91Ó°ÊÓ

Use Neville's method to obtain the approximations for Lagrange interpolating polynomials of degrees one, two, and three to approximate each of the following: a. \(f(8.4)\) if \(f(8.1)=16.94410, f(8.3)=17.56492, f(8.6)=18.50515, f(8.7)=18.82091\) b. \(f\left(-\frac{1}{3}\right)\) if \(f(-0.75)=-0.07181250, f(-0.5)=-0.02475000, f(-0.25)=0.33493750\), \(f(0)=1.10100000\) c. \(f(0.25)\) if \(f(0.1)=0.62049958, f(0.2)=-0.28398668, f(0.3)=0.00660095, f(0.4)=\) \(0.24842440\) d. \(f(0.9)\) if \(f(0.6)=-0.17694460, f(0.7)=0.01375227, f(0.8)=0.22363362, f(1.0)=\) \(0.65809197\)

Use Eq. (3.10) or Algorithm \(3.2\) to construct interpolating polynomials of degree one, two, and three for the following data. Approximate the specified value using each of the polynomials. a. \(f(0.43)\) if \(f(0)=1, f(0.25)=1.64872, f(0.5)=2.71828, f(0.75)=4.48169\) b. \(f(0)\) if \(f(-0.5)=1.93750, f(-0.25)=1.33203, f(0.25)=0.800781, f(0.5)=0.687500\)

Determine the clamped cubic spline \(s\) that interpolates the data \(f(0)=0, f(1)=1, f(2)=2\) and satisfies \(s^{\prime}(0)=s^{\prime}(2)=1\).

Construct the natural cubic spline for the following data. a. \begin{tabular}{c|c} \(x\) & \(f(x)\) \\ \hline \(8.3\) & \(17.56492\) \\ \(8.6\) & \(18.50515\) \end{tabular} b. \begin{tabular}{c|c} \(x\) & \(f(x)\) \\ \hline \(0.8\) & \(0.22363362\) \\ \(1.0\) & \(0.65809197\) \end{tabular} c. \begin{tabular}{c|c} \(x\) & \(f(x)\) \\ \hline\(-0.5\) & \(-0.0247500\) \\ \(-0.25\) & \(0.3349375\) \\ 0 & \(1.1010000\) \end{tabular} d. \begin{tabular}{c|r} 1\. \(x\) & \multicolumn{1}{|c}{\(f(x)\)} \\ \hline \(0.1\) & \(-0.62049958\) \\ \(0.2\) & \(-0.28398668\) \\ \(0.3\) & \(0.00660095\) \\ \(0.4\) & \(0.24842440\) \end{tabular}

Use Newton the forward-difference formula to construct interpolating polynomials of degree one, two, and three for the following data. Approximate the specified value using each of the polynomials. a. \(\quad f\left(-\frac{1}{3}\right)\) if \(f(-0.75)=-0.07181250, f(-0.5)=-0.02475000, f(-0.25)=0.33493750\), \(f(0)=1.10100000\) b. \(\quad f(0.25)\) if \(f(0.1)=-0.62049958, f(0.2)=-0.28398668, f(0.3)=0.00660095, f(0.4)=\) \(0.24842440\)

Use the Newton backward-difference formula to construct interpolating polynomials of degree one, two, and three for the following data. Approximate the specified value using each of the polynomials. a. \(f(-1 / 3)\) if \(f(-0.75)=-0.07181250, f(-0.5)=-0.02475000, f(-0.25)=0.33493750\), \(f(0)=1.10100000\) b. \(f(0.25)\) if \(f(0.1)=-0.62049958, f(0.2)=-0.28398668, f(0.3)=0.00660095, f(0.4)=\) \(0.24842440\)

Use appropriate Lagrange interpolating polynomials of degrees one, two, and three to approximate each of the following: a. \(f(8.4)\) if \(f(8.1)=16.94410, f(8.3)=17.56492, f(8.6)=18.50515, f(8.7)=18.82091\) b. \(f\left(-\frac{1}{3}\right)\) if \(f(-0.75)=-0.07181250, f(-0.5)=-0.02475000, f(-0.25)=0.33493750\), \(f(0)=1.10100000\) c. \(f(0.25)\) if \(f(0.1)=0.62049958, f(0.2)=-0.28398668, f(0.3)=0.00660095, f(0.4)=\) \(0.24842440\) d. \(f(0.9)\) if \(f(0.6)=-0.17694460, f(0.7)=0.01375227, f(0.8)=0.22363362, f(1.0)=\) \(0.65809197\)

Use appropriate Lagrange interpolating polynomials of degrees one, two, and three to approximate each of the following: a. \(f(0.43)\) if \(f(0)=1, f(0.25)=1.64872, f(0.5)=2.71828, f(0.75)=4.48169\) b. \(\quad f(0)\) if \(f(-0.5)=1.93750, f(-0.25)=1.33203, f(0.25)=0.800781, f(0.5)=0.687500\) c. \(f(0.18)\) if \(f(0.1)=-0.29004986, f(0.2)=-0.56079734, f(0.3)=-0.81401972, f(0.4)=\) \(-1.0526302\) d. \(f(0.25)\) if \(f(-1)=0.86199480, f(-0.5)=0.95802009, f(0)=1.0986123, f(0.5)=\) \(1.2943767\)

Use the Newton backward-difference formula to construct interpolating polynomials of degree one, two, and three for the following data. Approximate the specified value using each of the polynomials. a. \(f(0.43)\) if \(f(0)=1, f(0.25)=1.64872, f(0.5)=2.71828, f(0.75)=4.48169\) b. \(f(0.25)\) if \(f(-1)=0.86199480, f(-0.5)=0.95802009, f(0)=1.0986123, f(0.5)=\) \(1.2943767\)

$$ \begin{aligned} &\text { Show that the polynomial interpolating the following data has degree } 3 .\\\ &\begin{array}{c|r|r|r|r|r|r} x & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline f(x) & 1 & 4 & 11 & 16 & 13 & -4 \end{array} \end{aligned} $$