If a polynomial has degree n, it might possibly have n linear factors. The polynomial might, of course, have one or more irreducible quadratic factors, each of which contributes 2 to the degree of the polynomial and thereby reduces the possible number of linear factors by 2.
The number of linear factors could therefore be n, n-2, n-4, etc., but not n-1, n-3, n-5, etc..
Since each linear factor contributes a zero to the polynomial, and since there are no zeros except those contributed by the linear factors, the total number of zeros of a degree n polynomial is n, n-2, n-4, ....
If a polynomial has even degree, it is therefore possible that it could be made up entirely of irreducible quadratic factors and therefore have no zeros. On the other hand a polynomial of odd degree must have at least one linear factor and must therefore have at least one zero.
Note that some of the linear factors may be identical, so some of the zeros might appear more than once. Any set of identical zeros appears as one point on a graph, but the shape of the graph at the zeros is strongly dependent on the number of zeros represented by that point.
If a point on the x axis represents two identical zeros, corresponding to two identical linear factors, then we say that the point represents zero of multiplicity 2. The graph will have the shape of a parabola with its vertex at the zero point.
More generally, if a point on the x axis represents m identical zeros, corresponding to m identical linear factors, then we say that the point represents zero of multiplicity m. The graph at the zero point will have the shape of a y = a x^m function at the origin.
Degree 4
As seen in the previous section, a degree 4 polynomial could consist of
4 linear factors, or
2 linear factors and 1 irreducible quadratic factor, or
2 irreducible quadratic factors.
If there are 4 linear factors, they could
all be distinct, or
2 could be identical with the other two distinct, or
we could have two pairs of identical factors, or
we could have 3 identical factors and one distinct, or
we could have 4 identical factors.
If there are only two linear factors they are either distinct or identical.
The shape of the graph clearly depends on both the number of linear factors and the multiplicity of each.
The table below depicts the possibilities:
Number of linear factors Numbers of identical, distinct linear factors Shape of graph at zeros 4
4 distinct 2 identical, 2 distinct 3 identical, 1 distinct 4 identical 2 pairs of different identical factors
all linear 1 parabolic (mult. 2), 2 linear 1 cubic (mult. 3), 1 linear 1 quartic (mult. 4) 2 parabolic (each mult. 2) 2
2 distinct 2 identical
2 linear parabolic (mult. 2) 0 no linear factors no zeros
Degree 5
A degree 5 polynomial might have 5 linear factors, 3 linear factors or 1 linear factor.
If the polynomial has 5 linear factors, it is possible to have
2 identical and 3 distinct
3 identical and 2 distinct
2 pairs of identical factors and one distinct
4 identical and 1 distinct
3 identical and the other two identical to one another but not to the others
5 identical
A polynomial with 3 linear factors could have 3 distinct factors, 2 identical and 1 distinct or 3 identical factors.
You should make a table, like the one given for degree 4 polynomials, summarizing all these possibilities.
Exercises 1-21. List the possible numbers of linear and irreducible quadratic factors for polynomials of degree 6 and 7.
2. For a degree 6 polynomial with one irreducible quadratic factor and four linear factors list the possible numbers of repeated and distinct zeros.
Sketch a typical graph for each of these possibilities. Recall that when a zero is repeated twice the shape of the function near that zero is much like that of a parabola near its vertex. If a zero is repeated n times the shape of the graph near the zero is much like that of the y = x^p function near the origin. So if a zero is repeated three times the graph near the zero will have a shape similar to that of the y = x^3 function, while a zero repeated four times will exhibit behavior similar to at of the y = x^4 function.
Fitting polynomials to data pointsAs you saw very early in the course, if we have 3 data points we can use them to create 3 simultaneous linear equations using the form y = a2 x^2 + a1 x + a0. These three equations can be solved for the parameters a0, a1 and a2.
It should be easy to see then that if we have 4 data points we can use them to create 4 simultaneous linear equations using the form y = a3 x^3 + a2 x^2 + a1 x + a0. These four equations can be solved for the parameters a0, a1, a2 and a3.
The next logical statement is that if we have 5 data points we can use them to create 5 simultaneous linear equations using the form y = a4 x^4 + a3 x^3 + a2 x^2 + a1 x + a0. These five equations can be solved for the parameters a0, a1, a2, a3 and a4.
