$\begingroup$

I feel like an idiot having to ask but I am honestly stuck and can't find the solution in my old text books.

We know that the Taylor series expansion of the exponential function is given by $ e^z = \sum_{k=0}^{\infty}\frac{z^k}{k!}$. Similarly it follows that $ ze^z = \sum_{k=0}^{inf}k\frac{z^k}{k!}$.

I know that $\sum_{k=0}^{\infty}e^k\frac{z^k}{k!}=e^{ez}$ and $\sum_{k=0}^{\infty}e^{kr}\frac{z^k}{k!}=e^{e^rz}$, but honestly don't know how to get there. I am assuming that you have to take the log of the exponential of the function, but can't seem to get it right.

Any help would be appreciated!

$\endgroup$

1 Answer

$\begingroup$

If $c$ is any complex number, then replacing $z$ with $cz$ yields $$ e^{cz}=\sum_{k=0}^{\infty}c^k\frac{z^k}{k!}$$ Taking $c=e$ or $c=e^r$ leads to the formulas in your question.

$\endgroup$ 1

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy