Taylor Series

by Justin Skycak (@justinskycak) on

Many non-polynomial functions can be represented by infinite polynomials.

This post is part of the book Justin Math: Calculus. Suggested citation: Skycak, J. (2019). Taylor Series. In Justin Math: Calculus. https://justinmath.com/taylor-series/

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The sum formula for a geometric series is an example of representing a non-polynomial function as an infinite polynomial within a particular range of inputs.

$\begin{align*} \frac{x}{1-x} = x + x^2 + x^3 + \cdots \hspace{.5cm} (\mbox{provided } |x|<1) \end{align*}$


Many other non-polynomial functions can be represented by infinite polynomials called Taylor series. The general formula for the Taylor series of a function $f(x)$, centered about a point $x=c$, is

$\begin{align*} f(x) = \sum\limits_{n=0}^\infty \frac{f^{(n)}(c)}{n!}(x-c)^n. \end{align*}$


Just like for the geometric series sum formula, the Taylor series can only be used when it converges. The ratio test is particularly useful for finding the x-values for which the series converges.

For the sake of example, we will compute the Taylor series of several familiar functions: $e^x$, $\sin x$, and $\ln x$. To introduce some variety, we will center each series at a different x-value.

Taylor Series of the Exponential Function

For $f(x)=e^x$, we have $f’(x)=e^x$, $f’^\prime(x)=e^x$, and in general $f^{(n)}(x)=e^x$ for all values of $n$. The Taylor series of $f(x)=e^x$ centered at $x=0$ is then given by

$\begin{align*} f(x) &= \sum\limits_{n=0}^\infty \frac{f^{(n)}(0)}{n!}(x-0)^n \\ &= \sum\limits_{n=0}^\infty \frac{e^0}{n!}x^n \\ &= \sum\limits_{n=0}^\infty \frac{x^n}{n!} \, . \end{align*}$


Applying the ratio test, we see that the series converges when

$\begin{align*} \lim\limits_{n\to\infty} \left|\frac{ \frac{x^{n+1}}{(n+1)!} }{ \frac{x^n}{n!} } \right| &< 1 \\ \lim\limits_{n\to\infty} \left|\frac{x}{n+1} \right| &< 1 \\ 0 < 1\, . \end{align*}$


Thus, the series converges for all values of $x$.

Taylor Series of Sine

For $f(x)=\sin x$, we have $f’(x)=\cos x$, $f’^\prime(x)=-\sin x$, $f’’^\prime(x)=-\cos x$, $f^{(4)}(x)=\sin x$, and in general $f^{(2n)}(x)=(-1)^n \sin x$ and $f^{(2n+1)}(x)=(-1)^n \cos x$.

The Taylor series of $f(x)=\sin x$ centered at $x=\pi$ is then given by

$\begin{align*} f(x) &= \sum\limits_{n=0}^\infty \frac{f^{(n)}(\pi)}{n!}(x-\pi)^n \\ &= \sum\limits_{n=0}^\infty \frac{f^{(2n)}(\pi)}{(2n)!}(x-\pi)^{2n}+\frac{f^{(2n+1)}(\pi)}{(2n+1)!}(x-\pi)^{2n+1} \\ &= \sum\limits_{n=0}^\infty \frac{(-1)^n \sin (\pi)}{(2n)!}(x-\pi)^{2n}+\frac{(-1)^n \cos (\pi)}{(2n+1)!}(x-\pi)^{2n+1} \\ &= \sum\limits_{n=0}^\infty \frac{(-1)^n (0)}{(2n)!}(x-\pi)^{2n}+\frac{(-1)^n (-1)}{(2n+1)!}(x-\pi)^{2n+1} \\ &= \sum\limits_{n=0}^\infty \frac{(-1)^{n+1}}{(2n+1)!}(x-\pi)^{2n+1} \, . \end{align*}$


Applying the ratio test, we see that the series converges when

$\begin{align*} \lim\limits_{n\to\infty} \left|\frac{ \frac{ (-1)^{(n+1)+1} }{ (2(n+1)+1)! }(x-\pi)^{2(n+1)+1} }{ \frac{(-1)^{n+1}}{(2n+1)!}(x-\pi)^{2n+1} } \right| &< 1 \\ \lim\limits_{n\to\infty} \left|\frac{ \frac{(x-\pi)^{2n+3} }{ (2n+3)! } }{ \frac{ (x-\pi)^{2n+1} }{(2n+1)!} } \right| &< 1 \\ \lim\limits_{n\to\infty} \left| \frac{(x-\pi)^2}{(2n+2)(2n+3)} \right| &< 1 \\ 0 &< 1. \end{align*}$


Thus, the series converges for all values of $x$.

