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Equality of Series

  1. May 27, 2016 #1
    1. The problem statement, all variables and given/known data
    Here is my equation that I want to find bs values
    9595352200_1464358764.png
    2. Relevant equations


    3. The attempt at a solution

    I convert sin to cos.
    3790159100_1464358842.png

    for bs at s=0 I get
    1047470800_1464348618.png
    and if s is not zero I can't derive a clear answer for bs.

    but in electronic engineer book that I read it wrote
    7779146900_1464348841.png
     
    Last edited: May 27, 2016
  2. jcsd
  3. May 27, 2016 #2
    My advice will be to use the trigonometric identities ##sin(A)sin(B)=\frac{1}{2}(cos(A-B)-cos(A+B))## and ##cos(A)cos(B)=\frac{1}{2}(cos(A-B)+cos(A+B)## for ##A=B=s(\phi-\alpha)##.
     
  4. May 27, 2016 #3
    Thank you for your best suggestion
    I did it for both side so I have
    2729523900_1464359418.png

    but it doesn't the same or I can't derive book answer , what is my mistake?
     
  5. May 27, 2016 #4
    Ok sorry my suggestion doesn't seem to work, more specifically when you move from line 1 to line 2 and 3 of your last post, that is sufficient but not necessary . Try instead to transfer everything to one side and use the identity ##CcosAcosB-CsinAsinB=Ccos(A+B)##


    Sorry this doesn't seem to work either, I am all out of ideas for now.

    Cant think of anything else than to expand any term of the form ##sin(s(\phi-\alpha))=sin(s\phi)cos(s\alpha)-sin(s\alpha)cos(s\phi)## and similar for cosines and see if you can get anything in the end.​
     
    Last edited: May 27, 2016
  6. May 27, 2016 #5

    Ray Vickson

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    You need to tell us if the summation index ##s## goes over ##0,1,2, \ldots## or over ##\ldots -2, -1,0,1,2, \ldots## That can make a real difference.
     
  7. May 27, 2016 #6
    Well seems to me the only way to make it work is s to take values in the whole integer set from negative to positive. Then what he did in post #3 is pretty much valid with some additional correction and justification regarding line 3 and 5 of post #3.
     
  8. May 27, 2016 #7

    Ray Vickson

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    I was hoping to hear from the OP himself/herself.
     
  9. May 27, 2016 #8
    Well this thing puzzled me also, I thought my first suggestion should work because is the only straightforward way to transform different products (product of cosines and product of sines) to a sum of the same things. I don't know maybe I am still wrong and the solution is totally different, but seems to me its working (with some corrections) for s through all integers.
     
  10. May 27, 2016 #9
    Thanks for you response.
    m should be positive and m=0,1,2,3,....
     
  11. May 27, 2016 #10

    SammyS

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    Ray's question regarding indices was in regards to s, not m .
     
    Last edited: May 27, 2016
  12. May 27, 2016 #11

    Ray Vickson

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    What is 'm'? Your original post had no 'm' in it anywhere. It did have an 's', though, and the issue is about the range of 's'. If that range is infinite (either one-way or two-way) the right-hand-side in your original equation is a divergent infinite series. So, I am not sure your question makes any mathematical sense.
     
  13. May 27, 2016 #12
    However Baby1 if s is taking values in bounded integer sets of the form (-k,-k+1,...,-1,0,1,...,k-1,k) the solution can be ##b_s=\frac{I\omega\epsilon\pi}{2(\pi-\alpha)}## and satisfies both line 2 and line 3 of your post #3. For line 2 is obvious for line 3 you have to find out as I am sure you will .
     
    Last edited: May 28, 2016
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