Binomial Series to find a Maclaurin Series

**psu1024** · July 19th 2009, 11:04 PM

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!

**Opalg** · July 19th 2009, 11:45 PM

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 $y=x^2$ .

**psu1024** · July 20th 2009, 06:56 AM

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?

**Opalg** · July 21st 2009, 12:15 AM

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 $\frac1{\sqrt[3]{1+x^2}} = 1-\frac{x^2}{3}+\frac {2x^4}{3^22!}-\frac {28x^6}{3^33!}+\ldots$ .

**psu1024** · July 21st 2009, 07:15 AM

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.

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