In the notation now established we have $h_1=1$ , $h_2=1$ and $h_3=2$ . For a natural cubic spline we require $s^{″}$ to be zero at $x_1$ and $x_4$ . Values of $s^{″}$ at the other data points are found from the system of equations

$$\begin{array}{rcll}{h}_1\underset{=0}{\underset{\⏟}{s^{″}\left(x_1\right)}}+2\left(h_1+h_2\right)s^{″}\left(x_2\right)+h_2{s}^{″}\left(x_3\right)& =& 6\left(\frac{f_3-f_2}{h_2}-\frac{f_2-f_1}{h_1}\right)& \\ h_2{s}^{″}\left(x_2\right)+2\left(h_2+h_3\right)s^{″}\left(x_3\right)+h_3\underset{=0}{\underset{\⏟}{s^{″}\left(x_4\right)}}& =& 6\left(\frac{f_4-f_3}{h_3}-\frac{f_3-f_2}{h_2}\right)& \end{array}$$

In this case the equations become

$\left(\begin{array}{cc}\hfill 4\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 6\hfill \end{array}\right)\left(\begin{array}{c}\hfill s^{″}\left(x_2\right)\hfill \\ \hfill s^{″}\left(x_3\right)\hfill \end{array}\right)=\left(\begin{array}{c}\hfill 1.74\hfill \\ \hfill -1.86\hfill \end{array}\right)$

Solving the coupled pair of equations leads to $s^{″}\left(x_2\right)=0.534783s^{″}\left(x_3\right)=-0.399130$

We now find the coefficients $a_1$ , $b_1$ , etc. from the formulae and deduce that the spline is

$$\begin{array}{rcll}s\left(x\right)& =& 0.08913{\left(x-1\right)}^3+0.05087\left(x-1\right)+0.1\left(1<x<2\right)& \\ s\left(x\right)& =& -0.15565{\left(x-2\right)}^3+0.267391{\left(x-2\right)}^2+0.318261\left(x-2\right)+0.24\left(2<x<3\right)& \\ s\left(x\right)& =& 0.033261{\left(x-3\right)}^3-0.199565{\left(x-3\right)}^2+0.386087\left(x-3\right)+0.67\left(3<x<5\right)& \\ & & & \end{array}$$

1. We are interested in the Lagrange polynomials at the point $x=22$ so we consider

   $L_1\left(22\right)=\frac{\left(22-x_2\right)\left(22-x_3\right)\left(22-x_4\right)}{\left(x_1-x_2\right)\left(x_1-x_3\right)\left(x_1-x_4\right)}=\frac{\left(22-15\right)\left(22-20\right)\left(22-25\right)}{\left(10-15\right)\left(10-20\right)\left(10-25\right)}=0.056.$

   Similar calculations for the other Lagrange polynomials give

   $L_2\left(22\right)=-0.288,L_3\left(22\right)=1.008,L_4\left(22\right)=0.224,$

   and we find that our interpolated polynomial, evaluated at $x=22$ is

   $$\begin{array}{rcll}p\left(22\right)& =& f_1{L}_1\left(22\right)+f_2{L}_2\left(22\right)+f_3{L}_3\left(22\right)+f_4{L}_4\left(22\right)& \\ & =& 9.23\times 0.056+8.41\times -0.288+7.12\times 1.008+4.13\times 0.224& \\ & =& 6.197& \\ & =& 6.20,\text{to 2 decimal places,}& \end{array}$$

   which serves as the approximation to $f\left(22\right)$ .

2. $$\begin{array}{rcll}f\left(2\right)& =& \frac{\left(2-1\right)\left(2-3\right)}{\left(3.5-1\right)\left(3.5-3\right)}\times 342.9+\frac{\left(2-1\right)\left(2-3.5\right)}{\left(3-1\right)\left(3-3.5\right)}\times 295.5+\frac{\left(2-3\right)\left(2-3.5\right)}{\left(1-3\right)\left(1-3.5\right)}\times 99.8& \\ & =& -274.32+443.25+29.94& \\ & =& 198.87& \end{array}$$

   Estimate is $199$ (to $3$ sig. fig.)

