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Lecture 51: Expanding Functions into Power Series

Postgraduate Entrance Exam Mathematics study notes: Lecture 51: Expanding Functions into Power Series. Original formulas, diagrams, and examples are retained.

MathematicsRecord date4 min read
Graduate Entrance Examination MathematicsAdvanced MathematicsInfinite Series
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51.1 Expanding Functions into Power Series

51.1.1 Taylor Series

Why use a power-series expansion?

  • Some functions have no elementary antiderivative; $e^{x^2}$ is a standard example. After expansion, the series can be differentiated or integrated term by term within its interval of convergence.
  • A power series is also useful for numerical approximation. For example,
$$ e^{0.2}=1+0.2+\frac{0.2^2}{2!}+\frac{0.2^3}{3!}+\cdots. $$

Keeping finitely many terms gives an approximation at the required accuracy.

Theorem: power-series-expansion

If $f(x)$ can be represented on $(x_0-R,x_0+R)$ by a power series in $x-x_0$, that representation is unique and must be the Taylor series of $f$ about $x_0$:

$$ f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n. $$
Theorem: Taylor-series-convergence

Suppose that $f$ has the required derivatives in a neighborhood of $x_0$. Taylor's formula is

$$ f(x)=\sum_{k=0}^{n}\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k+R_n(x). $$

With the Lagrange remainder,

$$ R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1}, $$

where $\xi$ lies between $x$ and $x_0$. At a given $x$, the Taylor series converges to $f(x)$ if and only if $\lim_{n\to\infty}R_n(x)=0$.

51.1.2 Common Power-Series Expansions

$$ \frac{1}{1-x}=\sum_{n=0}^{\infty}x^n,\qquad |x|<1. $$
$$ \frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^n x^n,\qquad |x|<1. $$
$$ e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!},\qquad x\in\mathbb R. $$
$$ \sin x=\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!},\qquad x\in\mathbb R. $$
$$ \cos x=\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!},\qquad x\in\mathbb R. $$
$$ \ln(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}x^n,\qquad -1<x\le1. $$
$$ (1+x)^\alpha =1+\alpha x+\frac{\alpha(\alpha-1)}{2!}x^2+\cdots,\qquad |x|<1. $$
$$ \sum_{n=1}^{\infty}\frac{x^n}{n}=-\ln(1-x),\qquad -1\le x<1. $$

Endpoint convergence must be checked separately; it cannot be inferred from the open interval alone.

51.1.3 Two Expansion Methods

Method 1: Direct expansion

  1. Compute $f^{(n)}(x_0)$ and write the formal Taylor series
$$ \sum_{n=0}^{\infty}\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n. $$
  1. Verify that $\lim_{n\to\infty}R_n(x)=0$ and determine the interval of convergence.

Method 2: Indirect expansion

Start from a known expansion and use algebraic operations, term-by-term differentiation, term-by-term integration, or substitution. Track the resulting interval of convergence and check its endpoints after every transformation.

51.2 Common Question Types

Question Type 1: expanding-a-function

  • Try indirect expansion first, because repeated differentiation is often cumbersome.
  • For an expansion about $x=x_0$, rewrite the expression in terms of $x-x_0$.
  • For example, to expand $\sin x$ about $x_0=\frac{\pi}{4}$, let $u=x-\frac{\pi}{4}$:
$$ \sin x =\sin\left(\frac{\pi}{4}+u\right) =\frac{\sqrt2}{2}\left(\cos u+\sin u\right), $$

then substitute the Maclaurin series for $\cos u$ and $\sin u$.

Useful transformations

To apply the geometric series, rewrite an expression in the form

$$ \frac{1}{1-u(x)} \quad\text{or}\quad \frac{1}{1+u(x)}, $$

and first verify that $|u(x)|<1$.

Question Type 2: summing-a-series

A known expansion can be read in reverse to evaluate a series. For example,

$$ \frac{1}{1-x}=1+x+x^2+\cdots,\qquad |x|<1, $$

and

$$ \sum_{n=1}^{\infty}\frac{x^n}{n}=-\ln(1-x),\qquad -1\le x<1. $$

Typically, differentiate or integrate the target series term by term until it matches a known expansion, then use an initial value to determine the constant of integration.