Chapter 1: Lagrange Interpolation
- Page ID
- 209033
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Let \(\mathbb{T} \) be a time scale with forward jump operator and delta differentiation operator \(\sigma \)and \(\Delta\), respectively.
In this section we construct the Lagrange interpolation polynomial on an arbitrary time scale. We present the theoretical background of this construction and solve numerical examples. We assume that \(\mathbb{T} \)is a given time scale with forward jump operator \(\sigma\) delta derivative operator \(\Delta \)and graininess function \(\mu \).
Let \(\mathcal{P}_n \), \(n\in \mathbb{N}_0 \), denote the set of all polynomials of degree \(\leq n \)defined over the set \(\mathbb{R} \)of real numbers. Let \(n\in \mathbb{N}\) and \(x_i\in \mathbb{T} \), \(i\in \{0, 1, \ldots, n\} \), be distinct and \(y_i \), \(i\in \{0, 1, \ldots, n\} \), be given real numbers. We will find \(p_n\in \mathcal{P}_n \)such that \(p_n(x_i)=y_i \), \(i\in \{0, 1, \ldots, n\} \). Below we introduce the form of a polynomial taking the given values \(y_i \)at the points \(x_i \)for \(i\in \{0, 1, \ldots, n\} \).
Suppose that \(n\in \mathbb{N} \). Then there exist polynomials \(L_k\in \mathcal{P}_n \), \(k\in \{0, 1, \ldots, n\} \), such that
\[
L_k(x_i)= \left\{
\begin{array}{l}
1\quad \text{if}\quad i=k\\ \\
0\quad \text{if}\quad i\ne k,
\end{array}
\right.
\notag\]
\(i, k\in \{0, 1, \ldots, n\} \). Moreover,
\[
p_n(x)= \sum_{k=0}^n L_k(x) y_k, \quad x\in \mathbb{T},
\notag\]
satisfies the condition \(p_n(x_i)=y_i \), \(i\in \{0, 1, \ldots, n\} \), \(p_n\in \mathcal{P}_n \).
Proof
Define
\[
L_k(x)= C_k\sum_{i=0, i\ne k}^n (x-x_i),\quad x\in \mathbb{T},
\notag\]
where \(C_k\in \mathbb{R} \), \(k\in \{0, 1, \ldots, n\} \), will be determined below. We have \(L_k(x_i)=0 \), \(i\in \{0, 1, \ldots, n\} \), \(i\ne k \), and
\[\begin{aligned}
L_k(x_k)&=& C_k \prod_{i=0, i\ne k}^n(x_k-x_i)\\ \\
&=& 1,\quad k\in \{0, 1, \ldots, n\}.
\end{aligned}\notag\]
Thus,
\[
C_k= \frac{1}{\prod_{i=0, i\ne k}^n(x_k-x_i)},\quad k\in \{0, 1, \ldots, n\},
\notag\]
and
\begin{equation}
\label{1.4}
L_k(x)=\prod_{i=0, i\ne k}^n \frac{x-x_i}{x_k-x_i}, \quad x\in \mathbb{T}, \quad k\in \{0, 1, \ldots, n\}.
\end{equation}
We have that \(L_k\in \mathcal{P}_n \), \(k\in \{0, 1, \ldots, n\} \), and \(p_n\in \mathcal{P}_n \). This completes the proof.
The uniqueness of the polynomial given in the previous theorem is proved next.
Assume that \(n\in \mathbb{N}_0 \). Let \(x_i\in \mathbb{T} \), \(i\in \{0, 1, \ldots, n\} \), be distinct and \(y_i\in \mathbb{R} \), \(i\in \{0, 1, \ldots, n\} \). Then there exists a unique polynomial \(p_n\in \mathcal{P}_n \)such that
\[
p_n(x_i)=y_i,\quad i\in \{0, 1, \ldots, n\}.
\notag\]
Proof
The existence of the polynomial \(p_n \)follows by the previous theorem. Suppose that there exist two polynomials \(p_n, q_n \in \mathcal{P}_n \)such that
\[
p_n(x_i)=q_n(x_i)=y_i,\quad i\in \{0, 1, \ldots, n\}.
\notag\]
Then the polynomial \(h_n=p_n-q_n \)has \(n+1 \)distinct roots. Therefore \(h_n\equiv 0 \)or \(p_n\equiv q_n \). This completes the proof.
Now, we define formally the polynomial in the above theorems.