Taylor Series of Natural Log

For $f(x)=\ln x$, we have $f’(x)=\frac{1}{x}$, $f’^\prime(x)=-\frac{1}{x^2}$, $f’’^\prime(x)=\frac{2}{x^3}$, and in general $f^{(n)}(x)=(-1)^{n-1} \frac{(n-1)!}{x^n}$ for $n>1$. The Taylor series of $f(x)=\ln x$ centered at $x=1$ is then given by

$\begin{align*} f(x) &= \sum\limits_{n=0}^\infty \frac{f^{(n)}(1)}{n!}(x-1)^n \\ &= f(1) + \sum\limits_{n=1}^\infty \frac{f^{(n)}(1)}{n!}(x-1)^n \\ &= \ln 1 + \sum\limits_{n=1}^\infty \frac{ (-1)^{n-1}\frac{(n-1)!}{1^n} }{n!}(x-1)^n \\ &= \sum\limits_{n=1}^\infty \frac{ (-1)^{n-1} }{n}(x-1)^n \, . \end{align*}$


Applying the ratio test, we see that the series converges when

$\begin{align*} \lim\limits_{n\to\infty} \left|\frac{ \frac{(-1)^{(n+1)-1}}{n+1}(x-1)^{n+1} }{ \frac{(-1)^{n-1}}{n}(x-1)^n } \right| &< 1 \\ \lim\limits_{n\to\infty} \left|\frac{n}{n+1}(x-1) \right| &< 1 \\ |x-1| \lim\limits_{n\to\infty} \left|\frac{n}{n+1} \right| &< 1 \\ |x-1| &< 1 . \end{align*}$


Thus, the series converges for $0<x<2$.

Derivation

To see where the formula for the Taylor series comes from, we start by performing repeated integration on the function $f^{(n+1)}(x)$.

$\begin{align*} &\underbrace{\int\limits_c^x \cdots \int\limits_c^x}_{\text{n+1 integrals} } f^{(n+1)}(x) \, (dx)^{n+1} \\ &= \underbrace{\int\limits_c^x \cdots \int\limits_c^x}_{\text{n integrals} } f^{(n)}(x)-f^{(n)}(c) \, (dx)^{n} \\ &= \underbrace{\int\limits_c^x \cdots \int\limits_c^x}_{\text{n-1 integrals} } f^{(n-1)}(x)-f^{(n-1)}(c) - f^{(n)}(c)(x-c) \, (dx)^{n-1} \\ &= \underbrace{\int\limits_c^x \cdots \int\limits_c^x}_{\text{n-2 integrals} } f^{(n-2)}(x)-f^{(n-2)}(c) - f^{(n-1)}(c)(x-c) - \frac{f^{(n)}(c)}{2}(x-c)^2 \, (dx)^{n-2} \\ &= \hspace{.5cm} \cdots \\ &= f(x)-f(c)-f'(c)(x-c) - \frac{f''(c)}{2}(x-c)^2 - \cdots - \frac{f^{(n)}(c)}{n!}(x-c)^n \end{align*}$


Solving for $f(x)$, we find

$\begin{align*} f(x) = f(c) &+ f'(c)(x-c) + \frac{f''(c)}{2}(x-c)^2 + \cdots + \frac{f^{(n)}(c)}{n!} (x-c)^n \\ &+ \underbrace{\int\limits_c^x \cdots \int\limits_c^x}_{\text{n+1 integrals} } f^{(n+1)}(x) \, (dx)^{n+1} \, . \end{align*}$


Taking the limit as $n \to \infty$, we can express $f$ as the sum of its Taylor series and some remainder term.

$\begin{align*} f(x) = \sum\limits_{n=0}^\infty \frac{f^{(n)}(c)}{n!}(x-c)^n + \lim\limits_{n\to\infty} \underbrace{\int\limits_c^x \cdots \int\limits_c^x}_{\text{n+1 integrals} } f^{(n+1)}(x) \, (dx)^{n+1} \end{align*}$


For many familiar functions, with $x$ sufficiently close to $c$, it is often the case that the remainder decays to zero:

$\begin{align*} \lim\limits_{n\to\infty} \underbrace{\int\limits_c^x \cdots \int\limits_c^x}_{\text{n+1 integrals} } f^{(n+1)}(x) \, (dx)^{n+1} = 0 \end{align*}$


For example, the remainder decays to zero if $f$ is any polynomial, because differentiating an $\text{n}^\text{th}$ degree polynomial $n+1$ times always yields a result of $0$, and the integral of $0$ is always $0$. (But this is rather trivial since the Taylor series of a polynomial is the polynomial itself.)

