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$P(x)$ is a monic polynominal ( the highest coefficient is 1 ). $deg P(x) = n$. Prove that exist a constant $c$ that $P(x)+c.T_n(x)$ has all its roots real.

- $T_{n-1}(x)$ is the $n$th Chebyshev polynomial

Please help me :((

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Let $Q$ be *any* polynomial of degree $n$ with simple, real zeros. Let $x_0, x_1, \ldots, x_n$ be any set of points which are separated by the zeros of $Q$.

The polynomial $Q$ changes sign at each of its zeros because they are all simple. Thus the signs of the values $Q(x_0), Q(x_1), \ldots, Q(x_n)$ alternate. That is,

$$

Q(x_{k-1})Q(x_k) < 0, \quad k=1,2,\ldots,n.

\tag{1}

$$

Let $P$ be any polynomial of degree $n$ and define

$$

F_c(x) = P(x) + cQ(x).

$$

We will show that, for all $c$ large enough,

$$

F_c(x_{k-1})F_c(x_k) < 0, \quad k=1,2,\ldots,n.

\tag{2}

$$

This will mean that the function $F_c$ changes sign at least $n$ times, and since it is a polynomial of degree $n$ this will imply that all zeros of $F_c$ are simple and real.

For each $k \in \{1,2,\ldots,n\}$ the expression

$$

\begin{align*}

G_k(c) &:= F_c(x_{k-1})F_c(x_k) \\

&\,= Q(x_{k-1})Q(x_k)c^2 + [P(x_{k-1})Q(x_k) + P(x_k)Q(x_{k-1})]c + P(x_{k-1})P(x_k)

\end{align*}

$$

is a quadratic polynomial in the variable $c$. By equation $(1)$ the coefficient of $c^2$ in $G_k(c)$ is negative, and it follows that there is a $c_k$ such that $G_k(c) < 0$ for all $c > c_k$.

Taking $c > \max\{c_1,c_2,\ldots,c_n\}$ yields $(2)$, as desired.

Finally we note that the Chebyshev polynomial $T_n$ is a polynomial of degree $n$ with simple, real zeros, so the statement is indeed true in the special case $Q = T_n$.

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