# 3.3: Taylor Series

• • Contributed by Marcia Levitus

Before discussing more applications of Maclaurin series, let’s expand our discussion to the more general case where we expand a function around values different from zero. Let’s say that we want to expand a function around the number $$h$$. If $$h=0$$, we call the series a Maclaurin series, and if $$h\neq0$$ we call the series a Taylor series. Because Maclaurin series are a special case of the more general case, we can call all the series Taylor series and omit the distinction. The following is true for a function $$f(x)$$ as long as the function and all its derivatives are finite at $$h$$:

$\label{taylor} f(x)=a_0 + a_1(x-h)+a_2(x-h)^2+...+a_n(x-h)^n = \displaystyle\sum_{n=0}^{\infty}a_n(x-h)^n$

The coefficients are calculated as

$\label{taylorcoeff} a_n=\frac{1}{n!}\left( \frac{d^n f}{dx^n}\right)_h$

Notice that instead of evaluating the function and its derivatives at $$x=0$$ we now evaluate them at $$x=h$$, and that the basis set is now $$1, (x-h), (x-h)^2,...,(x-h)^n$$ instead of $$1, x, x^2,...,x^n$$. A Taylor series will be a good approximation of the function at values of $$x$$ close to $$h$$, in the same way Maclaurin series provide good approximations close to zero.

To see how this works let’s go back to the exponential function. Recall that the Maclaurin expansion of $$e^x$$ is shown in Equation $$3.1.3$$. We know what happens if we expand around zero, so to practice, let’s expand around $$h=1$$. The coefficient $$a_0$$ is $$f(1)= e^1=e$$. All the derivatives are $$e^x$$, so $$f'(1)=f''(1)=f'''(1)...=e.$$ Therefore, $$a_n=\frac{e}{n!}$$ and the series is therefore

$\label{taylorexp} e\left[ 1+(x-1)+\frac{1}{2}(x-1)^2+\frac{1}{6}(x-1)^3+... \right]=\displaystyle\sum_{n=0}^{\infty}\frac{e}{n!}(x-1)^n$

We can use the same arguments we used before to conclude that $$e^x\approx ex$$ if $$x\approx 1$$. If $$x\approx 1$$, $$(x-1)\approx 0$$, and the terms $$(x-1)^2, (x-1)^3$$ will be smaller and smaller and will contribute less and less to the sum. Therefore,

$e^x \approx e \left[ 1+(x-1) \right]=ex.$

This is the equation of a straight line with slope $$e$$ and $$y$$-intercept 0. In fact, from Equation $$3.1.7$$ we can see that all functions will look linear at values close to $$h$$. This is illustrated in Figure $$\PageIndex{1}$$, which shows the exponential function (red) together with the functions $$1+x$$ (magenta) and $$ex$$ (blue). Not surprisingly, the function $$1+x$$ provides a good approximation of $$e^x$$ at values close to zero (see Equation $$3.1.3$$) and the function $$ex$$ provides a good approximation around $$x=1$$ (Equation \ref{taylorexp}). Figure $$\PageIndex{1}$$: Two linear approximations of the exponential function. The function $$e^x$$ is plotted in red together with the function $$y = 1+x$$ (magenta) and $$y=ex$$ (blue). (CC BY-NC-SA; Marcia Levitus)

Example $$\PageIndex{1}$$:

Expand $$f(x)=\ln{x}$$ about $$x=1$$

Solution

$f(x)=a_0 + a_1(x-h)+a_2(x-h)^2+...+a_n(x-h)^n, a_n=\frac{1}{n!}\left( \frac{d^n f}{dx^n}\right)_h \nonumber$

$a_0=f(1)=\ln(1)=0 \nonumber$

The derivatives of $$\ln{x}$$ are:

$f'(x) = 1/x, f''(x)=-1/x^2, f'''(x) = 2/x^3, f^{(4)}(x)=-6/x^4, f^{(5)}(x)=24/x^5... \nonumber$

and therefore,

$f'(1) = 1, f''(1)=-1, f'''(1) = 2, f^{(4)}(1)=-6, f^{(5)}(1)=24.... \nonumber$

To calculate the coefficients, we need to divide by $$n!$$:

• $$a_1=f'(1)/1!=1$$
• $$a_2=f''(1)/2!=-1/2$$
• $$a_3=f'''(1)/3!=2/3!=1/3$$
• $$a_4=f^{(4)}(1)/4!=-6/4!=-1/4$$
• $$a_n=(-1)^{n+1}/n$$

The series is therefore:

$f(x)=0 + 1(x-1)-1/2 (x-1)^2+1/3 (x-1)^3...=\displaystyle{\color{Maroon}\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(x-1)^{n}} \nonumber$

Note that we start the sum at $$n=1$$ because $$a_0=0$$, so the term for $$n=0$$ does not have any contribution.

Need help? The links below contain solved examples.