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Quotient of Real Analytic Functions

  1. Apr 6, 2013 #1
    1. The problem statement, all variables and given/known data
    How can I show that [itex] \frac{x}{e^x-1} [/itex] is real analytic in a neighborhood of zero, excluding zero?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 7, 2013 #2

    Dick

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    Use the taylor series expansion of e^x. Then simplify a little and use that if f(x) and g(x) are real analytic, then f(x)/g(x) is real analytic at 0 if g(0)≠0.
     
  4. Apr 7, 2013 #3
    How do I prove that if f(x) and g(x) are real analytic, then f(x)/g(x) is real analytic where g(0)≠0?
     
  5. Apr 7, 2013 #4

    Dick

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    Doesn't your book prove ANY theorems about real analytic functions? And what I actually meant to say is that if f(x) and g(x) are real analytic in a neighborhood of a and g(a)≠0, then f(x)/g(x) is real analytic in a neighborhood of a.
     
  6. Apr 7, 2013 #5
    Well no, and I can't even find any mention of it online.
     
  7. Apr 7, 2013 #6

    Bacle2

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    Well,maybe this will work: you can also see this as a quotient. You can ignore the -1 for a start, in e^x-1 ,and find the series for e^(-x) =1/e^x

    then multiply by x , and play around to get the -1 back in .

    Maybe massaging the series for xe^-x , to get the -1 back in will work.
     
  8. Apr 7, 2013 #7

    SammyS

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  9. Apr 7, 2013 #8

    Bacle2

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    How about using Taylor's theorem ( in the right domain)?
     
  10. Apr 7, 2013 #9
    That won't work. Say the Taylor series for [itex] \frac{ x }{ e^x -1 } [/itex] about zero is [itex] \sum_{n=0}^{\infty} \frac{B_n}{n!} x^n [/itex]. We know nothing about [itex] \lim_{ n \to \infty } \sqrt[n]{ \frac{B_n}{n!}} [/itex] because we know nothing about [itex] \lim_{n \to \infty} \frac{B_{n+1}}{B_n} [/itex]. For all we know, the radius of convergence could be 0.

    In fact, Taylor's Theorem only applies when we know a function can be represented as a power series. That is what we are trying to prove here.

    Edit: On closer inspection, B_n seems to be less or equal to 1, but I can't prove this.
     
    Last edited: Apr 7, 2013
  11. Apr 7, 2013 #10
    Does anyone have a good way of proving this? Any help is appreciated.
     
  12. Apr 7, 2013 #11

    Bacle2

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    Why don't you give us Bn?
     
  13. Apr 7, 2013 #12
    I doubt it has a simple formula, just by calculating the first few coefficients. Is there not some general way of proving the analyticity of a quotient? Neither of SammyS's links are helpful in this regard.
     
  14. Apr 7, 2013 #13

    Bacle2

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    How about this ( change of approach):

    Do you accept that the composition and product of analytic functions is analytic (taking care of the need for working out the composition of the domains of analyticity)?

    If you do, then use :

    f(x)= 1/(x-1) = 1+x+x^2+...... ( in |x|<1 )

    g(x)=e^x ;analytic everywhere

    f(g(x))= 1/(e^x-1)

    h(x)=x -- its own series--then h(x)*[f(g(x))]= x/(e^x -1)

    Proof of closure under composition should be annoying , but not too difficult conceptually, I think .

    Basically if g is analytic at y, then g(f(x)) is analytic when f(x)=y.
     
  15. Apr 8, 2013 #14

    Bacle2

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    Just curious: do you agree with my last post?
     
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