Function with Euler's constant

[eaglechief]

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

 

[HallsofIvy]

I'm not sure what you mean by "display as a trigonometric function". Would writing e^x as cos(x)+ i sin(x) help?

 

[Orodruin]

I'm not sure what you mean by "display as a trigonometric function". Would writing e^x as cos(x)+ i sin(x) help?

Regardless of what it means, writing it this way would not help since it is not true. It is $e^x$, not $e^{ix}$ ...

 

[mfb]

Well, you can replace x by ix.

 

[Svein]

Turn it upside down: e^{x}-1=\frac{x}{1!}+\frac{x^{2}}{2!}+..., so \frac{e^{x}-1}{x}=\frac{1}{1!}+\frac{x}{2!}+\frac{x^{2}}{3!}.... Don't know if that helps.

 

[WWGD]

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.

 

[eaglechief]

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 !

 

[Svein]

Second, i wonder why they "appear" only with 2n index (except B#1).

This may possibly help: https://en.wikipedia.org/wiki/Riemann_zeta_function.