Fourier Cosine Series Coefficient Calculation: A_n = (2)(-1)^{n+1}(n≥1)

[TranscendArcu]

Homework Statement

The Attempt at a Solution

So after I have properly extended the series, I want to find the A_n coefficients. In the case of A_0 I get 1/2 -- good, great. That's what the book gets.

Now for the general A_n:

[itex]A_n = \frac{1}{2} \int^2_{-2} (1)cos(\frac{n\pi x}{2}) dx [/itex]
because, in the piecewise function, the portions from -2 to -1 and from 1 to 2 are zero, we can just write this as

[itex]A_n = \frac{1}{2} \int^1_{-1} (1)cos(\frac{n\pi x}{2}) dx [/itex]
[itex]= \frac{1}{2}(\frac{2}{n\pi} sin(\frac{n\pi x}{2}) |^1_{-1} = \frac{2}{n\pi} sin(\frac{n\pi}{2}) n ≥ 1[/itex]

and because integers n's will just cause the sin term to alternate between -1 and 1:

[itex] = \frac{(2)(-1)^{n+1}}{n\pi} n≥1[/itex], and this is my general term for the coefficients. But this isn't what the book gets, so what have I done wrong?

 

[Ray Vickson]

Homework Statement

The Attempt at a Solution

So after I have properly extended the series, I want to find the A_n coefficients. In the case of A_0 I get 1/2 -- good, great. That's what the book gets.

Now for the general A_n:

[itex]A_n = \frac{1}{2} \int^2_{-2} (1)cos(\frac{n\pi x}{2}) dx [/itex]
because, in the piecewise function, the portions from -2 to -1 and from 1 to 2 are zero, we can just write this as

[itex]A_n = \frac{1}{2} \int^1_{-1} (1)cos(\frac{n\pi x}{2}) dx [/itex]
[itex]= \frac{1}{2}(\frac{2}{n\pi} sin(\frac{n\pi x}{2}) |^1_{-1} = \frac{2}{n\pi} sin(\frac{n\pi}{2}) n ≥ 1[/itex]

and because integers n's will just cause the sin term to alternate between -1 and 1:

[itex] = \frac{(2)(-1)^{n+1}}{n\pi} n≥1[/itex], and this is my general term for the coefficients. But this isn't what the book gets, so what have I done wrong?

If the period is 4, we need to know the values of f(x) between x = 2 and x = 4 (or, perhaps, between x = -2 and x = 0). What are those values?

RGV

 

[TranscendArcu]

Since it says to use a cosine series, I presume that the question intends for us to use an "even" function extension for the piecewise. Thus, the values for f(x) = 0 between -2<x<-1 and f(x) = 1 for -1 < x < 0. Assigning f(x)'s extension in such a way makes the function even and thus permits us to find a cosine series.

 

[Ray Vickson]

Since it says to use a cosine series, I presume that the question intends for us to use an "even" function extension for the piecewise. Thus, the values for f(x) = 0 between -2<x<-1 and f(x) = 1 for -1 < x < 0. Assigning f(x)'s extension in such a way makes the function even and thus permits us to find a cosine series.

OK, that seems fair. You say your answer differs from that in the book. What is the book's answer? Is it, perhaps, just a matter of some different normalization convention? I don't see anything obviously wrong with your computation.

RGV

 

[TranscendArcu]

This is the books answer:

my answer will be [itex]f(x) = \frac{1}{2} + \sum_{n=1}^∞ ( \frac{(2)(-1)^{n+1}}{n\pi}) cos(\frac{n\pi x}{2})[/itex]

But they seem to think that we're skipping some terms (hence the 2n - 1 things), but I don't have any terms zeroing so I don't need to make that adjustment.

 

[TranscendArcu]

So I just tried to plot the first couple of terms in the series in MATLAB. I've attached the resulting graph in this post. As you can see, my answer isn't doing the right thing -- it's not correctly approximating the solution. Now, granted, this is only seven terms, but it doesn't seem to be doing very well. So I'm thinking that I must have a mistake somewhere in this work...

 

[Ray Vickson]

So I just tried to plot the first couple of terms in the series in MATLAB. I've attached the resulting graph in this post. As you can see, my answer isn't doing the right thing -- it's not correctly approximating the solution. Now, granted, this is only seven terms, but it doesn't seem to be doing very well. So I'm thinking that I must have a mistake somewhere in this work...

For even n ≥ 2 we will have [itex] \int_{-1}^1 \cos(n \pi x /2)\, dx = 0[/itex] because in each interval (-1,0) and (0,1) the positive and negative areas cancel. For example, when n = 2 the areas from x = 0 to x = 1/2 and x = 1/2 to x = 1 are opposite.

RGV