# $\displaystyle\pi$, a beautiful number

This note is incomplete

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

Before we indulge in the facts, let's munch on some pie. (pun intended)

$\small \downarrow\downarrow\downarrow See \ Below! \ \downarrow\downarrow\downarrow$

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{aligned} \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{aligned} 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{aligned} \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{aligned}

F. Bellard found the rapidly converging BBP-type formula

$\pi = \frac{1}{2^6} \sum_{n=0}^\infty \frac{(-1)^n}{2^{10n}} \left(-\frac{2^5}{4n+1}-\frac{1}{4n+3}+\frac{2^8}{10n+1}-\frac{2^6}{10n+3}-\frac{2^2}{10n+5}-\frac{2^2}{10n+7}+\frac{1}{10n+9}\right)$

A related integral is

$\pi = \frac{22}{7} - \int_{0}^{1} \frac{x^4 (1-x)^4}{1+x^2} \text{dx}$

Boros and Moll state that it is not clear if there exists a natural choice of a rational polynomial whose integral between 0 and 1 produces $\pi-\dfrac{333}{106}$, where $\dfrac{333}{106}$ is the next convergent. However, an integral exists for the fourth convergent, namely

$\pi=\dfrac{355}{113} - \frac{1}{3164} \int_{0}^{1} \frac{x^8 (1-x)^8 (25 + 816x^2)}{1+x^2} \text{dx}$

Backhouse used the identity

\begin{aligned} l_{mn} & = \int_{0}^{1} \frac{x^m (1-x)^n}{1+x^2} \text{dx} \\ &= 2^{-(m+n+1)} \sqrt{\pi} \Gamma (m+1) \Gamma (n+1) \times _3 F_2 \left(1,\frac{m+1}{2},\frac{m+2}{2}; \frac{m+n+2}{2},\frac{m+n+3}{2};-1\right) \\ &=a + b\pi +c \ln 2 \end{aligned} for positive integer $m$ and $n$ and where $a,b$ and $c$ are rational constant to generate a number of formulas for $\pi$. In particular, if $2m-n \equiv 0 \pmod 4$ then $c=0$

A similar formula was subsequently discovered by Ferguson, leading to a two-dimensional lattice of such formulas which can be generated by these two formulas given by $\pi = \sum_{k=0}^{\infty} \left(\frac{4+8r}{8k+1}-\frac{8r}{8k+2}-\frac{4r}{8k+3}-\frac{2+8r}{8k+4}-\frac{1+2r}{8k+5}-\frac{1+2r}{8k+6}+\frac{r}{8k+7} \right)\left(\frac{1}{16}\right)^k$

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

This note is incomplete

There is also an amazing $\pi$ song out there!

2 years, 5 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

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

- 2 years, 5 months ago

Will do!

- 2 years, 5 months ago

Thank you for contributing

- 2 years, 5 months ago

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

- 2 years, 5 months ago

Your LaTeX went wrong: \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.

- 2 years, 4 months ago

I know that. And tHank you

- 2 years, 4 months ago

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+...}}}}

- 2 years, 5 months ago

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

- 2 years, 5 months ago

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

- 2 years, 5 months ago

- 2 years, 5 months ago

$\pi=\cfrac{4}{1+\cfrac{1^2}{2+\cfrac{3^2}{2+\cfrac{5^2}{2+\cdots}}}}$

- 2 years, 5 months ago

Thank you for contributing

- 2 years, 5 months ago

X X, where is your comment?

- 2 years, 4 months ago

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

- 2 years, 4 months ago

Oh!

- 2 years, 4 months ago

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

- 2 years, 4 months ago

Can you give me your LaTeX?(Without the

- 2 years, 4 months ago

@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]

- 2 years, 4 months ago

\begin{aligned} \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{aligned}

- 2 years, 4 months ago

@X X Thank you

- 2 years, 4 months ago

@X X I need the raw latex

- 2 years, 4 months ago

\text{\begin{aligned} \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{aligned} }

- 2 years, 4 months ago

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

- 2 years, 4 months ago

@X X ok

- 2 years, 4 months ago

@X X Thank you

- 2 years, 4 months ago

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.

- 2 years, 4 months ago

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

- 2 years, 4 months ago

@X X Searching online what that is

- 2 years, 4 months ago

@X X got it

- 2 years, 4 months ago

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

- 2 years, 4 months ago

@X X But it does not include the last bit

- 2 years, 4 months ago

@X X X X, Are you online?

- 2 years, 4 months ago

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

- 2 years, 4 months ago

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

- 2 years, 4 months ago

OHHHH! Now I get It!

- 2 years, 4 months ago

Haha, were you asking about the \geq command?

- 2 years, 4 months ago

No

- 2 years, 4 months ago

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

- 2 years, 4 months ago

I will report it

- 2 years, 4 months ago

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.

- 2 years, 4 months ago

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

$>Stuff$

$>stuff$

appears as

Stuff

stuff

- 2 years, 4 months ago

Ok. Thanks Farhat.

- 2 years, 4 months ago

\begin{align} l{mn} & = \int{0}^{1} \frac{x^m (1-x)^n}{1+x^2} \text{dx} \ &= 2^{-(m+n+1) \sqrt{\pi} \Gamma (m+1) \Gamma (n+1) \times 3 F2 \left(1,\frac{m+1}{2},\frac{m+2}{2}; \frac{m+n+2}{2},\frac{m+n+3}{2};-1\right) \ &=a + b\pi +c \ln 2 \end{align}

why is this wrong?

(I took out the start and end brackets for you to inspect.)

- 1 year, 8 months ago

But I didn't notice anything wrong with the Backhouse identity.

- 1 year, 8 months ago

I already realised where the mistake was

- 1 year, 8 months ago

Add this series too which I found it, $\sum_{n=1}^{\infty}\frac{\Gamma(n-1/2)\Gamma(1/2+1)}{\Gamma(n-1)}=\pi$

- 8 months ago