3. We tabulate the data for convenience:

   $\begin{array}{ccccc}\hfill \hfill & \hfill x_n\hfill & \hfill f_n\hfill & \hfill x_n^2\hfill & \hfill x_n{f}_n\hfill \\ ̲& ̲& ̲& ̲& ̲\\ \hfill \hfill & \hfill 2\hfill & \hfill 2.2\hfill & \hfill 4\hfill & \hfill 4.4\hfill \\ \hfill \hfill & \hfill 3\hfill & \hfill 5.4\hfill & \hfill 9\hfill & \hfill 16.2\hfill \\ \hfill \hfill & \hfill 5\hfill & \hfill 6.5\hfill & \hfill 25\hfill & \hfill 32.5\hfill \\ \hfill \hfill & \hfill 7\hfill & \hfill 13.2\hfill & \hfill 49\hfill & \hfill 92.4\hfill \\ ̲& ̲& ̲& ̲& ̲\\ \hfill \sum \hfill & \hfill 17\hfill & \hfill 27.3\hfill & \hfill 87\hfill & \hfill 145.5\hfill \end{array}$

   The quantity $\sum 1$ counts the number of data points and in this case is equal to 4.
   It follows that the pair of equations for $m$ and $c$ are as follows:

   $$\begin{array}{rcll}4c+17m& =& 27.3& \\ 17c+87m& =& 145.5& \end{array}$$

   Solving these gives $c=-1.67$ and $m=2.00$ , to 2 decimal places, and we see that the least squares best fit straight line to the given data is

   $y=-1.67+2.00x$

4. In the notation now established we have $h_1=2$ , $h_2=1$ and $h_3=2$ . For a natural cubic spline we require $s^{″}$ to be zero at $x_1$ and $x_4$ . Values of $s^{″}$ at the other data points are found from the system of equations $$\begin{array}{rcll}{h}_1\underset{=0}{\underset{\&UnderBrace;}{s^{″}\left(x_1\right)}}+2\left(h_1+h_2\right)s^{″}\left(x_2\right)+h_2{s}^{″}\left(x_3\right)& =& 6\left(\frac{f_3-f_2}{h_2}-\frac{f_2-f_1}{h_1}\right)& \\ h_2{s}^{″}\left(x_2\right)+2\left(h_2+h_3\right)s^{″}\left(x_3\right)+h_3\underset{=0}{\underset{\&UnderBrace;}{s^{″}\left(x_4\right)}}& =& 6\left(\frac{f_4-f_3}{h_3}-\frac{f_3-f_2}{h_2}\right)& \end{array}$$

   In this case the equations become

   $\left(\begin{array}{cc}\hfill 6\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 6\hfill \end{array}\right)\left(\begin{array}{c}\hfill s^{″}\left(x_2\right)\hfill \\ \hfill s^{″}\left(x_3\right)\hfill \end{array}\right)=\left(\begin{array}{c}\hfill -5.82\hfill \\ \hfill 1.02\hfill \end{array}\right)$

   Solving the coupled pair of equations leads to

   $s^{″}\left(x_2\right)=-1.026857s^{″}\left(x_3\right)=0.341143$

   We now find the coefficients $a_1$ , $b_1$ , etc. from the formulae and deduce that the spline is given by

   $$\begin{array}{rcll}s\left(x\right)& =& -0.08557{\left(x-2\right)}^3+0.592286\left(x-2\right)+1.34\left(2<x<4\right)& \\ s\left(x\right)& =& 0.228{\left(x-4\right)}^3-0.513429{\left(x-4\right)}^2-0.434571\left(x-4\right)+1.84\left(4<x<5\right)& \\ s\left(x\right)& =& -0.02843{\left(x-5\right)}^3+0.170571{\left(x-5\right)}^2-0.777429\left(x-5\right)+1.12\left(5<x<7\right)& \\ & & & \end{array}$$