Taylor and Maclaurin Series

Letting a Taylor polynomial run on forever gives a power series. Where it converges to

f

, the function and the series are two descriptions of the same thing.

Taylor series

For

e^x

, every derivative is

e^x

, so each one equals 1 at

x = 0

and the coefficients are

\frac{1}{n!}

. For

\sin x

, the derivatives at 0 repeat the pattern

0, 1, 0, -1

, so only odd powers appear, with alternating signs. For

\cos x

the pattern is

1, 0, -1, 0

, which keeps the even powers. The geometric series supplies the fourth.

Four Maclaurin series to know

\begin{aligned} \frac{1}{1 - x} &= \sum_{n=0}^{\infty} x^n \\[4pt] &= 1 + x + x^2 + x^3 + \cdots \\[10pt] e^x &= \sum_{n=0}^{\infty} \frac{x^n}{n!} \\[4pt] &= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \\[10pt] \sin x &= \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n + 1)!} \\[4pt] &= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \\[10pt] \cos x &= \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \\[4pt] &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \end{aligned}

The first converges to

\frac{1}{1 - x}

for

-1 < x < 1

. The other three converge to their functions for every

x

New series from known ones

Computing

f^{(n)}(a)

for every

n

is rarely practical. It's usually faster to start from a known series and change it. Substitute an expression for

x

, multiply or divide by a power of

x

, add series, or differentiate or integrate term by term. The new series is the Taylor series of the new function. It converges at least on the open interval where the series it came from converges, after any substitution, and the endpoints have to be checked separately.

Example 1

Find the first four terms and the general term of the Maclaurin series for

f(x) = \frac{4}{5 - 2x}

, and find the interval on which it converges to

f

Factor out 5 to get the form

\frac{1}{1 - u}

, with

u = \frac{2x}{5}

\begin{aligned} \frac{4}{5 - 2x} &= \frac{4}{5}\cdot\frac{1}{1 - 2x/5} \\[4pt] &= \frac{4}{5}\sum_{n=0}^{\infty} \Big(\frac{2x}{5}\Big)^n \\[4pt] &= \frac{4}{5} + \frac{8x}{25} + \frac{16x^2}{125} \\[4pt] &\qquad + \frac{32x^3}{625} + \cdots \\[4pt] &\qquad + \frac{2^{n+2} x^n}{5^{n+1}} + \cdots \end{aligned}

The geometric series needs

|u| < 1

, so

\big|\frac{2x}{5}\big| < 1

, or

-\frac{5}{2} < x < \frac{5}{2}

. At

x = \pm\frac{5}{2}

the terms have size

\frac{4}{5}

and don't approach 0, so the series diverges there by the nth term test. It converges to

f

exactly on

-\frac{5}{2} < x < \frac{5}{2}

Example 2

Let

h(x) = \int_0^x \frac{e^{2t} - 1 - 2t}{t^2}\,dt

, where the integrand is defined to be 2 at

t = 0

. Find the first three nonzero terms and the general term of the Maclaurin series for

h

Substitute

2t

for

x

in the series for

e^x

. Subtracting

1 + 2t

removes the first two terms.

\begin{aligned} e^{2t} - 1 - 2t &= \frac{(2t)^2}{2!} + \frac{(2t)^3}{3!} + \cdots \\[4pt] &= \sum_{n=2}^{\infty} \frac{2^n t^n}{n!} \end{aligned}

Divide by

t^2

. The new series agrees with the integrand for

t \ne 0

, and its value at

t = 0

is 2, the value chosen there.

\begin{aligned} \frac{e^{2t} - 1 - 2t}{t^2} &= 2 + \frac{4t}{3} + \frac{2t^2}{3} + \cdots \\[4pt] &= \sum_{n=2}^{\infty} \frac{2^n t^{n-2}}{n!} \end{aligned}

Integrate term by term from 0 to

x

\begin{aligned} h(x) &= 2x + \frac{2x^2}{3} + \frac{2x^3}{9} + \cdots \\[4pt] &= \sum_{n=2}^{\infty} \frac{2^n x^{n-1}}{(n - 1)\,n!} \end{aligned}

The series for

e^x

converges for every

x

, and integrating keeps the radius of convergence, so this one converges for every

x

too.

Example 3

Find the Maclaurin series for

g(x) = \frac{6x}{(1 + 3x^2)^2}

and its radius of convergence.

The squared denominator is what the derivative of

\frac{1}{1 + 3x^2}

produces: that derivative is

-\frac{6x}{(1 + 3x^2)^2} = -g(x)

. Write

\frac{1}{1 + 3x^2}

as a geometric series first, with

u = -3x^2

\begin{aligned} \frac{1}{1 + 3x^2} &= \sum_{n=0}^{\infty} (-3x^2)^n \\[4pt] &= \sum_{n=0}^{\infty} (-1)^n 3^n x^{2n} \end{aligned}

Differentiate both sides. The

n = 0

term is a constant, which drops out. Then multiply by

-1

\begin{aligned} g(x) &= -\sum_{n=1}^{\infty} (-1)^n 3^n \cdot 2n\, x^{2n-1} \\[4pt] &= \sum_{n=1}^{\infty} (-1)^{n+1}\, 2n\, 3^n x^{2n-1} \\[4pt] &= 6x - 36x^3 + 162x^5 \\[4pt] &\qquad - 648x^7 + \cdots \end{aligned}

The geometric series needs

3x^2 < 1

, so

|x| < \frac{1}{\sqrt{3}}

. Differentiating keeps the radius, so it's

\frac{1}{\sqrt{3}}