Explicit Form for Coefficients of Extended Hyperbolic Secant Function

Question

Consider the function: $$\frac{3}{e^x+e^{{w_3}x}+e^{{w_3^2}x}}=\sum_{n=0}^\infty{E_{3,n}\frac{x^n}{n!}}$$ Note here that $w_3=e^{\frac{2i\pi}{3}}$ I am trying to get an explicit formula for the $E_{3,i}$. So I try and rewrite the LHS: $$\frac{3}{e^x+e^{{w_3}x}+e^{{w_3^2}x}}=\frac{3e^{-x}}{1-\left[-e^{(w_3-1)x}-e^{(w_3^2-1)x}\right]}$$ $$=3e^{-x}\sum_{n=0}^\infty(-1)^n\left[e^{(w_3-1)x}+e^{(w_3^2-1)x}\right]^n$$ $$=3e^{-x}\sum_{n=0}^\infty(-1)^n\left\{e^{(w_3-1)x}\left[1+e^{(w_3^2-w_3)x}\right]\right\}^n$$ $$=3e^{-x}\sum_{n=0}^\infty(-1)^ne^{(w_3-1)nx}\sum_{k=0}^n\binom{n}{k}e^{(w_3^2-w_3)kx}$$ $$=3\sum_{n=0}^\infty\sum_{k=0}^\infty\binom{n}{k}(-1)^ne^{[(w_3-1)n+(w_3^2-w_3)k-1]x}$$ $$=3\sum_{n=0}^\infty\sum_{k=0}^\infty\sum_{j=0}^\infty\binom{n}{k}(-1)^n[(w_3-1)n+(w_3^2-w_3)k-1]^j\frac{x^j}{j!}$$

It is here I get confused how to finish. How do i get it simpler and finalize so that I can extract the coefficient and get my formula?

EDIT:

Is better approach this? Consider Coefficient extraction:

$$E_{3,n}=n![x^n]\frac{3e^{-x}}{1-\left[-e^{(w_3-1)x}-e^{(w_3^2-1)x}\right]}$$ $$=n![x^n]3e^{-x}\sum_{r=0}^n(-1)^r\left[e^{(w_3-1)x}+e^{(w_3^2-1)x}\right]^r$$ $$...$$ $$=n![x^n]3\sum_{r=0}^n\sum_{k=0}^r\binom{r}{k}(-1)^re^{[(w_3-1)r+(w_3^2-w_3)k-1]x}$$ $$=n![x^n]3\sum_{r=0}^n\sum_{k=0}^r\sum_{j=0}^\infty\binom{r}{k}(-1)^r[(w_3-1)r+(w_3^2-w_3)k-1]^j\frac{x^j}{j!}$$ And since I"m considering the $n$-th coefficient, then i need the case when $j=n$? $$3\sum_{r=0}^n\sum_{k=0}^r\binom{r}{k}(-1)^r[(w_3-1)r+(w_3^2-w_3)k-1]^n$$ Then I can simplify at least one of the expressions in the trinomial above, since $w_3^2-w_3=-i\sqrt{3}$. So $$E_{3,n}=\sum_{r=0}^n\sum_{k=0}^r\binom{r}{k}(-3)^r[(w_3-1)r-i\sqrt{3}k-1]^n$$

I know this is not the right answer because I've put the formula into Mathematica and am not getting the correct coefficients. The coefficients are: $$1,0,0,-1,0,0,19,...$$

Answer 1

Here are some hints which could be helpful.

Composition of formal power series

Let's consider the following part of OPs calculation

\begin{align*} \frac{1}{1-\left[-e^{(w_3-1)x}-e^{(w_3^2-1)x}\right]}=\sum_{n=0}^{\infty}(-1)^n \left[e^{(w_3-1)x}+e^{(w_3^2-1)x}\right]^n\tag{1} \end{align*}

This representation is not valid when considering the ring of formal power series. The expression above is in fact the composition of the power series

\begin{align*} G(x)=\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n \end{align*} with \begin{align*} F(x)=-\left[e^{(w_3-1)x}+e^{(w_3^2-1)x}\right] \end{align*}

Note, that a composition

\begin{align*} G(F(x))=\sum_{n=0}^{\infty}\left(F(x)\right)^n \end{align*} is only valid, iff $F(0)=0$. If $F$ has no constant term it is assured, that a finite number of summands is used as contribution for the coefficients of the powers of $x$. (You might also look at this section of the referred Wiki page.)

