Engineering Mathematics: Reciprocals of Binomial Coefficients

$\frac{1}{\binom{n}{r}}=\frac{r!(n-r)!}{n!}=\frac{\Gamma (r+1)\Gamma (n-r+1)}{\Gamma(n+1) }= (n+1)\int _{0}^{1} t^{r}(1-t)^{r} dt$

(1)

RBC.ID.1: Trigonometric Identity

$\sum _{m\geqslant 1}\frac{(2x)^{2m}}{m\binom{2m}{m}} = \frac{2x}{\sqrt{1-x^{2}}}\sin^{-1}x$

(2)

proof

$\sum _{m\geqslant 1}\frac{(2x)^{2m}}{m\binom{2m}{m}} = \sum _{m\geqslant 1}(2x)^{2m} \beta (m+1,m)$

(3)

$\sum _{m\geqslant 1}\frac{(2x)^{2m}}{m\binom{2m}{m}} = \sum _{m\geqslant 1}(2x)^{2m} \int _{0}^{1} t^{m}(1-t)^{m-1} dt$

(4)

by geometric series

$\sum _{m\geqslant 1}\frac{(2x)^{2m}}{m\binom{2m}{m}} = \int _{0}^{1} \frac{4x^{2}t}{1-4x^{2}t(1-t)} dt$

(5)

replacing (5) by,

$s= x(2t-1)$

(6)

results

$\int _{x}^{-x}\frac{s}{s^{2}+ (1-x^{2})}ds + x\int _{x}^{-x}\frac{1}{s^{2}+ (1-x^{2})}ds$

(7)

guided by the following identities:

$\int \frac{1}{a^{2}+u^{2}}du = \frac{1}{a}\tan^{-1}(\frac{u}{a}) + C$

(8)

$\int \sin^{-1}(\frac{u}{a})du = u\sin^{-1}(\frac{u}{a} )+ \sqrt{a^{2}-u^{2}} + C$

(9)

$\tan^{-1}(\frac{x}{\sqrt{1-x^{2}}}) = \tan^{-1}(\tan (\sin^{-1}x))= \sin^{-1}x$

(10)

from (7) , (8) , (9) , and (10) , The first part of (7):

$\int _{-x}^{x}\frac{s}{s^{2}+(1-x^{2})}ds = \left | \frac{s}{\sqrt{1-x^{2}}}\tan^{-1}(\frac{s}{\sqrt{1-x^{2}}}) \right |_{-x}^{x} - \int _{-x}^{x}\frac{1}{\sqrt{1-x^{2}}}\tan^{-1}(\frac{s}{\sqrt{1-x^{2}}})$

(11)

The first part of (11) equals zero

$\int _{-x}^{x}\frac{s}{s^{2}+(1-x^{2})}ds = - \int _{-x}^{x}\frac{1}{\sqrt{1-x^{2}}}\tan^{-1}(\frac{s}{\sqrt{1-x^{2}}})$

(12)

$\int _{-x}^{x}\frac{s}{s^{2}+(1-x^{2})}ds = - \int _{-x}^{x}\frac{1}{\sqrt{1-x^{2}}}\sin^{-1}(s)= -\frac{1}{\sqrt{1-x^{2}}} \left | s\sin^{-1}(s) +\sqrt{s^{2}} \right |_{-x}^{x}$

(13)

$-\frac{1}{\sqrt{1-x^{2}}} \left | s\sin^{-1}(s) +\sqrt{s^{2}} \right |_{-x}^{x}=0$

(14)

from (7) , (8) , (9) , and (10) , The second part of (7):

$x\int _{-x}^{x}\frac{1}{s^{2}+(1-x^{2})} ds = \left | \frac{x}{\sqrt{1-x^{2}}}\tan^{-1}( \frac{s}{\sqrt{1-x^{2}}}) \right |_{-x}^{+x}$

(15)

$x\int _{-x}^{x}\frac{1}{s^{2}+(1-x^{2})} ds = \left | \frac{x}{\sqrt{1-x^{2}}}\tan^{-1}( \tan\sin^{-1}s ) \right |_{-x}^{+x}$

(16)

$x\int _{-x}^{x}\frac{1}{s^{2}+(1-x^{2})} ds = \left | \frac{x}{\sqrt{1-x^{2}}} \sin^{-1}s \right |_{-x}^{+x}$

(17)

$x\int _{-x}^{x}\frac{1}{s^{2}+(1-x^{2})} ds = \frac{2x}{\sqrt{1-x^{2}}}\sin^{-1}x$

(18)

from (14 and (18) ,

$\sum _{m\geqslant 1}\frac{(2x)^{2m}}{m\binom{2m}{m}} = \frac{2x}{\sqrt{1-x^{2}}}\sin^{-1}x$

(19)

if x=1/2 , then

$\sum _{m\geq 1}\frac{1}{m\binom{2m}{m}}=\frac{\pi \sqrt{3}}{9}$

(20)

RBC.ID.2: Derivative of Trigonometric Identity-1

$\sum _{m\geq 1}\frac{1}{\binom{2m}{m}}= \frac{2\pi }{9\sqrt{3}} + \frac{1}{3}$

(21)

Proof

differntiate (19) in respect to x

$\sum _{m\geq 1}\frac{2m\times 2(2x)^{2m-1}}{m\binom{2m}{m}}=\frac{d}{dx}(\frac{2x}{\sqrt{1-x^{2}}})\sin^{-1}(x)+ \frac{d}{dx}(\sin^{-1}(x))\frac{2x}{\sqrt{1-x^{2}}}$

