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Series Problem: Find the Sum


Date: 6/24/96 at 15:52:35
From: Anonymous
Subject: Series Problem: Sum

Find  sum[sin(nx)/(3^n),(n,0,oo)] if sin x=1/3 and x is in the 
first quadrant.


Date: 6/24/96 at 17:5:22
From: Doctor Anthony
Subject: Re: Series Problem: Sum

The series is sin(x)/3 + sin(2x)/3^2 + sin(3x)/3^3 + .. to infinity.

Let sin(x) be represented by Imaginary part of e^(ix) and in general
sin(nx) = Imaginary part of e^(nix).  So the series can be written:

Imag.part of e^(ix)/3 + e^(2ix)/3^2 + e^(3ix)/3^3 + ....to infinity.

This is a geometric series with common ratio e^(ix)/3

Sum to infinity = 1/(1-e^(ix)/3) and we must now take the imaginary 
part.

Sum = 3/(3-(cos(x)+isin(x)) = 3/(3-cos(x)-isin(x))

    = 3(3-cos(x)+i*sin(x))/{((3-cos(x))^2 + sin^2(x)}

    = (9-3cos(x)+3i*sin(x))/{9-6cos(x)+cos^2(x)+sin^2(x)}

    = (9-3cos(x)+3i*sin(x))/{10-6cos(x)}

Now taking the imaginary part we get

Sum = 3sin(x)/(10-6cos(x))

Now we have sin(x)=1/3 and therefore cos(x)= 2sqrt(2)/3

Sum = 3(1/3)/{10-6*2sqrt(2)/3}

    = 1/{10-4sqrt(2)} 

This can be changed to have the surd in the numerator, and so the sum 
is also given by:

      (5+2sqrt(2))/34 

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Sequences, Series

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