*$6\cdot20 \implies 13^k=13, 7^n=7$;
*$10\cdot12 \implies 13^k=11, 7^n=1$.
And clearly, only the third case yields a solution in integers with $k=n=1$.
Case 2: $c$ is odd: let $c=2m+1$. Then, $b!\equiv13^c≡-1\pmod7 \implies b=3$ or $b=6$.
Sub-case: $b=6$. We have $7^a+720=13^{2m+1}$. In modulo 8 this becomes $(-1)a\equiv5^{2m+1}≡5$ which is not possible.
Sub-case: $b=3$. We have $7^a+6=13^{2m+1}$ [In modulo 13 this becomes $7^a+6≡0$ which means $a=12q+1$ and then…].
In modulo 4, this reduces to $(-1)^a+2\equiv1$ which means that $a$ is odd: $a=2p+1$ and then we have
$$
\begin{aligned}
& 7^{2 p+1}+6=13^{2 m+1} \\
& \implies 7^{2 p+1}-7=13^{2 m+1}-13 \\
& \implies 7(7^{2 p}-1)=13(13^{2 m}-1) \\
& \implies 7(7^p-1)(7^p+1)=13(13^m-1)(13^m+1).
\end{aligned}
$$
I suspect $p=m=0$ is the only possible solution but cannot see a proof yet...
| The Dipohatine equation
$7^a+6=13^c$
has no solutions when $a\ge2$. We prove this with some fairly creative modular arithmetic.
Modulo 49
If $a\ge 2$, then $7^a\overset{\text{forced}}{\equiv}0\bmod49$ and so $13^c\equiv6$. From calculating powers of $13\bmod 49$ we then infer that $c\equiv11\bmod14$. The modulus on the exponent is reduced to $14$ because $13^{14}\equiv1\bmod49$.
With the modulus on $c$ reduced to $14$, we seek a different modulus whose Euler totient is divisible by $14$.
Modulo 29
Since $c\equiv11\bmod14$ and $13\equiv10^2$ is a quadratic residue $\bmod29$, we infer that $13^c\equiv13^{11}\equiv4\bmod29$. But then $7^a\equiv27$ whereas $7\equiv6^2$ is a quadratic residue and $27\equiv3×3^2$ is not. (The number $29$ is $\not\equiv\pm1\bmod12$.) $\rightarrow\leftarrow$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4624870",
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"source": "stackexchange",
"question_score": "1",
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Proof that $(x+y)^n = x^n + y^n$ iff. $x = 0 \lor y = 0 \lor x = -y$ for $n$ odd and $\geq 3$ I was going through Calculus by Spivak when in the first chapter I encountered problem 16, which in the end recites
you should know make a good guess as to when $\left(x+y\right)^n = x^n + y^n$
quoting the results of previous exercises with a fixed $n$, which had
this result iff $x = 0 \lor y = 0 \lor x = -y$
I started going on with the proof and immediately thought of using induction, but I was later discouraged about it when, seeing how I "proved" the cases with fixed $n$, I wasn't using an "induction-like" procedure, and I found myself lost (I saw Spivak proof and he uses Rolle's theorem on the function $f\left(x\right) = x^n + y^n - \left(x+y\right)^n$, but it isn't a proof that I find satisfying, as I think there's a proof that doesn't use such methods.
Here is a look into the procedure I used for $n=5$:
$$\left(x+y\right)^n = x^5 + 5 x^4 y + 10 x^3 y^2 + 10 x^2 y^3 + 5 x y^4 + y^5 = x^5 + y^5$$
$$5\left(x^4y + 2x^3y^2 + 2x^2y^3 + xy^4\right) = 0$$
$$xy\left(x^3 + 2x^2y + 2xy^2 + y^3\right) = 0$$
So either $x = 0 \lor y = 0$ gives us a solution, so now we consider the part inside parenthesis
$$x^3 + 2x^2y + 2xy^2 + y^3= x^3 + 3x^2y + 3xy^2 + y^3 - x^2y - xy^2 = 0$$
$$\left(x+y\right)^3 - xy\left(x + y\right) = 0$$
So either
$$xy = \left(x+y\right)^2 \iff 0 = x^2 + xy + y^2 \iff x = -y $$ or
$$\left(x+y\right)^3 = 0 \land \left(x+y\right) = 0 \iff x = -y$$
For a generic $n$, I went with:
$$\left(x+y\right)^n = \sum_{k = 0}^n \binom{n}{k}x^{n-k}y^k = x^n + y^n$$
Unpacking the first and last term of the sum that gives us
$$\sum_{k=1}^{n-1} \binom{n}{k}x^{n-k}y^k = 0$$
$$xy\left(\sum_{k=1}^{n-1} \binom{n}{k}x^{n-1-k}y^{k-1}\right) = 0$$
Which proves the condition $x = 0 \lor y = 0 \implies \left(x+y\right)^n = x^n + y^n$
Now I tried to find a way to transform the inner sum into the form
$$\left(x+y\right)^{n-2} - z\left(x+y\right)$$
Where $z$ is any possible polynomial that uses $x$ or $y$ or both
This way I should resolve $$\left(x+y\right)^{n-2} = \left(x+y\right) = 0$$ which yields $x = -y$ as the only result excluding $x = y = 0$
--- Update ---
Following Mark Bennet advise, I went on to prove that, with $|x| > |y|, z = y/x$ (I also tried with $z = x/y$), the sum
$$\sum_{i=1}^{n-1}\binom{n}{i}z^i = 0 \iff z = -1 \lor z = 0$$ which for $z < 0$ has the same number of positive and negative terms. Due to the properties of the binomial, we also know that the $i$-th and the $\left(n-i\right)$-th term of the sum have the same coefficient $\binom{n}{i}$. So we can write it as
$$\sum_{i=1}^{\left(n-1\right)/2}\binom{n}{i} \cdot z^i\left(1+z^{n-2i}\right)$$
My idea was to prove that such sum is never equal to $0$ except when $z = 0 \lor z = -1$. Do to prove such a thing, I had 3 main ideas:
*
*-prove that every element of the sum differs by values whose sum $\neq 0$
*-prove that the sum has only 2 real solutions, so that they can only be $z = 0 \lor z = 1$
*-prove that the sum is a parabola: this implies that it only has 2 real solutions
But again, I cannot find a way to prove such things
--- Update 2 ---
Thinking about the problem, I tried to pose it as
$$\left(1+z\right)^n < 1 + z^n$$
with $z = y/x \land -1 < z < 0$ (We already proved for $z$ > 0 that the claim is false)
So, listing down a couple of things:
$$1 + z < 1 \land z^n < z \implies \left(1+z\right)^n < 1 +z < 1 + z^n \forall n$$
This means that $$\forall z \in \left(-1, 0\right), \left(1+z\right)^n\neq 1 + z^n$$
As this is the only left range, it means that $z$ must either be $-1 \lor 0$, which implies $y = 0 \lor x = -y$, which proves our claim
| Like you mentioned, solving $(1+z)^n = 1+z^n$ for $0<z<1$ is easy by binomial or Bernoulli. For when $z<0,$ let $z\to -z$ and for $0<z<1$ and odd $n$, look at:
$$z^n+(1-z)^n = 1.$$ But this cannot happen since:
$$1 = z^n + (1-z)^n < z + (1-z) = 1.$$
| {
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"url": "https://math.stackexchange.com/questions/4627535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Solve $\log_{4}3x+1,+\log_{3}16x+1=\log_{2}12x+4$ Here is a basic logarithmic problem that I am struggling to solve:
$\log_{4}(3x+1)+\log_{3}(16x+1)=\log_{2}(12x+4)$
Here is my step:
$\log_{4}(3x+1)+\log_{4}(\frac{16x+1}{3})=\log_{2}(12x+4)$
Since the bases are the same now, we can multiply the arguments.
$\log_{4}(3x+1)*(\frac{16x+1}{3})=\log_{2}(12x+4)$
$\log_{4}\frac{48x^2+19x+1}{3}=\log_{2}(12x+4)$
The next thing to do is to change the base of 2 to base 4 on the right side.
Therefore,
$\log_{4}\frac{48x^2+19x+1}{3}=\log_{4}\frac{12x+4}{\frac{1}{2}}$
The right side can be simplified further, so
$\log_{4}\frac{48x^2+19x+1}{3}=\log_{4}24x+8$
Since the bases are the same on both sides, the arguments are also the same.
$\frac{48x^2+19x+1}{3}=24x+8$
$48x^2+19x+1=72x+24$
$48x^2-53x-23=0$
Using the quadratic formula, I get
$x=\frac{-(-53)\pm\sqrt{(-53)^2-4*48*-23}}{2*48}$
$x=\frac{53\pm\sqrt{2809+4116}}{96}$
$x=\frac{53\pm\sqrt{7225}}{96}$
$x=\frac{53\pm85}{96}$
$x=\frac{23}{16}$
or
$x=\frac{-1}{3}$
Since log argument cannot be negative, the second solution will not work. But when I checked the solution, $x=5$, which is not the answer that I have. A seemingly easy problem is driving me crazy. Can anyone point out what I'm doing wrong?
| The issue with your solution, as already noted in the comments, is when you converted $$\log_3(16x+1)$$ to $$\log_4\left(\frac{16x+1}3\right)$$
The log base rule states that:
$$\log_ba=\frac{\log_da}{\log_db}$$
In this case, $$\log_3(16x+1)$$
would convert to $$\frac{\log_4(16x+1)}{\log_43}$$
which is not equal to $$\log_4\left(\frac{16x+1}3\right)$$
Take, for example, $$\frac{\log_4{16}}{\log_44}=\frac21=2$$
But,
$$\log_4\left(\frac{16}4\right)=\log_44=1$$
Which is obviously not the same as $2$.
| {
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"source": "stackexchange",
"question_score": "1",
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} |
Evaluate $(g \circ f)^{-1}(x) = {\left( {\frac{{x - 7}}{2}} \right)^{\frac{1}{3}}}$
If $g\left( x \right) = ax + c$, $f\left( x \right) = {x^b} + 3$, and $(g \circ f)^{-1}(x) = {\left( {\frac{{x - 7}}{2}} \right)^{\frac{1}{3}}}$ what is the value of $a+b+c$?
My approach is as follows
Given $g\left( x \right) = ax + c$ & $f\left( x \right) = {x^b} + 3 \Rightarrow {\left( {f\left( x \right) - 3} \right)^{\frac{1}{b}}} = x$
$\frac{{g\left( x \right) - c}}{a} = x$
${\left( {f\left( x \right) - 3} \right)^{\frac{1}{b}}} = \frac{{g\left( x \right) - c}}{a}$
$g\left( x \right) = g\left( x \right) \Rightarrow {g^{ - 1}}\left( {g\left( x \right)} \right) = x$
I am not able to proceed.
| Notice that $$(g\circ f)(x) = 2x^3+7$$
But by definition of $g$ and $f$ we have also $$(g\circ f)(x) =ax^b+3a+c$$
Now if $x=0$ then we get $7 =3a+c$ and then $ax^b = 2x^3$. So if $x=1$ we get $a=2$ and $c=1$ and $x^b=x^3$. So if $x=2$ we get $2^b = 8$ and thus $b=3$.
Notice: Since we don't know apriori that $b$ is positive integer you can not do just the comparison of the coefficients.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
The most elementary proof of divisibility of sum of powers I am wondering how one can prove that for an arbitrary odd natural number $n$ and an arbitrary natural number $a$ the power-sum $S_{(a,n)}=1^n+2^n+\ldots +a^n $ is divisible by $S_{(a,1)}=1+2+\ldots + a$ in the most elementary way (possible) which comes to your mind. By an elementary way I mean the way which even middle school student could follow.
I traditionally tried to compute $S_{(a,n)}$ using the expansion of $(x-1)^{n+1}=\sum_{k=0}^{n+1}(-1)^k\binom{n+1}{k}x^{n+1-k}$.
I substituted $x=1,2,3,\ldots , a$ and then summed up results. After telescoping and cancelations I received $a^{n+1}=\sum_{k=1}^{n+1}(-1)^{k+1}\binom{n+1}{k}S_{(a,n+1-k)}$, from which we can express $S_{(a,n)}$.
Nevertheless this approach seems to me tough and not so easy to follow for a middle school student; A loads of technical re-writing and quite long symbolic computations needed to go through.
My question: I am missing something very evident and elegant which gives the desired divisibility without tears?
| Note that $r^n+(P-r)^n$ is divisible by $P$ by Binomial expansion and cancelling of the only terms without factors of $P$ (as $n$ is odd). If $a$ is odd, then all the terms in $S_{(a,n)}$ pair up in this way, with $P=a$, except for $a^n$, which is certainly a multiple of $a$, so $S_{(a,n)}$ is a multiple of $a$. If $a$ is even, then all the terms in $S_{(a,n)}$ pair up in this way, except for $a^n$ and $\Big(\dfrac{a}{2}\Big)^n$, so so $S_{(a,n)}$ is a multiple of $\dfrac{a}{2}$. We can make exactly the same argument with $P=a+1$, except that, if $a$ is odd, we find $S_{(a,n)}$ is a multiple of $\dfrac{a+1}{2}$ and if $a$ is even, $S_{(a,n)}$ is a multiple of $a+1$. Hence $S_{(a,n)}$ is always a multiple of $\dfrac{a(a+1)}{2}=S_{(a,1)}$.
Illustrative examples.
$1^n+2^n+3^n+4^n+5^n=(1^n+4^n)+(2^n+3^n) + 5^n$, so divisible by 5; also $1^n+2^n+3^n+4^n+5^n=(1^n+5^n)+(2^n+4^n) + 3^n$, so divisible by 3.
$1^n+2^n+3^n+4^n+5^n+6^n=(1^n+5^n)+(2^n+4^n) + 3^n +6^n$, so divisible by 3; also $1^n+2^n+3^n+4^n+5^n+6^n=(1^n+6^n)+(2^n+5^n) + (3^n+4^n)$, so divisible by 7.
| {
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Let $a$, $b$, $c$ be positive integers such that $\frac{bc}{b+c}$, $\frac{ca}{c+a}$, $\frac{ab}{a+b}$ are integers. Show that $\gcd(a,b,c) >1$. Let $a$, $b$, $c$ be positive integers such that $\frac{bc}{b+c}$, $\frac{ca}{c+a}$, $\frac{ab}{a+b}$ are integers. Show that $\gcd(a,b,c) >1$.
I can see that $\gcd(b,c)$, $\gcd(c,a)$, and $\gcd(a,b)$ are all greater than 1, but how can we show $\gcd(a,b,c)>1$? Thanks!
| Let's assume $(a,b,c)=1$. Since $\frac{ab}{a+b}$ is a positive integer, we must have $(a,b)>1$. The reason is clear. If $(a,b)=1$, then:
$$(a+b,a)=(a+b,b)=(a,b)=1 \implies (a+b, ab)=1.$$
Now, suppose $a=dm$ and $b=dn$, where $d>1$ and $(m,n)=1$. We will have:
$$\frac{ab}{a+b}=\frac{dmn}{m+n};$$
but $(m+n, mn)=1$ because $(m,n)=1$. Therefore:
$$m+n|d \implies d=s(m+n), $$
where $s$ is a positive integer. So $a=(m+n)sm$ and $b=(m+n)sn$.
Since we have assumed that $(a,b,c)=1$, we must have: $(c, (m+n)s)=1$.
Similarly, since $\frac{ac}{a+c}$ is a positive integer, we have:
$$(a,c)=((m+n)sm, c)=(m,c)>1.$$
Suppose $c=hc_1$ and $m=hm_1$, where $h$ is a positive integer and $(c_1,m_1)=1$. So,
$$\frac{ac}{a+c}=\frac{h((m+n)sm_1)c_1}{c_1+(m+n)sm_1},$$
but $(c_1, (m+n)sm_1)=1$; as a result, $((m+n)sm_1c_1, (m+n)sm_1+c_1)=1$. Thus, we must have:
$$c_1+(m+n)sm_1|h \\ \implies h \geq c_1+(m+n)sm_1 > m,$$
which is a contradiction because:
$$m=hm_1 \geq h.$$
| {
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For $c>b>a$ with $c,b,a\in\Bbb N,c-b=b-a=1$, prove $\frac{a^c}{b^b}+\frac{b^c}{c^b}+\frac{c^c}{a^b}\ge\frac{a^b}{b^a}+\frac{b^b}{c^a}+\frac{c^b}{a^a}$
Given $c>b>a$, also $c-b=b-a=1$ where $c,b,a$ are Natural numbers, prove that
$$
\frac{a^c}{b^b}+\frac{b^c}{c^b}+\frac{c^c}{a^b} \geqslant \frac{a^b}{b^a}+\frac{b^b}{c^a}+\frac{c^b}{a^a}
$$
The LHS becomes $$
\frac{a^{b+c} c^b+b^{b+c} a^b+c^{b+c} \cdot b^b}{(a b c)^b},
$$
and the RHS becomes
$$
\frac{a^{b+a} c^a+b^{a+b} a^a+c^{a+b} b^a}{(a b c)^a}.
$$
Now, $a^{b+c}>a^{b+a}$,$c^b>c^a$, $b^{b+c}>b^{b+a}$,$a^b>a^a$,
$c^{b+c}>c^{b+a}$,$b^b>b^a$. So the numerator on LHS is clearly bigger than the numerator on RHS. But $(abc)^{b}>(abc)^{a}$. How do I proceed.
$\textbf{Edit}$: added an important detail.
| Hint for $a,b,c>3$ we have :
$$\frac{a^c}{b^b}+\frac{b^c}{c^b}+\frac{c^c}{a^b}\geq \frac{(a+b+c)^{c-b}}{3^{c-b-1}}$$
And :
$$\frac{a^b}{b^a}+\frac{b^b}{c^a}+\frac{c^b}{a^a}<3\frac{c^b}{a^a}$$
We have for $c\geq 2(a+b)$:
$$\frac{(a+b+c)^{c-b}}{3^{c-b-1}}-3\frac{c^{b}}{a^{a}}>0$$
For the case $a+b\leq c\leq 2(a+b)$ :
We have :
$$\frac{c^{c}}{a^{b}}-\left(\frac{a^{b}}{b^{a}}+\frac{b^{b}}{c^{a}}+\frac{c^{b}}{a^{a}}\right)>0$$
Or :
$$\frac{c^{c}}{a^{b}}-3\frac{c^{b}}{a^{a}}>0$$
With your constraint we have :
$$f(x)=\frac{\left(x+2\right)^{\left(x+2\right)}}{x^{\left(x+1\right)}}-\frac{\left(x+2\right)^{\left(x+1\right)}}{x^{x}}\geq \lim_{x\to\infty}f(x)$$
$$g(x)=\frac{x^{\left(x+2\right)}}{\left(x+1\right)^{\left(x+1\right)}}-\frac{x^{\left(x+1\right)}}{\left(x+1\right)^{x}}\geq \lim_{x\to \infty}g(x)$$
$$h(x)=\frac{\left(x+1\right)^{\left(x+2\right)}}{\left(x+2\right)^{\left(x+1\right)}}-\frac{\left(x+1\right)^{\left(x+1\right)}}{\left(x+2\right)^{x}}\geq \lim_{x\to 0^+}h(x)$$
Reference :
https://arxiv.org/pdf/1504.05874.pdf
| {
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Solve $\frac{x-8}{x-10}+\frac{x-4}{x-6}=\frac{x-5}{x-7}+\frac{x-7}{x-9}$ Solve $\frac{x-8}{x-10}+\frac{x-4}{x-6}=\frac{x-5}{x-7}+\frac{x-7}{x-9}$
$\Rightarrow \frac{(x-10)+2}{x-10}+\frac{(x-6)+2}{x-6}=\frac{(x-7)+2}{x-7}+\frac{(x-9)+2}{x-9} \ \ \ ...(1)$
$\Rightarrow 1+\frac{2}{x-10}+1+\frac{2}{x-6}=1+\frac{2}{x-7}+1+\frac{2}{x-9}\ \ \ ...(2)$
$\Rightarrow \frac{2}{x-10}+\frac{2}{x-6}=\frac{2}{x-7}+\frac{2}{x-9}\ \ \ ...(3)$
$\Rightarrow \frac{1}{2}(\frac{2}{x-10}+\frac{2}{x-6})=\frac{1}{2}(\frac{2}{x-7}+\frac{2}{x-9})\ \ \ ...(4)$
$\Rightarrow \frac{1}{x-10}+\frac{1}{x-6}=\frac{1}{x-7}+\frac{1}{x-9}\ \ \ ...(5)$
$\Rightarrow \frac{x-6+x-10}{(x-10)(x-6)}=\frac{x-9+x-7}{(x-7)(x-9)}\ \ \ ...(6)$
$\Rightarrow \frac{2x-16}{(x-10)(x-6)}=\frac{2x-16}{(x-7)(x-9)}\ \ \ ...(7)$
$\Rightarrow (x-10)(x-6)=(x-7)(x-9)\ \ \ ...(8)$
$\Rightarrow x^2-16x+60=x^2-16x+63\ \ \ ...(9)$
$\Rightarrow 60=63$
I feel my calculations are all correct, but 60 cannot equal 63, so something went wrong, but I can't see it. Thanks for the help.
| Look closesly at step $(7)$
$$\frac{2x-16}{(x-10)(x-6)}=\frac{2x-16}{(x-9)(x-7)}$$
The numerators are equal for any value of $x$, but is the same true for the denominator too i.e. for any $x\in\mathbb{R}/\{6,7,9,10\}$? No!