Since $F(x)=-e^{(w_3-1)x}-e^{(w_3^2-1)x}=-2+\cdots$ has a constant term $-2\neq 0$, it does not fulfill this requirement and the representation (1) is not valid.

Observe, that on the other hand the correct representation provided in the answer of @amcalde

\begin{align*} \frac{1}{1+\sum_{n=1}^{\infty}\frac{x^{3n}}{(3n)!}} =\sum_{t=0}^{\infty}\left(-\sum_{n=1}^{\infty}\frac{x^{3n}}{(3n)!}\right)^t\tag{2} \end{align*}

is a valid composition of formal power series $G(x)=\frac{1}{1+x}$ and $H(x)=-\sum_{n=1}^{\infty}\frac{x^{3n}}{(3n)!}$ with $H(0)=0$.

Calculation of reciprocals of power series

The paper Composita and its properties by V.V. Kruchinin and D.V. Kruchinin presents in section 5 techniques to obtain the coefficients of reciprocals of formal power series. In order to obtain a valid composition of power series the authors consider also $xF(x)$ instead of $F(x)$.

But note, that it is not easy to find simple representations. Before trying to obtain the coefficients of the reciprocal of the rather complex expression \begin{align*} e^x+e^{{w_3}x}+e^{{w_3^2}x} \end{align*} you could try to start with the easier expression \begin{align*} e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}, \end{align*} Here we know the solution $e^{-x}=\sum_{n=0}^{\infty}(-1)^n\frac{x^n}{n!}$ and we can better see, how Kruchinins approach could be applied.

OEIS: A002115

Using the representation (2) to obtain the coefficients of small powers of $x$ we get \begin{align*} \frac{1}{1+\sum_{n=1}^{\infty}\frac{x^{3n}}{(3n)!}}=1-\frac{x^3}{3!}+\frac{19x^6}{6!}-\frac{1513x^9}{9!}+\cdots \end{align*} We can find this sequence in OEIS as A002115.

Since there is no simple representation of the coefficients of the series stated, this could indicate that no one is available.

Answer 2

Take a look at this sum: $e^x+e^{{w_3}x}+e^{{w_3^2}x}$. By the property of $w_3$ you can easily compute the power series. So you have: $$\frac{3}{e^x+e^{{w_3}x}+e^{{w_3^2}x}} = \frac{1}{\sum_{n=0}^\infty \frac{x^{3n}}{(3n)!}} = \frac{1}{1 + \sum_{n=1}^\infty \frac{x^{3n}}{(3n)!}}$$ We can compute this using the geometric series formula $$\frac{1}{1 + \sum_{n=1}^\infty \frac{x^{3n}}{(3n)!}} = 1 + (-\sum_{n=1}^\infty \frac{x^{3n}}{(3n)!}) + (-\sum_{n=1}^\infty \frac{x^{3n}}{(3n)!})^2 + (-\sum_{n=1}^\infty \frac{x^{3n}}{(3n)!})^3 + \cdots $$ It is clear that the only nonzero $E_{3,n}$ must have $n$ a multiple of three. Moreover: $$\frac{3}{e^x+e^{{w_3}x}+e^{{w_3^2}x}} = 1 - \frac{x^3}{3!} - \frac{x^6}{6!} + \frac{x^6}{3! 3!} + \cdots = 1 - \frac{x^3}{3!} - \frac{19 x^6}{6!} + \cdots $$ Which is what you got with Mathematica.

I hope this helps, or at least shows you how to repeat your Mathematica result.