Two Plus Two Newer Archives Geometric Series
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#1
11-11-2007, 09:53 PM
 yellowjack Senior Member Join Date: Nov 2004 Location: fighting -EV Posts: 2,152
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 )

#2
11-11-2007, 10:20 PM
 rufus Senior Member Join Date: Aug 2005 Posts: 425
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 )

[/ 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
?
#3
11-11-2007, 11:41 PM
 de Moivre Junior Member Join Date: Oct 2007 Posts: 5
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.
#4
11-11-2007, 11:51 PM
 yellowjack Senior Member Join Date: Nov 2004 Location: fighting -EV Posts: 2,152
Re: Geometric Series

thanks both for your quick replies, ill keep that site bookmarked from now on
#5
11-24-2007, 07:54 PM
 yellowjack Senior Member Join Date: Nov 2004 Location: fighting -EV Posts: 2,152
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
#6
11-27-2007, 02:55 AM
 tshort Senior Member Join Date: May 2005 Posts: 1,143
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

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.

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