Uniqueness of function approximation over three points?

Question

Given a function $f(x)$, we want to approximate $f$ using $P(x)$, such that: $P(x_0) = f(x_0)$, $P(x_2) = f(x_2)$, $P'(x_1) = f'(x_1)$. Prove that such a $P$ is unique $\iff$ $x_1 \neq \frac{1}{2}(x_0+x_2)$. I have no idea where to start with this, any help would be appreciated!

Answer 1

That statement is wrong without any further restrictions, because if $P$ satisfies $$ P(x_0) = f(x_0) \, , \quad P(x_2) = f(x_2) \,, \quad P'(x_1) = f'(x_1) \tag 1 $$ then for any differentiable function $g$ $$ \tilde P(x) = P(x) + g(x)(x-x_0)(x-x_1)^2(x-x_2) $$ satisfies $(1)$ as well.

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What you are probably looking for are quadratic polynomials $$ P(x) = a x^2 + b x + c $$ satisfying the interpolation conditions $(1)$. Then $$ \begin{aligned} a x_0^2 + b x_0 + c &= f(x_0) \\ a x_2^2 + b x_2 + c &= f(x_2) \\ 2 a x_1 + b &= f'(x_1) \end{aligned} $$ must hold. This is a linear system of 3 equations for the 3 variables $a, b, c$. It is well-known that the system has a unique solution if and only if the determinant of the coefficients $$ \begin{vmatrix} x_0^2 & x_0 & 1 \\ x_2^2 & x_2 & 1 \\ 2x_1 & 1 & 0 \end{vmatrix} $$ is not zero (and no or many solutions otherwise, depending on the values on the right-hand side of the equations).

This should lead you to the intended condition $x_1 \neq \frac{1}{2}(x_0+x_2)$ .

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A different way to look at it is the following: If both $P_1$ and $P_2$ are quadratic polynomials satisfying $(1)$, then $Q := P_1 - P_2$ is a polynomial of degree at most 2 and satisfies

$$ Q(x_0) = Q(x_2) = Q'(x_1) = 0 \, . $$ On the other hand, the vertex (or turning point) of a quadratic function is in the middle between the zeros, so that $$ Q'(\frac {x_0 + x_2}{2}) = 0\, . $$ It follows that $Q \equiv 0$ or $x_1 = \frac{1}{2}(x_0+x_2)$.