Analysis

Power series and Taylor series

Infinite polynomial-like sums that represent functions locally, such as eˣ = 1+x+x²/2!+…

IntuitionA polynomial that never stops

Some functions are hard to compute directly — square roots, exponentials, sines — but if you let a polynomial grow to infinitely many terms, it can match the function exactly near a point. The classic example is $\displaystyle e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$ : add up enough terms and you get a value indistinguishable from the true exponential.

This cubic

1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6}

is the degree-3 Taylor polynomial of

e^x

x=0

. Compare its graph with

e^x

: they nearly overlap near

x=0

and drift apart farther away.

SchoolWriting a function as an infinite sum

Definition: Power series

A power series centered at $c$ is an expression $\displaystyle\sum_{n=0}^{\infty} a_n (x-c)^n$ , where $a_n$ are fixed real coefficients and $x$ is the variable. It is really a "polynomial with infinitely many terms."

\sum_{n=0}^{\infty} a_n (x-c)^n = a_0 + a_1(x-c) + a_2(x-c)^2 + \cdots

A power series does not converge for every $x$ : it always converges on some interval centered at $c$ , called the interval of convergence, whose half-length is the radius of convergence $R$ . When $x$ is inside this interval, the infinite sum adds up to a finite number; when $x$ is outside, the terms blow up and the sum has no meaning.

a_n = \frac{f^{(n)}(c)}{n!}, \qquad f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x-c)^n

Behavior by position relative to R
Position	Behavior
$\|x-c\| < R$	Absolute convergence
$\|x-c\| > R$	Divergence
$\|x-c\| = R$	Must check endpoints separately

UndergraduateTheorems: radius of convergence and term-by-term differentiation

Theorem: Radius of convergence via the ratio test

For the power series $\displaystyle\sum_{n=0}^{\infty} a_n (x-c)^n$ , suppose $L = \displaystyle\lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|$ exists (as a finite number or $+\infty$ ). Then the radius of convergence is $R = \dfrac{1}{L}$ , with the convention that $R = \infty$ if $L=0$ and $R = 0$ if $L=\infty$ .

Why is it true?

The ratio test compares consecutive terms of the series to a geometric series: if the ratio of consecutive terms is eventually less than 1 in absolute value, the series behaves like a convergent geometric series and adds up to a finite number.

Proof

Fix $x \neq c$ and apply the ratio test to the terms $b_n = a_n(x-c)^n$ of the numerical series $\sum b_n$ . We compute

$\displaystyle \lim_{n\to\infty} \left| \frac{b_{n+1}}{b_n} \right| = \lim_{n\to\infty} \left| \frac{a_{n+1}(x-c)^{n+1}}{a_n(x-c)^n} \right| = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| \cdot |x-c| = L\,|x-c|$ .

By the ratio test for numerical series, $\sum b_n$ converges absolutely when $L|x-c| < 1$ , i.e. when $|x-c| < 1/L$ , and diverges when $L|x-c| > 1$ , i.e. when $|x-c| > 1/L$ . So the series converges absolutely for every $x$ with $|x-c| < 1/L$ and diverges for every $x$ with $|x-c| > 1/L$ .

This is exactly the definition of the radius of convergence: the largest $R$ such that the series converges for all $|x-c| < R$ . Hence $R = 1/L$ . The boundary cases $L=0$ (ratio always shrinks to 0, so the series converges for every $x$ , giving $R=\infty$ ) and $L=\infty$ (ratio blows up for any $x \neq c$ , so the series converges only at $x=c$ , giving $R=0$ ) follow the same argument taking the appropriate limits.

Theorem: Term-by-term differentiation

If $\displaystyle\sum_{n=0}^{\infty} a_n (x-c)^n$ has radius of convergence $R>0$ , then $f(x) = \sum_{n=0}^{\infty} a_n(x-c)^n$ is differentiable on $(c-R,\ c+R)$ , and its derivative can be computed term by term: $\displaystyle f'(x) = \sum_{n=1}^{\infty} n a_n (x-c)^{n-1}$ , with the same radius of convergence $R$ .

Why is it true?

A power series behaves like an infinite polynomial, and polynomials can be differentiated term by term; the theorem says this familiar rule survives the passage to infinitely many terms, as long as we stay strictly inside the radius of convergence.

