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I was doing a question on polynomials , where it was found that $P(x)=P(-x)$ in the interval $[-\sqrt2,\sqrt 2 ]$ It was then concluded that $P(x)=P(-x)$ holds for all values of $x$ ,"since it is a polynomial". Can someone help me understand why it could be generalized ?

Edit - $P(x)$ is a polynomial with real coefficients .

Tanya
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    Your title is misleading: it holds for all $x$ because of two things as the answers make clear: 1) $P(x)$ is a polynomial and 2) it is equal to some other polynomial at an infinite number of points - in that case, the two polynomials have to be equal at all points. – NickD May 31 '20 at 13:20
  • Thanks ,I will change it – Tanya Jun 01 '20 at 16:18

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Let $Q(x)=P(x)-P(-x)$, then $Q(x)$ is a real polynomial since $P(x)$ is (make sure you can show this!). By the assumption, $Q$ has infinitely many roots. But the only real polynomial with infinitely many roots is the zero polynomial. Hence $Q(x)=0$ for all real $x$, so $P(x)=P(-x)$ for all real $x$.

Minus One-Twelfth
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If polynomial $f(x)$ has degree less than $n$, then $n+1$ number of points on the polynomial determines the polynomial.

Since $P(x) - P(-x)$ is a polynomial, and there are infinite number of $t \in [-\sqrt 2, \sqrt2]$ satisfying $P(t) - P(-t) = 0$, we have $P(x) - P(-x) \equiv 0 $ as function.

dust05
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  • "If polynomial f(x) has degree less than n, then n+1 number of points on the polynomial determines the polynomial."-Why is this true? – Tanya May 31 '20 at 07:56
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    See https://math.stackexchange.com/questions/837902/prove-there-exists-a-unique-n-th-degree-polynomial-that-passes-through-n1-p – Minus One-Twelfth May 31 '20 at 07:57
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    @Grace500 Let $f(x) = a_0 + a_1 x + \cdots + a_n x^n$ and $(x_0, y_0), \cdots, (x_n, y_n)$ be the points of the polynomial, i.e. $f(x_i) = y_i$ for all $i \in {0, \cdots, n}$. Then we have the linear system of $a_0, \cdots, a_n$. $n+1$ number of variables, $n+1$ number of equations. And the system is linearly independent. https://en.wikipedia.org/wiki/Vandermonde_matrix – dust05 May 31 '20 at 07:59