Chapter 2: Sigma-Lagrange Interpolation
- Page ID
- 209034
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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 show that a given function can be approximated with so called \(\sigma\)-Lagrange polynomials. We will show that there are classes of time scales for which the Lagrange interpolation polynomials and the \(\sigma\)-Lagrange interpolation polynomials are different and there are classes of time scales for which the Lagrange interpolation polynomials and the \(\sigma\)-Lagrange interpolation polynomials coincide.
In the following, we define the \(\sigma\)-polynomials.
With \(\mathcal{P}_n^{\sigma} \), \(n\in \mathbb{N}_0 \), we will denote the set of all functions in the form
\[
a_n(\sigma(x))^n+a_{n-1}(\sigma(x))^{n-1}+\cdots+ a_1(x)\sigma(x)+a_0,\quad x\in \mathbb{T},
\notag\]
where \(a_j\in \mathbb{R} \), \(j\in \{0, 1, \ldots, n\} \). Let \(a, b\in \mathbb{T} \), \(a<b \). A function \(g\in \mathcal{P}_n^{\sigma} \)will be called
a \(\sigma\)-polynomial.
To define a \(\sigma\)-Lagrange polynomial we have a requirement that the points in the data set should be \(\sigma\)-distinct.
Let \(n\in \mathbb{N}_0 \). The points \(x_j\in [a, b) \), \(j\in \{0, 1, \ldots, n\} \), will be called \(\sigma\)-distinct if \(\sigma(x_n)\leq b \)and
\[
\sigma(x_0)<\sigma(x_1)<\ldots < \sigma(x_n).
\notag\]
Let \(\mathbb{T}=\{-1, 1\}\bigcup \left\{1+\left(\frac{1}{2}\right)^n: n\in \mathbb{N}_0\right\}\bigcup \{3, 4, 5\} \)and \(a=-1 \), \(b=5 \). Take the points
\[
x_0=-1,\quad x_1=1,\quad x_2=3.
\notag\]
Then
\[
\sigma(x_0)=1,\quad \sigma(x_1)=1,\quad \sigma(x_2)=4.
\notag\]
Thus, the points \(\{x_0, x_1, x_2\} \)are not \(\sigma\)-distinct.
Let \(\mathbb{T}=2^{\mathbb{N}_0} \), \(a=1 \), \(b=16 \). Take the points
\[
x_0=1,\quad x_1=2,\quad x_2=4.
\notag\]
Then
\[
\sigma(x_0)=2,\quad \sigma(x_1)=4,\quad \sigma(x_2)=8.
\notag\]
Therefore \(\{x_0, x_1, x_2\} \)are \(\sigma\)-distinct points.
As in the case of Lagrange interpolation, one can prove the following result.
Suppose that \(n\in \mathbb{N} \)and \(x_j\in \mathbb{T} \), \(j\in \{0, 1, \ldots, n\} \), are \(\sigma\)-distinct. Then there exist unique
\(\sigma\)-polynomials \(L_{\sigma k}\in \mathcal{P}_{n}^{\sigma} \), \(k\in \{0, 1, \ldots, n\} \), such that
\[
L_{\sigma k}(x_i)= \left\{
\begin{array}{l}
1\quad \text{if}\quad i=k\\ \\
0\quad \text{if}\quad i\ne k,
\end{array}
\right.
\notag\]
\(i, k\in \{0, 1, \ldots, n\} \). Moreover,
\[\begin{aligned}
p_{\sigma n}(x)&=& \sum_{k=0}^n L_{\sigma k}(x) y_k\\ \\
&=& \sum_{k=0}^n \left(\prod_{j=0, j\ne k}^n \frac{\sigma(x)-\sigma(x_j)}{\sigma(x_k)-\sigma(x_j)}\right)y_k,\quad x\in \mathbb{T}
\end{aligned}\notag\]
satisfies the condition \(p_{\sigma n}(x_i)=y_i \), \(i\in \{0, 1, \ldots, n\} \), \(p_{\sigma n}\in \mathcal{P}_n^{\sigma} \).
Based on the statement of the above theorem we define the \(\sigma\)-Lagrange polynomials as follows.
