Chapter 5: Delta Differentiation

Theorem

Suppose that \(n\geq 0 \), \(a, b\in \mathbb{T} \), \(a<b \), \(x_j\in [a, b] \), \(j\in \{0, 1, \ldots, n\} \), are distinct and \(f: [a, b]\to \mathbb{R} \), \(p_n: [a, b]\to \mathbb{R} \)is a polynomial of degree \(n \)interpolating \(f \)at the points \(x_j\in [a, b] \), \(j\in \{0, 1, \ldots, n\} \)and \(f^{\Delta^k}(x) \)exist for any \(x\in [a, b] \)and for any \(k\in \{1, \ldots, n+1\} \). Then for any \(x\in [a, b] \)there exists \(\xi=\xi(x)\in (a, b) \)and distinct points \(\eta_j \), \(j\in \{1, \ldots, n\} \), in \((a, b) \), such that
\[
f^{\Delta}(x)-p_n^{\Delta}(x)= \frac{f^{\Delta^{n+1}}(\xi)}{\pi_{n}^{*\Delta^{n+1}}(\xi)}\pi_{n}^*(x), \quad x\in [a, b],
\notag\]
or
\[
H_{min, n+1}(\xi)\leq \frac{f^{\Delta}(x)-p_n^{\Delta}(x)}{\pi_{n}^*(x)}\leq H_{max, n+1}(\xi), \quad x\in [a, b],
\notag\]
where
\[\begin{aligned}
H_{max, n+1}(\xi)&=& \max\left\{ \frac{f^{\Delta^{n+1}}(\xi)}{\pi_{n}^{*\Delta^{n+1}}(\xi)}, \frac{f^{\Delta^{n+1}}(\rho(\xi))}{\pi_{n}^{*\Delta^{n+1}}(\rho(\xi))}\right\},\\ \\
H_{min, n+1}(\xi)&=& \min\left\{ \frac{f^{\Delta^{n+1}}(\xi)}{\pi_{n}^{*\Delta^{n+1}}(\xi)}, \frac{f^{\Delta^{n+1}}(\rho(\xi))}{\pi_{n}^{*\Delta^{n+1}}(\rho(\xi))}\right\},\\ \\
\pi_n^*(x)&=& (x-\eta_1)\ldots (x-\eta_n), \quad x\in [a, b],.
\end{aligned}\notag\]

Proof

Let \(p_n \)be the Lagrange interpolation polynomial for the function \(f \)with interpolation points \(x_j \), \(j\in \{0, 1, \ldots, n\} \). Then the function \(f-p_n \)has at least \(n+1 \)GZs in \([a, b] \). Hence and the Rolle theorem, it follows that there exist \(\eta_j \), \(j\in \{1, \ldots, n\} \), in \((a, b) \)which are GZs of the function \(f^{\Delta}-p_n^{\Delta} \). Define the function
\[
\chi(t)= f^{\Delta}(t)-p_n^{\Delta}(t)-\frac{f^{\Delta}(x)-p_n^{\Delta}(x)}{\pi_{n}^*(x)}\pi_{n}^*(t),\quad t\in [a, b].
\notag\]
Then
\[\begin{aligned}
\chi(\eta_j)&=& f^{\Delta}(\eta_j)-p_n^{\Delta}(\eta_j)-\frac{f^{\Delta}(x)-p_n^{\Delta}(x)}{\pi_{n}^*(x)}\pi_{n}^*(\eta_j)\\ \\
&=& f^{\Delta}(\eta_j)-p_n^{\Delta}(\eta_j)\\ \\
&=& 0, \quad x\in [a, b], \quad j\in \{0, 1, \ldots, n\},
\end{aligned}\notag\]
and \(\chi(x)=0, \quad x\in [a, b] \).
Thus, \(\chi: [a, b]\to \mathbb{R} \)has at least \(n+1 \)GZs. Hence and the Rolle theorem, it follows that \(\chi^{\Delta^{n}} \)has at least one GZ on \((a, b) \). Therefore there exists an \(\xi=\xi(x)\in (a, b) \)such that
\[
\chi^{\Delta^{n}}(\xi)=0\quad \text{or}\quad \chi^{\Delta^{n}}(\rho(\xi))\chi^{\Delta^{n}}(\xi)<0.
\notag\]
Note that
\[
\chi^{\Delta^{n}}(t)= f^{\Delta^{n+1}}(t)-\frac{f^{\Delta}(x)-p_n^{\Delta}(x)}{\pi_{n}^*(x)}\pi_{n}^{*\Delta^{n}}(t),\quad t\in [a, b].
\notag\]

1. Let \(\chi^{\Delta^{n}}(\xi)=0 \). Then
\[
f^{\Delta^{n+1}}(\xi)= \frac{f^{\Delta}(x)-p_n^{\Delta}(x)}{\pi_{n}^*(x)}\pi_{n}^{*\Delta^{n}}(\xi), \quad x\in [a, b],
\notag\]
or
\[\begin{aligned}
f^{\Delta}(x)-p_n^{\Delta}(x)&=& \frac{f^{\Delta^{n+1}}(\xi)}{\pi_{n}^{*\Delta^{n}}(\xi)}\pi_{n}^*(x)\\ \\
&=& \frac{f^{\Delta^{n+1}}(\xi)}{\pi_{n}^{*\Delta^{n}}(\xi)}\pi_{n}^*(x), \quad x\in [a, b].
\end{aligned}\notag\]

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