Power Series representation

[jamesjohnes123]

Hello everyone.

I'm here with the following question about how to represent the following function as a power series.




Sorry about the image quality. From my understanding of how to represent a function as a power series, I should use the identity 1/1-x and try to make my function as close as possible to that identity. But, I have no clue on how to do that when I have the denominator elevated to a power.

Any help would be much appreciated. Hope you have a great day/night.

 

[Cubist]

You could start with...

[math] \frac{x^2}{\left(1 - x^2\right)^2} = x^2 \cdot \frac{1}{1-x^2} \cdot \, [/math]???

Then perhaps substitute u=? to make parts of the expression look like 1/(1-u).

Please post back with your work, and please use brackets so that we don't confuse "1/1-x" with (1/1) - x

 

[HallsofIvy]

You could also find the Taylor's series or MacLaurin series for \(\displaystyle f(x)= \frac{x^2}{(1- x^2)^2}\) or think of this as being \(\displaystyle \left(\frac{x}{1- x^2}\right)^2\) and recognize \(\displaystyle \frac{1}{1- x^2}\) as the sum of the geometric series \(\displaystyle \sum (r^2)^i\)

Conceptually, the simplest thing to do is to multiply out \(\displaystyle (1- x^2)^2= 1- 2x^2+ x^4\) and do the indicated division: divide \(\displaystyle 1- 2x^2+ x^4\) into \(\displaystyle x^2+ 0x^4+ 0x^6+ \cdot\cdot\cdot\). 1 divides into \(\displaystyle x^2\) \(\displaystyle x^2\) times and then we subtract \(\displaystyle (x^2+ 0x^4+ 0x^6)- (x^2- 2x^4+ x^6)= 2x^4- x^6\). divides into that \(\displaystyle 2x^4\) \(\displaystyle 2x^4\) times and then we subtract \(\displaystyle (2x^4- x^6+ 0x^8)- (2x^4- 4x^6+ 2x^8)= 3x^6- 2x^8\). So far we have the quotient \(\displaystyle x^2+ 2x^4\) and we can continue dividing until we see a pattern for the coefficients.