Chapter 4: Sigma-Hermite Interpolation
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In this section, we will construct the \(\sigma \)-Hermite interpolation polynomials. We will demonstrate the difference between the Hermite interpolation polynomial and \(\sigma \)-Hermite interpolation polynomial.
As was mentioned in the previous sections, the \(\sigma \)-interpolation polynomials provide an alternative way to interpolate a given set of data.
They may coincide with the interpolation polynomial in certain cases and differ in others. The numerical examples presented in this section demonstrate these situations.
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, the forward jump operator \(\sigma \)be \(\Delta \)-differentiable on \([a, b] \)and \(\sigma^{\Delta}(x_j)\ne 0 \), \(j\in\{0, 1, \ldots, n\} \).
Let also, \(y_j, z_j \in \mathbb{R} \), \(j\in \{0, 1, \ldots, n\} \). Then there exists a unique \(\sigma \)-polynomial \(p_{\sigma 2n+1}\in \mathcal{P}_{2n+1}^{\sigma} \)such that
\[
p_{\sigma 2n+1}(x_j)=y_j,\quad p_{\sigma 2n+1}^{\Delta}(x_j)=z_j,\quad j\in \{0, 1, \ldots, n\}.
\notag\]
Proof
Let \(M_k \), \(k\in \{0, 1, \ldots, n\} \), be the polynomials as in the proof of Theorem \ref{theorem1.12}.
We will find a polynomial \(p_{\sigma 2n+1}\in \mathcal{P}_{2n+1}^{\sigma} \)in the following manner.
\[
p_{\sigma 2n+1}(x)= \sum_{j=0}^n \left(y_j+(\sigma(x)-\sigma(x_j))(\alpha_j y_j+\beta_j z_j)\right)M_j(x) L_j(x),\quad x\in [a, b],
\notag\]
where \(\alpha_j \), \(\beta_j\in \mathbb{R} \), \(j\in \{0, 1, \ldots, n\} \), will be determined below. We have
\[\begin{aligned}
p_{\sigma 2n+1}(x_k)&=& \sum_{j=0}^n \left(y_j+(\sigma(x_k)-\sigma(x_j))(\alpha_j y_j+\beta_j z_j)\right)M_j(x_k) L_j(x_k)\\ \\
&=& y_k,\\ \\
p_{\sigma 2n+1}^{\Delta}(x)&=& \sum_{j=0}^n \sigma^{\Delta}(x)(\alpha_j y_j+\beta_j z_j)M_j(\sigma(x))L_j(\sigma(x))\\ \\
&&+ \sum_{j=0}^n \left(y_j+(\sigma(x)-\sigma(x_j))(\alpha_j y_j+\beta_j z_j)\right)\left(M_j^{\Delta}(x)L_j(x)+ M_j(\sigma(x))L_j^{\Delta}(x)\right),\\ \\
p_{\sigma 2n+1}^{\Delta}(x_k)&=& \sum_{j=0}^n \sigma^{\Delta}(x_k)(\alpha_j y_j+\beta_j z_j)M_j(\sigma(x_k))L_j(\sigma(x_k))\\ \\
&&+\sum_{j=0}^n \left(y_j+(\sigma(x_k)-\sigma(x_j))(\alpha_j y_j +\beta_j z_j)\right)\left(M_j^{\Delta}(x_k)L_j(x_k) +M_j(\sigma(x_k))L_j^{\Delta}(x_k)\right)\\ \\
&=& \sigma^{\Delta}(x_k)(\alpha_k y_k+\beta_k z_k)M_k(\sigma(x_k))L_k(\sigma(x_k))\\ \\
&&+ y_k\left(M_k^{\Delta}(x_k)+M_k(\sigma(x_k))L_k^{\Delta}(x_k)\right)\\ \\
&=& z_k,
\end{aligned}\notag\]
or
\[
\sigma^{\Delta}(x_k)\left(\alpha_k y_k+\beta_k z_k\right)M_k(\sigma(x_k))L_k(\sigma(x_k))=z_k -y_k\left(M_k^{\Delta}(x_k)+M_k(\sigma(x_k))L_k^{\Delta}(x_k)\right),
