# Function with Euler's constant

1. Sep 25, 2015

### eaglechief

Hello all,

is there an alternative way of displaying

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

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

2. Sep 26, 2015

### HallsofIvy

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

3. Sep 26, 2015

### Orodruin

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

4. Sep 26, 2015

### Staff: Mentor

Well, you can replace x by ix.

5. Sep 26, 2015

### 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.

6. Sep 26, 2015

### 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.

7. Sep 28, 2015

### 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 !

8. Sep 28, 2015