MAT159H1/Unit 07
Unit 07 · Analysis 1B

Power Series

Functions as infinite polynomials, now on rigorous footing: radius of convergence, and — thanks to uniform convergence on compact subintervals — term-by-term calculus inside that radius. The capstone of the course.

Why it matters & where it comes from
Why it exists

A power series turns a complicated function into something computable, differentiable, and integrable term by term — with the radius of convergence telling you exactly where that is valid.

In the real world

It is how computers evaluate transcendental functions, how differential equations are solved by series methods, and how machine learning and applied mathematics use truncated expansions to approximate and analyze functions.

Story & open questions

Madhava found power series for sine, cosine, and arctangent around 1400, well before Newton or Leibniz. From these same ideas comes Euler's identity $e^{i\pi}+1=0$, often called the most beautiful equation in mathematics — a single power series ($e^x$) evaluated at an imaginary argument.

By the end you can

Radius of convergence

Definition

A power series $\sum c_n(x-a)^n$ converges absolutely for $|x-a|<R$ and diverges for $|x-a|>R$, for some $R\in[0,\infty]$ called the radius of convergence; convergence at the two endpoints $x=a\pm R$ must be checked separately.

Cauchy–Hadamard formula

$\dfrac1R=\operatorname{limsup}_{n\to\infty}\sqrt[n]{|c_n|}$ (or, when the limit exists, $\dfrac1R=\lim\left|\dfrac{c_{n+1}}{c_n}\right|$ by the ratio test) — this is exactly the root/ratio test from Unit 5, applied with $x$ fixed and solving the resulting inequality $L<1$ for $|x-a|$.

Example Endpoint behaviour

$\sum\dfrac{x^n}{n}$ has $R=1$ by the ratio test. At $x=1$: $\sum1/n$ diverges. At $x=-1$: $\sum(-1)^n/n$ converges (alternating test). So the interval of convergence is $[-1,1)$ — asymmetric, because the two endpoints genuinely behave differently.

Term-by-term calculus

Theorem

Within radius $R$, a power series converges uniformly on every closed subinterval $[a-r,a+r]$ with $r<R$ (by the M-test, comparing with a convergent geometric-type series), so it may be differentiated and integrated term by term there, with the differentiated/integrated series having the same radius $R$.

Example

Integrating $\dfrac{1}{1+x}=\sum_{n\ge0}(-x)^n$ term by term gives $\ln(1+x)=\sum_{n\ge0}\dfrac{(-1)^n x^{n+1}}{n+1}$ for $|x|<1$.

Example Differentiating

Differentiating $\dfrac1{1-x}=\sum x^n$ term by term gives $\dfrac1{(1-x)^2}=\sum_{n\ge1}nx^{n-1}$, valid for $|x|<1$.

Uniqueness of the coefficients If a power series represents $f$ near $a$, its coefficients must be $c_n=f^{(n)}(a)/n!$ — differentiating term by term $n$ times and evaluating at $x=a$ kills every term except the $n$-th. So a function has at most one power series representation about a given centre: it must be its own Taylor series.

Common Maclaurin series

Reference list

$e^x=\displaystyle\sum_{n\ge0}\dfrac{x^n}{n!}$ ($R=\infty$); $\sin x=\sum_{n\ge0}\dfrac{(-1)^nx^{2n+1}}{(2n+1)!}$ ($R=\infty$); $\cos x=\sum_{n\ge0}\dfrac{(-1)^nx^{2n}}{(2n)!}$ ($R=\infty$); $\dfrac1{1-x}=\sum_{n\ge0}x^n$ ($R=1$); $\ln(1+x)=\sum_{n\ge1}\dfrac{(-1)^{n+1}x^n}{n}$ ($R=1$); $\arctan x=\sum_{n\ge0}\dfrac{(-1)^nx^{2n+1}}{2n+1}$ ($R=1$).

Example Building a new series

$e^{-x^2}=\displaystyle\sum_{n\ge0}\dfrac{(-x^2)^n}{n!}= \sum_{n\ge0}\dfrac{(-1)^nx^{2n}}{n!}$ — substitute directly into the known series for $e^x$, valid for all $x$ since the original has $R=\infty$.

Common pitfalls

• Endpoints need separate testing with the Unit 5 series tests; the radius alone never decides them.
• Term-by-term differentiation and integration are valid strictly inside the radius — the differentiated series can behave differently right at the boundary.
• A function need not equal its Taylor series everywhere the series converges (recall Cauchy's $e^{-1/x^2}$ example from Unit 4) — but if it does, the representation is unique.
• Substituting into a known series (as with $e^{-x^2}$) changes the radius according to the substitution — always recheck it.

Practice

Practice 1

Radius of convergence of $\sum\dfrac{x^n}{n!}$?

Answer
$R=\infty$ (ratio test: $\left|\tfrac{x}{n+1}\right|\to0$ for every $x$).
Practice 2

Maclaurin series of $e^{-x^2}$?

Answer
$\sum\dfrac{(-1)^n x^{2n}}{n!}$, $R=\infty$.
Practice 3

Interval of convergence of $\sum\dfrac{(x-2)^n}{n\,3^n}$?

Solution
Ratio test gives $R=3$, centred at $2$: candidate interval $(-1,5)$. At $x=5$: $\sum1/n$ diverges. At $x=-1$: $\sum(-1)^n/n$ converges. So $[-1,5)$.
Practice 4 · Cauchy–Hadamard

Find $R$ for $\sum n^n x^n$.

Answer
$\sqrt[n]{|c_n|}=n\to\infty$, so $1/R=\infty$, giving $R=0$ — the series converges only at $x=0$.
Practice 5 · Term-by-term integration

Use $\arctan x=\sum_{n\ge0}\dfrac{(-1)^nx^{2n+1}}{2n+1}$ at $x=1$ to justify $\dfrac{\pi}{4}=1-\dfrac13+\dfrac15-\cdots$ (you may assume the series' behaviour at the endpoint without proving it).

Answer
Substituting $x=1$ into the arctangent series gives $\arctan(1)=\pi/4$ on the left and the stated alternating series on the right — a classical series for $\pi$.
Practice 6 · Differentiating twice

Starting from $\dfrac1{1-x}=\sum_{n\ge0}x^n$, find a series for $\dfrac{2}{(1-x)^3}$.

Solution
Differentiate twice: $\dfrac{2}{(1-x)^3}=\sum_{n\ge2}n(n-1)x^{n-2}$, for $|x|<1$.
Practice 7 · Uniqueness

Two power series centred at $0$ both equal $\cos x$ on $(-1,1)$. What can you conclude about their coefficients, and why?

Answer
They must be identical, term by term — coefficients of a power series representation are forced to be $f^{(n)}(0)/n!$ by repeated term-by-term differentiation, so there is only one power series for $\cos x$ at a given centre.
Practice 8 · Radius after substitution

$\dfrac1{1-x}=\sum x^n$ has $R=1$. What is the radius of convergence of $\dfrac1{1-x^3}$ as a power series in $x$?

Answer
Substitute $x^3$ for $x$: $\sum(x^3)^n=\sum x^{3n}$, which needs $|x^3|<1$, i.e. $|x|<1$ — the radius is still $R=1$.

Expanded topic checklist

Course coverage. This unit is part of the complete MAT159H1 topic map. It should cover the following items before you move on:

Textbook/source alignment

Use the current UofT MAT159H1 calendar topics as the coverage spine, and pair lecture notes with a Spivak-style proof-calculus text for extra reading. The exact assigned textbook should still be checked against the instructor syllabus.

Core theorem map

The power-series spine is: radius first, endpoint behavior second, then algebraic operations within the open interval of convergence.

Study routine. When manipulating a power series, state the interval on which the manipulation is justified.

Deep practice extensions

Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.

Extension 1 · Radius

Find the radius of $\sum x^n/n$.

Guide
Ratio or root test gives radius $1$.
Extension 2 · Endpoints

Determine the interval of convergence for $\sum x^n/n$.

Guide
At $x=1$ it is harmonic and diverges; at $x=-1$ it is alternating harmonic and converges. Interval: $[-1,1)$.
Extension 3 · Termwise integration

Use $1/(1-x)=\sum x^n$ to get a series for $\ln(1-x)$.

Guide
Integrate termwise from $0$ to $x$: $-\ln(1-x)=\sum_{n=0}^\infty x^{n+1}/(n+1)$ for $|x|<1$.
Extension 4 · Differentiate

Differentiate $\sum x^n$ inside its radius and identify the sum.

Guide
Termwise derivative gives $\sum_{n=1}^\infty n x^{n-1}=1/(1-x)^2$ for $|x|<1$.