If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
021. `query 21
Question: `qWhat are the possible number of linear and irreducible quadratic factors for a polynomial of degree 6?
Your solution:
Confidence Assessment:
Given Solution:
`a** You can have as many as 6 linear and 3 irreducible quadratic factors for a polynomial of degree 6.
For a polynomial of degree 6:
If you have no irreducible quadratic factors then to have degree 6 you will need 6 linear factors.
If you have exactly one irreducible quadratic factor then this factor is of degree 2 and you will need 4 linear factors.
If you have exactly two irreducible quadratic factors then the product of these factors is of degree 4 and you will need 2
linear factors.
If you have three irreducible quadratic factors then the product of these factors is of degree 6 and you can have no linear
factors.
For a polynomial of degree 7:
If you have no irreducible quadratic factors then to have degree 7 you will need 7 linear factors.
If you have exactly one irreducible quadratic factor then this factor is of degree 2 and you will need 5 linear factors to give
you degree 7.
If you have exactly two irreducible quadratic factors then the product of these factors is of degree 4 and you will need 3
linear factors to give you degree 7.
If you have three irreducible quadratic factors then the product of these factors is of degree 6 and you will need 1 linear
factor to give you degree 7. **
Self-critique (if necessary):
Self-critique rating:
Question: `qFor a degree 6 polynomial with one irreducible quadratic factor and four linear factors list the possible numbers of repeated and distinct zeros.
Your solution:
Confidence Assessment:
Given Solution:
`a** there could be 1 root repeated 4 times, 2 roots with 1 repeated 3 times and the other distinct from it, 2 distinct roots
each repeated twice, three distinct roots with one of them repeated twice, or four distinct roots. Explanation:
You have one zero for every linear factor, so there will be four zeros.
Since the degree is even the far-left and far-right behaviors will be the same, either both increasing very rapidly toward
+infinity or both decreasing very rapidly toward -infinity.
You can have 4 distinct zeros, which will result in a graph passing straight thru the x axis at each zero, passing one way
(up or down) through one zero and the opposite way (down or up) through the next.
You can have 2 repeated and 2 distinct zeros. At the repeated zero the graph will just touch the x axis as does a
parabola at its vertex. The graph will pass straight through the x axis at the two distinct zeros.
You can have 3 repeated and 1 distinct zero. At the 3 repeated zeros the graph will level off at the instant it passes thru
the x axis, in the same way the y = x^3 graph levels off at x = 0. The graph will pass through the x axis at the one distinct
zero.
You can have two pairs of 2 repeated zeros. At each repeated zero the graph will just touch the x axis as does a
parabola at its vertex. Since there are no single zeros (or any other zeros repeated an odd number of times) the graph
will not pass through the x axis, so will remain either entirely above or below the x axis except at these two points.
You can have four repeated zeros. At the repeated zero the graph will just touch the x axis, much as does a parabola at
its vertex except that just as the y = x^4 function is somewhat flatter near its 'vertex' than the y = x^2 function, the graph
will be flatter near this zero than would be a parabolic graph. **
Self-critique (if necessary):
Self-critique rating:
Question: `qDescribe a typical graph for each of these possibilities. Describe by specifying the shape of the graph at each of its zeros, and describe the far-left and far-right belavior of the graph.
Your solution:
Confidence Assessment:
Given Solution:
The graph will pass through the x axis 4 times. Near each of
the 4 zeros the graph will appear to be a straight line, though of course as you
get away from the x axis the graph in between the zeros will need to curve in
order to return to the axis.
The graph will get increasingly steep at far right and far left, with the slopes
at left and right having opposite signs (i.e., if the slope at far right is
positive, with the graph moving up and to the right, then at far left the slope
must be negative, moving down as it moves to the right; similarly if the slope
at far left is positive, the slope at far right will be negative).
The graph might look like the figure below; it might also be 'upside down' compared to the graph in the figure.
ERRONEOUS STUDENT SOLUTION AND INSTRUCTOR CORRECTION
for 4 distincts, the graph curves up above the x axis then it crosses it 4 timeswithe the line retreating from the direction it
came from
for say 2 distincts and 2 repeats, line curves above x axis then it kisses the axis then it crosses it twice, retreating to the
side it came from
for 1 distinct and 3 it curves above x axis then it crosses once and kisses, finnaly heading off to the opposite side it came
from
INSTRUCTOR COMMENTS:
{The graph can't go off in th opposite direction. Since it is a product of four linear factors and any number of quadratic
factors its degree is even so its large-x behavior is the same for large positive as for large negative x. It doesn't kiss at a
degree-3 root, it acts like the y = x^3 polynomial, leveling off just for an instant as it passes through the zero and on to the
other side of the axis. **
problems 3-5
Self-critique (if necessary):
Self-critique rating:
Question: `qIt doesn't matter if you don't have a graphing utility--you can answer these questions based on what you know about the shape of each power function.
Why does a cubic polynomial, with is shape influenced by the y = x^3 power function, fit the first graph better than a quadratic or a linear polynomial?
