Taylor and Maclaurin Series Theorems, calculus lesson

A Taylor series can converge and still add up to the wrong function. So two questions are always separate. Does the series converge at $x$ ? Does it converge to $f(x)$ ? The answer to the second comes from the remainder, and once it is settled for a few basic functions, uniqueness of coefficients lets you get many more series by algebra alone.

Key terms

Equality requires the remainder to tend to zero

Definition Let $f$ have derivatives of all orders on an open interval $I$ containing $a$ , let $T_n$ be its Taylor polynomials at $a$ and $R_n(x) = f(x) - T_n(x)$ . For $x \in I$ ,

f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n \quad \text{if and only if} \quad \lim_{n \to \infty} R_n(x) = 0.

This holds because the partial sums of the Taylor series are exactly the $T_n(x)$ , and $T_n(x) \to f(x)$ means $R_n(x) \to 0$ .

In plain words The series equals the function at $x$ exactly when the error of the Taylor polynomials at $x$ shrinks to zero.

Example For $e^x$ at $a = 0$ and $x \neq 0$ the Lagrange form gives

R_n(x) = \frac{e^z}{(n + 1)!} x^{n+1}

with $z$ between $0$ and $x$ , so

\lvert R_n(x) \rvert \le e^{\lvert x \rvert} \frac{\lvert x \rvert^{n+1}}{(n + 1)!}.

For any fixed $x$ the factorial eventually outgrows the power, so

\lim_{n \to \infty} \frac{\lvert x \rvert^{n+1}}{(n + 1)!} = 0.

At $x = 0$ the remainder is $0$ anyway. So $e^x = \sum \frac{x^n}{n!}$ for every real $x$ .

Taylor’s inequality for series

Definition Let $f$ have derivatives of all orders on $[a - d, a + d]$ , where $d > 0$ , and suppose there are constants $M_n$ with $\bigl\lvert f^{(n+1)}(t) \bigr\rvert \le M_n$ for every $n \ge 0$ and every $t$ in that interval. Taylor’s inequality gives

\lvert R_n(x) \rvert \le \frac{M_n}{(n + 1)!} \lvert x - a \rvert^{n+1}, \qquad \lvert x - a \rvert \le d.

If the right side tends to $0$ as $n \to \infty$ , in particular if one constant $M$ bounds every derivative on that interval, then the Taylor series equals $f(x)$ for all $\lvert x - a \rvert \le d$ .

In plain words If you can bound all the derivatives on an interval well enough, the Taylor series is guaranteed to give the right answer there.

Example Every derivative of $\sin x$ is $\pm \sin x$ or $\pm \cos x$ , so $M_n = 1$ works on every interval. Then

\lvert R_n(x) \rvert \le \frac{\lvert x \rvert^{n+1}}{(n + 1)!} \to 0 \qquad \text{for every } x,

and the Maclaurin series of $\sin x$ equals $\sin x$ on all of $\mathbb{R}$ .

Convergence domain

Definition For a function $f$ with derivatives of all orders at $a$ , three sets can differ. They are the domain of $f$ , the interval of convergence of its Taylor series at $a$ , and the set of $x$ where the series converges to $f(x)$ . The third is contained in the second, and both can be much smaller than the first.

In plain words A function can be perfectly fine at a point where its series has already stopped working, and a series can converge to the wrong value.

Example The function $f(x) = \frac{1}{1 + x^2}$ is defined and smooth for every real $x$ . Its Maclaurin series $\sum (-1)^n x^{2n}$ has radius $1$ , so at $x = 2$ the function equals $\frac{1}{5}$ but the series diverges. For $e^{-1/x^2}$ (with value $0$ at $0$ ) the series converges everywhere and equals the function only at $0$ .

Uniqueness of coefficients

Definition If $f(x) = \sum c_n (x - a)^n$ for all $x$ with $\lvert x - a \rvert < R$ , where $R > 0$ , then

c_n = \frac{f^{(n)}(a)}{n!} \qquad \text{for every } n \ge 0.

So a function has at most one power series expansion about a given center, and any such expansion is its Taylor series. This follows by differentiating the series term by term $n$ times and setting $x = a$ .

In plain words However you find a power series for $f$ , by substitution, multiplication or any other valid method, it is the Taylor series. You can then read derivatives off the coefficients.

Example From $\frac{1}{1 + x^2} = \sum (-1)^n x^{2n}$ the coefficient of $x^{10}$ is $(-1)^5 = -1$ . So

\frac{f^{(10)}(0)}{10!} = -1, \qquad f^{(10)}(0) = -10! = -3628800,

with no differentiation needed.

Operations with valid convergence conditions

Definition Let $f(x) = \sum a_n (x - a)^n$ for $\lvert x - a \rvert < R_1$ and $g(x) = \sum b_n (x - a)^n$ for $\lvert x - a \rvert < R_2$ . Then the following hold.

Sums and Cauchy products give $f + g$ and $f g$ for $\lvert x - a \rvert < \min(R_1, R_2)$ .
Term-by-term differentiation and integration give $f'$ and antiderivatives of $f$ for $\lvert x - a \rvert < R_1$ .
For a series centered at $0$ , substituting $u = k x^m$ into $\sum a_n u^n$ is valid where $\lvert k x^m \rvert < R_1$ .

By uniqueness, each resulting series is the Taylor series of the new function. Endpoints must be checked separately each time.

In plain words Combine known series freely, but write down where each result holds. The answer is valid only where every ingredient was valid.

Example Substituting $u = -x^2$ into $e^u = \sum \frac{u^n}{n!}$ , valid for every $u$ , gives

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

for every real $x$ . The coefficient of $x^4$ is $\frac{1}{2}$ , so the fourth derivative of $e^{-x^2}$ at $0$ is $4! \cdot \frac{1}{2} = 12$ .

Check yourself

Prove that the Maclaurin series of $\cos x$ equals $\cos x$ for every real $x$ .

Show the answer

Every derivative of $\cos x$ is $\pm \sin x$ or $\pm \cos x$ , so it is bounded by $M = 1$ everywhere. Taylor’s inequality gives

\lvert R_n(x) \rvert \le \frac{\lvert x \rvert^{n+1}}{(n + 1)!},

which tends to $0$ for every $x$ .

Find $f^{(7)}(0)$ for $f(x) = x^2 e^x$ .

Show the answer

Multiply the series for $e^x$ by $x^2$ to get

x^2 e^x = \sum_{n=0}^{\infty} \frac{x^{n+2}}{n!}.

The coefficient of $x^7$ comes from $n = 5$ and is $\frac{1}{5!}$ . By uniqueness

f^{(7)}(0) = \frac{7!}{5!} = 42.

Find the Maclaurin series of $\ln(1 + x^2)$ and the set where it equals the function.

Show the answer

Substitute $u = x^2$ into

\ln(1 + u) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} u^n}{n},

which is valid for $-1 < u \le 1$ . This gives

\ln(1 + x^2) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^{2n}}{n},

valid when $x^2 \le 1$ , that is for $-1 \le x \le 1$ . For $\lvert x \rvert > 1$ the terms do not tend to $0$ .

Taylor and Maclaurin Series Theorems

Key terms

Equality requires the remainder to tend to zero

Taylor’s inequality for series

Convergence domain

Uniqueness of coefficients

Operations with valid convergence conditions

Check yourself

How this lesson connects

Builds on

Resources for this lesson

Taylor polynomial explorer