course mth 163 °úçfÈöy}ÑŒÎ×½W‡{÷RýèŠÕžÆ·Åûassignment #021
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22:50:07 What are the possible number of linear and irreducible quadratic factors for a polynomnial of degree 6?
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RESPONSE --> a degree 6 polynomial could have 6 linear factors, 4 linear factors, 2 linear factors and 1 irreduciable quadratic factors, or 2 irreduciable quadratic factors, or 3 irreduciable quadratic factors. confidence assessment: 1
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22:54:30 ** 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. **
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RESPONSE --> self critique assessment: 3
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23:15:26 For a degree 6 polynomial with one irreducible quadratic factor and four linear factors list the possible numbers of repeated and distinct zeros.
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RESPONSE --> There could be: all be distinct, 2 could be identical with the other two distinct, or two pairs of identical factors, or there could be 3 identical factors and one distinct, or 4 identical factors. confidence assessment: 1
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23:28:00 ** 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. **
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RESPONSE --> I got the correct amounts, I didn't talk about the graph because it was not asked for. self critique assessment: 2
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23:44:54 Describe 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.
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RESPONSE --> As mentioned in the pervious explaination and my graph: There will be four zeros. The degree is even, so the far-left and far-right sides will be the same, either both increasing very rapidly toward +infinity or both decreasing very rapidly toward -infinity. There can be 4 distinct zeros, which will caue the graph to go through the x axis going up or down through one zero and the other way down or up through the next. There is a possibility of 2 repeated and 2 distinct zeros. At the repeated zero, the graph will just touch the x axis. The graph will pass through the x axis at the two distinct zeros. If there are 3 repeating and 1 distinct zero. At the zeros the graph will level off where it passes thru the x axis. The graph will pass through the x axis at the one distinct zero. If there are two pairs of 2 repeated zeros, at each repeated zero the graph will touch the x axis. The graph will not pass through the x axis. With four repeated zeros the graph will touch the x axis. confidence assessment: 2
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23:45:42 ERRONEOUS STUDENT SOLUTION AND INSTRUCTOR CORRESION 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. **
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RESPONSE --> Mine was a little longer detail. self critique assessment: 1
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23:54:13 It 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?
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RESPONSE --> The quadratic polynomial would have a parabloa shape and a liner polynomial would be more of a straight line. When plotted out, these data points fit the cubic graph, which begins below the x-axis and then at a point begins to curve upward like half of an inverted parabola. confidence assessment: 2
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23:54:34 ** 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. **
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RESPONSE --> self critique assessment: 3
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00:08:46 On 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?
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RESPONSE --> Both the 4th and 6th degree polynomial has a wider vertex, but thinner than the 2nd degree polynomial. When you look at the graph, it has a wider ""base"" or vertex than a 2nd degree polynomial would have. confidence assessment: 2
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00:08:52 ** higher even degrees flatten out more near their 'vertices' **
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RESPONSE --> self critique assessment: 3
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00:09:17 On 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?
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RESPONSE --> This is the same question.... confidence assessment: 3
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00:09:25 STUDENT 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.
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RESPONSE --> self critique assessment: 3
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00:59:26 What is the degree 2 Taylor approximation for f(t) = e^(2t), and what is your approximation to f(.5)? How close is your approximation to the actual value of e^(2t)?
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RESPONSE --> The degree 2 Taylor approximation for 2t is 1 + 2t + 2t^2 / 2!. When you substitute in .5 for t you get: 1 + 2(.5) + [(2*.5)^2 / (1*2)] 1 + 1 + [(1^2) / 2] 2 + (1/2) 2.5 When you do e^(2t) with t = .5 e^(2*.5) e^1 2.71828 confidence assessment: 2
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01:04:21 ** The degree 2 Taylor polynomial is f(t) = 1 + 2t + (2t)^2 / 2! = 1 + 2 t + 4 t^2 / 2 = 1 + 2 t + 2 t^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. **
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RESPONSE --> self critique assessment: 3
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01:25:12 By how much does your accuracy improve when you make the same estimate using the degree 3 Taylor approximation?
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RESPONSE --> 1 + 2t + (2t)^2 / 2! + (2t)^3 / 3! 1 + 2(.5) + [2*(.5^2)] + [3*(.5^3)] 1 + 1 + (2*.25) + (3*.125) 2 + (.5) + (.375) 2.875 The difference is .157 Accuracy improves greatly. confidence assessment: 3
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01:26:47 ** The degree 3 Taylor polynomial is f(t) = 1 + 2t + (2t)^2 / 2! + (2t)^3 / 3! = 1 + 2 t + 4 t^2 / 2 + 8 t^3 / 6 = 1 + 2 t + 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. **
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RESPONSE --> self critique assessment: 3
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02:08:37 Describe 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.
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RESPONSE --> The graph will be a line that is decreasing at a decreasing rate. confidence assessment: 1
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02:12:25 ** The errors, rounded to the nearest thousandth, are: degree-2 error: -.218 degree-3 error: -.051 degree-4 error: -.010 degree-5 error: -.002 A graph of error vs. degree decreases rapidly toward the horizontal axis, showing that the error decreases rapidly toward zero as the degree increases. **
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RESPONSE --> I should have said at an increaseing rate. self critique assessment: 1
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02:36:29 What 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.