What can a cubic polynomial do with this data that a quadratic can't?
Your solution:
Confidence Assessment:
Given Solution:
`a** the concavity (i.e., the direction of curvature) of a cubic can change. Linear graphs don't curve, quadratic graphs
can be concave either upward or downward but not both on the same graph. Cubics can change concavity from upward
to downward. **
Self-critique (if necessary):
Self-critique rating:
Question: `qOn problem 5, how do the shapes of 4th-degree polynomials and 6th-degree polynomials progressively differ from the shape of a 2d-degree polynomial in such a way as to permit a better and better fit to the graph?
Your solution:
Confidence Assessment:
Given Solution:
`a** higher even degrees flatten out more near their 'vertices' **
Self-critique (if necessary):
Self-critique rating:
Question: `qOn problem 5, how do the shapes of 4th-degree polynomials and 6th-degree polynomials progressively differ from the shape of a 2d-degree polynomial in such a way as to permit a better and better fit to the graph?
Your solution:
Confidence Assessment:
Given Solution:
`aSTUDENT RESPONSE: progressively more flexing, because more curves, and fit graph better the that of a lesser
degree
INSTRUCTOR COMMENT: On a degree-2 polynomial there is only one change of direction, which occurs at the
vertex. For degrees 4 and 6, respectively, there can be as many as 3 and 5 changes of direction, respectively. For higher
degrees the graph has more ability to 'wobble around' to follow the data points.
Self-critique (if necessary):
Self-critique rating:
Question: `qWhat is the degree 2
Your solution:
Confidence Assessment:
Given Solution:
`a** The degree 2
Therefore we have
T(.5) = 1 + 2 * .5 + 2 * .5^2 = 1 + 1 + 2 * .25 = 1 + 1 + .5 = 2.5.
The actual value of e^(2t) at t = .5 is
f(.5) = e^(2 * .5) = e^1 = e = 2.718, approx..
The approximation is .218 less than the actual function. **
Self-critique (if necessary):
Self-critique rating:
Question: `qBy how much does your accuracy
improve when you make the same estimate using the degree 3
Your solution:
Confidence Assessment:
Given Solution:
`a** The degree 3
+ 2 t^2 + 4 t^3 / 3.
Therefore we have
T(.5) = 1 + 2 * .5 + 2 * .5^2 + 4 * .5^3 / 3 = 1 + 1 + 2 * .25 + 4 * .125 / 3 = 1 + 1 + .5 + .167 = 2.667.
The actual value of e^(2t) at t = .5 is
f(.5) = e^(2 * .5) = e^1 = e = 2.718, approx..
The approximation is .051 less than the actual function. This is about 4 times closer than the approximation we obtained
from the degree-2 polynomial. **
Self-critique (if necessary):
Self-critique rating:
Question: `qDescribe your graph of the error vs. the degree of the approximation for degree 2, degree 3, degree 4 and degree 5 approximations to e^.5.
Your solution:
Confidence Assessment:
Given Solution:
`a**
The value of e^.5 to 10 significant figures is 1.648721270. Each approximation should be rounded to a number of significant figures great enough to give at least a 2-significant-figure difference.
The Taylor approximation for e^x is 1 + x + x^2 / 2! + x^3 / 3! + ... + x^n / n! + ...
Thus for degree 2 the approximation is 1 + .5 + .5^2/2 = 1.625
F0r degree 3 the approximation is 1 + .5 + .5^2/2 + .5^3/6 = 1.6458
For degree 4 the approximation is 1 + .5 + .5^2/2 + .5^3/6 + .5^4/24 = 1.64838 approx
For degree 5 the approximation is 1 + .5 + .5^2/2 + .5^3/6 + .5^4/24 + .5^5/120 = 1.648697917
The errors, if calculations are done accurately enough to yield a difference with 3 significant figures, are as follows:
ê^.5 - (1 + .5 + .5^2/2 ) = 0.0237
ê^.5 - (1 + .5 + .5^2/2 + .5^3/6 ) = 0.00288
ê^.5 - (1 + .5 + .5^2/2 + .5^3/6 + .5^4/24 ) = 0.000283
e^.5 - (1 + .5 + .5^2/2 + .5^3/6 + .5^4/24 + .5^5/120) =
.000023354
Self-critique (if necessary):
Self-critique rating:
Question: `qWhat are your degree four approximations for e^.2, e^.4, e^.6 e^.8 and e^1? Describe the graph of the approximation error vs. x.