(22)

$\sum _{m\geq 1}\frac{4(2x)^{2m-1}}{\binom{2m}{m}}=\frac{d}{dx}(\frac{2x}{\sqrt{1-x^{2}}})\sin^{-1}(x)+ \frac{d}{dx}(\sin^{-1}(x))\frac{2x}{\sqrt{1-x^{2}}}$

(23)

$\sum _{m\geq 1}\frac{4(2x)^{2m-1}}{\binom{2m}{m}}=(\frac{2\sqrt{1-x^{2}}+2x^{2}(\sqrt{1-x^{2}})^{\frac{-1}{2}}}{(\sqrt{1-x^{2}})^{2}})\sin^{-1}(x)+ \frac{d}{dx}(\sin^{-1}(x))\frac{2x}{\sqrt{1-x^{2}}}$

(24)

multiply and divide the first part of right-hand side by $(\sqrt{1-x^{2}})^{\frac{1}{2}}$

$\sum _{m\geq 1}\frac{4(2x)^{2m-1}}{\binom{2m}{m}}=(\frac{2(1-x^{2})+2x^{2}}{(\sqrt{1-x^{2}})(1-x^{2})})\sin^{-1}(x)+ \frac{d}{dx}(\sin^{-1}(x))\frac{2x}{\sqrt{1-x^{2}}}$

(25)

$\frac{d}{dx}(\sin^{-1}(x)) = \frac{1}{\sqrt{1-x^{2}}}$

(26)

from (25) and (26)

$\sum _{m\geq 1}\frac{4(2x)^{2m-1}}{\binom{2m}{m}}=(\frac{2(1-x^{2})+2x^{2}}{(\sqrt{1-x^{2}})(1-x^{2})})\sin^{-1}(x)+ \frac{1}{\sqrt{1-x^{2}}} \frac{2x}{\sqrt{1-x^{2}}}$

(27)

$\sum _{m\geq 1}\frac{4(2x)^{2m-1}}{\binom{2m}{m}}=(\frac{2(1-x^{2})+2x^{2}}{(\sqrt{1-x^{2}})(1-x^{2})})\sin^{-1}(x)+ \frac{2x}{1-x^{2}}$

(28)

if x=1/2

$\sum _{m\geq 1}\frac{4}{\binom{2m}{m}}= \frac{\pi }{6} (\frac{2(3/4)+1/2}{3/4 \sqrt{3/4}}) + 4/3$

(29)

divide both sides by 4

$\sum _{m\geq 1}\frac{1}{\binom{2m}{m}}= \frac{2\pi }{9\sqrt{3}} + \frac{1}{3}$

(30)

RBC.ID.3: Derivative of Trigonometric Identity-2

$\sum_{n=0}^{\infty } \frac{1}{(2n+1)\binom{2n}{n}} = \frac{2\pi }{3\sqrt{3}}$

Proof

$\int _{0}^{1}x^{n}(1-x)^{n} dx = \frac{(n!)^{2}}{(2n+1)!} = \frac{1}{(2n+1)\binom{2n}{n}}$

(31)

$\sum _{n=0}^{\infty }\frac{1}{(2n+1)\binom{2n}{n}}=\sum _{n=0}^{\infty }\int _{0}^{1}x^{n}(1-x)^{n} dx$

(32)

$\sum _{n=0}^{\infty }\frac{1}{(2n+1)\binom{2n}{n}}=\int _{0}^{1}\sum _{n=0}^{\infty }x^{n}(1-x)^{n} dx$

(33)

Using geometric mean

$\int _{0}^{1}\sum _{n=0}^{\infty }x^{n}(1-x)^{n}dx = \int _{0}^{1} \frac{1}{1-x(1-x)} dx$

(34)

$\int _{0}^{1} \frac{1}{1-x(1-x)} dx = \int _{0}^{1} \frac{1}{x^{2}-x+1} dx = \int _{0}^{1} \frac{1}{(x-\frac{1}{2})^{2}+(\frac{\sqrt{3}}{2})^{2}} dx$

(35)

Replacing

$x-\frac{1}{2} = \frac{\sqrt{3}}{2}\tan \theta$

$dx = \frac{\sqrt{3}}{2} \sec ^{2} \theta \, \, d\theta = \frac{\sqrt{3}}{2\cos ^{2} \theta } \, \, d\theta$

(36)

$\int _{0}^{1} \frac{1}{(x-\frac{1}{2})^{2}+(\frac{\sqrt{3}}{2})^{2}} dx = \int _{\frac{\pi }{6}}^{\frac{-\pi }{6}} \frac{1}{(\frac{\sqrt{3}\tan \theta }{2})^{2}+(\frac{\sqrt{3}}{2})^{2}} \frac{\sqrt{3}}{2\cos ^{2} \theta } \, \, d\theta$

(37)

$\int _{0}^{1} \frac{1}{(x-\frac{1}{2})^{2}+(\frac{\sqrt{3}}{2})^{2}} dx = \frac{2}{\sqrt{3}} \int _{\frac{\pi }{6}}^{\frac{-\pi }{6}} \, \, d\theta = \frac{2\pi }{3\sqrt{3}}$

(38)

From (31) and (38) , Therefore ,

$\sum_{n=0}^{\infty } \frac{1}{(2n+1)\binom{2n}{n}} = \frac{2\pi }{3\sqrt{3}}$

(39)

Engineering Mathematics

Saturday, October 22, 2011

Reciprocals of Binomial Coefficients

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