Can we write $e^{-\alpha x}$ as $\sum_{n=0}^\infty c_n\left(\alpha\right)\gamma\left(x\right)^n$ such that $\lim_{x\to\infty}\gamma\left(x\right)=0$

Do there exist continuous functions $$c_n\colon\mathbb R^+\to\mathbb R$$ and $$\gamma\colon\mathbb R^+\to\mathbb R$$ such that $$\lim_{x\to\infty}\gamma\left(x\right)=0$$ and the following equation is true for all $$\alpha,x>0$$ $$e^{-\alpha x}=\sum_{n=1}^{\infty}c_n\left(\alpha\right)\gamma\left(x\right)^n$$ If yes, how regular can we make $$\gamma$$, can it be smooth? Analytic? What about the $$c_n$$?

Clarifying edit: all the $$x$$-dependence is in $$\gamma$$ and all the $$\alpha$$ dependence is in the $$c_n$$

Suppose that $$c_n(\alpha)$$ and $$\gamma(x)$$ with the desired properties exist. Necessarily $$\gamma$$ is one-to-one, and hence $$e^{-\alpha \gamma^{-1}(x)} = \sum_{n = 0}^\infty c_n(\alpha) x^n.$$ Since the right-hand side converges for some $$x > 0$$, it defines a function $$f_\alpha(x) = \sum_{n = 0}^\infty c_n(\alpha) x^n$$ which is real-analytic near $$0$$ and not identically equal to zero. In particular, $$f_\alpha(x) \sim c_{k(\alpha)}(\alpha) x^{k(\alpha)}$$ as $$x \to 0$$, where $$k(\alpha)$$ is the index of the first non-zero coefficient $$c_n(\alpha)$$. This means that $$e^{-\gamma^{-1}(x)} \sim c_{k(\alpha)}(\alpha) x^{k(\alpha)/\alpha},$$ and therefore $$k(\alpha) / \alpha$$ is constant. Since $$k(\alpha)$$ is integer-valued, this is not possible, unless $$k(\alpha) = 0$$. But this means that $$e^{-\gamma^{-1}(x)}$$ has a positive right limit at $$0$$, contrary to the assumption that $$\gamma^{-1}(x) \to \infty$$ as $$x \to 0^+$$.
• If $\gamma(x) = \gamma(y)$, then $e^{-\alpha x} = e^{-\alpha y}$, and hence $x = y$. Jan 31, 2022 at 12:56