Assume that \(n\in \mathbb{N}_0 \). Let \(x_i\in \mathbb{T} \), \(i\in \{0, 1, \ldots, n\} \), be distinct and \(y_i\in \mathbb{R} \), \(i\in \{0, 1, \ldots, n\} \). The polynomial
\[
p_n(x)= \sum_{k=0}^n L_k(x) y_k, \quad x\in \mathbb{T},
\notag\]
where \(L_k \), \(k\in \{0, 1, \ldots, n\} \), are defined with \eqref{1.4}, will be called the Lagrange interpolation polynomial of degree \(n \)with 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 distinct and \(f: [a, b]\to \mathbb{R} \)be a given function. The polynomial
\[
p_n(x)= \sum_{k=0}^n L_k(x) f(x_k),\quad x\in \mathbb{T},
\notag\]
where \(L_k \), \(k\in \{0, 1, \ldots, n\} \), are defined with \eqref{1.4}, will be called the Lagrange interpolation polynomial of degree \(n \)with interpolation points \(x_i \), \(i\in \{0, 1, \ldots, n\} \), for the function \(f \).
Let \(\mathbb{T}=\mathbb{Z} \). We will construct the Lagrange interpolation polynomial for the set
\[
\{(-3, 0), (-2, 2), (0, 1), (1, 0)\}.
\notag\]
Here
\[\begin{aligned}
x_0&=& -3,\quad x_1= -2,\quad x_2= 0,\quad x_3= 1,\\ \\
y_0&=& 0,\quad y_1= 2,\quad y_2= 1, \quad y_3=1,\quad n=3.
\end{aligned}\notag\]
Then
\[\begin{aligned}
L_0(x)&=& \prod_{i=1}^3\frac{x-x_i}{x_0-x_i}\\ \\
&=& \frac{(x+2)x(x-1)}{-1\cdot (-3)\cdot (-4)}\\ \\
&=& -\frac{(x-1) x (x+2)}{12},\\ \\
L_1(x)&=& \prod_{i=0, i\ne 1}^3 \frac{x-x_i}{x_1-x_i}\\ \\
&=& \frac{(x+3)x(x-1)}{1\cdot (-2)\cdot (-3)}\\ \\
&=& \frac{(x-1) x(x+3)}{6},\\ \\
L_2(x)&=& \prod_{i=0, i\ne 2}^3 \frac{x-x_i}{x_2-x_i}\\ \\
&=& \frac{(x+3)(x+2)(x-1)}{3\cdot 2\cdot (-1)}\\ \\
&=& -\frac{(x-1)(x+2)(x+3)}{6},\\ \\
L_3(x)&=& \prod_{i=0}^2 \frac{x-x_i}{x_3-x_i}\\ \\
&=& \frac{(x+3)(x+2)x}{4\cdot 3\cdot 1}\\ \\
&=& \frac{x(x+2)(x+3)}{12},\quad x\in \mathbb{T}.
\end{aligned}\notag\]
Hence,
\[\begin{aligned}
p_3(x)&=& y_0 L_0(x)+y_1 L_1(x)+y_2 L_2(x)+y_3 L_3(x)\\ \\
&=& 2\left(\frac{x(x-1)(x+3)}{6}\right)-\frac{(x-1)(x+2)(x+3)}{6}\\ \\
&&+ \frac{x(x+2)(x+3)}{12}\\ \\
&=& \frac{(x-2)(x-1)(x+3)}{6}+\frac{x(x+2)(x+3)}{12}\\ \\
&=& \frac{(x+3)(2(x-1)(x-2)+x(x+2))}{12}\\ \\
&=& \frac{(x+3)(3x^2-4x+4)}{12},\quad x\in \mathbb{T}.
\end{aligned}\notag\]
Let \(\mathbb{T}=2^{\mathbb{N}_0} \). Construct the Lagrange interpolation polynomial for the set
\[
\{(1, -1), (4, 0), (8, 1), (16, 2)\}.
\notag\]
Suppose that \(n\in \mathbb{N}_0 \)and \(x_j\in \mathbb{T} \), \(j\in \{0, 1, \ldots, n\} \), are distinct points. For \(x\in \mathbb{T} \)we define the polynomials
\[
\pi_{n+1}(x)=\prod_{j=0}^n (x-x_j),\quad \Pi_{n+1}^k(x)=\pi_{n+1}^{\Delta^k}(x),\quad k\in \mathbb{N}_0.
\notag\]
Let \(\mathbb{T}=2^{\mathbb{N}_0} \), \(x_0=1 \), \(x_1=2 \). Here
\[
n=1,\quad \sigma(x)=2x,\quad x\in \mathbb{T}.
\notag\]
Then
\[\begin{aligned}
\pi_2(x)&=&(x-1)(x-2)\\ \\
&=& x^2-3x+2,\quad x\in \mathbb{T},
\end{aligned}\notag\]
and
\[\begin{aligned}
\Pi_2^1(x)&=& \pi_2^{\Delta}(x)\\ \\
&=& \sigma(x)+x-3\\ \\
&=& 3x-3, \\ \\
\Pi_2^2(x)&=& \pi_2^{\Delta^2}(x)\\ \\
&=& 3,\quad x\in\mathbb{T}.
\end{aligned}\notag\]
Let \(\mathbb{T}=\mathbb{Z} \)and
\[
x_0=1,\quad x_1=2,\quad x_2=3.
\notag\]
Then
\[\begin{aligned}
\pi_3(x)&=& (x-1)(x-2)(x-3)\\ \\
&=& (x^2-3x+2)(x-3)\\ \\
&=& x^3-6x^2+11x-6,\quad x\in \mathbb{T}.
\end{aligned}\notag\]
We have
\[
\sigma(x)= x+1,\quad x\in \mathbb{T},
\notag\]
and
\[\begin{aligned}
\Pi_3^1(x)&=& \pi_3^{\Delta}(x)\\ \\
&=&(\sigma(x))^2+x \sigma(x)+x^2-6(\sigma(x)+x)+11\\ \\
&=& (x+1)^2+x(x+1)+x^2-6(x+1+x)+11\\ \\
&=& 3x^2-9x+6,\\ \\
\Pi_3^2(x)&=& \pi_3^{\Delta^2}(x)\\ \\
&=& 3(\sigma(x)+x)-9\\ \\
&=& 6x-6,\\ \\
\Pi_3^3(x)&=& \pi_3^{\Delta^3}(x)\\ \\
&=& 6,\quad x\in \mathbb{T}.
\end{aligned}\notag\]
Let \(\mathbb{T}=\left(\frac{1}{4}\right)^{\mathbb{N}_0} \),
\[
x_0=\frac{1}{64},\quad x_1=\frac{1}{16},\quad x_2=\frac{1}{4},\quad x_3= 1.
\notag\]
Find \(\pi_4(x) \), \(\Pi_4^1(x) \), \(\Pi_4^2(x) \), \(\Pi_4^3(x) \), \(\Pi_4^4(x) \), \(x\in \mathbb{T} \).
The following theorem gives the error in approximating a function \(f \)by a Lagrange polynomial.
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 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
\[
f(x)-p_n(x)= \frac{f^{\Delta^{n+1}}(\xi)}{\Pi_{n+1}^{n+1}(\xi)}\pi_{n+1}(x),\quad x\in [a, b],
\notag\]
or
\[
F_{min, n+1}(\xi)\leq \frac{f(x)-p_n(x)}{\pi_{n+1}(x)}\leq F_{max, n+1}(\xi),\quad x\in [a, b],
\notag\]
where
\[\begin{aligned}
F_{max, n+1}(\xi)&=& \max\left\{ \frac{f^{\Delta^{n+1}}(\xi)}{\Pi_{n+1}^{n+1}(\xi)}, \frac{f^{\Delta^{n+1}}(\rho(\xi))}{\Pi_{n+1}^{n+1}(\rho(\xi))}\right\},\\ \\
F_{min, n+1}(\xi)&=& \min\left\{ \frac{f^{\Delta^{n+1}}(\xi)}{\Pi_{n+1}^{n+1}(\xi)}, \frac{f^{\Delta^{n+1}}(\rho(\xi))}{\Pi_{n+1}^{n+1}(\rho(\xi))}\right\}.
\end{aligned}\notag\]
Proof
Let \(p_n \)be the Lagrange interpolation polynomial for the function \(f \)with interpolation points \(x_j \), \(j\in \{0, 1, \ldots, n\} \). Define the function
\[
\phi(t)= f(t)-p_n(t)-\frac{f(x)-p_n(x)}{\pi_{n+1}(x)}\pi_{n+1}(t),\quad t\in [a, b].
