Series Problem: Find the SumDate: 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/ |
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