What can you conclude from this? One way to look at it is equating this to $0$, as mentioned in the comments. But the question is why? Let's cross-multiply and see where it gets us:
$$(2x-16)(x-9)(x-7)=(2x-16)(x-10)(x-6)$$
You have to consider two cases now (why so will be clear by the cases).
Case 1. $(2x-16)\ne0$
Then, you can divide both LHS and RHS by $(2x-16)$ and it gets cancelled out and spirals down to your steps 8 onward. However, this leads to an impossibility. You must conclude that this case is not true.
Case 2. $(2x-16)=0$
How is this going to be any different from the previous case? That lies in what we did first in case 1. Divided both sides by $(2x-16)$. However, now this would be equivalent to division by $0$, which is undefined. So, how do we get the solution?
The trick is to look at the case itself!
$$2x-16=0\Rightarrow x=8$$
Aaand.. these are exhaustive cases, so there is a sinlge solution to the original equation.
| {
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Show that $a^2+b^2+c^2=x^2+y^2+z^2$
Let $a,b,c,x,y,z$ be real numbers such that
$$ a^2+x^2=b^2+y^2=c^2+z^2=(a+b)^2+(x+y)^2=(b+c)^2+(y+z)^2=(c+a)^2+(z+x)^2 $$
Show that $a^2+b^2+c^2=x^2+y^2+z^2$.
Progress: Note that $$0=\sum_{\text{cyc}} (a + b)^2 + (x + y)^2 - \sum_{\text{cyc}} a^2 + x^2 = (a + b + c)^2 + (x + y + z)^2\implies a+b+c=x+y+z=0.$$
So we need to show $$ab+bc+ca=xy+yz+zx.$$ Also, taking $c=-(a+b),z=-(x+y)$, we get $$a^2+x^2=b^2+y^2=(a+b)^2+(x+y)^2.$$
So we get $$y^2+b^2+2ab+2yx=0\implies \sum_{cyc}y^2+b^2+2ab+2yx=0\implies (a+b)^2+(b+c)^2+(c+a)^2+(x+y)^2+(y+x)^2+(x+z)^2=0\implies a=b=c=x=y=z=0. $$
I am not sure of the last part. So can someone check? Can someone also give an alternate solution to this problem?
| Writing this using complex numbers makes it a little easier. The equations mean that these following complex numbers have the same distance from the origin of the complex plane.
$$a+ix=re^{i\theta_1}\mbox{ with } a=r\cos(\theta_1),x=r\sin(\theta_1).$$
$$b+iy=re^{i \theta_2}\mbox{ with } b=r\cos(\theta_2),y=r\sin(\theta_2).$$
$$c+iz=re^{i \theta_3}\mbox{ with } c=r\cos(\theta_3),z=r\sin(\theta_3).$$
Now, $$r=(a+b)^2+(x+y)^2=|(a+ix)+(b+iy)|=r|(e^{i\theta_1}+e^{i\theta_2})|$$
$$\implies |e^{i\theta_1}+e^{i\theta_2}|=1$$
For this to happen, $\theta_1=\frac{2\pi}{3}+\theta_2$. Similarly, $\theta_2=\frac{2\pi}{3}+\theta_3$ and $\theta_3=\frac{2\pi}{3}+\theta_1$.
The three points lie on the vertices of an equilateral triangle.
$$a^2+b^2+c^2=r^2(\cos^2(\theta_1)+\cos^2(\frac{2\pi}{3}+\theta_1)+\cos^2(\frac{2\pi}{3}-\theta_1))=\frac {3r^2}{2}\tag 1$$
$$x^2+y^2+z^2=r^2(\sin^2(\theta_1)+\sin^2(\frac{2\pi}{3}-\theta_1)+\sin^2(\frac{2\pi}{3}+\theta_1))=\frac{3r^2}{2}\tag 2$$
which completes our proof.
I have used $re^{i\theta}=r(\cos(\theta)+i\sin(\theta))$ . The proof that $\sin^2(\theta_1)+\sin^2(\frac{2\pi}{3}-\theta_1)+\sin^2(\frac{2\pi}{3}+\theta_1)=\cos^2(\theta_1)+\cos^2(\frac{2\pi}{3}+\theta_1)+\cos^2(\frac{2\pi}{3}-\theta_1)=3/2$ can be found online.
| {
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How do I solve the equation $(x - 1)(x - 2)(x - 3) = (x - 2)(x - 3)(x - 4)$? Problem: $(x - 1)(x - 2)(x - 3) = (x - 2)(x - 3)(x - 4)$
Heres my question with this problem: why do I end up with a wrong answer when I divide both sides by $(x-2)(x-3)$ to cancel out the $(x-2)(x-3)$ on both sides. Is this not allowed and why? Please provide the explanation to this question. You do not have to solve the problem.
Thank you very much.
| You cannot divide both sides by $(x-2)(x-3)$ if it equals $0$. When you divide both sides by it, you are ignoring the solutions $x=2,3$
$$(x-1)(x-2)(x-3)=(x-2)(x-3)(x-4)$$
A proper way to solve this is as follows
$$(x-1)(x-2)(x-3)-(x-4)(x-2)(x-3)=0$$$$[(x-1)-(x-4)](x-2)(x-3)=0$$
$$3(x-2)(x-3)=0$$
| {
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Hard geometry problem: given a triangle with sides 16, 30, 34, find the area of the triangle introduced by the inscribed circle You are given a triangle $\triangle PQR$ with sides $16, 30, 34$. Let the incircle touch the sides of $\triangle PQR$ at $X,Y,$ and $Z$. Given that the ratio $[XYZ]/[PQR]$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers, find $m+n$.
What I did first was to draw the diagram and I noticed that $PX=PZ, QX=QY, RY=RZ$ due to two intercepting lines both being tangent to a circle. Since triangles $PXZ, QXY, RYZ$ are isosceles I bisected them and got another right triangle after I tried to do similar triangles but got a little stuck.
My other friend used "barycentric coordinates", which I thought was completely unnecessary.
Any help would be appreciated.
| John Omeilan has provided an excellent solution. However the answer can be simpler if we observe that $\Delta PQR$ is a right angled triangle.
Since $16^2+30^2=34^2$, $\Delta PQR$ is a right angled triangle and hence $ [PQR] = \frac{16 \times 30}{2}=240$
$s=\frac{p+q+r}{2}=\frac{30+16+34}{2}=40$
Let the radius of the inscribed circle be $x$
From $xs=[PQR] =240$, we have
$$x=\frac{240}{40}=6$$
Notice that $\sin R=\frac{16}{34}=\frac{8}{17},
\sin P=\frac{15}{17}$
and $\sin Q=\frac{34}{34}=1$
Also note that $\angle YOZ=180^o-R$.
Hence $ [YOZ]=\frac{1}{2}x^2\sin(180^o-R)=\frac{1}{2}\times 6^2 \times \sin R=18 \times \frac{8}{17}$
Similarly $[ZOX]=\frac{1}{2}x^2 \times \frac{15}{17}=18 \times \frac{15}{17}$
and $[XOY]=\frac{1}{2}x^2=\frac{1}{2}\times 6^2=18$
Hence $[XYZ]=18 \times \left(\frac{8}{17}+\frac{15}{17}+1 \right)=\frac{720}{17}$
Thus $\frac{[XYZ]}{[PQR]}=\frac{\frac{720}{17}}{240}=\frac{3}{17}$
| {
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How can I parametrize the intersection curve of cone $z=\sqrt{(x^2+y^2)}$ and plane $z=1+x+y$ Usually, I equalize both equation, then complete the square in order to find something that looks like $\cos^2 t+\sin^2 t=1$ to get a value of $x(t)$ and $y(t)$. I then substitute to get $z(t)$. I am unable to apply that technique with this question. I think it has something to do with the square root and the fact that the intersection curve is split in two.
| We have $z^2=x^2+y^2$ and $z^2=(1+x+y)^2$, which gives $$x^2+y^2=1+2x+2y+2xy+x^2+y^2$$ which is equivalent to $1+2x+y(2+2x)=0$.
Now we can set $x=t$ and $y=-\frac{1+2x}{2+2x}=-\frac{1+2t}{2+2t}$ and $z=1+x+y=1+t-\frac12\frac{1+2t}{1+t}$.
Furthermore, $z=\sqrt{x^2+y^2}$ gives $z>0$, removing the problem that the curve is split in two. This means that we need $t>-1$.
| {
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Prove that $ a, b, c $ are in arithmetic sequence for the given condition. A question on my mathematics textbook, Mathematics—Textbook for Class XI, goes thus:
If $a\left(\frac{1}{b} + \frac{1}{c}\right), b \left(\frac{1}{c} + \frac{1}{a}\right)$ and $c \left(\frac{1}{a} + \frac{1}{b}\right)$ are in arithmetic sequence, prove that $a, b, c$ are also in arithmetic sequence.
I've tried solving the question by proceeding with this as the first step:
\begin{equation}
b \left(\frac{1}{c} + \frac{1}{a}\right) - a \left(\frac{1}{b} + \frac{1}{c}\right) = c \left(\frac{1}{a} + \frac{1}{b}\right) - b \left(\frac{1}{c} + \frac{1}{a}\right)
\end{equation}
However, I've been unable to reach this step:
\begin{equation}
b - a = c - b
\end{equation}
which, I believe, is required to arrive at to prove that $a, b, c$ are indeed in arithmetic sequence.
A guidance towards the right steps would be helpful.
| Hint: Let $x_1=a\bigg(\frac{1}{b}+\frac{1}{c}\bigg)$, $x_2=b\bigg(\frac{1}{a}+\frac{1}{c}\bigg)$ and
$x_3=c\bigg(\frac{1}{a}+\frac{1}{b}\bigg)$.
Then
$$
x_3-2x_2+x_1=\frac{(ab+ac+bc)(a-2b+c)}{abc}
$$
Can you finish from here ?
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Closed form expression for $\psi_{e^{\pi}}^{(3)}(1-i)$ Let $\psi_q(z)$ be the q-DiGamma function defined for a real variable $\Re(z)>0$ as $$\psi_q(z)=\frac{1}{\Gamma_q(z)}\frac{\partial}{\partial z} (\Gamma_q(z))$$
where $\Gamma_q(z)$ is the q-Gamma function defined as $$\Gamma_q(z)=(1-q)^{1-z}\prod_{n=0}^{\infty}\frac{1-q^{n+1}}{1-q^{n+z}}$$
Question I am looking for a closed form for $$\psi_{e^{\pi}}^{(3)}(1-i)$$ where $i=\sqrt{-1}$
Here is a beautiful answer for calculating $$\psi_{e^{\pi}}^{(3)}(1)$$
Wolfram Alpha gives the expansion at $x=\infty$:
$$\psi_x^{(3)}(1)=\ln^4(x)\left(x^{-1}+9x^{-2}+\dots\right)$$
and these match Oeis A$001158$ with divisor $\sigma_v(n)$ and various theta functions after plugging the sum back in here. Use $\vartheta_v(0,x)=\vartheta_v(x)$:
$$\psi_x^{(3)}(1)=\ln^4(x)\sum_{n=1}^\infty\frac{\sigma_3(n)}{x^n}=\frac{\ln^4(x)}{480}\left(\vartheta_2\left(\frac1{\sqrt x}\right)^8+ \vartheta_3\left(\frac1{\sqrt x}\right)^8+ \vartheta_4\left(\frac1{\sqrt x}\right)^8-2\right)$$
Therefore:
$$\psi_{e^\pi}^{(3)}(1)=\frac{\pi^4}{480}\left(\vartheta_2^8\left(e^{-\frac\pi2}\right)+ \vartheta_3^8\left(e^{-\frac\pi2}\right)+ \vartheta_4^8\left(e^{-\frac\pi2}\right)-2\right)$$
Clicking “more digits” here shows a smaller error each time implying the result is true.
Now use Dedekind $\eta(z)$ identities for $\vartheta_v\left(e^{-\frac\pi2}\right)$ when $v=2$, $v=3$, and $v=4$
$$\psi_{e^\pi}^{(3)}(1)= \frac{\pi^4}{480}\left(\left(2\frac{\eta^2(i)}{\eta\left(\frac i2\right)}\right)^8+\left(\frac{\eta^5\left(\frac i2\right)}{\eta^2\left(i\right)\eta^2\left(\frac i4\right)}\right)^8+\left(\frac{\eta^2\left(\frac i4\right)}{\eta\left(\frac i2\right)}\right)^8-2\right)$$
Using special values in terms of $\Gamma\left(\frac14\right)$:
$\eta\left(\frac i4\right)=2\eta(4i)=\frac{\sqrt[4]{\sqrt2-1} \Gamma\left(\frac14\right)}{2^\frac{13}{16}\pi^\frac34},\eta\left(\frac i2\right)=\frac{\Gamma\left(\frac14\right)}{2^\frac 78\pi^\frac34},\eta(i)=\frac{\Gamma\left(\frac14\right)}{2\pi^\frac34}$
Finally, substitute and have a form in terms of $\Gamma\left(\frac14\right)$ which has no elementary closed form. Therefore:
$$\boxed{\psi_{e^\pi}^{(3)}(1)=\frac{11\Gamma\left(\frac14\right)^8}{5120\pi^2}-\frac{\pi^4}{240}}$$
shown here
If anyone could please solve this question by hand or mathematica or sage math. I would be highly indebted to you all.
| I am not an expert in such sums, this derivation can probably be done in a simpler way. From the link provided by Tyma Gaidash, we have
\begin{align*}
\psi _{{\rm e}^\pi }^{(3)} (1 - {\rm i}) = & - \pi ^4 \sum\limits_{n = 1}^\infty {\frac{1}{{{\rm e}^{\pi n} + 1}}} + 7\pi ^4 \sum\limits_{n = 1}^\infty {\frac{1}{{({\rm e}^{\pi n} + 1)^2 }}} \\ & - 12\pi ^4 \sum\limits_{n = 1}^\infty {\frac{1}{{({\rm e}^{\pi n} + 1)^3 }}} + 6\pi ^4 \sum\limits_{n = 1}^\infty {\frac{1}{{({\rm e}^{\pi n} + 1)^4 }}} .
\end{align*}
Re-expanding the fractions in powers of ${\rm e}^{-\pi n}$, changing the order of summation and re-summing again leads to the identities
\begin{align*}
& \sum\limits_{n = 1}^\infty {\frac{1}{{{\rm e}^{\pi n} + 1}}} = -\sum\limits_{k = 1}^\infty {\frac{{( - 1)^k }}{{{\rm e}^{\pi k} - 1}}} ,
\\ &\sum\limits_{n = 1}^\infty {\frac{1}{{({\rm e}^{\pi n} + 1)^2 }}} = \sum\limits_{k = 1}^\infty {\frac{{( - 1)^k k}}{{{\rm e}^{\pi k} - 1}}} - \sum\limits_{k = 1}^\infty {\frac{{( - 1)^k }}{{{\rm e}^{\pi k} - 1}}} ,
\\ & \sum\limits_{n = 1}^\infty {\frac{1}{{({\rm e}^{\pi n} + 1)^3 }}} = -\frac{1}{2}\sum\limits_{k = 1}^\infty {\frac{{( - 1)^k k^2 }}{{{\rm e}^{\pi k} - 1}}} + \frac{3}{2}\sum\limits_{k = 1}^\infty {\frac{{( - 1)^k k}}{{{\rm e}^{\pi k} - 1}}} - \sum\limits_{k = 1}^\infty {\frac{{( - 1)^k }}{{{\rm e}^{\pi k} - 1}}} ,
\\ &\sum\limits_{n = 1}^\infty {\frac{1}{{({\rm e}^{\pi n} + 1)^4 }}} = \frac{1}{6}\sum\limits_{k = 1}^\infty {\frac{{( - 1)^k k^3 }}{{{\rm e}^{\pi k} - 1}}} - \sum\limits_{k = 1}^\infty {\frac{{( - 1)^k k^2 }}{{{\rm e}^{\pi k} - 1}}} + \frac{{11}}{6}\sum\limits_{k = 1}^\infty {\frac{{( - 1)^k k}}{{{\rm e}^{\pi k} - 1}}} - \sum\limits_{k = 1}^\infty {\frac{{( - 1)^k }}{{{\rm e}^{\pi k} - 1}}} .
\end{align*}
Substituting back to the original formula yields the simpler form
$$
\psi _{{\rm e}^\pi }^{(3)} (1 - {\rm i}) = \pi ^4 \sum\limits_{k = 1}^\infty {\frac{{( - 1)^k k^3 }}{{{\rm e}^{\pi k} - 1}}} .
$$
Now we have
$$
\sum\limits_{k = 1}^\infty {\frac{{( - 1)^k k^3 q^k }}{{1 - q^k }}} = \frac{{\theta _4^8 (0,q) - 1}}{{16}}
$$
for $|q|<1$ with $\theta_4$ being one of the theta functions. Hence,
$$
\psi _{{\rm e}^\pi }^{(3)} (1 - {\rm i}) = \pi ^4 \frac{{\theta _4^8 (0,{\rm e}^{ - \pi } ) - 1}}{{16}}.
$$
Using the specific value
$$
\theta _4 (0,{\rm e}^{ - \pi } ) = \frac{{\pi ^{1/4} }}{{2^{1/4} \Gamma (3/4)}},
$$
(see equation $(45)$ here) we finally have
$$\boxed{
\psi _{{\rm e}^\pi }^{(3)} (1 - {\rm i}) = \frac{{\pi ^6 }}{{64\Gamma ^8 (3/4)}} - \frac{{\pi ^4 }}{{16}}.}
$$
Numerical check.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4642004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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} |
$\int \frac{x^2}{\sqrt{5-x^2}}dx$ using substitution $u = \sqrt{5-x^2}$? I was trying to solve $$\int\frac{x^2}{\sqrt{5-4x^2 }} dx$$ by using the substitution
$$
u=\sqrt{5-4x^2}\\
dx = \frac{\sqrt{5-4x^2}}{-4x}du$$ So the integral can be written as $$\int \left(-\frac{x}{4}\right) du$$ and
$$
x= \begin{cases}
\sqrt{(5-u^2)/4}, & x >0\\
-\sqrt{(5-u^2)/4}, &x<0
\end{cases}
$$
I've rewritten the integral as
$$-\frac18\int\sqrt{5-u^2}du$$
using trig substitution with $$u= \sqrt5\sin(t)$$ we have
$$-\frac5{16}t - \frac5{16} \sin(t)\cos(t)$$ $$t=\arcsin\left(\frac{u^2}{\sqrt5}\right)$$
If we now reverse the substitutions and simplify, I got the final answer as $$-\frac5{16} \arcsin\left(\sqrt{\frac{5-4x^2}5}\right) - \frac{x}8\sqrt{5-4x^2} $$However this differs from the answer in my textbook which is:
$$-\frac{x}{8}\sqrt{5-4x^2}+\frac5{16}\arcsin\left(\frac{2x}{\sqrt5}\right)+C$$
My textbook used different methods that are simpler and easier, but I want to know where I made a mistake.
| Note that
$$\arcsin\sqrt{\frac{5-4x^2}5}= \arccos\sqrt{1-\frac{5-4x^2}5}
=\arccos\frac{2x}{\sqrt5}=\frac\pi2-\arcsin\frac{2x}{\sqrt5}
$$
Therefore, the two results differs by the constant $-\frac\pi2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4644624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Derivative of $y = \sin^3(\frac\pi 3(\cos(\frac\pi{3\sqrt2}(-4x^3 + 5x^2 + 1)^{3/2})))$ at $x=1$ Today I came across a problem:
If $y = \sin^3\left(\frac\pi 3\left(\cos\left(\frac\pi{3\sqrt2}\left(-4x^3 + 5x^2 + 1\right)^{3/2}\right)\right)\right)$, then at $x=1$ which of the following option is correct?