\notag\]
Then
\[\begin{aligned}
\phi(x_j)&=& f(x_j)-p_n(x_j)-\frac{f(x)-p_n(x)}{\pi_{n+1}(x)}\pi_{n+1}(x_j)\\ \\
&=& f(x_j)-f(x_j)\\ \\
&=& 0,\quad j\in \{0, 1, \ldots, n\},
\end{aligned}\notag\]
and \(\phi(x)=0 \). Thus, \(\phi: [a, b]\to \mathbb{R} \)has at least \(n+2 \) generalized zeros (GZs). Hence and the Rolle theorem, it follows that \(\phi^{\Delta^{n+1}} \)has at least one GZ on \((a, b) \). Therefore there exists an \(\xi=\xi(x)\in (a, b) \)such that
\[
\phi^{\Delta^{n+1}}(\xi)=0\quad \text{or}\quad \phi^{\Delta^{n+1}}(\rho(\xi))\phi^{\Delta^{n+1}}(\xi)<0.
\notag\]
Note that
\[
\phi^{\Delta^{n+1}}(t)= f^{\Delta^{n+1}}(t)-\frac{f(x)-p_n(x)}{\pi_{n+1}(x)}\pi_{n+1}^{\Delta^{n+1}}(t),\quad t\in [a, b].
\notag\]
1. Let \(\phi^{\Delta^{n+1}}(\xi)=0 \). Then
\[
f^{\Delta^{n+1}}(\xi)= \frac{f(x)-p_n(x)}{\pi_{n+1}(x)}\pi_{n+1}^{\Delta^{n+1}}(\xi)
\notag\]
or
\[\begin{aligned}
f(x)-p_n(x)&=& \frac{f^{\Delta^{n+1}}(\xi)}{\pi_{n+1}^{\Delta^{n+1}}(\xi)}\pi_{n+1}(x)\\ \\
&=& \frac{f^{\Delta^{n+1}}(\xi)}{\Pi_{n+1}^{n+1}(\xi)}\pi_{n+1}(x).
\end{aligned}\notag\]
2. Let
\[
\phi^{\Delta^{n+1}}(\rho(\xi))\phi^{\Delta^{n+1}}(\xi)<0.
\notag\]
Then
\[\begin{aligned}
\phi^{\Delta^{n+1}}(\rho(\xi))&=& f^{\Delta^{n+1}}(\rho(\xi))-\frac{f(x)-p_n(x)}{\pi_{n+1}(x)}\pi_{n+1}^{\Delta^{n+1}}(\rho(\xi))\\ \\
&=& f^{\Delta^{n+1}}(\rho(\xi))-\frac{f(x)-p_n(x)}{\pi_{n+1}(x)}\Pi_{n+1}^{n+1}(\rho(\xi)),
\end{aligned}\notag\]
and
\[
\phi^{\Delta^{n+1}}(\xi)= f^{\Delta^{n+1}}(\xi)-\frac{f(x)-p_n(x)}{\pi_{n+1}(x)}\Pi_{n+1}^{n+1}(\xi).
\notag\]
Hence,
\[\begin{aligned}
0&>& \phi^{\Delta^{n+1}}(\rho(\xi))\phi^{\Delta^{n+1}}(\xi)\\ \\
&=& \left(f^{\Delta^{n+1}}(\rho(\xi))-\frac{f(x)-p_n(x)}{\pi_{n+1}(x)}\Pi_{n+1}^{n+1}(\rho(\xi))\right)\\ \\
&&\times \left(f^{\Delta^{n+1}}(\xi)-\frac{f(x)-p_n(x)}{\pi_{n+1}(x)}\Pi_{n+1}^{n+1}(\xi)\right)\\ \\
&=& \left(\frac{f(x)-p_n(x)}{\pi_{n+1}(x)}\right)^2\Pi_{n+1}^{n+1}(\rho(\xi))\Pi_{n+1}^{n+1}(\xi)\\ \\
&&-\frac{f(x)-p_n(x)}{\pi_{n+1}(x)}\left(\Pi_{n+1}^{n+1}(\rho(\xi))f^{\Delta^{n+1}}(\xi)+\Pi_{n+1}^{n+1}(\xi) f^{\Delta^{n+1}}(\rho(\xi))\right)\\ \\
&&+f^{\Delta^{n+1}}(\rho(\xi))f^{\Delta^{n+1}}(\xi).
\end{aligned}\notag\]
Hence,
\[
F_{min, n+1}(\xi)\leq \frac{f(x)-p_n(x)}{\pi_{n+1}(x)}\leq F_{max, n+1}(\xi).
