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Math Help - series proof

  1. #1
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    series proof

    hello I am given prove that

    \frac{sin\theta}{2} +   \frac{sin2\theta}{2^2} + \frac{sin3\theta}{2^3} +...........=  \frac{2sin\theta}{5-4cos\theta}

    what I have done
    this is a Geometric progression therefore sum to infinity = \frac{a}{1-r}

    a=  \frac{sin\theta}{2}

    r =  \frac{sin2\theta}{2^2}  / \frac{sin\theta}{2}
    r =  \frac{sin2\theta}{2sin\theta}

    substituting back into the equation I get

    sum to infinity =  \frac{2sin^2\theta}{2-2cos\theta}


    can someone please show me where I am going wrong, thanks.
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  2. #2
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    Re: series proof

    Quote Originally Posted by sigma1 View Post
    hello I am given prove that

    \frac{sin\theta}{2} +   \frac{sin2\theta}{2^2} + \frac{sin3\theta}{2^3} +...........=  \frac{2sin\theta}{5-4cos\theta}

    what I have done
    this is a Geometric progression therefore sum to infinity = \frac{a}{1-r}

    a=  \frac{sin\theta}{2}

    r =  \frac{sin2\theta}{2^2}  / \frac{sin\theta}{2}
    r =  \frac{sin2\theta}{2sin\theta}

    substituting back into the equation I get

    sum to infinity =  \frac{2sin^2\theta}{2-2cos\theta}


    can someone please show me where I am going wrong, thanks.
    It's not geometric, simplify the 2nd and 3rd terms using sin(A + B) = sinAcosB + cosAsinB to compare with 1st term sine function

     \frac{sin \theta}{2} \ + \ \frac{sin \theta cos \theta}{2} \ + \ \frac{sin \theta (4cos^2 \theta -1)}{8} \ + \ ...

    You don't have a common ratio here

    Last edited by agentmulder; April 9th 2013 at 11:18 PM.
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  3. #3
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    Re: series proof

    Quote Originally Posted by agentmulder View Post
    It's not geometric, simplify the 2nd and 3rd terms using sin(A + B) = sinAcosB + cosAsinB to compare with 1st term sine function

     \frac{sin \theta}{2} \ + \ \frac{sin \theta cos \theta}{2} \ + \ \frac{sin \theta (4cos^2 \theta -1)}{8} \ + \ ...

    You don't have a common ratio here

    thank you for pointing that out. it would be a great help if you could show me how you would attempt to further solve this.
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  4. #4
    Member agentmulder's Avatar
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    Re: series proof

    Your problem requires cleverness i do not have at this time so i am looking forward to a solution posted by another MHF member. The series is interesting.



    WIA is also stumped.

    series(sin(nx))/(2^n&# 41; , n = 1 to infinity) - Wolfram|Alpha
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  5. #5
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    Re: series proof

    Quote Originally Posted by sigma1 View Post
    hello I am given prove that

    \frac{sin\theta}{2} +   \frac{sin2\theta}{2^2} + \frac{sin3\theta}{2^3} +...........=  \frac{2sin\theta}{5-4cos\theta}

    what I have done
    this is a Geometric progression therefore sum to infinity = \frac{a}{1-r}

    a=  \frac{sin\theta}{2}

    r =  \frac{sin2\theta}{2^2}  / \frac{sin\theta}{2}
    r =  \frac{sin2\theta}{2sin\theta}

    substituting back into the equation I get

    sum to infinity =  \frac{2sin^2\theta}{2-2cos\theta}


    can someone please show me where I am going wrong, thanks.
    You could probably use induction.
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  6. #6
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    Re: series proof

    I'll replace \theta with x for convenience.

    Using the identity \sin(x)=i\frac{e^{-ix}-e^{ix}}{2}, your series may be rewritten as

    \displaystyle{\sum_{n=1}^{\infty}\frac{i}{2^{n+1}}  e^{-inx}}-\displaystyle{\sum_{n=1}^{\infty}\frac{i}{2^{n+1}}  e^{inx}}.

    Individually, these two sums are convergent geometric series (do you know why?). Calculate the sum for each term and add them up. I have verified that you do indeed get the desired result (after some algebraic manipulations), but I have no desire to type it out. You should be able to work it out yourself.

    You may need to justify why you are able to break the sum into two infinite series, but this is a straightforward application of limit theorems since both your series converge.
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