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Final sum

  1. May 20, 2008 #1
    1. The problem statement, all variables and given/known data

    Hi :smile:

    Find the (finite,ultimate,definitive,peremptory,eventua,conclusive) sum:

    [tex]a) cos^2x+cos^22x+...+cos^2nx ; b) sin^2x+sin^22x+...+sin^2nx [/tex]

    2. Relevant equations

    z=A+Bi

    3. The attempt at a solution

    [tex]A=cos^2x+cos^22x+...+cos^2nx[/tex]

    [tex]B=sin^2x+sin^22x+...+sin^2nx[/tex]

    [tex]z=cos^2x+cos^22x+...+cos^2nx+i(sin^2x+sin^22x+...+sin^2nx)=(cos^2x+isin^2x)+(cos^22x+isin^22x)+...+(cos^2nx+isin^2nx)=(cos^2x+isin^2x)+[/tex]
    [tex](cos^2x+isin^2x)^2+...+(cos^2x+isin^2x)^n[/tex]

    How will I continue?
     
    Last edited: May 20, 2008
  2. jcsd
  3. May 20, 2008 #2

    Borek

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    Staff: Mentor

    Not sure what you are trying to do.

    First, you defined two sums - a and b.

    Is your task to find sum of those two sums? a+b is what you are looking for?

    If so, why do use i?

    Do you know what Pythagorean trigonometric identity is?
     
  4. May 20, 2008 #3
    I don't know if they are separate or together... I should use "i", since that's my task, to find it through i.
     
  5. May 21, 2008 #4
    Any help?
     
  6. May 21, 2008 #5

    Borek

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    You won't get more help without giving more information. At present - at least IMHO - question as posted, and the idea of using i for calculations, doesn't make sense.
     
  7. May 21, 2008 #6

    Dick

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    The real part of exp(i*n*x)^2 is cos(n*x)^2-sin(n*x)^2. So summing the geometric series exp(i*2*n*x) and taking the real part will give you the difference A-B. What's the sum A+B? That's one way to do it.
     
  8. May 23, 2008 #7
    [tex](\frac{1+cos2x}{2}+i\frac{1-cos2x}{2})+(\frac{1+cos4x}{2}+i\frac{1-cos4x}{2})+(\frac{1+cosnkx}{2}+i\frac{1-cosnkx}{2})[/tex]

    [tex]\frac{1}{2}(1+cos2x+i(1-cos2x))+\frac{1}{2}(1+cos2x+i(1-cos2x))+\frac{1}{2}(1+cos4x+i(1-cos4x))+\frac{1}{2}(1+cosnkx+i(1-cosnkx))[/tex]

    How will I get rid of "1"?

    How will I go next?
     
  9. May 23, 2008 #8
    Dick, someone?
     
  10. May 23, 2008 #9

    Dick

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    Why did you ignore my last message? It had some reasonable suggestions in it.
     
  11. May 23, 2008 #10
    I have never learn about it. Is mine correct? How will I continue solving?
     
  12. May 23, 2008 #11

    Dick

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    What you wrote is a version of A+Bi. Do you know how to sum the series
    cos(2n)+cos(4n)+cos(6n)+...?
     
  13. May 23, 2008 #12
    Actully, I have never learned. So I don't know.
     
  14. May 23, 2008 #13

    Dick

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    Can you sum a geometric series?
     
  15. May 23, 2008 #14
    I don't know how.
     
  16. May 23, 2008 #15

    Dick

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    If you haven't been given the sum of the finite series sin(nx) and cos(nx), or been taught how to derive them by summing geometric series like exp(i*n*x), then I don't know how you are supposed to do this problem.
     
  17. May 23, 2008 #16
    I know something...
    [tex]a=cosx+cos2x+...+cosnx[/tex]

    [tex]B=sinx+sin2x+...+sinnx[/tex]

    [tex]A=Re(z) ; B=Im(z)[/tex]

    [tex]z=A+Bi[/tex]

    [tex]z=A+iB=(cosx+cos2x+...+cosnx)+i(sinx+sin2x+...+sinnx)= (cosx+isinx)+[/tex]

    [tex]+(cos2x+isin2x)+...+(cosnx+isinnx)= (cosx+isinx)+(cosx+isinx)^2+...+(cosx+isinx)^n=[/tex]

    I don't know where to go out of here.
     
  18. May 23, 2008 #17

    Dick

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    That's a geometric series with a common ratio of exp(ix). You should know how to write down it's sum. http://en.wikipedia.org/wiki/Geometric_progression
    Once you do that the real part of the sum is A and the imaginary part is B.
     
  19. May 23, 2008 #18
    For (cosx+isinx) is simpler, but still got no clue how to continue.
     
  20. May 23, 2008 #19

    Dick

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    I just TOLD you. Review geometric series.
     
  21. May 23, 2008 #20
    Yes, yes, but I don't understand this step.
    Now I have
    [tex](cosx+isinx)(1+(cosx+isinx)+...+(cosx+sinx)^n^-^1)[/tex]
    I understand this. But don't understand where the next step comes from?
     
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