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Complex sum

  1. May 6, 2009 #1
    1. The problem statement, all variables and given/known data

    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)

    2. Relevant equations

    ratio test

    3. 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
     
  2. jcsd
  3. May 6, 2009 #2

    Dick

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    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.
     
  4. May 6, 2009 #3
    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) ]
     
  5. May 6, 2009 #4

    Dick

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    All the terms vanish except for n=3. I'll say it once more. (z-6)^(3-3)=1.
     
  6. May 6, 2009 #5
    so f'''(6) = (-1)^3 * (3)^2 / 2^3 * 3 * 2 * 1 * 1 ?

    I'm sorry, was my radius of convergence ok?
     
  7. May 6, 2009 #6

    Dick

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    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?
     
  8. May 6, 2009 #7
    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!
     
  9. May 6, 2009 #8

    gabbagabbahey

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

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