Assume that \(n\in \mathbb{N}_0 \). Let \(x_i\in \mathbb{T} \), \(i\in \{0, 1, \ldots, n\} \), be \(\sigma\)-distinct and \(y_i\in \mathbb{R} \), \(i\in \{0, 1, \ldots, n\} \). The \(\sigma\)-polynomial
\[
p_{\sigma n}(x)= \sum_{k=0}^n L_{\sigma k}(x) y_k, \quad x\in \mathbb{T}
\notag\]
will be called the \(\sigma\)-Lagrange interpolation polynomial of degree \(n \)with \(\sigma\)-interpolation points \(x_i \), \(i\in \{0, 1, \ldots, n\} \).
Assume that \(n\in \mathbb{N}_0 \). Let \(x_i\in [a, b]\subset \mathbb{T} \), \(i\in \{0, 1, \ldots, n\} \), be \(\sigma\)-distinct and \(f: [a, b]\to \mathbb{R} \)be a given function. The \(\sigma\)-polynomial
\[
p_{\sigma n}(x)= \sum_{k=0}^n L_{\sigma k}(x) f(x_k), \quad x\in \mathbb{T},
\notag\]
will be called the \(\sigma\)-Lagrange interpolation polynomial of degree \(n \)with \(\sigma\)-interpolation points \(x_i \), \(i\in \{0, 1, \ldots, n\} \), for the function \(f \).
In the following, we will compute the Lagrange and the \(\sigma\)-Lagrange polynomials for a given set of data on a time scale and compare them.
Let \(\mathbb{T}=\{-2, -1, 0, 3, 7\} \),
\[
a=-2,\quad b=7,\quad x_0=-2,\quad x_1=0,
\notag\]
\(f: \mathbb{T}\to \mathbb{R} \)is defined by
\[
f(t)=t+3,\quad t\in \mathbb{T}.
\notag\]
We will find the \(\sigma\)-Lagrange interpolation polynomial for the function \(f \)with \(\sigma\)-interpolation points \(\{x_0, x_1\} \). We have
\[\begin{aligned}
\sigma(x_0)&=& \sigma(-2)=-1,\\ \\
\sigma(x_1)&=& \sigma(0)= 3,\\ \\
L_{\sigma 0}(x)&=& \frac{\sigma(x)-\sigma(x_1)}{\sigma(x_0)-\sigma(x_1)}\\ \\
&=& -\frac{1}{4}(\sigma(x)-3),\\ \\
L_{\sigma 1}(x)&=& \frac{\sigma(x)-\sigma(x_0)}{\sigma(x_1)-\sigma(x_0)}\\ \\
&=& \frac{1}{4}(\sigma(x)+1),\quad x\in \mathbb{T}, \\ \\
f(x_0)&=& f(-2)\\ \\
&=& 1,\\ \\
&=& 3.
\end{aligned}\notag\]
Thus,
\[\begin{aligned}
p_{\sigma 1}(x)&=& f(x_0)L_{\sigma 0}(x)+f(x_1) L_{\sigma 1}(x)\\ \\
&=& -\frac{1}{4}(\sigma(x)-3)+\frac{3}{4}(\sigma(x)+1)\\ \\
&=& \frac{1}{2}(\sigma(x)+3),\quad x\in [-2, 7].
\end{aligned}\notag\]
Note that,
\[\begin{aligned}
p_{\sigma 1}(0)&=& \frac{1}{2}(\sigma(0)+3)\\ \\
&=& 3=f(0),\\ \\
p_{\sigma 1}(-2)&=& \frac{1}{2}(\sigma(-2)+3)\\ \\
&=& 1=f(-2).
\end{aligned}\notag\]
Also, we have
\[\begin{aligned}
p_{\sigma 1}(-1)&=& \frac{1}{2}(\sigma(-1)+3)\\ \\
&=& \frac{3}{2}.
\end{aligned}\notag\]
Now, we will find the Lagrange interpolation polynomial for the function \(f \)with interpolation points \(\{ x_0, x_1\} \). We have
\[\begin{aligned}
L_0(x)&=& \frac{x-x_1}{x_0-x_1}\\ \\
&=& -\frac{1}{2}x,\\ \\
L_1(x)&=& \frac{x-x_0}{x_1-x_0}\\ \\
&=& \frac{x+2}{2},\quad x\in [-2, 7].
\end{aligned}\notag\]
Therefore
\[\begin{aligned}
p_1(x)&=& f(x_0)L_0(x)+f_1(x)L_1(x)\\ \\
&=& -\frac{1}{2}x+3\frac{x+2}{2}\\ \\
&=& x+3,\quad x\in [-2, 7].
