Taylor Series Introduction

Section: 
10.2
Date: 
Monday, October 18, 2010 - 14:00 - 15:00
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fall2010math2300_10-2-taylor-series-notes.pdf66.67 KB

Math 2300 Section 005 – Calculus II – Fall 2010

Notes on Taylor Series – October 18, 2010

Definition of a Taylor Series

The Taylor series for f(x) centered at x=0 is

f(x)=f(0)+f^{{\prime}}(0)x+\frac{f^{{\prime\prime}}(0)}{2!}x^{2}+\frac{f^{{(3)}}(0)}{3!}x^{3}+\cdots=\sum _{{k=0}}^{\infty}\frac{f^{{(k)}}(0)}{k!}x^{k}.

Similarly, the Taylor series for f(x) centered at x=a is

f(x)=f(a)+f^{{\prime}}(a)(x-a)+\frac{f^{{\prime\prime}}(a)}{2!}(x-a)^{2}+\cdots=\sum _{{k=0}}^{\infty}\frac{f^{{(k)}}(a)}{k!}(x-a)^{k}.

Examples

Since a Taylor series is the same as a Taylor polynomial, but is taken to infinite degree, we already know many Taylor series. From the last section, we know the following.

\displaystyle\sin x \displaystyle=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+\cdots=\sum _{{k=0}}^{\infty}\frac{(-1)^{k}}{(2k+1)!}x^{{2k+1}}
\displaystyle\cos x \displaystyle=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+\cdots=\sum _{{k=0}}^{\infty}\frac{(-1)^{k}}{(2k)!}x^{{2k}}
\displaystyle e^{x} \displaystyle=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots=\sum _{{k=0}}^{\infty}\frac{x^{k}}{k!}
\displaystyle(1+x)^{p} \displaystyle=1+px+\frac{p(p-1)}{2!}x^{2}+\frac{p(p-1)(p-2)}{3!}x^{3}+\cdots=\sum _{{k=0}}^{\infty}\binom{p}{k}x^{k}
\displaystyle\ln(x+1) \displaystyle=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\cdots=\sum _{{k=1}}^{\infty}\frac{(-1)^{{k+1}}}{k}x^{k}

where

\binom{p}{k}=\frac{p!}{k!(p-k)!}.

There is a bit of a technicality here. In section 10.1, we wrote \approx between the functions and their Taylor polynomials. Now, we're writing = instead. (\approx means “approximately equal to,” while = means “equal to.”) The main point to understand here is detailed in the next portion of these notes.

Intervals of Convergence of Taylor Series

Here is the main point: Taylor series are power series. Remember that power series are given by a series in the form

\sum _{{k=1}}^{\infty}c_{k}x^{k}=\sum _{{k=1}}^{\infty}c_{k}(x-0)^{k}\qquad\text{or}\qquad\sum _{{k=1}}^{\infty}c_{k}(x-a)^{k}

where the c_{k} are coefficients. Thus, for Taylor series, the derivative and factorial parts of each term account for the c_{k}. But, more importantly, power series have a radius of convergence. This means that for some radius r, the interval around where the series is centered will cause the series to converge. I.e., if a is the center of the series then any x in the interval (a-r,a+r) plugged in to the series for x will cause the series to converge. As to whether or not the specific numbers a\pm r cause the series to converge, we have to explicitly test them and determine that for ourselves.

Like power series in general, Taylor series may have an infinite radius of convergence. This simply means that any real number value for x will cause the series to converge.

Let's look at two examples, one with an infinite radius of convergence and one with a finite radius of convergence.

Examples of Radius/Interval of Convergence

  1. Find the radius and interval of convergence of the Taylor series for e^{x} centered at x=0.

    Solution: We know that the Taylor series for e^{x} at x=0 is

    1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+\cdots=\sum _{{k=0}}^{\infty}\frac{x^{k}}{k!}.

