Question

Find the sum of the following series:
$\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+.......$

Answer 1

$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\frac{x^6}{6!}+\frac{x^7}{7!}+....$ --- ( 1 )

$e^{-x}=1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}-\frac{x^5}{5!}+\frac{x^6}{6!}-\frac{x^7}{7!}+....$ --- ( 2 )

Subtracting ( 2 ) from ( 1 ) we get,

$e^x-e^{-x}=[1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\frac{x^6}{6!}+\frac{x^7}{7!}+..]-[1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}-\frac{x^5}{5!}+\frac{x^6}{6!}-\frac{x^7}{7!}+..]$

$e^x-e^{-x}=2x+\frac{2x^3}{3!}+\frac{2x^5}{5!}+\frac{2x^7}{7!}+..$

$\frac{e^x-e^{-x}}{2}=x+\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+..$

$\therefore$ $\frac{e^x-e^{-x}}{2}-x=\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+..$