Functions Expressed as Power Series

A power series is an infinite sum of the form \(\sum_{n=0}^{\infty} c_n (x-a)^n\). When a function \(f(x)\) can be written exactly as one of these sums (for the values of \(x\) where the sum converges), we say \(f(x)\) has a power series representation. This is a powerful idea: it lets us treat complicated functions like polynomials with infinitely many terms, which is useful for approximation, integration, and solving differential equations.

Rather than guessing coefficients from scratch every time, most power series representations are built from one starting point: the geometric series.

Recall that a geometric series with first term 1 and common ratio \(x\) is

\( \dfrac{1}{1-x} = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots \)

This equation holds only when \(|x| < 1\), which you can confirm using the ideas behind geometric series convergence. Outside that interval, the series diverges even though the left-hand side, \(\frac{1}{1-x}\), is still defined for every \(x \ne 1\). This is the key idea to keep in mind throughout this topic: a power series represents its function only inside its interval of convergence, not everywhere the function itself exists.

The graph above shows \(f(x) = \frac{1}{1-x}\). Near \(x = 0\), this smooth curve is exactly matched by the polynomial-like sum \(1 + x + x^2 + x^3 + \cdots\), one more term at a time.

Once you know the series for \(\frac{1}{1-x}\), you can substitute an expression in place of \(x\) to generate power series for many other functions, without starting the derivation over.

Example 1: Find a power series representation for \(\dfrac{1}{1+x}\).

Since \(\frac{1}{1+x} = \frac{1}{1-(-x)}\), replace \(x\) with \(-x\) in the geometric series:

\( \dfrac{1}{1+x} = \sum_{n=0}^{\infty} (-x)^n = \sum_{n=0}^{\infty} (-1)^n x^n = 1 - x + x^2 - x^3 + \cdots \)

This series converges for \(|-x| < 1\), which is the same as \(|x| < 1\).

Example 2: Find a power series representation for \(\dfrac{1}{1+x^2}\).

Replace \(x\) with \(-x^2\) instead:

\( \dfrac{1}{1+x^2} = \sum_{n=0}^{\infty} (-x^2)^n = \sum_{n=0}^{\infty} (-1)^n x^{2n} = 1 - x^2 + x^4 - x^6 + \cdots \)

Since \(|-x^2| < 1\) is equivalent to \(|x| < 1\), this representation is valid on \((-1, 1)\).

Inside its interval of convergence, a power series can be differentiated or integrated term by term, and the result is the power series for the derivative or antiderivative of the original function. This opens the door to functions that don't come directly from a substitution.

Example 3: Find a power series representation for \(\ln(1+x)\).

Start from the series found in Example 1:

\( \dfrac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^n x^n \)

Since \(\displaystyle\int \frac{1}{1+x}\,dx = \ln(1+x) + C\), integrate both sides term by term:

\( \ln(1+x) = \sum_{n=0}^{\infty} (-1)^n \dfrac{x^{n+1}}{n+1} + C = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dfrac{x^4}{4} + \cdots \)

Setting \(x = 0\) gives \(\ln(1) = 0\), so \(C = 0\). The series is valid for \(-1 < x \le 1\); the right endpoint can be checked with the alternating series test conditions, since at \(x = 1\) the series becomes the alternating harmonic series.

Substitution and calculus operations can shrink, shift, or leave unchanged the interval where a power series matches its function, so it's important to re-check convergence after every manipulation rather than assuming it carries over automatically. A quick check at the endpoints often relies on the same reasoning used in the Divergence test: if the terms of the series don't shrink to zero at an endpoint, the series diverges there immediately, and that endpoint is excluded from the interval.

Building power series from the geometric series through substitution, differentiation, and integration is the foundation for representing functions like \(e^x\), \(\sin x\), and \(\ln(1+x)\) as infinite sums, and it prepares you for the more general Taylor and Maclaurin series methods, where coefficients are found directly from derivatives instead of algebraic manipulation. Practicing these substitution and calculus tricks first makes that later, more general method much easier to understand.

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