\end{aligned}\notag\]
Then
\[
p_1(0)=3=f(0),\quad p_1(-2)=1=f(-2).
\notag\]
We also have
\[
p_1(-1)= -1+3=2\neq p_{\sigma 1}(-1)=\frac{3}{2}.
\notag\]
In the above example we see that in general, the Lagrange interpolation polynomial and the \(\sigma\)-Lagrange interpolation polynomial for a function \(f \)are different.
Let \(\mathbb{T}=\left\{ -1, -\frac{1}{4}, -\frac{1}{8}, 0, 2, 3, 7\right\} \),
\[
a=x_0=-1,\quad b=7,\quad x_1=-\frac{1}{8},\quad x_2=3.
\notag\]
Find the \(\sigma\)-Lagrange and Lagrange interpolation polynomials for the function \(f: \mathbb{T}\to \mathbb{R} \)defined by
\[
f(t)=\frac{t+1}{t^2-t+1}+3t,\quad t\in [-1, 7],
\notag\]
with \(\sigma\)-interpolation and interpolation points \(x_0 \), \(x_1 \), \(x_2 \).
In the following example we see that on some time scales, the Lagrange and \(\sigma\)-Lagrange polynomials are the same.
Let \(\mathbb{T}=\mathbb{Z} \),
\[
a=x_0=-1,\quad x_1=1,\quad x_2=3,
\notag\]
\(f: \mathbb{T}\to \mathbb{R} \)is defined by
\[
f(t)= t^2+t+1,\quad t\in \mathbb{T}.
\notag\]
We will find the \(\sigma\)-Lagrange interpolation polynomial with \(\sigma\)-interpolation points \(x_0, x_1, x_2 \). We have
\[\begin{aligned}
\sigma(x_0)&=& \sigma(-1)=0,\\ \\
\sigma(x_1)&=& \sigma(1)= 2,\\ \\
\sigma(x_2)&=& \sigma(3)=4.
\end{aligned}\notag\]
Then
\[
\sigma(x_0)<\sigma(x_1)<\sigma(x_2),
\notag\]
i.e., \(x_0, x_1, x_2 \)are \(\sigma\)-distinct points. We have
\[\begin{aligned}
L_{\sigma 0}(x)&=& \frac{(\sigma(x)-\sigma(x_1))(\sigma(x)-\sigma(x_2))}{(\sigma(x_0)-\sigma(x_1))(\sigma(x_0)-\sigma(x_2))}\\ \\
&=& \frac{(x+1-2)(x+1-4)}{(0-2)(0-4)}\\ \\
&=& \frac{(x-1)(x-3)}{8},\\ \\
L_{\sigma 1}(x)&=& \frac{(\sigma(x)-\sigma(x_0))(\sigma(x)-\sigma(x_2))}{(\sigma(x_1)-\sigma(x_0))(\sigma(x_1)-\sigma(x_2))}\\ \\
&=& \frac{(x+1-0)(x+1-4)}{(2-0)(2-4)}\\ \\
&=& -\frac{(x-3)(x+1)}{4},\\ \\
L_{\sigma 2}(x)&=& \frac{(\sigma(x)-\sigma(x_0))(\sigma(x)-\sigma(x_1))}{(\sigma(x_2)-\sigma(x_0))(\sigma(x_2)-\sigma(x_1))}\\ \\
&=& \frac{(x+1-0)(x+1-2)}{(4-0)(4-2)}\\ \\
&=& \frac{(x-1)(x+1)}{8},\quad x\in [-1, 4],\\ \\
f(x_0)&=& f(-1)\\ \\
&=& 1,\\ \\
f(x_1)&=& f(1)\\ \\
&=& 3,\\ \\
f(x_2)&=& f(3)\\ \\
&=& 13.
\end{aligned}\notag\]
Hence,
\[\begin{aligned}
p_{\sigma 2}(x)&=& f(x_0)L_{\sigma 0}(x)+f(x_1) L_{\sigma 1}(x)+f(x_2) L_{\sigma 2}(x)\\ \\
&=& \frac{(x-1)(x-3)}{8}-3\frac{(x-3)(x+1)}{4}+13\frac{(x-1)(x+1)}{8}\\ \\
&=& \frac{x^2-4x+3-6x^2+12x +18+13x^2-13}{8}\\ \\
&=& x^2+x+1,\quad x\in [-1, 4],
\end{aligned}\notag\]
is the \(\sigma\)-Lagrange interpolation polynomial for the function \(f \).
