\(\displaystyle\pi\), a beautiful number

This note is incomplete


\(\huge\pi \ \mathcal{can \ be \ expressed \ in \ many \ beautiful \ patterns \ and \ ways.}\)

\(\small \downarrow\downarrow\downarrow See \ Below! \ \downarrow\downarrow\downarrow\)


Take The Nilakantha Series and The Gregory-Leibniz Series.

Nilakantha Series: \(\displaystyle \pi = 3 + \frac{4}{2 \times 3 \times 4} - \frac{4}{4 \times 5 \times 6} + \frac{4}{6 \times 7 \times 8} - \frac{4}{8 \times 9 \times 10} \cdots\)

Gregory-Leibniz Series: \(\displaystyle \frac{\pi}{4} = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{2k-1} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} \cdots \)

\(\displaystyle \textbf{But the Gregory-Leibniz Series takes MORE than 300 terms to be correct to 2 decimal places!}\) It can be transformed into

\(\displaystyle \pi = \sum_{k=1}^\infty \frac{3^k-1}{4^k} \zeta(k+1)\) where \(\displaystyle \zeta\) is the Riemann Zeta function and so that the error after \(k\) terms is \(\displaystyle \approx \left(\frac{3}{4}\right)^k\)


There is also Machin's Formula: \(\displaystyle \frac{1}{4} \pi = 4 \tan^{-1} \left(\frac{1}{5}\right) - \tan^{-1} \left(\frac{1}{239}\right)\)

Abraham Sharp gave the infinite sum series, \(\displaystyle \pi = \sum_{k=0}^\infty \frac{2 (-1)^k 3^{\frac{1}{2}-k}}{2k+1}\)


Simple series' of infinite sums:

\(\displaystyle \frac{1}{4} \pi \sqrt{2} = \sum_{k=1}^\infty \left[\frac{(-1)^{k+1}}{4k-1} + \frac{(-1)^{k+1}}{4k-3}\right] = 1 + \frac{1}{3} - \frac{1}{5} + \frac{1}{7} - \frac{1}{9} + \cdots \) \(\quad\quad\quad\quad\quad\small(\text{related to the Gregory-Leibniz series})\)

\(\displaystyle \frac{1}{4} (\pi-3) = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{2k(2k+1)(2k+2)} = \frac{1}{2 \times 3 \times 4} - \frac{1}{4 \times 5 \times 6} + \frac{1}{6 \times 7 \times 8} - \cdots \)

\(\displaystyle \frac{1}{6} \pi^2 = \sum_{k=1}^\infty \frac{1}{k^2} = 1 + \frac{1}{4} + \frac{1}{9} +\frac{1}{16} \cdots \)

\(\displaystyle \frac{1}{8} \pi^2 = \sum_{k=1}^\infty \frac{1}{(2k-1)^2} = 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \dots \)

There is also \(\displaystyle \frac{\pi^4}{90} = \zeta(4)\)


In 1666 (Newton's miracle year) Newton used a geometric construction to derive the formula

\(\displaystyle \pi = \frac{3}{4} \sqrt{3} + 24 \int_{0}^{\frac{1}{4}} \sqrt{x-x^2} \text{dx} = \frac{3 \sqrt{3}}{4} + 24 \left(\frac{1}{12} - \frac{1}{5 \times 2^5} - \frac{1}{28 \times 2^7} - \frac{1}{72 \times 2^9} \cdots \right) \)

The coefficients can be found in

\(\displaystyle I(x) = \int \sqrt{x-x^2} \text{dx} = \frac{1}{4} (2x-1) \sqrt{x-x^2} - \frac{1}{8} \sin^{-1} (1-2x) \)

by taking the series expansion of \(I(x)-I(0)\) about 0 obtaining

\(I(x) = \dfrac{2}{3} x^{\frac{3}{2}} - \dfrac{1}{5} x^{\frac{3}{2}} - \dfrac{1}{28} x^{\frac{7}{2}} - \dfrac{1}{72} x^{\frac{9}{2}} - \dfrac{5}{704} x^{\frac{11}{2}} \cdots \)


