Chapter 2, Problem 9TP

Solution

A polynomial function that fits the data in each table, using finite differences and a system of equations; and $f\left(x\right)+g\left(x\right)$ .

It has been determined that $f\left(x\right)=\frac{1}{3}{x}^3+4x^2-\frac{28}{3}x+9$ and $g\left(x\right)=2x^3-5x^2+x$ .

It has further been evaluated that $f\left(x\right)+g\left(x\right)=\frac{7}{3}{x}^3-x^2-\frac{25}{3}x+9$ .

Given:

The following tables:

  $\begin{array}{l}\begin{array}{ccccccc}x& 1& 2& 3& 4& 5& 6\\ f\left(x\right)& 4& 9& 26& 57& 104& 169\end{array}\\ \begin{array}{ccccccc}x& 1& 2& 3& 4& 5& 6\\ g\left(x\right)& -2& -2& 12& 52& 130& 258\end{array}\end{array}$

Concept used:

The degree of a polynomial that fits a given data set can be obtained by repeatedly analyzing the differences between the data values corresponding to equally spaced values of the independent variable.

Calculation:

Repeatedly calculating the difference until the differences are constant, for the first table, it follows that:

$x$	$f\left(x\right)$	$\Delta f$	${\Delta }^{2}f$	${\Delta }^{3}f$
1	4	$9-4=5$	$17-5=12$	$14-12=2$
2	9	$26-9=17$	$31-17=14$	$16-14=2$
3	26	$57-26=31$	$47-31=16$	$18-16=2$
4	57	$104-57=47$	$65-47=18$
5	104	$169-104=65$
6	169

It can be seen from the above table that constant finite difference is obtained after three rounds of finding differences.

This implies that $f\left(x\right)$ is a polynomial of degree $3$ .

Then, it can be written that,

  $f\left(x\right)=a{x}^3+b{x}^2+cx+d$

From the given table,

  $f\left(1\right)=4$ ,

  $f\left(2\right)=9$ ,

  $f\left(3\right)=26$ ,

  $f\left(4\right)=57$

Put $x=1,2,3,4$ in $f\left(x\right)=a{x}^3+b{x}^2+cx+d$ to get,

  $f\left(1\right)=a+b+c+d$ ,

  $f\left(2\right)=8a+4b+2c+d$ ,

  $f\left(3\right)=27a+9b+3c+d$ ,

  $f\left(4\right)=64a+16b+4c+d$

Put the previously obtained values in these expressions to get,

  $a+b+c+d=4$ … (1)

  $8a+4b+2c+d=9$ … (2)

  $27a+9b+3c+d=26$ … (3)

  $64a+16b+4c+d=57$ … (4)

Subtracting equation (1) from equation (2),

  $7a+3b+c=5$ … (5)

Subtracting equation (2) from equation (3),

  $19a+5b+c=17$ … (6)

Subtracting equation (3) from equation (4),

  $37a+7b+c=31$ … (7)

Subtracting equation (5) from equation (6),

  $12a+2b=12$ … (8)

Subtracting equation (6) from equation (7),

  $18a+2b=14$ … (9)

Subtracting equation (8) from equation (9),

  $6a=2$

Simplifying,

  $a=\frac{1}{3}$

Put $a=\frac{1}{3}$ in equation (8) to get,

  $4+2b=12$

Simplifying,

  $2b=8$

On further simplification,

  $b=4$

Put $a=\frac{1}{3}$ and $b=4$ in equation (5) to get,

  $\frac{7}{3}+12+c=5$

Simplifying,

  $c=5-\frac{7}{3}-12$

On further simplification,

  $c=-\frac{28}{3}$

Put $a=\frac{1}{3}$ , $b=4$ and $c=-\frac{28}{3}$ in equation (1) to get,

  $\frac{1}{3}+4-\frac{28}{3}+d=4$

Simplifying,

  $d=\frac{28}{3}-\frac{1}{3}$

On further simplification,

  $d=9$

Put $a=\frac{1}{3}$ , $b=4$ , $c=-\frac{28}{3}$ and $d=9$ in $f\left(x\right)=a{x}^3+b{x}^2+cx+d$ to get,

  $f\left(x\right)=\frac{1}{3}{x}^3+4x^2-\frac{28}{3}x+9$

This is the required polynomial function.

