<s>[INST] Calculate a function with the use of a development point and geometric series Assuming I have this function :
$$ f(x) = \frac{1 + x^3}{2-x}  $$
The question was to calculate this with the use of the geometric series and  the development point $x_{0} = 0$.
Well setting first the function to a geometric series is quite clear here .
I got $ f(x)=$ $\sum_{n=1}^\infty \frac{x^n}{2^{n+1}}(1+x^3) $ as a result . What's the meaning of calculating an infinite series here ? By calculating does this mean that I can choose to calculate the first terms of the series (For example from n=0 to n=3 and then   $  '+\ldots '$ ) ?
Plus why do I need to use the development point $x_{0}$ . I can already calculate the function with the geometric series only . Won't that be enough or is there another idea to implement this?
 [/INST] We are looking for a representation
\begin{align*}
f(x)=\sum_{n=0}^\infty a_n(x-x_0)^n=\sum_{n=0}^\infty a_nx^n\tag{1}
\end{align*}
where the series expansion is at the point $x_0=0$.

We obtain
\begin{align*}
\color{blue}{f(x)}&\color{blue}{=\frac{1+x^3}{2-x}}
=\frac{1}{2}\cdot\frac{1+x^3}{1-\frac{1}{2}x}\\
&=\frac{1}{2}\sum_{n=0}^\infty\frac{1}{2^{n}}x^n(1+x^3)\\
&=\sum_{n=0}^\infty\frac{1}{2^{n+1}}x^n+\sum_{n=0}^\infty\frac{1}{2^{n+1}}x^{n+3}\\
&=\sum_{n=0}^\infty\frac{1}{2^{n+1}}x^n+\sum_{n=3}^\infty\frac{1}{2^{n-2}}x^{n}\\
&\,\,\color{blue}{=\frac{1}{2}+\frac{1}{4}x+\frac{1}{8}x^2+\sum_{n=3}^\infty\frac{9}{2^{n+1}}x^n}
\end{align*}
with the last line being a representation according to (1).

</s>