I am stucked at this problem:

If $f\in C^1[a,b]$ and $x_0,...,x_n$ are $n+1$ distinct points in $[a,b]$, Then there exist unique polynomial $H_{2n+1}$ of degree at most $2n+1$ that satisfies the conditions $H_{2n+1}(x_k)=f(x_k)$ and $H_{2n+1}'(x_k)=f'(x_k)$ for each $k\in\{0,...,n\}$ (This is the Hermite interpolation polynomial for $f$).

We can prove it as follows:

Suppose that there is another polynomial with $P(x_k)=f(x_k)$ and $P'(x_k)=f'(x_k)$ for $k = 0,...,n$ and that the degree of $P$ is at most $2n+1$.

Let $D(x)=H_{2n+1}(x)-P(x)$

Then $D$ is a polynomial of degree at most $2n+1$ with $D(x_k)=0$ and $D'(x_k)=0$ for each $k=0,1,...,n$. Thus $D$ has zeros of multiplicity $2$ at each $x_k$, so

$D(x)=(x-x_0)^2(x-x_1)^2...(x-x_n)^2 Q(x)$

Either, $D(x)$ is of degree $2n$ or more which should be a contradiction, or $Q(x)=0$, which implies that $D(x)=0$, This implies that $P(x)$ is $H_{2n+1}(x)$, so this polynomial is unique. Q.E.D.

Now suppose that instead of requiring $f\in C^1[a,b]$ we require that $f\in C^2[a,b]$ and instead of requiring $H_{2n+1}'(x_k)=f'(x_k)$ we require that $H_{2n+1}''(x_k)= f''(x_k)$ for each $k=0,1,...,n$. Explain why the uniqueness proof fails in this case. (Do not give counter-examples, just explain why proving the uniqueness fails)

I've tried to explain it using systems of linear equations but I am not sure how good it was. (We got $n$ linear equations in $n$ variables and additional $n$ linear equations in $n-2$ variables of the $n$ variables from the first system, I am not sure how to continue)

Thanks for any hint/help.

  • 2
    $\begingroup$ It's not just the uniqueness. Even the existence is not guaranteed (without changing the restrictions on the degree of the polynomial, at least.$$ $$ For instance, if $n=0$, then if the polynomial is of degree $2*0+1$, then the second derivative is $0$ and we can't satisfy the new requirement. $\endgroup$ – Kitegi Aug 7 '15 at 8:32

From the given data you can approximate f'(x_k). Now you make a new paired data for f'(X). We know the pairs (f'(x_k),f"(x_k))=(g(x_k),g'(x_k)). Using Hermite again we can find g.


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