Unit 10 Topic 10.14

Taylor and Maclaurin Series

every AB topic, then series

4 min readfreeNot practiced

A Taylor polynomial has a last term. If you never stop adding terms, you get a power series, the Taylor series of the function. For the functions in this course (the four series below and every series built from them), that series does more than approximate: on its interval of convergence it equals the function. That is a property of these functions, not of every function: a few functions have a Taylor series that converges to something else, and the course does not use them.

From polynomials to a series

The Taylor series of ff about x=ax = a is

∑n=0∞f(n)(a)n!(x−a)n=f(a)+f′(a)(x−a)+f″(a)2!(x−a)2+⋯\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x - a)^n = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \cdots

About a=0a = 0 it is called the Maclaurin series. The Taylor polynomial PnP_n is its partial sum through the (x−a)n(x - a)^n term. So the questions of the last three topics fit together: Pn(x)P_n(x) is a partial sum, the error bounds measure how far a partial sum is from the sum, and the interval of convergence says where the partial sums settle at all.

The graph of y = e to the x from x = -3 to 2.5 with the first three Taylor polynomials about 0: the line 1 + x, the parabola 1 + x + x squared over 2, and the cubic that adds x cubed over 6. Each new term brings the polynomial closer to e to the x over a wider stretch; at x = 2 their values are 3, 5 and about 6.333, approaching e squared, about 7.389.
Figure 1. The partial sums of the Maclaurin series for exe^x close in on the curve. LevlPrep original diagram.

For exe^x, every derivative at 0 is 1, so the Maclaurin series is ∑xnn!\sum \frac{x^n}{n!}. The figure shows its first partial sums closing in on exe^x. The ratio test gives |x|n+1→0\frac{|x|}{n + 1} \to 0 for every xx, so the series converges everywhere, and it converges to exe^x.

Four series to know by heart

FunctionMaclaurin seriesGeneral termConverges to the function for
11−x\frac{1}{1 - x}1+x+x2+x3+⋯1 + x + x^2 + x^3 + \cdotsxnx^n−1<x<1-1 < x < 1
exe^x1+x+x22!+x33!+⋯1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdotsxnn!\frac{x^n}{n!}every xx
sin⁡x\sin xx−x33!+x55!−⋯x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots(−1)nx2n+1(2n+1)!\frac{(-1)^n x^{2n+1}}{(2n+1)!}every xx
cos⁡x\cos x1−x22!+x44!−⋯1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots(−1)nx2n(2n)!\frac{(-1)^n x^{2n}}{(2n)!}every xx

Patterns help you remember them. sin⁡x\sin x is odd, so only odd powers appear; cos⁡x\cos x is even, so only even powers. Both alternate in sign because their derivatives cycle through sin⁡,cos⁡,−sin⁡,−cos⁡\sin, \cos, -\sin, -\cos. The series for 11−x\frac{1}{1 - x} is just the geometric series, which is why it works only for |x|<1|x| < 1.

Building a Taylor series from derivatives

Worked example 1. Find the Taylor series for f(x)=exf(x) = e^x about x=2x = 2.

Every derivative of exe^x is exe^x, so f(n)(2)=e2f^{(n)}(2) = e^2 for every nn.

The coefficients are e2n!\frac{e^2}{n!}, and the series is ∑n=0∞e2(x−2)nn!\sum_{n=0}^{\infty} \frac{e^2 (x - 2)^n}{n!}=e2= e^2+e2(x−2)+ e^2(x - 2)+e22!(x−2)2+ \frac{e^2}{2!}(x - 2)^2+⋯+ \cdots

When a question gives a formula for f(n)(a)f^{(n)}(a) instead of the function, the same recipe applies: divide by n!n! and simplify. If f(n)(0)=n!3nf^{(n)}(0) = \frac{n!}{3^n}, the coefficients are 13n\frac{1}{3^n} and the series is the geometric series ∑(x3)n\sum \left(\frac{x}{3}\right)^n.

Reading coefficients backward

If a function is given by a power series ∑cn(x−a)n\sum c_n (x - a)^n with a positive radius, then that series is its Taylor series, so cn=f(n)(a)n!c_n = \frac{f^{(n)}(a)}{n!} and

f(n)(a)=n!⋅cn.f^{(n)}(a) = n! \cdot c_n.

Worked example 2. g(x)g(x)=∑n=0∞(−1)nxnn+2= \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{n + 2}. Find g″(0)g''(0).

The coefficient of x2x^2 is c2c_2=(−1)22+2= \frac{(-1)^2}{2 + 2}=14= \frac{1}{4}, so g″(0)=2!⋅14=12g''(0) = 2! \cdot \frac{1}{4} = \frac{1}{2}.

Using known series to find sums and limits

Worked example 3. Find ∑n=0∞(−1)n(2n+1)!(π6)2n+1\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n + 1)!}\left(\frac{\pi}{6}\right)^{2n+1}.

This is the series for sin⁡x\sin x with x=π6x = \frac{\pi}{6}, so the sum is sin⁡π6=12\sin\frac{\pi}{6} = \frac{1}{2}.

Worked example 4. Find limx→01−cos⁡xx2\lim_{x \to 0} \frac{1 - \cos x}{x^2}.

1−cos⁡x1 - \cos x=x22!−x44!+⋯= \frac{x^2}{2!} - \frac{x^4}{4!} + \cdots, so 1−cos⁡xx2\frac{1 - \cos x}{x^2}=12−x224= \frac{1}{2} - \frac{x^2}{24}+⋯→12+ \cdots \to \frac{1}{2}.

Series or polynomial?

When a question asks for a Taylor polynomial, write a finite sum that ends at the stated degree. When it asks for a series, write the first few terms followed by "+⋯+ \cdots", or use sigma notation with the general term. Mixing the two, a polynomial with "+⋯+ \cdots" or a series with no general term when one is asked for, costs points on the exam.

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