Similarly, repeatedly calculating the difference until the differences are constant, for the second table, it follows that:

$x$	$g\left(x\right)$	$\Delta g$	${\Delta }^{2}g$	${\Delta }^{3}g$
1	-2	$-2-\left(-2\right)=0$	$14-0=14$	$26-14=12$
2	-2	$12-\left(-2\right)=14$	$40-14=26$	$38-26=12$
3	12	$52-12=40$	$78-40=38$	$50-38=12$
4	52	$130-52=78$	$128-78=50$
5	130	$258-130=128$
6	258

It can be seen from the above table that constant finite difference is obtained after three rounds of finding differences.

This implies that $g\left(x\right)$ is a polynomial of degree $3$ .

Then, it can be written that,

  $g\left(x\right)=a{x}^3+b{x}^2+cx+d$

From the given table,

  $g\left(1\right)=-2$ ,

  $g\left(2\right)=-2$ ,

  $g\left(3\right)=12$ ,

  $g\left(4\right)=52$

Put $x=1,2,3,4$ in $g\left(x\right)=a{x}^3+b{x}^2+cx+d$ to get,

  $g\left(1\right)=a+b+c+d$ ,

  $g\left(2\right)=8a+4b+2c+d$ ,

  $g\left(3\right)=27a+9b+3c+d$ ,

  $g\left(4\right)=64a+16b+4c+d$

Put the previously obtained values in these expressions to get,

  $a+b+c+d=-2$ … (10)

  $8a+4b+2c+d=-2$ … (11)

  $27a+9b+3c+d=12$ … (12)

  $64a+16b+4c+d=52$ … (13)

Subtracting equation (10) from equation (11),

  $7a+3b+c=0$ … (14)

Subtracting equation (11) from equation (12),

  $19a+5b+c=14$ … (15)

Subtracting equation (12) from equation (13),

  $37a+7b+c=40$ … (16)

Subtracting equation (14) from equation (15),

  $12a+2b=14$ … (17)

Subtracting equation (15) from equation (16),

  $18a+2b=26$ … (18)

Subtracting equation (17) from equation (18),

  $6a=12$

Simplifying,

  $a=2$

Put $a=2$ in equation (17) to get,

  $24+2b=14$

Simplifying,

  $2b=-10$

On further simplification,

  $b=-5$

Put $a=2$ and $b=-5$ in equation (14) to get,

  $14-15+c=0$

Simplifying,

  $c=15-14$

On further simplification,

  $c=1$

Put $a=2$ , $b=-5$ and $c=1$ in equation (10) to get,

  $2-5+1+d=-2$

Simplifying,

  $d=-2-2+5-1$

On further simplification,

  $d=0$

Put $a=2$ , $b=-5$ , $c=1$ and $d=0$ in $g\left(x\right)=a{x}^3+b{x}^2+cx+d$ to get,

  $g\left(x\right)=2x^3-5x^2+x$

This is the required polynomial function.

It has been determined that $f\left(x\right)=\frac{1}{3}{x}^3+4x^2-\frac{28}{3}x+9$ and $g\left(x\right)=2x^3-5x^2+x$ .

Then,

  $\begin{array}{c}f\left(x\right)+g\left(x\right)=\left(\frac{1}{3}{x}^3+4x^2-\frac{28}{3}x+9\right)+\left(2x^3-5x^2+x\right)\\ =\left(\frac{1}{3}+2\right)x^3+\left(4-5\right)x^2+\left(-\frac{28}{3}+1\right)x+9\\ =\left(\frac{1+6}{3}\right)x^3+\left(-1\right)x^2+\left(\frac{-28+3}{3}\right)x+9\\ =\frac{7}{3}{x}^3-x^2-\frac{25}{3}x+9\end{array}$

So, $f\left(x\right)+g\left(x\right)=\frac{7}{3}{x}^3-x^2-\frac{25}{3}x+9$ .

This is the required expression.

Conclusion:

It has been determined that $f\left(x\right)=\frac{1}{3}{x}^3+4x^2-\frac{28}{3}x+9$ and $g\left(x\right)=2x^3-5x^2+x$ .

It has further been evaluated that $f\left(x\right)+g\left(x\right)=\frac{7}{3}{x}^3-x^2-\frac{25}{3}x+9$ .