Derivative of Geometric Series

[Saladsamurai]

Homework Statement

I am having trouble following what is going on in this solution. We are looking to find the expectation value of:

f(x,y)=\frac{1}{4^{x+y}}\cdot\frac{9}{16}

I have gotten it down to:

E(X) = \frac{3}{4}\sum_{x=0}^\infty x\cdot\left(\frac{1}{4}\right)^x\qquad(1)

We know that for a geometric series with an initial value of 1 we can write for 0 < r < 1:

\sum_{x=0}^\infty r^x = \frac{1}{1-r}

taking the derivative of both sides wrt 'r' yields:

\sum_{x=1}^\infty r^{x-1} = \frac{1}{(1-r)^2}\qquad(2)Here is where I get confused:

I thought it was a simple matter of plugging in:

\frac{3}{4}\cdot\frac{1}{(1-1/4)^2} = 4/3

However, the solution gives:\frac{3}{4}\cdot\underbrace{\frac{1}{4}}\cdot\frac{1}{(1-1/4)^2} = 1/3

I am a little confused as to where the factor of 1/4 is coming from. I am having a feeling that it has something to do with the fact that (1) runs from 0 to infinite and (2) runs from 1 to infinite.

Any thoughts?

 

[hbweb500]

Ah, you forgot to divide out a (1/4) in order to get the summand into the form x r^{x-1}. In particular:

\sum_{x=0}^\infty x r^{x}

Divide out an r, dropping the power in the summand down by 1:

r \sum_{x=0}^\infty x r^{x-1}

Recognize that this is the derivative of the series with respect to r:

= r \sum_{x=0}^\infty \frac{d}{dr} \left(r^x\right)

Take the derivative outside of the sum and apply your knowledge about the geometric series:

= r \frac{d}{dr} \left( \sum_{x=0}^\infty r^x \right) = r \frac{d}{dr} \left( \frac{1}{1-r} \right) = r \frac{1}{(1-r)^2}

So that r out in front is where the extra (1/4) factor comes from.

 

[Saladsamurai]

Ah, you forgot to divide out a (1/4) in order to get the summand into the form x r^{x-1}. In particular:

\sum_{x=0}^\infty x r^{x}

Divide out an r, dropping the power in the summand down by 1:

r \sum_{x=0}^\infty x r^{x-1}

Recognize that this is the derivative of the series with respect to r:

= r \sum_{x=0}^\infty \frac{d}{dr} \left(r^x\right)

Take the derivative outside of the sum and apply your knowledge about the geometric series:

= r \frac{d}{dr} \left( \sum_{x=0}^\infty r^x \right) = r \frac{d}{dr} \left( \frac{1}{1-r} \right) = r \frac{1}{(1-r)^2}

So that r out in front is where the extra (1/4) factor comes from.

Wow, great explanation hbweb500! Thanks for the help