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Math Help - To sum n terms

  1. #1
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    To sum n terms

    Hi,

    iS THERE Any one out there who can provide an expression for the sum of n terms:

    i*X* COS i*X

    where X is in radians, and i varies from 1 to n?

    Thanks in advance for your help.

    Cheers
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  2. #2
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    Dear Almo,

    you can put out the x front of a bracket, I mean:
    sum i*x*cos(i*x) = x*sum i*cos(i*x)

    It can be interesting eventuell the derivation of sin(i*x) is.
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  3. #3
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    Quote Originally Posted by Almo View Post
    Hi,

    iS THERE Any one out there who can provide an expression for the sum of n terms:

    i*X* COS i*X

    where X is in radians, and i varies from 1 to n?

    Thanks in advance for your help.

    Cheers
    I take it that i in this case is your counter, not the imaginary number i = \sqrt{-1}?


    So you're asking what does

    \sum_{i = 1}^n{ix\cos{ix}} equal right?
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  4. #4
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    tHANKS FOR YOUR HELP

    G'day


    Many thanks for your interest.

    Yes you are correct that i is NOT an imaginary number i.e. sqrt(-1).

    It is as you say the sum of the terms whereby i is initialy 1 then increases by 1 to the limit n.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    How about this use the following lemma

    \sum_{n=1}^{m}n\cdot x^n=\frac{x^{m+1}(mx-m-1)+x}{(x-1)^2}

    So your sum of \sum_{n=1}^{m}xn\cos(nx)=x\Re\left\{\sum_{n=1}^{m}  ne^{inx}\right\} and apply the above lemma with x=e^{ix}
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  6. #6
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    Thanks

    Thanks for your suggestion. I'm afraid I don't understand if your last line is my desired answer. Would you please rephrase your solution in terms of my i and X?

    Cheers
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