Self-critique (if necessary):
Self-critique rating:
Question: `qThe following are the approximations and errors:
The degree-4 approximation to the exponential function e^x is 1 + x + x^2 / 2! + x^3 / 3! + x^4 / 4!. We evaluate this expression for x = .2, .4, 6., .8, 1.0:
For x = 0.2 the degree 4 approximation is 1.2214
This differs from e^.2 = 1.221402758 by 2.75816*10^(-06) = .000002758
For x = 0.4 the degree 4 approximation is 1.49173333
This differs from e^.4 = 1.491824698 by 9.13643*10^(-05) = .00009136
For x = 0.6 the degree 4 approximation is 1.8214
This differs from e^.6 = 1.8221188 by 0.0007188
For x = 0.8 the degree 4 approximation is 2.2224
This differs from e^.8 = 2.225540928 by 0.003140928
For x = 1 the degree 4 approximation is 2.70833333
This differs from e^1 = 2.718281828 by 0.009948495
The errors can be written to 2 significant figures as .0000027, .000091,
.000071, .0031 and .0099.
A graph of approximation error vs. x increases exponentially, with over a
10-fold increase with every increment of 0.2.
This graph is shown below:
Self-critique (if necessary):
Self-critique rating:
Question: `qWhat is the function which gives the quadratic approximation to the natural log function?
Your solution:
Confidence Assessment:
Given Solution:
`a** The function is P2(x) = (x-1) - (x-1)^2/2.
A table of values of ln(x), P2(x) and P2(x) - ln(x):
x ln(x) P2(x) P2(x) - ln(x)
.6 -0.5108256237 -0.48 0.03082562376
.8 -0.2231435513 -0.22 0.003143551314
1.2 0.1823215567 0.18 -0.002321556793
1.4 0.3364722366 0.32 -0.01647223662
At x = 1 we have ln(x) = ln(1) = 0, and P2(x) = P2(1) = (1-1) - (1-1)^2 / 2 = 0. There is no difference in values at x =
1.
As we move away from x = 1 the approximation becomes less and less accurate. **
Self-critique (if necessary):
Self-critique rating:
Question: `qWhat is the error in the degree 2 approximation to ln(x) for x = .6, .8, 1.2 and 1.4? Why does the approximation get better as x approaches 1?
Your solution:
Confidence Assessment:
Given Solution:
`a** The respective errors are .03, .00314, .00232, .016472.
There is no error at x = 1, since both the function and the approximation give us 0. As we move away from 1 the
approximation becomes less and less accurate. **
Self-critique (if necessary):
Self-critique rating:
Question: `qproblem 12. What does the 1/x graph do than no quadratic function can do?
Your solution:
Confidence Assessment:
Given Solution:
`a** The y = 1/x graph has vertical asymptotes at the y axis and horizontal asymptotes at the x axis. The parabolas we
get from quadratic functions do have neither vertical nor horizontal asymptotes. **
Self-critique (if necessary):
Self-critique rating:
Question: `qWhat are the errors in the quadratic approximation to 1/x at x = .6, .8, 1, 1.2, and 1.4?
Describe a graph of the approximation error vs. x.
Your solution:
Confidence Assessment:
Given Solution:
`a** The quadratic approximation to 1/x is the second-degree
P2(x) = 1 - (x - 1) + (x - 1)^2.
A table of values of 1/x, P2(x) and P2(x) - 1/x:
| x | 1/x | P2(x) | P2(x) - 1/x |
| .6 | 1.666... | 1.56 | -.1066... |
| .8 | 1.25 | 1.24 | -0.01 |
| 1.2 | 0.8333.... | 0.84 | 0.006666 |
| 1.4 | 0.714285... | 0.76 | 0.457 |
A graph of approximation error vs. x decreases at a decreasing rate to 0 at x = 1, then increases at an increasing rate for x
> 1.
This shows how the accuracy of the approximation decreases as we move away from x = 1.
The graph of approximation error vs. x gets greater as we move away from x = 1.**
Self-critique (if necessary):
Self-critique rating:
If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.
Question: What is the maximum number of zeros possible for a polynomial of degree 5 which contains one irreducible quadratic factor?
Your solution:
Confidence Assessment:
Question: What are the zeros of each of the following polynomials, and what is the multiplicity of each zero?
y = (x - 3) * (x + 2) * (x^2 + 4 x + 3)
y = (x - 3) * (x + 2) * (x^2 + 4 x + 4)
y = (x - 3) * (x + 2) * (x^2 + 4 x + 5)
Your solution:
Confidence Assessment:
Question: What are the possible number of linear and irreducible quadratic factors for a polynomnial of degree 5?
Your solution:
Confidence Assessment:
Question: What are the degree 2 and degree 3 Taylor approximations of the function f(t) = e^(t/2)?
The approximate value of e^(t / 2) for t = 0.2 is 1.1052, accurate to five significant figures.
By how much does the value of the degree-2 Taylor polynomial, evaluated at t = 0.2, differ from this value?
By how much does the value of the degree-3 Taylor polynomial, evaluated at t = 0.2, differ from this value?
Optional moderately challenging question: Accurate to within +- 0.1, what is the smallest value of t for which the degree-3 Taylor polynomial differs from the value of the exponential function by more than your answer to the first question (i.e., by more than the value of the degree-2 polynomial evaluated at t = 0.2 differed from that of the given function at t = 0.2)?
Your solution:
Confidence Assessment:
Self-critique rating: