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.
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.
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.
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.
- Find the radius and interval of convergence, including testing the endpoints separately.
- Differentiate and integrate power series term by term, and justify why this is valid.
- Recognize and use the common Maclaurin series.
- Explain why a power series's coefficients are uniquely determined by the function it represents.
Radius of convergence
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.
$\dfrac1R=\operatorname{limsup}_{n\to\infty}\sqrt[n]{|c_n|}$ (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 $|x-a|$.
$\sum\dfrac{x^n}{n}$ 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
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 $R$.
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 $|x|<1$.
Differentiating $\dfrac1{1-x}=\sum x^n$ term by term gives $\dfrac1{(1-x)^2}=\sum_{n\ge1}nx^{n-1}$, valid for $|x|<1$.
Common Maclaurin series
$e^x=\displaystyle\sum_{n\ge0}\dfrac{x^n}{n!}$ ($R=\infty$); $\sin x=\sum_{n\ge0}\dfrac{(-1)^nx^{2n+1}}{(2n+1)!}$ ($R=\infty$); $\cos x=\sum_{n\ge0}\dfrac{(-1)^nx^{2n}}{(2n)!}$ ($R=\infty$); $\dfrac1{1-x}=\sum_{n\ge0}x^n$ ($R=1$); $\ln(1+x)=\sum_{n\ge1}\dfrac{(-1)^{n+1}x^n}{n}$ ($R=1$); $\arctan x=\sum_{n\ge0}\dfrac{(-1)^nx^{2n+1}}{2n+1}$ ($R=1$).
$e^{-x^2}=\displaystyle\sum_{n\ge0}\dfrac{(-x^2)^n}{n!}= \sum_{n\ge0}\dfrac{(-1)^nx^{2n}}{n!}$ — substitute directly into the known series for $e^x$, valid for all $x$ since the original has $R=\infty$.
• 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
Radius of convergence of $\sum\dfrac{x^n}{n!}$?
Answer
Maclaurin series of $e^{-x^2}$?
Answer
Interval of convergence of $\sum\dfrac{(x-2)^n}{n\,3^n}$?
Solution
Find $R$ for $\sum n^n x^n$.
Answer
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
Starting from $\dfrac1{1-x}=\sum_{n\ge0}x^n$, find a series for $\dfrac{2}{(1-x)^3}$.
Solution
Two power series centred at $0$ both equal $\cos x$ on $(-1,1)$. What can you conclude about their coefficients, and why?
Answer
$\dfrac1{1-x}=\sum x^n$ has $R=1$. What is the radius of convergence of $\dfrac1{1-x^3}$ as a power series in $x$?
Answer
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:
- Radius and interval. Find radius of convergence and test endpoints separately.
- Termwise operations. Differentiate and integrate power series inside the radius of convergence.
- Taylor series. Connect Taylor coefficients to function derivatives and convergence to the original function.
- Uniform convergence on compact subintervals. Use power-series convergence to justify termwise operations away from endpoints.
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.
The power-series spine is: radius first, endpoint behavior second, then algebraic operations within the open interval of convergence.
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.
Find the radius of $\sum x^n/n$.
Guide
Determine the interval of convergence for $\sum x^n/n$.
Guide
Use $1/(1-x)=\sum x^n$ to get a series for $\ln(1-x)$.
Guide
Differentiate $\sum x^n$ inside its radius and identify the sum.