Solutions to Selected Practice Exercises

Solution to Exercise One Practice Problems

Exercise 1a)

ABC Children's Party Company

Maximum children attending the party Cost per Child Total Cost of Party
10 $37 $370
25 $28 $700
50 $22 $1100
100 $15 $1500

The four equations in four unknowns:

[latex]a(10^3) + b(10^2) + c(10) + d = 37[/latex]

[latex]a(25^3) + b(25^2) + c(25) + d = 28[/latex]

[latex]a(50^3) + b(50^2) + c(50) + d = 22[/latex]

[latex]a(100^3) + b(100^2) + c(100) + d = 15[/latex]

 

Equations in Table Form

a b c d cost
1000 100 10 1 37
15625 625 25 1 28
125000 2500 50 1 22
1000000 10000 100 1 15

 

This provides the flow of the solution matrix math used in a spreadsheet.
Figure 8.1 Matrix Setup for Exercise 1a

Long Description

 


Resulting Pricing Polynomial

[latex]cost = (-8.52 * 10^{-5})x^3 + (1.62 * 10^{-2})x^2 + (-1.09)x + (4.63 * 10^1)[/latex]

 

The matrix math and the graph of the solution illustrate the actual equation line compared to the interpolation polynomial l.
Figure 8.2 Exercise 1b

Long Description

 


Solution to Exercise Two Practice Problems

 

2a)

Newton’s Divided Difference Table is populated as follows:

Newton’s Divided Difference Table

x y [latex]b_0[/latex] Linear [latex]b_1(x-10)[/latex] Quadratic [latex]b_2(x-10)(x-25)[/latex] Cubic [latex]b_3(x-10)(x-25)(x-50)[/latex]
10 37 37
[latex]\frac{37-28}{10-25}=-0.6[/latex]
25 28 28 [latex]\frac{-0.6-(-.24)}{-40}=0.009[/latex]
[latex]\frac{28-22}{25-50}=-0.24[/latex] [latex]\frac{0.009-0.001}{-90}=0.0000888[/latex]
50 22 22 [latex]\frac{-0.24-(-0.14)}{-75}=0.001[/latex]
[latex]\frac{22-15}{50-100}=-0.14[/latex]
100 15 15

 

Simplifies to: [latex]-0.0000888x^3 + 0.016548x^2 - 1.0926x + 46.36[/latex]

2b)

2b Table

x y or f(x)
-6.2 -8
-3 -7
-1.5 -2.2
1 0.7
3.5 3
4.25 5
7.9 8

 

2b Difference Table

x f(x) 1st Divided Difference 2nd Divided Difference
- [latex]b_0[/latex] [latex]b_1(x - x_0)[/latex] [latex]b_2(x - x_0)(x - x_1)[/latex]
-3 -7 - -
- - [latex]\large \frac {-7 - 0.7}{-3 - 1} = \frac {-7.7}{-4} = 1.925[/latex] -
1 0.7 - [latex]\large \frac {1.925 - 1.493}{-3 - 7.9} = \frac {0.432}{-10.9} = -0.040[/latex]
- - [latex]\large \frac {0.7 - 11}{1 - 7.9} = \frac {-10.3}{-6.9} = 1.493[/latex] -
7.9 11 - -

 

Simplifies to: -0.040x2 + 1.845x – 1.105


Solution to Chapter Three Practice Exercises

Exercise 3b)

 

The flow of the matrix multiplication and the graph compare the measured velocity to the Direct Interpolation demonstrating that direct method interpolation provides an excellent estimate within the interval of the four closest points. Note the dotted line is the cubic polynomial generated by the graphing app in the spreadsheet.
Figure 8.3 Exercise 3b

Long Description

 


Solution to Chapter Four Practice Exercise

4a)

The Setup:

Abbreviated List of weekly Dow Jones closing averages:

