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Let $n \in \mathbb N$, let $p:\mathbb R \to \mathbb R$ be a real polynomial, and let $\lambda_1< \lambda_2 <\dots < \lambda_n$. Now let $$ f \in \mathrm{span} \left \{ p(x - \lambda_k)e^{\lambda_k x} : k=1,\dots, n \right \}, \quad f \neq 0. $$ It is well-known that if $p =1$ then $f$ has at most $n-1$ distinct (real) zeros in $[0,1]$. What about the case where $p \neq 0$? Can I find assumptions on the $\lambda_j$'s so that $f$ has again at most $n-1$ distinct zeros in $[0,1]$? I'm in particular interested in the case where the $\lambda_k$'s are of the form $\lambda_k = C + \alpha k$ for some fixed $\alpha,C>0$.

Thank you for any help!

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  • $\begingroup$ You certainly need some restrictions on $p$. If you want it to be true for an arbitrary polynomial $p$ with some restrictions on $\lambda$'s alone (which seems to me to be the formal meaning of what is written now), then you are doomed (just recall that $p$ can approximate absolutely any continuous function on any compact set with any precision). $\endgroup$
    – fedja
    Commented Jan 28, 2023 at 21:12

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To estimate the number of real zeros from above, you need some assumption on the degree of $p$ (otherwise no such estimate is possible, even for $n=1$, see @fedja's comment). Then there is a general theorem: If $P_j$ are real polynomials of degrees $d_j$ then $$P_1(x)e^{\lambda_1 x}+\ldots +P_n(x)e^{\lambda_nx}$$ has at most $d_1+\ldots+d_n+n-1$ real zeros.

See, for example, Polya Szego, Part V, Chap I, Sect. 5, probl. 75.

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  • $\begingroup$ Thank you very much for the answer! Do you think one can gain anything in the setting where $P_j=P(\cdot - \lambda_j)$? Or does this seem unrealistic ... $\endgroup$
    – r_l
    Commented Jan 29, 2023 at 9:06

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