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To find an entire function

  1. Sep 2, 2012 #1
    1. The problem statement, all variables and given/known data

    To find entire functions which satisfy g([itex]\frac{1}{n}[/itex]) = g(-[itex]\frac{1}{n}[/itex]) = [itex]\frac{1}{n^{2}}[/itex]

    2. Relevant equations

    How many functions can be found?

    3. The attempt at a solution

    Because the function is entire, it can be expanded in the Taylor series. But how can I work out the question?
     
  2. jcsd
  3. Sep 2, 2012 #2

    Zondrina

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    Hint :

    This condition here : g([itex]\frac{1}{n}[/itex]) = g(-[itex]\frac{1}{n}[/itex])

    Should look oddly familiar to : [itex]g(x) = g(-x)[/itex] which is the definition of an EVEN function.

    For example, consider these functions :

    f(x) = x^2
    f(x) = x^4
    ...
    f(x) = x^2n

    ;)
     
  4. Sep 2, 2012 #3

    Dick

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    You should be able to easily find one entire function that satisfies that. A hint might be, 'don't think too hard'.
     
  5. Sep 2, 2012 #4
    So n can only be even numbers with the Ʃa[itex]_{n}[/itex]z[itex]^{n}[/itex]=[itex]\frac{1}{n^{2}}[/itex]. And then?
     
  6. Sep 2, 2012 #5

    Zondrina

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    Not necessarily, consider : cos(x), cosh(x), |x|

    Those are all even functions as well.
     
    Last edited: Sep 2, 2012
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