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| 10 Theory slides |
| 8 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
Consider the following table of values.
x | f(x) |
---|---|
1 | 2 |
2 | 3 |
3 | 7 |
4 | 16 |
5 | 32 |
6 | 57 |
Looking at the x-values, it can be noted that the values are equally spaced. It is possible to find the first finite differences by subtracting two consecutive values. Consider the following example. f(2) - f(1) = 3 - 2 ⇓ f(2) - f(1) = 1
The first differences for each pair of consecutive terms can be found by doing the same process. Then, the second differences are found by subtracting consecutive first differences, third differences by subtracting consecutive second differences, and so on. Find every first, second, and third difference. Is there something that can be noted about the differences?As part of an entrepreneur unit in school, Zosia is learning some interesting math. During her lesson, Zosia's teacher explained that modeling data is useful to understanding business. He explained that there are various methods to write a polynomial function that models data.
As an example, the teacher told the students about a certain company that has experienced ups and downs in earnings due to economic downturns.
The teacher showed the students the following graph, where y represents the profit, in millions of dollars, and x represents the years before and after 2000. Here, negative values of y represent losses.
Notice that three of the points are x-intercepts.
x= 0, f(x)= 7
Identity Property of Addition
Multiply
.LHS /42.=.RHS /42.
Simplify quotient
Rearrange equation
Sometimes it can be useful to examine the differences between function values when the input values are separated by a known quantity. These differences are known as finite differences.
A finite difference is a mathematical expression that can be written in the following form.
f(x+c_1)-f(x+c_2)
In this expression, c_1 and c_2 are constants. There are three basic finite differences that are especially useful.
These differences are shown below.
Finite differences can be used to determine the degree of a polynomial function that models a data set. This is because finite differences have some really useful properties.
The finite differences of polynomial functions satisfy the following two properties.
To justify these properties, the minor cases will be examined first.
Substitute expressions
Distribute -1
(a+b)^2=a^2+2ab+b^2
Distribute a
Distribute b
Subtract terms
Distribute -1
Subtract terms
Substitute expressions
(a+b)^2=a^2+2ab+b^2
Distribute a
Distribute b
Distribute -2
Subtract terms
It can be concluded from the previous cases that, for a polynomial of degree n, its nth differences are the last constant non-zero differences. Therefore, the nth differences of a polynomial of degree n are constant and nonzero. Note that this is an informal justification and should not be taken as a proof.
Interested about Zosia's recent entrepreneur lessons, her uncle told her about his own experience. Some years ago, Zosia's uncle started a local business selling art supplies.
Her uncle showed Zosia how his sales went the first few years of business. At the end of the first year, he was $1000 behind because of a starting loan. Then he started seeing returns.
Year | 1 | 2 | 3 | 4 | 5 |
---|---|---|---|---|---|
Income | -$1000 | $1000 | $6000 | $14 000 | $25 000 |
Recalling her math lessons, Zosia decided to help her uncle by modeling the income with a polynomial function. Help Zosia do the following.
x | 1 | 2 | 3 | 4 | 5 |
---|---|---|---|---|---|
y | -1 | 1 | 6 | 14 | 25 |
By the Properties of Finite Differences, the constant differences can help determine the degree of the polynomial that best fits the data. The first differences are shown below.
The first differences are not constant. This means that the polynomial is not of degree 1 and that other differences need to be examined. Now it is time to look at the second differences.
Looking at the differences, it can be noted that the second differences are constant and non-zero. By the Properties of Finite Differences, the degree of the polynomial is 2.
To find a polynomial function of degree 2 to fit the data, the quadratic regression function of the calculator will be used. To view this function, press STAT, scroll to the right to view the CALC options, and choose the fifth option in the list, QuadReg.
This function is similar to a line of best fit function, but instead of fitting the data with a line, the quadratic regressions fits the data using a parabola. Using the information given, it is possible to write the function. y = 1.5x^2 - 2.5x Remember that at the beginning the numbers were divided by 1000. Therefore, to obtain the real income, each coefficient must be multiplied by 1000. y = 1500x^2 - 2500x It should be noted that the function can also be found by setting and solving a system of equations, where the coefficients are the variables. This method is explained later in the lesson.
Indicate the degree of the polynomial function that best fits the data in each table of values.
When given a set of points, it is possible to determine a polynomial that perfectly fits these points. To do so, it is important to consider a principle called the (n+1) Point Principle.
Given a set of n + 1 points in the coordinate plane, where no pair of points lies on a vertical line, there is a unique polynomial of degree at most n that fits the points perfectly.
To justify this principle, the smaller cases will be examined to come to a conclusion.
In general, n+1 points generate a system of n+1 equations with n+1 variables, where each variable is a coefficient of a polynomial of degree n. y_1 = a_1 x_1^n + a_2 x_1^(n-1) + ⋯ + a_(n-1)x_1 + a_n y_2 = a_1 x_2^n + a_2 x_2^(n-1) + ⋯ + a_(n-1)x_2 + a_n ... y_n = a_1 x_n^n + a_2 x_n^(n-1) + ⋯ + a_(n-1)x_n + a_n A solution can result in a polynomial with a degree less than n if the leading coefficient a_1 is zero. It should be noted that this is an informal justification and should not be taken as a proof.
