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Complex Fourier Series into a Cosine Series

  • #1

Homework Statement



a. Represent f(x)=|x| in -2<x<2 with a complex Fourier series

b. Show that the complex Fourier Series can be rearranged into a cosine series

c. Take the derivative of that cosine series. What function does the resulting series represent?
[/B]


Homework Equations



L%7D%2C%20-L%3Cx%3CL.gif


L%7Ddx.gif
[/B]


The Attempt at a Solution


[/B]
Ok, so the first part seems pretty straight forward, I plugged the variables in, and due to the absolute value broke the integral up from -2 to 0 and then from 0 to -2 and added them together.

Providing I made no errors (I hope...), I get

%20%5Cfrac%7B1%7D%7Bin%5Cpi%7D%20%28%7Be%5E%7Bin%5Cpi%7D%7D%20-%20%7Be%5E%7B-in%5Cpi%7D%29%7D%5D.gif


Using Euler's, I simplify this to:

1%7D%7Bin%5Cpi%7D%5B2cos%28n%5Cpi%29%20-%20%5Cfrac%7B1%7D%7Bin%5Cpi%7D%202i%20sin%28n%5Cpi%29%5D.gif


which, since for each n the sin (n pi) will become 0, my assumption is the above simply becomes (...?):

gif.latex?C_n%3D%5Cfrac%7B1%7D%7Bin%5Cpi%7D%5B2cos%28n%5Cpi%29%20%5D.gif


I am completely befuddled as to how to move into part b though... I also don’t know what to do with the n’s of the summation. So far, all the cosine series I've done so far have been summed from 0 to ∞ - I don’t know what to do with the -∞ to 0 part of the summation.

Any help would be greatly appreciated. Thanks for looking...
 
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Answers and Replies

  • #2
blue_leaf77
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I suggest that you check again your calculation of ##C_n##. The integral of ##C_n## can be simplified to
$$
C_n = \frac{1}{2L} \int_{-L}^L f(x) e^{in\pi x/L} dx = \frac{1}{2L} \int_{-L}^L f(x) \cos\frac{n\pi x}{L} dx
$$
The integral with ##i\sin\frac{n\pi x}{L}## vanishes because the product ##f(x)\sin\frac{n\pi x}{L}## forms an odd function. In this regard, I would expect ##C_n## to be real for any ##n##.
 
  • #3
BvU
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I don’t know what to do with the -∞ to 0 part of the summation.
Note that a summation index is just a place holder that takes on a value: $$\sum_{i=-\infty}^0 g(i) = \sum_{i=0}^{+\infty} g(-i)$$

[edit] should have used ##n## instead of ##i## to avoid confusion, sorry. I mean a sum over integer numbers
 
  • #4
Thank you two so much for your insight! I was making a silly error when integrating the absolute value.

I'm still getting a little stuck (very stuck) on b: "Show that the complex Fourier Series can be rearranged into a cosine series"

So, for part A: Represent /x/ with a complex Fourier Series, after correcting my C_n calculations, I got the answer of:

[itex]
C_n = \frac{2 cos (n\pi ) -2}{(n\pi)^2}

[/itex]

[itex]f(x) = \sum_{-\infty}^{\infty} C_ne^{-in\pi x /2}[/itex] , -2 < x < 2

Now, for part B, I have to rearrange this into a cosine series....I was trying to start by using Euler's identity to start:

[itex]f(x) = \sum_{-\infty}^{\infty} C_ne^{-in\pi x /2}[/itex]

[itex]= C_n [cos (\frac{n\pi x}{2})-isin(\frac{n\pi x}{2})][/itex]

....and this is as far as I get. I think (if my understanding of the material is correct) that I could make a conceptual argument as to why it can be just represented with the cosine series using some graphs that have been shown to me, but I don't know what I need to do to show it via formulas....

Thanks very much for reading and any help you can offer...
 
