MHB Help with Maclaurin series representation

[fernlund]

Hello! So, I'm having a bit of a problem with an exercise in my Calculus book. I'm supposed to find the Maclaurin series representation of

$$ \frac{1+x^3}{1+x^2} $$

and then express it as a sum. Am I really supposed to differentiate this expression a bunch of times..? That will be very complicated quickly.

I've tried to solve this using Wolfram Mathematica as a help, to find $$ f'(0) $$, $$ f''(0) $$ etc, but of course I want to do it by myself.

Is there any kind of trick that I'm missing? A substitution or another way of writing this?

 

[Prove It]

Hello! So, I'm having a bit of a problem with an exercise in my Calculus book. I'm supposed to find the Maclaurin series representation of

$$ \frac{1+x^3}{1+x^2} $$

and then express it as a sum. Am I really supposed to differentiate this expression a bunch of times..? That will be very complicated quickly.

I've tried to solve this using Wolfram Mathematica as a help, to find $$ f'(0) $$, $$ f''(0) $$ etc, but of course I want to do it by myself.

Is there any kind of trick that I'm missing? A substitution or another way of writing this?

$\displaystyle \begin{align*} \frac{1 +x^3}{1 + x^2} = \left( 1 + x^3 \right) \, \frac{1}{1 - \left( -x^2 \right) } \end{align*}$

Now notice that $\displaystyle \begin{align*} \frac{1}{1 - \left( -x^2 \right) } \end{align*}$ is the closed form of a geometric series with $\displaystyle \begin{align*} r = -x^2 \end{align*}$.

 

[fernlund]

$\displaystyle \begin{align*} \frac{1 +x^3}{1 + x^2} = \left( 1 + x^3 \right) \, \frac{1}{1 - \left( -x^2 \right) } \end{align*}$

Now notice that $\displaystyle \begin{align*} \frac{1}{1 - \left( -x^2 \right) } \end{align*}$ is the closed form of a geometric series with $\displaystyle \begin{align*} r = -x^2 \end{align*}$.

Ah, thank you very much! I've got it from here :)