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Math Help - Second derivative of infinite series gives 0^0 situation

  1. #1
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    Second derivative of infinite series gives 0^0 situation

    What is the correct way to handle this problem?

    Evaluate the second derivative wrt x of the summation of [(2x+1)^(n+1)]/(n!) from n=0..inf at x=-1/2

    My solution was to take the derivative twice, giving:

    summation of [4(n+1)(2x+1)^(n-1)]/[(n-1)!] from n=1..inf

    then substituting x=-1/2 giving

    summation of [4(n+1)*0^(n-1)]/[(n-1)!] from n=1..inf

    Now the mistake I originally made here was to assume 0^(n-1) will cancel to zero making the solution zero, but the first term of the series creates a 0^0 situation which evidently should be treated as 1 -- resulting in the final solution being 8. I'm not very well versed on the proper way to handle 0^0 situations and was wondering if I should always treat 0^0 = 1 in problems like this?

    I recognize that I could just break out the first two terms of the original series before taking the two derivatives to avoid getting to a 0^0 situation, but I wasn't sure if this was necessarily the best way to do it. Thanks.
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  2. #2
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    Quote Originally Posted by drumist View Post
    I recognize that I could just break out the first two terms of the original series before taking the two derivatives to avoid getting to a 0^0 situation, but I wasn't sure if this was necessarily the best way to do it. Thanks.
    This is the wisest way to handle this kind of problem
    I personally take a great care of the first terms of a series in order to be sure of what I am doing
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by drumist View Post
    What is the correct way to handle this problem?

    Evaluate the second derivative wrt x of the summation of [(2x+1)^(n+1)]/(n!) from n=0..inf at x=-1/2

    My solution was to take the derivative twice, giving:

    summation of [4(n+1)(2x+1)^(n-1)]/[(n-1)!] from n=1..inf

    then substituting x=-1/2 giving

    summation of [4(n+1)*0^(n-1)]/[(n-1)!] from n=1..inf

    Now the mistake I originally made here was to assume 0^(n-1) will cancel to zero making the solution zero, but the first term of the series creates a 0^0 situation which evidently should be treated as 1 -- resulting in the final solution being 8. I'm not very well versed on the proper way to handle 0^0 situations and was wondering if I should always treat 0^0 = 1 in problems like this?

    I recognize that I could just break out the first two terms of the original series before taking the two derivatives to avoid getting to a 0^0 situation, but I wasn't sure if this was necessarily the best way to do it. Thanks.
    Quote Originally Posted by running-gag View Post
    This is the wisest way to handle this kind of problem
    I personally take a great care of the first terms of a series in order to be sure of what I am doing
    Let's give a little stronger hint here. Your second derivative is



    There is a problem with simplifying the coefficient the way you did. Compare what happens to these two "identical" expressions when n = 0:
    <--The unsimplified version

    <--The simplification

    The first expression is 0. The second expression is undefined because (-1)! is undefined. In order to correctly calculate the series at n = 0 you have to use the first expression. What does this do when you plug x = -1/2 into your series?

    -Dan
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