Function with Euler's constant

  • Thread starter eaglechief
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  • #1
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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
 

Answers and Replies

  • #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?
 
  • #3
Orodruin
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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?
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
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.
 
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  • #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.
 
  • #7
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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 !
 

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