Euler's convergence improvement transformation gives

\(\displaystyle \frac{\pi}{2} = \frac{1}{2} \sum_{n=0}^\infty \frac{(n!)^2 2^{n+1}}{(2n+1)!} = \sum_{n=0}^\infty \frac{n!}{(2n+1)!!} = 1 + \frac{1}{3} + \frac{1 \times 2}{3 \times 5} + \frac{1 \times 2 \times 3}{3 \times 5 \times 7} + \cdots = 1 + \frac{1}{3}\left(1 + \frac{2}{5}\left(1 + \frac{3}{7}\left(1 + \frac{4}{9}(1 + \cdots )\right)\right)\right)\)

This corresponds to plugging in \(\displaystyle x = \frac{1}{\sqrt{2}}\) into the power series for the hypergeometric function \(\displaystyle_2 F_1 (a,b;c;x),\)

\(\displaystyle \frac{\sin^{-1} x}{\sqrt{1-x^2}} = \sum_{i=0}^\infty \frac{(2x)^{2i+1} i!^2 }{2 (2i+1)!} = _2 F_1 (1,1;\frac{3}{2};x^2)x\)

Despite the convergence improvement, series \(\left(\displaystyle \diamond\right)\) converges at only one bit/term. At the cost of a square root, Gosper has noted that \(\displaystyle x = \frac{1}{2}\) gives 2 bits/term,

\(\displaystyle \frac{1}{9} \sqrt{3} \pi = \frac{1}{2} \sum_{i=0}^\infty \frac{(i!)^2}{(2i+1)!} \)

and \(\displaystyle x = \sin\left(\frac{\pi}{10}\right)\) gives almost 3.39 bits/term

\(\displaystyle \frac{\pi}{5 + \sqrt{\phi+2}} = \frac{1}{2} \sum_{i=0}^\infty \frac{(i!)^2}{\phi^{2i+1} (2i+1)!} \)

where \(\displaystyle\phi\) is the golden ratio (not Euler's Totient function). Gosper also obtained

\(\displaystyle \pi = 3 + \frac{1}{60}\left(8 + \frac{2 \times 3}{7 \times 8 \times 3}\left(13 + \frac{3 \times 5}{10 \times 11 \times 3}\left(18 + \frac{4 \times 7}{13 \times 14 \times 3}(23 + \cdots)\right)\right)\right)\)


A spigot algorithm for \(\displaystyle \pi\) is given by Rabinowitz and Wagon.

More amazingly still, a closed-form expression giving a digit-extraction algorithm which produces digits of \(\displaystyle\pi\) (or \(\displaystyle \pi^2 )\) in base-16 was discovered by Bailey.

\(\displaystyle \pi = \sum_{n=0}^\infty \left( \frac{4}{8n+1} - \frac{2}{8n+4} - \frac{1}{8n+5} - \frac{1}{8n+6}\right) \left(\frac{1}{16}\right)^n \)

This formula, known as the BBP formula, was discovered using the PSLQ algorithm and is equivalent to

\(\displaystyle \pi = \int_{0}^{1} \frac{16y - 16}{y^4 - 2y^3 + 4y - 4} \ d \ y \)

There is a series of BBP-type formulas for \(\displaystyle\pi\) in powers of \(\displaystyle (-1)^k\), the first few independent formulas of which are

