Chapter 3: Hermite Interpolation

Theorem (Hermite Interpolation Theorem)

Let \(n\in \mathbb{N}_0 \)and let \(a \), \(b\in \mathbb{T} \), \(a<b \), and \(x_j\in \mathbb{T} \), \(j\in \{0, 1, \ldots, n\} \), be \(\sigma \)-distinct and \(x_j \neq \sigma(x_{k}) \)for all \(j,k\in \{0,1, \ldots, n\} \).
Let also, \(y_j, z_j \in \mathbb{R} \), \(j\in \{0, 1, \ldots, n\} \). Then there exists a unique polynomial \(p_{2n+1}\in \mathcal{P}_{2n+1} \)such that
\begin{equation}
\label{1.13} p_{2n+1}(x_j)=y_j,\quad p_{2n+1}^{\Delta}(x_j)=z_j,\quad j\in \{0, 1, \ldots, n\}.
\end{equation}

Proof

For \(x\in [a, b] \), define the polynomial
\[
M_k(x)= \prod_{j=0, j\ne k}^n \frac{x-\sigma(x_j)}{x_k-\sigma(x_j)},\quad k\in \{0, 1, \ldots, n\}.
\notag\]
We have
\[
M_k(\sigma(x_j))=0,\quad M_k(x_k)=1,\quad j, k\in \{0, 1, \ldots, n\},\quad j\ne k.
\notag\]
We will search a polynomial \(p_{2n+1}\in \mathcal{P}_{2n+1} \)in the following form
\[
p_{2n+1}(x)= \sum_{j=0}^n \left(y_j+(x-x_j)(\alpha_j y_j+\beta_j z_j)\right)M_j(x) L_j(x),\quad x\in [a, b],
\notag\]
where \(\alpha_j \), \(\beta_j\in \mathbb{R} \), \(j\in \{0, 1, \ldots, n\} \), will be determined by the conditions \eqref{1.13}. We have
\[\begin{aligned}
p_{2n+1}(x_k)&=& \sum_{j=0}^n \left(y_j+(x_k-x_j)(\alpha_j y_j+\beta_j z_j)\right)M_j(x_k) L_j(x_k)\\ \\
&=& y_k,\\ \\
p_{2n+1}^{\Delta}(x)&=& \sum_{j=0}^n (\alpha_j y_j+\beta_j z_j)M_j(\sigma(x))L_j(\sigma(x))\\ \\
&&+ \sum_{j=0}^n \left(y_j+(x-x_j)(\alpha_j y_j+\beta_j z_j)\right)\left(M_j^{\Delta}(x)L_j(x)+ M_j(\sigma(x))L_j^{\Delta}(x)\right),\\ \\
p_{2n+1}^{\Delta}(x_k)&=& \sum_{j=0}^n (\alpha_j y_j+\beta_j z_j)M_j(\sigma(x_k))L_j(\sigma(x_k))\\ \\
&&+\sum_{j=0}^n \left(y_j+(x_k-x_j)(\alpha_j y_j +\beta_j z_j)\right)\left(M_j^{\Delta}(x_k)L_j(x_k) +M_j(\sigma(x_k))L_j^{\Delta}(x_k)\right)\\ \\
&=& (\alpha_k y_k+\beta_k z_k)M_k(\sigma(x_k))L_k(\sigma(x_k))\\ \\
&&+ y_k\left(M_k^{\Delta}(x_k)+M_k(\sigma(x_k))L_k^{\Delta}(x_k)\right)\\ \\
&=& z_k
\end{aligned}\notag\]
or
\[
\left(\alpha_k y_k+\beta_k z_k\right)M_k(\sigma(x_k))L_k(\sigma(x_k))=z_k -y_k\left(M_k^{\Delta}(x_k)+M_k(\sigma(x_k))L_k^{\Delta}(x_k)\right),
\notag\]
or
\[
\alpha_k y_k+\beta_k z_k= \frac{z_k -y_k\left(M_k^{\Delta}(x_k)+M_k(\sigma(x_k))L_k^{\Delta}(x_k)\right)}{M_k(\sigma(x_k))L_k(\sigma(x_k))},
\notag\]
and
\[\begin{aligned}
p_{2n+1}(x)&=& \sum_{j=0}^n \left(y_j+\frac{z_j-y_j\left(M_j^{\Delta}(x_j)+M_j(\sigma(x_j))L_j^{\Delta}(x_j)\right)}{M_j(\sigma(x_j))L_j(\sigma(x_j))}(x-x_j)\right)M_j(x)L_j(x)\\ \\
&=& \sum_{j=0}^n \bigg( \left(1-\frac{M_j^{\Delta}(x_j)+M_j(\sigma(x_j))L_j^{\Delta}(x_j)}{M_j(\sigma(x_j))L_j(\sigma(x_j))}(x-x_j)\right)y_j\\ \\
&&+ \frac{z_j}{M_j(\sigma(x_j))L_j(\sigma(x_j))}(x-x_j)\bigg) M_j(x) L_j(x),
\end{aligned}\notag\]
\(x\in [a, b] \). Now, suppose that there are two polynomials such that \(p_{2n+1}, q_{2n+1}\in \mathcal{P}_{2n+1} \)and
\[
p_{2n+1}(x_k)=q_{2n+1}(x_k)=y_k,\quad p_{2n+1}^{\Delta}(x_k)=q_{2n+1}^{\Delta}(x_k)=z_k,\quad k\in \{0, 1, \ldots, n\}.
\notag\]
Let \(h_{2n+1}=p_{2n+1}-q_{2n+1} \). Then \(h_{2n+1}\in \mathcal{P}_{2n+1} \)and it has at least \(2n+2 \)GZs. Thus,
\[
h_{2n+1}\equiv 0\quad \text{or}\quad p_{2n+1}\equiv q_{2n+1}\quad \text{on}\quad [a, b].
\notag\]
This completes the proof.