After seeing her enthusiasm about his experience, Zosia's uncle invited her out to visit and get some first hand experience with how a business works. Nervous about her first flight, Zosia watched a documentary about airplanes and their speeds at takeoff.
In the documentary, the horizontal distance traveled by the plane during the first few minutes after takeoff was analyzed.
Zosia thought she could model the distances by finding a polynomial that passes through the following four points. (1,5), (2,9) (3,15), (4,24) Here, the x-coordinates represent the minutes after takeoff and the y-coordinates represent the horizontal distance between the plane and the airport.
y = ax^3 + bx^2 + cx + d To find the values of the coefficients, the points are substituted into the equation above to generate a system of four equations. 5 = a(1)^3 + b(1)^2 + c(1) + d 9 = a(2)^3 + b(2)^2 + c(2) + d 15 = a(3)^3 + b(3)^2 + c(3) + d 24 = a(4)^3 + b(4)^2 + c(4) + d These equations can be rewritten by calculating the powers and rearranging the terms. 1a + 1b + 1c + 1d = 5 8a + 4b + 2c + 1d = 9 27a + 9b + 3c + 1d=15 64a + 16b + 4c + 1d = 24 Since the system is big, solving it by elimination or substitution can get messy. Because of this, it is preferred to use a graphing calculator to solve the system. To input a system of equations as a matrix, press 2ND, then x^(-1). Then, scroll to the right to reach the EDIT menu.
Selecting one of the matrices in the menu opens a window to modify its elements. To input a system of four equations with four variables, a 4*5 matrix will be needed. The last column stores the results at the right-hand side of each equation.
To solve the system of equations, press 2ND and x^(-1) once more. However, this time, select the MATH menu and scroll down until the rref(
option is selected.
Finally, input matrix [A] to the rref(
function. To do so, push 2ND and x^(-1) to bring up all available matrices. In the NAMES menu, choose matrix [A].
The solutions can be read vertically in the last column, using the remaining coefficients as reference points for the variables. [ cccc 1&0&0&0&0.166... 0&1&0&0&0 0&0&1&0&2.833... 0&0&0&1&2 ] ⇕ 1a + 0b + 0c + 0d = 0.166... 0a + 1b + 0c + 0d = 0 0a + 0b + 1c + 0d= 2.833... 0a + 0b + 0c + 1d = 2 Now that the coefficients have been calculated, it is possible to write the polynomial function. Remember to round each coefficient to two decimal places. y = 0.166... x^3 + 0x^2 + 2.833... x + 2 ⇓ y = 0.17x^3 + 2.83x + 2
x= 5
Calculate power
Multiply
Add terms
Zosia is really into environmentalism. After watching the documentary about planes, she became increasingly worried about the atmospheric carbon dioxide levels on Earth and decided to do some research.
In her research, Zosia found data about the average levels, in parts per million, of carbon dioxide in the atmosphere per year. Some results are listed below. The years are counted from the year 1995, so 0 is the year 1995, 5 is 2000, and so on.
Year | Carbon Dioxide Levels (ppm) |
---|---|
0 | 360 |
5 | 369 |
10 | 379 |
15 | 390 |
20 | 401 |
25 | 414 |
Recalling her recent introduction to polynomial functions, Zosia decided to do the following.
QuartRegon a graphing calculator.
To find a polynomial of degree four that fits the points, press STAT again and move to the CALC menu. Then, scroll down to select the function QuartReg.
Now that the coefficients are known, it is possible to write an function for the carbon levels. It is important to remember that the symbol E^(-5) is equivalent to multiplying the number by 10^(-5). C(t) = 6.667*10^(-5)t^4 -0.003t^3 + 0.063t^2 - 1.538t + 360.0
t= 30
DiagnosticOnand press ENTER. Finally, reopen the window from Part A.
The closer the value of R^2 is to 1, the better the polynomial function fits the points. Since R^2 is very close to 1, the function fits the points really well. Therefore, it is possible that the prediction is reliable, but more points would clarify if that is the case.
The following table of values was presented earlier in the lesson.
x | f(x) |
---|---|
1 | 2 |
2 | 3 |
3 | 7 |
4 | 16 |
5 | 32 |
6 | 57 |
Below are the first, second, and third differences of these values.
Note that the third differences are constant and different from zero. By the Properties of Finite Differences, the function values are represented by a cubic polynomial function. f(x) = ax^3 + bx^2 + cx + d Since many points are given, the best way to find the coefficients is to use the cubic regression function on a graphing calculator. The first thing to do is to input the data into the calculator. Push STAT, choose Edit, and then enter the values in the first two columns.
To find a polynomial of degree three that fits the points, press STAT again and move to the CALC menu. Then, scroll down to select the function CubicReg.
