Geometric Series
 

I realize geometric series isn't exactly probability, but attracts the mathematically-inclined posters, so here goes. I also posted in the SMP forum.

I need to show that:

2*sum[ (( 1/3)^t) * cos(wt) ]
where we are summing from t=1 to inf, is equal to:
( 1 - 3cos(w) ) / ( 3cos(w) - 5 )

Any links/tips would be helpful. I haven't been able to find anything via google.

Re: Geometric Series
 

[ QUOTE ]
I realize geometric series isn't exactly probability, but attracts the mathematically-inclined posters, so here goes. I also posted in the SMP forum.

I need to show that:

2*sum[ (( 1/3)^t) * cos(wt) ]
where we are summing from t=1 to inf, is equal to:
( 1 - 3cos(w) ) / ( 3cos(w) - 5 )

Any links/tips would be helpful. I haven't been able to find anything via google.

[/ QUOTE ]

You're going to have better results for a question like this in the homework section of http://www.physicsforums.com/
(They also have latex support there so you can make your post more legibly.)

Have you seen:
e^(i theta)=cos theta + i sin theta
?

Re: Geometric Series
 

Rufus has the right idea.

((1/3)^t)*cos(wt) is the real part of [(1/3)e^(iw)]^t. So the sum is the real part of (1/3)e^(iw)/[1-(1/3)e^(iw)], using the formula for the sum of a geometric series.

Now substitute e^(iw)=cos w + i sin w, and use the fact that the real part of (a+ib)/(c+id) is (ac+bd)/(c^2+d^2).

When I do the algebra, I get

(1-3cos(w))/(3cos(w)-5),

which is what you wanted to show.

Re: Geometric Series
 

thanks both for your quick replies, ill keep that site bookmarked from now on

Re: Geometric Series
 

Hi again, quick question regarding (a+bi)/(c+di)

What is the justification for dropping the imaginary part? i.e. Only using (ac+bd)/(c^2+d^2) and dropping the i[(ac+bd)/(c^2+d^2)] component

Re: Geometric Series
 

[ QUOTE ]
Hi again, quick question regarding (a+bi)/(c+di)

What is the justification for dropping the imaginary part? i.e. Only using (ac+bd)/(c^2+d^2) and dropping the i[(ac+bd)/(c^2+d^2)] component

[/ QUOTE ]

Original equation:

2 * Sum[ (1/3)^t cos(wt) ]

You then perform a manipulation trick adding in (1/3)^t sin(wt) so it is summable by geometric series (2 * Sum[ ((1/3) e^iw)^t]). This equation is a new equation different from the original one. You have added an imaginary part to the equation so you may sum the series. You then drop the imaginary part after you sum the series and are left with the equal real parts.

On the other hand, you may use the identity:

cos(z) = (e^(iz)+e^(-iz)) / 2

Then your series becomes:

Sum[(1/3*e^(iw))^t] + Sum[(1/3*e^(-iw))^t]

Which sums to:

-(e^iw)/((e^iw)-3) + 1/(3e^(iw)-1)

Algebraic manipulations will then yield your desired equation (using (e^iw = cos w + i sin w):

1-3Cos(w)/(3Cos(w)-5)

In this second method you don't have to worry about adding in an imaginary part and later dropping it.