# Thread: Binomial Series to find a Maclaurin Series

1. ## Binomial Series to find a Maclaurin Series

I'm having a tough time finding the Maclaurin series (we have to use the binomial series) for
$
f(x)=\frac{1}{\sqrt [3]{1+x^2}}
$

Am I supposed to try to make it fit the form $(1+x)^k$? If so, I'm confused about how to do so. Any help walking me through this step by step would be greatly appreciated. Thanks!

2. Originally Posted by psu1024
I'm having a tough time finding the Maclaurin series (we have to use the binomial series) for
$
f(x)=\frac{1}{\sqrt [3]{1+x^2}}
$

Am I supposed to try to make it fit the form $(1+x)^k$? If so, I'm confused about how to do so. Any help walking me through this step by step would be greatly appreciated. Thanks!
Use the formula for $(1+y)^{-1/3}$ and then substitute $y=x^2$.

3. So does the equation become this:
$
(1+y)^{-1/3} = 1-\frac{y}{3}+\frac {2y^2}{3^22!}-\frac {28y^3}{3^33!}+...
$

How do I substitute the $x^2$ back into the equation though?

4. Originally Posted by psu1024
So does the equation become this:
$
(1+y)^{-1/3} = 1-\frac{y}{3}+\frac {2y^2}{3^22!}-\frac {28y^3}{3^33!}+...
$

How do I substitute the $x^2$ back into the equation though?
That's correct. Now replace y by x^2, and you get $\frac1{\sqrt[3]{1+x^2}} = 1-\frac{x^2}{3}+\frac {2x^4}{3^22!}-\frac {28x^6}{3^33!}+\ldots$.

5. Thanks! That's how I ended up doing the problem. I just wasn't sure if there was some weird substitution technique or something. Thanks for your help.