The question:
$\left(\frac{1+t}{1-t}\right)^{\frac{1}{2}}\:=\:\sum _{n=0}^{\infty }\:a_nt^n$
The question asks for what is $a_n$
These are the steps I've done:
$\left(\frac{1+t}{1-t}\right)^{\frac{1}{2}}\:=\:\left(\frac{1+t}{1-t}\right)^{\frac{1}{2}}\left(\frac{1+t}{1+t}\right)^{\frac{1}{2}}\:=\:\frac{1+t}{\left(1-t^2\right)^{\frac{1}{2}}}\:=\:\left(1+t\right)\left(1-t^2\right)^{-\frac{1}{2}}$
I know I need to expand $\left(1-t^2\right)^{-\frac{1}{2}}$ to then multiply it with $\left(1+t\right)$ formally and get a result. However, I do not know how to get from $\frac{1}{1-t}\:=\:\sum _{n=0}^{\infty }\:t^n\:\rightarrow \:\frac{1}{1-t^2}\:=\:\sum \:_{n=0}^{\infty \:}\:t^{2n}\:\rightarrow \:\frac{1}{\left(1-t^2\right)^{\frac{1}{2}}}\:=\:?$ or even if I am in the right track.
I appreciate any help! :)
Expand $x\mapsto \sqrt{1-x}$ into
$$\sqrt{1-x} = \sum_{n=0}^{\infty} (-x)^n \binom{1/2}{n}.$$ It is not very hard to prove that this is the correct Taylor expansion. Then, differentiate both sides of the equation and finally substitute $ x = t^2$.