\[ \begin{align} \pi & = 4 \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} \\ &= 3 \sum_{k=0}^\infty (-1)^k \left[\frac{1}{6k+1} + \frac{1}{6k+5} \right] \\ &= 4 \sum_{k=0}^\infty (-1)^k \left[\frac{1}{10k+1} - \frac{1}{10k+3} + \frac{1}{10k+5} - \frac{1}{10k+7} + \frac{1}{10k+9}\right] \\ &= \sum_{k=0}^\infty (-1)^k \left[\frac{3}{14k+1} - \frac{3}{14k+3} + \frac{3}{14k+5} - \frac{4}{14k+7} + \frac{4}{14k+9} - \frac{4}{14k+11} + \frac{4}{14k+13}\right] \\ &= \sum_{k=0}^\infty (-1)^k \left[\frac{2}{18k+1}+\frac{3}{18k+3}+\frac{2}{18k+5}-\frac{2}{18k+7}-\frac{2}{18k+11}+\frac{2}{18k+13}+\frac{3}{18k+15}+\frac{2}{18k+17}\right] \\ &= \sum_{k=0}^\infty (-1)^k \left[\frac{3}{22k+1}-\frac{3}{22k+3}+\frac{3}{22k+5}-\frac{3}{22k+7}+\frac{3}{22k+9}+\frac{8}{22k+11}+\frac{3}{22k+13}-\frac{3}{22k+15}+\frac{3}{22k+17}-\frac{3}{22k+19}+\frac{1}{22k+21}\right]\end{align} \] Similarly, there are a series of BBP-type formulas for \(\pi\) in powers of \(2^k\), the first few independent formulas of which are

\[ \begin{align} \pi & = \sum_{k=0}^\infty \frac{1}{16^k} \left[\frac{4}{8k+1} - \frac{2}{8k+4} - \frac{1}{8k+5} - \frac{1}{8k+6}\right] \\ &= \frac{1}{2} \sum_{k=0}^\infty \frac{1}{16^k} \left[\frac{8}{8k+2} + \frac{4}{8k+3} + \frac{4}{8k+4} - \frac{1}{8k+7} \right] \\ &= \frac{1}{16} \sum_{k=0}^\infty \frac{1}{256^k} \left[\frac{64}{16k+1} - \frac{32}{16k+4} - \frac{16}{16k+5} - \frac{16}{16k+6} + \frac{4}{16k+9} - \frac{2}{16k+12} - \frac{1}{16k+13} - \frac{1}{16k+14} \right] \\ &= \frac{1}{32} \sum_{k=0}^\infty \frac{1}{256^k} \left[\frac{128}{1k+2} + \frac{64}{16k+3}+\frac{64}{16k+4}-\frac{16}{16k+7} + \frac{8}{16k+10}+\frac{4}{16k+11}+\frac{4}{16k+12}-\frac{1}{16k+15}\right] \\ &= \frac{1}{32} \sum_{k=0}^\infty \frac{1}{4096^k} \left[\frac{256}{24k+2}+\frac{192}{24k+3}-\frac{256}{24k+4}-\frac{96}{24k+6}-\frac{96}{24k+8}+\frac{16}{24k+10}-\frac{4}{24k+12}-\frac{3}{24k+15}-\frac{6}{24k+16}-\frac{2}{24k+18}-\frac{1}{24k+20}\right] \\ &= \frac{1}{64} \sum_{k=0}^\infty \frac{1}{4096^k} \left[\frac{256}{24k+1}+\frac{256}{24k+2}-\frac{384}{24k+3}-\frac{256}{24k+4}-\frac{64}{24k+5}+\frac{96}{24k+8}+\frac{64}{24k+9}+\frac{16}{24k+10}+\frac{8}{24k+12}-\frac{4}{24k+13}+\frac{6}{24k+15}+\frac{6}{24k+16}+\frac{1}{24k+17}+\frac{1}{24k+18}-\frac{1}{24k+20}-\frac{1}{24k+20}\right] \\ &= \frac{1}{96} \sum_{k=0}^\infty \frac{1}{4096^k}\left[\frac{256}{24k+2}+\frac{64}{24k+3}+\frac{128}{24k+5}+\frac{352}{24k+6}+\frac{64}{24k+7}+\frac{288}{24k+8}+\frac{128}{24k+9}+\frac{80}{24k+10}+\frac{20}{24k+12}-\frac{16}{24k+14}-\frac{1}{24k+15}+\frac{6}{24k+16}-\frac{2}{23k+17}-\frac{1}{24k+19}+\frac{1}{24k+20}-\frac{2}{24k+21}\right] \\ &= \frac{1}{96} \sum_{k=0}^\infty \frac{1}{4096^k} \left[\frac{256}{24k+1} + \frac{320}{24k+3} + \frac{256}{24k+4} - \frac{192}{24k+5}-\frac{224}{24k+6}-\frac{64}{24k+7}-\frac{192}{24k+8}-\frac{64}{24k+9}-\frac{64}{24k+10}-\frac{28}{24k+12}-\frac{4}{24k+13}-\frac{5}{24k+15}+\frac{3}{24k+17}+\frac{1}{24k+18}+\frac{1}{24k+19}+\frac{1}{24k+21}-\frac{1}{24k+22}\right] \\ & = \frac{1}{96} \sum_{k=0}^\infty \frac{1}{4096^k} \left[\frac{512}{24k+1}-\frac{256}{24k+2}+\frac{64}{24k+3}-\frac{512}{24k+4}-\frac{32}{24k+6}+\frac{64}{24k+7}+\frac{96}{24k+8}+\frac{64}{24k+9}+\frac{48}{24k+10}-\frac{12}{24k+12}-\frac{8}{24k+13}-\frac{16}{24k+14}-\frac{1}{24k+15}-\frac{6}{24k+16}-\frac{2}{24k+18}-\frac{1}{24k+19}-\frac{1}{24k+20}-\frac{1}{24k+21}\right] \\ &=\frac{1}{4096} \sum_{k=0}^\infty \frac{1}{65536^k} \left[\frac{16384}{32k+1}-\frac{8192}{32k+4}-\frac{4096}{32k+5}-\frac{4096}{32k+6}+\frac{1024}{32k+9}-\frac{512}{32k+12}-\frac{256}{32k+13}-\frac{256}{32k+14}+\frac{64}{32k+17}-\frac{32}{32k+20}-\frac{16}{32k+21}-\frac{16}{32k+22}+\frac{4}{32k+25}-\frac{2}{32k+28}-\frac{1}{32k+29}-\frac{1}{32k+30}\right] \end{align}\]


