# geometric power series

• Dec 7th 2009, 09:24 PM
jlmills5
geometric power series
My questions require me to put functions into the a/(1-r) form. I don't understand how to do this.

The question- Find a power series for the function, centered at c, and determine the interval of convergence.

f(x)= 1/(2-x), c=5

How do I put this into the a/(1-r) form?
• Dec 8th 2009, 03:40 AM
HallsofIvy
Quote:

Originally Posted by jlmills5
My questions require me to put functions into the a/(1-r) form. I don't understand how to do this.

The question- Find a power series for the function, centered at c, and determine the interval of convergence.

f(x)= 1/(2-x), c=5

How do I put this into the a/(1-r) form?

Divide both numerator and denominator by 2!
• Dec 8th 2009, 10:05 AM
jlmills5
I can see that, but what about the c=5? I don't understand how to incorporate that and create the a/(1-r) form.

My key shows 1/(-3-(x-5)) = (-1/3)/(1+(1/3)(x-5))

I can see how they derived the 2nd part from the first, but I have no clue how they even got the first part! Help!
• Dec 8th 2009, 10:21 AM
nehme007
If you want a series centered at c = 0, you let r be something times x. If you want a series centered at c = 5, you let r be something times x-5. To get an x-5 in the denominator, you do the following:
$2 - x = 2 - x + 5 - 5 \\
= -3 - x + 5 \\
= -3 - (x-5)$
.
Then proceed to divide the numerator and denominator by -3. Then you'll have the appropriate form with a = -1/3 and r = (1/3)(x-5).
• Dec 8th 2009, 10:30 AM
jlmills5
Thank you, thank you, thank you!