    Recall that finding the radius of convergence of a power series requires us to use the ratio test. Specifically, the ratio test says that, in order for the series a_{n} to converge, we must have

    \lim _{{k\rightarrow\infty}}\frac{|a_{{k+1}}|}{|a_{k}|}<1.

    How this relates to the radius of a power series isn't immediately obvious, but let's work through this calculation for the above Taylor series for e^{x} and it will become more obvious along the way. Notice that in our situation we have a_{k}=x^{k}/k!. Thus, we need the following limit to be less than 1 in order for the series to converge.

    \lim _{{k\rightarrow\infty}}\frac{\left|\displaystyle\frac{x^{{k+1}}}{(k+1)!}\right|}{\left|\displaystyle\frac{x^{k}}{k!}\right|}=\lim _{{k\rightarrow\infty}}\frac{\displaystyle\frac{|x^{{k+1}}|}{(k+1)!}}{\displaystyle\frac{|x^{k}|}{k!}}=\lim _{{k\rightarrow\infty}}\frac{k!|x^{{k+1}}|}{(k+1)!|x^{k}|}=|x|\lim _{{k\rightarrow\infty}}\frac{k!}{(k+1)!}.

    The last step follows since \frac{|x^{{k+1}}|}{|x^{k}|}=\left|\frac{x^{{k+1}}}{x^{k}}\right|=|x| and this has nothing to do with k (it is essentially a constant as far as k is concerned) and we may pull it to the front of the limit. Now, we're considering the limit

    |x|\lim _{{k\rightarrow\infty}}\frac{k!}{(k+1)!}=|x|\lim _{{k\rightarrow\infty}}\frac{1}{k+1}=|x-0|\lim _{{k\rightarrow\infty}}\frac{1}{k+1}

    where we have replaced x by x-0 since they are obviously equal. (Remember that |a-b|=|b-a| is the distance between a and b, and so |x|=|x-0| is a nice mathematical way to represent how far x is from zero.) What we have shown is that for x satisfying

    |x-0|\lim _{{k\rightarrow\infty}}\frac{1}{k+1}<1

    the Taylor (power) series for e^{x} evaluated at that x will converge. Since the limit is zero, the inequality is true for all x. This means that the Taylor series for e^{x} converges for all real numbers x. Thus, we have an infinite radius of convergence and the interval of convergence is (-\infty,\infty).

  2. Find the radius and interval of convergence of the Taylor series of \ln(x+1) centered at x=0.

    Solution: Since the Taylor series is

    \ln(x+1)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\cdots=\sum _{{k=1}}^{\infty}\frac{(-1)^{{k+1}}}{k}x^{k}

    we have the following limit calculation given by the ratio test. Again, we just pick everything apart inside the limit until we can take the limit.

    \displaystyle\lim _{{k\rightarrow\infty}}\frac{|a_{{k+1}}|}{|a_{k}|} \displaystyle=\lim _{{k\rightarrow\infty}}\frac{\displaystyle\left|\frac{(-1)^{{k+2}}x^{{k+1}}}{(k+1)}\right|}{\left|\displaystyle\frac{(-1)^{{k+1}}x^{k}}{k}\right|}
    \displaystyle=\lim _{{k\rightarrow\infty}}\frac{\displaystyle\frac{|(-1)^{{k+2}}|\cdot|x^{{k+1}}|}{|k+1|}}{\displaystyle\frac{|(-1)^{{k+1}}|\cdot|x^{k}|}{|k|}}
    \displaystyle=\lim _{{k\rightarrow\infty}}\frac{\displaystyle\frac{|x^{{k+1}}|}{|k+1|}}{\displaystyle\frac{|x^{k}|}{|k|}}
    \displaystyle=\lim _{{k\rightarrow\infty}}\frac{|x^{{k+1}}|\cdot|k|}{|x^{k}|\cdot|k+1|}
    \displaystyle=|x|\lim _{{k\rightarrow\infty}}\frac{k}{k+1}
    \displaystyle=|x|
    \displaystyle=|x-0|

