# Misunderstanding on Taylor series expansion

Sometimes, I try to find the series expansion of some random functions but I am usually wrong, especially when using already known formulas such as $$\frac{1}{1-X} = 1 + X + X^2 + X^3 + ... \quad (1)$$ Now imagine I want to find the series expansion of $$\frac{1}{2+x^2}$$, here is what I do :

$$\frac{1}{2+x^2}=\frac{1}{1-(-x^2-1)}=1+(-x^2-1)+(-x^2-1)^3+... \quad (2)$$

But the result is not the same as the results in the internet that finds : $$\frac{1}{2+x^2}= \frac{1}{2}- \frac{1}{4}x^{2}+\frac{1}{8}x^{4}- \frac{1}{16}x^{6}$$.

My guess is that the $$X$$ in (1) must verify $$\lvert X \rvert < 1$$ and when I take $$X = (-x^2-1)$$ in (2) then this condition is not verified. I also think that to find the right answer, computers do $$\frac{1}{2+x^2} = \frac{1}{2}\frac{1}{1+(\frac{x^2}{2})}$$

• Didn't you answer your own question in the question itself? Jan 9 at 16:05
• It was a guess, I do not know if my guess is right or wrong
– John
Jan 9 at 16:06
• In a Taylor series abut $0$, the terms are of the form "constant times power of $x$". More generally, in a Taylor series about $a$, the terms are of the form "constant times power of $x-a$. In any case, they are not of the form "constant times power of $-x^2-1$. So even if your series converged, it wouldn't be a Taylor series. Jan 9 at 18:05

The expansion $$\frac{1}{1-x} = 1 + x + x^2 + x^3 + ...$$ only works if $$|x|<1$$ as you pointed out yourself. Because $$2+x^2>1$$ the substitution you make is guaranteed to give you a divergent sum.
Now, by factoring out $$\frac{1}{2}$$ you solve that problem, since now it works whenever $$|x|<\sqrt{2}$$. So the substitution is fine (that is, it results in a convergent series).