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.

  • $\begingroup$ Of course the part that you can integrate and differentiate a series needs to be justified, infinite series integration and differentiation is different than finite cases. $\endgroup$ – jimjim Feb 3 '11 at 3:57

Are these the notes you took in class? ie.. do you just want some explanation on some of the details? Well for your first example, in the last step the $$1\over 5$$ is multiplied into the summation giving the term $$x^{n+1}\over {5^{n+1}(n+1)}$$. Then the power series is shifted by replacing n with n-1 (which explains the term you have written in the last step). However, in doing this, the series now starts at $n=1$ not $n=0$. Was that a typo? Clearly the term to the right of the last summation is undefined at $n=0$

For the second example (and I guess the first one too) the term $C$ is an arbitrary constant of integration. So to evaluate the definite integral in your second question, you can choose $C$ to be zero as any antiderivative can be used in the fundamental theorem of calculus.

In general, power series are nice because polynomials are by far the easiest functions to work with, and a power series allows you approximate a much harder function as a polynomial. There are only a few well known power series in elementary calculus (namely, $sin(x)$, $cos(x)$, $e^x$, and $1\over {x-1}$ ) but by differentiating and integrating you can produce power series for many, many functions. That's the whole point of this.

  • $\begingroup$ Thanks Joe. Not notes, just me trying to understand what's going on in the problems. I was just wondering if my line of thinking was on the right path or if I was just completely out of line. When I first was introduced to this section, I didn't understand why we were integrating/differentiating functions and how they produced the power series we wanted. $\endgroup$ – ShrimpCrackers Feb 3 '11 at 2:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.