It is clear that we can continue this reasoning to any number of data points.
In general then, if we have n data points we can use them to create n simultaneous linear equations using the form a(n) x^n + x(n-1) x^(n-1) + ... + a2 x^2 + a1 x + a0. These n equations can be solved for the parameters a0, a1, a2, ..., a(n-1), a(n).
The polynomial model we get won't necessarily be a good model of our data. This is because the model will tend to 'wobble' around in order to reach each point, as opposed to our usual expectation that the true model is a smooth curve without 'wobbles'.
We are usually better off using a best-fit with many data points. A linear or a quadratic model is often good enough, but in many cases cubic or even quartic best-fit model might be more appropriate.
The decision about when to use each kind of model requires a good knowledge of calculus and of the phenomena we are attempting to model.
Exercises 3-53. The data table and corresponding graph shown below are to be fit with a linear, a quadratic and a cubic best-fit function.
-1.6
-0.95283
-1.2
-0.89621
-0.8
-0.72863
-0.4
-0.42074
0 -0.0456
0.4
0.346737
0.8
0.762977
1.2
0.897338
1.6
1.02358
Use DERIVE or another utility to plot a graph of the data and of the linear, quadratic and cubic best fits to this data.
4. In terms of the shapes of the y = x^3 and y = x^2 graphs, why does the x^3 term of a cubic polynomial allow the cubic best-fit function to achieve a much better fit of this data than the best-fit quadratic polynomial?
5. Fit the data shown below with a quadratic polynomial, then with a quartic (degree 4) polynomial, using DERIVE or an equivalent utility.
-2
11
-1.5
6
-1
1
-0.5
1
0
1
0.5
1
1
1
1.5
6
2
11
Which polynomial comes closer to filling in the jagged corner of the graph? Would you expect a degree 6 polynomial to do an even better job?
Approximating with polynomialsMost of the functions we use to model the real world can be closely approximated by polynomials. Of the functions studied in this course, the linear, quadratic and positive-p power functions are in fact polynomials. The negative-power functions, exponential functions and logarithmic functions, on the other hand, are not polynomials. The sine, cosine and other trigonometric functions studied during the second semester of this course are also not polynomials. All these non-polynomial functions can be approximated as closely as desired, over a specified interval of x values, by polynomials.
A polynomial approximation of y = e^xThe exponential function y = f(x) = e^x is very close in value to 1 + x + x^2 / 2, as long as x is close to 0. The table below gives values of e^x and 1 + x + x^2 / 2 for x values between -1 and 1:
e^x
-1 .0367879 0.5 -0.8
0.449329
0.52
-0.6
0.548812
0.58
-0.4
0.67032
0.68
-0.2
0.818731
0.82
0
1
1
0.2
1.221403
1.22
0.4
1.491825
1.48
0.6
1.822119
1.78
0.8
2.225541
2.12
1
2.718282
2.5
In fact the exponential function y=e^x is even closer in value to 1 + x + x^2 / 2 + x^3 / 6, as the following table demonstrates:
x e^x 1+x+x^2/2+x^3/6 -1
0.367879
0.333333
-0.8
0.449329
0.434667
-0.6
0.548812
0.544
-0.4
0.67032
0.669333
-0.2
0.818731
0.818667
0 1
1
0.2
1.221403
1.221333
0.4
1.491825
1.490667
0.6
1.822119
1.816
0.8
2.225541
2.205333
1
2.718282
2.666667
A polynomial which approximates the exponential function as closely as we might wish can be built from the following pattern:
e^x = 1 + x + x^2 / 2 + x^3 / 6 + x^4 / 24 + x^5 / 120 + x^6 / 720 + x^7 / 5040 + ....
The pattern should be very clear to you, except perhaps for the denominators 2, 6, 24, 120, 720, 5040, .... Actually these denominators are easily constructed. If we look at the ratios of the denominator sequence we see that they are 3, 4, 5, 6, 7, .... This should make the following scheme obvious:
We get 6 if we multiply 2 by 3
We get 24 by multiplying 6 by 4
We get 120 by multiplying 24 by 5
We get 720 by multiplying 120 by 6
We get 5040 by multiplying 720 by 7.
