$\sqrt[3]{31}$ is about $3.14138$. Why is this so close to $\pi$?

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    $\begingroup$ You could have just said $\sqrt[3]{31}\approx3.1414$. $\endgroup$ – blue Mar 12 '14 at 4:32
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    $\begingroup$ 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 Martin Mar 12 '14 at 4:35
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    $\begingroup$ @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? $\endgroup$ – wckronholm Mar 12 '14 at 4:46
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    $\begingroup$ "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 '14 at 16:17
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    $\begingroup$ This strikes me as an odd question. It's like asking "why is 3.15 so close to 3.14?" $\endgroup$ – Alec Feb 23 '16 at 13:45

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


is small we just drop it.


I don't believe you . Can you prove it or provide a reference?

First off you asked for some references



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.

  • $\begingroup$ I don't believe you . Can you prove it or provide a reference? $\endgroup$ – chubakueno Mar 13 '14 at 0:27
  • $\begingroup$ @chubakueno references provided as you ask. I can maybe dig up a few more if you need them. $\endgroup$ – bobbym Mar 13 '14 at 8:41
  • $\begingroup$ Never knew about this series. interesting.. $\endgroup$ – user85798 Mar 13 '14 at 8:50
  • $\begingroup$ How did you get that? $\endgroup$ – Agnishom Chattopadhyay Dec 26 '16 at 5:50
  • $\begingroup$ 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. $\endgroup$ – bobbym Dec 26 '16 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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    $\begingroup$ $243 = 3^5$ by the way $\endgroup$ – Mr Pie May 3 '18 at 14:00

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 :


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 $$\pi^3\approx 32$$

  • $\begingroup$ but the question is why $\pi^3\approx 31$ $\endgroup$ – Jaume Oliver Lafont Jun 10 '17 at 8:26
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    $\begingroup$ @JaumeOliverLafont $32\approx 31$. $\endgroup$ – Mr Pie Feb 11 '18 at 1:49

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