10
$\begingroup$

What is the origin of the Ramanujan's approximate identity $$\pi^4\approx 2143/22,\;\;\tag 1$$ which is valid with $10^{-9}$ relative accuracy? For comparison, the relative accuracy of the well known $\pi\approx 22/7$ is only $4\cdot10^{-4}$ and in this case we have the identity $$\frac{22}{7} - \pi = \int_0^1 (x-x^2)^4 \frac{dx}{1+x^2}, \tag{2}$$ which explains why the difference is small (concerning this identity, see Source and context of $\frac{22}{7} - \pi = \int_0^1 (x-x^2)^4 dx/(1+x^2)$?).

Of course, (1) can be rewritten in the form $$\zeta(4)\approx 2143/1980,$$ so maybe some fast convergent series for $\zeta(4)$ can be used to get this approximate identity (in the case of $\frac{22}{7}-\pi$, a series counterpart of (2) is $$\sum_{k=0}^\infty \frac{240}{(4k+5)(4k+6)(4k+7)(4k+9)(4k+10)(4k+11)}=\frac{22}{7}-\pi$$ - see Source and context of $\frac{22}{7} - \pi = \int_0^1 (x-x^2)^4 dx/(1+x^2)$?).

P.S. I just discovered that this question was discussed in https://math.stackexchange.com/questions/1359015/is-there-an-integral-for-pi4-frac214322 and in https://math.stackexchange.com/questions/1649890/is-there-a-series-to-show-22-pi42143 is anything to add to the answers given there?

$\endgroup$
  • 2
    $\begingroup$ This is convergent of the continued fraction starting $[97; 2, 2, 3, 1, 16539, 1, 6, 3]$. The large $16539$ might explain it. $\endgroup$ – joro Feb 26 '16 at 8:51
  • 16
    $\begingroup$ Large 16539 is a question, not an answer. $\endgroup$ – Fedor Petrov Feb 26 '16 at 9:20
15
$\begingroup$

I think Ramanujan's thought was very simple. He calculated the decimal expansion of $\pi^4$ and he got: $$\pi^4 = 97.409091034... \approx 97.4090909...= 97.4 +1/110$$
And then: $$ 97.4 + 1/110 = 10715/110 = 2143/22$$

$\endgroup$
  • 3
    $\begingroup$ This is a very nice observation and leads to a question why $10\pi^4-1/11\approx 974.0000012$ is a near integer. $\endgroup$ – Zurab Silagadze Feb 26 '16 at 11:22
  • 1
    $\begingroup$ How do you know he looked at patterns in the decimal expansion instead of finding rational approximations in the usual way (as joro said in a comment to the question, $2143/22$ is just a convergent to the continued fraction)? $\endgroup$ – Jeppe Stig Nielsen Feb 26 '16 at 17:23
  • 2
    $\begingroup$ I think it's the most natural way. I found very good article : Ramanujan for lowbrows. Berndt and Bhargava discussed about OP and other problems. $\endgroup$ – user68208 Feb 27 '16 at 0:16
  • $\begingroup$ We may want to add another unit fraction to obtain $5\pi^4-\frac{1}{2·11}-\frac{1}{2^9 5^5} \approx 486.99999999955$ $\endgroup$ – Jaume Oliver Lafont Mar 10 '16 at 16:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.