Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What are the proofs of existence and uniqueness of Hermite interpolation polynomial? suppose $x_{0},...,x_{n}$ are distinct nodes and $i=1 , ... ,n$ and $m_{i}$ are in Natural numbers. prove exist uniqueness polynomial $H_{N}$ with degree N=$m_{1}+...+m_{n}$-1 satisfying $H_{N}^{(k)}$=$y_{i}^{(k)}$ k=0,1,...,$m_{i}$ & i=$0,1,\ldots,n$ ?

share|improve this question
add comment

1 Answer

I think you've got your indices mixed up a bit; they're sometimes starting at $0$ and sometimes at $1$. I'll assume that the nodes are labeled from $1$ to $n$ and the first $m_i$ derivatives at $x_i$ are determined, that is, the derivatives from $0$ to $m_i-1$.

A straightforward proof consists in showing how to construct a basis of polynomials $P_{ik}$ that have non-zero $k$-th derivative at $x_i$ and zero for all other derivatives and nodes. For given $i$, start with $k=m_i-1$ and set

$$P_{i,m_i-1}=(x-x_i)^{m_i-1}\prod_{j\ne i}(x-x_j)^{m_j}\;.$$

Then decrement $k$ in each step. Start with

$$Q_{i,k}=(x-x_i)^k\prod_{j\ne i}(x-x_j)^{m_j}\;,$$

which has zero derivatives up to $k$ at $x_i$, and subtract out multiples of the $P_{i,k'}$ with $k'\gt k$, which have already been constructed, to make the $k'$-th derivatives at $x_i$ with $k'\gt k$ zero. Doing this for all $i$ yields a basis whose linear combinations can have any given values for the derivatives.

Uniqueness follows from the fact that the number of these polynomials is equal to the dimension $d=\sum_i m_i$ of the vector space of polynomials of degree up to $d-1$. Since the $P_{ik}$ are linearly independent, there's no more room for one more that also satisfies one of the conditions, since it would have to be linearly independent of all the others.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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