The Wallis Product is another magnificent way of expressing \(\pi\):

\(\displaystyle \prod_{n=1}^\infty \left(\frac{2n}{2n-1} \times \frac{2n}{2n+1}\right) = \frac{2}{1} \times \frac{2}{3} \times \frac{4}{3} \times \frac{4}{5} \times \frac{6}{5} \times \frac{6}{7} \times \frac{8}{7} \times \frac{8}{9} \cdots = \frac{\pi}{2}\)

PROOF:

1)Euler's infinite product for the sine function

\[\frac{\sin x}{x} = \prod_{n=1}^\infty \left(1-\frac{x^2}{n^2 \pi^2}\right)\]

Let \(\displaystyle x = \frac{\pi}{2}\)

\[\longrightarrow \frac{2}{\pi} = \prod_{n=1}^\infty \left(1-\frac{1}{4n^2}\right)\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\]

\[\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\longrightarrow \frac{\pi}{2} = \prod_{n=1}^\infty \left(\frac{4n^2}{4n^2-1}\right) = \prod_{n=1}^\infty \left(\frac{2n}{2n-1} \times \frac{2n}{2n+1}\right) = \frac{2}{1} \times \frac{2}{3} \times \frac{4}{3} \times \frac{4}{5} \times \frac{6}{5} \times \frac{6}{7} \times \frac{8}{7} \times \frac{8}{9} \cdots\]

2)Proof using Integration

Let: \(\displaystyle I(n) = \int_{0}^{\pi} \sin^n x \ dx \)

\(\small \text{(a form of Wallis' Integrals)}\)

Integrate by parts:

\(\quad\quad u=\sin^{n-1} x\)

\(\Rrightarrow du=(n-1) \sin^{n-2} x \ \cos x \ dx\)

\(\quad dv = \sin x \ dx\)

\(\Rrightarrow v = - \cos \ x\)

\(\displaystyle \Rrightarrow I(n) = \int_{0}^{\pi} - \sin^{n-1} x \cos x |_{0}^{\pi}\)


There is also a continued fraction form : \[\pi=\cfrac{4}{1+\cfrac{1^2}{2+\cfrac{3^2}{2+\cfrac{5^2}{2+\ddots}}}} \]

There is also a curious identity where \(314 \equiv 159 + 265 \pmod{10}\) involving the first \(9\) digits of \(\pi\)


I am going to update this in the future and in the meantime, we can comment about the magical number, \(\pi\)

Watch Out! There is also Golden ratio in another note!