*
*$2y'+\sqrt{3} \pi^2 y = 0$
*$2y' +3 \pi^2 y = 0$
*$\sqrt2 y' - 3\pi^2 y = 0 $
*$y' + 3\pi^2 y = 0$
This question is from JEE Main 2023 examination.
My attempt:
Firstly I used the identity $\sin^3(x) = \frac34 \sin(x) - \frac14 \sin(3x)$ and then by applying a brute force chain rule, I got that $f'(1) = \frac3{16} \pi^2$ and $f(1) = -\frac18$.
So, second option i.e. $2y' + 3\pi^2 y = 0$ is correct.
But I don't think that this question is designed to be solved in this manner. Is there something which I'm missing?
| It is good to know the answer and solve. You never see your mistakes. I tried to be quick:
$y=\sin^3A$ and $A=\frac\pi 3\cos B$ and $B=\frac{\pi}{3\sqrt 2}C^{3/2}$ and $C=-4x^3+5x^2+1$.
$y'=3\sin^2A\cos A\frac\pi 3(-\sin B)\frac{\pi}{3\sqrt2}\frac32C^{1/2}(-12x^2+10x).$
At $x=1$, $C=2$, $B=\frac{2\pi}3$, $A=-\frac\pi 6$.
At $x=1$, $y'=3(-\frac12)^2\frac{\sqrt{3}}{2}\frac{\pi}3(-\frac{\sqrt 3}{2})\frac{\pi}{3\sqrt2}\frac32\sqrt 2(-2)=\frac{3\pi^2}{16}$.
At $x=1$, $y=\sin^3(-\frac\pi 6)=(-\frac12)^2=-\frac18$.
At $x=1$, $\large\frac{y'}y=\large\frac{\frac{3\pi^2}{16}}{-\frac18}=-\frac{3\pi^2}2\implies 2y'+3\pi^2y=0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4646083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Computing the limit of $ \frac{1^3+2^3+\cdots+n^3}{\sqrt{4n^8 +1}} $ I had this exercise:
Compute the limit
$$ \lim_{n\to\infty} \frac{1^3+2^3+\cdots+n^3}{\sqrt{4n^8 +1}} $$
I tried two different approaches and got different answers.
Approach 1:
$$\begin{split}
\lim_{n\to\infty} \frac{1^3+2^3+\cdots+n^3}{n^4\sqrt{4 +1/n^8}}&=\lim_{n\to\infty} \frac{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\cdots+\frac{n^3}{n^4}}{\sqrt{4 +1/n^8}}\\
&=\lim_{n\to\infty} \frac{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\cdots+\frac{1}{n}}{\sqrt{4 +1/n^8}}\\
&= \frac{0}{\sqrt{4+0}}\\
&= 0\\
\end{split}$$
Approach 2:
We substitute sum of cubes of $n$ natural numbers in the numerator and get
$$\begin{split}
\lim_{n\to\infty} \frac{n^2(n+1)^2}{4n^4\sqrt{4 +1/n^8}}&= \lim_{n\to\infty} \frac{n^2(n+1)^2}{4n^4\sqrt{4 +1/n^8}}\\
&= \lim_{n\to\infty} \frac{n^4(1+\frac{1}{n})^2}{4n^4\sqrt{4 +1/n^8}}\\
&= \lim_{n\to\infty} \frac{(1+\frac{1}{n})^2}{4\sqrt{4 +1/n^8}}\\
&= \lim_{n\to\infty} \frac{1}{4\sqrt{4}}\\
&= \frac{1}{8}\\
\end{split}$$
I'm not sure why the two methods are giving me different answers and which one is correct?
| One more method is to see the Laurent series of the expression. It is equal to $$\frac{1}{8}+\frac{1}{4n}+\frac{1}{8n^2}-\frac{1}{64n^8}+O\left(\left(\frac{1}{n}\right)^9\right)$$
You can see that it confirms your second approach.
| {
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"answer_count": 2,
"answer_id": 1
} |
Prove $\dfrac{n^2}{\ln n^2-1} - \dfrac{\dfrac{n^2}{2}}{\ln \dfrac{n^2}{2}-1.1} > n$ for $n \ge 347$ Prove $\dfrac{n^2}{\ln n^2-1} - \dfrac{\dfrac{n^2}{2}}{\ln \dfrac{n^2}{2}-1.1} > n$ for $n \ge 347$. It seems to be true for all $n \ge 11$, but I only need it to be true for $n \ge 347$. I tried to manipulate the equation in all sorts of ways but I couldn't get it to work. These are the difference of prime bounds if that helps. $\dfrac{x}{\ln x-1} < \pi(x)$ for $x \ge 5393$ and $\pi(x) < \dfrac{x}{\ln x-1.1}$ for $x \ge 60184$ where $\pi(x)$ is counts the number of primes less than or equal to $x$.
| I tried to manipulate the equation in all sorts of ways but I couldn't get it to work
In fact, there is an exact solution to the equation
$$\dfrac{n^2}{\log( n^2)-1} - \dfrac{\dfrac{n^2}{2}}{\log \left(\frac{n^2}{2}\right)-a} = n$$ which can rewrite
$$n=4\frac{\left(\log (n)-\frac{1}{2}\right) (\log
(n)-\alpha )}{\log (n)-\beta }$$ where
$$\alpha=\frac{1}{2} (a+\log (2)) \qquad \beta=\frac{1}{2} (2 a+2 \log (2)-1)$$
Let $n=e^x$ and write
$$e^{-x}=\frac 14 \frac{x-\beta }{\left(x-\frac{1}{2}\right) (x-\alpha )}$$ which shows an exact explicit solution in terms of the generalized Lambert function (have a look at equation $(4)$).
Since we just look for an estimate, using series
$$e^x=4 x+(2 a-4+2\log
(2))+O\left(\frac{1}{x}\right) $$ $$x= -\left(\frac{a}{2}+\frac{\log (2)}{2}-1\right)-W_{-1}\left(-\frac{e^{1-\frac{a}{2}}}{4 \sqrt{2}}\right)$$ giving, for $a=1.1$, $x_0=2.05486$ corresponding to $n_0=7.80572$
Newton iterates are
$$\left(
\begin{array}{cc}
k & x_k & n_k\\
0 & 2.05486 & 7.80572 \\
1 & 2.33134 & 10.2917 \\
2 & 2.30879 & 10.0623 \\
3 & 2.30852 & 10.0595 \\
\end{array}
\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4647326",
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"source": "stackexchange",
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"answer_count": 2,
"answer_id": 1
} |
How do I integrate $\int_0^{\infty} \frac{\log(t+1)}{t^2+a^2} \mathrm{d}t$? How do I integrate the following integral:
$$\int_0^{\infty} \frac{\log(t+1)}{t^2+a^2} \mathrm{d}t$$
Where $a$ is some parameter?
I know that the solution includes Lerch Transcendents and logs (which is what I'm trying to arrive at); however, I've tried integrating this function, but failed.
I've already tried using simplifying it using series, which yielded a bunch of integrals as follows:
$$-\sum_{k \geq 1} \frac{(-1)^k}{k} \int_0^1 \frac{t^k} {t^2+a^2} \mathrm{d}t - \frac{1}{a^2} \sum_{k \geq 1} \frac{(-1)^k}{k} \int_0^1 \frac{t^k} {t^2+\frac{1}{a^2}} \mathrm{d}t + \frac{1}{a^2} \sum_{k \geq 1} \frac{(-1)^k}{k} \int_0^1 \frac{\log(t)} {t^2+\frac{1}{a^2}} \mathrm{d}t$$
However, I don't know how I'd proceed forwards from here.
| let's generalize your integral
$$ \mathcal{I}(b) = \int_0^\infty \frac{\ln(1+b \, x)}{a^2+x^2} \, \mathrm{d}x \hspace{5mm} \cdots (1) $$
Using Feynmann's Trick we get (differentiating w.r.t b)
\begin{align*}
\mathcal{I}'(b) &= \int_0^\infty \frac{x}{(1+b\,x) \, (x^2 + a^2)} \, \mathrm{d}x \\
&= \frac{1}{1+a^2 \, b^2} \int_0^\infty \left( \frac{-b}{1+bx} + \frac{x}{x^2+a^2} + \frac{a^2 b}{x^2 + a^2} \right) \, \mathrm{d}x \\
&= \frac{1}{1+a^2 \, b^2} \left( - \ln(1+b \, x) + \frac{1}{2} \, \ln(x^2+a^2) + ab \tan^{-1} \left(\frac{x}{a} \right) \right) \bigg{|}_0^\infty \\
&= \frac{1}{1+a^2 \, b^2} \left( \ln \left( \frac{\sqrt{x^2+a^2}}{1+b x} \right)\bigg{|}_0^\infty + \frac{a b \pi}{2} \right) \\
&= \frac{1}{1+a^2 \, b^2} \left( - \ln(a) + \frac{a b \pi}{2} \right)
\end{align*}
Now we have
$$ \mathcal{I}'(b) = \frac{1}{1+a^2 \, b^2} \left( - \ln(a) + \frac{a b \pi}{2}
\right) \hspace{5mm} \cdots (2)$$
After integrating (2) w.r.t b we get
$$ \mathcal{I}(b) = \frac{- \ln(a)}{a} \, \tan^{-1}(a b) + \frac{\pi}{4a} \ln(1 + a^2 b^2) + \mathrm{C} \hspace{5mm} \cdots (3) $$
Notice in (1) $$ \mathcal{I}(0) = 0 $$
So putting $b = 0$ in (3) we get
$$ \mathcal{I}(0) = 0 = 0 + 0 + \mathrm{C} \implies \mathrm{C} = 0 $$
So from (3)
$$ \mathcal{I}(b) = \frac{- \ln(a)}{a} \, \tan^{-1}(a b) + \frac{\pi}{4a} \ln(1 + a^2 b^2) \hspace{5mm} \cdots (4) $$
Now putting $b = 1$ in (4) we get desired answer
$$ \boxed{ \int_0^\infty \frac{\ln(1+x)}{a^2 + x^2} \, \mathrm{d}x = \frac{- \ln(a)}{a} \, \tan^{-1}(a) + \frac{\pi}{4a} \ln(1 + a^2) }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4648635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
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Distance between the point of touching in three touching circles $\textbf{Question : }$ Say you have three touching circles $\Gamma_1,\Gamma_2,\Gamma_3$ with radii $x,y,z$ and centers $A,B,C$ as per the diagram, then prove the following $$|DE|=\frac{2}{\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x}+\dfrac{1}{z}\right)}}$$
$\textbf{My Attempt}$
Let $\angle{DAE} = \alpha, \angle{FBE}=\beta , \angle{FCD}=\pi-\alpha-\beta$ then through sine rule in the $\Delta ABC$ we can say
$$\frac{\sin{\alpha}}{y+z}=\frac{\sin{\beta}}{z+x}=\frac{\sin{(\alpha+\beta)}}{x+y} $$
As we know
$$|DE|=2x\sin\frac{\alpha}{2}$$
We just need to prove
$$\sin\frac{\alpha}{2} = \dfrac{1}{x}\dfrac{1}{\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x}+\dfrac{1}{z}\right)}}$$
So using the sine rule relation
$$\sin\beta = \frac{z+x}{y+z} \sin \alpha$$
And again from the sine rule relation
$$\sin \alpha = \frac{y+z}{x+y} \sin (\alpha+\beta)$$
Now substituting $\sin \beta$ in terms of $\sin \alpha$
$$\sin \alpha = \sin \alpha \cdot \sqrt{1-\left( \frac{z+x}{y+z}\sin \alpha \right)^2}+\sqrt{1-\sin^2 \alpha}\cdot \frac{z+x}{y+z}\sin \alpha $$
$$1 = \sqrt{1-\left( \frac{z+x}{y+z}\sin \alpha \right)^2}+\sqrt{1-\sin^2 \alpha}\cdot \frac{z+x}{y+z}$$
Now this is the step where I get stuck, it is simply too complicated to solve by hand for me at least. Maybe someone can suggest a way to solve this in reasonable time or even better a different approach...
| Partial solution
Set $\gamma=\frac{z+x}{y+z}$ and $\mu=\sin^2\alpha$. You end up with $$1=\sqrt{1-\gamma^2\mu}+\sqrt{\gamma^2-\gamma^2\mu}$$
Squaring gives $$1=(1+\gamma^2)-2\gamma^2\mu+2\sqrt{(1-\gamma^2\mu)(\gamma^2-\gamma^2\mu)}$$
Which can be re-written to $$-\gamma^2+2\gamma^2\mu=2\sqrt{\gamma^2-(1+\gamma^2)\mu+\gamma^4\mu^2}$$
Squaring again gives $$\gamma^4-4\gamma^4\mu+4\gamma^4\mu^2=4\gamma^2-4(1+\gamma^2)\mu+4\gamma^4\mu^2$$
Which simplifies to a linear equation in $\mu$, hurray! We get:
$$\mu=\frac{4(\gamma^2-\gamma^4)}{4(1+\gamma^2-\gamma^4)}=\frac{1-\gamma^2}{\gamma^{-2}+1-\gamma^2}$$
Now we only need to solve $\sin\frac\alpha2$ from $\mu=\sin^2\alpha$.
Set $\sin^2\frac\alpha2=\nu$. We get
$$\mu=\sin^2\alpha=4\sin^2\frac\alpha2(1-\sin^2\frac\alpha2)=4\nu-(2\nu)^2=-((2\nu-1)^2-1)=1-(2\nu-1)^2,$$ so $\nu=\frac{1\pm\sqrt{1-\mu}}2$ and $$\sin\frac\alpha2=\sqrt{\frac{1-\sqrt{1-\mu}}2}=\sqrt{\frac{1-\sqrt{1-\frac{1-\gamma^2}{\gamma^{-2}+1-\gamma^2}}}2} = \sqrt{\frac{1-\frac{\gamma^{-1}}{\sqrt{\gamma^{-2}+1-\gamma^2}}}2}$$
Can you take over from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4648790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find the coefficients of product Given the following product,
$$(1+ax)(1+a^2x)(1+a^3x)\cdots (1+a^mx) $$
where $a$ is some real number which will be taken to be unity in the end. I want to know the coefficient of general term of expansion $x^p$. I know that the coefficient of $x^p$ comes from choosing $p$ terms from $m$, but the terms will reduce because some kinds of products have the same coefficients. How to write the coefficient step by step? Thanks in advance.
| Products of this type are known to be generalisations of the binomial
\begin{align*}
(1+x)^m=\sum_{k=0}^m\binom{m}{k}x^k
\end{align*}
Often such generalisations are written with $q$ instead of $a$ and here I'll follow this convention. We denote with $[x^p]$ the coefficient of $x^p$ in a series and want to find
\begin{align*}
[x^p]\left(1+qx\right)\left(1+q^2x\right)\cdots\left(1+q^mx\right)=[x^p]\prod_{k=1}^m\left(1+q^kx\right)\tag{1}
\end{align*}
According to the q-binomial theorem the following is valid
\begin{align*}
\color{blue}{\prod_{k=0}^{m-1}\left(1+q^kx\right)=\sum_{k=0}^mq^{\binom{k}{2}}\binom{m}{k}_q x^k}\tag{2}
\end{align*}
with $\binom{m}{k}_{q}$ the so-called q-binomial coefficients. The q-binomial coefficients follow a recurrence relation similar to the binomial coefficients, namely
\begin{align*}
\binom{m}{k}_q&=q^k\binom{m-1}{k}+\binom{m-1}{k-1}_q\qquad\qquad 1\leq k\leq m\tag{3.1}\\
\binom{m}{0}_q&=\binom{m}{m}_q=1\tag{3.2}\\
\binom{m}{1}_q&=\binom{m}{m-1}_q=1+q+\cdots+q^{k-1}\tag{3.3}
\end{align*}
We can now transform (1) to use the sum formula in (2) so that we can extract the coefficient of $x^p$.
We obtain from (1) and (2)
\begin{align*}
\color{blue}{[x^p]}\color{blue}{\prod_{k=1}^m\left(1+q^kx\right)}
&=[x^p]\prod_{k=1}^m\left(1+q^{k-1}(qx)\right)\tag{4.1}\\
&=[x^p]\prod_{k=0}^{m-1}\left(1+q^k(qx)\right)\tag{4.2}\\
&=[x^p]\sum_{k=0}^mq^{\binom{k}{2}}\binom{m}{k}_q (qx)^k\tag{4.3}\\
&=[x^p]\sum_{k=0}^mq^{\binom{k+1}{2}}\binom{m}{k}_q x^k\tag{4.4}\\
&\,\,\color{blue}{=q^{\binom{p+1}{2}}\binom{m}{p}_q}\tag{4.5}
\end{align*}
Comment:
*
*In (4.1) we write $q^k=q^{k-1}q$.
*In (4.2) we shift the index to start with $k=0$ and are now able to apply (2).
*In (4.3) we apply (2), replacing $x$ with $qx$.
*In (4.4) we use $\binom{q}{2}+q=\frac{1}{2}q(q-1)+q=\frac{1}{2}q(q+1)=\binom{q+1}{2}$.
*In (4.5) we select the coefficient of $x^p$.
Example $m=5, p=3$: Let's calculate a small example. We obtain with $m=5$ and $p=3$:
\begin{align*}
\color{blue}{[}&\color{blue}{x^3]\prod_{k=1}^5\left(1+q^kx\right)}=q^{\binom{4}{2}}\binom{5}{2}_q\tag{$\rightarrow 4.1, 4.5$}\\
&=q^6\left(q^2\binom{4}{2}_q+\binom{4}{1}_q\right)\tag{$\rightarrow 3.1$}\\
&=q^6\left(q^2\left(q^2\binom{3}{2}_q+\binom{3}{1}_q\right)+\binom{4}{1}_q\right)\tag{$\rightarrow 3.1$}\\
&=q^6\left(q^4\binom{3}{1}_q+q^2\binom{3}{1}_q+\binom{4}{1}_q\right)\tag{$\rightarrow 3.3$}\\
&=q^6\left(q^4\left(1+q+q^2\right)+q^2\left(1+q+q^2\right)+\left(1+q+q^2+q^3\right)\right)\\
&\,\,\color{blue}{=q^6\left(1+q+2q^2+2q^3+2q^4+q^5+q^6\right)}
\end{align*}
in accordance with a plausibility check by WolframAlpha.
| {
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"source": "stackexchange",
"question_score": "1",
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Is there an integral that proves $\pi > 333/106$? The following integral,
$$ \int_0^1 \frac{x^4(1-x)^4}{x^2 + 1} \mathrm{d}x = \frac{22}{7} - \pi $$
is clearly positive, which proves that $\pi < 22/7$.
Is there a similar integral which proves $\pi > 333/106$?
| From the relationship
$$\frac{333}{106}=\frac{377-2·22}{120-2·7}$$
and integrals
$$\pi = \frac{22}{7} - \int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx$$
and
$$\pi = \frac{377}{120} - \frac{1}{2}\int_0^1 \frac{x^5(1-x)^6}{1+x^2}dx$$
we obtain
$$\int_0^1 \frac{14x^4(1-x)^4-60x^5(1-x)^6}{1+x^2}dx = \int_0^1 \frac{2x^4(1-x)^4(7-30x(1-x)^2)}{1+x^2}dx = \int_0^1 \frac{2x^4(1-x)^4(7-30x+60x^2-30x^3)}{1+x^2}dx = 106\pi-333$$
Therefore,
$$\frac{1}{53} \int_0^1 \frac{x^4(1-x)^4(7-30x+60x^2-30x^3)}{1+x^2}dx = \pi-\frac{333}{106}$$
The numerator can be shown to be nonnegative for $0\leq x\leq 1$.