\notag\]
or
\[
\alpha_k y_k+\beta_k z_k= \frac{z_k -y_k\left(M_k^{\Delta}(x_k)+M_k(\sigma(x_k))L_k^{\Delta}(x_k)\right)}{\sigma^{\Delta}(x_k)M_k(\sigma(x_k))L_k(\sigma(x_k))},
\notag\]
and
\[\begin{aligned}
p_{\sigma 2n+1}(x)&=& \sum_{j=0}^n \left(y_j+\frac{z_j-y_j\left(M_j^{\Delta}(x_j)+M_j(\sigma(x_j))L_j^{\Delta}(x_j)\right)}{\sigma^{\Delta}(x_j)M_j(\sigma(x_j))L_j(\sigma(x_j))}(\sigma(x)-\sigma(x_j))\right)M_j(x)L_j(x)\\ \\
&=& \sum_{j=0}^n \bigg( \left(1-\frac{M_j^{\Delta}(x_j)+M_j(\sigma(x_j))L_j^{\Delta}(x_j)}{\sigma^{\Delta}(x_j)M_j(\sigma(x_j))L_j(\sigma(x_j))}(\sigma(x)-\sigma(x_j))\right)y_j\\ \\
&&+ \frac{z_j}{\sigma^{\Delta}(x_j)M_j(\sigma(x_j))L_j(\sigma(x_j))}(\sigma(x)-\sigma(x_j))\bigg) M_j(x) L_j(x),
\end{aligned}\notag\]
\(x\in [a, b] \). Now, suppose that there are two \(\sigma \)-polynomials such that \(p_{\sigma 2n+1}, q_{\sigma 2n+1}\in \mathcal{P}_{2n+1}^{\sigma} \)and
\[
p_{\sigma 2n+1}(x_k)=q_{\sigma 2n+1}(x_k)=y_k,\quad p_{\sigma 2n+1}^{\Delta}(x_k)=q_{\sigma 2n+1}^{\Delta}(x_k)=z_k,\quad k\in \{0, 1, \ldots, n\}.
\notag\]
Let \(h_{\sigma 2n+1}=p_{\sigma 2n+1}-q_{\sigma 2n+1} \). Then \(h_{\sigma 2n+1}\in \mathcal{P}_{2n+1}^{\sigma} \)and it has at least \(2n+2 \)GZs. Thus,
\[
h_{\sigma 2n+1}\equiv 0\quad \text{or}\quad p_{\sigma 2n+1}\equiv q_{\sigma 2n+1}\quad \text{on}\quad [a, b].
\notag\]
This completes the proof.
Taking into account the last theorem, the \(\sigma \)-Hermite interpolation polynomials for a given data set on a time scale and for a given function on an
arbitrary time scale are defined as follows.
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, the forward jump operator \(\sigma \)be \(\Delta \)-differentiable on \([a, b] \)and \(\sigma^{\Delta}(x_j)\ne 0 \), \(j\in\{0, 1, \ldots, n\} \).
Let also, \(y_j, z_j \in \mathbb{R} \), \(j\in \{0, 1, \ldots, n\} \). The polynomial
\[\begin{aligned}
p_{\sigma 2n+1}(x)&=&
\sum_{j=0}^n \bigg( \left(1-\frac{M_j^{\Delta}(x_j)+M_j(\sigma(x_j))L_j^{\Delta}(x_j)}{\sigma^{\Delta}(x_j)M_j(\sigma(x_j))L_j(\sigma(x_j))}(\sigma(x)-\sigma(x_j))\right)y_j\\ \\
&&+ \frac{z_j}{\sigma^{\Delta}(x_j)M_j(\sigma(x_j))L_j(\sigma(x_j))}(\sigma(x)-\sigma(x_j))\bigg) M_j(x) L_j(x),
\end{aligned}\notag\]
\(x\in [a, b] \), will be called \(\sigma \)-Hermite interpolation polynomial for the set \(\{(x_j, y_j, z_j): j\in \{0, 1, \ldots, n\}\} \).
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, the forward jump operator \(\sigma \)be \(\Delta \)-differentiable on \([a, b] \)and \(\sigma^{\Delta}(x_j)\ne 0 \), \(j\in\{0, 1, \ldots, n\} \).