Sources: Mathworld Wolfram and Wikipedia

Help from Members: Agnishom Chattopadhyay, X X, Andrei Li


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Note by Mohmmad Farhan
3 months ago

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  Easy Math Editor

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Your LaTeX went wrong:\(\text{ \displaystyle \rightarrow I(n) = \int_{0}^{\pi} - \sin^{n-1} x \cos x |_{0}^{\pi} }\).

And we cannot see your last line for BBP-type formulas.

X X - 1 month, 3 weeks ago

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I know that. And tHank you

Mohmmad Farhan - 1 month, 3 weeks ago

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Very beautiful indeed. Great thanks to you Farhat; especially if you are in grade 4! :D

Syed Hamza Khalid - 2 months, 2 weeks ago

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Yes; how do you know?

Syed Hamza Khalid - 1 month, 3 weeks ago

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@Mohmmad Farhan Actually I used to play Tankionline now I feel its trash so I play tanki X or Fortnite or PUBG mobil

Syed Hamza Khalid - 1 month, 3 weeks ago

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@Mohmmad Farhan Oh okay

Syed Hamza Khalid - 1 month, 3 weeks ago

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@Mohmmad Farhan Arre!!! that is surprising to hear!!! Mujhe laga thaa that you were from Singapore??

Aaghaz Mahajan - 1 month, 3 weeks ago

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@Mohmmad Farhan Well I didnt write Farhat........I knew its Farhan!!! :)

Aaghaz Mahajan - 1 month, 3 weeks ago

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@Mohmmad Farhan That is soo cool!!! Then you shifted to Singapore I guess???

Aaghaz Mahajan - 1 month, 3 weeks ago

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@Mohmmad Farhan HAHA!!! Ok bro!!!

Aaghaz Mahajan - 1 month, 3 weeks ago

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@Mohmmad Farhan Also, if you are on gmail, then share your email ID..........I'll send you the hangouts invite.....!!!

Aaghaz Mahajan - 1 month, 3 weeks ago

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@Mohmmad Farhan Ok listen, I got an idea.......instead of chatting over here on the community, we can chat on Slack or Google Hangouts.....Are you on either of them???

Aaghaz Mahajan - 1 month, 3 weeks ago

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@Aaghaz Mahajan Wait... I thought Brilliant Slack has been removed?

X X - 1 month, 3 weeks ago

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@X X That's why I was going for a hangouts invite.......where we can share stuff and all!!!

Aaghaz Mahajan - 1 month, 3 weeks ago

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@Mohmmad Farhan Ok chillax buddy!!! No problems!!!! We can then still talk here...... :)

Aaghaz Mahajan - 1 month, 3 weeks ago

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@X X Just gotta find a random community where both people are available and then send a personal link!!! :)

Aaghaz Mahajan - 1 month, 3 weeks ago

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@X X Well yeah, but slack is still an online community........we can still communicate with people over there!!!

Aaghaz Mahajan - 1 month, 3 weeks ago

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@Aaghaz Mahajan Yeah, that's true. But why did they remove Brilliant Slack?

X X - 1 month, 3 weeks ago

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@X X I think it was because RARELY people used to come online!!! Also, it was silent for about 1 year and the maintenance wasn't worth it I guess.......

Aaghaz Mahajan - 1 month, 3 weeks ago

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@Mohmmad Farhan Why so???

Aaghaz Mahajan - 1 month, 3 weeks ago

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You can add \(\dfrac{\pi^4}{90}=\zeta(4)\) also

X X - 3 months ago

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Thank you for contributing

Mohmmad Farhan - 3 months ago

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Will do!