\notag\]
This completes the proof.
Note that as stated in the next remark, the error vanishes if the number of data points increases to infinity.
Suppose that all conditions of the previous theorem hold. If
\[
\lim_{n\to\infty} \max_{x\in [a, b]}\left(\frac{f^{\Delta^{n+1}}(\xi)}{\Pi_{n+1}^{n+1}(\xi)}\pi_{n+1}(x)\right)=0
\notag\]
and
\[
\lim_{n\to\infty} \max_{x\in [a, b]}\left(\frac{f^{\Delta^{n+1}}(\rho(\xi))}{\Pi_{n+1}^{n+1}(\rho(\xi))}\pi_{n+1}(x)\right)=0,
\notag\]
then
\[
\lim_{n\to\infty}\max_{x\in [a, b]}|f(x)-p_n(x)|=0.
\notag\]
Let \(\mathbb{T}=\displaystyle \left\{0,\frac{1}{6},\frac{1}{4},\frac{1}{2},1,2,3,4\right\} \). Let
\[
a=x_0=0,\quad x_1=\frac{1}{4},\quad x_2=1,\quad x_3=3,\quad b=4,
\notag\]
and
\[
f(x)=\frac{x+1}{x^2+x+6},\quad x\in \mathbb{T}.
\notag\]
We will construct the Lagrange interpolation polynomial \(p_3 \)of \(f \)for the points \(x_0, x_1, x_2, x_3 \)and compare the graphs of
\(p_3 \)and \(f \).
First, note that
\[\begin{aligned}
f(x_0)&=& f(0)=\frac{1}{6},\\
f(x_1)&=&f(\frac{1}{4})=\frac{20}{101},\\
f(x_2)&=&f(1)=\frac{1}{4},\\
f(x_3)&=&f(3)=\frac{2}{9}.
\end{aligned}\notag\]
Also, we compute
\[\begin{aligned}
L_0(x)&=&\displaystyle \frac{(x-\frac{1}{4})(x-1)(x-3)}{(0-\frac{1}{4})(0-1)(0-3)}\\
&=& \displaystyle -\frac{4}{3}(x-\frac{1}{4})(x-1)(x-3),\\\\
L_1(x)&=&\displaystyle \frac{(x-0)(x-1)(x-3)}{(\frac{1}{4}-0)(\frac{1}{4}-1)(\frac{1}{4}-3)}\\
&=& \displaystyle \frac{64}{33}x(x-1)(x-3,)\\\\
L_2(x)&=&\displaystyle \frac{(x-0)(x-\frac{1}{4})(x-3)}{(1-0)(1-\frac{1}{4})(1-3)}\\
&=& \displaystyle -\frac{2}{3}x(x-\frac{1}{4})(x-3),\\\\
L_3(x)&=&\displaystyle \frac{(x-0)(x-\frac{1}{4})(x-1)}{(3-0)(3-\frac{1}{4})(3-1)}\\
&=& \displaystyle \frac{2}{33}x(x-\frac{1}{4})(x-1),\quad x\in \mathbb{T}.
\end{aligned}\notag\]
Then the Lagrange interpolation polynomial \(p_3 \)of \(f \)for the points \(\displaystyle 0, \frac{1}{4}, 1, 3 \) is
\[\begin{aligned}
p_3(x)&=&\displaystyle f(0)L_0(x)+f(\frac{1}{4})L_1(x)+f(1)L_2(x)+f(3)L_3(x) \\
&=& \displaystyle\frac{1}{6}\left(-\frac{4}{3}(x-\frac{1}{4})(x-1)(x-3)\right) \\
&+& \displaystyle\frac{20}{101}\left(\frac{64}{33}x(x-1)(x-3)\right) \\
&+& \displaystyle\frac{1}{4}\left(-\frac{2}{3}x(x-\frac{1}{4})(x-3)\right)\\
&+& \displaystyle\frac{2}{9}\left(\frac{2}{33}x(x-\frac{1}{4})(x-1)\right)\\
&=& \displaystyle-\frac{2}{9}(x-\frac{1}{4})(x-1)(x-3)\\
&+&\frac{1280}{3333}x(x-1)(x-3)\\
&-& \displaystyle \frac{1}{6}x(x-\frac{1}{4})(x-3)\\
&+&\frac{4}{297}x(x-\frac{1}{4})(x-1),\quad x\in \mathbb{T}.
\end{aligned}\notag\]