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Function with Euler's constant

  1. Sep 25, 2015 #1
    Hello all,

    is there an alternative way of displaying

    ##\frac{x}{e^{x}-1}##

    as a trigonometric function, not using the bernoulli-numbers ?

    Thanks in advance
     
  2. jcsd
  3. Sep 26, 2015 #2

    HallsofIvy

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    I'm not sure what you mean by "display as a trigonometric function". Would writing [itex]e^x[/itex] as [itex]cos(x)+ i sin(x)[/itex] help?
     
  4. Sep 26, 2015 #3

    Orodruin

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    Regardless of what it means, writing it this way would not help since it is not true. It is ##e^x##, not ##e^{ix}## ...
     
  5. Sep 26, 2015 #4

    mfb

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    Well, you can replace x by ix.
     
  6. Sep 26, 2015 #5

    Svein

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    Turn it upside down: [itex]e^{x}-1=\frac{x}{1!}+\frac{x^{2}}{2!}+... [/itex], so [itex] \frac{e^{x}-1}{x}=\frac{1}{1!}+\frac{x}{2!}+\frac{x^{2}}{3!}...[/itex]. Don't know if that helps.
     
  7. Sep 26, 2015 #6

    WWGD

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    Maybe set up a triangle with one side ## f(x)=x ## and the other side a local linearization of ## e^x -1 ##, then use Taylor series for arcsin or arc cos? Just an idea, Ihave not worked it out.
     
  8. Sep 28, 2015 #7
    Thanks for your answers, I do think the answers of Svein and WWGD do help me the best.

    Basically, i am trying to understand, what the Bernoulli Numbers "do" and why they can be developed in a series expansion leading to the simple result x/(e^x-1).
    I started by checking that Faulhaber-formulas, where the bernoulli-numbers appear in the last term while summarising x^2n with escalating x. Second, i wonder why they "appear" only with 2n index (except B#1).

    thx for any hint !
     
  9. Sep 28, 2015 #8

    Svein

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