The uniqueness of Hermite's interpolating polynomial in the case of $n=2$ nodes

Question

The problem is : I have tried to prove the problem using proof by contradiction. However, I got stuck. I am not sure which role the derivatives of the function $p$ and $f$ should play, or how we should use the points $x_0$ and $x_1$.

Answer 1

I thought it would interesting to give a proof of uniqueness by contradiction.

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Suppose there are two polynomials $p$ and $q$ of degree at most $3$ such that $$ f(x_i) = p(x_i) = q(x_i), \quad i = 0, 1,$$ and $$ f'(x_i) = p'(x_i) = q'(x_i), \quad i = 0, 1,$$ I claim that $r = p - q$ is the zero polynomial. It is clear that $r$ has degree at most $3$, i.e., $$r (x) = a x^3 + bx^2 + cx + d$$ for suitable $a$, $b$, $c$ and $d$. Moreover, we have $$ r(x_i) = p(x_i) - q(x_i) = 0, \quad i = 0, 1,$$ and $$ r'(x_i) = p'(x_i) - q'(x_i), \quad i = 0, 1.$$ We will now show that $a=0$. This will be done by repeated application of Rolle's theorem. It can be helpful to draw a line and mark and name the points as they are discovered. By Rolle's theorem, $r'$ has at least one zero $\xi$ between $x_0$ and $x_1$. Moreover, since $r'(x_0) = r'(x_1) =0$, Rolles's theorem show that $r''$ has at least one zero between $x_0$ and $\xi$ and at least one zero between $\xi$ and $x_1$. In particular, $r''$ has at least two zeros between $x_0$ and $x_1$. By another application of Rolle's theorem, we conclude that $r'''$ has at least one zero. However, $$r'''(x) = 6a,$$ so $a=0$ as claimed. We now know that $$r(x) = bx^2 + cx + d.$$ This implies $$r'(x) = 2bx + c.$$ Since $r'$ has two distinct zeros, we have immediately $b=c=0$. This reduces $r$ to the tantalizing form $$r(x) = d.$$ Since $r$ has at least one zero, we conclude that $r=0$. It follows that $p=q$.

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In reality, I have merely adapted the proof of the error formula, i.e., $$\forall x \: \exists \xi \: : \: f(x) - p(x) = \frac{f^{(4)}(\xi)}{4!}(x-x_0)^2(x-x_1)^2,$$ where $p$ is the polynomial given in the text, i.e., Hermite's interpolation polynomial of degree at most $3$. It is also possible to give a formula for $p$ directly in terms of the given values of $f$ and $f'$, but this is a subject for a separate investigation.

Answer 2

Well, it seems like the polynomial has to be cubic. So, it would be given by: $p(x)=a_0+a_1x+a_2x^2+a_3x^3$. So, if we know what the values of $a_0$ to $a_3$ are, we've fully specified the polynomial. The conditions given involve four conditions that translate to four equations. These equations are:

\begin{align} a_0 +a_1 x_0 +a_2 x_0 ^2 + a_3 x_0^3 = f(x_0)\\ a_0 +a_1 x_1 +a_2 x_1 ^2 + a_3 x_1^3 = f(x_1)\\ a_1+ a_2 2x_0+a_3 3x_0^2=f'(x_0)\\ a_1+ a_2 2x_1+a_3 3x_1^2=f'(x_1) \end{align}

Four equations in four unknowns means there is a unique solution, unless the system is under-specified.

The system becoming under-specified can only happen if one of the equations can be constructed from a linear combination of the other three. For a general system, $Au=b$, this happens if the determinant, $|A|=0$.

For your particular system, the matrix $A$ of the LHS becomes -

$$ A= \begin{bmatrix} 1 & -h/2 & h^2/4 & -h^3/8 \\ 1 & h/2 & h^2/4 & h^3/8 \\ 0 & 1 & -h & 3h^2/4\\ 0 & 1 & h & 3h^2/4 \end{bmatrix} $$

The Determinant of this matrix turns out to be $|A| = \frac{h^4}{4}$.

This determinant can only be $0$ if $h=0$. Otherwise, the system will have one unique solution.

You might also find the Taylor series expansion of a smooth function instructive.