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=$0,1,\ldots,n$ ?
Asked
Active
Viewed 3,319 times
1 Answers
5
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.
joriki
- 238,052