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Find the function for this Taylor series

  1. Jul 23, 2012 #1
    [itex]\sum_{m=0}^\infty \frac{(m-1)^{m-1}x^{m}}{m!}[/itex]


    Interesting result...
     
    Last edited: Jul 23, 2012
  2. jcsd
  3. Jul 23, 2012 #2
    I doubt there is a closed form for it. Plus, it only converges if |x|<1/e. After looking at the graph, it's similar to the Lambert W function when |x|<1/e: [tex]-W(-x)=\sum_{m=1}^{\infty}{\frac{m^{m-1}x^{m}}{m!}}[/tex] Subtracting 1 from both sides will make it approximately equal to your sum. However, there is no way to manipulate my series to put yours in terms of the W function.
     
    Last edited: Jul 23, 2012
  4. Jul 23, 2012 #3
    Another interesting thing of note, if [tex]f(x)=\sum_{m=0}^\infty \frac{(m-1)^{m-1}x^{m}}{m!}[/tex] then [tex]f'(x)=\sum_{n=0}^\infty \frac{n^{n}x^{n}}{n!}[/tex] as long as |x|<1/e.
     
    Last edited: Jul 24, 2012
  5. Jul 24, 2012 #4
    so...
    :-)
     
  6. Jul 24, 2012 #5
    Lagrange inversion...

    1/(1- ...) = ...

    voilá
     
  7. Jul 25, 2012 #6
    The Lagrange Inversion Theorem is used to find the inverse of a function. You didn't ask for the inverse in your original post :P
     
  8. Jul 25, 2012 #7

    haruspex

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    How about this? Integrate W wrt x:
    [tex]\sum_{m=1}^{\infty}{\frac{m^{m-1}x^{m+1}}{(m+1)!}} = \sum_{m=2}^{\infty}{\frac{(m-1)^{m-2}x^{m}}{m!}} = x\sum_{m=2}^{\infty}{\frac{(m-1)^{m-2}x^{m-1}}{m!}}[/tex]
    Now divide by x and differentiate:
    [tex]\sum_{m=2}^{\infty}{\frac{(m-1)^{m-1}x^{m-2}}{m!}}[/tex]
    .. and finally multiply by x2

    [edit] ... finally finally, add in the m = 1 term
     
    Last edited: Jul 25, 2012
  9. Jul 25, 2012 #8
    Well, it would seem I don't belong posting in this forum. I shouldn't have missed that.
     
  10. Jul 27, 2012 #9
    No, that`s wrong.

    Yours is shorter and better. From what you stated, you just need one step further ...
    and voilá, 1/ 1-...
     
  11. Jul 27, 2012 #10

    micromass

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    Did you actually know the answer to this problem??
     
  12. Jul 27, 2012 #11

    haruspex

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    Is this right?
    [itex]f(x) = \sum_{m=0}^\infty \frac{(m-1)^{m-1}x^{m}}{m!} = xG(x) - ∫G(x)[/itex]
    where G(x) = -W(-x).
    Hence [itex]f(x) = ∫x.dG(x)[/itex]
    But I don't know where we're trying to get to.
     
  13. Jul 27, 2012 #12
    yes, I know













    p.3, 14
     
    Last edited: Jul 27, 2012
  14. Jul 28, 2012 #13
    Even though I showed I really am a fool for missing something so obvious, I decided to finish this.

    [tex]y=\sum_{m=0}^{\infty}\frac{(m-1)^{m-1}x^{m}}{m!}[/tex]
    [tex]y'=\sum_{m=0}^{\infty}\frac{m^{m}x^{m}}{m!}[/tex]
    Move on to the W function.
    [tex]-W(-x)=\sum_{m=1}^{\infty}\frac{m^{m-1}x^{m}}{m!}[/tex]
    Differentiate.
    [tex]\frac{-W(-x)}{x(W(-x)+1)}=\sum_{m=1}^{\infty}\frac{m^{m}x^{m-1}}{m!}[/tex]
    [tex]\frac{-W(-x)}{W(-x)+1}=\sum_{m=1}^{\infty}\frac{m^{m}x^{m}}{m!}[/tex]
    Take note the subscript in that sum is m=1 instead of m=0.
    [tex]y'=1-\frac{W(-x)}{W(-x)+1}[/tex]
    [tex]y=\frac{x}{W(-x)}+C[/tex]
    In another form.
    [tex]y=-e^{W(-x)}+C[/tex]
    Both are true when |x|<1/e.
     
    Last edited: Jul 28, 2012
  15. Jul 28, 2012 #14
    Congratulations FOOL.
    There are other ways to get the function, but yours is great and different to others I have seen in the literature.

    Also congratulations to haruspex who did a great job, indeed, I wonder how you could posibly have found the function from your first post where you handled the sums, their indexes and the x's and x^2's...


    Just another note:
    FOOL, you say |x|<1/e

    however W(-1/e)=-1 and
    y={1/e}/{W(-1/e)}+C= -1/e +C

    What do you say about that?
     
    Last edited: Jul 28, 2012
  16. Jul 28, 2012 #15
    I say that |x|<1/e since that is the radius of convergence of [tex]y=\sum_{m=0}^{\infty}\frac{(m-1)^{m-1}x^{m}}{m!}[/tex]
     
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