Now, we will find the Lagrange interpolation polynomial for the function \(f \)with interpolation points \(x_0 \), \(x_1 \), \(x_2 \). We have
\[\begin{aligned}
L_0(x)&=& \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}\\ \\
&=& \frac{(x-1)(x-3)}{8},\\ \\
L_1(x)&=& \frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}\\ \\
&=& -\frac{(x+1)(x-3)}{4},\\ \\
L_2(x)&=& \frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}\\ \\
&=& \frac{(x+1)(x-1)}{8},\quad x\in [-1, 4].
\end{aligned}\notag\]
Hence,
\[\begin{aligned}
p_2(x)&=& f(x_0)L_0(x)+f(x_1)L_1(x)+f(x_2)L_2(x)\\ \\
&=& \frac{(x-1)(x-3)}{8}-3\frac{(x+1)(x-3)}{4}+13\frac{(x+1)(x-1)}{8}\\ \\
&=& \frac{x^2-4x+3-6x^2+12x+18+13x^2-13}{8}\\ \\
&=& x^2+x+1, \quad x\in [-1, 4].
\end{aligned}\notag\]
Moreover,
\[
p_2(x)=p_{\sigma 2}(x)=f(x), \quad x\in [-1, 4].
\notag\]
In the above example we see that that there are cases for which the Lagrange interpolation polynomial and the \(\sigma\)-Lagrange interpolation polynomial for a function \(f \)coincide.
Let \(\mathbb{T}=2^{\mathbb{N}_0} \),
\[
a=1=x_0,\quad x_1=2,\quad x_2=8,\quad x_3=32=b,
\notag\]
\(f: \mathbb{T}\to \mathbb{R} \)is defined by
\[
f(t)=t^3+t^2-t+4,\quad t\in \mathbb{T}.
\notag\]
Find the \(\sigma\)-Lagrange interpolation polynomial for the function \(f \)with \(\sigma\)-interpolation points \(x_0 \), \(x_1 \), \(x_2 \).
Now, we will describe the classes of time scales for which the \(\sigma\)-Lagrange and Lagrange interpolation polynomials coincide.
Let \(\mathbb{T} \)be a time scale such that \(\sigma(t)=ct+d \) for any \(t\in \mathbb{T} \)and some constants \(c,d \), Let also, \(n\in \mathbb{N} \)and
\[
a=x_0<x_1<\ldots <x_n=b,\quad x_j\in \mathbb{T},\quad j\in \{0, 1, \ldots, n\},
\notag\]
are \(\sigma\)-interpolation and interpolation points. Then
\[
L_k(x)=L_{\sigma k}(x),\quad x\in [a, b],\quad k\in \{0, 1, \ldots, n\}.
\notag\]
Proof
Since \(\sigma(t)=ct+d \)for any \(t\in \mathbb{T} \), we get
\[\begin{aligned}
L_{\sigma k}(x)&=& \prod_{j=0, j\ne k}^n \frac{\sigma(x)-\sigma(x_j)}{\sigma(x_k)-\sigma(x_j)}\\ \\
&=& \prod_{j=0, j\ne k}^n \frac{(cx+d)-(cx_j+d)}{(cx_k+d)-(cx_j+d)}\\ \\
&=& \prod_{j=0, j\ne k}^n \frac{x-x_j}{x_k-x_j}\\ \\
&=& L_k(x),\quad k\in \{0, 1, \ldots, n\},\quad x\in [a, b].
\end{aligned}\notag\]
This completes the proof.
It is clear that the uniqueness of interpolating polynomial of degree \(n \)is not violated. Indeed, if \(\sigma \)is not a linear function, the \(\sigma \)-Lagrange interpolation polynomial of a function \(f \)is not a polynomial of degree \(n \)or not a polynomial at all. For instance, if \(\mathbb{T}=\mathbb{N}_0^2 \), then \(\sigma(t)(\sqrt{t}+1)^2 \), \(t\in \mathbb{T} \), and the \(\sigma \)-Lagrange interpolation polynomial is not a polynomial. Therefore, for any function \(f \), defined on a time scale \(\mathbb{T} \), there is a unique interpolation polynomial of degree \(n \).
