Chapter 18, Problem 14P

Develop cubic splines for the data in Prob. 18.6 and (a) predict f (4) and f (2.5) and (b) verify that $f_2\left(3\right)$ and $f_3\left(3\right)=19$.

To determine

To calculate: The value of $f\left(4\right)$, and $f\left(2.5\right)$ by developing the cubic splines for the data points mentioned Prob 18.6.

Answer to Problem 14P

Solution:

$f_3\left(4\right)=48.41157$ and $f_2\left(2.5\right)=10.78398$

Explanation of Solution

Given Information:

Write the data points mentioned in the Prob. 18.6.

$\text{x}$	1	2	3	5	7	8
$f\left(\text{x}\right)$	3	6	19	99	291	444

Formula Used:

A third order polynomial is derived for each interval between data points in cubic splines.

Write the expression of polynomial for each interval from Eq.18.36.

$f_i\left(x\right)=\left(\begin{array}{l}\frac{{f^{″}}_i\left(x_{i-1}\right){\left(x_{í}-x\right)}^3}{6\left(x_{í}-x_{i-1}\right)}+\frac{{f^{″}}_i\left(x_i\right)}{6\left(x_{í}-x_{i-1}\right)}\times {\left(x-x_{i-1}\right)}^3\\ +\left[\frac{f\left(x_{i-1}\right)}{\left(x_{í}-x_{i-1}\right)}-\frac{f^{″}\left(x_{i-1}\right)\times \left(x_{í}-x_{i-1}\right)}{6}\right]\times \left(x_{í}-x\right)\\ +\left[\frac{f\left(x_i\right)}{\left(x_{í}-x_{i-1}\right)}-\frac{f^{″}\left(x_i\right)\times \left(x_{í}-x_{i-1}\right)}{6}\right]\times \left(x-x_{i-1}\right)\end{array}\right)$

Here, 4n conditions are required to evaluate 4n unknowns.

Write the equations to calculate the unknowns,

$\left(\begin{array}{l}\left(x_{í}-x_{i-1}\right)f^{″}\left(x_{i-1}\right)+2\left(x_{i+1}-x_{i-1}\right)f^{″}\left(x_i\right)\\ +\left(x_{i+1}-x_i\right)f^{″}\left(x_{i+1}\right)\end{array}\right)=\left(\frac{6\left[f\left(x_{i+1}\right)-f\left(x_i\right)\right]}{\left(x_{i+1}-x_i\right)}+\frac{6\left[f\left(x_{i-1}\right)-f\left(x_i\right)\right]}{\left(x_i-x_{i-1}\right)}\right)$

Calculation:

Recall the table mentioned in the Prob. 18.6.

$\text{x}$	1	2	3	5	7	8
$f\left(\text{x}\right)$	3	6	19	99	291	444

From the table the value is,

$\begin{array}{c}\text{For }{x}_0=1,f\left(x_0\right)=3\\ \text{For }{x}_1=2,f\left(x_1\right)=6\\ \text{For }{x}_2=3,f\left(x_2\right)=19\\ \text{For }{x}_3=5,f\left(x_3\right)=99\\ \text{For }{x}_5=7,f\left(x_5\right)=291\end{array}$

Recall the equations to calculate the unknowns,

$\left(\begin{array}{l}\left(x_{í}-x_{i-1}\right)f^{″}\left(x_{i-1}\right)+2\left(x_{i+1}-x_{i-1}\right)f^{″}\left(x_i\right)\\ +\left(x_{i+1}-x_i\right)f^{″}\left(x_{i+1}\right)\end{array}\right)=\left(\begin{array}{l}\frac{6\left[f\left(x_{i+1}\right)-f\left(x_i\right)\right]}{\left(x_{i+1}-x_i\right)}\\ +\frac{6\left[f\left(x_{i-1}\right)-f\left(x_i\right)\right]}{\left(x_i-x_{i-1}\right)}\end{array}\right)$ …… (1)

Write the equation(1) for $i=1$.

$\left(\begin{array}{l}\left(x_1-x_0\right)f^{″}\left(x_0\right)+2\left(x_2-x_0\right)f^{″}\left(x_1\right)\\ +\left(x_2-x_1\right)f^{″}\left(x_2\right)\end{array}\right)=\left(\frac{6\left[f\left(x_2\right)-f\left(x_1\right)\right]}{\left(x_2-x_1\right)}+\frac{6\left[f\left(x_0\right)-f\left(x_1\right)\right]}{\left(x_1-x_0\right)}\right)$

Substitute all the above values.