Let also, \(f: [a, b]\to \mathbb{R} \)be \(\Delta \)-differentiable. The polynomial
\[\begin{aligned}
p_{\sigma 2n+1}(x)&=& \sum_{j=0}^n \bigg( \left(1-\frac{M_j^{\Delta}(x_j)+M_j(\sigma(x_j))L_j^{\Delta}(x_j)}{\sigma^{\Delta}(x_j)M_j(\sigma(x_j))L_j(\sigma(x_j))}(\sigma(x)-\sigma(x_j))\right)f(x_j)\\ \\
&&+ \frac{f^{\Delta}(x_j)}{\sigma^{\Delta}(x_j)M_j(\sigma(x_j))L_j(\sigma(x_j))}(\sigma(x)-\sigma(x_j))\bigg) M_j(x) L_j(x),
\end{aligned}\notag\]
\(x\in [a, b] \), will be called \(\sigma \)-Hermite interpolation polynomial for the function \(f \).
The next example demonstrates that on a general time scale the Hermite and \(\sigma \)-Hermite polynomials may be different.
Let \(\mathbb{T}=\{-1, 1, 2, 5, 9, 10\} \),
\[
a=-1,\quad b=9,\quad n=1,\quad x_0=-1,\quad x_1=5,\quad y_0=y_1=0,\quad z_0=z_1=1.
\notag\]
We have
\[\begin{aligned}
\sigma(x_0)&=& \sigma(-1)=1,\\ \\
\sigma(x_1)&=& \sigma(5)= 9,\\ \\
L_0(x)&=& \frac{x-x_1}{x_0-x_1}=-\frac{1}{6}(x-5),\\ \\
L_0(\sigma(x_0))&=& L_0(1)=\frac{2}{3},\\ \\
L_0^{\Delta}(x)&=& -\frac{1}{6},\\ \\
L_1(x)&=& \frac{x-x_0}{x_1-x_0}=\frac{1}{6}(x+1),\\ \\
L_1(\sigma(x_1))&=& L_1(9)= \frac{5}{3},\\ \\
L_1^{\Delta}(x)&=& \frac{1}{6},\\ \\
M_0(x)&=& \frac{x-\sigma(x_1)}{x_0-\sigma(x_1)}= -\frac{1}{10}(x-9),\\ \\
M_0(\sigma(x_0))&=& M_0(1)=\frac{4}{5},\\ \\
M_0^{\Delta}(x)&=& -\frac{1}{10},\\ \\
M_1(x)&=& \frac{x-\sigma(x_0)}{x_1-\sigma(x_0)}= \frac{1}{4}(x-1),\\ \\
M_1(\sigma(x_1))&=& M_1(9)= 2,\\ \\
M_1^{\Delta}(x)&=& \frac{1}{4},\\ \\
\sigma^{\Delta}(x_0)&=& \frac{\sigma(\sigma(x_0))-\sigma(x_0)}{\sigma(x_0)-x_0}\\ \\
&=& \frac{\sigma(1)-1}{1-(-1)}= \frac{2-1}{2}= \frac{1}{2},\\ \\
\sigma^{\Delta}(x_1)&=& \frac{\sigma(\sigma(x_1))-\sigma(x_1)}{\sigma(x_1)-x_1}\\ \\
&=& \frac{\sigma(9)-9}{9-5}= \frac{1}{4}.