Proof

First we show the differentiated series has the same radius of convergence. Let $c_n = n a_n$ be the coefficients of the term-by-term derivative (shifted index). Since $n^{1/n} \to 1$ as $n \to \infty$ , we have $\limsup_n |n a_n|^{1/n} = \limsup_n |a_n|^{1/n}$ , and by the Cauchy–Hadamard formula the two series $\sum a_n(x-c)^n$ and $\sum n a_n (x-c)^{n-1}$ have the same radius of convergence $R$ .

Next, fix any closed subinterval $[c-\rho, c+\rho]$ with $\rho < R$ . On this subinterval the series of derivatives $\sum n a_n (x-c)^{n-1}$ converges uniformly, because its terms are dominated (for $n$ large) by $M \rho^{n-1}$ for constants coming from a slightly larger radius $\rho' \in (\rho, R)$ , and the Weierstrass M-test applies.

Uniform convergence of the derivative series on $[c-\rho, c+\rho]$ together with pointwise convergence of the original series $\sum a_n(x-c)^n$ lets us invoke the standard theorem on differentiating a series of functions term by term: the sum $f(x)$ is differentiable on $[c-\rho,c+\rho]$ and $f'(x)$ equals the sum of the derivative series there. Since $\rho < R$ was arbitrary, this holds on all of $(c-R,\ c+R)$ .

UndergraduateReal-World Applications and Worked Examples

Engineers and scientists rarely evaluate $\sin$ , $\cos$ , $e^x$ or $\sqrt{\ }$ by hand; calculators and computer chips use truncated power series (with a controlled error term) to compute them. Physicists use the small-angle approximation $\sin x \approx x - \dfrac{x^3}{6}$ to linearize pendulum and oscillator equations, and financial models use exponential series to approximate continuous compounding over short time steps.

Example: Radius and interval of convergence with an asymmetric endpoint

Find the radius and interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{n\cdot 4^n}(x-2)^n$ .

Solution

Here $c=2$ and $a_n = \frac{(-1)^n}{n\cdot 4^n}$ . By the ratio test theorem above, $L = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n\to\infty} \frac{n}{4(n+1)} = \frac{1}{4}$ , so $R = 4$ .

This gives the open interval $(-2,6)$ where the series converges absolutely. We must check the two endpoints separately, since the ratio test is inconclusive there.

At $x=-2$ : $(x-2)^n = (-4)^n$ , so the term becomes $\frac{(-1)^n}{n\cdot 4^n}(-4)^n = \frac{(-1)^n(-1)^n 4^n}{n \cdot 4^n} = \frac{1}{n}$ , giving the harmonic series $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}$ , which diverges.

At $x=6$ : $(x-2)^n = 4^n$ , so the term becomes $\frac{(-1)^n}{n\cdot 4^n}\cdot 4^n = \frac{(-1)^n}{n}$ , the alternating harmonic series $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$ , which converges by the alternating series test.

So the interval of convergence is $(-2,\ 6]$ : closed on the right, open on the left.

Example: Estimating cos(0.2) for a pendulum correction

An engineer needs $\cos(0.2)$ (radians) to a precision of $10^{-6}$ when modeling the restoring torque of a pendulum. Use the Maclaurin series of $\cos x$ truncated at the $x^4$ term to estimate it, and bound the error.

Solution

The Maclaurin series of $\cos x$ is $\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots$ , a special case of $\displaystyle f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x-c)^n$ with $c=0$ .

Truncating at the $x^4$ term with $x=0.2$ : $1 - \dfrac{(0.2)^2}{2} + \dfrac{(0.2)^4}{24} = 1 - 0.02 + 0.0000667 \approx 0.980067$ .

Because $\cos x$ 's Maclaurin series is alternating with decreasing terms for small $x$ , the error is bounded by the first omitted term: $\dfrac{(0.2)^6}{720} \approx 8.9\times 10^{-8}$ , which is already far below the required $10^{-6}$ tolerance.

So the engineer can safely use $\approx 0.980067$ instead of evaluating a transcendental cosine directly — exactly what a calculator's internal algorithm does.

For a power series with $L = \lim_{n\to\infty} |a_{n+1}/a_n| = 5$ , what is the radius of convergence?

What does the Taylor series formula $\displaystyle f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x-c)^n$ require of $f$ at $c$ ?

If $\displaystyle\sum_{n=0}^{\infty} a_n (x-c)^n$ has radius of convergence $R$ , what is the radius of convergence of the differentiated series $\displaystyle f'(x) = \sum_{n=1}^{\infty} n a_n (x-c)^{n-1}$ ?

Using degree-4 Maclaurin truncation, engineers approximate $\cos(0.2)$ with error bounded by which quantity?