Suppose that \(n\in \mathbb{N}_0 \)and \(x_j\in \mathbb{T} \), \(j\in \{0, 1, \ldots, n\} \), are \(\sigma \)-distinct points. For \(x\in \mathbb{T} \), define the \(\sigma \)-polynomials
\[
\pi_{\sigma n+1}(x)=\prod_{j=0}^n (\sigma(x)-\sigma(x_j)),\quad \Pi_{\sigma n+1}^k(x)=\pi_{\sigma n+1}^{\Delta^k}(x),\quad x\in \mathbb{T},\quad k\in \mathbb{N}_0.
\notag\]
These functions are employed in the error estimate for the \(\sigma \)-Lagrange interpolation.
Let \(\mathbb{T}=\left\{ 0, \frac{1}{6}, \frac{1}{3}, \frac{1}{2}, 1+\frac{1}{2^n}:n\in \mathbb{N}_0\right\} \),
\[
a=0,\quad b=1,\quad x_0=0,\quad x_1=\frac{1}{3}.
\notag\]
We will compute \(\displaystyle\pi_{\sigma 2}(x) \), \(\displaystyle\pi_{\sigma 2}\left(\frac{1}{2}\right) \)and \(\displaystyle\Pi_{\sigma 2}^1\left(\frac{1}{2}\right) \).
We have
\[\begin{aligned}
\sigma(x_0)&=& \frac{1}{6},\\ \\
\sigma(x_1)&=& \frac{1}{2},\\ \\
\sigma\left(\frac{1}{2}\right)&=& 1,\\ \\
\sigma(1)&=& 1.
\end{aligned}\notag\]
Hence,
\[\begin{aligned}
\pi_{\sigma 2}(x)&=& (\sigma(x)-\sigma(x_0))(\sigma(x)-\sigma(x_1))\\ \\
&=& \left(\sigma(x)-\frac{1}{6}\right)\left(\sigma(x)-\frac{1}{2}\right),\quad x\in \mathbb{T},\\ \\
\pi_{\sigma 2}\left(\frac{1}{2}\right)&=& \left(\sigma\left(\frac{1}{2}\right)-\frac{1}{6}\right)\left(\sigma\left(\frac{1}{2}\right)-\frac{1}{2}\right)\\ \\
&=& \left(1-\frac{1}{6}\right)\left(1-\frac{1}{2}\right)\\ \\
&=& \frac{5}{12},\\ \\
\Pi_{\sigma 2}^1\left(\frac{1}{2}\right)&=& \pi_{\sigma 2}^{\Delta}\left(\frac{1}{2}\right)\\ \\
&=& \frac{\left(\sigma^2\left(\frac{1}{2}\right)-\frac{1}{6}\right)\left(\sigma^2\left(\frac{1}{2}\right)-\frac{1}{2}\right)
-\left(\sigma\left(\frac{1}{2}\right)-\frac{1}{6}\right)\left(\sigma\left(\frac{1}{2}\right)-\frac{1}{2}\right)}{\sigma\left(\frac{1}{2}\right)-\frac{1}{2}}\\ \\
&=& \frac{\left(\sigma\left( 1\right)-\frac{1}{6}\right)\left(\sigma\left( 1\right)-\frac{1}{2}\right)
-\left(\sigma\left(\frac{1}{2}\right)-\frac{1}{6}\right)\left(\sigma\left(\frac{1}{2}\right)-\frac{1}{2}\right)}{\sigma\left(\frac{1}{2}\right)-\frac{1}{2}}\\ \\
&=& \frac{\left(1-\frac{1}{6}\right)\left(1-\frac{1}{2}\right)-\left(1-\frac{1}{6}\right)\left(1-\frac{1}{2}\right)}{1-\frac{1}{2}}\\ \\
&=& 0.
\end{aligned}\notag\]
Let \(\mathbb{T}=\{0, 2, 3, 5, 9, 16, 18\} \),
\[
a=0, \quad b=18, \quad x_0=0,\quad x_1=3,\quad x_2=9.
\notag\]
Find \(\Pi_{\sigma 2}^2\left(3\right) \).