\end{aligned}\notag\]
Hence, using that \(y_0=y_1=0 \), \(z_0=z_1=1 \), we get
\[\begin{aligned}
p_{\sigma 3}(x)&=& \frac{z_0}{\sigma^{\Delta}(x_0)M_0(\sigma(x_0))L_0(\sigma(x_0))}(\sigma(x)-\sigma(x_0))M_0(x)L_0(x)\\ \\
&&+\frac{z_1}{\sigma^{\Delta}(x_1)M_1(\sigma(x_1))L_1(\sigma(x_1))}(\sigma(x)-\sigma(x_1))M_1(x)L_1(x)\\ \\
&=& \frac{1}{\frac{1}{2}\cdot \frac{4}{5}\cdot \frac{2}{3}}(\sigma(x)-1)\left(-\frac{1}{10}(x-9)\right)\left(-\frac{1}{6}(x-5)\right)\\ \\
&&+ \frac{1}{\frac{1}{4} \cdot 2\cdot \frac{5}{3}}(\sigma(x)-9)\left(\frac{1}{6}(x+1)\right)\left(\frac{1}{4}(x-1)\right)\\ \\
&=& \frac{1}{16}(\sigma(x)-1)(x-9)(x-5)+\frac{1}{20}(\sigma(x)-9)(x+1)(x-1),\quad x\in \mathbb{T},\\ \\
p_{\sigma 3}(2)&=& \frac{1}{16}(\sigma(2)-1)(2-9)(2-5)+\frac{1}{20}(\sigma(2)-9)(2+1)(2-1)\\ \\
&=& \frac{1}{16}\cdot (5-1)\cdot (-7)\cdot (-3)+\frac{1}{20}\cdot(5-9)\cdot 3\\ \\
&=& \frac{21}{4}-\frac{3}{5}\\ \\
&=& \frac{105-12}{20}\\ \\
&=& \frac{93}{20},
\end{aligned}\notag\]
and
\[\begin{aligned}
p_3(x)&=& \frac{z_0}{M_0(\sigma(x_0))L_0(\sigma(x_0))}(x-x_0)M_0(x) L_0(x)\\ \\
&&+ \frac{z_1}{M_1(\sigma(x_1))L_1(\sigma(x_1))}(x-x_1)M_1(x) L_1(x)\\ \\
&=& \frac{1}{\frac{2}{3}\cdot \frac{4}{5}}(x+1)\left(-\frac{1}{10}(x-9)\right)\left(-\frac{1}{6}(x-5)\right)\\ \\
&&+\frac{1}{2\cdot \frac{5}{3}}(x-5)\left(\frac{1}{6}(x+1)\right)\left(\frac{1}{4}(x-1)\right)\\ \\
&=& \frac{1}{32}(x+1)(x-9)(x-5)+\frac{1}{80}(x-5)(x+1)(x-1), \quad x\in \mathbb{T}, \\ \\
p_3(2)&=& \frac{1}{32}(2+1)(2-9)(2-5)+\frac{1}{80}(2-5)(2+1)(2-1)\\ \\
&=& \frac{63}{32}-\frac{9}{80}\\ \\
&=& \frac{315-18}{160}\\ \\
&=& \frac{297}{160}.
\end{aligned}\notag\]
Consequently we have
\[
p_3(2)\ne p_{\sigma 3}(2).
\notag\]
In the above example we see that the Hermite and \(\sigma \)-Hermite interpolation polynomials may be different.
Let \(\mathbb{T}=\left\{-4, -1, 0, \frac{1}{2}, \frac{5}{6}, 1, \frac{4}{3}, 2, 7\right\} \),
\[
a=-4, \quad b-7, \quad x_0=-4,\quad x_1=0,\quad x_2=\frac{5}{6},
\notag\]
the function \(f: \mathbb{T}\to \mathbb{R} \)is defined by
\[
f(t)= \frac{t+1}{t^2+4t+7}+1,\quad t\in \mathbb{T}.
\notag\]
Find the \(\sigma \)-Hermite interpolation polynomial for the function \(f \).
In the next theorem we will give some criteria for coincidence of the Hermite and \(\sigma \)-Hermite interpolation polynomials.
Assume that \(\mathbb{T} \)is a time scale such that \(\sigma(t)=ct+d \)for any \(t\in \mathbb{T} \)and for some real constants \(c,d \).
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.
Let also, \(y_j, z_j \in \mathbb{R} \), \(j\in \{0, 1, \ldots, n\} \).
Then
\[
p_{\sigma 2n+1}\equiv p_{2n+1}.