Mohmmad Farhan - 3 months ago

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@Agnishom Chattopadhyay, How to input the greater than symbol?

Mohmmad Farhan - 1 month, 3 weeks ago

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Mohammad Farhan, can you tell me the latex code of how you got the matter completely in a box type manner from "There is a series of BBP-type formulas" and also at some other places.

I know that we can keep " > " symbol to achieve it but it works oly for a continuous string without any break.

Ram Mohith - 1 month, 3 weeks ago

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My name is Farhat. You just simply add an another > below it without any break. For example,

\(>Stuff\)

\(>stuff\)

appears as

Stuff

stuff

Mohmmad Farhan - 1 month, 3 weeks ago

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@Mohmmad Farhan Ok. Thanks Farhat.

Ram Mohith - 1 month, 3 weeks ago

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Umm, you press the relevant key on your keyboard, I guess?

Agnishom Chattopadhyay Staff - 1 month, 3 weeks ago

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@Agnishom Chattopadhyay, I saw an 'algebra' problem named 'hej' that is not written in english

Mohmmad Farhan - 1 month, 3 weeks ago

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@Mohmmad Farhan I will report it

Mohmmad Farhan - 1 month, 3 weeks ago

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OHHHH! Now I get It!

Mohmmad Farhan - 1 month, 3 weeks ago

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@Mohmmad Farhan Haha, were you asking about the \geq command?

Agnishom Chattopadhyay Staff - 1 month, 3 weeks ago

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@Agnishom Chattopadhyay No

Mohmmad Farhan - 1 month, 3 weeks ago

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X X, where is your comment?

Mohmmad Farhan - 1 month, 4 weeks ago

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Sorry, I didn't notice you add that to the above, so I deleted my coment.

X X - 1 month, 4 weeks ago

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Can you help align my equal signs for the BBP-type formulas

Mohmmad Farhan - 1 month, 4 weeks ago

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@Mohmmad Farhan Can you give me your LaTeX?(Without the

X X - 1 month, 4 weeks ago

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@X X \displaystyle \pi = 4 \sum{k=0}^\infty \frac{(-1)^k}{2k+1} = 3 \sum{k=0}^\infty (-1)^k \left[\frac{1}{6k+1} + \frac{1}{6k+5} \right] = 4 \sum{k=0}^\infty (-1)^k \left[\frac{1}{10k+1} - \frac{1}{10k+3} + \frac{1}{10k+5} - \frac{1}{10k+7} + \frac{1}{10k+9}\right] = \sum{k=0}^\infty (-1)^k \left[\frac{3}{14k+1} - \frac{3}{14k+3} + \frac{3}{14k+5} - \frac{4}{14k+7} + \frac{4}{14k+9} - \frac{4}{14k+11} + \frac{4}{14k+13}\right] = \sum_{k=0}^\infty (-1)^k \left[\frac{2}{18k+1}+\frac{3}{18k+3}+\frac{2}{18k+5}-\frac{2}{18k+7}-\frac{2}{18k+11}+\frac{2}{18k+13}+\frac{3}{18k+15}+\frac{2}{18k+17}\right]

Mohmmad Farhan - 1 month, 4 weeks ago

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@Mohmmad Farhan \[\begin{align} \pi &= 4 \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} \\ &= 3 \sum_{k=0}^\infty (-1)^k \left[\frac{1}{6k+1} + \frac{1}{6k+5} \right] \\ &= 4 \sum_{k=0}^\infty (-1)^k \left[\frac{1}{10k+1} - \frac{1}{10k+3} + \frac{1}{10k+5} - \frac{1}{10k+7} + \frac{1}{10k+9}\right] \\ &= \sum_{k=0}^\infty (-1)^k \left[\frac{3}{14k+1} - \frac{3}{14k+3} + \frac{3}{14k+5} - \frac{4}{14k+7} + \frac{4}{14k+9} - \frac{4}{14k+11} + \frac{4}{14k+13}\right] \\ &= \sum_{k=0}^\infty (-1)^k \left[\frac{2}{18k+1}+\frac{3}{18k+3}+\frac{2}{18k+5}-\frac{2}{18k+7}-\frac{2}{18k+11}+\frac{2}{18k+13}+\frac{3}{18k+15}+\frac{2}{18k+17}\right] \end{align}\]

X X - 1 month, 4 weeks ago

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@X X X X, Are you online?