$\left[\left(2-1\right)f^{″}\left(1\right)+2\left(3-1\right)f^{″}\left(2\right)+\left(3-2\right)f^{″}\left(3\right)\right]=\left(\frac{6\times \left[19-6\right]}{\left(3-2\right)}+\frac{6\times \left[3-6\right]}{\left(2-1\right)}\right)$

Because of natural spline condition $f^{″}\left(1\right)=0$ and $f^{″}\left(8\right)=0$ thus,

$4f^{″}\left(2\right)+f^{″}\left(3\right)=60$

Similarly, equation(1) can be write for all interior knot $i=2,3,5,7$.

$f^{″}\left(2\right)+6f^{″}\left(3\right)+2f^{″}\left(5\right)=162$

$2f^{″}\left(3\right)+8f^{″}\left(5\right)+2f^{″}\left(7\right)=336$

$2f^{″}\left(5\right)+6f^{″}\left(7\right)=342$

Write all the above equations in matrix form.

$\left[\begin{array}{cccc}4& 1& 0& 0\\ 1& 6& 2& 0\\ 0& 2& 8& 2\\ 0& 0& 2& 6\end{array}\right]\left[\begin{array}{c}{f}^{″}\left(2\right)\\ f^{″}\left(3\right)\\ f^{″}\left(5\right)\\ f^{″}\left(7\right)\end{array}\right]=\left[\begin{array}{c}60\\ 162\\ 336\\ 342\end{array}\right]$

Solve the equations by using the MATLAB,

Write the following codes in MATLAB.

A=zeros(4,4);

A(1,1)=4;

A(1,2)=1;

A(2,1)=1;

A(2,2)=6;

A(2,3)=2;

A(3,2)=2;

A(3,3)=8;

A(3,4)=2;

A(3,5)=1;

A(4,3)=2;

A(4,4)=6;

%

B= [60;162;336;342];

Sol = A\B

The values are,

$\begin{array}{l}{f}^{″}\left(2\right)=10.84716\\ f^{″}\left(3\right)=16.61135\\ f^{″}\left(5\right)=25.74236\\ f^{″}\left(7\right)=48.41921\end{array}$

Recall the polynomial expression for each interval,

$f_i\left(x\right)=\left(\begin{array}{l}\frac{{f^{″}}_i\left(x_{i-1}\right){\left(x_{í}-x\right)}^3}{6\left(x_{í}-x_{i-1}\right)}+\frac{{f^{″}}_i\left(x_i\right)}{6\left(x_{í}-x_{i-1}\right)}\times {\left(x-x_{i-1}\right)}^3\\ +\left[\frac{f\left(x_{i-1}\right)}{\left(x_{í}-x_{i-1}\right)}-\frac{f^{″}\left(x_{i-1}\right)\times \left(x_{í}-x_{i-1}\right)}{6}\right]\times \left(x_{í}-x\right)\\ +\left[\frac{f\left(x_i\right)}{\left(x_{í}-x_{i-1}\right)}-\frac{f^{″}\left(x_i\right)\times \left(x_{í}-x_{i-1}\right)}{6}\right]\times \left(x-x_{i-1}\right)\end{array}\right)$ …… (2)

Write the equation (2) for $i=1$.

$f_1\left(x\right)=\left(\begin{array}{l}\frac{{f^{″}}_1\left(x_0\right){\left(x_1-x\right)}^3}{6\left(x_1-x_0\right)}+\frac{{f^{″}}_1\left(x_1\right)}{6\left(x_1-x_0\right)}\times {\left(x-x_0\right)}^3\\ +\left[\frac{f\left(x_0\right)}{\left(x_1-x_0\right)}-\frac{f^{″}\left(x_0\right)\times \left(x_1-x_0\right)}{6}\right]\times \left(x_1-x\right)\\ +\left[\frac{f\left(x_1\right)}{\left(x_1-x_0\right)}-\frac{f^{″}\left(x_1\right)\times \left(x_1-x_0\right)}{6}\right]\times \left(x-x_0\right)\end{array}\right)$

Substitute the all values.

$f_1\left(x\right)=\left(\begin{array}{l}\frac{{f^{″}}_1\left(1\right){\left(2-x\right)}^3}{6\left(2-1\right)}+\frac{{f^{″}}_1\left(2\right)}{6\left(2-1\right)}\times {\left(x-1\right)}^3\\ +\left[\frac{f\left(1\right)}{\left(2-1\right)}-\frac{f^{″}\left(1\right)\times \left(2-1\right)}{6}\right]\times \left(2-x\right)\\ +\left[\frac{f\left(2\right)}{\left(2-1\right)}-\frac{f^{″}\left(2\right)\times \left(2-1\right)}{6}\right]\times \left(x-1\right)\end{array}\right)$

Again, substitute all the values.

