Answer to Question #122708 in Differential Equations for Prathibha Rose C S

Question

Express sin inverse x in the form of a power series summation an x^n by solving y'= (1-x^2) ^(1/2) in two ways use this result to obtain the formula pi/6=1/2+ 1/2 .1/(3.(2)^3)+1/2.3/4.1/5.(2)^5)+ etc ......

Answer 1

$y'=\sqrt{1-x^2}$

$y=\int\sqrt{1-x^2}dx$

$x=sinu, u=arcsinx,dx=cosudu$

$y=\int cos^2udu=\int \frac {cos2u+1}{2}du=\frac {cosusinu+u}{2}$

$y=\frac {arcsinx+x\sqrt{1-x^2}}{2}+C$

$y=\sum^{\infin}_{n=0} a_nx^n$

$y'=\sum^{\infin}_{n=1} na_nx^{n-1}=\sqrt{1-x^2}$

$c+2\sum^{\infin}_{n=0} a_nx^n=arcsinx+x\sum^{\infin}_{n=1} na_nx^{n-1}$

$c+2\sum^{\infin}_{n=0} a_nx^n=arcsinx+\sum^{\infin}_{n=1} na_nx^n$

$arcsinx=c+2\sum^{\infin}_{n=0} a_nx^n-\sum^{\infin}_{n=1} na_nx^n$

$arcsinx=c+2a_0+\sum^{\infin}_{n=1}(2-n)a_nx^n$

$arcsin0=0\implies c+2a_0=0$

Answer:

$arcsinx=\sum^{\infin}_{n=1}(2-n)a_nx^n$

$(arcsinx)'=\sum ^{\infin}_{n=1}n(2-n)a_nx^{n-1}=\frac {1}{\sqrt{1-x^2}}$

$(arcsin0)'=1=a_1$

$(arcsin0)'''=1\implies a_3=-1/6$

By same way we can get $a_4,a_5,...$

So:

$\frac {\pi}{6}=arcsin(1/2)=\frac {1}{2}+\frac {1}{6}\cdot\frac {1}{2^3}+...$