Theorem

Suppose that \(n\in \mathbb{N}_0 \), \(a, b\in \mathbb{T} \), \(a<b \), \(x_j\in [a, b] \), \(j\in \{0, 1, \ldots, n\} \), are \(\sigma \)-distinct and \(f: [a, b]\to \mathbb{R} \), \(f^{\Delta^k}(x) \)exist for any \(x\in [a, b] \)and for any \(k\in \{1, \ldots, 2n+2\} \). Then for any \(x\in [a, b] \)there exists \(\xi=\xi(x)\in (a, b) \)such that
\[
f(x)-p_{2n+1}(x)= \frac{f^{\Delta^{2n+2}}(\xi)}{\pi_{n+1}(x)\zeta_{n+1}(x)}\left(\pi_{n+1}\zeta_{n+1}\right)^{\Delta^{2n+2}}(\xi), \quad x\in [a, b]
\notag\]
or
\[
G_{min, 2n+2}(\xi)\leq \frac{f(x)-p_{2n+1}(x)}{\pi_{n+1}(x)\zeta_{n+1}(x)}\leq G_{max, 2n+2}(\xi), \quad x\in [a, b]
\notag\]
where
\[\begin{aligned}
G_{max, 2n+2}(\xi)&=& \max\left\{ \frac{f^{\Delta^{2n+2}}(\xi)}{\left(\pi_{n+1}\zeta_{n+1}\right)^{\Delta^{2n+2}}(\xi)}, \frac{f^{\Delta^{2n+2}}(\rho(\xi))}{\left(\pi_{n+1}\zeta_{n+1}\right)^{\Delta^{2n+2}}(\rho(\xi))}\right\},\\ \\
G_{min, 2n+2}(\xi)&=& \min\left\{ \frac{f^{\Delta^{2n+2}}(\xi)}{\left(\pi_{n+1}\zeta_{n+1}\right)^{\Delta^{2n+2}}(\xi)}, \frac{f^{\Delta^{2n+2}}(\rho(\xi))}{\left(\pi_{n+1}\zeta_{n+1}\right)^{\Delta^{2n+2}}(\rho(\xi))}\right\}.
\end{aligned}\notag\]