In fact, we can get a smaller numerator that is still nonnegative, which leads to a closer approximation to $\pi$
$$\pi=\frac{21991}{7000} +\frac{1}{50} \int_0^1 \frac{x^4 (1 - x)^4 \left(4 - 27 x (1 - x)^2\right)}{\left(1 + x^2\right)} dx$$
Here the numerator has been adjusted to have a double zero in $(0,1)$ while not crossing the axis WA link
The resulting fraction is
$$\frac{21991}{7000} = \frac{22}{7}-\frac{9}{7000}$$
which allows for writing the double inequality
$$\frac{22}{7}-\frac{9}{7000}<\pi<\frac{22}{7}$$
or, equivalently, an upper bound for the error in Archimedes' approximation
$$\frac{22}{7}-\pi<\frac{9}{7000}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Lateral surface area by hand I need to find the exact solution for the lateral surface area of the solid generated by revolving the region bounded by $y=x^2$, $y=0$, $x=0$, and $x=\sqrt 2$ about the X axis.
I have come up with my own solution, but I'm not sure if it's right.
Before substituting values back in, I got
$$ \frac\pi{16} \left( (1+4x^2)^{3/2} 2x - \frac32 \ln\sqrt{1+4x^2+2x} + x\sqrt{1+4x^2} \right) \text{ from }0\text{ to }\sqrt2$$
Can someone please verify this? For my integration by parts, I set $u = \sec^3\theta - \sec\theta$ and $dv = \sec^2\theta d\theta$
Thanks so much in advance.
| I assume you have in your earlier steps:
\[ \sigma = 2\pi \int_0^\sqrt{2} x^2 \sqrt{1+4x^2} dx \]
\[ \tan \theta = 2x \; \mathrm{and} \; \sec \theta = \sqrt{1+4x^2} \]
\[ \sigma = \frac{\pi}2 \int_{x=0}^{x=\sqrt{2}} \sec^3\theta \tan^2\theta d\theta. \]
So you have a messy antiderivative you worked out, and want to check whether it's right? Why not see what happens if you take the derivative of your expression above? If everything is fine, you should get back to $2\pi x^2 \sqrt{1+4x^2}$. I get:
\[ d \sigma = \frac{\pi}{16} \left[ 24x^2\sqrt{1+4x^2}+2(1+4x^2)^{\frac 32} - \frac{3(8x+2)}{4(1+4x^2+2x)} + \sqrt{1+4x^2} + 4x^2(1+4x^2)^{-\frac 12} \right] dx. \]
Yes, there are a few problems here. More things should be cancelling out. The middle term (from $\ln$) is conspicuous for involving $\sqrt{1+4x^2+2x}$ instead of $\sqrt{1+4x^2}$. I'm pretty sure the $2x$ snuck under the radical by accident, and the term should really be $\ln | \sqrt{1+4x^2}+2x |$. If I fix that in your expression and derive again, I get
\[ d \sigma = \frac{\pi}{16} \left[ 24x\sqrt{1+4x^2}+2(1+4x^2)^{\frac 32} - 3(1+4x^2)^{-\frac 12} + \sqrt{1+4x^2} + 4x^2(1+4x^2)^{-\frac 12} \right] dx. \]
Now there are some like terms and related terms, but they won't cancel out far enough, because some of your coefficients are wrong. Maybe from here you can find your mistakes?
Other hints:
*
*The correct antiderivative in terms of $x$ never involves a factor of 3, unless you count the term $(1+4x^2)^{\frac 32}$.
*If you have a good antiderivative in terms of $\theta$, you might want to plug in to that equation instead of converting back to $x$, by finding values for $\tan \theta$ and $\sec \theta$ at $x=0$ and $x=\sqrt{2}$. (At $x=\sqrt{2}$, $\sec \theta$ is rational!)
*When you have a final answer, compare to my result $\frac{\pi}{32}\left[102\sqrt{2} - \ln\left(3+2\sqrt{2}\right)\right]$.
| {
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Funny identities Here is a funny exercise
$$\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$$
(If you prove it don't publish it here please).
Do you have similar examples?
| $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3) = \pi$
(using the principal value),
but if you blindly use the addition formula
$\tan^{-1}(x) + \tan^{-1}(y)
= \tan^{-1}\dfrac{x+y}{1-x y}$
twice, you get zero:
$\tan^{-1}(1) + \tan^{-1}(2) =
\tan^{-1}\dfrac{1+2}{1-1*2}
=\tan^{-1}(-3)$;
$\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3)
=\tan^{-1}(-3) + \tan^{-1}(3)
=\tan^{-1}\dfrac{-3+3}{1-(-3)(3)}
= 0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inequality of three variable polynomial I read that one can prove by AM-GM-inequality that for all $a,b,c\in\mathbb{R}_+$ we have that
$$11(a^6 + b^6 + c^6) + 40abc(ab^2 + bc^2 + ca^2) \ge 51abc(a^2b + b^2c + c^2a)$$
How this can be done? Is it possible to prove stronger inequalities like
$$10(a^6 + b^6 + c^6) + 41abc(ab^2 + bc^2 + ca^2) \ge 51abc(a^2b + b^2c + c^2a),$$
or even
$$a^6 + b^6 + c^6 + 50abc(ab^2 + bc^2 + ca^2) \ge 51abc(a^2b + b^2c + c^2a)$$
without computer?
| Each of the inequalities below is easily verified by AM-GM
\begin{eqnarray*}
&29(& a^6 & & & + a^3bc^2 & +3a^2b^3c & & & & & \geq & 5a^3b^2c & & &) \\
&29(& & b^6 & & & + a^2b^3c & +3ab^2c^3 & & & & \geq & & 5ab^3c^2 & &) \\
&29(& & & c^6 & +3a^2b^3c & & + ab^2c^3 & & & & \geq & & & 5a^2bc^3 &) \\
&4 (& a^6 & & & & + a^2b^3c & & & +ab^3c^2 & & \geq & 3a^3b^2c & & &) \\
&4 (& & b^6 & & & & + ab^2c^3 & & & +a^2bc^3 & \geq & & 3ab^3c^2 & &) \\
&4 (& & & c^6 & + a^3bc^2 & & & +a^3b^2c & & & \geq & & & 3a^2bc^3 &). \\
\end{eqnarray*}
Now add these together, cancel some terms, divide by $3$ and we are dumb.
| {
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Proof of the formula $1+x+x^2+x^3+ \cdots +x^n =\frac{x^{n+1}-1}{x-1}$
Possible Duplicate:
Value of $\sum x^n$
Proof to the formula
$$1+x+x^2+x^3+\cdots+x^n = \frac{x^{n+1}-1}{x-1}.$$
| Since $$\frac1{1-x}=1+x+x^2+x^3+\cdots,$$ we have
$$
\frac{1-x^{n+1}}{1-x}=(1+x+x^2+x^3+\cdots)-(x^{n+1}+x^{n+2}+x^{n+3}+x^{n+4}+\cdots)
$$
And on the right hand side every thing cancels except $1+x+x^2+\cdots+x^n$.
(This argument is probably circular! :) )
| {
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"source": "stackexchange",
"question_score": "5",
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Nice expression for minimum of three variables? As we saw here, the minimum of two quantities can be written using elementary functions and the absolute value function.
$\min(a,b)=\frac{a+b}{2} - \frac{|a-b|}{2}$
There's even a nice intuitive explanation to go along with this: If we go to the point half way between two numbers, then going down by half their difference will take us to the smaller one. So my question is: "Is there a similar formula for three numbers?"
Obviously $\min(a,\min(b,c))$ will work, but this gives us the expression:
$$\frac{a+\left(\frac{b+c}{2} - \frac{|b-c|}{2}\right)}{2} - \frac{\left|a-\left(\frac{b+c}{2} - \frac{|b-c|}{2}\right)\right|}{2},$$
which isn't intuitively the minimum of three numbers, and isn't even symmetrical in the variables, even though its output is. Is there some nicer way of expressing this function?
| Based on Christian Blatter's answer and the trigonometric solution of the cubic equation, we can derive the following unusual solution.
Let $a$, $b$ and $c$ be real numbers. Then they are the roots of the equation
$$(x-a)(x-b)(x-c)=0$$
which can also be written as
$$x^3-(a+b+c)x^2+(ab+bc+ca)x-abc=0.$$
The trigonometric formula gives the solutions of a cubic equation in terms of its coefficients. If we substitute in the above coefficients and simplify we get an expression for the three roots, but now their size order is clear. Let $M$ be the average of $a$, $b$ and $c$, and let $P$ and $Q$ be the quadratic and geometric means of the quantities $a+b-2c$, $a-2b+c$ and $-2a+b+c$. That is:
$$M=\frac{a+b+c}3,$$
$$P=\sqrt{\frac{(a+b-2c)^2+(a-2b+c)^2+(-2a+b+c)^2}3},$$
$$Q=\sqrt[3]{(a+b-2c)(a-2b+c)(-2a+b+c)}.$$
Then we have
$$\max(a,b,c)=\frac{\sqrt 2}3P\cos\left(\frac 13\arccos\left(\sqrt 2\left(\frac QP\right)^3\right)\right)+M,$$
$$\mathrm{median}(a,b,c)=\frac{\sqrt 2}3P\cos\left(\frac 13\arccos\left(\sqrt 2\left(\frac QP\right)^3\right)+\frac{2\pi}3\right)+M,$$
$$\min(a,b,c)=\frac{\sqrt 2}3P\cos\left(\frac 13\arccos\left(\sqrt 2\left(\frac QP\right)^3\right)+\frac{4\pi}3\right)+M.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
A general formula for $\sum (k-1)(k-2)(k-3)$? What is a "simpler" formula for
$$\sum_{3}^{n} \frac{(k-1)(k-2)(k-3)}{6}$$
| Perhaps a messy and boring way, we can use the generating function.
$$\sum_{k=3}^{n}\frac{(k-1)(k-2)(k-3)}{6}=\frac{1}{6}\sum_{k=2}^{n-1}k(k-1)(k-2)$$
In addition, the generating function for $k(k-1)(k-2)$ is $x^3\left(\frac{1}{1-x}\right)^{(3)}.$
Hence, the sum is the coefficient of $x^{n-1}$ in $\frac{1}{6}\frac{1}{1-x}x^3\left(\frac{1}{1-x}\right)^{(3)}=\frac{x^3}{(1-x)^5}$, which is
$$\binom{n-1-3+5-1}{5-1}=\binom{n}{4},\quad n\geqslant3.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Algebra Problem The expression $x^2-4x+5$ is a factor of $ax^3+bx^2+25$. Express the sum a+b as an integer.
Please give an explanation of how the answer
| Since $x^2-4x+5=0$ has the roots $2 \pm i$ they must also be roots of $ax^3+bx^2+25\,$ therefore:
$$
0 = a(2+i)^3+b(2+i)^2+25=8a -6a + 4b - b + 25 + i\,(12 a -a + 4b) \\
\iff
\begin{cases}
\begin{align}
2a + 3b + 25 = 0 \\
11 a + 4b = 0
\end{align}
\end{cases}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Finding Value of the Infinite Product $\prod \Bigl(1-\frac{1}{n^{2}}\Bigr)$ While trying some problems along with my friends we had difficulty in this question.
*
*True or False: The value of the infinite product $$\prod\limits_{n=2}^{\infty} \biggl(1-\frac{1}{n^{2}}\biggr)$$ is $1$.
I couldn't do it and my friend justified it by saying that since the terms in the product have values less than $1$, so the value of the product can never be $1$. I don't know whether this justification is correct or not. But i referred to Tom Apostol's Mathematical Analysis book and found a theorem which states, that
*
*The infinite product $\prod(1-a_{n})$ converges if the series $\sum a_{n}$ converges.
This assures that the above product converges. Could anyone help me in finding out where it converges to? And,
*
*Does there exist a function $f$ in $\mathbb{N}$ ( like $n^{2}$, $n^{3}$) such that $\displaystyle \prod\limits_{n=1}^{\infty} \Bigl(1-\frac{1}{f(n)}\Bigr)$ has the value $1$?
| We have
\begin{align*}
p_{k} &= \prod_{n = 2}^{k} \left( 1 - \frac{1}{n^{2}} \right) = \prod_{n=2}^{k} \frac{(n-1)(n+1)}{n^{2}} \\ & = \frac{1 \cdot 3}{2 \cdot 2} \cdot \frac{2\cdot 4}{3 \cdot 3} \cdot \frac{3\cdot 5}{4 \cdot 4} \cdot \cdots \cdot \frac{(k-2)\cdot k}{(k-1)\cdot(k-1)} \cdot \frac{(k-1)(k+1)}{k\cdot k}\\ & = \frac{1}{2}\left(1 + \frac{1}{k}\right)
\end{align*}
because all but the first and last numerators and denominators cancel. Therefore
\[
\prod_{n=2}^{\infty} \left( 1 - \frac{1}{n^{2}} \right) =
\lim_{k\to\infty} p_{k} = \frac{1}{2}.
\]
To put this into a little context: Euler has shown that
\[
\sin{(\pi z)} = \pi z \prod_{n=1}^{\infty} \left( 1 - \frac{z^{2}}{n^{2}} \right)
\]
for all $z \in \mathbb{C}$.
We would like to plug in $z = 1$ on both sides. This doesn't work because we get $0 = 0$. But a simple trick yields what we want:
\[
\pi \prod_{n=2}^{\infty} \left( 1 - \frac{1}{n^{2}} \right) = \lim_{z \to 1} \frac{\sin{(\pi z)}}{(1- z^2)} = \lim_{z \to 1} \frac{\pi \cos{(\pi z)}}{-2z} = \frac{\pi}{2}
\]
and cancelling on both sides with $\pi$ gives $\frac{1}{2}$ for the infinite product.
There is a lot of fun that you can have with Euler's product. For instance $z = \frac{1}{2}$ yields the Wallis product formula
\[
\frac{\pi}{2} = \frac{2 \cdot 2}{1 \cdot 3} \cdot \frac{4 \cdot 4}{3 \cdot 5} \cdot \frac{6 \cdot 6}{5 \cdot 7} \cdot \cdots = \prod_{n=1}^{\infty} \frac{2n}{2n -1}\frac{2n}{2n +1}
\]
and $z = i$ yields
\[
\frac{\sin{(i\pi)}}{i \pi} = \frac{e^{\pi} - e^{-\pi}}{2\pi} =
\prod_{n = 1}^{\infty} \left( 1 + \frac{1}{n^{2}} \right).
\]
A thorough treatment of these and many other topics involving infinite products can be found in chapters 1 and 2 of Remmert, Classical topics in complex function theory, Springer GTM 172, 1998. The title of the German original is simply Funktionentheorie 2.
| {
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"timestamp": "2023-03-29T00:00:00",
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Recurrence equation: $u_n = 4u_{n−1} + 4u_{n−2}$ ; is $4x+4 = 4$ the characteristic equation? Given this recurrence equation:
$u_1 = 0, u_2 = 1$
$u_n = 4u_{n−1} + 4u_{n−2}$
Is the correct characteristic equation:
$4x+4 = 4$
EDIT:
Complete solve:
The characteristic equation is
$x^2-4x-4=0$
We solve the quadratic equation...
$\alpha = 5$
$\beta=-1$
So:
$u_n = c_1 \alpha^n + c_2 \beta^n$
We solve the equation...
$c_1 = 1/30$
$c_2 = 1/6$
Finally:
$u_n = \dfrac{5^n}{30} + \dfrac{(-1)^n}{6}$
| Give Wilf's techniques a try... define $U(z) = \sum_{n \ge 0} u_{n + 1} u^z$,
and write the recurrence as:
$$
u_{n + 2} = 4 u_{n + 1} + 4 u_n \qquad u_1 = 0, u_2 = 1
$$
By properties of generating functions:
$$
\frac{U(z) - u_1 - u_2 z}{z^2} = 4 \cdot \frac{U(z) - u_1}{z} + 4 \cdot U(z)
$$
Maxima solves this as:
$$
U(z) = \frac{z}{1 - 4 z - 4 z^2}
= \frac{1}{2^{5/2}} \cdot \frac{1}{1 - (2^{3/2} + 2) z}
- \frac{1}{2^{5/2}} \cdot \frac{1}{1 + (2^{3/2} - 2) z}
$$
Two geometric series:
$$
u_{n + 1} = \frac{1}{2^{5/2}} \cdot
\left(
(2^{3/2} - 2)^n - (-1)^n (2^{3/2} + 2)^n
\right)
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\int \cos^3 x\;\sin^2 xdx$ Is this correct? I thought it would be but when I entered it into wolfram alpha, I got a different answer.
$$\int (\cos^3x)(\sin^2x)dx = \int(\cos x)(\cos^2x)(\sin^2x)dx
= \int (\cos x)(1-\sin^2x)(\sin^2x)dx.$$
let $u = \sin x$, $du = \cos xdx$
$$\int(1-u^2)u^2du = \int(u^2-u^4)du
= \frac{u^3}{3} - \frac{u^5}{5} +C$$
Plugging in back $u$, we get $\displaystyle\frac{\sin^3 x}{3} - \frac{\sin^5 x}{5}$ + C
| Take $\sin x=t, \cos x dx=dt$
\begin{align*}
\therefore \int \cos^{3}x\sin^{2}x dx &=\int \cos^{2}x\sin^{2} x\cos x dx\
&=\int (1-\sin^{2}x) \sin^{2} x \cos x dx\
&= \int (1-t^{2})t^{2} dt\
&=\int (t^{2}-t^{4})dt\
&=\frac{t^{3}}{3}-\frac{t^{5}}{5}+c\
&= \frac{\sin^{3}x}{3}-\frac{\sin^{5}x}{5}+c
\end{align*}
| {
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Counting orbits under $\operatorname{Sym}(3)$ I'd like a closed form expression $x(n)$ for the number of orbits of the symmetric group on $3$ points acting on the triples in $\{ (a,b,c) \mid a,b,c \in \Bbb{Z}, 1 \leq a,b,c \leq n, c = 2n−a−b \}$.
I feel like this should be a really basic problem, but my standard method of attack fails: look it up in OEIS and prove the known formula. My backup plan of "think about it" has failed: I don't know how to deal with the restriction on $c$. (Without the restriction, I happen to have learned this is the same thing as counting $1 \leq a \leq b \leq c \leq n$, and I happen to have learned this is Binomial($n+2$, $3$) because you need to place two bars between $n$ stars, but I have no general context for this.)
I suspect this is a pretty standard counting problem even with the restriction, but I never really learned how to count (fish, fish, fish, …, fish, fish, …, fish).
The counts $x(n)$ for $n=1$ to $30$ are: $0, 1, 2, 3, 4, 6, 7, 9, 11, 13, 15, 18, 20, 23, 26$, $29, 32, 36, 39, 43, 47, 51, 55, 60, 64, 69, 74, 79, 84, 90$.
I think there are $(n-1)\cdot(n+4)/2$ triples, but even that is a little fuzzy (increase increases by $1$ each time). I have no idea how many of them have $2$ equal components.
| This one can also be done using the Polya Enumeration Theorem. Recall the cycle index of the symmetric group on three elements which can be computed using pen and paper by factorizing the six constituent permutations and has the value
$$Z(S_3) = \frac{1}{6}\left(a_1^3 + 3 a_1 a_2 + 2 a_3 \right).$$
Then the desired value is given by
$$[z^{2n}] Z(S_3)(z+z^2+\cdots+z^n).$$
Actually doing the substitution we get
$$[z^{2n}]\frac{1}{6}(z+z^2+\cdots+z^n)^3 \\+
[z^{2n}]\frac{1}{2} (z+z^2+\cdots+z^n) (z^2+z^4+\cdots+z^{2n}) \\+
[z^{2n}]\frac{1}{3} (z^3+z^6+\cdots+z^{3n}).$$
Now the contribution from the first term is
$$[z^{2n}]\frac{z^3}{6}(1+z+\cdots+z^{n-1})^3
= [z^{2n-3}]\frac{1}{6} \frac{(1-z^n)^3}{(1-z)^3}.$$
This is in fact (use the Newton binomial)
$$[z^{2n-3}]\frac{1}{6} \frac{1-3z^n}{(1-z)^3}
=\frac{1}{6}{2n-3+2\choose 2}
-\frac{1}{2}{n-3+2\choose 2}
\\= \frac{1}{6}{2n-1\choose 2}
-\frac{1}{2}{n-1\choose 2}
\\ = \frac{1}{12} n^2 +\frac{1}{4} n - \frac{1}{3}.$$
For the second term we see by inspection that for those terms from $z+z^2+\cdots+z^n$ that are even there exists a matching term from $z^2+z^4+\cdots+z^{2n}$ that gives an exponent sum of $2n$ making for a contribution of
$$\frac{1}{2}\bigg\lfloor\frac{n}{2}\bigg\rfloor
= \frac{1}{4} n - \frac{1}{4} (n\bmod 2).$$
Finally the third term contributes the value $1/3$ if $n$ is divisible by three.
