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Analytic Functions and Intervals of Convergence

  1. Nov 29, 2012 #1
    Working out of Boas' Mathematical Methods in the Physical Sciences; Chapter 14, section 2, problem 42...

    I'm supposed to write the power series of the following function, then find the disk of convergence for the series. Boas goes on to state, "What you are looking for is the point nearest the origin, at which the function does not have a derivative. Then the disk of convergence has center at the origin and extends to that point. The series converges inside the disk."

    f(z)= sinh(z)


    So I derived (rather than looked up) the series expansion for hyperbolic sine:
    [tex]
    x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!}+...
    [/tex]

    so that breaks down to
    [tex]
    \frac{x^{2n+1}}{2n+1}
    [/tex]

    By the ratio test I have come to a final expression of
    [tex]
    \frac{x^2 (2n+1)}{2n+3}
    [/tex]

    Not sure if I did that correctly. Anyway, I know I'm supposed to take the limit as n -> Infinity but that doesn't work here because this (above) expression will just "blow up" but the answer in the back tells me it converges "for all z". So I know I'm doing something incorrectly. Just looking for some pointers. Thanks.
     
  2. jcsd
  3. Nov 29, 2012 #2

    Dick

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    What happened to the factorials? (2n+1) isn't the same (2n+1)!.
     
  4. Nov 29, 2012 #3
    True. Completely overlooked that omission.

    Allow me to retry and return with questions.


    (Thanks)
     
  5. Nov 29, 2012 #4

    Dick

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    (Welcome). Allowed.
     
  6. Nov 30, 2012 #5
    Ugh, FYI don't casually read a political blog in the middle of trying a math problem, complete tangent.

    Anyway, it's completely possible that my factorial simplifying is off but I've come to this final expression

    [tex]
    \frac{x^2}{(2n+1)(2n+2)}
    [/tex]

    Does that seem correct?

    *I've noticed that I've replaced z by x, just noting that. It's supposed to be z.
     
  7. Nov 30, 2012 #6

    Dick

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    It's in the right ballpark. But I would say (2n+1)!/(2n+3)!=1/((2n+2)(2n+3)). How did you get that? By reading the political blog?
     
  8. Nov 30, 2012 #7
    You're correct, and by looking at my notes I was originally as well. I guess I shouldn't pay attention to politics in matters of numbers, huh?
     
  9. Nov 30, 2012 #8
    I still don't see how it converges for "all x" though. If I try to expand the denominator and then take the limit I just get zero.
     
  10. Nov 30, 2012 #9

    Mute

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    Is there any value of x you can pick such that the ratio is not less than 1 as you take n to infinity?
     
  11. Nov 30, 2012 #10
    Just the initial x at the beginning of the series expansion if I get your question correctly.
     
  12. Nov 30, 2012 #11

    Mute

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    By "beginning of the series expansion" do you mean n =1, 2 or so? That's not what I asking about. Let me try to be clearer: You calculated the ratio of the (n+1)th term in the series to the nth term in the series. If you take the limit as n goes to infinity of that expression, you are finding that the expression goes to zero, yes? My question was, is there any value of x that you can plug into that ratio such that when you take the limit as n goes to infinity, the ratio is more than 1? (Remember that the ratio test tells you a series diverges if the ratio of the (n+1)th term to the nth term is greater than 1 as you take the limit as n goes to infinity).
     
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