## 17Calculus - Fourier Series

1st Order

2nd/Higher Order

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This page covers two areas related to Fourier Series. First, we present an introduction to Fourier Series, then we discuss how to solve differential equations using Fourier Series. If you are just learning about Fourier Series, you can go through the introduction and practice problems and skip the section related to solving differential equations.

What is a Fourier Series?

The main idea of Fourier Series is that we want to build an infinite series, using the basic trig functions sine and cosine, that is equivalent to a more complicated function. The series can then be manipulated more easily than the original function.

Here is a great video to get you started. He explains why we need to build these functions, goes through an example and then explains the big picture.

### Dr Chris Tisdell - building functions [13min-52secs]

The solution to the practice problem at the very end of this video can be found in his free workbook found here.

video by Dr Chris Tisdell

How to Calculate Fourier Series

As you saw in that video, there are some basic equations required to calculate the Fourier Series. To build a Fourier Series for a function $$f(t)$$ with period $$2L$$, it is required that $$f(t)$$ and it's derivative $$f'(t)$$ be piecewise continuous on the interval $$[-L,L]$$.

Fourier Series Equations

Fourier Series

$$\displaystyle{ f(t) = a_0 + \sum_{n=1}^{\infty}{ \left[ a_n \cos \frac{n \pi t}{L} + b_n \sin \frac{n \pi t}{L} \right] } }$$

constants

$$\displaystyle{ a_0 = \frac{1}{2L} \int_{-L}^{L}{f(t)~dt} }$$

$$\displaystyle{ a_n = \frac{1}{L} \int_{-L}^{L}{f(t)\cos \frac{n\pi t}{L} ~dt} }$$

$$\displaystyle{ b_n = \frac{1}{L} \int_{-L}^{L}{f(t)\sin \frac{n\pi t}{L} ~dt} }$$

Knowing if the original $$f(t)$$ is either even or odd can help us a lot when finding the Fourier Series. Of course, we do not require that $$f(t)$$ be even or odd, but you remember from precalculus that cosine is an even function and sine is odd. So, for even functions $$b_n=0$$ and for odd functions $$a_n=0$$.

Practice

Unless otherwise instructed, find the Fourier Series for these functions.

$$\displaystyle{ f(t) = \left\{\begin{array}{rr} -1 & -\pi < t < 0 \\ 0 & t = 0, \pm \pi \\ 1 & 0 < t < \pi \end{array} \right. }$$
$$f(t) = f(t+2\pi)$$ for all $$t$$

Problem Statement

Find the Fourier Series for $$\displaystyle{ f(t) = \left\{\begin{array}{rr} -1 & -\pi < t < 0 \\ 0 & t = 0, \pm \pi \\ 1 & 0 < t < \pi \end{array} \right. }$$
$$f(t) = f(t+2\pi)$$ for all $$t$$

Solution

### 1296 video

video by Dr Chris Tisdell

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$$\displaystyle{f(x) = \left\{\begin{array}{rr} 0 & -1 \leq x \leq 0 \\ 1 & 0 < x < 1 \end{array} \right. }$$
with period $$2$$

Problem Statement

Find the Fourier Series for $$\displaystyle{f(x) = \left\{\begin{array}{rr} 0 & -1 \leq x \leq 0 \\ 1 & 0 < x < 1 \end{array} \right. }$$
with period $$2$$

Solution

### 1297 video

video by PatrickJMT

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$$\displaystyle{ f(x) = \left\{\begin{array}{rr} -3 & -1 < x < 0 \\ 3 & 0 < x < 1 \end{array} \right. }$$
$$f(x) = f(x+2)$$

Problem Statement

Find the Fourier Series for $$\displaystyle{ f(x) = \left\{\begin{array}{rr} -3 & -1 < x < 0 \\ 3 & 0 < x < 1 \end{array} \right. }$$ where $$f(x) = f(x+2)$$

Solution

### 1298 video

video by Dr Chris Tisdell

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Solving Differential Equations

These videos show how to use Fourier Series to solve differential equations.

video by MIT OCW

video by MIT OCW

### MIT OCW - Finding Particular Solutions via Fourier Series; Resonant Terms; Hearing Musical Sounds [45min-46secs]

video by MIT OCW

Practice

Unless otherwise instructed, solve these differential equations using Fourier Series.

Solve $$y'' + 3y' + 3y = f(x)$$ for the square wave of period 2: $$\displaystyle{f(x) = \left\{\begin{array}{rr} 1 & 0 \lt x \lt 1 \\ -1 & -1 \lt x \lt 0 \end{array} \right. }$$

Problem Statement

Solve $$y'' + 3y' + 3y = f(x)$$ for the square wave of period 2: $$\displaystyle{f(x) = \left\{\begin{array}{rr} 1 & 0 \lt x \lt 1 \\ -1 & -1 \lt x \lt 0 \end{array} \right. }$$

Solution

### 3548 video

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Solve $$y'' + \omega^2y = f(x)$$, $$\omega \neq n \pi$$ for the square wave of period 2: $$\displaystyle{f(x) = \left\{\begin{array}{rr} 1 & 0 \lt x \lt 1 \\ -1 & -1 \lt x \lt 0 \end{array} \right. }$$

Problem Statement

Solve $$y'' + \omega^2y = f(x)$$, $$\omega \neq n \pi$$ for the square wave of period 2: $$\displaystyle{f(x) = \left\{\begin{array}{rr} 1 & 0 \lt x \lt 1 \\ -1 & -1 \lt x \lt 0 \end{array} \right. }$$

Solution

### 3549 video

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You CAN Ace Differential Equations

### Topics You Need To Understand For This Page

 precalculus - even and odd functions infinite series basics of differential equations

### Related Topics and Links

external links you may find helpful

fourier series youtube playlist

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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### Topics Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

Differential Equations

Precalculus

Engineering

Circuits

Semiconductors

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