The error in the \(\sigma \)-Lagrange interpolation is given int the next 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 \(k\in \{1, \ldots, n+1\} \). Then for any \(x\in [a, b] \)there exists \(\xi=\xi(x)\in (a, b) \)such that
\[
f(x)-p_{\sigma n}(x)= \frac{f^{\Delta^{n+1}}(\xi)}{\Pi_{\sigma n+1}^{n+1}(\xi)}\pi_{\sigma n+1}(x), \quad x\in [a, b],
\notag\]
or
\[
F_{\sigma min, n+1}(\xi)\leq \frac{f(x)-p_{\sigma n}(x)}{\pi_{\sigma n+1}(x)}\leq F_{\sigma max, n+1}(\xi), \quad x\in [a, b],
\notag\]
where
\[\begin{aligned}
F_{\sigma max, n+1}(\xi)&=& \max\left\{ \frac{f^{\Delta^{n+1}}(\xi)}{\Pi_{\sigma n+1}^{n+1}(\xi)}, \frac{f^{\Delta^{n+1}}(\rho(\xi))}{\Pi_{\sigma n+1}^{n+1}(\rho(\xi))}\right\},\\ \\
F_{\sigma min, n+1}(\xi)&=& \min\left\{ \frac{f^{\Delta^{n+1}}(\xi)}{\Pi_{\sigma n+1}^{n+1}(\xi)}, \frac{f^{\Delta^{n+1}}(\rho(\xi))}{\Pi_{\sigma n+1}^{n+1}(\rho(\xi))}\right\}.
\end{aligned}\notag\]
Proof
Let \(p_{\sigma n} \)be the \(\sigma \)-Lagrange interpolation polynomial for the function \(f \)with \(\sigma \)-interpolation points \(x_j \), \(j\in \{0, 1, \ldots, n\} \). Define the function
\[
\phi(t)= f(t)-p_{\sigma n}(t)-\frac{f(x)-p_{\sigma n}(x)}{\pi_{\sigma n+1}(x)}\pi_{\sigma n+1}(t),\quad t\in [a, b].
\notag\]
From here, the proof repeats the corresponding proof for the Lagrange Polynomials. This completes the proof.
If
\[
\lim_{n\to\infty} \max_{x\in [a, b]}\left(\frac{f^{\Delta^{n+1}}(\xi)}{\Pi_{\sigma n+1}^{n+1}(\xi)}\pi_{\sigma n+1}(x)\right)=0
\notag\]
and
\[
\lim_{n\to\infty} \max_{x\in [a, b]}\left(\frac{f^{\Delta^{n+1}}(\rho(\xi))}{\Pi_{\sigma n+1}^{n+1}(\rho(\xi))}\pi_{\sigma n+1}(x)\right)=0,
\notag\]
then
\[
\lim_{n\to\infty}\max_{x\in [a, b]}|f(x)-p_{\sigma n}(x)|=0.
\notag\]
The last example shows clearly the difference of Lagrange and \(\sigma \)-Lagrange polynomials on a time scale whose forward jump operator is not a linear function.
\begin{example}
Let \(\mathbb{T}=\left\{\sqrt{2n+1},n\in\mathbb{N}_0\right\}=\{1,\sqrt{3},\sqrt{5},\ldots\} \). Take
\[
a=x_0=\sqrt{3},\quad x_1=\sqrt{7},\quad x_2=\sqrt{11},\quad x_3=\sqrt{17},\quad b=\sqrt{23},
\notag\]
and
\[
f(x)=\frac{x^2+1}{2x^2+5}, \quad x\in \mathbb{T}.
\notag\]
We will write the Lagrange interpolation polynomial \(p_3 \)and the \(\sigma \)-Lagrange interpolation polynomial \(p_{\sigma 3} \)interpolating the function
\(f \)at the points \(x_0, x_1, x_2, x_3 \). We will also compare the graphs of the two types interpolating polynomials with the function \(f \).
Notice that on the given time scale we have
\[
\sigma(x)=\sqrt{x^2+2}, \quad x\in \mathbb{T}.
\notag\]
Also
\[\begin{aligned}
f(x_0)&=&f(\sqrt{3})=\displaystyle \frac{3+1}{6+5}=\frac{4}{11}, \\\\
f(x_1)&=&f(\sqrt{7})=\displaystyle \frac{7+1}{14+5}=\frac{8}{19}, \\\\
f(x_2)&=&f(\sqrt{11})=\displaystyle \frac{11+1}{22+5}=\frac{4}{9}, \\\\
f(x_3)&=&f(\sqrt{17})=\displaystyle \frac{17+1}{34+5}=\frac{6}{13}.