\notag\]
Proof
Let \(\sigma(t)=ct+d \)for any \(t\in \mathbb{T} \)and for some real constants \(c \)and \(d \). Then \(\sigma^\Delta(t)=c, \quad t\in \mathbb{T} \)and
\[\begin{aligned}
p_{\sigma 2n+1}(x)&=&
\sum_{j=0}^n \bigg( \left(1-\frac{M_j^{\Delta}(x_j)+M_j(\sigma(x_j))L_j^{\Delta}(x_j)}{\sigma^{\Delta}(x_j)M_j(\sigma(x_j))L_j(\sigma(x_j))}(\sigma(x)-\sigma(x_j))\right)y_j\\
&&+ \frac{z_j}{\sigma^{\Delta}(x_j)M_j(\sigma(x_j))L_j(\sigma(x_j))}(\sigma(x)-\sigma(x_j))\bigg) M_j(x) L_j(x)\\
&=& \sum_{j=0}^n \bigg( \left(1-\frac{M_j^{\Delta}(x_j)+M_j(\sigma(x_j))L_j^{\Delta}(x_j)}{cM_j(\sigma(x_j))L_j(\sigma(x_j))}(cx+d-(cx_j+d))\right)y_j\\
&&+ \frac{z_j}{cM_j(\sigma(x_j))L_j(\sigma(x_j))}(cx+d-(cx_j+d))\bigg) M_j(x) L_j(x)\\
&=& \sum_{j=0}^n \bigg( \left(1-\frac{M_j^{\Delta}(x_j)+M_j(\sigma(x_j))L_j^{\Delta}(x_j)}{M_j(\sigma(x_j))L_j(\sigma(x_j))}(x-x_j)\right)y_j\\
&&+ \frac{z_j}{M_j(\sigma(x_j))L_j(\sigma(x_j))}(x-x_j)\bigg) M_j(x) L_j(x)\\
&=& p_{2n+1}(x),\quad x\in [a, b].
\end{aligned}\notag\]
This completes the proof.
As we have proved the above theorem, one can deduct the following result regarding the error in \(\sigma \)-Hermite interpolation.
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
\[
f(x)-p_{\sigma 2n+1}(x)= \frac{f^{\Delta^{2n+2}}(\xi)}{\pi_{n+1}(x)\zeta_{n+1}(x)}\left(\pi_{n+1}\zeta_{n+1}\right)^{\Delta^{2n+2}}(\xi), \quad x\in [a,b],
\notag\]
or
\[
G_{min, 2n+2}(\xi)\leq \frac{f(x)-p_{\sigma 2n+1}(x)}{\pi_{n+1}(x)\zeta_{n+1}(x)}\leq G_{max, 2n+2}(\xi), \quad x\in [a,b].
\notag\]
In the following example we consider a time scale with a nonlinear forward jump operator. We show the difference between Hermite and \(\sigma \)-Hermite polynomials graphically.
Let \(\mathbb{T}=\mathbb{N}_0^2=\{0,1,4,9,16,\ldots\} \)and \([a,b]=[1,49] \).
Let
\[x_0=1, x_1=9, x_2=25.\notag\]
We will find the Hermite interpolation polynomial \(p_5(x) \)and the \(\sigma\)-Hermite interpolation polynomial
\(p_{\sigma 5}(x) \)for a function \(f \)satisfying
\[f(1)=2,\quad f(9)=4, \quad f(25)=10,\notag\]
and
\[f^\Delta(1)=3, \quad f^\Delta(9)=6, \quad f^\Delta(25)=7.\notag\]
On this time scale we have
\[\begin{aligned}
\sigma(x)&=&(\sqrt{x}+1)^2, \\
\sigma^\Delta(x)&=& \frac{\sigma(\sigma(x))-\sigma(x)}{\sigma(x)-x}=\frac{2\sqrt{x}+3}{2\sqrt{x}+1},\quad x\in \mathbb{T}.
\end{aligned}\notag\]
We first compute the polynomials \(L_k \)and \(M_k \)for \(k=0,1,2 \).
\[\begin{aligned}
L_0(x)&=&\frac{(x-9)(x-25)}{(-8)(-24)}=\frac{x^2-34x+225}{192},\\
L_1(x)&=&\frac{(x-1)(x-25)}{(8)(-16)}=-\frac{x^2-26x+25}{128},\\
L_2(x)&=&\frac{(x-1)(x-9)}{(24)(16)}=\frac{x^2-10x+9}{384},\\
M_0(x)&=&\frac{(x-16)(x-36)}{(-15)(-35)}=\frac{x^2-52x+576}{525},\\
M_1(x)&=&\frac{(x-4)(x-36)}{(5)(-27)}=-\frac{x^2-40x+144}{135},\\
M_2(x)&=&\frac{(x-4)(x-16)}{(21)(9)}=\frac{x^2-20x+64}{189},\quad x\in [1,49].
