Power Series and Functions

Representing a function as a power series simply means finding an endless polynomial that equals that function, at least for $x$ near the center. All we're really saying in this lesson is that once you know one such series, you can get lots more for free by doing ordinary algebra to it. The starting point $\frac{1}{1-x} = 1 + x + x^2 + \cdots$ is just the geometric series, where each term is the last one times $x$. "Substitute" means swap $x$ for something else (like $-x^2$), and "term by term" means you differentiate or integrate each power separately, exactly like a polynomial ($x^3$ becomes $3x^2$ or $\frac{x^4}{4}$). The formulas look busy, but every step is something you already know how to do.

Why bother? Because a lot of functions, like $\ln(1 + x)$ and $\arctan x$, have derivatives that look like $\frac{1}{1 - \text{something}}$. If we can write that derivative as a geometric series, integrating term by term hands us a series for the original function.

Substitute. Replace $x$ by anything: $-x$, $x^2$, $3x$. The condition $|x| < 1$ changes with it, since it now applies to whatever you put in.

• $\dfrac{1}{1+x^2} = \dfrac{1}{1 - (-x^2)} = \sum (-1)^n x^{2n}$, for $|x| < 1$.
• $\dfrac{1}{5 - x} = \dfrac15\cdot\dfrac{1}{1 - x/5} = \sum \dfrac{x^n}{5^{n+1}}$, for $|x| < 5$.

Multiply by powers of $x$. $\dfrac{x}{1-x} = \sum x^{n+1}$.

Differentiate or integrate. Inside the radius of convergence you can treat the series like a long polynomial:

• Differentiating $\sum x^n$ gives $\dfrac{1}{(1-x)^2} = \sum n x^{n-1}$.
• Integrating $\sum (-1)^n x^n$ gives $\ln(1 + x) = \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1} x^n}{n}$.
• Integrating $\sum (-1)^n x^{2n}$ gives $\arctan x = \sum \dfrac{(-1)^n x^{2n+1}}{2n+1}$.

The radius of convergence never changes when you differentiate or integrate. That makes sense: multiplying or dividing the $n$th term by $n$ is tiny next to the $|x|^n$ that decides whether the terms shrink.

A common mistake. It's tempting to assume the whole interval carries over too. It doesn't: the endpoints can change. $\sum x^n$ diverges at $x = -1$, but its integral, the series $\sum \frac{x^{n+1}}{n+1}$ for $-\ln(1 - x)$, converges there. So after differentiating or integrating, keep the same $R$ but recheck the endpoints.

Free sums. Plugging numbers into these series evaluates series you couldn't otherwise sum. In the last unit we could prove the alternating harmonic series converges, but not what it converges to. Now we can: $x = 1$ in the $\ln(1+x)$ series gives $1 - \frac12 + \frac13 - \cdots = \ln 2$, and $x = 1$ in the $\arctan$ series gives $1 - \frac13 + \frac15 - \cdots = \frac\pi4$.

Probability generating functions and the z-transforms used in digital filters are handled exactly this way: engineers multiply, differentiate and integrate the series instead of the function.