More generally, we can place an upper bound on the size of the $\text{n}^\text{th}$ remainder:

$\begin{align*} \left| \underbrace{\int\limits_c^x \cdots \int\limits_c^x}_{\text{n+1 integrals} } f^{(n+1)}(x) \, (dx)^{n+1} \right| \leq \left( \max\limits_{[c,x]} \left| f^{(n+1)} \right| \right) \left| \underbrace{\int\limits_c^x \cdots \int\limits_c^x}_{\text{n+1 integrals} } 1 \, (dx)^{n+1} \right| \end{align*}$


Then, since

$\begin{align*} \underbrace{\int\limits_c^x \cdots \int\limits_c^x}_{\text{n+1 integrals} } 1 \, (dx)^{n+1} &= \frac{(x-c)^{n+1}}{(n+1)!} \end{align*}$


we must have that

$\begin{align*} \left| \lim\limits_{n\to\infty} \underbrace{\int\limits_c^x \cdots \int\limits_c^x}_{\text{n+1 integrals} } f^{(n+1)}(x) \, (dx)^{n+1} \right| \leq \lim\limits_{n\to\infty} \left( \max\limits_{[c,x]} \left| f^{(n+1)} \right| \right) \frac{\left| x-c \right|^{n+1}}{(n+1)!} \, . \end{align*}$


Provided that the $(n+1)^\text{st}$ derivative doesn’t grow large enough to overpower the $(n+1)!$ term in the denominator as $n \to \infty$, the remainder will decay to zero. Then the function will be equal to its Taylor series, provided that the series converges.

$\begin{align*} f(x) = \sum\limits_{n=0}^\infty \frac{f^{(n)}(c)}{n!}(x-c)^n \end{align*}$


Exercises

Compute the Taylor series for the following functions, centered at the given points. Also compute the interval of convergence. (You can view the solution by clicking on the problem.)

$\begin{align*} 1) \hspace{.5cm} &f(x)= \ln (1+x) \\ &\mbox{at } x=0 \end{align*}$
Solution:
$\begin{align*} &\sum\limits_{n=1}^\infty \frac{(-1)^{n+1} }{n}x^n \\ & -1 < x \leq 1 \end{align*}$


$\begin{align*} 2) \hspace{.5cm} &f(x)=\frac{1}{(1+x)^2} \\ &\mbox{at } x=0 \end{align*}$
Solution:
$\begin{align*} &\sum\limits_{n=0}^\infty (-1)^n(1+n) x^n \\ & -1 < x < 1 \end{align*}$


$\begin{align*} 3) \hspace{.5cm} &f(x)=\cos x \\ &\mbox{at } x=0 \end{align*}$
Solution:
$\begin{align*} &\sum\limits_{n=0}^\infty \frac{(-1)^n}{(2n)!} x^{2n} \\ & -\infty < x < \infty \end{align*}$


$\begin{align*} 4) \hspace{.5cm} &f(x)= \cos x \\ &\mbox{at } x=\pi \end{align*}$
Solution:
$\begin{align*} &\sum\limits_{n=0}^\infty \frac{(-1)^{n+1} }{(2n)! } (x-\pi)^{2n} \\ & -\infty < x < \infty \end{align*}$


$\begin{align*} 5) \hspace{.5cm} &f(x)= \arctan x \\ &\mbox{at } x=0 \end{align*}$
Solution:
$\begin{align*} &\sum\limits_{n=0}^\infty \frac{(-1)^n}{2n+1} x^{2n+1} \\ & -1 \leq x \leq 1 \end{align*}$


$\begin{align*} 6) \hspace{.5cm} &f(x)= 2^x \\ &\mbox{at } x=-1 \end{align*}$
Solution:
$\begin{align*} &\sum\limits_{n=0}^\infty \frac{(\ln 2)^n}{(2n)!} (x+1)^n \\ & -\infty < x < \infty \end{align*}$


$\begin{align*} 7) \hspace{.5cm} &f(x)= \frac{1}{x} \\ &\mbox{at } x=2 \end{align*}$
Solution:
$\begin{align*} &\sum\limits_{n=0}^\infty \frac{(-1)^n}{2^{n+1} } (x-2)^n \\ & 0 < x < 4 \end{align*}$


$\begin{align*} 8) \hspace{.5cm} &f(x)= \frac{1}{x} \\ &\mbox{at } x=-10 \end{align*}$
Solution:
$\begin{align*} &\sum\limits_{n=0}^\infty - \frac{1}{10^{n+1} }(x+10)^n \\ & -20 < x < 0 \end{align*}$


This post is part of the book Justin Math: Calculus. Suggested citation: Skycak, J. (2019). Taylor Series. In Justin Math: Calculus. https://justinmath.com/taylor-series/


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