Mohmmad Farhan - 1 month, 4 weeks ago

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@X X I need the raw latex

Mohmmad Farhan - 1 month, 4 weeks ago

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@Mohmmad Farhan \(\text{\begin{align} \pi &= 4 \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} \\ &= 3 \sum_{k=0}^\infty (-1)^k \left[\frac{1}{6k+1} + \frac{1}{6k+5} \right] \\ &= 4 \sum_{k=0}^\infty (-1)^k \left[\frac{1}{10k+1} - \frac{1}{10k+3} + \frac{1}{10k+5} - \frac{1}{10k+7} + \frac{1}{10k+9}\right] \\ &= \sum_{k=0}^\infty (-1)^k \left[\frac{3}{14k+1} - \frac{3}{14k+3} + \frac{3}{14k+5} - \frac{4}{14k+7} + \frac{4}{14k+9} - \frac{4}{14k+11} + \frac{4}{14k+13}\right] \\ &= \sum_{k=0}^\infty (-1)^k \left[\frac{2}{18k+1}+\frac{3}{18k+3}+\frac{2}{18k+5}-\frac{2}{18k+7}-\frac{2}{18k+11}+\frac{2}{18k+13}+\frac{3}{18k+15}+\frac{2}{18k+17}\right] \end{align} }\)

X X - 1 month, 4 weeks ago

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@X X Your BBP-type formula's LaTeX must went wrong. Perhaps you should copy my LaTeX again.

X X - 1 month, 3 weeks ago

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@X X But it does not include the last bit

Mohmmad Farhan - 1 month, 3 weeks ago

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@X X Also, I think you can add Wallis Product, too.

X X - 1 month, 4 weeks ago

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@X X got it

Mohmmad Farhan - 1 month, 4 weeks ago

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@X X Searching online what that is

Mohmmad Farhan - 1 month, 4 weeks ago

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@X X Thank you

Mohmmad Farhan - 1 month, 4 weeks ago

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@Mohmmad Farhan I noticed you editted the expression, but I think you should copy my raw LaTeX again, because sometimes words like "\, *, _, #" will disappear when we type it.(And obviously, the alignment went wrong) To avoid this, I added \text{...}, but I think you can still copy it.

X X - 1 month, 4 weeks ago

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@X X Add the \(\text{\[ \]}\) one, and remember the sum, the _ is missing.

X X - 1 month, 4 weeks ago

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@X X ok

Mohmmad Farhan - 1 month, 4 weeks ago

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@X X Thank you

Mohmmad Farhan - 1 month, 4 weeks ago

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Oh!

Mohmmad Farhan - 1 month, 4 weeks ago

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Nice!

There are also the continued fractions, one of them being: \[\pi=\frac{4}{1+\frac{1^2}{2+\frac{2+\frac{3^2}{2+\frac{5^2}{2+...}}}}\]

Andrei Li - 3 months ago

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Thank you for contributing

Mohmmad Farhan - 3 months ago

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\(\pi=\cfrac{4}{1+\cfrac{1^2}{2+\cfrac{2+\cfrac{3^2}{2+\cfrac{5^2}{2+\ddots}}}}\)

Mohmmad Farhan - 3 months ago

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Strangely enough, my computer refuses to accept the continued fraction LaTeX...

Andrei Li - 3 months ago

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@Andrei Li me too.. (so sad)

Mohmmad Farhan - 3 months ago

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@Mohmmad Farhan \(\pi=\cfrac{4}{1+\cfrac{1^2}{2+\cfrac{3^2}{2+\cfrac{5^2}{2+\cdots}}}} \)

X X - 3 months ago

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