# Complex sum

## Homework Statement

a) determine where f(z) is analytic.

f(z) = sum from n to inf. [ ( (-1)^n * n^2 ) / ( 2^n ) * (z - 6)^n ]

b) evaluate f'''(6)

ratio test

## The Attempt at a Solution

a) I used the ratio test and found that f(z) is analytic on the circle of convergence C: | z - 6 | < 1/2.

b) I found the 3rd derivative of the sum as

f'''(z) = sum from 0 to inf [ ( (-1)^n * n^2 ) / ( 2^n ) * n * (n-1) * (n-2) * (z - 6)^(n-3) ]

then when i plug in 6 into this i get

f'''(6) = 0

any suggestions or comments are much apperciated. thanks

Dick
Homework Helper
The third derivative of your sum is going to come only from the n=3 term in the sum. It's not zero. (z-6)^(3-3)=1, not zero. And the sum for your third derivative should be the sum from 3 to inf.

ok so

f'''(z) = sum from 3 to inf [ ( (-1)^n * n^2 ) / ( 2^n ) * n * (n-1) * (n-2) * (z - 6)^(n-3) ]

f'''(6) = sum from 3 to inf [ ( (-1)^n * n^2 ) / ( 2^n ) * n * (n-1) * (n-2) * (6 - 6)^(n-3) ]

f'''(6) = sum from 3 to inf [ ( (-1)^n * n^2 ) / ( 2^n ) * n * (n-1) * (n-2) * (0)^(n-3) ]

Dick
Homework Helper
All the terms vanish except for n=3. I'll say it once more. (z-6)^(3-3)=1.

so f'''(6) = (-1)^3 * (3)^2 / 2^3 * 3 * 2 * 1 * 1 ?

I'm sorry, was my radius of convergence ok?

Dick
Homework Helper
so f'''(6) = (-1)^3 * (3)^2 / 2^3 * 3 * 2 * 1 * 1 ?

I'm sorry, was my radius of convergence ok?

I didn't look closely at it the first time but no it's not. Why do you think it's 1/2? What did you get from the ratio test for the limit a_(n+1)/a_n?

I did

| (-1)^(n+1) * (n+1)^2 * 1/( 2^(n+1)) * (z-6)^(n+1) * 2^n/( (-1)^n * n^2 * (z-6)^n ) |

= (n+1)^2 * 1 / (2*n^2) * |z-6|

lim n->inf (n+1)^2 * 1 / (2*n^2) = 1/2

so 1/2|z-6| < 1

oh.. did I make the mistake there

|z-6| = 2 ?

also, is this the circle where f(z) is analytic?

thanks!

gabbagabbahey
Homework Helper
Gold Member
I did

| (-1)^(n+1) * (n+1)^2 * 1/( 2^(n+1)) * (z-6)^(n+1) * 2^n/( (-1)^n * n^2 * (z-6)^n ) |

= (n+1)^2 * 1 / (2*n^2) * |z-6|

lim n->inf (n+1)^2 * 1 / (2*n^2) = 1/2

so 1/2|z-6| < 1

This last line should be $0<\frac{1}{2}|z-6|<1$ and hence the sum converges on the disk $0<|z-6|<2$...

also, is this the circle where f(z) is analytic?

The ratio test tells you where the sum converges. It so happens that each individual term in the sum is a polynomial in $z$; hence each term is analytic. That means that the sum will be analytic so long as it exists/converges, and therefor $f(z)$ is analytic on the disk $0<|z-6|<2$