Question

If $y=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...\infty$ and $|x|<1$, then $x=y+\frac{y^2}{a!}+\frac{y^3}{b!}+\dots \dots +\infty$. Find $a^2+b$.

Answer 1

Since,

$y=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...\infty$

$y={\log }_e(1+x)$

$\therefore$$e^y=1+x$

$1+y+\frac{y^2}{2!}+\frac{y^3}{3!}+...=1+x$

$\therefore$$x=y+\frac{y^2}{2!}+\frac{y^3}{3!}+...,\infty$

$\Rightarrow a=2,b=3\therefore a^2+b=7$