Question

Power series expansion

Answer 1

\[\int\limits_{0}^{4} \ln(1+x^4)dx\] also list first 3 terms.

So we always have to use \[\frac{ 1 }{ 1+x }\] as a starting point?

Answer 2

@SithsAndGiggles @myininaya

So the best approach for a such question is by first using a different variable right?

\[\frac{ 1 }{ 1+t } = \frac{ 1 }{ 1-(-t) }\]

But, I'm not exactly sure how to find a power series representation of this

Answer 3

Well there's no single right way to approach this. If you memorize the power series for \(\ln(1+x)\) or \(\ln(1-x)\) or any other variation, you can replace the argument with whatever may be appropriate.

What I would do is, considering I don't remember the power series for every single function out there, would be to work with the given logarithm's derivative.
\[\frac{d}{dx}\ln(1+x^4)=\frac{4x^3}{1+x^4}=4x^3\sum_{n=0}^\infty(-x^4)^n=4\sum_{n=0}^\infty(-1)^nx^{4n+3}\]
Integrating gives the power series for the logarithm:
\[\ln(1+x^4)=4\sum_{k=0}^\infty\frac{(-1)^nx^{4n+4}}{4n+4}=\sum_{k=0}^\infty\frac{(-1)^nx^{4n+4}}{n+1}\]

Answer 4

As far as your approach is concerned, you're on the right track in terms of trying to find the geometric series expression for the function.
\[\frac{1}{1-x}=\sum x^n~~\implies~~\frac{1}{1-f(x)}=\sum [f(x)]^n\]

Answer 5

Very cool, I will have to go over this a few times to fully understand this process, thank you.

Haha, I was just writing the power series representation for \[\frac{ 1 }{ 1-x }\] to confirm it and you answered that as well.

Thank you so much!

Answer 6

You're welcome!

Answer 7

Cool! I understood the process, was a little hesitant on the \[4 \sum_{n=0}^{\infty} (-1)^n x^{4n+3}\] thinking where did the x^3 go but I realized it's in x^4n+3, very cool stuff, thanks again.