π\displaystyle\pi, a beautiful number

This note is incomplete

π can be expressed in many beautiful patterns and ways.\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)

↓↓↓See Below! ↓↓↓\small \downarrow\downarrow\downarrow See \ Below! \ \downarrow\downarrow\downarrow

Take The Nilakantha Series and The Gregory-Leibniz Series.

Nilakantha Series: π=3+42×3×4−44×5×6+46×7×8−48×9×10⋯\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: π4=∑k=1∞(−1)k+12k−1=1−13+15−17+19⋯\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

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

π=∑k=1∞3k−14kζ(k+1)\displaystyle \pi = \sum_{k=1}^\infty \frac{3^k-1}{4^k} \zeta(k+1) where ζ\displaystyle \zeta is the Riemann Zeta function and so that the error after kk terms is ≈(34)k\displaystyle \approx \left(\frac{3}{4}\right)^k

There is also Machin's Formula: 14π=4tan⁡−1(15)−tan⁡−1(1239)\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, π=∑k=0∞2(−1)k312−k2k+1\displaystyle \pi = \sum_{k=0}^\infty \frac{2 (-1)^k 3^{\frac{1}{2}-k}}{2k+1}

Simple series' of infinite sums:

14π2=∑k=1∞[(−1)k+14k−1+(−1)k+14k−3]=1+13−15+17−19+⋯\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 (related to the Gregory-Leibniz series)\quad\quad\quad\quad\quad\small(\text{related to the Gregory-Leibniz series})

14(π−3)=∑k=1∞(−1)k+12k(2k+1)(2k+2)=12×3×4−14×5×6+16×7×8−⋯\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

16π2=∑k=1∞1k2=1+14+19+116⋯\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

18π2=∑k=1∞1(2k−1)2=1+132+152+172+…\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 π490=ζ(4)\displaystyle \frac{\pi^4}{90} = \zeta(4)

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

π=343+24∫014x−x2dx=334+24(112−15×25−128×27−172×29⋯ )\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

I(x)=∫x−x2dx=14(2x−1)x−x2−18sin⁡−1(1−2x)\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)I(x)-I(0) about 0 obtaining

I(x)=23x32−15x32−128x72−172x92−5704x112⋯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

π2=12∑n=0∞(n!)22n+1(2n+1)!=∑n=0∞n!(2n+1)!!=1+13+1×23×5+1×2×33×5×7+⋯=1+13(1+25(1+37(1+49(1+⋯ ))))\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 x=12\displaystyle x = \frac{1}{\sqrt{2}} into the power series for the hypergeometric function 2F1(a,b;c;x),\displaystyle_2 F_1 (a,b;c;x),

sin⁡−1x1−x2=∑i=0∞(2x)2i+1i!22(2i+1)!=2F1(1,1;32;x2)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 x=12\displaystyle x = \frac{1}{2} gives 2 bits/term,

193π=12∑i=0∞(i!)2(2i+1)!\displaystyle \frac{1}{9} \sqrt{3} \pi = \frac{1}{2} \sum_{i=0}^\infty \frac{(i!)^2}{(2i+1)!}

and x=sin⁡(π10)\displaystyle x = \sin\left(\frac{\pi}{10}\right) gives almost 3.39 bits/term

π5+ϕ+2=12∑i=0∞(i!)2ϕ2i+1(2i+1)!\displaystyle \frac{\pi}{5 + \sqrt{\phi+2}} = \frac{1}{2} \sum_{i=0}^\infty \frac{(i!)^2}{\phi^{2i+1} (2i+1)!}

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

π=3+160(8+2×37×8×3(13+3×510×11×3(18+4×713×14×3(23+⋯ ))))\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 π\displaystyle \pi is given by Rabinowitz and Wagon.

More amazingly still, a closed-form expression giving a digit-extraction algorithm which produces digits of π\displaystyle\pi (or π2)\displaystyle \pi^2 ) in base-16 was discovered by Bailey.