This gives the final result
$$\frac{1}{12} n^2 +\frac{1}{2} n
+ \begin{cases} 0 & \quad\text{if}\quad n \bmod 6 = 0,\\
-\frac{7}{12} & \quad\text{if}\quad n \bmod 6 = 1,5 \\
-\frac{1}{3} & \quad\text{if}\quad n \bmod 6 = 2,4 \\
-\frac{1}{4} & \quad\text{if}\quad n \bmod 6 = 3.
\end{cases}.$$
The computation at this MSE link is somewhat similar, as is the one at this MSE link II which also uses the Newton binomial.
| {
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Prove inequality: When $n > 2$, $n! < {\left(\frac{n+2}{\sqrt{6}}\right)}^n$ Prove: When $n > 2$,
$$n! < {\left(\frac{n+2}{\sqrt{6}}\right)}^n$$
PS: please do not use mathematical induction method.
EDIT: sorry, I forget another constraint, this problem should be solved by
algebraic mean inequality.
Thanks.
| For $a_1, a_2, \ldots, a_n \geqslant 0$, we have:$$\sqrt [ n ]{ { a }_{ 1 }\cdot { a }_{ 2 }\cdots { a }_{ n } } \leqslant \frac { { a }_{ 1 }+{ a }_{ 2 }+\cdots +{ a }_{ n } }{ n } $$
Consider ${ \left( n! \right) }^{ 2 }=\left( n\cdot 1 \right) \left( \left( n-1 \right) \cdot 2 \right) \cdots \left( 1\cdot n \right) $, for $n>1$:
$$\sqrt [ n ]{ { \left( n! \right) }^{ 2 } } =\sqrt [ n ]{ \left( n\cdot 1 \right) \left( \left( n-1 \right) \cdot 2 \right) \cdots \left( 1\cdot n \right) } \leqslant \frac { n\cdot 1+\left( n-1 \right) \cdot 2+\cdots +\left( n-\left( n-1 \right) \right) \cdot n }{ n } \\ =\frac { n\left( 1+2+\cdots +n \right) -1\times 2-2\times 3-\cdots -\left( n-1 \right) \times n }{ n } \\ =\frac { n\left( \frac { n\left( n+1 \right) }{ 2 } \right) -\frac { n\left( { n }^{ 2 }-1 \right) }{ 3 } }{ n } \\ =\frac { \left( n+1 \right) \left( n+2 \right) }{ 6 } <\frac { { \left( n+2 \right) }^{ 2 } }{ 6 } $$
That's the conclusion, which we can get using high school math not the Stirlings.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/28084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Nice proofs of $\zeta(4) = \frac{\pi^4}{90}$? I know some nice ways to prove that $\zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \pi^2/6$. For example, see Robin Chapman's list or the answers to the question "Different methods to compute $\sum_{n=1}^{\infty} \frac{1}{n^2}$?"
Are there any nice ways to prove that $$\zeta(4) = \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}?$$
I already know some proofs that give all values of $\zeta(n)$ for positive even integers $n$ (like #7 on Robin Chapman's list or Qiaochu Yuan's answer in the linked question). I'm not so much interested in those kinds of proofs as I am those that are specifically for $\zeta(4)$.
I would be particularly interested in a proof that isn't an adaption of one that $\zeta(2) = \pi^2/6$.
| Let $ f(x)=x^2-\frac{1}{12} $ defined on $ [-\frac{1}{2},-\frac{1}{2}) $, and extend $ f $ to be periodic on $ \mathbb{R} $.
We compute that
\begin{align*}
\hat f(0)=\int_{-1/2}^{1/2}f(x)dx=\int_{-1/2}^{1/2}\left(x^2-\frac{1}{12}\right)dx=0.
\end{align*}
And for $ k\ne 0 $:
\begin{align*}
\hat f(k)&=\int_{-1/2}^{1/2}f(x)e^{-2\pi ikx}dx=\int_{-1/2}^{1/2}\left(x^2-\frac{1}{12}\right)e^{-2\pi ikx}dx=\int_{-1/2}^{1/2}x^2e^{-2\pi ikx}dx\\
&=-\left.\frac{x^2e^{-2\pi i kx}}{2\pi ik}\right|_{-1/2}^{1/2}+\frac{1}{\pi ik}\int_{-1/2}^{1/2}xe^{-2\pi ikx}dx=\frac{1}{\pi ik}\int_{-1/2}^{1/2}xe^{-2\pi ikx}dx\\
&=\frac{1}{2\pi^2k^2 }\int_{-1/2}^{1/2}xd(e^{-2\pi ikx})=\left.\frac{1}{2\pi^2k^2}xe^{-2\pi ikx}\right|_{-1/2}^{1/2}+\frac{1}{2\pi^2k^2}\int_{-1/2}^{1/2} e^{-2\pi ikx}dx=\frac{(-1)^k}{2\pi^2k^2}.
\end{align*}
By the Parseval identity
\begin{align*}
\int_{-1/2}^{1/2}|f(x)|^2dx=\sum_{k=-\infty}^{\infty}|\hat{f}(k)|^2=|\hat{f}(0)|^2+2\sum_{k=1}^{\infty}|\hat{f}(k)|^2=\sum_{k=1}^{\infty}\frac{1}{2\pi^4 k^4}.
\end{align*}
On the other hand,
\begin{align*}
\int_{-1/2}^{1/2}|f(x)|^2dx&=\frac 1{180}.
\end{align*}
Hence, we have $$ \sum_{k=1}^{\infty}\frac{1}{k^4}=\frac{\pi^4}{90}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/28329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "119",
"answer_count": 15,
"answer_id": 12
} |
Finding the first 5 terms of the Maclaurin Series of $\frac{\sin x}{e^x}$ Question.
Find the first 5 terms of the Maclaurin Series of $\frac{\sin x}{e^x}$.
I'm trying to find it using division. Is there possibly a shorter way because every time I try to do it, I get the wrong answer:
$$
x-2x^3/3+\cdots$$
I don't really like polynomial long division.
This is how I set it up:
$$
\frac{x - x^3/3! + x^5/5! - x^7/7!+\cdots}{ 1 + x + x^2/2! + x^3/3! + x^4/4!+\cdots}
$$
I'm still getting the same answer.
| Here are the first two terms:
$x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} + \cdots$
$=x\left(1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!}+ \dfrac{x^5}{5!}+ \dfrac{x^6}{6!}\cdots\right) -x^2 -\dfrac{2x^3}{3}-\dfrac{x^4}{6}-\dfrac{x^5}{30}-\dfrac{x^6}{120}-\dfrac{x^7}{5040}-\cdots$
$= (x-x^2)\left(1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!}+ \dfrac{x^5}{5!}+ \dfrac{x^6}{6!}\cdots\right) +\dfrac{x^3}{3}+\dfrac{x^4}{3}+\dfrac{2x^5}{15}+\dfrac{x^6}{30}+\dfrac{17x^7}{2520}+\cdots$
and you can continue from there, taking the first term of the residual of the second part as the next term to put into the multiplication in the first part.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/29416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Can this integral $\int_0^{2\pi} \frac{d\theta}{(a^2 \cos^2 \theta +b^2\sin^2\theta)^{3/2}}$ be written in the form of a elliptic integral I am trying to find the magnetic field due to an elliptic loop of wire. How to do integrals of the type $$\int_0^{2\pi} \frac{d\theta}{(a^2 \cos^2 \theta +b^2\sin^2\theta)^{3/2}}$$ Where a and b are the axes of the ellipse. From dimensional considerations in the problem, I have deduced that the value of the integral must be must reduced to $$\frac{ L}{(ab)^2}$$ where $L$ is the circumference of the ellipse. How do I show it.
OK. So now I know from wikipedia that the circumference of ellipse involves something called elliptic integrals, so thankfully, I wouldnt have to be looking for a closed form. But does this reduce to the elliptic integral? Or have I done something wrong?
| Wow, this really worked out incredibly nicely!
The elliptic integral which gives $L$ is
$$
\frac{L}{4} = J(a,b) := \int_0^{\pi/2} \sqrt{a^2 \cos^2 \theta + b^2 \sin^2 \theta} \, d\theta
$$
and it is clearly symmetric, $J(a,b)=J(b,a)$.
The change of variables $t=b \tan\theta$ gives
$$
\sqrt{a^2 \cos^2 \theta + b^2 \sin^2 \theta} = \cos \theta \, \sqrt{a^2 + t^2},
\qquad
\cos \theta = \frac{b}{\sqrt{b^2 + t^2}},
\qquad
d\theta = \frac{b}{b^2 + t^2} \, dt;$$
thus
$$
J(a,b) = \int_0^\infty \frac{b^2 \, \sqrt{a^2 + t^2}}{(b^2 + t^2)^{3/2}} \, dt.
$$
Applying the same change of variables to $1/4$ of your integral (that is, with integration from $0$ to $\pi/2$ instead of to $2\pi$) results in
$$
\frac{1}{b^2} \int_0^\infty \frac{\sqrt{b^2 + t^2}}{(a^2 + t^2)^{3/2}} \, dt,
$$
which, from comparison with the above,
clearly equals $J(b,a)/(ab)^2$.
Using the symmetry of $J$, and multiplying by 4, we conclude that your integral indeed equals $L/(ab)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/34516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 2
} |
How to find roots of $X^5 - 1$? How to find roots of $X^5 - 1$? (Or any polynomial of that form where $X$ has an odd power.)
| Edit 2: Since the equation $x^n-1=0$ is equivalent to $x=\sqrt[n]{1}$, it has the following $n$ solutions:
$$e^{i\dfrac{2\pi (1+k)}{n}}=\cos\left( \dfrac{2\pi (1+k)}{n}\right)+i\sin\left( \dfrac{2\pi (1+k)}{n}\right),\qquad k=0,1,\ldots ,n-1,$$
where $n$ is a positive integer, odd or even.
See comment by Gerry Myerson indicating how the Galois group for an irreducible quartic determines whether the method exposed below works.
Although late I present another algebraic solution, less elegant but more "automatic", in the sense that no clever trick is necessary. [Edit in response to Gerry Myerson's comment] We are going to factor it into two factors, one being linear and the other quadratic. This method works for quintic polynomials, as long as we are able to find one solution and solve the quadratic polynomial factor. If it fails, in principle we can solve the quadratic polynomial by means of a resolvent cubic equation (also in this post in Portuguese) [end of edit]. For the general polynomial of 7th degree or higher it is not applicable, because it depends on finding one root by inspection (or numerically) and the remaining ones algebraically.
By inspection we see that $x=1$ is a zero of $x^{5}-1$. By long division or Ruffini's rule, we find
$$x^{5}-1=\left( x-1\right) \left( x^{4}+x^{3}+x^{2}+x+1\right) .$$
Now we factor $x^{4}+x^{3}+x^{2}+x+1$ into two quadratic polynomials
$$x^{4}+x^{3}+x^{2}+x+1=\left( x^{2}+bx+c\right) \left( x^{2}+Bx+C\right) ,$$
whose (real) coefficients have to be found by comparing LHS with the expanded RHS. Since
$$\left( x^{2}+bx+c\right) \left( x^{2}+Bx+C\right)$$
$$ =x^{4}+\left( B+b\right) x^{3}+\left( C+bB+c\right) x^{2}+\left(bC+cB\right) x+cC,$$
the coefficients must satisfy
$$\left\{
\begin{array}{c}
B+b=1 \\
C+bB+c=1 \\
bC+cB=1 \\
cC=1%
\end{array}%
\right. \Leftrightarrow \left\{
\begin{array}{c}
B=1-b \\
1/c+b\left( 1-b\right) +c=1 \\
b/c+c\left( 1-b\right) =1 \\
C=1/c%
\end{array}%
\right. $$
One of the solutions of $b/c+c\left( 1-b\right) =1$ is $c=1$. By
substitution we find the remaining coefficients:
$$\left\{
\begin{array}{c}
B=\frac{1}{2}\mp \frac{1}{2}\sqrt{5} \\
b=\frac{1}{2}\pm \frac{1}{2}\sqrt{5} \\
c=1 \\
C=1%
\end{array}%
\right. $$
Thus
$$x^{4}+x^{3}+x^{2}+x+1=\left( x^{2}+\left( \frac{1}{2}+\frac{1}{2}\sqrt{5}%
\right) x+1\right) \left( x^{2}+\left( \frac{1}{2}-\frac{1}{2}\sqrt{5}%
\right) x+1\right) $$
and finally
$$x^{5}-1=\left( x-1\right) \left( x^{2}+\left( \frac{1}{2}+\frac{1}{2}\sqrt{5%
}\right) x+1\right) \left( x^{2}+\left( \frac{1}{2}-\frac{1}{2}\sqrt{5}%
\right) x+1\right) .$$
Hence the roots of $x^{5}-1$ are the roots of these three factors. They are, respectively, $x_1=1$ and
$$x_{2,3}=-\frac{1}{4}-\frac{1}{4}\sqrt{5}\pm \frac{1}{4}\sqrt{-10+2\sqrt{5}},\quad x_{4,5}=-\frac{1}{4}+\frac{1}{4}\sqrt{5}\pm \frac{1}{4}\sqrt{-10-2\sqrt{5}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/39202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
} |
Solving $\sqrt{x+5} = x - 1$ I'm currently learning about radicals and simplifying them, and I came across this problem on the internet and tried to solve it:
$$\sqrt{x+5} = x - 1$$
So I used this logic:
$$
\begin{align}
\sqrt{x+5} &= x - 1 \\
x + 5 &= (x-1)^2 \\
x + 5 &= (x-1)(x-1) \\
x + 5 &= x^2 - 2x + 1 \\
0 &= x^2 - 3x - 4 \\
0 &= (x-4)(x+1) \\
\end{align}
$$
Therefore, $x = -1$ and $x = 4$ satisfy the equation $0 = (x-4)(x+1)$.
But then I tried to plug them in the original problem $\sqrt{x+5}=x-1$:
$$
\begin{align}
\sqrt{4 + 5} &= 4 - 1 \\
\sqrt{9} &= 3 \\
3 &= 3
\end{align}
$$
So using 4 works as expected, but when using $-1$:
$$
\begin{align}
\sqrt{-1 + 5} &= -1 - 1 \\
\sqrt{4} &= -2 \\
2 &\ne -2
\end{align}
$$
At what stage am I going wrong?
And according the WolframAlpha, the solution is $x = 4$.
| *
*Function on the left hand side is strictly increasing (i.e always increasing).
*Function on the right hand side is strictly increasing (i.e always increasing) too.
*We draw a graph to find out where too look for the roots. You can use this or any other tool to draw two graphs for left and right hand side functions. We'll see that there is one interception at $x=4$ and there will be no other interceptions because functions striclty increase.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/41152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 7,
"answer_id": 6
} |
Asymptotic approximation for confluent hypergeometric function I have the following nasty expression that I would like to expand in powers of $\frac{1}{N}$:
\begin{align}
\frac{2^{\frac{3}{2}} 3^{\frac{1}{2}} \Biggl[ \sqrt{u} \cdot \Gamma\left(\frac{2+N}{4}\right)
\cdot {}_1F_1 \left( \frac{2+N}{4},\frac{1}{2},\frac{3r^2}{2u} \right)
-\sqrt{6} r \cdot \Gamma \left( \frac{4+N}{4} \right) \cdot {}_1F_1 \left( \frac{4+N}{4},\frac{3}{2},\frac{3r^2}{2u} \right) \Biggr]
}{N \cdot u^{\frac{1}{2}} \Biggl[ \sqrt{u} \cdot \Gamma\left(\frac{N}{4}\right) \cdot {}_1F_1 \left( \frac{N}{4},\frac{1}{2},\frac{3r^2}{2u} \right) -\sqrt{6} r \cdot \Gamma \left( \frac{2+N}{4} \right) \cdot {}_1F_1 \left( \frac{2+N}{4},\frac{3}{2},\frac{3r^2}{2u} \right) \Biggr]}
\end{align}
where you can assume that $N \in \mathbb{N}$ (but could be analytically continued to $\mathbb{R}^+$), $u \in \mathbb{R}^+$, and $r \in \mathbb{R}^+$. Furthermore, ${}_1F_1$ is the confluent hypergeometric function sometimes written as $M(a,b,z)$.
Using a different route I have obtained a value for the limit $N \to \infty$, but I'd like to a) reproduce this result using the above expression and b) find the $O\left(\frac{1}{N}\right)$ corrections. So far I have tried numerous identities from the NIST Handbook of Mathematical Functions, but I simply seem to lack the experience to make real progress.
If anyone knows a solution or has an idea of how to proceed next, I'd greatly appreciate their help.
With best regards,
Jan
| Perhaps not a big help: For ${}_1F_1((2+N)/4,3/2,3r^2/2u)$ we want asymptotics of ${}_1F_1(1/2+x,3/2,a)$ as $x \to 0$. Using the series, I get
$$
{}_1F_1\left(\frac{1}{2}+x,\frac{3}{2},a\right)=
\frac{\sqrt{\pi} \mathrm{erf} (i \sqrt{a})}{2 \sqrt{-a}} + \sum_{k = 0}^{\infty} \frac{\Bigl(\Psi \Bigl(\frac{1}{2} + k\Bigr) + \gamma + 2 \operatorname{ln} (2)\Bigr) a^{k}}{(1 + 2 k) k!} x + \operatorname{O} \bigl(x^{2}\bigr)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/47121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Getting the inverse of a lower/upper triangular matrix For a lower triangular matrix, the inverse of itself should be easy to find because that's the idea of the LU decomposition, am I right? For many of the lower or upper triangular matrices, often I could just flip the signs to get its inverse. For eg: $$\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
-1.5 & 0 & 1
\end{bmatrix}^{-1}=
\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
1.5 & 0 & 1
\end{bmatrix}$$
I just flipped from -1.5 to 1.5 and I got the inverse.
But this apparently doesn't work all the time. Say in this matrix:
$$\begin{bmatrix}
1 & 0 & 0\\
-2 & 1 & 0\\
3.5 & -2.5 & 1
\end{bmatrix}^{-1}\neq
\begin{bmatrix}
1 & 0 & 0\\
2 & 1 & 0\\
-3.5 & 2.5 & 1
\end{bmatrix}$$
By flipping the signs, the inverse is wrong.
But if I go through the whole tedious step of gauss-jordan elimination, I would get its correct inverse like this: $\begin{bmatrix}
1 & 0 & 0\\
-2 & 1 & 0\\
3.5 & -2.5 & 1
\end{bmatrix}^{-1}=
\begin{bmatrix}
1 & 0 & 0\\
2 & 1 & 0\\
1.5 & 2.5 & 1
\end{bmatrix}$
And it looks like some entries could just flip its signs but not for others.
Then this is kind of weird because I thought the whole idea of getting the lower and upper triangular matrices is to avoid the need to go through the tedious process of gauss-jordan elimination and can get the inverse quickly by flipping signs? Maybe I have missed something out here. How should I get an inverse of a lower or an upper matrix quickly?
| Given that the diagonal elements of a triangular matrix $A$ are all $1$. It is immediate to see that the characteristic polynomial of $A$ is
\begin{align*}
f(\lambda) = \det(\lambda I_{(n)} - A) = (\lambda - 1)^n =
\lambda^n - \binom{n}{1}\lambda^{n - 1} + \binom{n}{2}\lambda^{n - 2} + \cdots + (-1)^n.