\end{aligned}\notag\]
Let
\[\begin{aligned}
g_1(x)&=& x^2,\\ \\
g_2(x)&=& x,\quad x\in \mathbb{T}.
\end{aligned}\notag\]
Then
\[\begin{aligned}
g_1^\Delta(x)&=& x+\sigma(x)=2x+2\sqrt{x}+1,\\ \\
g_2^\Delta(x)&=& 1,\quad x\in \mathbb{T}.
\end{aligned}\notag\]
We compute
\[\begin{aligned}
L_0^\Delta(x)&=& \frac{2x+2\sqrt{x}-33}{192},\\
L_1^\Delta(x)&=&-\frac{2x+2\sqrt{x}-25}{128},\\
L_2^\Delta(x)&=& \frac{2x+2\sqrt{x}-9}{384},\\
M_0^\Delta(x)&=&\frac{2x+2\sqrt{x}-51}{525},\\
M_1^\Delta(x)&=&-\frac{2x+2\sqrt{x}-39}{135},\\
M_2^\Delta(x)&=&\frac{2x+2\sqrt{x}-19}{189},\quad x\in[1,49].
\end{aligned}\notag\]
Then we have
\[\begin{aligned}
L_0(\sigma(x_0))&=& L_0(4)=\frac{35}{64},\quad L_0^\Delta(x_0)=L_0^\Delta(1)=-\frac{29}{192},\\
L_1(\sigma(x_1))&=& L_1(16)=\frac{135}{128},\quad L_1^\Delta(x_1)=L_1^\Delta(9)=\frac{1}{128},\\
L_2(\sigma(x_2))&=& L_2(36)=\frac{315}{128},\quad L_2^\Delta(x_2)=L_2^\Delta(25)=\frac{17}{128},
\end{aligned}\notag\]
\[\begin{aligned}
M_0(\sigma(x_0))&=& M_0(4)=\frac{128}{175},\quad M_0^\Delta(x_0)=M_0^\Delta(1)=-\frac{47}{525},\\
M_1(\sigma(x_1))&=& M_1(16)=\frac{16}{6},\quad M_1^\Delta(x_1)=M_1^\Delta(9)=\frac{1}{9},\\
M_2(\sigma(x_2))&=& M_2(36)=\frac{640}{189},\quad M_2^\Delta(x_2)=M_2^\Delta(25)=\frac{41}{189}.
\end{aligned}\notag\]
Using these values we compute
\[\begin{aligned}
p_5(x)&=&\left(2+\frac{17}{2}(x-1)\right)\left(\frac{x^2-34x+225}{192}\right)\left(\frac{x^2-52x+576}{525}\right)\\
&+&\left(4+\frac{44}{15}(x-9)\right)\left(\frac{x^2-26x+25}{128}\right)\left(\frac{x^2-40x+144}{135}\right)\\
&+&\left(10+\frac{1}{25}(x-25)\right)\left(\frac{x^2-10x+9}{384}\right)\left(\frac{x^2-20x+64}{189}\right), \quad x\in[1,49],
\end{aligned}\notag\]
and
\[\begin{aligned}
p_{\sigma 5}(x)&=&\left(2+\frac{51}{10}((\sqrt{x}+1)^2-4)\right)\left(\frac{x^2-34x+225}{192}\right)\left(\frac{x^2-52x+576}{525}\right)\\
&+&\left(4+\frac{308}{135}((\sqrt{x}+1)^2-16)\right)\left(\frac{x^2-26x+25}{128}\right)\left(\frac{x^2-40x+144}{135}\right)\\
&+&\left(10+\frac{11}{325}((\sqrt{x}+1)^2-36)\right)\left(\frac{x^2-10x+9}{384}\right)\left(\frac{x^2-20x+64}{189}\right),\quad x\in[1,49].
\end{aligned}\notag\]
The example above shows that the Hermite interpolation polynomial and \(\sigma\)-Hermite interpolation polynomial can be different and moreover,
the \(\sigma\)-Hermite interpolation polynomial may not be a polynomial in the classical sense.