Introduction

Welcome to this guide on Finding Taylor series, this article will guide you through how to find the Taylor or Maclaurin series of a function.

Taylor series

Intuitive Understanding

Consider someone walking, the distance he walks cannot be modeled using limited polynomials, if we know where the person started, his velocity, his acceleration at his starting point, we can approximate his path using kinematic equations. (assuming constant acceleration)

$r=r_0+v_{0}t+\frac{1}{2}{a}_0{t}^2$

If we know the jerk (derivative of acceleration) at a point, we can further approximate its path:

$r=r_0+v_{0}t+\frac{1}{2}{a}_0{t}^2+\frac{1}{6}{j}_0{t}^3$

We can do this further if we know the snap (derivative of jerk), crackle (derivative of snap) and pop (derivative of crackle), we can approximate his path as this:

$r=r_0+v_{0}t+\frac{1}{2}{a}_0{t}^2+\frac{1}{6}{j}_0{t}^3+\frac{1}{24}{s}_0{t}^4+\frac{1}{120}{c}_0{t}^5+\frac{1}{720}{p}_0{t}^6$

(These are all real terms)

This is essentially the Taylor Series, it finds the change of a function at one point, then the change of change of function at one point, and so on.

Formula

Assume we can approximate the function like this:

$P_n(x)=a_0+a_1(x-c)+a_2(x-c{)}^2+...+a_n(x-c{)}^n+...$

We require that $P_n(c)=f(c)$, $P_n^{\prime }(c)=f^{\prime }(c)$, $P_n^{\prime \prime }(c)=f^{\prime \prime }(c)$ and so on.

Thus, the Taylor Series is:

$f(x)=\sum _{n=0}^{\infty }\frac{f^{(n)}(a)}{n!}(x-a{)}^n$

If $a=0$, then the series is called the Maclaurin Series.

Example 1:

Find the Maclaurin series of $f(x)=e^x$

To find the Maclaurin series of this function, first we need to find $f^n(0)$, it is not hard to see $f^n(x)=e^x$, thus $f^n(0)=e^0=1$, hence the Maclaurin series is 

$f(x)=\sum _{n=0}^{\infty }\frac{f^n(0)}{n!}(x-0{)}^n$

$=\sum _{n=0}^{\infty }\frac{x^n}{n!}$

$=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...+\frac{x^n}{n!}$

$=\sum _{n=0}^{\infty }\frac{x^n}{n!}$

Example 2:

Find the Maclaurin series of $f(x)=\sin x$

To find the Maclaurin of this function, we need to find $f^n(0)$ first. We shall first find the derivatives of $f(x)$:

It is not hard to notice there is a cycle of $0$, $1$, $0$, $-1$ going on, thus the Maclaurin series turns into

$f(x)=\sum _{n=0}^{\infty }\frac{f^n(0)}{n!}(x-0{)}^n$

$=0\cdot 1+1\cdot x+0\cdot \frac{x^2}{2!}-1\cdot \frac{x^3}{3!}+...+\frac{(-1{)}^n{x}^{2n+1}}{(2n+1)!}$

$=\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n+1}}{(2n+1)!}$

Example 3:

Find the Maclaurin series of $f(x)=\frac{1}{1-x}$

One can definitely repeat the method we used before to obtain the Maclaurin series of this function, I encourage you to work the series out in this way, but there are faster approaches, recall that for a convergent geometric series, the following equation holds true:

$\sum _{n=0}^{\infty }{a}_0\cdot r^n=\frac{a_0}{1-r}$

Let $a_0=1$ and $r=x$, we have

$\sum _{n=0}^{\infty }{x}^n=\frac{1}{1-x}$

The left side of the equality is merely an infinite sum of power functions, which fits the form of Maclaurin series, thus we can conclude

$\frac{1}{1-x}=1+x+x^2+x^3+...+x^n=\sum _{n=0}^{\infty }{x}^n$

Before we continue to do more question, here are some important Maclaurin series you should memorize:

Find the Maclaurin series of $f(x)=x\sin x$

Using the old method for this function will be too time consuming, here is a faster approach. We already worked out the Maclaurin series for $\sin x$, we can directly substitute this into the function

$x\sin x=x\cdot \sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n+1}}{(2n+1)!}$

$=\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n+1}}{(2n+1)!}$

Practice

Find the Taylor/Maclaurin series of the following functions