We note that 6 = 3*2*1, 24 = 4*3*2*1, 120 = 5*4*3*2*1, etc.. These products are called factorials:
When we multiply all the counting numbers from n down to 1, we call the result n!. We read this as 'n factorial'. Thus for example, 8! = 8*7*6*5*4*3*2*1 = 40,320.
We therefore write e^x in the more explicit form
e^x = 1 + x + x^2 / 2! + x^3 / 3! + x^4 / 4! + . . . + x^n / n! + . . .
where the last . . . indicates that the pattern continues forever.
The meaning of the phrase 'continues forever' might not be completely clear to you; this concept requires the concepts and techniques of calculus for clarification.
The infinite polynomial given above is called the Taylor series of e^x. All the functions studied in this course have Taylor series.
The degree n Taylor polynomial for a function is just the polynomial obtained from the Taylor series, up to and including the degree n term of the series.
The degree 2 Taylor series for e^x is 1 + x + x^2 / 2!.
The degree 3 series for this function is 1 + x + x^2 / 2! + x^3 / 3!, while the degree 4 series is 1 + x + x^2 / 2! + x^3 / 3! + x^4 / 4!.
The degree n approximation corresponding to a Taylor series is just the degree n Taylor polynomial.
Exercises 6-86. Using the Taylor Series given above, find the degree 2 approximation of e^(2t). You will simply substitute 2t for x in the degree 2 expression e^x = 1 + x + x^2 / 2, then expand the squared term. You will obtain 1 + 2t + 4 t^2 / 2 = 1 + 2t + 2 t^2.
Use this approximation to determine an approximate value for f(t) = e ^ (2t) when t = .5. Compare with the actual value of f(.5), as given by your calculator.
Repeat for the degree 3 approximation. By how much does the approximation improve when you go from degree 2 to degree 3?
7. Find the degree 2, degree 3, degree 4 and degree 5 approximations to e ^ .5. For each of these approximations find the error in the approximation. (The error is the difference between the approximation and the actual value as given by your calculator).
Sketch a graph of the error vs. the degree of the approximation.
8. Find the degree 4 approximations of e^x for x = .2, .4, .6, .8 and 1. For each approximation find the error. Sketch a graph of the error vs. the value of x.
Other seriesA Taylor Series expression for the natural log function y = ln(x) is
ln(x) = (x-1) - (x-1)^2 / 2 + 2(x-1)^3 / 6 - 3(x-1)^3 / 24 + 4(x-1)^3 / 120 ....
This polynomial is defined in terms of powers of (x-1), for reasons that will be clarified in calculus. These reasons have to do with the fact that the natural logarithm of 0 is undefined.
Other simply expressed polynomial approximations include the following:
sin(x) = x - x^3 / 6 + x^5 / 120 - x^7 / 5040 ... + (-1)^(2n+1) x^(2n+1) / (2n+1) ! + ...
cos(x) = 1 - x^2/2 + x^4 / 24 - x^6 / 720 + ... + (-1)^(2n) x^2n / (2n) ! + ...
1/x = 1 - (x-1) + 2(x-1)^2 / 2 - 3(x-1)^2 / 6 + 4(x-1)^3 / 24 ...
1/x^2 = 1 - 2(x-1) + 3(x-1)^2 / 2 - 4(x-1)^2 / 6 + 5(x-1)^3 / 24 ...
1/x^3 = 1 - 3(x-1) + 4(x-1)^2 / 2 - 5(x-1)^2 / 6 + 6(x-1)^3 / 24 ...
You should note the similarity between these series and the series for the exponential function. Each of these series uses every other term of the series for the exponential function, with the sin(x) function using the odd-numbered terms of the series and the cos(x) function using the even-numbered terms. These series also alternate positive terms with negative terms.
The sin(x) and cos(x) functions are easily evaluated on your calculator. These functions have not yet been studied, though you might know something about them from previous courses. For the purposes of this course they are used only as examples; it is not necessary to know anything about the meanings of the functions.
