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Is there a formula which helps in approximation of $\pi$ as $\dfrac{p}{q}$ where $p,q \in \mathbb{Z}$?

I got this site though : [http://qin.laya.com/tech_projects_approxpi.html ] which shows the various fractions from $\pi \approx \frac{3}{2}$ to $\approx\frac{2646693125139304345}{842468587426513207}$

So is there any relation/formula/easy trick/anything which can help to find these fractions??

Thanks!!

NeilRoy
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  • One way would be to simply take truncations of Wallis product https://en.wikipedia.org/wiki/Wallis_product This will give a series of rational approximations which is getting closer and closer to $\pi$, but at very slow rate. – Wojowu Oct 03 '15 at 06:49
  • Maybe a useful reference: https://en.wikipedia.org/wiki/Proof_that_22/7_exceeds_%CF%80 – Moya Oct 03 '15 at 06:49
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    Try https://en.wikipedia.org/wiki/Approximations_of_%CF%80 for a listing of various methods of approximating $\pi$. It notes that the continued fraction representation can be used to generate successive best rational approximations. – amd Oct 03 '15 at 07:25

5 Answers5

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There are infinitely many ways to approximate pi with rational numbers, and there is a whole industry devoted to doing this to huge degrees of accuracy. Perhaps the simplest rational approximations to pi are found in its continued-fraction expansion. The first two are $3$ and $\frac{22}7$, and already the fourth, $\frac{355}{113}$, is accurate to six decimal places.

John Bentin
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This series converges to $\pi$ from above, starting at $\frac{22}{7}$

$$\pi=\frac{22}{7}-\sum_{k=0}^\infty \frac{96 (160 k^2+422 k+405)}{(4 k+3) (4 k+5) (4 k+7) (4 k+9) (4 k+11) (4 k+13) (4 k+15) (4 k+17)}$$

A simpler one is $$\pi=\frac{22}{7}-\sum_{k=1}^{\infty} \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}$$

A series to prove $\frac{22}{7}-\pi>0$

Community
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Jaume Oliver Lafont
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  • It is interesting to compare above summation with integration per $$\pi = \frac{22}{7} - \int_{0}^{1}\frac{x^4(1-x)^4}{1+x^2},\mathrm{d}x$$ – Alex Feb 14 '16 at 02:28
  • As a follow up to previous comment, in the summation case the denominator is 8-th degree polynomial and the numerator is 2-nd degree polynomial (both with regards to "k"), while in the integration case the denominator is 2-nd degree polynomial and the numerator is 8-th degree polynomial (both with regards to "x"). – Alex Feb 14 '16 at 02:47
  • In fact, it was the other way: the series has eight factors in the denominator in an attempt to match Dalzell's integral symmetry. However, the numerator is not that symmetric, which is the reason for the linked question... – Jaume Oliver Lafont Feb 14 '16 at 05:37
  • @Alex Related: http://math.stackexchange.com/questions/1654104/series-and-integrals-for-inequalities-and-approximations-to-pi – Jaume Oliver Lafont Feb 14 '16 at 06:51
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Yes. There are several. The best know is the Leibniz formula:

$$1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots = \frac{\pi}{4}$$

This is an infinite series but just do the first few terms to get: $$\pi = 4\frac{3\cdot5\cdot7 - 5\cdot7 + 3\cdot7 - 3\cdot5}{3\cdot5\cdot17} = \frac{304}{105} \approx 2.8952380952380\ldots$$ Okay, I never said it would be a good approximation, but each successive term gets closer. Add $4/9$ to that and we get $3.3396825396825\ldots$

EDIT
Here's a better one: $$\frac{\pi}{4} = \sum_{n=0}^\infty\frac{2}{(4n+1)(4n + 3)}$$ So for the first three terms we get: $$\pi = 8\left(\frac13 + 7\frac15 + 11\frac19 + 15\frac1{13}\right) \approx 3.017071817071817\ldots$$

rubik
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fleablood
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    The best known?! The convergence of the Leibniz series is (as you note) awful! – preferred_anon Oct 03 '15 at 07:29
  • @DanielLittlewood I think he meant the "best-known" (though that is arguable in itself) and not "(the) best (formula) known". Ah, the travails of the English Language. :) – Deepak Feb 14 '16 at 01:53
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    Also, the Leibniz formula should properly be called the Madhava-Leibniz formula, as the Indian mathematician Madhava is now known to have discovered it centuries before Leibniz. – Deepak Feb 14 '16 at 02:00
  • I definitely meant the most known. I never really paid attention to methods (as long as I know someone can calculate pi I don't really care about the details). It's the one I see mentioned the most. – fleablood Feb 14 '16 at 06:48
  • Ah, that makes more sense. Not sure why I was so aggressive when I posted that comment! – preferred_anon Feb 14 '16 at 12:51
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The best way to obtain rational approximants of $\pi$ is by using continued fractions.

But the simple continued fraction for $\pi$ has no pattern - you will never be able to guess it without knowing the value of $\pi$ first.

So, you can use one of the better, regular continued fractions for $\pi$, like this one:

$$\pi=3+\dfrac{1}{6+\dfrac{3^2}{6+\dfrac{5^2}{6+...}}}$$

It converges much better than simple series, for example:

$$\pi=3+\dfrac{1}{6+\dfrac{3^2}{6+\dfrac{5^2}{6}}}=\frac{1321}{420}=3.14524$$

The next fraction (with $7^2$) looks nicer:

$$\frac{989}{315}=3.13968$$

See here for another three regular fractions for $\pi$.

You can also use infinite products instead of series, because you will immediately get a fraction at every step. Like Wallis product for example:

$$\prod^{\infty}_{n=1}\frac{2n}{2n-1}\frac{2n}{2n+1}=\frac{2}{1}\frac{2}{3}\frac{4}{3}\frac{4}{5}\frac{6}{5}\frac{6}{7}...$$

But it converges very slowly and gives huge fractions, for example:

$$\prod^{20}_{n=1}\frac{2n}{2n-1}\frac{2n}{2n+1}=\frac{151115727451828646838272}{48691767863540419643025}=3.10352$$

Yuriy S
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There is one made by Ramanujan in 1914 that aproximates $\frac{1}{\pi}$.

$$\frac{4}{\pi}=\frac{1}{882}\sum_{n=0}^{\infty}\frac{(-1)^{n}(4n)!}{(4^n n!)^4}\frac{1123+21460 n}{882^{2n}}$$

Paramanand Singh
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Ivan S. Guerra
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