I am looking for results on the probabilities of common roots of $f(x)$ and $f'(x)$ for $f(x) = \sum\_{s \in S\_f} u\_s x^s$ and $u\_s$ i.i.d $\mathcal{N}(0,1)$-distributed, for $x \in [0,1]$. Or, put differently, how likely is it for a root of $f(x)$ to have multiplicity larger than 1?


Thank you for any ideas or reading suggestions.



Multiple roots of Random Polynomials
You did not let us know what $S\_f$ is. Since the title of your post is "Multiple roots of Random Polynomials", it seems reasonable to assume that $S\_f$ is the set of integers $n\_0,\dots,n\_k$ such that $0\le n\_0<\dots<n\_k$.


One of the following cases must occur:


*Case 1: $n\_0\ge2$.* Then $0$ is a multiple root of $f$ with probability $1$.


*Case 2: $n\_0=1$.* Then, for $g(x):=f(x)/x$, the multiple roots of $f$ are the same as those of $g$ (except for the zero-probability event $\{u\_1=0\}$) -- which reduces the consideration to


*Case 3: $n\_0=0$.* Then $f(x)=v\_0+h\_v(x)$, where $h\_v(x):=v\_1 x^{n\_1}+\dots+v\_k x^{n\_k}$, $v:=(v\_1,\dots,v\_k)$, and $v\_j:=u\_{n\_j}$ for $j=0,\dots,k$. The random finite set  

$$X\_v:=\{x\in\mathbb R\colon f'(x)=0\}=\{x\in\mathbb R\colon h'\_v(x)=0\}$$
depends only on $v$, but not on $v\_0$.


Next, $f$ has multiple roots iff $-v\_0$ is in the random finite set $h\_v(X\_v)=\{h\_v(x)\colon x\in X\_v\}$. Therefore and because the distribution of $v\_0$ is continuous and because $v\_0$ is independent of $v$, we conclude that the probability that $f$ has multiple roots is
$$P(-v\_0\in h\_v(X\_v))=EP(-v\_0\in h\_v(X\_v)|v)=E0=0.$$