Proof

Let \(p_{2n+1} \)be the Hermite interpolation polynomial for the function \(f \)with interpolation points \(x_j \), \(j\in \{0, 1, \ldots, n\} \). Define the function
\[
\psi(t)= f(t)-p_{2n+1}(t)-\frac{f(x)-p_{2n+1}(x)}{\pi_{n+1}(x)\zeta_{n+1}(x)}\pi_{n+1}(t)\zeta_{n+1}(t),\quad t\in [a, b].
\notag\]
Then
\[\begin{aligned}
\psi(x_j)&=& f(x_j)-p_{2n+1}(x_j)-\frac{f(x)-p_{2n+1}(x)}{\pi_{n+1}(x)\zeta_{n+1}(x)}\pi_{n+1}(x_j)\zeta_{n+1}(x_j)\\ \\
&=& f(x_j)-f(x_j)\\ \\
&=& 0, \quad x\in [a, b], \quad j\in \{0, 1, \ldots, n\},
\end{aligned}\notag\]
and \(\psi(x)=0, \quad x\in [a, b] \). Thus, \(\psi: [a, b]\to \mathbb{R} \)has at least \(n+2 \)GZs. Hence and the Rolle theorem (see the Appendix of this book), it follows that \(\psi^{\Delta} \)has at least \(n+1 \)GZs on \((a, b) \)that do not coincide with \(x_j \), \(j\in \{0, 1, \ldots, n\} \). Next,
\[\begin{aligned}
\psi^{\Delta}(t)&=& f^{\Delta}(t)- p_{2n+1}^{\Delta}(t)-\frac{f(x)-p_{2n+1}(x)}{\pi_{n+1}(x)\zeta_{n+1}(x)}(\pi_{n+1}\zeta_{n+1})^{\Delta}(t)\\ \\
&=& f^{\Delta}(t)-p_{2n+1}^{\Delta}(t)\\ \\
&&- \frac{f(x)-p_{2n+1}(x)}{\pi_{n+1}(x)\zeta_{n+1}(x)}\left(\pi_{n+1}^{\Delta}(t)\zeta_{n+1}(\sigma(t))+\pi_{n+1}(t)\zeta_{n+1}^{\Delta}(t)\right),
\end{aligned}\notag\]
\(t\in [a, b] \). Hence,
\[\begin{aligned}
\psi^{\Delta}(x_j)&=& f^{\Delta}(x_j)-p_{2n+1}^{\Delta}(x_j)\\ \\
&&-\frac{f(x)-p_{2n+1}(x)}{\pi_{n+1}(x)\zeta_{n+1}(x)}\left(\pi_{n+1}^{\Delta}(x_j) \zeta_{n+1}(\sigma(x_j))+\pi_{n+1}(x_j)\zeta_{n+1}^{\Delta}(x_j)\right)\\ \\
&=& 0,\quad j\in \{0, 1, \ldots, n\},
\end{aligned}\notag\]
i.e., \(\psi^{\Delta} \)has at least \(n+1 \)GZs at \(x_j \), \(j\in \{0, 1, \ldots, n\} \). Therefore \(\psi^{\Delta} \)has at least \(2n+2 \)GZs in \([a, b] \).
Then \(\psi^{\Delta^{2n+2}} \)has at least one GZ in \((a, b) \)and there exists a \(\xi=\xi(x)\in (a, b) \)such that
\[
\psi^{\Delta^{2n+2}}(\xi)=0\quad \text{or}\quad \psi^{\Delta^{2n+2}}(\rho(\xi))\psi^{\Delta^{2n+2}}(\xi)<0.
\notag\]
Observe that
\[
\psi^{\Delta^{2n+2}}(t)= f^{\Delta^{2n+2}}(t)-\frac{f(x)-p_{2n+1}(x)}{\pi_{n+1}(x)\zeta_{n+1}(x)}\left(\pi_{n+1}\zeta_{n+1}\right)^{\Delta^{2n+2}}(t),\quad t\in [a, b].
\notag\]

