$\sqrt[3]{31}$ is about $3.14138$. Why is this so close to $\pi$?
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5$\begingroup$ You could have just said $\sqrt[3]{31}\approx3.1414$. $\endgroup$– blueMar 12, 2014 at 4:32
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21$\begingroup$ because there are a hundred and seven bazillion constants in the world and only so much room for all of them. a bunch of random numbers is going to contain some that are close to each other. $\endgroup$– Greg MartinMar 12, 2014 at 4:35
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2$\begingroup$ @LuciusTarquiniusSuperbus But what was your process? Were you just plugging random numbers in and seeing what comes out? Or was there some more intentional approach? $\endgroup$– wckronholmMar 12, 2014 at 4:46
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3$\begingroup$ "Close" is a matter of scale. Why is $\sqrt{1776}$ so close to $42$? Is the American revolution really is "The Answer"? $\endgroup$– Asaf Karagila ♦Mar 13, 2014 at 16:17
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2$\begingroup$ This strikes me as an odd question. It's like asking "why is 3.15 so close to 3.14?" $\endgroup$– AlecFeb 23, 2016 at 13:45
4 Answers
This series is the reason:
$$ \frac{\pi^3}{32} = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^3} $$
Now just truncate the series at the third term and multiply both sides by 32.
$$\pi^3\approx 32-\frac{32}{27}+\frac{32}{125}=31 + \left(\frac{32}{125}-\frac{5}{27}\right)$$
Now because
$$\frac{32}{125}-\frac{5}{27}=0.0708148$$
is small we just drop it.
@chubakueno
I don't believe you . Can you prove it or provide a reference?
First off you asked for some references
http://en.wikipedia.org/wiki/List_of_formulae_involving_%CF%80
http://www.dansmath.com/pages/pipage.html
Enter this at Alpha "Sum[(-1)^n / (2n+1)^3, {n,0,infinity}]//FullSimplify"
From the book Integrals and Series Vol 1 by Prudnikov, Brychkov, Marichev. p653 #2
I do not have a proof but suspect it might be possible using a Fourier series. Anyway, it does not belong in this thread so maybe you should open up another thread and ask the question about whether the series quoted sums to what the references say.
Castellano gives:
$$\pi^3 \approx \left ( 31+\frac{62^2+14}{28^4} \right )$$
An amazing approximation and appears to be done empirically. Here the fraction is 10 times smaller then in the other example. Again, we can just drop it. It appears we can come up with lots of these.
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$\begingroup$ I don't believe you . Can you prove it or provide a reference? $\endgroup$ Mar 13, 2014 at 0:27
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$\begingroup$ @chubakueno references provided as you ask. I can maybe dig up a few more if you need them. $\endgroup$– bobbymMar 13, 2014 at 8:41
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$\begingroup$ Never knew about this series. interesting.. $\endgroup$ Mar 13, 2014 at 8:50
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$\begingroup$ Hi; Was just looking around for some approximations of pi and remembered this one. We could discuss it in the chatroom because comments are not for discussions. $\endgroup$– bobbymDec 26, 2016 at 6:30
We have the following series $$\pi^6-31^2=3\sum_{k=0}^\infty \left(\frac{77}{(2k+3)^6}+\frac{243}{(2k+5)^6}\right)$$
(see https://math.stackexchange.com/a/1651175/134791)
The difference is close to zero because the terms of the summation are small.
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1
Disclaimer: Not totally relevant answer.
Even better approximation $$ \sqrt[10]{93648}=3.14159248\ldots $$ due to the fact that $$ \frac{\pi^{10}}{93555}=\sum_{n=1}^\infty\frac{1}{n^6}=\frac{1025}{1024}\sum_{n=1}^\infty\frac{1}{(2k+1)^{10}}, $$ and hence $$ \pi^{10}=93555\cdot\frac{1025}{1024}\left(1+\frac{1}{3^{10}}+\cdots\right)=93648.047\ldots $$
Say that you want to approximate $\Gamma\left(\frac14\right)$, you know that
$$\int_0^1 t^{x-1}(1-t)^{y-1}\;\mathrm{d}x=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$
and / also
$$\int_0^{\pi/2} \sin^{2n-1}\theta\cos^{2m-1}\theta\;\mathrm{d}\theta=\frac12\frac{\Gamma(n)\Gamma(m)}{\Gamma(n+m)}$$
Specially for $m=1/2$:
$$\int_0^\pi \sin^{2n-1}\theta\;\mathrm{d}\theta=\frac{\Gamma(n)\Gamma(1/2)}{\Gamma(n+1/2)}=\sqrt{\pi}\frac{\Gamma(n)}{\Gamma(n+1/2)}$$
Say you want to aproximate the function $\sin\theta$ on $(0,\pi)$ with the function : $a\,\theta(\pi-\theta)$ so that the latter integral is exact for some chosen $n$ (you want to find $a$) As we can see, the relation is satysfied for all $a$ for $n=1/2$ trivially. By substitution $\theta = \pi t$ :
$$\int_0^\pi \left[a\,\theta(\pi-\theta)\right]^{2n-1}\;\mathrm{d}\theta=a^{2n-1}\pi^{4n-1}\frac{\Gamma^2(n)}{\Gamma(2n)}$$
Lets say the integral is exact for $n=1$ i.e.
$$a\pi^{3}\frac{\Gamma^2(2)}{\Gamma(4)} = \sqrt{\pi}\frac{\Gamma(1)}{\Gamma(1+1/2)}$$
Solving for $a$ gives us :
$$a=\frac{2}{\pi^3}$$
So, the aproximate relation now is :
$$\sqrt{\pi}\frac{\Gamma(n)}{\Gamma(n+1/2)} \approx \frac{2^{2n-1}}{\pi^{2n-2}}\frac{\Gamma^2(n)}{\Gamma(2n)} $$
Which turns into equality when $n=1$ or trivially when $n=1/2$
When we insert $n=3/4$ and $5/4$ to the approximate formula and with the help of the Euler reflection formula we get $$\pi^3\approx 32$$
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$\begingroup$ but the question is why $\pi^3\approx 31$ $\endgroup$ Jun 10, 2017 at 8:26
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