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Let $λ_1,\ldots,λ_m$ real numbers pairwise distinct and $μ_1,\ldots,μ_m$ real numbers all nonzero. We know from polynomial interpolation that for a given $r$ such that $1\leq r\leq m$, there exists an unique polynomial $R$ of degree less than $(m+r+1)$ such that $R(λ_i)=μ_i,1≤i≤m$ and $R'(λ_i)=0,1≤i≤r$ .

I want to prove the existence of a real polynomial $S$ such that the polynomial $$P(x)=R(x)+S(x)\displaystyle\prod_{i=1}^{r}(x-\lambda_i)^2\displaystyle\prod_{i=r+1}^{m}(x-\lambda_i)$$ admits exactly $(m+r+\mathrm{deg}(S))$ real roots pairwise distinct.

In other words I am looking for a real polynomial $P$ which satisfies $$P(λ_i)=μ_i,1≤i≤m \mbox{ and } P'(λ_i)=0,1≤i≤r,$$ and whose roots are all real and simple.

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1. As I already told you in my comment to your previous problem, mathoverflow.net/questions/111492/… you misstate the interpolation theorem: to interpolate at m points you need a polynomial of degree m-1. Otherwise your polynomial is not unique. 2. If you do not restrict the degree of S, the answer is probably "yes", but what is the purpose of such result?? –  Alexandre Eremenko Nov 5 '12 at 21:14
    
First, I am sorry, I mean deg(R)<m+r. Secondly, this result will very usefull to prove a result concerning the controllability of discrete-time bilinear system. –  driss-alamilouati Nov 5 '12 at 21:28
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1 Answer

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If there are no restrictions on $S$, one can apply the following construction. Let $$Q(x)=S(x)\prod_{j=1}^r(x-\lambda_j)^2\prod_{j=r+1}^m(x-\lambda_j).$$ We want to construct $S$. Consider first the case when $R$ does not change sign on the real line. Suppose it is positive. Then choose $S$ so that

a) All double zeros of $Q$ are maxima. This is achieved by inserting zeros of $S$ at properly chosen places.

b) That the minima of $Q$ on each interval between its zeros are less (more negative) than $-\max R$ (the maximum of $R$ is taken over some big interval that contains all zeros of $Q$). This can be achieved just by multiplying $S$ on a sufficiently large constant.

Now when you graph $-R$ and $Q$ on the same picture, you see that the graphs intersect as many times as the degree of $Q$ is. So $R+Q$ has all zeros real. You also make them simple, by another multiplication of $Q$ on a constant.

The general case is similar. Consider the real zeros of $R$, say $x_1,...,x_k$. On the intervals $(x_k,x_{k+1})$ where $R(x)>0$, you want all double zeros of $Q$ to be maxima. On the intervals $(x_k,x_{k+1})$ where $R(x)<0$, you want all double zeros of $Q$ to be minima. This is achieved by placing the zeros of $S$ at appropriate places. Each new zero of $S$ switches the sign of $Q$ at this place.

Now after you achieved this (that all double roots of $Q$ are maxima where $R$ is positive, and minima where it is negative), it only remains to multiply $Q$ by a very large positive constant, to make sure that the graphs of $-R$ and $Q$ intersect as many times as needed. Finally, by slightly perturbing the constant multiple in $S$, you achieve that all these real roots are simple.

But I don't now how useful this polynomial can be, as its degree can be larger than the number of interpolation nodes. You can try to minimize the degree of $S$, the answer will depend on the relative position of the double roots $\lambda_j$ among all $\lambda_j$. But roughly speaking the degree of $Q$ constructed with this method, can be about twice as large as that of $R$.

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