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Solving series with complex exponentials.

  1. Dec 18, 2013 #1
    1. The problem statement, all variables and given/known data
    I don't know how to come up with this final solution:
    [tex]1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}+\frac{re^{-j(\theta-\phi)}}{a-re^{-j(\theta-\phi)}}=\frac{a^2-r^2}{a^2-2at\cos(\theta-\phi)+r^2}[/tex]
    2. Relevant equations

    3. The attempt at a solution
    For [itex]\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}[/itex], multiply by [itex]\frac{a-re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}[/itex]

    [tex]\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}(a-re^{j(\theta-\phi)})[/tex]
    [tex]=\frac{ar^n e^{jn(\theta-\phi)}}{a^n}-\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}+\frac{ar^{n-1} e^{j(n-1)(\theta-\phi)}}{a^{n-1}}-\frac{r^{n}e^{j(n(\theta-\phi)}}{a^{n-1}}\cdot \cdot \cdot \cdot\cdot \cdot+\frac{are^{j(\theta-\phi)}}{a}-\frac{r^2e^{j2(\theta-\phi)}}{a}[/tex]
    [tex]=re^{j(\theta-\phi)}-\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}[/tex]
    [tex]\Rightarrow\;\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}=\frac{re^{j(\theta-\phi)}-\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}}{a-re^{j(\theta-\phi)}}[/tex]

    The same steps are use:
    [tex]\Rightarrow\;\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=\frac{re^{-j(\theta-\phi)}-\frac{r^{n+1}e^{-j(n+1)(\theta-\phi)}}{a^n}}{a-re^{-j(\theta-\phi)}}[/tex]


    I expand the series out. You can see most cancel each other and leaving only two terms which factored out into 4 simpler terms. You can see two of the terms match the answer:
    [tex]1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}-\frac{\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}} {a-re^{j(\theta-\phi)}} +\frac{re^{-j(\theta-\phi)}}{a-re^{-j(\theta-\phi)}}-\frac{\frac{r^{n+1}e^{-j(n+1)(\theta-\phi)}}{a^n}} {a-re^{-j(\theta-\phi)}}[/tex]

    Compare with [itex]1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}+\frac{re^{-j(\theta-\phi)}}{a-re^{-j(\theta-\phi)}}[/itex]

    [tex]\Rightarrow\;\frac{\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}} {a-re^{j(\theta-\phi)}}+\frac{\frac{r^{n+1}e^{-j(n+1)(\theta-\phi)}}{a^n}} {a-re^{-j(\theta-\phi)}}=0[/tex]

    I have no idea how to make this zero, please help.

    Thanks
     
    Last edited: Dec 18, 2013
  2. jcsd
  3. Dec 18, 2013 #2
    I can't follow at all what you have done, so I can't comment on it.

    Perhaps the best approach would be to use the commonly known result for geometric series: [tex] \sum_{n=0}^\infty x^n = \frac{1}{1-x} [/tex] for [itex] |x| < 1 [/itex]
     
  4. Dec 18, 2013 #3
    Thanks for the reply.
    It is given [itex]r\leq a[/itex] and also [itex]|e^{j(\theta-\phi)}|\leq 1[/itex]. So that fits [itex]|x|\leq 1[/itex]

    [tex]\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}\;\Rightarrow\;x=\left(\frac{r}{a}\right) e^{j(\theta-\phi)}\;\Rightarrow\;\frac{1}{1-x}=\frac{a}{a-re^{j(\theta-\phi)}}[/tex]


    The same steps are use:
    [tex]\Rightarrow\;\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}\;\Rightarrow\;x=\left(\frac{r}{a}\right) e^{-j(\theta-\phi)} \;\Rightarrow\;\frac{1}{1-x}=\frac{a}{a-re^{-j(\theta-\phi)}}[/tex]

    But that will give:
    [tex]1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{a}{a-re^{j(\theta-\phi)}}+\frac{a}{a-re^{-j(\theta-\phi)}}[/tex]

    That's does not give the right answer. Please help.
     
  5. Dec 18, 2013 #4
    Notice that the summation in clamtrox' post starts at ##0## and yours starts at ##1##.
     
  6. Dec 18, 2013 #5
    Thanks, I missed that. I have to see whether it's making any difference.
     
  7. Dec 18, 2013 #6
    So with the extra term of [itex]n=0[/itex], both will give a 1 for [itex]n=0[/itex].

    [tex]1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{a}{a-re^{j(\theta-\phi)}}+\frac{a}{a-re^{-j(\theta-\phi)}}+2[/tex]

    I then factor 1 into one of the fraction:

    [tex]1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{a}{a-re^{j(\theta-\phi)}}+\frac{a}{a-re^{-j(\theta-\phi)}}+2=1+\frac{2a-re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}+\frac{2a-re^{-j(\theta-\phi)}}{a-re^{-j(\theta-\phi)}}[/tex]

    I still cannot get the right answer.
     
  8. Dec 18, 2013 #7
    Don't you think that should be -2 instead of +2?

    You can then give -1 to each of the fraction to get the right answer.
     
  9. Dec 18, 2013 #8
    Can you explain how you can make it to -1. If you look at post #3, it's all positive result. n=0 then the result is +1. I don't understand why you can assign -1 to it. Please help.

    Thanks
     
  10. Dec 19, 2013 #9
    You agree that
    $$\sum_{i=0}^{\infty} x^i=\frac{1}{1-x}$$

    for ##|x| < 1##.

    The above series is:
    $$1+x+x^2+x^3+.......=\frac{1}{1-x}$$
    Subtract 1 from both the sides to get
    $$\sum_{i=1}^{\infty} x^i =\frac{1}{1-x}-1$$

    Do you agree and does this clear your doubt?
     
    Last edited: Dec 19, 2013
  11. Dec 19, 2013 #10
    [tex] \sum_{n=1}^\infty x^n = x \sum_{n=0}^\infty x^n [/tex]
     
  12. Dec 19, 2013 #11

    Intrastellar

    User Avatar
    Gold Member

    [tex] \sum_{k=m}^\infty ar^k=\frac{ar^m}{1-r} [/tex]

    Here is the more general formula, it should be easier to go from here.
     
  13. Dec 19, 2013 #12
    Yes, I made a mistake, it's -1 instead.

    The book suggested the method in my original post and I just don't see how that work!!
     
  14. Dec 19, 2013 #13
    Do you have a link that explain a little more on this? What is the limit of [itex]r[/itex] in this?
     
  15. Dec 19, 2013 #14
  16. Dec 19, 2013 #15
    Thanks for your help. Lastly I just want to verify this:

    [tex]\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}\;\Rightarrow\;x=\left(\frac{r}{a}\right) e^{j(\theta-\phi)}[/tex]

    It is given ##r\leq a## so ##\left(\frac{r}{a}\right)^n\leq 1##

    [tex]|e^{jn(\theta-\phi)}|=e^{jn(\theta-\phi)}e^{-jn(\theta-\phi)}=1[/tex]

    [tex]\Rightarrow\;|x|=\left|\left(\frac{r}{a}\right) e^{j(\theta-\phi)}\right|\leq 1[/tex]
    Thanks
     
  17. Dec 19, 2013 #16
    I just read through the article. I don't think this can work for my problem. As posted, the ##r\leq 1## in my case. This does not work for ##r=1## as the denominator become zero and the solution is unbounded.

    So I have to go back to my original post #1. Maybe that's the reason the book used a complicated formula.

    Please help.

    Thanks
     
  18. Dec 19, 2013 #17
    You are right, and that is also the reason I edited my previous post.

    Can you perhaps post the complete problem statement? I don't see the bounds stated anywhere in the OP.
     
  19. Dec 19, 2013 #18
    This is part of the derivation of Laplace equation on a disk. The books show these three steps. I am trying to verify this by deriving from beginning to the end. The important thing is the boundary of the disk is ##r=a##. So the solution has to be finite when ##r=a## which rules out the easy solution of the few suggestions above.

    It is given by the book:
    [tex]1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}+\frac{re^{-j(\theta-\phi)}}{a-re^{-j(\theta-\phi)}}=\frac{a^2-r^2}{a^2-2at\cos(\theta-\phi)+r^2}[/tex]


    This is my attempt:
    For [itex]\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}[/itex], multiply by [itex]\frac{a-re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}[/itex]

    [tex]\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}(a-re^{j(\theta-\phi)})[/tex]

    [tex]=\frac{r^n e^{jn(\theta-\phi)}}{a^{n-1}}+\frac{r^{n-1} e^{j(n-1)(\theta-\phi)}}{a^{n-2}}+\frac{r^{n-2} e^{j(n-2)(\theta-\phi)}}{a^{n-3}}\cdot \cdot \cdot \cdot\cdot \cdot+\frac{r^2e^{j2(\theta-\phi)}}{a}+re^{j(\theta-\phi)}[/tex]

    [tex]-\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}-\frac{r^{n}e^{j(n(\theta-\phi)}}{a^{n-1}}-\frac{r^{n-1}e^{j(n-1)(\theta-\phi)}}{a^{n-2}}\cdot \cdot \cdot \cdot\cdot \cdot-\frac{r^3e^{-j3(\theta-\phi)}}{a^2}-\frac{r^2e^{j2(\theta-\phi)}}{a}[/tex]
    As you can see, all except two terms get cancelled out giving:
    [tex]=re^{j(\theta-\phi)}-\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}[/tex]
    [tex]\Rightarrow\;\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}=\frac{re^{j(\theta-\phi)}-\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}}{a-re^{j(\theta-\phi)}}[/tex]

    The same steps are use:
    [tex]\Rightarrow\;\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=\frac{re^{-j(\theta-\phi)}-\frac{r^{n+1}e^{-j(n+1)(\theta-\phi)}}{a^n}}{a-re^{-j(\theta-\phi)}}[/tex]


    I expand the series out. You can see most cancel each other and leaving only two terms which factored out into 4 simpler terms. You can see two of the terms match the answer:
    [tex]1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}-\frac{\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}} {a-re^{j(\theta-\phi)}} +\frac{re^{-j(\theta-\phi)}}{a-re^{-j(\theta-\phi)}}-\frac{\frac{r^{n+1}e^{-j(n+1)(\theta-\phi)}}{a^n}} {a-re^{-j(\theta-\phi)}}[/tex]

    Compare with [itex]1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}+\frac{re^{-j(\theta-\phi)}}{a-re^{-j(\theta-\phi)}}[/itex]

    [tex]\Rightarrow\;\frac{\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}} {a-re^{j(\theta-\phi)}}+\frac{\frac{r^{n+1}e^{-j(n+1)(\theta-\phi)}}{a^n}} {a-re^{-j(\theta-\phi)}}=0[/tex]

    I have no idea how to make this zero, please help.
     
    Last edited: Dec 19, 2013
  20. Dec 19, 2013 #19

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    I cannot figure out what you are trying to do, but it does seem you are making the problem 100 times more difficult than it is. You want to evaluate
    [tex] 1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}
    = 1 + 2\, \text{Re}\left[ \sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)} \right] [/tex]
    The summation can be easily computed using
    [tex] \sum_{n=1}^{\infty} x^n = \frac{x}{1-x}, \;\; x = (r/a)\, e^{j(\theta-\phi)}. [/tex]
     
  21. Dec 19, 2013 #20
    Yes, you are correct, but this formula cannot accommodate when ##r=a##. This is a Laplace equation on a disk where boundary condition is ##r=a##. So this formula cannot be used.
     
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