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RESPONSE --> The degree four approximation is: 1 + 2t + 2t^2 + 4t^3/3 + 2t^4/3 SO: 1 + 2(.2) + 2(.2)^2 + 4(.2)^3/3 + 2(.2)^4/3 1 + .4 + .08 + .011 + .00107 1.492 e^.2 = 1.221 1 + 2(.4) + 2(.4)^2 + 4(.4)^3/3 + 2(.4)^4/3 1 + .8 + .32 + .0853 + .0171 2.222 e^.4 = 1.492 1 + 2(.6) + 2(.6)^2 + 4(.6)^3/3 + 2(.6)^4/3 1 + 1.2 + .72 + .288 + .0864 3.29 e^.6 = 1.822 1 + 2(.8) + 2(.8)^2 + 4(.8)^3/3 + 2(.8)^4/3 1 + 1.6 + 1.28 + .683 + .273 4.836 e^.8 = 2.226 1 + 2(1) + 2(1)^2 + 4(1)^3/3 + 2(1)^4/3 1 + 2 + 2 + 1.3 + .667 6.967 e^.2 = 2.718 The graph is increasing. confidence assessment: 1
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02:37:02 The following are the approximations and errors: 0.2 1.2214 1.221402758 2.75816E-06 0.4 1.4917 1.491824698 9.13643E-05 0.6 1.8214 1.8221188 0.0007188 0.8 2.2224 2.225540928 0.003140928 1 2.7083 2.718281828 0.009948495 The errors can be written 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 os .2.
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RESPONSE --> I think somehow I got the wrong numbers in my approximations. I will go back and check them. confidence assessment: 1
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10:28:17 ** 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. **
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RESPONSE --> It clicked before I had a chance to answer. I understand from the notes where the P2(x) = (x-1) - (x-1)^2/2 came from. What did we need to talk about the table. I missed that somewhere in the question? And whre did the number come from? A table is not mentioned until question 10. Just a bit comfused. self critique assessment: 1
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23:00:20 What 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?
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RESPONSE --> The equation for these would be: ln (x) = (x-1) - (x-1)^2/2 ln (.6) = (.6-1) - (.6-1)^2/2 ln (.6) = -.4 - (-.4)^2/2 ln (.6) = -.4 - (.16)/2 ln (.6) = -.4 - .08 ln (.6) = -.48 actual: -.511 difference: -.031 ln (.8) = (.8-1) - (.8-1)^2/2 ln (.8) = -.2 - (-.2)^2/2 ln (.8) = -.2 - (.04)/2 ln (.8) = -.2 - .02 ln (.8) = -.22 actual: -.223 difference: -.003 ln (1.2) = (1.2-1) - (1.2-1)^2/2 ln (1.2) = .2 - (.2)^2/2 ln (1.2) = .2 - (.04)/2 ln (1.2) = .2 - .02 ln (1.2) = .18 actual: .182 difference: .002 ln (1.4) = (1.4-1) - (1.4-1)^2/2 ln (1.4) = .4 - (.4)^2/2 ln (1.4) = .4 - (.16)/2 ln (1.4) = .4 - .08 ln (1.4) = .32 actual: .34 difference: .02 The approximation gets better as x approaches 1 because 1 is when the natural log becomes positive. confidence assessment: 2
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23:07:13 ** 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. **
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RESPONSE --> I got a different error for 1.4. My reason for being more accurate as it approaches 1 was wrong, however I do understand the explaination. self critique assessment: 3
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23:22:07 problem 12. What does the 1/x graph do than no quadratic function can do?
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RESPONSE --> The 1/x graph is a mirror image on the positive sides of the y and x axis to the negaitve side of the x and y axis. confidence assessment: 2
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23:28:02 ** 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. **
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RESPONSE --> I was not totally sure how to word the fact that it had asymptotes along the y axis and the x axis. I think I got fairly close with what I was trying to say. self critique assessment: 2
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00:00:52 What 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.
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RESPONSE --> 1/x = 1 - (x-1) + 2(x-1)^2 / 2 1/(.6) = 1 - (.6-1) + 2(.6-1)^2 / 2 1/(.6) = 1 - (-.4) + (-.4)^2 1/(.6) = 1.4 + .16 1/(.6) = 1.56 1/(.6) = 1.67 error: .11 1/(.8) = 1 - (.8-1) + 2(.8-1)^2 / 2 1/(.8) = 1 - (-.2) + (-.2)^2 1/(.8) = 1.2 + .04 1/(.8) = 1.24 1/(.8) = 1.25 error: .01 1/(1.2) = 1 - (1.2-1) + 2(1.2-1)^2 / 2 1/(1.2) = 1 - (.2) + (.2)^2 1/(1.2) = .8 + .04 1/(1.2) = .84 1/(1.2) = .83 error: -.01 1/(1.4) = 1 - (1.4-1) + 2(1.4-1)^2 / 2 1/(1.4) = 1 - (.4) + (.4)^2 1/(1.4) = .6 + .16 1/(.6) = .76 1/(.6) = .71 error: -.05 At this time, my graph makes a straight line. confidence assessment: 2
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00:01:56 ** The quadratic approximation to 1/x is the second-degree Taylor polynomial 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 1.666666666 1.56 - 0.1066666666 .8 1.25 1.24 - 0.01 1.2 0.8333333333 0.84, 0.006666666666 1.4 0.7142857142 0.76 0.04571428571. A graph of appoximation 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.**
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RESPONSE --> I think my graph was off. I will go back and look at it again. self critique assessment: 2
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