Fourier Series - Will someone walk me through this

In summary, the conversation discussed a particular book for a class on applied mathematics and the difficulties the student was having with a specific chapter. They shared an example from the book and expressed the need for more guidance in understanding the steps. This led to a discussion on solving Laplace's equation in polar coordinates and finding a Fourier series solution. The student was able to gain a better understanding and solve the problem with the help of the expert's explanation.
  • #1
FrogPad
810
0
The book for the class that I'm currently taking is "Introduction to Applied Mathematics" by Gilbert Strang. Things have been good with this class until this chapter. If you have used this book, you will understand what I mean when I say it is different. Things have been ok, because I've been understanding it. But I seriously need someone to hand hold me through an example. Here is one of the two worked examples on this material. (pg 278)

EXAMPLE 2 [itex] u_0(\theta) = \theta,\,\,\,-\pi<\theta<\pi.[/itex]

The Fourier series for [itex] u_0 [/itex] was calculated in equation (12);
[tex]\theta = 2\left(\frac{\sin\theta}{1}-\frac{\sin 2\theta}{2}+\frac{\sin 3\theta}{3}- \ldots \right). [/tex]

Therefore the Fourier series (21) for [itex] u [/itex] is
[tex] u(r,\theta)=2\left(\frac{r \sin\theta}{1}-r^2 \frac{\sin2\theta}{2}+r^3 \frac{\sin3\theta}{3}-\ldots \right). [/tex]

On the circle [itex] u [/itex] has a sudden jump at [itex] \theta = \pi [/itex], but inside the circle that jump is gone. The powers [itex] r^k [/itex] damp the high frequencies and Laplace's equation always has smooth solutions.If someone would please help me fill in the missing steps, I would be relieved.

I feel like someone has showed me: [tex] 5x^2 + 30x +2 =0[/tex] and asked me to solve it for [itex] x [/itex] but I've never heard of the quadratic equation, or completing the square :)

I mean yes it would be nice if I could figure out those "in between steps". But, it is not coming to me... and we are moving quickly into the next chapter. So I need to know this. Thank you.
 
Last edited:
Physics news on Phys.org
  • #2
What you have given here, simply doesn't make sense. Given the Fourier series for [itex]u_0(\theta)[/itex] does not automatically give a Fourier series for [itex]u(r,\theta)[/itex] without some more information- for example the fact that [itex]u(r,\theta)[/itex] satisfies some partial differential equation inside the circle r= 1.

I suspect, although you didn't say it, that this [itex]u(r,\theta)[/itex] satisfies Laplace's[\b] equation inside the circle. Certainly any (reasonable) function of [itex]\theta[/itex] can be written as a Fourier series. If it is also a function of r, then the coefficients may be functions or r. Assume a solution of the form
[tex]u(r,\theta)= u_1(r)\frac{sin \theta}{1}- u_2(r)\frac{sin 2\theta}{2}+ u_3(r)\frac{sin 3\theta}{3}- ... [/tex]
, plug it into the partial differential equation and see what differential equations you get for un(r).
 
  • #3
Sorry HallsofIvy. This example was under the section of Solution of Laplaces Equation.

Thank you by the way! It's how you worded it, but something clicked (I think, or hope anyways).

[tex] \nabla^2 = 0 [/tex]

Laplaces equation in polar coordinates:
[tex] \frac{1}{r} \frac{\partial}{\partial r}\left( r \frac{\partial u}{\partial r}\right)+\frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2}=0 [/tex]

Assume:
[tex] u(r,\theta) = a_0 + \sum_{n=1}^{\infty}a_n r^n \cos n\theta + \sum_{n=1}^{\infty} b_n r^n \sin n\theta [/tex]

When [itex] r=1 [/itex] then [itex] u(r=1,\theta) = u_0(\theta) [/itex]

Thus:
[tex] u_0 = \theta [/tex]

We rewrite this as a Fourier series:
[tex] u_0(\theta) = 2\left( \frac{\sin \theta}{1}-\frac{\sin 2\theta}{2}+\frac{\sin 3 \theta}{3} - \ldots \right) =\ldots[/tex]

[tex] \ldots = a_0 + (a_1 r \cos \theta + a_2 r^2 \cos 2 \theta + \ldots) + (b_1 r \sin \theta + b_2 r^2 \sin 2\theta+\ldots) [/tex]

Now [itex] u_0 = u [/itex] when r=1 so:
[tex] a_n =0 [/tex] and [tex] r=1 [/tex] yields:

[tex] (b_1 r \sin \theta + b_2 r^2 \sin 2\theta + b_3 r^3 \sin 3 \theta + \ldots) [/tex] which is close to [tex] u_0 [/tex] but not it yet so, we have to solve for [itex] b_n [/itex]

by inspection:
[tex] b_n = \frac{2\alpha}{n}\,\,\left|\,\,\alpha = \left\{\begin{array}{c} 1\,\,\,\,n=\{1,3,5,\ldots\} \\ -1\,\,\,\,n=\{2,4,6,\ldots\} \end{array} [/tex]

Thus:
[tex] u(r,\theta) = \sum_{n=1}^{\infty} \frac{2 \alpha}{n}(r^n \sin n\theta ) [/tex]

plugging [itex] u_n(r,\theta) = \frac{2 \alpha}{n}(r^n \sin n\theta ) [/itex] into Laplaces equation:
[tex] \nabla^2(u_n(r,\theta)) = 0 [/tex]

Thus satisfying the boundary condition, and Laplaces equation.
 
Last edited:

FAQ: Fourier Series - Will someone walk me through this

What is a Fourier series?

A Fourier series is a mathematical representation of a periodic function as a sum of sine and cosine functions. It is named after Joseph Fourier, a French mathematician who first introduced the concept in the early 19th century.

How is a Fourier series calculated?

A Fourier series is calculated using a formula that involves integration of the given function over one period. This results in coefficients for each sine and cosine term, which are then used to create the series.

What is the purpose of a Fourier series?

A Fourier series is used to represent a periodic function in terms of simpler trigonometric functions, making it easier to analyze and manipulate. It has applications in fields such as signal processing, image processing, and physics.

What are the key components of a Fourier series?

The key components of a Fourier series are the trigonometric functions (sine and cosine), the coefficients for each term, and the period of the function. These components are used to create the series, which can be used for various mathematical operations.

How is a Fourier series different from a Fourier transform?

A Fourier series represents a periodic function as a sum of trigonometric functions, while a Fourier transform represents a non-periodic function as a sum of complex exponential functions. Additionally, a Fourier series deals with continuous functions, while a Fourier transform can handle both continuous and discrete functions.

Similar threads

Back
Top