$\begin{array}{c}{f}_1\left(x\right)=\left(\begin{array}{c}\frac{\left(0\right){\left(2-x\right)}^3}{6\left(2-1\right)}+\frac{\left(10.84716\right)}{6\left(2-1\right)}\times {\left(x-1\right)}^3\\ +\left[\frac{3}{\left(2-1\right)}-\frac{\left(0\right)\times \left(2-1\right)}{6}\right]\times \left(2-x\right)\\ +\left[\frac{6}{\left(2-1\right)}-\frac{\left(10.84716\right)\times \left(2-1\right)}{6}\right]\times \left(x-1\right)\end{array}\right)\\ =\frac{\left(10.84716\right)}{6}{\left(x-1\right)}^3+\left[3\right]\left(2-x\right)+\left[6-\frac{\left(10.84716\right)}{6}\right]\left(x-1\right)\\ =1.80786{\left(x-1\right)}^3+3\left(2-x\right)+4.19241\left(x-1\right)\end{array}$

Similarly solve for the $i=2,3,5,7$.

$f_2\left(x\right)=1.80786{\left(3-x\right)}^3+2.768559{\left(x-2\right)}^3+4.19241\left(3-x\right)+16.23144\left(x-2\right)$

$f_3\left(x\right)=1.384279{\left(5-x\right)}^3+2.145197{\left(x-3\right)}^3+3.962882\left(5-x\right)+40.91921\left(x-3\right)$

$f_5\left(x\right)=2.145179{\left(7-x\right)}^3+4.034934{\left(x-5\right)}^3+40.91921\left(7-x\right)+129.3603\left(x-3\right)$

$f_7\left(x\right)=8.069869{\left(8-x\right)}^3+282.9301\left(8-x\right)+444\left(x-7\right)$

Calculate the value of $f\left(4\right)$.

Substitute $4$ for $x$ in $f_3\left(x\right)$.

$\begin{array}{c}{f}_3\left(4\right)=1.384279{\left(5-x\right)}^3+2.145197{\left(4-3\right)}^3+3.962882\left(5-4\right)+40.91921\left(4-3\right)\\ =1.384279+2.145197+3.962882+40.91921\\ =48.41157\end{array}$

Calculate the value of $f\left(2.5\right)$.

Substitute $2.5$ for $x$ in $f_2\left(x\right)$.

$\begin{array}{c}{f}_2\left(2.5\right)=1.80786{\left(3-2.5\right)}^3+2.768559{\left(2.5-2\right)}^3+4.19241\left(3-2.5\right)+16.23144\left(2.5-2\right)\\ =1.80786{\left(0.5\right)}^3+2.768559{\left(0.5\right)}^3+4.19241\left(0.5\right)+16.23144\left(0.5\right)\\ =10.78398\end{array}$

Therefore, the value $f\left(4\right)=48.41157$ and $f\left(2.5\right)=10.78398$.

To determine

To show: The value of $f_2\left(3\right)$, and $f_3\left(3\right)=19$.

Answer to Problem 14P

Solution:

$f_2\left(3\right)=f_3\left(3\right)=19$ verified.

Explanation of Solution

Given Information:

$f_2\left(x\right)=1.80786{\left(3-x\right)}^3+2.768559{\left(x-2\right)}^3+4.19241\left(3-x\right)+16.23144\left(x-2\right)$

$f_3\left(x\right)=1.384279{\left(5-x\right)}^3+2.145197{\left(x-3\right)}^3+3.962882\left(5-x\right)+40.91921\left(x-3\right)$

Calculation:

Calculate the value of $f_2\left(3\right)$.

Recall the expression of $f_2\left(x\right)$.

$f_2\left(x\right)=1.80786{\left(3-x\right)}^3+2.768559{\left(x-2\right)}^3+4.19241\left(3-x\right)+16.23144\left(x-2\right)$

Substitute $3$ for $x$ in $f_2\left(x\right)$.

$\begin{array}{c}{f}_2\left(3\right)=1.80786{\left(3-3\right)}^3+2.768559{\left(3-2\right)}^3+4.19241\left(3-3\right)+16.23144\left(3-2\right)\\ =2.768559+16.23144\\ =19\end{array}$

Calculate the value of $f_3\left(3\right)$.

Substitute $3$ for $x$ in $f_3\left(x\right)$.

$\begin{array}{c}{f}_3\left(x\right)=1.384279{\left(5-3\right)}^3+2.145197{\left(3-3\right)}^3+3.962882\left(5-3\right)+40.91921\left(3-3\right)\\ =1.384279{\left(2\right)}^3+3.962882\left(2\right)\\ =19\end{array}$

Thus, $f_2\left(3\right)=f_3\left(3\right)=19$ proved.