why is $\sqrt[3]{31}$ so close to $\pi$?

Question

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

Answer 1

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.

Answer 2

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 $$ \pi^{10}=93555\cdot\frac{1025}{1024}\left(1+\frac{1}{3^{10}}+\cdots\right)=93648.047\ldots $$

Answer 3

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.

Answer 4

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 $$\pi^3\approx 32$$