1. Let \(\psi^{\Delta^{2n+2}}(\xi)=0 \). Then
\[
f^{\Delta^{2n+2}}(\xi)= \frac{f(x)-p_{2n+1}(x)}{\pi_{n+1}(x)\zeta_{n+1}(x)}\left(\pi_{n+1}\zeta_{n+1}\right)^{\Delta^{2n+2}}(\xi), \quad x\in [a, b],
\notag\]
or
\[\begin{aligned}
f(x)-p_{2n+1}(x)&=& \frac{f^{\Delta^{2n+2}}(\xi)}{\left(\pi_{n+1}\zeta_{n+1}\right)^{\Delta^{2n+2}}(\xi)}\pi_{n+1}(x)\zeta_{n+1}(x), \quad x\in [a, b].
\end{aligned}\notag\]

2. Let
\[
\psi^{\Delta^{2n+2}}(\rho(\xi))\psi^{\Delta^{2n+2}}(\xi)<0.
\notag\]
Then
\[\begin{aligned}
\psi^{\Delta^{2n+2}}(\rho(\xi))&=& f^{\Delta^{2n+2}}(\rho(\xi))-\frac{f(x)-p_{2n+1}(x)}{\pi_{n+1}(x)\zeta_{n+1}(x)}\left(\pi_{n+1}\zeta_{n+1}\right)^{\Delta^{2n+2}}(\rho(\xi)), \quad x\in [a, b],
\end{aligned}\notag\]
and
\[
\psi^{\Delta^{2n+2}}(\xi)= f^{\Delta^{2n+2}}(\xi)-\frac{f(x)-p_{2n+1}(x)}{\pi_{n+1}(x)\zeta_{n+1}(x)}\left(\pi_{n+1}\zeta_{n+1}\right)^{\Delta^{2n+2}}(\xi,
\quad x\in [a, b].
\notag\]
Then we have,
\[\begin{aligned}
0&>& \psi^{\Delta^{2n+2}}(\rho(\xi))\psi^{\Delta^{2n+2}}(\xi)\\ \\
&=& \left(f^{\Delta^{2n+2}}(\rho(\xi))-\frac{f(x)-p_{2n+1}(x)}{\pi_{n+1}(x)\zeta_{n+1}(x)}\left(\pi_{n+1}\zeta_{n+1}\right)^{\Delta^{2n+2}}(\rho(\xi))\right)\\ \\
&&\times \left(f^{\Delta^{2n+2}}(\xi)-\frac{f(x)-p_{2n+1}(x)}{\pi_{n+1}(x)\zeta_{n+1}(x)}\left(\pi_{n+1}\zeta_{n+1}\right)^{\Delta^{2n+2}}(\xi)\right)\\ \\
&=& \left(\frac{f(x)-p_{2n+1}(x)}{\pi_{n+1}(x)\zeta_{n+1}(x)}\right)^2 \left(\pi_{n+1}\zeta_{n+1}\right)^{\Delta^{2n+2}}(\xi)\left(\pi_{n+1}\zeta_{n+1}\right)^{\Delta^{2n+2}}(\rho(\xi))\\ \\
&&-\frac{f(x)-p_{2n+1}(x)}{\pi_{n+1}(x)\zeta_{n+1}(x)}\left(\left(\pi_{n+1}\zeta_{n+1}\right)^{\Delta^{2n+2}}(\rho(\xi))
f^{\Delta^{2n+2}}(\xi)+\left(\pi_{n+1}\zeta_{n+1}\right)^{\Delta^{2n+2}}(\xi) f^{\Delta^{2n+2}}(\rho(\xi))\right)\\ \\
&&+f^{\Delta^{2n+2}}(\rho(\xi))f^{\Delta^{2n+2}}(\xi), \quad x\in [a, b].
\end{aligned}\notag\]
Hence,
\[
G_{min, 2n+2}(\xi)\leq \frac{f(x)-p_{2n+1}(x)}{\pi_{n+1}(x)\zeta_{n+1}(x)}\leq G_{max, 2n+2}(\xi), \quad x\in [a, b].
\notag\]
This completes the proof.