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Taylor series

  1. Jul 23, 2007 #1
    1. The problem statement, all variables and given/known data

    For which real numbers c is [tex](e^x+e^{-x})/2 \leq e^{cx^2}[/tex] for all real x?

    2. Relevant equations

    3. The attempt at a solution
    I think you can expand both sides into series and term by term compare them. The left side is
    Can someone help me with the right side?
  2. jcsd
  3. Jul 23, 2007 #2
    Try x=iy. LHS = cos(y), RHS=exp(-cy^2)

    Hmmm... maybe not. x is real.




    c>0.5 means that each term is +ve.
    Last edited: Jul 24, 2007
  4. Jul 24, 2007 #3


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    I guess you mean [tex]\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}[/tex]

    ( 2n! really means 2 (n!))

    Why don't you expand the RHS also to get
    [tex]\sum_{n=0}^{\infty}\frac{(cx^2)^{n}}{n!} =
    \sum_{n=0}^{\infty}\frac{c^n \times x^{2n}}{n!}[/tex]

    and then simply compare
  5. Jul 24, 2007 #4
  6. Jul 24, 2007 #5
    Sorry--my calculus is a little fuzzy.

    Are you allowed to use the same Taylor series expansion of exp(x) with exp(cx^2)? Because you generate the Taylor series expansion by taking the derivatives of exp(x) at 0, and the derivatives of exp(x) at 0 are different from the derivatives of exp(cx^2) at 0, it seems like you might need some variant of the chain rule to make that work. I mean instead of just plugging in cx^2 for x.
  7. Jul 24, 2007 #6
    Nah, you can just substitute x'=cx^2 on both LHS and RHS.
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