Now it is possible to write a polynomial function of degree three that fits the given data. y = 0.33x^3 - 0.5x^2 + 0.17x + 2
Since the value of R^2 is exactly 1, the polynomial function fits the points perfectly. This confirms the conclusion made considering the Properties of Finite Differences. In general, when encountering a data set, it can be useful to find a polynomial function to model the data, as it might lead to more accurate predictions or conclusions.
Find the second differences of the following data set.
x | 2 | 4 | 6 | 8 | 10 |
---|---|---|---|---|---|
f(x) | 8 | 76 | 256 | 596 | 1144 |
Notice that the x-values in the table are equally spaced. To find the second differences, we begin by finding the first differences. We do this by subtracting each pair of consecutive function values.
Now that we know the first differences, we can proceed to finding the second differences. To do this, we subtract each pair of consecutive first differences.
As we can see, the second differences are 112, 160, and 208.
We will follow the same procedure as we did in Part A. However, notice that the given points are not in order. Let's first organize the points so that the x-values are in increasing order.
(1,-15), (3,169), (5,3001),
(7,16 689) (9,59 041), (11,161 305)
With the points organized, we can see that the x-values are equally spaced. Now let's begin by finding the first differences. As before, we will subtract each pair of consecutive y-values.
We are ready to find the second differences by subtracting the values we found for the first differences.
Repeating the same process one more time will give us the third differences.
The third differences for the given set of points are 8208, 17 808, and 31 248.
Indicate the degree of the polynomial function that best fits each of the following data sets.
x | 3 | 4 | 5 | 6 | 7 | 8 |
---|---|---|---|---|---|---|
g(x) | -11 | -35 | -79 | -149 | -251 | -391 |
According to the Properties of Finite Differences, if the nth differences of equally spaced values are constant and different from zero, then the values can be represented by a polynomial function of degree n. Let's begin by finding the first differences of the given function values.
As we can see, the first differences are not constant — the values are not equal. This means that a linear polynomial does not fit the data well. Let's continue and find the second differences.
Once more, we see that the second differences are not constant either. This means that a quadratic polynomial is not the best fit. Let's find the third differences.
This time, all the third differences are equal and different from zero — they are all equal to -6. This implies that the degree of polynomial function that best fits the data set is 3.
As in Part A, we will find the finite differences of the data set to determine the degree of the required polynomial. However, notice that the given points are not ordered. Therefore, let's organize them so that the x-values are in increasing order.
(2,-16),(3,-9),(4,0),(5,11),(6,24)
We can see that the x-values are equally spaced. Let's continue our process by finding the first differences. To do this, we subtract each pair of consecutive y-values.
Since the first differences are not constant, let's continue to find the second differences.
As we can see, the second differences are constant and different from zero. Therefore, the degree of the polynomial function that best fits the given data set is 2.
Consider the following statement.
If the nth differences of equally spaced values are and zero, then the values can be represented by a polynomial function of degree n. |
When we want to determine the degree of the polynomial function that best fits a set of values, we study the finite differences of y-values. In fact, if the nth differences of equally space values are constant and different from 0, the values can be represented by a polynomial function of degree n.
If the nth differences of equally spaced values are constant and different from zero, then the values can be represented by a polynomial function of degree n.
Consequently, the missing parts of the given statement are constant and different from.
Davontay is interested in finding a cubic polynomial that passes through the points (3,0), (-2,0), (1,0), and (4,72). In the process, he first wrote a factored function and then performed a certain substitution to find the value of the leading coefficient.
However, Davontay's teacher said that he made a mistake.
Let's check Davontay's computations carefully to determine where he made a mistake. As we can see, he first wrote the points through which the cubic function passes. (3,0), (-2,0), (1,0), (4,72) A general cubic polynomial has the form ax^3+bx^2+cx+d. Since three of the given points are x-intercepts, we can use the factored form of a cubic polynomial, which is what Davontay did. ( 3,0), ( -2,0), ( 1,0) [0.75em] f(x) = a(x + 3)(x - 2)(x + 1) However, the factor corresponding to an x-intercept of the form (k,0) is x-k. In other words, to form the factor, the sign of the x-intercept is changed. This is where Davontay made his mistake — he forgot to change the signs. ( 3,0), ( -2,0), ( 1,0) [0.75em] f(x) = a(x + 3)(x - 2)(x + 1) * [0.75em] f(x) = a(x- 3)(x+ 2)(x- 1) ✓ Therefore, Davontay's mistake is that the sign on each factor is wrong.
To write the correct function rule, we can follow Davontay's steps but correct his mistake. Let's write the points through which the polynomial passes.
(3,0), (-2,0), (1,0), (4,72)
Since three of the given points are x-intercepts, we can use the factored form of a cubic polynomial.
f(x) = a(x-k_1)(x-k_2)(x-k_3)
Here, k_1, k_2, and k_3 are the x-intercepts. Let's substitute the corresponding values.
Our next step is to find the value of a. To do so, we can use the fact that the polynomial also passes through ( 4, 72), which implies that f( 4)= 72. Let's evaluate f(x) at x= 4.
We got that the value of a is 4. We are ready to write the required cubic polynomial. f(x) = 4(x-3)(x+2)(x-1)