# Weird Power Series Expansion

When I do the power series expansion of $\displaystyle f(x)={1\over(4-2x^3)}$ and find the interval of convergence, I get two different answers depending on the way that I solve it.

If I break $f(x)$ up into $\displaystyle \frac14\left({1\over (1-(1/2)x^3)}\right)$ and then use the geometric series $${1\over(1-x)}= 1 + x +x^2 ... \text{if } |x|<1$$ I get $-(2)^{1/3} < x < 2^{1/3}$

If I break $f(x)$ up into $\displaystyle {1\over(1-(2x^3-3))}$ I end up getting an interval of convergence of $1 < x < 2^{1/3}$

What am I doing wrong?

• Is my edit fine ?
– user312097
Apr 27 '17 at 18:51
• These are expansions at different points. No wonder they converge on different intervals. There's nothing wrong about it. Apr 27 '17 at 19:32
• Thanks! I appreciate the edit! Apr 27 '17 at 21:43

You are not doing anything wrong. You have developed two different representations of the same function $f(x)=\frac{1}{4-2x^3}$.

The first is a power series in powers of $x^3$

$$f(x)=\frac14 \sum_{n=0}^\infty \left(\frac{x^3}{2}\right)^n=\frac14\sum_{n=0}^\infty \frac{x^{3n}}{2^{n}}\tag 1$$

for $|x|<2^{1/3}$.

The second is a power series in powers of $2x^3-3$

$$f(x)= \sum_{n=0}^\infty \left(2x^3-3\right)^n=\sum_{n=0}^\infty 2^n(x^3-3/2)^n\tag 2$$

for $|2x^3-3|<1\implies 1<x<2^{1/3}$.

Note that $(1)$ is a power series in $x$ centered at $x=0$, which $(2)$ is a power series in $x^3$ centered at $x^3=3/2$.

• You're welcome! My pleasure. -Mark Apr 27 '17 at 21:45
• Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark May 1 '17 at 3:48