I'm looking at functions in the form $$ f(x) = \sum_{i=1}^n a_i e^{b_i x} $$ with $a_i, b_i \in \mathbb{R}$ and $b_i < 0$.
I'm interested in the maximum value of this function at certain intervals and I tried several values for $a_i$ and $b_i$. I did an observation and wonder if this is generally true. And, if true, how to prove this.
Does $f(x)$ always have at most one local maximum?
Some properties: $$ \begin{aligned} \lim_{x \to -\infty} f(x) &= \pm \infty \\ \lim_{x \to \infty} f(x) &= 0 \\ f^{(k)}(x) &= \sum_{i=1}^n a_i {b_i}^k e^{b_i x} \end{aligned} $$
We can split these functions into some cases by looking at the behaviour of the function at $-\infty$ and $\infty$: $$ l(f) = \lim_{x \to -\infty} \operatorname{sgn}(f(x)) = \operatorname{sgn}(a_{\operatorname{arg min}_i(b_i)}) = \pm 1 \\ r(f) = \lim_{x \to \infty} \operatorname{sgn}(f(x)) = \operatorname{sgn}(a_{\operatorname{arg max}_i(b_i)}) = \pm 1 $$
Some examples:
$$
\begin{aligned}
f_1(x) &= e^{-x} & l(f_1) = 1 \quad r(f_1) = 1 \\
f_2(x) &= e^{-x} - e^{-2x} & l(f_2) = -1 \quad r(f_2) = 1 \\
f_3(x) &= e^{-x} - 2e^{-2x} - 3e^{-3x} + 4 e^{-4x} & l(f_3) = 1 \quad r(f_3) = 1
\end{aligned}
$$
So we have:
- $f_1$ - No local minimum/maximum
- $f_2$ - One local maximum
- $f_3$ - First a local minimum, then a local maximum
- $-f_1$ - No local minimum/maximum
- $-f_2$ - One local minimum
- $-f_3$ - First a local maximum, then a local minimum
Are there any other cases?
No. Let $f(x)=P(\exp(-x))$ where $P$ is a polynomial. Then $f$ is of the desired form, and has as many local maxima in $(0,\infty)$ as $P$ does in $[0,1]$. Which can easily exceed $1$.