# Function with Euler's constant

Hello all,

is there an alternative way of displaying

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

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

HallsofIvy
Homework Helper
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
Staff Emeritus
Homework Helper
Gold Member
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
Mentor
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.

aikismos
WWGD
Gold Member
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.

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