# The origin of the Ramanujan's $\pi^4\approx 2143/22$ identity

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?

• This is convergent of the continued fraction starting $[97; 2, 2, 3, 1, 16539, 1, 6, 3]$. The large $16539$ might explain it.
– joro
Commented Feb 26, 2016 at 8:51
• Large 16539 is a question, not an answer. Commented Feb 26, 2016 at 9:20
• You get an even larger second term in the continued fraction for 22 pi^4 = [2143, 363893, 1, ...].
– Max
Commented Jun 21, 2022 at 16:04

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: $$97.4 + 1/110 = 10715/110 = 2143/22$$

• This is a very nice observation and leads to a question why $10\pi^4-1/11\approx 974.0000012$ is a near integer. Commented Feb 26, 2016 at 11:22
• 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)? Commented Feb 26, 2016 at 17:23
• 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.
– user68208
Commented Feb 27, 2016 at 0:16
• We may want to add another unit fraction to obtain $5\pi^4-\frac{1}{2·11}-\frac{1}{2^9 5^5} \approx 486.99999999955$ Commented Mar 10, 2016 at 16:41
• that looks very probable but actually I found a source which explains the different way Ramanujan found it, see my answer
– Max
Commented Jun 21, 2022 at 16:14

Perhaps the most intuitive way of explaining this identity is with the help of a continued fraction $$97+\frac{1}{2+\frac{1}{2+\frac{1}{3+\frac{1}{1+\frac{1}{16539+\frac{1}{\ddots}}}}}}$$ The unexpectedly large partial quotient of 16539 means that using all partial quotients up to (but not including) this value gives an incredibly close approximation of $$\pi^4$$ given the number of terms used. What this large number means essentially is that it takes a denominator of that many times more than the previous in order to obtain a better approximation, implying that the previous denominator is a really close approximation. The same method can be used to show that 355/113 is a very close approximation of $$\pi$$ for the number of partial quotients used.

Evaluating the truncation $$97+\frac{1}{2+\frac{1}{2+\frac{1}{3+\frac{1}{1}}}}$$ gives the famous approximation of $$\frac{2143}{22}$$ for $$\pi^4$$.

FWIW, I found a source according to which

He found this by first squaring the square of π which gives 97.40909103…, then by subtracting 9^2 = 81 he got 16.40909103…, multiplying this by 22 he got 361.0000027…, 361 being the square of 19.

(Thereafter, the source compares to the alternate "method" mentioned in the earlier answer, π^4 = 97.40909103... ≈ 97.409090909...)

The Wikipedia page "Approximations of π" cites his "Lost notebook page 16"

from Ramanujan, who claimed the Goddess of Namagiri appeared to him in a dream and told him the true value of π.