0
$\begingroup$

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.

$\endgroup$
  • 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.