\end{align*}
By Cayley-Hamilton theorem, this ensues
\begin{align*}
0 = A^n - \binom{n}{1}A^{n - 1} + \binom{n}{2}A^{n - 2} + \cdots + (-1)^nI_{(n)}.
\end{align*}
Rearranging terms yields
\begin{align*}
A \left((-1)^nA^{n - 1} + (-1)^{n + 1}\binom{n}{1}A^{n - 2} + (-1)^{n + 2}\binom{n}{2}A^{n - 3} + \cdots + (-1)^{2n - 1}\binom{n}{n - 1}I_{(n)}\right) = I_{(n)}.
\end{align*}
By definition, this shows
\begin{align*}
A^{-1} = (-1)^nA^{n - 1} + (-1)^{n + 1}\binom{n}{1}A^{n - 2} + (-1)^{n + 2}\binom{n}{2}A^{n - 3} + \cdots + (-1)^{2n - 1}\binom{n}{n - 1}I_{(n)}.
\end{align*}
When $n$ is small, this formula should be easy to apply.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "51",
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Inequality: $(x + y + z)^3 \geq 27 xyz$ Edit: $a,b,c$ and $x,y,z$ are positive, real numbers.
Since $(a-b)^2 \geq 0~$, $a^2 + b^2 - 2ab\geq0~$ and $a^2 + b^2 \geq 2ab~$. Similarly, $a^2 + c^2 \geq 2ac~$ and $b^2 + c^2 \geq 2bc~$.
Adding these inequalities together, $2(a^2 +b^2 + c^2) \geq 2(ab + ac +bc)~$ and accordingly, $a^2 +b^2 + c^2 \geq ab + ac +bc~$
Multiplying both sides by $(a + b + c)$:
$(a^2 +b^2 + c^2)(a+b +c) \geq (ab + ac +bc)(a + b + c)~~$ and simplifying this, $ a^3 + b^3 + c^3 + \Sigma a^2 b \geq 3abc + \Sigma a^2 b $
Therefore, it follows that $a^3 + b^3 + c^3 \geq 3abc~$, and letting $a^3 = x~$, $b^3 = y~$, $c^3 = z~$: $x + y + z\geq 3\sqrt[3]{xyz}$
Cubing both sides, $(x + y + z)^3 \geq 27 xyz~$ which was to be proven.
I was wondering if there are alternative approaches to solve this problem (possibly using higher-level maths), and is my proof entirely correct?
| Here are two one-line-formula proofs:
*
*By AM-GM:
$$
\frac{x+y+z}{3} \geq \sqrt[3]{x y z}
$$
Now take the cubic value on both sides of this inequality.
*Use the following identity which also gives you the exact deviation in positive terms from $27 x y z$: (from which you can derive tighter bounds of the LHS)
$$
(x+y+z)^3 = 27 x y z + 3 (z-y)^2 x + 3 (x-z)^2 y+ 3 (y-x)^2 z +\\
+ \frac12 (x+y+z)((x-y)^2 + (y-z)^2 + (z-x)^2)
$$
All terms on the RHS are positive, so you can take lower bounds of the LHS by any term on the RHS or weighted sum of terms on the RHS, with weights between 0 and 1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/48621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 8,
"answer_id": 5
} |
Prove that $3(a^3+b^3+c^3) > (a+b+c)(a^2+b^2+c^2)$ Let $a,b,c$ be positive, not equal. Prove that $3(a^3+b^3+c^3) > (a+b+c)(a^2+b^2+c^2)$.
I know the proof by subtracting LHS by RHS and then doing some arrangement.
But isn't there any inequality which can be used in $(1+1+1)(a^3+b^3+c^3) >(a+b+c)(a^2+b^2+c^2)$, or an easy proof?
| There is a fairly general way to prove inequalities like this using rearrangement, but just for fun, let me show you how you can prove a large class of inequalities like this by repeated use of AM-GM. First subtract $a^3, b^3, c^3$ from the RHS, then symmetrically sum the inequalities
$$a^3 + a^3 + b^3 \ge 3 a^2 b.$$
Essentially I am using weighted AM-GM with rational coefficients. Here are some exercises that can be solved using this idea.
Exercise 1. Prove that if $n > 1$ is a positive integer and $a > 0$, then
$$\frac{1 + a + ... + a^n}{a + a^2 + ... + a^{n-1}} \ge \frac{n+1}{n-1}.$$
Exercise 2. Let $a, b, c \ge 0$. Prove that
$$a^5 + b^5 + c^5 \ge 5abc(b^2 - ac).$$
(Weighted AM-GM isn't necessarily the best way to solve either of these problems; it's just the argument I came up with first.)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 1
} |
Find all pairs of positive integers $(x,y)$ such that $x^2-y^2=a^3$ and $x^3-y^3=b^2$ for integral $a, b$? How to find ALL pairs of positive integers $(x,y)$ such that the difference in their squares is a perfect cube and the difference in their cubes is a perfect square.
i.e.,
Positive integers $(x,y)$ such that
$x^2-y^2=a^3$ and $x^3-y^3=b^2$ for integral $a, b$?
Finding infinite number of pairs is no problem, as in:
$( 2^{6j+1} \cdot 3^{6k} \cdot 5 , 2^{6j+1} \cdot 3^{6k+1} )$ for any integral $j,k \geq 0$
But how would you determine the exhaustive list?
| Not an answer, just thinking out loud about the question of whether we can ever take $x,y$ relatively prime.
We solve $x^2-y^2=a^3$ by writing $a^3=a_1a_2$ with $a_1,a_2$ of the same parity, then $x-y=a_1$, $x+y=a_2$, $2x=a_2+a_1$, $2y=a_2-a_1$, and we see any common divisor of $a_1,a_2$ other than perhaps $2$ is a common divisor of $x,y$. So we have $a_1=c^3$, $a_2=d^3$ with $c,d$ relatively prime, or else $a_1=2c^3$, $a_2=4d^3$, or else $a_1=4c^3$, $a_2=2d^3$. I'll just look at the first case.
We have $2x=d^3+c^3$, $2y=d^3-c^3$, $c,d$ relatively prime. Then $$8b^2=(2x)^3-(2y)^3=(d^3+c^3)^3-(d^3-c^3)^3=6d^6c^3+2c^9$$ so $(2b)^2=c^3(3d^6+c^6)$. Since $c,d$ are coprime, the only possibilities for $\gcd(c^3,3d^6+c^6)$ are $1$ and $3$. I'll just look at the first case.
Now $c^3$ and $3d^6+c^6$ must be squares, so $c=e^2$ and $3d^6+e^{12}=f^2$. So $$(f-e^6)(f+e^6)=3d^6$$ If $f-e^6,f+e^6$ are coprime (and they must be pretty nearly coprime, if you trace back through to the coprimality of $c,d$), then $f-e^6=g^6$, $f+e^6=3h^6$, or else $f-e^3=3g^6$, $f+e^6=h^6$. Taking, as usual, just the first case, we get $$g^6+2e^6=3h^6$$
That's as far as I go. There is the trivial solution, $e=g=h=1$. That rules out using congruences to show there aren't any solutions, but I suspect there aren't any other coprime solutions. Maybe someone can actually prove this, and clean up the cases I haven't discussed.
| {
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How to divide polynomial matrices If I am given two $2\times2$ polynomial matrices and I need to divide them, what are the steps I need to follow?
I know I need to do right division and left division, and that the answer will have the right quotient and remainder and the left quotient and remainder.
Please show me how to determine these values with an example.
I don't want to put up the exact question I need to answer, I'd like to do that myself. I just want an example please.
| Dividing matrices is multiplying by the inverse, i.e if you want to divide $A$ by $B$ then you calculate $A\cdot B^{-1}$ (right) or $B^{-1}\cdot A$ (left). In order to find the inverse you use the Gauss reduction or the adjoint formula. For $2\times2$ matrices you have:
$$\left( \begin{array}{cc} a&b \\ c&d \end{array}\right)^{-1} = \frac{1}{ad-bc}\left( \begin{array}{cc} d&-b \\ -c&a \end{array}\right)$$
It doesn't matter whether the entries are numbers or polynomials.
To elborate on the comments: For example, suppose $A=\left( \begin{array}{cc} x^2&x+1 \\ x+2&x^2+1 \end{array}\right)$ and $B=\left( \begin{array}{cc} x&x \\ x+1&2x \end{array}\right)$. Then you have $B^{-1}= \frac{1}{2x^2-x^2-x}\left( \begin{array}{cc} 2x&-x \\ -x-1&x \end{array}\right)$. Suppose you want to calculate the right quotient.
$$AB^{-1}=\frac{1}{x^2-x}\left( \begin{array}{cc} 2x^3-(x+1)^2&-x^3+x^2+x\\ 2x^2+4x-(x+1)(x^2+1)&x^3+x-x^2-2x \end{array}\right)$$
Now divide each entry by the determinant, to get:
$$AB^{-1}=\left( \begin{array}{cc} 2x+1&-x \\ -x&x \end{array}\right)+\frac{1}{x^2-x}\left( \begin{array}{cc} -x-1&x \\ 3x-1&-x \end{array}\right)$$
Multiplying both sides by $B$, you get:
$$A=\left( \begin{array}{cc} 2x+1&-x \\ -x&x \end{array}\right)\cdot B+\left( \begin{array}{cc} 0&1 \\ 2&1 \end{array}\right)$$
So the right quotient is $\left( \begin{array}{cc} 2x+1&-x \\ -x&x \end{array}\right)$ and the right reminder is $\left( \begin{array}{cc} 0&1 \\ 2&1 \end{array}\right)$.
| {
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Determining the Jordan form of a matrix given the invariant factors I am trying to recover the Jordan normal form of a matrix given a list of invariant factors and was wondering if I am proceeding correctly in constructing the Jordan blocks.
Let $F = \mathbb{C}$ and let $V$ be a finite dimensional vector space over $F$. Let $T:V\to V$ be a linear operator and give $V$ the structure of a module over the polynomial ring $F[x]$ by defining $x \alpha = T(\alpha) \alpha \in V$
let
$$
A = \left( \begin{array}{ccc}
x^2(x-1)^2 & 0 & 0 \\
0 & x(x-1)(x-2)^2 & 0 \\
0 & 0 & x(x-2)^3 \end{array} \right)
$$
be a relation matrix for V with respect to $\{v_1, v_2, v_3\}$ generators of $V$.
Then $d_1 = x$, $d_2 = x(x-1)(x-2)^2$ and $d_3 = x^2(x-1)^2(x-2)^3$ are the invariant factors of $T$. Then we know $ V = F[x] / (x) \oplus F[x]/ (x(x-1)(x-2)^2) \oplus F[x]/(x^2(x-1)^2(x-3)^3)$. Further we know that the minimal polynomial of $T$ is the largest of the invariant factors so that $m_T(x) = (x^2(x-1)^2(x-2)^3)$ and the characteristic polynomial will be the product of $d_1 d_2 d_3$.
Question: what is the appropriate Jordan normal form of T?
Since 0, 1 and are repeated roots and 2 is repeated 3 times.
Does that give me Jordan blocks $$ J_1 = \begin{pmatrix}0 & 1 \\0 & 0 \end{pmatrix}$$
$$ J_2 = \begin{pmatrix}1 & 1 \\0 & 1 \end{pmatrix}$$
and $$ J_3 = \begin{pmatrix}2 & 1 \\0 & 2 \end{pmatrix}$$
| Edit: I'm sorry, but my first answer was definitely incorrect and I really hope I didn't cause any confusion. Won't speedread problems in the future :)
Since we have $V = F[x] / (x) \oplus F[x]/ (x(x-1)(x-2)^2) \oplus F[x]/(x^2(x-1)^2(x-3)^3)$, we can look at our three summands separately.
Our first summand, $F[x]/(x)$, has a single eigenvalue of zero (of multiplicity one), so our first Jordan block is simply $$J_1 = \begin{pmatrix}0 \end{pmatrix}$$
Next, we look at our second summand, $F[x]/(x(x-1)(x-2)^2)$, which has eigenvalues 0, 1, and 2 of multiplicities 1, 1, and 2 (respectively), so our Jordan blocks are now $$ J_2 = \begin{pmatrix}0 \end{pmatrix}, J_3 = \begin{pmatrix} 1 \end{pmatrix}, J_4 = \begin{pmatrix}2 & 1 \\0 & 2 \end{pmatrix}$$
For our final summand of $F[x]/(x^2(x-1)^2(x-3)^3)$ the eigenvalues are 0, 1, and 2 (with multiplicities 2, 2, and 3 respectively), so the Jordan blocks will be of the form $$ J_5 = \begin{pmatrix}0 & 1 \\0 & 0 \end{pmatrix}, J_6 = \begin{pmatrix}1 & 1 \\0 & 1 \end{pmatrix}, J_7 = \begin{pmatrix}2 & 1 & 0 \\0 & 2 & 1 \\0 & 0 & 2\end{pmatrix}$$
Putting the 7 blocks together gives our answer.
| {
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Rational parameterization of surface The surface $$ (x^{2} + y^{2} - 2 y + 1) \cdot (x^{2} + y^{2} + 2 y + 1) \cdot (x^{2} + y^{2} - 2 x + 1) \cdot (x^{2} + y^{2} + 2 x + 1) - z^2 = 0$$
has the points $\left(\dfrac{t}{s},\dfrac{t}{s},\dfrac{s^{4} + 4 t^{4}}{s^{4}}\right)$, $ \left(\dfrac{- s^{2} + t^{2}}{s^{2} + t^{2}},\dfrac{2 s t}{s^{2} + t^{2}} ,\dfrac{8 s^{3} t - 8 s t^{3}}{s^{4} + 2 s^{2} t^{2} + t^{4}}\right) $ (up to sign and swapping $x,y$)
Can the rest of the points be parameterized (or at least some of them)? (such points exist).
What software can do this?
The surface is related to Euler bricks.
EDIT
Some empirical observation about the points of interest. Of the 280 points found 246 have denominator of x that is sum of two squares, 250 have denominator of y that is sum of two squares and 252 have denominator of z that is a square.
Explaining the relation with Euler bricks per John's request (almost sure this have been done before).
So I naïvely wasted my time with perfect Euler bricks.
$$\begin{align*}a^2+b^2&=s_1^2\qquad \qquad \rm(1)\\
a^2+c^2&=s_2^2\qquad \qquad \rm(2)\\
b^2+c^2&=s_3^2\qquad \qquad \rm(3)\\
a^2+b^2+c^2&=s_4^2\qquad \qquad \rm(4)\end{align*}$$
(4) can be parameterized by $(a,b,c) = (2s,2t,1-s^2-t^2)$
To eliminate (1) further parameterize (1) with $(s,t) = (x^2-y^2,2xy)$. Substituting leads to $a=2 x^{2} - 2 y^{2}, b=4 x y ,c= -x^{4} - 2 x^{2} y^{2} - y^{4} + 1$ and leaves only equations (2) and (3). (3) is the surface in question with $z=s_3$. The known parameterization make one of $a,b,c$ $0$.
Without the restricting factor of $2$ in $2s,2t$ the surface becomes
$$ (x^{2} + y^{2} - 2 x - 2 y + 2) \cdot (x^{2} + y^{2} - 2 x + 2 y + 2) \cdot (x^{2} + y^{2} + 2 x - 2 y + 2) \cdot (x^{2} + y^{2} + 2 x + 2 y + 2) - z^2=0$$
with known solutions $(t,t)$ and $x^2+y^2=2$ (up to sign)
| I may have some results on this in due course. So bookmark this page and check back in a week or two.
But meanwhile, I'm curious to know how this relates to Euler bricks, unless you'd rather keep that under your hat.
Your equation is clearly equivalent over $\mathbb{Q}$ to the pair:
$ p^2 + q^2 + r^2 = 1 $
$ (1 - q^2) (1 - r^2) = t^2 $
and also (with some more work) to :
$ 1 + X^2 = Y^2 = (1 + Z^2) (1 + T^2) $
which mocks one by its apparent simplicity, in particular the seeming flexibility in choosing $Z$ and $T$, but which in truth is as intractable as the previous forms (and many others) !
These, especially the first, suggest you are investigating the "two face diagonals and body diagonal" case, or perhaps a slightly weaker verstion of that with the product of two face diagonals an integer. Not a bad idea.
Aside from one-parameter solutions, such as the ones you gave (in homogenous form), I think I've shown that this equation has a set of extra rational points derived from the Weierstrass curve $ X^3 - 13 X - 18 = Y^2 $, which has Mordell-Weil rank 1 with torsion point $X, Y = -2, 0$ and generator:
$X, Y = \dfrac{17}{4}, \dfrac{15}{8}$
But the proof is quite intricate, and needs careful checking, and also of course typing here! So it probably won't appear (if at all) before next week. I'll post again whether or not it holds up.
I am fairly certain there is no general, two-parameter solution. In other words this is not a rational surface. It's most likely what is called a K3 surface.
EDIT a week later:
On checking I found there was an error in what led to that Weierstrass equation. Why is it that every damned error in dealing with this brute of a problem seems to yield an intriguing or even definitive result?! It's not wishful thinking either, because one can rarely see the solution at the point where the error is made.
jojo, if you ping me your email, to jhnrmsdn at yahoo dot co dot uk, I can email you some more results and see if I can get you on the private Euler Brick Yahoo group (no promises, because I don't own the group).
Meanwhile here is an observation relating to your result. With suitable scaling, the two face diagonal and body diagonal case is equivalent to the set:
$x^2 + u^2 = 1$
$y^2 + v^2 = 1$
$x^2 + y^2 + z^2 = 1$
by the simple observation that the first two with the third imply respectively :
$u^2 = 1 - x^2 = y^2 + z^2$
$v^2 = 1 - y^2 = x^2 + z^2$
So plugging $x$ and $y$ of the following general parametrization of the third:
$x = \dfrac{2 a}{a^2 + b^2 + 1}$
$y = \dfrac{2 b}{a^2 + b^2 + 1}$
$z = \dfrac{a^2 + b^2 -1}{a^2 + b^2 + 1}$
into the first two and clearing denominators gives :
$(a^2 + b^2 + 1)^2 - 4 a^2 = U^2$
$(a^2 + b^2 + 1)^2 - 4 b^2 = V^2$
which is a stronger version of your result, split into two factors which must each be square rather than simply their product being a square.
Of course that isn't to say the corresponding factors in your result, which is derived in a different way, need be square. But it illustrates a characteristic feature of this problem - It's like a vast algebraic hall of mirrors, where similar forms show up, which may be weaker or stronger or even identical (the latter amounting to addition formulae which allow one to leapfrog between solutions a la chord-tangent process).
| {
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Solving an Exponential Equation $$3^{x-1}+3^{x-2}+3^{x-3}=3159$$
Another exponential equation I'm having a hard time with, the answer is given and equals to : $8$. I'm absolutely sure I'm making a wrong step somewhere along the way. Any help is appreciated.
EDIT:
$$3^{x-3}(\frac{1}{8}+\frac{1}{243}+1)=3159$$
This is where I'm now. If this is correct, it should pretty simple from here on. But I just don't seem to reach 8 as the final answer.
| I would try factoring $3^x$ first, and isolating:
$$\begin{align*}
3^{x-1} + 3^{x-2} + 3^{x-3} &= 3159\\
3^x\left(3^{-1} + 3^{-2} + 3^{-3}\right) &= 3159\\
3^x &= \frac{3159}{3^{-1}+3^{-2}+3^{-3}}
\end{align*}$$
For an extra flourish,
$$\begin{align*}
\frac{3159}{3^{-1}+3^{-2}+3^{-3}} &= \frac{3159}{\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3}}\\
&= \frac{3159}{\quad\frac{9 + 3 + 1}{3^3}\quad}\\
&= \frac{3^3(3159)}{13}\\
&= \frac{3^3(3^5)(13)}{13}\\
&= 3^8.
