1
$\begingroup$

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 $$\frac{1}{2+x^2} = \frac{1}{2}\frac{1}{1+(\frac{x^2}{2})}$$

$\endgroup$
3
  • 1
    $\begingroup$ Didn't you answer your own question in the question itself? $\endgroup$ Jan 9 at 16:05
  • $\begingroup$ It was a guess, I do not know if my guess is right or wrong $\endgroup$
    – John
    Jan 9 at 16:06
  • $\begingroup$ 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. $\endgroup$ Jan 9 at 18:05
4
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.