Approximating π \pi

Number Theory Level 4

Thanks to @Naren Bhandari for this inspiration .

The sum S = ∑ n = 1 ∞ 1 n 5 ( n + 1 ) 5 S=\sum_{n=1}^{\infty} \frac{1}{n^5 (n+1)^5} can be written in the form S = A B π 4 + C D π 2 + E S=\frac{A}{B}\pi^4+\frac{C}{D}\pi^2+E for integers A , B , C , D , E A,B,C,D,E .

Solving this biquadratic equation , gives a formula for π \pi in terms of S S .

If instead we approximate S S by its partial sum S N = ∑ n = 1 N 1 n 5 ( n + 1 ) 5 S_N=\sum_{n=1}^{N} \frac{1}{n^5 (n+1)^5} we can get an approximation for π \pi ; the more terms we take in the sum, the better the approximations.

If p N p_N is the approximation to π \pi obtained from N N terms of the sum, what is the smallest N N such that ∣ p n − π ∣ < 1 0 − 9 |p_n-\pi|<10^{-9} ?

Bonus: can you find a taxicab along the way?


The answer is 4.

Chris Lewis by Chris Lewis , 40, United Kingdom

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Mark Hennings
Jul 5, 2020

Taking partial fractions, 1 n 5 ( n + 1 ) 5 = ( 1 n 5 − 1 ( n + 1 ) 5 ) − 5 ( 1 n 4 + 1 ( n + 1 ) 4 ) + 15 ( 1 n 3 − 1 ( n + 1 ) 3 ) − 35 ( 1 n 2 + 1 ( n + 1 ) 2 ) + 70 ( 1 n − 1 n + 1 ) \frac{1}{n^5(n+1)^5} \; = \; \left(\frac{1}{n^5} - \frac{1}{(n+1)^5}\right) - 5\left(\frac{1}{n^4 } + \frac{1}{(n+1)^4}\right) + 15\left(\frac{1}{n^3} - \frac{1}{(n+1)^3}\right) - 35\left(\frac{1}{n^2} + \frac{1}{(n+1)^2}\right) + 70\left(\frac{1}{n} - \frac{1}{n+1}\right) and hence S = 1 − 5 ( 2 ζ ( 4 ) − 1 ) + 15 − 35 ( 2 ζ ( 2 ) − 1 ) + 70 = 126 − 35 3 π 2 − 1 9 π 4 S \; = \; 1 - 5(2\zeta(4)-1) + 15 - 35(2\zeta(2) - 1) + 70 \; = \; 126 - \tfrac{35}{3}\pi^2 - \tfrac{1}{9}\pi^4 Solving the equation S N = 126 − 35 3 p N 2 − 1 9 p N 4 S_N \; = \; 126 - \tfrac{35}{3}p_N^2 - \tfrac{1}{9}p_N^4 gives the approximation for π \pi p N = 3 2 [ 1729 − 4 S N − 35 ] p_N \; = \; \sqrt{\frac32\left[\sqrt{{\red{1729}} - 4S_N} - 35\right]} (behold the t a x i c a b \red{\mathrm{taxicab}} ). The first four values of ∣ p N − π ∣ |p_N - \pi| are 1.52706 × 1 0 − 6 5.03269 × 1 0 − 8 4.17886 × 1 0 − 9 5.90394 × 1 0 − 10 1.52706\times10^{-6} \hspace{1cm} 5.03269\times10^{-8}\hspace{1cm} 4.17886\times10^{-9} \hspace{1cm} 5.90394\times10^{-10} which makes the answer 4 \boxed{4} .

3 Helpful 2 Interesting 3 Brilliant 0 Confused

Great solution, thanks for taking the time to write it up.

Any thoughts about improving the approximation? One interesting point is that the telescoping only ever leaves rational terms and terms in ζ ( 2 m ) \zeta(2m) - ie rational multiples of powers of π \pi ; if, for example, we define S ( k ) = ∑ n = 1 ∞ 1 n k ( n + 1 ) k S(k)=\sum_{n=1}^{\infty} \frac{1}{n^k (n+1)^k} we find

k k S ( k ) S(k)
1 1 1 1
2 2 − 3 + 1 3 π 2 -3+\frac13 \pi^2
3 3 10 − π 2 10-\pi^2
4 4 − 35 + 10 3 π 2 + 1 45 π 4 -35+\frac{10}{3}\pi^2+\frac{1}{45}\pi^4
5 5 126 − 35 3 π 2 − 1 9 π 4 126-\frac{35}{3}\pi^2-\frac{1}{9}\pi^4
6 6 − 462 + 42 π 2 + 7 15 π 4 + 2 945 π 6 -462+42\pi^2+\frac{7}{15}\pi^4+\frac{2}{945}\pi^6

and so on. Combining these seems helpful for improving accuracy.

Another idea is to estimate the error in the partial sums by integration. This can be done directly from the partial fraction decomposition. The only snag is the logarithmic terms from integrating 1 n \frac{1}{n} and 1 n + 1 \frac{1}{n+1} , but again these can be eliminated by combining the sums.

I'm not sure how much mileage there is here - the sums converge quickly, but not that quickly (the early rush of correct decimal places soon dries up). Any other ideas on how to improve the accuracy?

Chris Lewis - 11 months, 1 week ago

Try looking online for (Ramanujan’s) approximations for π \pi , and their extensions.

Mark Hennings - 11 months, 1 week ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...