I did fine in Calculus 1 because everything could be drawn up in diagrams, easily derived, and pretty much thought of in terms of logic. Limits make perfect sense, asymptotes make perfect sense, derivatives and second derivatives make sense, and even integrals make sense to me. I can diagram all of these things and see them visually and convince myself that they're true.

But now in Calc 2, I'm struggling like never before in my life. I feel depressed because of the hard time I am having with power series. More and more, I feel that I am just memorizing the stuff our teacher/book present and regurgitating it so that I can get my homework done and get a good grade, but it's not working because the day immediately after I work on some problems and watch the lecture, I've completely forgotten everything and the rationale behind it all. I don't understand differentiation/integration/substitution with these darn power series. I just don't get it. It seriously seems like wizardry to me.

Are there any good guides for thoroughly understanding this stuff?

  • $\begingroup$ What do you mean by power series ? Please provide an example. $\endgroup$
    – Furrane
    Oct 27 '16 at 22:57
  • $\begingroup$ representing functions with power series $\endgroup$
    – AleksandrH
    Oct 27 '16 at 23:04

Polynomials are nice. It would be really nice if every function was a polynomial. However that is not the case. What if we "stretch" the definition of polynomial a little bit, and allow infinite sums?

What I mean is that a polynomial is a finite sum $$a_0+a_1x+a_2x^2+\cdots+a_n x^n$$ What if we were to consider infinite sums $$a_0+a_1x+a_2x^2+a_3x^3+\cdots=\sum_{n=0}^\infty a_nx^n?$$ Since a series is simply a limit of finite sums, a series as above can be thought of as being approximated by polynomials.

Let's call such a series $\sum_{n=0}^\infty a_nx^n$ a power series. Let's try to see which kind of function can be described by a power series: Suppose $f(x)$ is such a function: there are $a_0,a_1,\ldots$ such that $$f(x)=\sum_{n=0}^\infty a_nx^n=a_0+a_1x+a_2x^2+\cdots$$ The natural question is *how do we calculate $a_n$ from $f$"? Wel, if we evaluate both sides at $x=0$, we obtain $$f(0)=a_0+a_10+a_20^2+\cdots=a_0$$ so we found the value of $a_0$. If we take derivatives, we obtain $$f'(x)=a_1+2a_2x+3a_3x^2+4a_4x^3+\cdots$$ so again evaluating at $0$ gives us $a_1=f'(0)$. Taking derivatives again, we obtain $$f''(x)=2a_2+3\cdot 2a_3x+4\cdot 3x^2+\cdots$$ so evaluating again at $x=0$ yields $2a_2=f''(0)$.

More generally, when we take the $n$-th derivative, the exponents of $x$ will decrease by $n$, and the coeficcient $a_n$ will be multiplied by $n!$. The terms $a_ix^i$, with $i<n$ will become $0$, and terms $a_ix^i$ with $i>n$ will become $b_ix^{i-n}$ for some $b_i$ (you can verify that $b_i=\frac{i!}{(i-n)!}a_i$, but that doesn't matter). Anyway, evaluating at $0$ will kill those guys, and we will have $$f^{(n)}(0)=n!a_n$$ where $f^{(n)}$ means that we take derivatives $n$ times. This gives us formulas for $a_n$, so that $$f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n.$$

Again, not every function admits such a representation, but a large class of functions does.

However, we were evaluating at $0$, and some functions are not defined at $0$ ($\log$, for example). We can make a similar analysis and look more generally at functiosn which can be represented by $$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(c)}{n!}(x-c)^n,$$ where $c$ is some element of the domain of $f$. Then $\log$ can be represented in this form: First we need to verify that $\log^{(n)}(1)=(-1)^{n+1}(n-1)!$, so $$\log(x)=\sum_{n=1}^\infty(-1)^{n+1}\frac{(x-1)^n}{n}$$.

There are some natural questions: In what sense does $\sum a_nx^n$ converges? Poitwise of uniformly? This is dealt with in the study of power/Taylor series.


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