# $$\displaystyle\pi$$, 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

This note is incomplete Note by Mohmmad Farhan
8 months, 2 weeks ago

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## Comments

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

- 7 months 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.

- 7 months 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

- 7 months ago

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Ok. Thanks Farhat.

- 7 months ago

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

Staff - 7 months ago

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

- 7 months ago

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I will report it

- 7 months ago

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OHHHH! Now I get It!

- 7 months ago

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Haha, were you asking about the \geq command?

Staff - 7 months ago

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No

- 7 months ago

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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.

- 7 months ago

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I know that. And tHank you

- 7 months ago

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X X, where is your comment?

- 7 months, 1 week ago

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

- 7 months, 1 week ago

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

- 7 months, 1 week ago

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Can you give me your LaTeX?(Without the

- 7 months, 1 week 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]

- 7 months, 1 week ago

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\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}

- 7 months, 1 week ago

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@X X X X, Are you online?

- 7 months, 1 week ago

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@X X I need the raw latex

- 7 months, 1 week ago

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\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} }

- 7 months, 1 week ago

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

- 7 months ago

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@X X But it does not include the last bit

- 7 months ago

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

- 7 months, 1 week ago

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@X X got it

- 7 months, 1 week ago

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@X X Searching online what that is

- 7 months, 1 week ago

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@X X Thank you

- 7 months, 1 week ago

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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.

- 7 months, 1 week ago

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

- 7 months, 1 week ago

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@X X ok

- 7 months, 1 week ago

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@X X Thank you

- 7 months, 1 week ago

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Oh!

- 7 months, 1 week ago

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

- 7 months, 3 weeks ago

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Nice!

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

- 8 months, 1 week ago

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Thank you for contributing

- 8 months, 1 week ago

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

- 8 months, 1 week ago

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

- 8 months, 1 week ago

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me too.. (so sad)

- 8 months, 1 week ago

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

- 8 months, 1 week ago

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

- 8 months, 1 week ago

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Thank you for contributing

- 8 months, 1 week ago

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Will do!

- 8 months, 1 week ago

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