Chapter One – Elimination (Substitution) Interpolation

A common method for solving the resulting system of equations is using linear algebra and matrix math. However, neither are necessary to illustrate this technique and apply to a practical problem. We will use elimination to solve the example below. While I think it is important students experience how basic algebra works for interpolation, they will quickly see that unless the numbers are small and simple this particular technique quickly becomes unwieldy for large values generated during the process.

For example:

Sample Vehicle Fuel Efficiency Measurements

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Apply expertise and experience to create a polynomial that will reasonably predict the fuel efficiency of the particular vehicle used to gather the above data.

Step one: Deciding that a quadratic equation looks like the best fit, we select the first, second and fourth points to construct a second degree (quadratic) polynomial.

Step Two: Even though the result will be a quadratic equation we are able to use straightforward linear techniques of elimination and substitution. The reason for this is that we are not trying to find x and y. The three points we selected already give us those. Instead, we are trying to create the quadratic in standard form  by solving for the unknown constants a, b and c:

[latex]ax^2 + bx + c = y[/latex]

Step Three: Lets create three quadratic equations with the same three unknowns a, b, c and replacing x, y in each with the actual data point values.

Eq. one:    [latex]a(45)^2 + b(45) + c = 43[/latex] ———–> [latex]2025a + 45b + c = 43[/latex]

Eq. two:    [latex]a(55)^2 + b(55) + c = 42[/latex] ———–> [latex]3025a + 55b + c = 42[/latex]

Eq. three: [latex]a(75)^2 + b(75) + c = 32[/latex]  ———–> [latex]5625a + 75b + c = 32[/latex]

Step Four: The elimination process:

45(Eq. two) – 55( Eq one):

[ [latex]136125a + 2475b + 45c = 1890[/latex] ] – [ [latex]111375a + 2475b + 55c = 2365[/latex] ]

Eq. four:  [latex]24750a - 10c = -475[/latex]   b is eliminated

55(Eq. three) – 75(Eq. two):

[ [latex]309375a + 4125b + 55c = 1760[/latex] ]   –  [ [latex]226875a + 4125b + 75c = 3150[/latex] ]

Eq. five:    [latex]82500a - 20c = -1390[/latex]  b is eliminated

Conduct elimination on the resulting two equations with two unknowns to eliminate c.

2(Eq. four) – Eq. five:

  [latex]49500a - 20c = - 950[/latex]

  [latex]82500a - 20c = -1390[/latex]

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         [latex]-33000a = 440[/latex]

Eq. six:      [latex]a = -0.013[/latex] c is eliminated

Plugging the resulting value of a into Eq. 4 allows us to solve for c:

 [latex]24750(-0.013) - 10c = -475[/latex]

 [latex]c = 14.508[/latex]

Step Five: Substitute a and c into any of the original equations to find b:

 [latex]2025(-0.013) + 45b + 14.508 = 43[/latex]

[latex]b = 1.233[/latex]

Our interpolated polynomial is:

                                                [latex]P(x) = -0.013x^2 + 1.233x + 14.508[/latex]

For students looking for a less manual process here is the setup using matrix math to run the calculations in a spreadsheet.

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Let’s look between 45 and 55 at  [latex]x = 50[/latex] and see how well our polynomial estimates a reasonable value:

                                           [latex]P(50) = -0.013x(50)^2 + 1.233(50) + 14.508[/latex]

It is recommended that the original points also be plugged into the equation as a check.

[latex]P(50) = 43.7[/latex] Plotting on our graph shows that this is indeed a very good estimate.

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