Is there a way to simplify $\sum_{n=0}^L \binom{L}{n} e^{p^n q^{L-n}}$?

Question

I have a nasty sum I've been working with that can be reduced to the following form $$ \sum_{n=0}^L \binom{L}{n} e^{-c p^n q^{L-n}}$$ where $c > 0$ is constant. However, I do not know any way to simplify/reduce this. Is there any nice way to do it, or do I need to start approximating the sum? Doing a bit of research suggests no, but I still have hope for it.

EDIT: Forgot to mention $p+q =1, 0<p<1.$

Answer 1

It's not much, but we can write

\begin{align} \sum_{n=0}^L \binom{L}{n} \exp\Big(-c \left(p^n q^{L-n}\right)\Big) &= \sum_{n=0}^L\, \binom{L}{n}\, \sum_{k\geq 0}\frac{{\left(-c \left(p^n q^{L-n}\right)\right)}^k}{k!} \\&= \sum_{k\geq 0}\,(-1)^k\,\frac{c^k}{k!} \, \left(\sum_{n=0}^L\, \binom{L}n\,{\left(p^k\right)}^n{\left(q^k\right)}^{L-n}\right) \\&= \sum_{k\geq 0}(-1)^k\,\frac{c^k\,{\left(p^k+q^k\right)}^L}{k!}. \end{align}

This looks somewhat easier to make estimates on, or else might point someone in a better direction.

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We can also attempt some calculus.

\begin{align} f(p, L, c) &= \sum_{k\geq 0}\,{\left(p^k+(1-p)^k\right)}^L\,\frac{(-c)^k}{k!} \\&= 2^L-c+ \sum_{k\geq 2}\,{\left(p^k+(1-p)^k\right)}^L\,\frac{(-c)^k}{k!} \end{align}

Then $\frac{\partial f}{\partial p}(p,L,c)$ equals

\begin{align} &\hphantom{=} \sum_{k\geq 2}\,L\,{\left(p^k+(1-p)^k\right)}^{L-1}\left(kp^{k-1}-k(1-p)^{k-1}\right)\,\frac{(-c)^k}{k!} \\&= \sum_{k\geq 1}\,L\,{\left(p^k+(1-p)^k\right)}^{L-1}\left(kp^{k-1}-k(1-p)^{k-1}\right)\,\frac{(-c)^k}{k!} \\&= -Lc\left[\left(\sum_{k\geq 1}\,{\left(p^k+(1-p)^k\right)}^{L-1}\,\frac{(-pc)^{k-1}}{(k-1)!}\right) - \left(\sum_{k\geq 1}\,{\left(p^k+(1-p)^k\right)}^{L-1}\,\frac{((1-p)c)^{k-1}}{(k-1)!}\right)\right] \end{align}

On the other hand,

\begin{align} \frac{\partial f}{\partial c}(p,L, c) &= -1-\sum_{k\geq 2}\,{\left(p^k+(1-p)^k\right)}^L\,\frac{(-c)^{k-1}}{(k-1)!} \\&= -\sum_{k\geq 1}\,{\left(p^k+(1-p)^k\right)}^L\,\frac{(-c)^{k-1}}{(k-1)!} \end{align}

Hence, for $L>1$ we have

\begin{align} \frac{\partial f}{\partial p}(p,L,c) &= -Lc\left[ \left(-\frac{\partial f}{\partial c}(p,L-1,pc)\right) - \left(-\frac{\partial f}{\partial c}(p,L-1,(1-p)c)\right) \right] \\&= Lc\left[ \frac{\partial f}{\partial c}(p,L-1,pc) - \frac{\partial f}{\partial c}(p,L-1,(1-p)c)\right] \end{align}

While this does not look like a simple DE to solve, it at least shows that $p=\frac12$ is a critical point for fixed $L>1$, $c>0$.