I'm having a hard time understanding what exact processes I have to take to represent a function as a power-series.
I understand that $\frac{1}{1-x} = 1 + x + x^2 + x^3 + ... = \sum_{0}^\infty x^n$
So, in words, $\frac{1}{1-x}$ can be represented as the power series $\sum_{0}^\infty x^n$.
Jumping to the problem:
Represent $f(x) = ln(5-x)$ as a power series.
$ln(5-x) = -\int\frac{1}{5-x}dx = -\frac{1}{5}\int\frac{1}{1-\frac{x}{5}} = -\frac{1}{5}\int[\sum_{0}^\infty (\frac{x}{5})^n]dx = C - \frac{1}{5}\sum_{0}^\infty \frac{x^{n+1}}{5^n(n+1)} = C - \sum_{0}^\infty \frac{x^n}{n5^n}$
In this problem, it's difficult to get ln(5-x) in some sort of geometric form to turn into a power series...but we can differentiate it to get it into the convenient geometric form $\frac{1}{5-x}$. The constant is factored out and put into power series form with $r = (\frac{x}{5})^n$. Then it is integrated. But because we're integrating a derivative of $ln(5-x)$ in the form of a power series, we're actually getting ln(5-x) in the form of a power series...which is what we wanted.
is this correct (btw, what happened to that 1/5 in that last step?)? This is one reason why we integrate/differentiate power series, right? To get some function in the form of power series where it can be difficult to do so?
Another problem:
This problem's solution confused me a bit (the original instruction says to use a power series to approximate the definite integral to six decimal places the limits of integration being 0 to 0.2, but I don't care about that right now): $\int xarctan(3x) = \int x \sum_{0}^\infty (-1)^n\frac{(3x)^{2n+1}}{2n+1} dx = \int\sum_{0}^\infty(-1)^n\frac{(3)^{2n+1}x^{2n+2}}{2n+1}dx = C + \sum_{0}^\infty(-1)^n\frac{(3)^{2n+1}x^{2n+3}}{(2n+1)(2n+3)}$
In this case, we have a function $xarctan(3x)$ we want to turn into a power series (so we can approximate). (After skipping alot of steps) The arctan is put into geometric form, and then into a power series. The power series, arctan(3x) is then integrated...with the constant x factored out for the moment. Then it is factored in and then the x itself is integrated, giving us $C + ...power series$ so essentially, we now finally have xarctan(3x) in power series form. Is this correct?
Just wanted a confirmation that I was seeing this correctly cause I had a hard time following in class what exactly the point of integrating/differentiating power series was.