OEIS A229580 mini challenge (RPL)

[Gerson W. Barbosa]

(05-03-2018 05:11 PM)Valentin Albillo Wrote:  

Gerson W. Barbosa Wrote:However I think you were right in the first place, a recursive formula could have indeed been of help for my particular application. In the program below I have used a closed-form formula to compute the last denominator of the continued fraction, but I might as well have computed it using a recursive formula, every term being computed at each step of the first FOR-NEXT loop

A few questions if you don't mind:

What's the origin of the hybrid formula (series + continued fraction) you're evaluating ? Is it original research on your part ?

Hello, Valentin,

No, I don't mind at all. That's a well known series, but I derived it in early 1983 when I was a Physics student who had just learned about Fourier series and decided to use them for an out-of-the-textbook application:

\(\pi = \int_{0}^{\pi }\left [ \frac{1}{2}+\frac{2}{\pi } \left ( \sin x+\frac{1}{3}\sin 3x+\frac{1}{5}\sin 5 x+\frac{1}{7}\sin 7x+\cdots \right )\right ]\)dx
\(\vdots\)
\(\pi =2\sqrt{2\sum_{n=1}^{\infty}\frac{1}{\left (2n-1 \right )^{2}} }\)

It worked, but would converge very slowly. Then I did a little more work and obtained

\(\pi =2\sqrt{1+2\sum_{n=1}^{\infty}\left ( \frac{1}{\left ( 2n-1 \right )^{2}}-\frac{1}{\left ( 2n-1 \right )\left ( 2n+1 \right )} \right )}\)

It didn't help much, but at least I could have \(\pi\) between two bounds.

I was bit disappointed two years later when I read in a Calculus book something like "in 1792 a French mathematician obtained this series for \(\pi\)", followed by one of the series above, I don't remember which.

I though I had lost them, but a few days ago I found the two sheets of papers I had written back then hiding between the pages of my Portuguese version of "The Elements of the Differential and Integral Calculus", by W. A. Granville, P. F. Smith and W. R. Longley. I know it is a useless series for computing \(\pi\), due to its extremely slow convergence rate, but I decided to play with it again just the same.

(05-03-2018 05:11 PM)Valentin Albillo Wrote:  Does it actually converge to Pi\(^2\)/8 and do the coefficients of the continued fraction part actually belong to the OEIS series (divided by 2) or it's just a tentative conjecture ?

The first part does converge, as we have seen, but the continued-fraction terms work only for the denominators up to 112*n. Incidentally 3, 20 and 112, when multiplied by two, match the second, third and fourth terms of the OEIS sequence, but that's all. For n = 1000, for instance, the first few denominators of the continued fraction (considering all numerators equal to 1), should be 4000, 3000, 1250, 1382, 1, 2, 1, 1, 5, 1. I don't think this continued fraction improvement should work for non-alternating series, though. For other alternating series I tried, like the one for the Reciprocal Fibonacci Constant, the denominators cycle between two values, for instance, 4n and 16n, while the numerators are second powers of consecutive integers. Since the numerators of this one appears to follow a fourth power pattern, I would guess the denominators would cycle between four values, but then again the original series is non-alternating, so this may be doomed to failure.

(05-03-2018 05:11 PM)Valentin Albillo Wrote:  I could check the convergence myself but if you already know I'd rather ask and avoid wasting time rediscovering your knowledge. Also, you begin evaluating the continued fraction part starting with the last denominator and working iteratively upwards to the first. I'd do it the opposite way, starting at the top and going downwards to the last denominator because this way it's easier to visualize the alleged convergence as each new denominator is considered.

Thank you very much for your offer. Please feel free to do any experiments if you think they're worthwhile. I hope my lack of formalism hasn't been a problem for the explanation you've requested.

Best regards,

Gerson.

P.S.:

The \(\pi\)/4 formula here is the example I meant, not the RFC formula. Sorry! Also, contrary to what I had stated, the latter is not an alternating series, so I may be wrong and perhaps there is a continued-fraction adjustment for the original \(\pi\)^2/8 series as well.