\end{align*}$$
Now you see that your equation is equivalent to
$$3^x = 3^8.$$
| {
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Is this asymptotic equation correct? Is this equation correct?
$$ \frac {1 + \Theta(\frac 1 {2n})} {(1 + \Theta(1/n))^2} = 1 + O(1 / n) $$
I need this equation to prove that
$$ \binom {2n} n = \frac {2 ^ {2n}} {\sqrt {\pi n}} (1 + O(1 / n)) $$
which could be calculated from Stirling's Approximation:
$$ n! = \sqrt {2\pi n} \left(\frac n e\right)^n \left(1 + \Theta({\frac 1 n})\right).$$
| Suppose $Bx\le\Theta(x)\le Cx$ for each $\Theta$.
$$
\begin{align}
\frac{1+\Theta(\frac{1}{2n})}{(1+\Theta(\frac{1}{n}))^2}
&\le\frac{1+\frac{C}{2n}}{(1+\frac{B}{n})^2}\\
&=(1+\frac{C}{2n})(1-2\frac{B}{n}+O(\frac{1}{n^2}))\\
&=(1+(\frac{C}{2}-2B)\frac{1}{n}+O(\frac{1}{n^2}))\\
&=1+O(\frac{1}{n})
\end{align}
$$
$$
\begin{align}
\frac{1+\Theta(\frac{1}{2n})}{(1+\Theta(\frac{1}{n}))^2}
&\ge\frac{1+\frac{B}{2n}}{(1+\frac{C}{n})^2}\\
&=(1+\frac{B}{2n})(1-2\frac{C}{n}+O(\frac{1}{n^2}))\\
&=(1+(\frac{B}{2}-2C)\frac{1}{n}+O(\frac{1}{n^2}))\\
&=1+O(\frac{1}{n})
\end{align}
$$
| {
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Difficult Integral: $\int\frac{x^n}{\sqrt{1+x^2}}dx$ How to calculate this difficult integral: $\int\frac{x^2}{\sqrt{1+x^2}}dx$?
The answer is $\frac{x}{2}\sqrt{x^2\pm{a^2}}\mp\frac{a^2}{2}\log(x+\sqrt{x^2\pm{a^2}})$.
And how about $\int\frac{x^3}{\sqrt{1+x^2}}dx$?
| I would first try the substitution $x=\tan(\theta)$, so that $\sqrt{1+x^2}=\sec(\theta)$. That gives
$$
\begin{align}
\int\frac{x^n}{\sqrt{1+x^2}}\;\mathrm{d}x
&=\int \tan^n(\theta)\sec(\theta)\;\mathrm{d}\theta\\
&=\tan^{n-1}(\theta)\sec(\theta)-(n-1)\int\tan^{n-2}(\theta)\;\sec^3(\theta)\;\mathrm{d}\theta\\
&=\tan^{n-1}(\theta)\sec(\theta)-(n-1)\int(\tan^n(\theta)+\tan^{n-2}(\theta))\;\sec(\theta)\;\mathrm{d}\theta\\
&=\frac{1}{n}\tan^{n-1}(\theta)\sec(\theta)-\frac{n-1}{n}\int\tan^{n-2}(\theta)\;\sec(\theta)\;\mathrm{d}\theta
\end{align}
$$
If $n$ is odd, this reduces to
$$
\int\tan(\theta)\sec(\theta)\;\mathrm{d}\theta=\sec(\theta)+C
$$
If $n$ is even, this reduces to
$$
\begin{align}
\int\sec(\theta)\;\mathrm{d}\theta&=\int\sec^2(\theta)\;\mathrm{d}\sin(\theta)\\
&=\int\frac{1}{2}\left(\frac{1}{1-\sin(\theta)}+\frac{1}{1+\sin(\theta)}\right)\;\mathrm{d}\sin(\theta)\\
&=\frac{1}{2}\log\left(\frac{1+\sin(\theta)}{1-\sin(\theta)}\right)+C\\
&=\log(\sec(\theta)+\tan(\theta))+C
\end{align}
$$
| {
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How to prove $a^2 + b^2 + c^2 \ge ab + bc + ca$? How can the following inequation be proven?
$$a^2 + b^2 + c^2 \ge ab + bc + ca$$
| From Cauchy-Schwarz
$ab+bc+ac=\sqrt{a^2}\sqrt{b^2}+\sqrt{b^2}\sqrt{c^2}+\sqrt{a^2}\sqrt{c^2} \leq \sqrt{a^2+b^2+c^2}\sqrt{a^2+b^2+c^2}$
Moving on;
$ab+bc+ac \leq a^2+b^2+c^2$
Done!
| {
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Solving Radical Equations $x-7= \sqrt{x-5}$ This the Pre-Calculus Problem:
$x-7= \sqrt{x-5}$
So far I did it like this and I'm not understanding If I did it wrong.
$(x-7)^2=\sqrt{x-5}^2$ - The Square root would cancel, leaving:
$(x-7)^2=x-5$ Then I F.O.I.L'ed the problem.
$(x-7)(x-7)=x-5$
$x^2-7x-7x+14=x-5$
$x^2-14x+14=x-5$
$x^2-14x-x+14=x-x-5$
$x^2-15x+14=-5$
$x^2-15x+14+5=-5+5$
$x^2-15x+19=0$
$(x-1)(x-19)=0$
Now this is where I'm stuck because when I tried to see if I got the right numbers in the parentheses I got this....
$x^2-19x-1x+19=0$
$x^2-20x+19=0$
As you may see I'm doing something bad because I don't get $x^2-15x+19$
Could anyone please help me and tell me what I'm doing wrong?
| We can avoid squaring both sides. Let $x-5=u^2$, where $u \ge 0$. Then $\sqrt{x-5}=u$. Also, $x=u^2+5$, so $x-7=u^2-2$. Thus our equation can be rewritten as
$$u^2-2=u, \quad\text{or equivalently}\quad u^2-u-2=0.$$
But
$$u^2-u-2=(u-2)(u+1).$$
Thus the solutions of $u^2-u-2=0$ are $u=2$ and $u=-1$. Since $u \ge 0$, we reject the solution $u=-1$.
We conclude that $u=2$, and therefore $x=u^2+5=9$.
| {
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Convergence of Series, problems with intermediate steps I have problems with two exercises. I know the answer of those two but both of them have a step which I don't understand.
1) I have to prove that $\prod_{k=2}^n \frac{k^3-1}{k^3+1}$ is convergent.
My first steps were:
$\prod_{k=2}^n \frac{k^3-1}{k^3+1}= \prod_{k=2}^n \frac{k-1}{k+1}\cdot \prod_{k=2}^n \frac{k^2+k+1}{k^2-k+1}=\frac{2(n-1)!}{(n+1)!}\prod_{k=2}^n \frac{k^2+k+1}{k^2-k+1}$
And then I got stuck. The author of the exercise does that:
$\frac{2(n-1)!}{(n+1)!}\cdot\prod_{k=2}^n \frac{k^2+k+1}{k^2-k+1} = \frac{2}{n(n+1)}\cdot\prod_{k=2}^n((k+1)^2-(k+1)+1)\cdot\prod_{k=2}^n\frac{1}{k^2-k+1}=$
$\frac{2}{n(n+1)}\cdot\prod_{k=3}^{n+1}(k^2-k+1)\cdot\prod_{k=2}^n\frac{1}{k^2-k+1}=\frac{2}{n(n+1)}\cdot\frac{(n+1)^2-(n+1)+1}{3}=\frac{2}{3}\cdot\frac{n^2+n+1}{n^2+n}\rightarrow\frac{2}{3}$
I understand what he does, but I don't know how he knew it. How can I use it for other exercises, so does somebody know how to remember this trick and for which kind of exercises this works?
2) I have to prove that $a_n:=\sqrt{n^2+n}-n$ is convergent and that a constant $A$ exist with $|a_n-a|<\frac{A}{n}.$
I proved that $a_n$ is convergent to $\frac{1}{2}$. For finding the constant $A$ the author says
It's easy to see, that: $|a_n -\frac{1}{2}|=|\frac{n}{\sqrt{n^2+n}+n}-\frac{1}{2}|<\frac{1}{8n}$, therefore $A:=\frac{1}{8n}$.
Unfortunately I don't see it. Can somebody please give a hint how he gets $\frac{1}{8}$?
thanks
| 1) The author uses a trick to compute the exact value of the infinite product but one does not need it to show the convergence only.
Namely, use that $\frac{k^3-1}{k^3+1}=1-\frac2{k^3+1}$, that $\frac2{k^3+1}<1$ for every $k\ge2$ and that the series $\sum\limits_{k}\frac2{k^3+1}$ converges absolutely.
2) Use the trick of the conjugate quantity, that is, $\sqrt{x}-\sqrt{y}=\frac{x-y}{\sqrt{x}+\sqrt{y}}$ for positive $x$ and $y$.
Thus, $a_n=\frac{n}{\sqrt{n^2+n}+n}$ (choose $x=n^2+n$ and $y=n^2$) hence $1-2a_n=\frac{\sqrt{n^2+n}-n}{\sqrt{n^2+n}+n}$, which is also $1-2a_n=\frac{n}{(\sqrt{n^2+n}+n)^2}$ (same $x$ and $y$).
Since $\sqrt{n^2+n}+n\ge2n$, this shows that $0\le 1-2a_n\le\frac1{4n}$, hence $2a_n\to1$ and, as the author writes, $\frac12-\frac1{8n}\le a_n\le\frac12$.
| {
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Solution to a system of linear equations in GF(2) Denote the system in $GF(2)$ as $Ax=b$, where:
$$
\begin{align}
A=&(A_{ij})_{m\times m}\\
A_{ij}=&
\begin{cases}
(1)_{n\times n}&\text{if }i=j\quad\text{(a matrix where entries are all 1's)}\\
I_n&\text{if }i\ne j\quad\text{(the identity matrix)}
\end{cases}
\end{align}
$$
that is, $A$ is a square matrix of order $m\times n$. And $b$ is a 0-1 vector with length $m\times n$. Now what is the solution of this system, if any, for a general pair of $m$ and $n$?
Example: For $m=2,n=3$ and $b=(0, 1, 0, 0, 1, 0)^T$, we have
$$
A=
\begin{pmatrix}
1 & 1 & 1 & 1 & 0 & 0 \\
1 & 1 & 1 & 0 & 1 & 0 \\
1 & 1 & 1 & 0 & 0 & 1 \\
1 & 0 & 0 & 1 & 1 & 1 \\
0 & 1 & 0 & 1 & 1 & 1 \\
0 & 0 & 1 & 1 & 1 & 1
\end{pmatrix}
$$
then one solution is $x=(1, 0, 1, 0, 1, 0)^T$
I know Gaussian elimination. I am trying but find it not very easy when dealing with a general case.
| Some observations, too long for a comment:
If $n$ is odd, your matrix is not invertible, and so there is no solution for arbitrary $b$ (and a solution will not be unique if it exists). First, do some row operations to rewrite the constituent blocks to $$\pmatrix{1&1&1&1\\0&0&0&0\\0&0&0&0\\0&0&0&0}, \pmatrix{1&0&0&0\\1&1&0&0\\1&0&1&0\\1&0&0&1}$$
Then do some column operations to rewrite the blocks to
$$\pmatrix{n\bmod 2&1&1&1\\0&0&0&0\\0&0&0&0\\0&0&0&0}, \pmatrix{1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1}$$
But if $n$ is odd, then columns $1$, $n+1$, $2n+1$, .. $(m-1)n+1$ are now all identical.
On the other hand, if $m$ is odd, then permuting the indices will turn the $m\times n$ problem into an $n\times n$ problem from the same family, and that will not be invertible either.
Some further progress in the even case (before I noticed Robert's elegant solution): We have the blocks
$$\pmatrix{0&1&1&1\\0&0&0&0\\0&0&0&0\\0&0&0&0}, \pmatrix{1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1}$$
Further column operations give
$$\pmatrix{0&1&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0}, \pmatrix{1&0&0&0\\0&1&1&1\\0&0&1&0\\0&0&0&1}$$
and by row operations we can blank out the new 1's to the right:
$$\pmatrix{0&1&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0}, \pmatrix{1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1}$$
Now everything decouples into one $2m\times 2m$ problem (containing the first two rows and columns of each block and $n-2$ separate $m\times m$ problems.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Linear diophantine equation $100x - 23y = -19$ I need help with this equation: $$100x - 23y = -19.$$ When I plug this into Wolfram|Alpha, one of the integer solutions is $x = 23n + 12$ where $n$ is a subset of all the integers, but I can't seem to figure out how they got to that answer.
| The continued fraction solution goes as follows:
Expand 100/23 into a continued fraction (I'm essentially using the GCD algorithm):
\begin{align*}
\frac{100}{23} & = 4 + \frac{8}{23}\\
& = 4 + \frac{1}{\frac{23}{8}}\\
& = 4 + \cfrac{1}{2 + \cfrac{7}{8}}\\
& = 4 + \cfrac{1}{2 + \cfrac{1}{\frac{8}{7}}}\\
& = 4 + \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}{7}}}
\end{align*}
Now that all the numerators are $1$, you're done getting the continued fraction, often written as the list of partial quotients like this:
$100/23 = [4,2,1,7]$
You can of course check: $4 + 1/(2 + 1/(1 + 1/7)) = 100/23$, the last equation above. You can change the $7$ at the end to $6 + 1/1$ to get an odd number of partial quotients but you don't have to.
Now if you look at $[4,2,1]$, which is the next to last convergent (the last being $[4,2,1,7]=100/23$) you get
$$4 + \frac{1}{2 + \frac{1}{1}} = \frac{13}{3}$$
The difference of cross product of the numerators and denominators of successive convergents is always $+1$ or $-1$, i.e:
$$100 \cdot 3 - 13 \cdot 23 = 1$$
If we multiply though by $-19$ we get:
$$100 \cdot (3 \cdot -19) - 23 \cdot (13 \cdot -19) = -19$$
so a particular solution $x_0,y_0$ is
\begin{align*}
x_0 & = 3 \cdot -19 = -57\\
y_0 & = 13 \cdot -19 = -247
\end{align*}
Since for all integer $t$
$$100 \cdot 23t - 23 \cdot 100t = 0$$
we can add that equation to
$$100 \cdot -57 - 23 \cdot -247 = -19$$
and get
$$100(23t-57) - 23(100t-247) = -19$$
so the general solution is
\begin{align*}
x & = 23t - 57\\
y & = 100t - 247
\end{align*}
To get the exact Wolfram answer change variables to $n = t-3$
$$x = 23(n+3)-57 = 23n+69-57 = 23n + 12$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The $3 = 2$ trick on Google+ I found out this on Google+ yesterday and I was thinking about what's the trick. Can you tell?
How can you prove $3=2$?
This seems to be an anomaly or whatever you call in mathematics. Or maybe I'm just plain dense.
See this illustration:
$$ -6 = -6 $$
$$ 9-15 = 4-10 $$
Adding $\frac{25}{4}$ to both sides:
$$ 9-15+ \frac{25}{4} = 4-10+ \frac{25}{4} $$
Changing the order
$$ 9+\frac{25}{4}-15 = 4+\frac{25}{4}-10 $$
This is just like $a^2 + b^2 - 2a b = (a-b)^2$. Here $a_1 = 3, b_1=\frac{5}{2}$ for L.H.S, and $a_2 =2, b_2=\frac{5}{2}$ for R.H.S. So it can be expressed as follows:
$$ \left(3-\frac{5}{2} \right) \left(3-\frac{5}{2} \right) =
\left(2-\frac{5}{2} \right) \left( 2-\frac{5}{2} \right) $$
Taking positive square root on both sides:
$$ 3 - \frac{5}{2} = 2 - \frac{5}{2} $$
$$ 3 = 2 .$$
I think it's something near the root.
| $2-(5/2)$ is not a positive square root.
| {
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Probability of collecting all 4 different items while picking 1 random item from the set a.) There are $4$ distinct items in the set. What is the probability of picking all $4$ items after picking $n\ge4$.
b.) How many items do you need to pick to collect all four with a probability of at least $.9$.
My answer for part a. (which I think is wrong):
The sample space is the all the possible ways of picking $n$ items from the set of $4$ items.
That number is equal to the number of non negative integer-valued solutions to this problem:
$a_1+a_2+a_3+a_4=n$
that number is $\binom{n+4-1}{4-1}=\binom{n+3}{3}$
All possible ways of picking $n$ items and having at least 1 of each of the $4$ items is the number of positive integer-valued solutions to this problem:
$a_1+a_2+a_3+a_4=n$
That number is $\binom{n-1}{4-1}=\binom{n-1}{3}$
So the probability of picking all $4$ items after picking $n$ items is:
$\frac{\binom{n-1}{3}}{\binom{n+3}{3}}$
Using this answer, I find that the probability of picking all $4$ items after picking $4$ random items is:
~$.02857\ne\frac{4}{4}\frac{3}{4}\frac{2}{4}\frac{1}{4}$
Please explain where I went wrong.
| The original problem is incompletely specified. And one problem with a "stars and bars" approach is that not all solutions are equally likely, making the calculation of probabilities very difficult.
We consider the following equivalent situation. An experiment consists of picking one of the digits $1$, $2$, $3$, $4$, with each choice equally likely. (a) If we perform this experiment $n$ times, what is the probability that we have chosen each of $1$, $2$, $3$, $4$ at least once? (b) What is the smallest $n$ such that if we perform the experiment $n$ times, the probability of having picked each digit at least once is at least $0.9$?
For (a), represent the result of the $n$ experiments as a word of length $n$ over the alphabet $\{1,2,3,4\}$. There are $4^n$ such words, all equally likely. Now we count the words that have all four digitss. As is often the case, it is a little easier to count the complement, the words in which not all of $1$, $2$, $3$, and $4$ appear.
To do this, we will use Inclusion-Exclusion. The number of words that do not have a $1$ is $3^n$. So is the number of words that do not have a $2$, and so on. Add up, for a total of $4\cdot 3^n$. This doesn't quite work, since we have double counted, for example, the words that have neither a $1$ nor a $2$. There are $2^n$ of these. But there are $\binom{4}{2}=6$ double counted patterns. This gives us the estimate $4\cdot 3^n -6\cdot 2^n$ for the number of words with at least one missing digit. However, we have subtracted once too many times the words with just the digit $1$, or just $2$, and so on. So the total number of words with at least one missing digit is
$$4\cdot 3^n -6\cdot 2^n +4\cdot 1^n.$$
It follows that the required probability is
$$1 -\frac{4\cdot 3^n -6\cdot 2^n +4\cdot 1^n}{4^n}.$$
(b) We want the smallest integer $n \ge 4$ such that
$$\frac{4\cdot 3^n -6\cdot 2^n +4\cdot 1^n}{4^n} \lt 0.1.$$
The dominant term is $(4\cdot 3^n)/4^n$. A little calculator work shows that the smallest $n$ that works is given by $n=13$.
| {
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How to prove those "curious identities"? How to prove
$$ \prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right) = \frac{n}{2^{n-1}}$$
and
$$ \prod_{k=1}^{n-1} \cos\left(\frac{k\pi}{n}\right) = \frac{\sin(\pi n/2)}{2^{n-1}}$$
| Denote $w = e^{i \pi/n}$. We have
$$\prod_{k = 1}^{n-1} \sin \left(\frac{k\pi}{n}\right)= \prod_{k = 1}^{n-1} \frac{w^k - w^{-k}}{2i} = \frac{1}{2^{n-1}} \prod_{k = 1}^{n-1} \frac{w^k}{i} (1-w^{-2k})$$
Since we have
$$\sum_{k = 0}^{n-1} x^k = \prod_{k = 1}^{n-1} (x-w^{2k})$$
Setting $x=1$ yields
$$\prod_{k = 1}^{n-1} (1-w^{2k}) = n$$
So we get
$$\prod_{k = 1}^{n-1} \sin \left(\frac{k\pi}{n}\right)= \frac{n}{2^{n-1}} \frac{w^{n(n-1)/2}}{i^{n-1}} = \frac{i^{n-1}}{i^{n-1}} \frac{n}{2^{n-1}} = \frac{n}{2^{n-1}}$$
I guess (but did not check) that the same kind of reasoning gives the one with $\cos$.
| {
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Wrong sign on the final answer, where did I go wrong? Algebra $12x^2-12x = 0$
This was suppose to be 18, but it is -18.
I think it is because of how I modified one of the factors.
$\sqrt{x^2+x^2}$
$\sqrt{2x^2}$
$\sqrt{2}\cdot \sqrt{x^2}$
$x\sqrt{2}$
If I let wolfram do this, it gives the same but it says "assuming X is positive". Isn't that redundant? Any negative number squared is positive? The problem could be with the other factor that I modified.
