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Homework Help: 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
    [tex]\sum_{n=0}^{\infty}\frac{x^{2n}}{2n!}[/tex]
    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.

    cosh(x)=1+x^2/2!+x^4/4!+x^6/6!+...

    exp(cx^2)=1+cx^2+c^2x^4/2!+c^3x^6/3!+...

    exp(cx^2)-cosh(x)=(c-1/2!)x^2+(c^2/2!-1/4!)x^4+(c^3/3!-1/6!)x^6+...

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

    nrqed

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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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