π=∑n=0∞(48n+1−28n+4−18n+5−18n+6)(116)n\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

π=∫0116y−16y4−2y3+4y−4 d y\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 π\displaystyle\pi in powers of (−1)k\displaystyle (-1)^k, the first few independent formulas of which are

π=4∑k=0∞(−1)k2k+1=3∑k=0∞(−1)k[16k+1+16k+5]=4∑k=0∞(−1)k[110k+1−110k+3+110k+5−110k+7+110k+9]=∑k=0∞(−1)k[314k+1−314k+3+314k+5−414k+7+414k+9−414k+11+414k+13]=∑k=0∞(−1)k[218k+1+318k+3+218k+5−218k+7−218k+11+218k+13+318k+15+218k+17]=∑k=0∞(−1)k[322k+1−322k+3+322k+5−322k+7+322k+9+822k+11+322k+13−322k+15+322k+17−322k+19+122k+21] \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 π\pi in powers of 2k2^k, the first few independent formulas of which are

π=∑k=0∞116k[48k+1−28k+4−18k+5−18k+6]=12∑k=0∞116k[88k+2+48k+3+48k+4−18k+7]=116∑k=0∞1256k[6416k+1−3216k+4−1616k+5−1616k+6+416k+9−216k+12−116k+13−116k+14]=132∑k=0∞1256k[1281k+2+6416k+3+6416k+4−1616k+7+816k+10+416k+11+416k+12−116k+15]=132∑k=0∞14096k[25624k+2+19224k+3−25624k+4−9624k+6−9624k+8+1624k+10−424k+12−324k+15−624k+16−224k+18−124k+20]=164∑k=0∞14096k[25624k+1+25624k+2−38424k+3−25624k+4−6424k+5+9624k+8+6424k+9+1624k+10+824k+12−424k+13+624k+15+624k+16+124k+17+124k+18−124k+20−124k+20]=196∑k=0∞14096k[25624k+2+6424k+3+12824k+5+35224k+6+6424k+7+28824k+8+12824k+9+8024k+10+2024k+12−1624k+14−124k+15+624k+16−223k+17−124k+19+124k+20−224k+21]=196∑k=0∞14096k[25624k+1+32024k+3+25624k+4−19224k+5−22424k+6−6424k+7−19224k+8−6424k+9−6424k+10−2824k+12−424k+13−524k+15+324k+17+124k+18+124k+19+124k+21−124k+22]=196∑k=0∞14096k[51224k+1−25624k+2+6424k+3−51224k+4−3224k+6+6424k+7+9624k+8+6424k+9+4824k+10−1224k+12−824k+13−1624k+14−124k+15−624k+16−224k+18−124k+19−124k+20−124k+21]=14096∑k=0∞165536k[1638432k+1−819232k+4−409632k+5−409632k+6+102432k+9−51232k+12−25632k+13−25632k+14+6432k+17−3232k+20−1632k+21−1632k+22+432k+25−232k+28−132k+29−132k+30] \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

π=126∑n=0∞(−1)n210n(−254n+1−14n+3+2810n+1−2610n+3−2210n+5−2210n+7+110n+9) \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

π=227−∫01x4(1−x)41+x2dx\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 π−333106\pi-\dfrac{333}{106}, where 333106\dfrac{333}{106} is the next convergent. However, an integral exists for the fourth convergent, namely

π=355113−13164∫01x8(1−x)8(25+816x2)1+x2dx\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

lmn=∫01xm(1−x)n1+x2dx=2−(m+n+1)πΓ(m+1)Γ(n+1)×3F2(1,m+12,m+22;m+n+22,m+n+32;−1)=a+bπ+cln⁡2\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 mm and nn and where a,ba,b and cc are rational constant to generate a number of formulas for π\pi. In particular, if 2m−n≡0(mod4)2m-n \equiv 0 \pmod 4 then c=0c=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 π=∑k=0∞(4+8r8k+1−8r8k+2−4r8k+3−2+8r8k+4−1+2r8k+5−1+2r8k+6+r8k+7)(116)k\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 π\pi:

∏n=1∞(2n2n−1×2n2n+1)=21×23×43×45×65×67×87×89⋯=π2\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

sin⁡xx=∏n=1∞(1−x2n2π2)\frac{\sin x}{x} = \prod_{n=1}^\infty \left(1-\frac{x^2}{n^2 \pi^2}\right)

Let x=π2\displaystyle x = \frac{\pi}{2}

⟶2π=∏n=1∞(1−14n2)\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

⟶π2=∏n=1∞(4n24n2−1)=∏n=1∞(2n2n−1×2n2n+1)=21×23×43×45×65×67×87×89⋯\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: I(n)=∫0πsin⁡nx dx\displaystyle I(n) = \int_{0}^{\pi} \sin^n x \ dx

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

Integrate by parts:

u=sin⁡n−1x\quad\quad u=\sin^{n-1} x

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

dv=sin⁡x dx\quad dv = \sin x \ dx

⇛v=−cos⁡ x\Rrightarrow v = - \cos \ x

⇛I(n)=∫0π−sin⁡n−1xcos⁡x∣0π\displaystyle \Rrightarrow I(n) = \int_{0}^{\pi} - \sin^{n-1} x \cos x |_{0}^{\pi}

There is also a continued fraction form : π=41+122+322+522+⋱\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≡159+265(mod10)314 \equiv 159 + 265 \pmod{10} involving the first 99 digits of π\pi

I am going to update this in the future and in the meantime, we can comment about the magical number, π\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 π\pi song out there!