$\sqrt{(6-x)^2+(6-x)^2}$
$x\sqrt2 -6\sqrt2$
To sum it up:
$x\sqrt2(x\sqrt2-6\sqrt2) = 2x^2-12x$
It should be $-2x^2+12x$, because that would indeed give positive 18 as it's answer.
Update! With all the info!
The task is to find x such that the area of the rectangle would be as big as possible. I used the Pythagorean theorem and wrote out the following equation.
$\sqrt{(6-x)^2+(6-x)^2} \cdot \sqrt{x^2+x^2} = A$ of the rectangle.
The left factor would be the long side of the rectangle and the right factor would be the short side. I need to write this out nicely and then find when the derivative is 0. For max area.
| As you point out, by the Pythagorean Theorem, the area of the rectangle can be written as
$$\sqrt{2x^2}\sqrt{2(6-x)^2}.$$
Note that from the geometry, we have $0 < x < 6$.
It follows that $\sqrt{2x^2}=\sqrt{2}\;x$ and $\sqrt{2(6-x)^2}=\sqrt{2}\;(6-x)$.
Thus, if $A(x)$ is the area, then
$$A(x)=2x(6-x).$$
(Incidentally, we can find the area in a more basic way. The two triangles at the Northeast and Southwest have combined area $x^2$, and the other two triangles have combined area $(6-x)^2$, for a total of $x^2+(6-x)^2=2x^2-12x+36$. We can now minimize this, or subtract from $6^2$ and conclude that $A(x)=12x-2x^2$.)
Maximize, using any of the standard tools. We could use the calculus, but in part because of the precalculus tag, we complete the square.
We have
$$A(x)=-2(x^2-6x)=-2((x-3)^2+9=18-2(x-3)^2.$$
The maximum is reached when $(x-3)^2=0$, that is, when $x=3$. The value of the area at $x=3$ is $18$.
Comment on error: The mistake in the post was in the simplification of $\sqrt{(6-x)^2+(6-x)^2}$.
You obtained $x\sqrt{2} -6\sqrt{2}$, and the correct value is $6\sqrt{2}-x\sqrt{2}$. The expression $x\sqrt{2}-6\sqrt{2}$ cannot be right, since for $\;0<x <6$, it gives a negative number. Maybe if you had written it as $\sqrt{2}(x-6)$ the mistake would have become obvious.
We are trying to find a certain length, the square root of $2(6-x)^2$. The square root of any (non-negative) number is non-negative.
A rule that one might want to remember is that $\sqrt{a^2}=|a|$. So our square root is $\sqrt{2}|6-x|$. Since $0 \le x \le 6$ by the geometry, $|6-x|=6-x$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Did I tackle this implicit differentiation problem correctly? $$2x^3 + x^2y-xy^3 = 2$$
$$\frac{\mathrm{d}}{\mathrm{d}x} [2x^3+x^2y -xy^3 ] = \frac{\mathrm{d}}{\mathrm{d}x}(2)$$
$$6x^2 + \left(2xy + x^2\frac{\mathrm{d}y}{\mathrm{d}x}\right) - \left( 1 y^3 + 3y6^2 \frac{\mathrm{d}y}{\mathrm{d}x}\right ) = 0$$
$$\frac{\mathrm{d}y}{\mathrm{d}x} (x^2 +3y^2)(6x^2+2x-y^3) = 0$$
$$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{6x^2+2x-y^3}{x^2+3y^2} $$
Did I tackle this question correctly?
| $2x^3 + x^2y-xy^3 = 2$
$$\frac{\mathrm{d}}{\mathrm{d}x} [2x^3+x^2y -xy^3 ] = \frac{\mathrm{d}}{\mathrm{d}x}(2)$$
$$6x^2 + \left(2xy + x^2\frac{\mathrm{d}y}{\mathrm{d}x}\right) - \left( 1 y^3 + 3y^2x \frac{\mathrm{d}y}{\mathrm{d}x}\right ) = 0$$
$$\frac{\mathrm{d}y}{\mathrm{d}x} (x^2 - 3y^2x) + (6x^2+2xy-y^3) = 0$$
$$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{6x^2+2xy-y^3}{3y^2x -x^2} $$
| {
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"url": "https://math.stackexchange.com/questions/75596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A "fast" way for computing $ \prod \limits_{i=1}^{45}(1+\tan i^\circ) $?
Which is the fastest paper-pencil approach to compute the product $$ \prod
\limits_{i=1}^{45}(1+\tan i^\circ) $$
| Using
$$
1+\tan x = \frac{\sin x + \cos x}{\cos x} = \frac{\sqrt{2} \cos (45^{\circ} - x)}{\cos x},
$$
the product can be written as:
$$
\prod_{x=1}^{45}(1+\tan x^\circ) = 2^{45/2} \prod_{x=1}^{45} \frac{\cos (45 - x)^{\circ}}{\cos x^{\circ}} \stackrel{(1)}{=} 2^{45/2} \cdot \frac{\prod\limits_{x=0}^{44} \cos x^{\circ}}{\prod\limits_{x=1}^{45} \cos x^{\circ}} \stackrel{(2)}{=} 2^{45/2} \cdot \frac{\cos 0}{\cos 45^{\circ}} = 2^{23},
$$
where we
*
*reindexed the product in the numerator, and
*cancelled the common factors.
Another approach. If $x+y = 45^{\circ}$, then
$$
1 = \tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y},
$$
which rearranges to
$$
\tan x \tan y + \tan x + \tan y = 1 \quad \implies \quad (1+\tan x)(1+\tan y) = 2.
$$
Now plug in $x = 0^{\circ}, 1^{\circ}, 2^{\circ}, \ldots, 45^{\circ}$, so that $y$ takes the same values but in the opposite order. Multiplying all these equations, we get
$$
\left[ \prod_{x=0}^{45} (1+\tan x^\circ) \right]^2 = 2^{46}.
$$
Taking square-roots and noting that $1+\tan 0^\circ = 1$, we get the answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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Problem with generating functions and binary recurrences I am considering the following recurrence:
$a_0 = 1$;
$a_1 = 2$
$a_{n} = 2 (a_{n - 1} + a_{n - 2})$
Then I proceeded with the generating function:
$F(x) = \displaystyle\sum_{n = 0}^\infty a_n x^n = 1 + 2x + \displaystyle\sum_{n = 2}^\infty a_{n} x^{n} = 1 + 2x + \displaystyle\sum_{n = 2}^\infty 2x^n(a_{n - 1} + a_{n - 2})$
$F(x) = 1 + 2x + \displaystyle\sum_{n = 2}^{\infty} 2x^{n} a_{n - 1} + \displaystyle\sum_{n = 2}^{\infty} 2x^{n} a_{n - 2}$
$F(x) = 1 + 2x + (2x \displaystyle\sum_{n = 2}^{\infty} x^{n - 1} a_{n - 1}) + (2x^{2} \displaystyle\sum_{n = 2}^{\infty} x^{n - 2} a_{n - 2})$
$F(x) = 1 + 2x + 2x(F(x) - 1) + 2x^{2}F(x)$
$F(x) = \frac{1}{1 - 2x - 2x^{2}}$ Let a, b be the roots of the quadratic.
$F(x) = \frac{1}{(x - a)(x - b)} = \displaystyle\sum_{n = 0}^{\infty} \frac{x^{n}(b^{-1 - n} - a^{-1 - n})}{\sqrt{3}}$
We should then have $a_{n} = \frac{b^{-1 - n} - a^{-1 - n}}{\sqrt{3}}$, but I know that this is false. Where have I gone wrong?
| Let be $a(n)$ a geometric progression, or $a(n)=q^n$
So,
$a(n)=2a(n-1)+2a(n-2)$ becomes
$q^n=2q^{n-1}+2q^{n-2}$
so $q=0$ or
$q^2=2q+2$
So, $q_1=\frac{2+\sqrt{12}}{2}=1+\sqrt{3}$ or $q_2=\frac{2-\sqrt{12}}{2}=1-\sqrt{3}$.
and
$a(n)=bq_1^n+cq_2^n$
$a(0)=1$, $b+c=1$
$a(1)=2$
| {
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Finding smallest root of equation By trial and modification, I want to find the smallest root of this equation:
$-2\tan\left(\frac{a}{2}\right)=\tan(a)$
and the following as well:
$-\tan\left(\frac{a}{\sqrt2}\right)=\sqrt2\tan({a/2})$
| Hint:
If $\frac{a}{2}=b$, then you equation became:
$-2\tan(b)-\tan(2b)=0$
But,
$\tan(2b)=\frac{2\tan(b)}{1-\tan^2(b)}$
so we have:
$-2\tan(b)-\frac{2\tan(b)}{1-\tan^2(b)}=0 \iff$
$\frac{-2tan(b)(1-tan^2(b))}{1-tan^2(b)}-\frac{2tan(b)}{1-tan^2(b)}=0$
The denominator couldn't be 0, so:
$1-\tan^2(b) \neq 0 \iff \tan^2(b) \neq 1 \iff (\tan(b) \neq -1 \quad or \quad \tan(b) \neq -1)$
Then the equation become:
$\tan^3(b)=2\tan(b) \iff \tan(b)=0 \quad or \quad \tan(b)=-\sqrt{2} \quad or \quad \tan(b)=\sqrt{2}$
But $a=2b$, so we get:
$\tan(2b)=\frac{2\tan(b)}{1-\tan^2(b)}$
$\tan(a)=\tan(2b)=0 \quad or -2\sqrt{2} \quad or \quad 2\sqrt{2}$
$\cdots$
| {
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Tricky congruence question Let $p>5$ be a prime,
$$n\triangleq\frac{4^p+1}{5}\;\quad\text{and}\;\quad b\triangleq\frac{n-1}{4}\quad.$$
It can be shown that both are integers, and also that $n$ is composite, $b$ is odd and $p$ divides $b$.
I'd like to prove that
$$4^b\equiv-1\pmod n\;\quad.$$
I tried rephrasing the congruence in terms of the other variables to see if I could find a way out by applying Fermat's — or even Euler's — Theorem, but I'm not getting anywhere.
| We will show the slightly stronger $$ 4^b \equiv -1 \pmod{4^p + 1}. $$ Fermat's theorem implies that $4^{p-1} \equiv 1 \pmod{p}$. Since $b = (4^{p-1}-1)/5$, it follows that $p|b$ (since $p > 5$). Suppose that $b = Xp$. Clearly $b$ is odd, therefore so is $X$. We have
$$\begin{align*} 4^{Xp} + 1 &= (4^{Xp} + 4^{(X-1)p}) - (4^{(X-1)p} - 4^{(X-2)p}) + \cdots + (4^p + 1) \\ &= (4^{(X-1)p} - 4^{(X-2)p} + \cdots + 1)(4^p + 1). \end{align*} $$ Here we used the fact that $X$ is odd. It follows that $4^p + 1$ divides $4^b + 1$.
| {
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How to prove a fact about the sum of three squares? How would I go about proving the following?
If $a$, $b$, $c$, $n$ are positive integers, then
$a^2+b^2+c^2 \neq 2^nabc$
I tried doing something similar to the proof for Adrien-Marie Legendre's Three Square theorem: $a^2+b^2+c^2=n$ iff there are not integers $k$, and $m$ so that $n=4^k(8m+7)$. It didn't quite work out...
$2^nabc$ is always even. So if $a^2+b^2+c^2 = 2^nabc$, then $a^2+b^2+c^2$ must be even.
That means there is $a_1$, $b_1$, $c_1$ so that
$a = 2a_1$, $b = 2b_1$, and $c = 2c_1$
So
$(2a_1)^2+(2b_1)^2+(2c_1)^2 = 2^nabc \rightarrow 2(2a_1^2+2b_1^2+2c_1^2)= 2^nabc$
and we get $2a_1^2+2b_1^2+2c_1^2= 2^{n-1}abc$
We can continue to do this procedure with $a_2$, $b_2$, $c_2$ then $a_3$, $b_3$, $c_3$ then ... $a_n$, $b_n$, $c_n$.
With $a_n$, $b_n$, $c_n$ we'd get
$2^na_n^2+2^nb_n^2+2^nc_n^2= 2^{n-n}abc=abc$
Since $a_n=2a_{n-1}$ and $a_0=a$,
$a_n = \frac{a}{2^n}$
and we get
$2^n(\frac{a}{2^n})^2+2^n(\frac{b}{2^n})^2+2^n(\frac{c}{2^n})^2=abc$
This just becomes the original equation.
$a^2+b^2+c^2 = 2^nabc$
| Since the right-hand side is even, either exactly one or all three of $a,b,c$ must be even.
The former case is impossible, as you can easily see by taking both sides mod 4.
In the latter case, let $2^k$ be the greatest power of 2 in the GCD of $a,b,c$. Then
$$\left(\frac{a}{2^k}\right)^2 + \left(\frac{b}{2^k}\right)^2 + \left(\frac{c}{2^k}\right)^2 = 2^{n+k} \frac{a}{2^k} \frac{b}{2^k} \frac{c}{2^k},$$
with at least one of the terms on the left-hand side odd, and we are back in case 1.
EDIT: Note that $n>0$ is essential. When $n=0$,
$$3^2 + 3^2 + 6^2 = 54 = 3\cdot3\cdot6.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/78122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Finding the slope of the tangent line to $\frac{8}{\sqrt{4+3x}}$ at $(4,2)$ In order to find the slope of the tangent line at the point $(4,2)$ belong to the function $\frac{8}{\sqrt{4+3x}}$, I choose the derivative at a given point formula.
$\begin{align*}
\lim_{x \to 4} \frac{f(x)-f(4)}{x-4} &=
\lim_{x \mapsto 4} \frac{1}{x-4} \cdot \left (\frac{8}{\sqrt{4+3x}}-\frac{8}{\sqrt{4+3 \cdot 4}} \right )
\\ \\ & = \lim_{x \to 4} \frac{1}{x-4} \cdot \left ( \frac{8}{\sqrt{4+3x}}-\frac{8}{\sqrt{16}} \right ) \\ \\ & = \lim_{ x \to 4} \frac{1}{x-4} \cdot \left ( \frac{8}{\sqrt{4+3x}}-2\right)
\end{align*}$
But now I can't figure it out, how to end this limit.
I know that the derivative formula for this function is $-\frac{12}{(4+3x)\sqrt{4+3x}}$.
Thanks for the help.
| HINT:
$$
\begin{eqnarray}
\frac{1}{x-4} \left ( \frac{8}{\sqrt{4+3x}}-2\right) &=&\frac{1}{x-4} \left ( \frac{8}{\sqrt{4+3x}}-2\right) \left ( \frac{8}{\sqrt{4+3x}}+2\right) \left ( \frac{8}{\sqrt{4+3x}}+2\right)^{-1} \\
&=& \frac{1}{x-4} \left( \frac{64}{4+3x} -4 \right) \left ( \frac{8}{\sqrt{4+3x}}+2\right)^{-1} \\
&=& \frac{1}{x-4} \left( \frac{-12(x-4)}{4+3x} \right) \left ( \frac{8}{\sqrt{4+3x}}+2\right)^{-1} \\ &=& \left( \frac{-12}{4+3x} \right) \left ( \frac{8}{\sqrt{4+3x}}+2\right)^{-1}
\end{eqnarray}
$$
Can you find the limit now ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/80010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
Why this limit can't be computed like this? This is the limit:
$$\lim_{x \to 2} \frac{3x^2-x-10}{x^2-4}$$
The solution manual says it is: $\frac{11}{4}$
I've tried to solve it like a polynomial like this:
$$ \frac{(\frac{3x^2}{3x^2}-\frac{x}{3x^2}-\frac{10}{3x^2})*3x^2}{(\frac{x^2}{x^2}-\frac{4}{x^2})*x^2}=$$
$$= \frac{(1-0-0)*3x^2}{(1+0)*x^2}=3$$
Can you please tell me where am I doing wrong? Or why these kind of operation doesn't work here? Thank you
| For this problem, you'll want to factor the numerator and the denominator, (which both tend to $0$ as $ x \to 2$), cancel a factor of $x-2$, and then try direct substitution again.
$\frac{3x^2 -x - 10}{x^2-4} = \frac{(3x +5)(x - 2)}{(x + 2)(x - 2)} = \frac{3x + 5}{x + 2}$ (when $x \neq 2$)
Now letting $x \to 2$, we get $\frac{3 \cdot 2 + 5}{2 + 4} = \frac{11}{4}$, as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/81145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding Complex Number $z$ in $\frac{z+2i}{z-2i}=\frac{7-6i}{5}$
What I did: Cross Multiply, try to expand out the mod and args, but they all seem to lead to dead end (probably I am not seeing something)
| All your equations are correct but apparently your main difficulty was how
to find $z$ in the algebraic form $a+bi$ from the given equation, which you
have shown is equivalent to the linear equation in $z$
$$
2z-12-(24+6z)i=0.
$$
To solve this equation add the terms in $z$ and separately the independent
terms.
$$
\left( 2-6i\right) z-(12+24i)=0.
$$
Move the independent term to the RHS.
$$
\left( 2-6i\right) z=12+24i.
$$
To find $z$ divide both sides by the coefficient of $z$, multiply both
numerator and denominator by the conjugate of the denominator and simplify
$$
\begin{eqnarray*}
z &=&\frac{12+24i}{2-6i}=\frac{6+12i}{1-3i}=\frac{\left( 6+12i\right) \left(
1+3i\right) }{\left( 1-3i\right) \left( 1+3i\right) } \\
&=&\frac{-30+30i}{10}=-3+3i.
\end{eqnarray*}
$$
So $a=-3,b=3$. You can easily find the modulus of $z$
$$
\left\vert z\right\vert =\sqrt{a^{2}+b^{2}}=\sqrt{(-3)^{2}+3^{2}}=3\sqrt{2}.
$$
Its argument $\theta =\arg z$ is the angle in the 2nd quadrant$^1$ such that $\tan \theta =b/a$. The range of the arctangent function is $]−\pi/2, \pi[$. Then
$$
\arg z=\pi+\arctan \left( \frac{b}{a}\right) =\pi+\arctan \left( \frac{3}{-3}\right)
=\pi+\arctan \left( -1\right) =\pi-\frac{1}{4}\pi=\frac{3}{4}\pi .
$$
As for $w$ you do not need to find its algebraic form. You can apply the
properties of the modulus and argument of a complex fraction. Since
$$
\left\vert \frac{2z}{w}\right\vert =\frac{\left\vert 2z\right\vert }{
\left\vert w\right\vert }=\frac{2\left\vert z\right\vert }{\left\vert
w\right\vert }=\frac{6\sqrt{2}}{\left\vert w\right\vert }=3,
$$
solving for $\left\vert w\right\vert $ you get $\left\vert w\right\vert =
\frac{6\sqrt{2}}{3}=2\sqrt{2}$, while from the following equation
$$
\arg \left( \frac{w}{z}\right) =\arg \left( w\right) -\arg \left( z\right)
=\arg \left( w\right) -\frac{3}{4}\pi =-\frac{1}{8}\pi
$$
you find $\arg \left( w\right) =-\pi /8+3\pi /4=5\pi /8$.
$^1$ added and corrected
| {
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"url": "https://math.stackexchange.com/questions/84947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Two Taylor expansions of $\frac{1}{1+\sqrt{2-z}}$ about $z=0$
How do you start expanding this function $$f(z)= \frac{1}{1+\sqrt{2-z}}$$ into two Taylor expansions about $z=0$?
The best I came up is to let $u=\sqrt{2-z}$ and then expand $f(z)$ as a geometric series.
| Alternatively, with some algebraic manipulations you get at:
$$
f(z) = \frac{1}{1+\sqrt{2-z}} = \frac{1-\sqrt{2-z}}{\left(1+\sqrt{2-z}\right)\left(1-\sqrt{2-z}\right)} = \frac{1-\sqrt{2-z}}{z-1} = \frac{\sqrt{2-z}-1}{1-z}
$$
Suppose, you worked out $\sqrt{2-z}-1 = \sum_{n=0}^\infty c_n z^n$, then
$$
\frac{\sqrt{2-z}-1}{1-z} = \sum_{n=0}^\infty \left( \sum_{m=0}^n c_m \right) z^n
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/85524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
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