\end{aligned}\notag\]
First, we compute \(L_0(x) \), \(L_1(x) \), \(L_2(x) \)and \(L_3(x) \)as follows,
\[\begin{aligned}
L_0(x)&=&\displaystyle \frac{(x-\sqrt{7})(x-\sqrt{11})(x-\sqrt{17})}{(\sqrt{3}-\sqrt{7})(\sqrt{3}-\sqrt{11})(\sqrt{3}-\sqrt{17})}, \\\\
L_1(x)&=&\displaystyle \frac{(x-\sqrt{3})(x-\sqrt{11})(x-\sqrt{17})}{(\sqrt{7}-\sqrt{3})(\sqrt{7}-\sqrt{11})(\sqrt{7}-\sqrt{17})}, \\\\
L_2(x)&=&\displaystyle \frac{(x-\sqrt{3})(x-\sqrt{7})(x-\sqrt{17})}{(\sqrt{11}-\sqrt{3})(\sqrt{11}-\sqrt{7})(\sqrt{11}-\sqrt{17})}, \\\\
L_3(x)&=&\displaystyle \frac{(x-\sqrt{3})(x-\sqrt{7})(x-\sqrt{11})}{(\sqrt{17}-\sqrt{3})(\sqrt{17}-\sqrt{7})(\sqrt{17}-\sqrt{11})}, \quad x\in \mathbb{T}.
\end{aligned}\notag\]
Then the Lagrange interpolation polynomial \(p_3 \)interpolating the function \(f \)at the points \(\sqrt{3}, \sqrt{7}, \sqrt{11}, \sqrt{17} \)is obtained as
\[\begin{aligned}
p_3(x)&=&f(\sqrt{3})L_0(x)+f(\sqrt{7})L_1(x)+f(\sqrt{11})L_2(x)+f(\sqrt{17})L_3(x)\\
&=&\displaystyle\frac{4}{11}\frac{(x-\sqrt{7})(x-\sqrt{11})(x-\sqrt{17})}{(\sqrt{3}-\sqrt{7})(\sqrt{3}-\sqrt{11})(\sqrt{3}-\sqrt{17})}\\\\
&+&\displaystyle \frac{8}{19}\frac{(x-\sqrt{3})(x-\sqrt{11})(x-\sqrt{17})}{(\sqrt{7}-\sqrt{3})(\sqrt{7}-\sqrt{11})(\sqrt{7}-\sqrt{17})}\\\\
&+&\displaystyle \frac{4}{9}\frac{(x-\sqrt{3})(x-\sqrt{7})(x-\sqrt{17})}{(\sqrt{11}-\sqrt{3})(\sqrt{11}-\sqrt{7})(\sqrt{11}-\sqrt{17})}\\\\
&+&\displaystyle \frac{6}{13}\frac{(x-\sqrt{3})(x-\sqrt{7})(x-\sqrt{11})}{(\sqrt{17}-\sqrt{3})(\sqrt{17}-\sqrt{7})(\sqrt{17}-\sqrt{11})}, \quad x\in \mathbb{T}.
\end{aligned}\notag\]
Next, we compute \(L_{\sigma 0}(x) \), \(L_{\sigma 1}(x) \), \(L_{\sigma 2}(x) \)and \(L_{\sigma 3}(x) \)as follows,
\[\begin{aligned}
L_{\sigma 0}(x)&=&\displaystyle \frac{(\sigma(x)-\sigma(\sqrt{7}))(\sigma(x)-\sigma(\sqrt{11}))(\sigma(x)-\sigma(\sqrt{17}))}
{(\sigma(\sqrt{3})-\sigma(\sqrt{7}))(\sigma(\sqrt{3})-\sigma(\sqrt{11}))(\sigma(\sqrt{3})-\sigma(\sqrt{17}))} \\\\
&=&\displaystyle \frac{(\sqrt{x^2+2}-3)(\sqrt{x^2+2}-\sqrt{13})(\sqrt{x^2+2}-\sqrt{19})}
{(\sqrt{5}-3)(\sqrt{5}-\sqrt{13})(\sqrt{5}-\sqrt{19})},\\\\
L_{\sigma 1}(x)&=&\displaystyle \frac{(\sigma(x)-\sigma(\sqrt{3}))(\sigma(x)-\sigma(\sqrt{11}))(\sigma(x)-\sigma(\sqrt{17}))}
