11.5: Taylor Series

A power series is a series of the form

$$\sum_{n=0}^\infty a_n x^n$$

where each $a_n$ is a number and $x$ is a variable.

A power series defines a function $f(x) = \sum_{n=0}^\infty a_n x^n$ where we substitute numbers for $x$.

Note: The function $f$ is only defined for those $x$ with $\sum_{n=0}^\infty a_n x^n$ convergent.

Geometric series as a power series

For $\vert x\vert < 1$ we computed

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

Taylor Series

If $f(x)$ is an infinitely differentiable function, then the Taylor series of $f(x)$ at $a$ is the series

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

Example

Compute the Taylor series of $f(x) = e^x$ at $a = 0$.

Solution

We know $f^{(n)}(x) = e^x$ for all $n \geq 0$. So $f^{(n)}(0) = 1$ and the Taylor series of $f(x) = e^x$ at $a = 0$ is

$$\sum_{n=0}^\infty \frac{1}{n!} x^n$$

Example

Compute the Taylor series at $a = 1$ of $f(x) = \sqrt{x}$.

Solution

Write $f(x) = x^\frac{1}{2}$. Then $f'(x) = \frac{1}{2} x^\frac{-1}{2}$, $f''(x) = \frac{-1}{4} x^\frac{-3}{2}$, $f'''(x) = \frac{3}{8} x^\frac{-5}{2}$, $f^{(4)}(x) = \frac{-5}{16} x^\frac{-7}{2} = \frac{(-1)^{4+1} 5 \cdot 3 \cdot 1}{2^4} x^\frac{-7}{2}$. In general, $f^{(n)}(x) = \frac{(-1)^{n+1} (2n - 1) \cdot (2n - 3) \cdots 3 \cdot 1}{2^n} x^\frac{-2n-1}{2}$ so that $f^{(n)}(1) = \frac{(-1)^{n+1} (2n - 1) \cdot (2n - 3) \cdots 3 \cdot 1}{2^n}$ and the Taylor series of $f(x) = \sqrt{x}$ at $a = 1$ is

$$1 + \sum_{n=1}^\infty (-1)^{n+1} \frac{(2n - 1) (2n -3) \cdots 3 \cdot 1}{n! 2^n} (x - 1)^n$$

Convergence of Taylor series

Given an infinitely differentiable function $f(x)$ with Taylor series (at $a$) $\sum_{n=0}^\infty b_n (x - a)^n$ either $\sum_{n=0}^\infty b_n (x - a)^n$ converges and is equal to $f(x)$ for every number $x$ or there is a number $R$ (called the radius of convergence) for which $\sum_{n=0}^\infty b_n (x - a)^n$ converges and is equal to $f(x)$ for $\vert x - a\vert < R$ while $\sum_{n=0}^\infty b_n (x - a)^n$ diverges for $\vert x - a\vert > R$.

Examples

• 
  
  $$e^x = \sum_{n=0}^\infty \frac{1}{n!} x^n$$
  
  
  for all $x$.
• 
  
  $$\frac{1}{1-x} = \sum_{n=0}^\infty x^n$$
  
  
  for $\vert x\vert < 1$ and diverges for $\vert x\vert > 1$

Operations on Taylor series

Differentiation: If $f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n = \sum_{n=0}^\infty a_n x^n$, then $f'(x) = \sum_{n=0}^\infty \frac{f^{(n+1)}(0)}{n!} x^n = \sum_{n=0}^\infty n a_n x^{n-1}$.

Integration: If $f(x) = \sum_{n=0}^\infty a_n x^n$, then $\int f(x) dx = C + \sum_{n=0}^\infty \frac{a_n}{n+1} x^{n+1}$.

Products: If $f(x) = \sum_{n=0}^\infty a_n x^n$ and $g(x) = \sum_{n=0}^\infty b_n x^n$, then $f(x) \cdot g(x) = \sum_{n=0}^\infty (\sum_{i=0}^n a_i b_{n-i}) x^n$.

Composition (monomial case): If $f(x) = \sum_{n=0}^\infty a_n x^n$ and $m$ is a positive integer, then $f(x^m) = \sum_{n=0}^\infty a_n x^{nm}$.

Example

Compute the Taylor series at $a = 0$ of $\int e^{x^2} dx$.

Solution

We know $e^x = \sum_{n=0}^\infty \frac{1}{n!} x^n$. So, $e^{x^2} = \sum_{n=0}^\infty \frac{1}{n!} x^{2n}$. Integrating, $\int e^{x^2} dx = \int \sum_{n=0}^\infty \frac{1}{n!} x^{2n} dx = C + \sum_{n=0}^\infty \frac{1}{n! (2n+1)} x^{2n+1}$.

That is, $\int e^{x^2} dx = C + x + \frac{1}{6} x^3 + \frac{1}{30} x^5 + \frac{1}{98} x^7 + \cdots$.

Example

Find the Taylor series at zero of $\frac{1 + x^2}{1 - x^3}$.

Solution

We know $\frac{1}{1 - u} = \sum_{n=0}^\infty u^n$ so that $\frac{1}{1 - x^3} = \sum_{n=0}^\infty x^{3n}$.

Multiplying, $\frac{1 + x^2}{1 - x^3} = (1 + x^2) \sum_{n=0}^\infty x^{3n} = \sum_{n=0}^\infty (x^{3n} + x^{3n+2}) = 1 + x^2 + x^3 + x^5 + x^6 + x^8 + x^9 + x^{11} + \cdots$