Solutions to Selected Practice Exercises

Solution to Exercise One Practice Problems

Exercise 1a)

ABC Children's Party Company

The four equations in four unknowns:

$a({10}^3)+b({10}^2)+c(10)+d=37$

$a({25}^3)+b({25}^2)+c(25)+d=28$

$a({50}^3)+b({50}^2)+c(50)+d=22$

$a({100}^3)+b({100}^2)+c(100)+d=15$

Equations in Table Form

Long Description

Resulting Pricing Polynomial

$cost=(-8.52\ast {10}^{-5})x^3+(1.62\ast {10}^{-2})x^2+(-1.09)x+(4.63\ast {10}^1)$

Long Description

Solution to Exercise Two Practice Problems

2a)

Newton’s Divided Difference Table is populated as follows:

Newton’s Divided Difference Table

Simplifies to: $-0.0000888x^3+0.016548x^2-1.0926x+46.36$

2b)

2b Table

2b Difference Table

Simplifies to: -0.040x² + 1.845x – 1.105

Solution to Chapter Three Practice Exercises

Exercise 3b)

Long Description

Solution to Chapter Four Practice Exercise

4a)

The Setup:

Abbreviated List of weekly Dow Jones closing averages:

Weekly Closing Averages

Long Description

Long Description

Solution to Chapter Five Practice Exercises

Step One:

1a) Find the difference between each actual value and its associated value generated by the interpolative polynomial. Square the result.

1b) Find the difference between each actual value and the Mean of the actual values. Square the result.

Step Two:

2a) Sum the results from 1a

2b) Sum the results from 1b

Step Three:

Divide 2a by 2b subtracting the result from 1.

Answer:  $R^2=0.882707285\approxeq 88.3\mathrm{\%}$

Solution to Chapter Six Practice Exercises

6a)

Select the Function to be approximated. Cos function centered at x=0

Derivatives of cos

$f(0)=cos(0)=1$

$f^{(1)}(0)=-sin(0)=0$

$f^{(2)}(0)=-cos(0)=-1$

$f^{(3)}(0)=sin(0)=0$

$f^{(4)}(0)=cos(0)=1$

$f^{(5)}(0)=-sin(0)=0$

$f^{(6)}(0)=-cos(0)=-1$

$f^{(7)}(0)=sin(0)=0$

$f^{(8)}(0)=cos(0)=1$

$f^{(9)}(0)=-sin(0)=0$

Plug derivatives into the general form of the Taylor polynomial:

$p(0)=\frac{1}{0!}+\frac{0}{1!}(x-a)+\frac{-1}{2!}(x-a{)}^2+\frac{0}{3!}(x-a{)}^3+\frac{1}{4!}(x-a{)}^4+\frac{0}{5!}(x-a{)}^5+\frac{-1}{6!}(x-a{)}^6+\frac{0}{7!}(x-a{)}^7+\frac{1}{8!}(x-a{)}^8+\frac{0}{9!}(x-a{)}^9$

Every other term has zero in the numerator so we can drop these and condense p(0). Further since a = 0 we can simplify the binomials.

$p(0)=\frac{1}{0!}+\frac{-1}{2!}(x{)}^2+\frac{1}{4!}(x{)}^4+\frac{-1}{6!}(x{)}^6+\frac{1}{8!}(x{)}^8$

f(x) is generated from an app precise to 15 decimal positions. p(x) is the Taylor approximation for Cosine.

Taylor Approximation for Cosine

Solution to Chapter Seven Practice Exercise

7a)

$r_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a{)}^{n+1}$ where c is between a and x

Solving the remainder twice for 0 and $\frac{\pi }{10}$

This will provide a range of possible values between 0 and $\frac{\pi }{10}$

$f^{(9)}(c)=-sin(c)$

for $c=0$     $r_8(0)=\frac{-sin(0)}{(9)!}(0-0{)}^9=0$

for $c=\frac{\pi }{10}$       $r_8(\frac{\pi }{10})=\frac{-sin(\frac{\pi }{10})}{(9)!}(\frac{\pi }{10}-0{)}^9=-0.000000000025384$

Drop negative as it is a matter of distance not direction.

Gives us an error possibility $0\le r_{10}\le 0.000000000025384$   $(2.54\ast {10}^{-11})$