{(\sigma(\sqrt{7})-\sigma(\sqrt{3}))(\sigma(\sqrt{7})-\sigma(\sqrt{11}))(\sigma(\sqrt{7})-\sigma(\sqrt{17}))} \\\\
&=&\displaystyle \frac{(\sqrt{x^2+2}-\sqrt{5})(\sqrt{x^2+2}-\sqrt{13})(\sqrt{x^2+2}-\sqrt{19})}
{(3-\sqrt{5})(3-\sqrt{13})(3-\sqrt{19})},\\\\
L_{\sigma 2}(x)&=&\displaystyle \frac{(\sigma(x)-\sigma(\sqrt{3}))(\sigma(x)-\sigma(\sqrt{7}))(\sigma(x)-\sigma(\sqrt{17}))}
{(\sigma(\sqrt{11})-\sigma(\sqrt{3}))(\sigma(\sqrt{11})-\sigma(\sqrt{7}))(\sigma(\sqrt{11})-\sigma(\sqrt{17}))} \\\\
&=&\displaystyle \frac{(\sqrt{x^2+2}-\sqrt{5})(\sqrt{x^2+2}-3)(\sqrt{x^2+2}-\sqrt{19})}
{(\sqrt{13}-\sqrt{5})(\sqrt{13}-3)(\sqrt{13}-\sqrt{19})},\\\\
L_{\sigma 3}(x)&=&\displaystyle \frac{(\sigma(x)-\sigma(\sqrt{3}))(\sigma(x)-\sigma(\sqrt{7}))(\sigma(x)-\sigma(\sqrt{11}))}
{(\sigma(\sqrt{17})-\sigma(\sqrt{3}))(\sigma(\sqrt{17})-\sigma(\sqrt{7}))(\sigma(\sqrt{17})-\sigma(\sqrt{11}))} \\\\
&=&\displaystyle \frac{(\sqrt{x^2+2}-\sqrt{5})(\sqrt{x^2+2}-3)(\sqrt{x^2+2}-\sqrt{13})}
{(\sqrt{19}-\sqrt{5})(\sqrt{19}-3)(\sqrt{19}-\sqrt{13})},\quad x\in \mathbb{T}.
\end{aligned}\notag\]
Hence, the \(\sigma \)-Lagrange polynomial \(p_{\sigma 3} \)interpolating the function \(f \)at the points \(\sqrt{3}, \sqrt{7}, \sqrt{11}, \sqrt{17} \)is obtained as
\[\begin{aligned}
p_{\sigma 3}(x)&=&f(\sqrt{3})L_{\sigma 0}(x)+f(\sqrt{7})L_{\sigma 1}(x)+f(\sqrt{11})L_{\sigma 2}(x)+f(\sqrt{17})L_{\sigma 3}(x)\\ \\
&=&\displaystyle\frac{4}{11}\frac{(\sqrt{x^2+2}-3)(\sqrt{x^2+2}-\sqrt{13})(\sqrt{x^2+2}-\sqrt{19})}
{(\sqrt{5}-3)(\sqrt{5}-\sqrt{13})(\sqrt{5}-\sqrt{19})}\\\\
&+& \displaystyle \frac{8}{19}\frac{(\sqrt{x^2+2}-\sqrt{5})(\sqrt{x^2+2}-\sqrt{13})(\sqrt{x^2+2}-\sqrt{19})}
{(3-\sqrt{5})(3-\sqrt{13})(3-\sqrt{19})}\\\\
&+& \displaystyle \frac{4}{9}\frac{(\sqrt{x^2+2}-\sqrt{5})(\sqrt{x^2+2}-3)(\sqrt{x^2+2}-\sqrt{19})}
{(\sqrt{13}-\sqrt{5})(\sqrt{13}-3)(\sqrt{13}-\sqrt{19})}\\\\
&+& \displaystyle \frac{6}{13}\frac{(\sqrt{x^2+2}-\sqrt{5})(\sqrt{x^2+2}-3)(\sqrt{x^2+2}-\sqrt{13})}
{(\sqrt{19}-\sqrt{5})(\sqrt{19}-3)(\sqrt{19}-\sqrt{13})}, \quad x\in \mathbb{T}.
\end{aligned}\notag\]
It is clear that \(p_3(x)\neq p_{\sigma 3}(x),\quad x\in \mathbb{T} \). Moreover, \(p_{\sigma 3} \)is not a polynomial.