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I'm working on the following question from a past Masters' exam in Analysis:

Prove the following:
(a) If $f:\mathbb R\to\mathbb R$ is continuous and differentiable everywhere and $f'$ has at most $n$ distinct zeros, then $f$ has at most $n+1$ distinct zeros.
(b) If $f:\mathbb R\to\mathbb R$ is defined by $f(x)=a_0+\sum_{i=1}^n a_i e^{b_i x}$ (where $a_i$ and $b_i$ are real numbers), then either $f$ is identically zero or it has at most $n$ distinct zeros.

Part (a) has already been answered here: Proving that if $f'$ has at most $n-1$ zeros, then $f$ has at most $n$ zeros. I'm trying to use induction to prove part (b) using part (a). Here's what I tried so far:

In the base case $n=1$ we have that $f(x)=a_0+a_1e^{b_1x}$ for $a_0,a_1,b_1\in\mathbb R$. Suppose $f$ is not identically zero. The only case where $f$ is nonconstant is when $a_1\neq 0$ and $b_1\neq 0$, and in this case it's easy to show $f$ is injective so that it has at most one zero.

So suppose the statement holds for $k$. I want to show it holds for $k+1$, so I let $f(x)=a_0+\sum_{i=1}^{k+1}a_ie^{b_ix}$. Then we have that $f'(x)=\sum_{i=1}^{k+1}a_ib_ie^{b_ix}$. I get stuck trying to show that $f'$ has at most $k$ zeros. I know by our assumption that $\sum_{i=1}^k a_ib_ie^{b_ix}$ has at most $k$ zeros, but I'm not sure how that helps. Any hints here?

blakedylanmusic
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    Hint: if you divide your formula for $f'(x)$ through by the nowhere zero function $e^{b_1x}$, you get an instance of the inductive hypothesis (which you should take to say that what you are trying to prove is true for any $a_i$ and $b_i$). – Rob Arthan Sep 13 '23 at 20:08
  • @RobArthan OMG I cannot believe I didn't think of that. Thank you!!!! – blakedylanmusic Sep 13 '23 at 21:03
  • @RobArthan So dividing through by say, $e^{b_{k+1}x}$ gives us $a_{k+1}b_{k+1}+\sum_{i=1}^k a_ib_ie^{(b_i-b_{k+1})x}$, which has at most $k$ zeros by hypothesis, so then $f'$ must also have at most $k$ zeros because $e^{b_{k+1}x}$ is nonzero. – blakedylanmusic Sep 13 '23 at 21:05
  • It's not very easy to read the superscripts in the comments on my laptop. You should divide through by $\exp(b_1 \cdot x)$ (for example: any $b_i$ would do). – Rob Arthan Sep 13 '23 at 22:29
  • @RobArthan yes, I divided by $\exp(b_{k+1}x)$ – blakedylanmusic Sep 13 '23 at 23:37
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    What you have done is fine: you can divide by $exp(b_{i} \cdot x)$ for any $i$. I chose $i = 1$ because I think it makes the algebra and notation easier, but $i = k +1$ is fine too. – Rob Arthan Sep 14 '23 at 21:33

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