Weekly Closing Averages

Week Actual [latex]a_1x^3[/latex] [latex]a_2x^2[/latex] [latex]a_3x[/latex] [latex]a_4[/latex] Interpolation
1 28,583.68 1.00 1.00 1.00 1 28,416.89149
2 28,939.67 8.00 4.00 2.00 1 28,034.20169
3 29,196.04 27.00 9.00 3.00 1 27,677.60694
4 28,722.85 64.00 16.00 4.00 1 27,346.5493
- - - - - - -
- - - - - - -
- - - - - - -
78 34,292.29 474,552.00 6,084.00 78.00 1 34,491.0287
79 34,577.37 493,039.00 6,241.00 79.00 1 34,485.14307
80 34,888.79 512,000.00 6,400.00 80.00 1 34,462.39157
81 34,511.99 531,441.00 6,561.00 81.00 1 34,422.21628
82 35,058.52 551,368.00 6.724.00 82.00 1 34,364.05925
83 35,084.53 571,787.00 6,889.00 83.00 1 34,287.36256

 

 

This matrix math solution to practice problem 4a).
Figure 8.4 Matrix Solution 4a

Long Description

 

 

The graph illustrates the results of the interpolation polynomial compared to the actual Dow Jones closing averages for between January 2020 and July 2021.
Figure 8.5 Graph of Weekly DJIA

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:  [latex]R^2 = 0.882707285 \approxeq 88.3\%[/latex]


Solution to Chapter Six Practice Exercises

6a)

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

Derivatives of cos

[latex]f(0) = cos (0) = 1[/latex]

[latex]f^{(1)}(0) = -sin (0) = 0[/latex]

[latex]f^{(2)}(0) = -cos (0) = -1[/latex]

[latex]f^{(3)}(0) = sin (0) = 0[/latex]

[latex]f^{(4)}(0) = cos (0) = 1[/latex]

[latex]f^{(5)}(0) = -sin (0) = 0[/latex]

[latex]f^{(6)}(0) = -cos (0) = -1[/latex]

[latex]f^{(7)}(0) = sin (0) = 0[/latex]

[latex]f^{(8)}(0) = cos (0) = 1[/latex]

[latex]f^{(9)}(0) = -sin (0) = 0[/latex]

Plug derivatives into the general form of the Taylor polynomial:

 

[latex]\large 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[/latex]

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.

 

[latex]\large 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[/latex]

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

 

Points students to the table of data encouraging them to graph a solution using their favorite app.
Figure 8.6 Speech Bubble

Taylor Approximation for Cosine

Degrees Radians f(x) p(x)
0 0 1.000000000000000 1.000000000000000
18 [latex]\Large \frac {\pi}{10}[/latex] 0.951056516295154 0.951056516297732
22.5 [latex]\Large \frac {\pi}{8}[/latex] 0.923879532511287 0.923879532535293
30 [latex]\Large \frac {\pi}{6}[/latex] 0.866025403784439 0.866025404210352
45 [latex]\Large \frac {\pi}{4}[/latex] 0.707106781186548 0.707106805683294
72 [latex]\Large \frac {2\pi}{5}[/latex] 0.309016994374947 0.309019668329804
90 [latex]\Large \frac {\pi}{2}[/latex] 0.000000000000000  0.000000000000000


Solution to Chapter Seven Practice Exercise

7a)

 

[latex]\large r_n(x) = \frac {f^{(n+1)}(c)}{(n+1)!}(x - a)^{n+1}[/latex] where c is between a and x

Solving the remainder twice for 0 and [latex]\large \frac {\pi}{10}[/latex]

This will provide a range of possible values between 0 and [latex]\large \frac {\pi}{10}[/latex]

[latex]\large f^{(9)}(c) = -sin(c)[/latex]

for [latex]\large c = 0[/latex]     [latex]\large r_8(0) = \frac {-sin(0)}{(9)!}(0 - 0)^9 = 0[/latex]

for [latex]\large c = \frac {\pi}{10}[/latex]       [latex]\large r_8(\frac {\pi}{10}) = \frac {-sin(\frac {\pi}{10})}{(9)!}(\frac {\pi}{10} - 0)^9 = -0.000000000025384[/latex]

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

Gives us an error possibility [latex]\large 0 \le r_{10} \le 0.000000000025384[/latex]   [latex](2.54 * 10^{-11})[/latex]

 

 

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The Art of Polynomial Interpolation Copyright © 2022 by Stuart Murphy is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License, except where otherwise noted.

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