Solving a Two-Equation Fourier Series

[Shomy]

[SOLVED] Fourier Series

Homework Statement

I know this may be a simple problem but I am just beginning to understand this subject and this question has confused me.

Expand the following as a whole-range Fourier series:

f(x) = 1 -pi < x < 0
f(x) = x 0 < x < pi

I can solve FS with one eqn, but what do you do with two, do you integrate twice or something?

Homework Equations

The Attempt at a Solution

 

[Hootenanny]

Welcome to PF Shomy,

The first thing to decide is whether the function is even, odd or neither.

 

[Shomy]

Thanks for the welcome!
The first part is even and the second is odd

 

[Hootenanny]

The first part is even and the second is odd

I suspect that the function is written thus,

$$f(x) = \left\{\begin{array}{cr}1 & -\pi<x<0 \\ x & 0<x<\pi\end{array}\right.$$

Which means it isn't actually two functions, but one piecewise function.

 

[Shomy]

Yeah

 

[Hootenanny]

Yeah

So now, is the function odd or even?

 

[Shomy]

The first part is even and the second is odd

 

[Hootenanny]

The first part is even and the second is odd

So overall the function is...?

 

[Shomy]

Odd?

 

[Hootenanny]

Odd?

Correct! What does this tell you about the Fourier series coefficients?

 

[Shomy]

One of the coefficients (cos) is zero

 

[Hootenanny]

One of the coefficients (cos) is zero

Correct, all the coefficients of cosine are zero. So what is the Euler formula for calculating the coefficients of sine?

 

[Shomy]

Well i meant that one of the coefficients (ie the cos ) is zero because i am using sum notation.

an = 1/pi integral f(x)cosmx dx from pi to -pi. Do I integrate both functions and sum them or what?

 

[Hootenanny]

Haven't we just agreed that the cosine coefficients are zero?

 

[Shomy]

Yeah sorry

 

[Hootenanny]

Yeah sorry

No problem . So what about the coefficients of sine?

 

[Shomy]

bm = 1/pi integral f(x)sinmx dx

So do you integrate the first part from pi to 0 and the second part from 0 to -pi

 

[Hootenanny]

So do you integrate the first part from pi to 0 and the second part from 0 to -pi

I believe it's the other way around. Overall the coefficient is given by,

$$b_n=\frac{1}{\pi}\int_{0}^{2\pi}f(x)\sin(nx)dx$$

However, since f(x) is defined piecewise we must write,

$$b_n = \frac{1}{\pi}\left[\int_{-\pi}^{0}1\cdot\sin(nx)dx + \int_{0}^{\pi}x\cdot\sin(nx)dx\right]$$

Is that what you meant?

 

[Shomy]

Yeah that's what was holding up. Thanks so much!

 

[Hootenanny]

Yeah that's what was holding up. Thanks so much!

No problem

 

[Redbelly98]

> "Odd?"

Correct! What does this tell you about the Fourier series coefficients?

If I understand the function definition, I have a problem with this answer. For example consider x = 2:

f(2) = 2
f(-2) = 1

So for x = 2,
f(x) is not equal to f(-x)
and
f(x) is not equal to -f(-x)

What does this tell us about odd vs. even vs. neither?

 

[Hootenanny]

Good catch Redbelly! Of course the function is neither odd nor even, I really don't know what I was thinking ! I'll PM Shomy now and hopefully they haven't handed the assignment in.

I've been making a few daft mistakes recently

 

[Shomy]

Thank You!

 

[Redbelly98]

I've been making a few daft mistakes recently

It only gets worse as you get older