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For each $n$, $a=(a_1,...,a_n)\in \mathbb{R}^n$, and $b=(b_1,...,b_n)\in \mathbb{R}^n$, set $$f_{a,b}(t)=\sum_{k=1}^{n}b_ke^{a_kt}, t\in \mathbb{R}.$$ $f_{a,b}$ is said to have a repeated zero at $t_0$ if $f_{a,b}(t_0)=f'_{a,b}(t_0)=0$. Is it true that for any natural $n$ and pairwise distinct $(a_1,...,a_n)\in \mathbb{R}^n$, the set {$b\in \mathbb{R}^n:f_{a,b}$ has a repeated zero} has Lebesgue measure zero in $\mathbb{R}^n$?

P.S. If $a_k$ are integers (or rationals), $f_{a,b}$ is, up to a change of variable, a polynomial, for which the answer is affirmative: Conditions on coefficients of polynomial to guarantee simple roots ; Do almost all polynomials of degree $n$ in $\mathbb R[x]$ have exactly $n$ distinct roots in $\mathbb C$?

Thanks.

Mikhail
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1 Answers1

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Yes.

Claim: there are smooth functions $e_3, \ldots, e_n: \mathbb{R} \rightarrow \mathbb{R}^n$ such that for any $t \in \mathbb{R}$, $(e_3(t),\ldots,e_n(t))$ is a basis of the space of vectors $b$ such that $\sum_{k=1}^n{b_ke^{a_kt}}=\sum_{k=1}^n{b_ka_ke^{a_kt}}=0$.

Suppose the claim is true.

Let $F:\mathbb{R}^{n-1} \rightarrow \mathbb{R}^n$ be given by $F(t,u_3,\ldots,u_n)=\sum_{i=3}^n{u_ie_i(t)}$. Then, the set of vectors $b\in \mathbb{R}^n$ for which $f_{a,b}$ has a repeated root is exactly the image of $F$.

Since $F$ is smooth from $\mathbb{R}^{n-1}$ to $\mathbb{R}^n$, its image has measure zero by Sard’s theorem.

Proof of the claim: fix some $t \in \mathbb{R}$ and let $f_i(t) \in \mathbb{R}^n$ be the vector $(a_k^ie^{a_kt})_{1 \leq k \leq n}$ for every $1 \leq i \leq n$.

A direct computation with the Vandermonde determinant shows that for any $1 \leq t \leq n$, $(f_i(t))_{1 \leq i \leq n}$ is a basis of $\mathbb{R}^n$.

Let $(e_1(t),\ldots,e_n(t))$ be the Gram-Schmidt orthonormalisation of the basis $(f_1(t),\ldots,f_n(t))$. Then it is a basis of $\mathbb{R}^n$, and the vector subspace spanned by $e_3(t),\ldots,e_n(t)$ is exactly the orthogonal of the vector subspace spanned by $e_1(t)$ and $e_2(t)$.

Finally, there are explicit formulae for the $e_i(t)$ as a function of the $e_j(t)$ for $j<i$ and $f_i(t)$, so, since every $f_i$ is smooth, so is every $e_i$, so we are done.

Aphelli
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