#Algebra

Note by Mohammad Farhat
2 years, 9 months ago

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Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai−1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 ∑i=13 \sum_{i=1}^3
\sin \theta sin⁡θ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

You can add π490=ζ(4)\dfrac{\pi^4}{90}=\zeta(4) also

X X - 2 years, 9 months ago

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Will do!

Mohammad Farhat - 2 years, 9 months ago

Thank you for contributing

Mohammad Farhat - 2 years, 9 months ago

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

Syed Hamza Khalid - 2 years, 8 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.

X X - 2 years, 8 months ago

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I know that. And tHank you

Mohammad Farhat - 2 years, 8 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+...}}}}

Andrei Li - 2 years, 9 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}}}}

Mohammad Farhat - 2 years, 9 months ago

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

Andrei Li - 2 years, 9 months ago

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@Andrei Li – me too.. (so sad)

Mohammad Farhat - 2 years, 9 months ago

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@Mohammad Farhat – π=41+122+322+522+⋯\pi=\cfrac{4}{1+\cfrac{1^2}{2+\cfrac{3^2}{2+\cfrac{5^2}{2+\cdots}}}}

X X - 2 years, 9 months ago

Thank you for contributing

Mohammad Farhat - 2 years, 9 months ago

X X, where is your comment?

Mohammad Farhat - 2 years, 8 months ago

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

X X - 2 years, 8 months ago

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Oh!

Mohammad Farhat - 2 years, 8 months ago

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

Mohammad Farhat - 2 years, 8 months ago

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@Mohammad Farhat – Can you give me your LaTeX?(Without the

X X - 2 years, 8 months 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]

Mohammad Farhat - 2 years, 8 months ago

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@Mohammad Farhat – π=4∑k=0∞(−1)k2k+1=3∑k=0∞(−1)k[16k+1+16k+5]=4∑k=0∞(−1)k[110k+1−110k+3+110k+5−110k+7+110k+9]=∑k=0∞(−1)k[314k+1−314k+3+314k+5−414k+7+414k+9−414k+11+414k+13]=∑k=0∞(−1)k[218k+1+318k+3+218k+5−218k+7−218k+11+218k+13+318k+15+218k+17]\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}

X X - 2 years, 8 months ago

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@X X – Thank you

Mohammad Farhat - 2 years, 8 months ago

@X X – I need the raw latex

Mohammad Farhat - 2 years, 8 months ago

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@Mohammad Farhat – \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} }

X X - 2 years, 8 months ago

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

X X - 2 years, 8 months ago

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@X X – ok

Mohammad Farhat - 2 years, 8 months ago

@X X – Thank you

Mohammad Farhat - 2 years, 8 months ago

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@Mohammad Farhat – 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 - 2 years, 8 months ago

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

X X - 2 years, 8 months ago

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@X X – Searching online what that is

Mohammad Farhat - 2 years, 8 months ago

@X X – got it

Mohammad Farhat - 2 years, 8 months ago

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

X X - 2 years, 8 months ago

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@X X – But it does not include the last bit

Mohammad Farhat - 2 years, 8 months ago

@X X – X X, Are you online?

Mohammad Farhat - 2 years, 8 months ago

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

Mohammad Farhat - 2 years, 8 months ago

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

Agnishom Chattopadhyay - 2 years, 8 months ago

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OHHHH! Now I get It!

Mohammad Farhat - 2 years, 8 months ago

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@Mohammad Farhat – Haha, were you asking about the \geq command?

Agnishom Chattopadhyay - 2 years, 8 months ago

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@Agnishom Chattopadhyay – No

Mohammad Farhat - 2 years, 8 months ago

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

Mohammad Farhat - 2 years, 8 months ago

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@Mohammad Farhat – I will report it

Mohammad Farhat - 2 years, 8 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.

Ram Mohith - 2 years, 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

>stuff>stuff

appears as

Stuff

stuff

Mohammad Farhat - 2 years, 7 months ago

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@Mohammad Farhat – Ok. Thanks Farhat.

Ram Mohith - 2 years, 7 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.)

Mohammad Farhat - 2 years ago

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

X X - 2 years ago

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I already realised where the mistake was

Mohammad Farhat - 2 years ago

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

Aruna Yumlembam - 11 months, 3 weeks ago
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