  • #5
blue_leaf77
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....and this is as far as I get. I think (if my understanding of the material is correct) that I could make a conceptual argument as to why it can be just represented with the cosine series using some graphs that have been shown to me, but I don't know what I need to do to show it via formulas....
Except for ##n=0##, for each ##n=m##, you also have ##n=-m## in the Fourier series. You have figured out that the Fourier coefficient looks like
$$
C_n = \frac{2 \cos (n\pi ) -2}{(n\pi)^2}
$$
Now compare this coefficient for ##n=m## and ##n=-m##, how do you conclude about the relation between them? Then consider the two terms out of the series which correspond to ##n=\pm m##
$$
C_m e^{-im\pi x/L} + C_{-m} e^{im\pi x/L}
$$
Taking the relation between ##C_m## and ##C_{-m}## you have found before, how can you simplify them?
 
  • #6
BvU
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Very good you fixed your intergration issue. ##1\over n^2## sounds a lot better than ##1\over n##.
You are way overthinking things: part a asked for the complex coefficients ##C_n## in $$f(x) = \sum_{n = -\infty}^\infty C_n e^{-in\pi/L}$$ and you found them.

Drat, blueleaf was faster (a lot faster!) and he's good.
 
  • #7
blue_leaf77
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Drat, blueleaf was faster (a lot faster!) and he's good.
May be, the timezone also has effect? I live in GMT+07.00 where it's currently approaching the dusk, if you are in the same zone as the US then I imagine it must be around breakfast time right now.
 
  • #8
BvU
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GMT+1 -- after lunch dip :smile:

Even the GMT+1 isn't good enough/unambiguous nowadays. They cheat and call it CET
 
  • #9
4
0
So I casually stumbled upon this problem and tried solving the problem on my own just for fun, but doesn't quite get the answer to part B and C. For B, do you just ignored the sine portion of the Euler's identity since f(x) is an even function?
 
Last edited:
  • #10
blue_leaf77
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For B, do you just ignored the sine portion of the Euler's identity since f(x) is an even function?
If you read through this post at least until post #6, you should get the hint why the sine terms do not appear.
 
  • #11
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0
Except for ##n=0##, for each ##n=m##, you also have ##n=-m## in the Fourier series. You have figured out that the Fourier coefficient looks like
$$
C_n = \frac{2 \cos (n\pi ) -2}{(n\pi)^2}
$$
Now compare this coefficient for ##n=m## and ##n=-m##, how do you conclude about the relation between them? Then consider the two terms out of the series which correspond to ##n=\pm m##
$$
C_m e^{-im\pi x/L} + C_{-m} e^{im\pi x/L}
$$
Taking the relation between ##C_m## and ##C_{-m}## you have found before, how can you simplify them?
Could you elaborate a bit on how this is supposed to rearrange the complex series to cosine series? Sorry, just really lost haha
 
  • #12
blue_leaf77
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Take a look at this.
 
  • #13
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0
Take a look at this.
So this is the work i've done and attempt to solve the last part. Could you take a look and see if there are mistakes? Thanks a lot.
https://scontent.fsnc1-1.fna.fbcdn.net/hphotos-xlf1/v/t34.0-12/12966620_10153752402479331_217279662_n.jpg?oh=9654213d445a9df020ce5fbf28c18fad&oe=5715E168
 
  • #14
BvU
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Well, it doesn't represent the sine function...
beware the PF rules: the OP has to have something to do left over ...
 
  • #15
Thanks for all your help, all. The posts through #6 clued me in as to how to actually show the sines do indeed disappear, and now I understand WHY, instead of just taking it as gospel that it does. The work shown above helped me confirm my understanding (and the work I turned in) was correct. And, lastly, that derivative represents f'(x) = d/dx /x/.

Really, really appreciate it! I wasn't grasping some of the concepts at play here; thanks so much!
 
  • #16
vela
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And, lastly, that derivative represents f'(x) = d/dx /x/.
Also known as a square wave, which is probably what they were looking for.
 

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