    What the ratio test tells us is that the series will converge if the value of this limit calculation is less than 1. That is, the series converges when |x-0|<1. Thus, the radius of convergence is 1. Therefore, we know that the series at least converges for x within 1 of 0, i.e., on (-1,1). What this doesn't tell us, unfortunately, is what happens at the endpoints of that interval. So, if we plug in x=-1 or x=1, we need to figure out whether the series will converge. This usually ends up being pretty easy. In this case, plugging in x=-1 probably won't work, since the function \ln(x+1) isn't defined there. We'll plug it in to the series anyway and see what happens. We get

    \sum _{{k=1}}^{\infty}\frac{(-1)^{{k+1}}}{k}x^{k}\Big|_{{x=-1}}=\sum _{{k=1}}^{\infty}\frac{(-1)^{{k+1}}}{k}(-1)^{k}=\sum _{{k=1}}^{\infty}\frac{1}{k}.

    Since plugging in x=-1 leads to a clearly non-convergent series, we don't add x=-1 to the interval of convergence and the interval therefore stays open at x=-1. However, plugging in x=1 we find that

    \sum _{{k=1}}^{\infty}\frac{(-1)^{{k+1}}}{k}x^{k}\Big|_{{x=1}}=\sum _{{k=1}}^{\infty}\frac{(-1)^{{k+1}}}{k}

    and this converges by the alternating series test. Therefore, we add x=1 to the interval of convergence and our final interval of convergence is (-1,1].

Binomial Series

Not all Taylor series are infinitely long. For instance, you know that the Taylor polynomials of any polynomial can only have up to the degree of the original polynomial. If you understand why Taylor polynomials of polynomials are limited by degree in this way, then you will understand that Taylor series of polynomials are the same.

Binomial series are Taylor series with a specific form. They are defined by

(1+x)^{p}=1+px+\frac{p(p-1)}{2!}x^{2}+\frac{p(p-1)(p-2)}{3!}x^{3}+\frac{p(p-1)(p-2)(p-3)}{4!}x^{4}+\cdots

and when p is some positive integer this just becomes

(1+x)^{p}=\sum _{{k=0}}^{\infty}\binom{p}{k}x^{k}\quad\text{where }\binom{p}{k}=\frac{p!}{k!(p-k)!}.

Examples of Binomial Series

  1. Find the Taylor series of (1+x)^{{10}} around x=0.

    Solution: This is really just asking us to expand the polynomial (1+x)^{{10}}. We could actually expand it out, but that would take a really long time. Instead, using the Binomial series with p=10 we have

    \displaystyle(1+x)^{{10}} \displaystyle=1+10x+\frac{90}{2!}x^{2}+\frac{720}{3!}x^{3}+\cdots+\frac{10!}{10!}x^{{10}}
    \displaystyle=1+10x+45x^{2}+120x^{3}+210x^{4}+252x^{5}+210x^{6}+120x^{7}+45x^{8}+10x^{9}+x^{{10}}
  2. Find the Taylor series for \frac{1}{1+x} about x=0.

    Solution: This is the Binomial series with p=-1. We find

    \displaystyle\frac{1}{1+x} \displaystyle=(1+x)^{{-1}}
    \displaystyle=1+(-1)x+\frac{(-1)(-2)}{2!}x^{2}+\frac{(-1)(-2)(-3)}{3!}x^{3}+\frac{(-1)(-2)(-3)(-4)}{4!}x^{4}+\cdots
    \displaystyle=1-x+\frac{(-1)^{2}2!}{2!}x^{2}+\frac{(-1)^{3}3!}{3!}x^{3}+\frac{(-1)^{4}4!}{4!}x^{4}+\cdots
    \displaystyle=1-x+x^{2}-x^{3}+x^{4}-x^{5}+\cdots
    \displaystyle=\sum _{{k=0}}^{\infty}(-1)^{k}x^{k}

© 2011 Jason B. Hill. All Rights Reserved.