The table below summarizes Taylor series polynomials for a few common functions.
| Function | Taylor polynomial of degree 2 or 3 | Taylor series |
| e^x | 1 + x + x^2 / 2 + x^3 / 6 (degree 3; degrees of other listed polynomials are 2) | 1+x+x^2/2+x^3/3! + x^4/4! + ... + x^n/n! + ... |
| ln(x) | (x-1) - (x-1)^2/2 + 2(x-1)^2 / 6 | (x-1) - (x-1)^2/2 + 2(x-1)^2 / 3! - 3(x-1)^4 / 4! ... + (-1)^(n-1) (x-1) ^ (n-1) / n! + ... |
| sin(x) | x - x^3 / 6 | x - x^3 / 3! + x^5 / 5! ... + (-1)^(2n+1) x^(2n+1) ... |
| cos(x) | 1 - x^2 | 1 - x^2 + x^4/4! + ... + (-1)^(2n) x^(2n) ... |
| 1/x | 1 - (x-1) + 2(x-1)^2 | 1 - (x-1) + 2(x-1)^2 / 2 - 3(x-1)^3 / 6 + 4(x-1)^4 / 24 ... + (-1)^n (n) (x-1)^n |
| 1/x^2 | 1 - 2(x-1) + 3(x-1)^2 / 2 | 1 - 2(x-1) + 3(x-1)^2 / 2 -43(x-1)^3 / 6 + 5(x-1)^4 / 24 .. + (-1)^n (n+1) (x-1)^n |
| 1/x^3 | 1 - 3(x-1) + 4(x-1)^2 / 2 | 1 - 3(x-1) + 4(x-1)^2 / 2 - 5(x-1)^3 / 6 + 6(x-1)^4 / 24 ... + (-1)^n (n+2) (x-1)^n |
9. Sketch a graph of y = ln(x) for 0 < x < 10. What aspect or aspects of this graph cannot be duplicated by the graph of a quadratic function?
Sketch on the same set of coordinate axes the parabolic graph of the degree 2 Taylor polynomial approximating y = ln(x).
Find the degree 2 approximation to ln(x) for x = .6, .8, 1, 1.2 and 1.4.
The function y = ln(x) is easily evaluated using your calculator. Determine the error of the approximation at each x value.
10. The function y = f(x) = cos(x) can be evaluated by your calculator or by DERIVE. If you use your calculator you must put it in radian mode before evaluating the function. An alternative to using radian mode is to multiply every x value by 180 / `pi (approximately 57.3) before evaluating the function.
A table of values for this function for x = 0 to `pi, and a corresponding graph, are given below for reference, in the event that you are unsure of your calculator's values:
x y = cos(x) 0
1
0.3925
0.923956
0.785
0.707388
1.1775
0.383235
1.57
0
1.9625
-0.38176
2.355
-0.70626
2.7475
-0.92335
3.14
-1
Verify that the cos(x) function starts with value 1 at x = 0, decreases until x = `pi/2 (approximately 1.57), where it is zero, then decreases to -1 by the time x = 3.14.
Sketch a graph of the cos(x) function for x from 0 to `pi (approx. 3.14).
What aspect of the shape of the graph of y = cos(x) can never be duplicated by the parabolic graph of a quadratic function?
Sketch on the same set of coordinate axis the parabolic graph of the degree 2 polynomial 1 - x^2 / 2, which is a quadratic function, approximating cos(x).
The two graphs start out pretty much together. At approximately what value of x does the polynomial first different from the cos(x) function by more than 10% of its value?
11. Part of the graph of the function y = sin(x) is shown below. The graph continues in the same manner for all positive and negative values of x.
How many zeros does the part of the y = sin(x) graph shown below have? What degree polynomial would be required in order to achieve this many zeros? What Taylor polynomial would you therefore start with in an attempt to approximate this graph?
Using trial and error, determine the value of this polynomial for the x = 3.14 zero of the sin(x) function. How close is the Taylor polynomial to 0?
12. Sketch a graph of y = 1/x, for x = 0 to 5.
What aspect(s) of the y = 1/x graph cannot be duplicated by any quadratic polynomial?
Use a degree 2 Taylor polynomial to approximate y = 1/x for x = .6, .8, 1, 1.2 and 1.4. For each x value find the error of your approximation. Plot the error vs. x.
