1. ## Taylor theorem

Prove the given equation using Taylor series theorem.

$\displaystyle sin^{-1}\frac{2x}{1+x^{2}}=2\left \{ x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+... \right \}$

2. ## Re: Taylor theorem

Originally Posted by rishilaish

Prove the given equation using Taylor series theorem.

$\displaystyle sin^{-1}\frac{2x}{1+x^{2}}=2\left \{ x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+... \right \}$
What have you tried? Where are you stuck?

3. ## Re: Taylor theorem

Originally Posted by rishilaish

Prove the given equation using Taylor series theorem.

$\displaystyle sin^{-1}\frac{2x}{1+x^{2}}=2\left \{ x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+... \right \}$
My 'proof' doesn't use the 'Taylor series theorem' but avoids tedious computation of derivatives...

Let's strart with the hypothesis that it exists the McLaurin expansion...

$\displaystyle \sin^{-1} \frac{2x}{1+x^{2}}= \sum_{n=0}^{\infty} a_{n}\ x^{n}$ (1)

...and we want to compute the $\displaystyle a_{n}$. Now, considering that is...

$\displaystyle \int \sin^{-1} t\ dt= t\ \sin^{-1} t + \sqrt{1-t^{2}}+c$ (2)

... setting in (2) $\displaystyle t=\frac{2 x}{1+x^{2}}$ we derive using (1) and a little of patience...

$\displaystyle \int_{0}^{x} \sin^{-1} t\ dt = \frac{2 x}{1+x^{2}}\ \sin^{-1} \frac{2x}{1+x^{2}} - \frac{2\ x^{2}}{1+x^{2}} =$

$\displaystyle = \frac{2 x}{1+x^{2}}\ (\sum_{n=0}^{\infty} a_{n}\ x^{n}-x)= \sum_{n=0}^{\infty} \frac{a_{n}}{n+1}\ x^{n+1}$ (3)

... and the identity (3) permits the computation of the $\displaystyle a_{n}$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$