course Mth 163 zу~kԐ^Ұ}߶assignment #021
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20:40:55 What are the possible number of linear and irreducible quadratic factors for a polynomnial of degree 6?
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RESPONSE --> polynomial of degree 6: 6 linear and no irreducible no linear and 6 irreducible 4 linear and 1 irreducible 2 linear and 2 irreducible the irreducibles consist of a degree of 2
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20:41:43 ** 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 --> i put 6 irreducibles, but of all had a degree of 2 then only three would be necessary.
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20:43:18 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 --> 2 zeros repeated twice 4 zeros 1 zero repeated three times and another 1 zero repeated four times
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20:43:45 ** 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 --> is there a difference between a root and a zero?
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20:48:47 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 --> 4 distinct zeros will go straight through the x axis at every zero moving up or down in an opposing pattern. 2 zeros repeated, the graph touches the x axis at it's vertex, is of a parabola shape. 1 repeated 3 times and the other disting from it would be a variaition of a cubic graph which goes up or down then passes straight through the x axis at another point. 1 zero repeated four times. would be a variation of a x^2 graph only fatter (horizontally stretched)?
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20:49:10 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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20:49:12 problems 3-5
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20:51:09 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 direction (up or down) that the graph of x^3 curves can change unlike a linear or quadratic graph.
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20:51:16 ** 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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20:52:00 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 --> the higher the degree the better the graph fits.
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20:52:09 11-09-2006 20:52:09 ** higher even degrees flatten out more near their 'vertices' **
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NOTES ------->
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20:54:10 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 --> question was just asked.
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20:54:42 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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20:58:04 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 --> f(t) = 1+2t+(2t)^2/2! = 1+2t+4t^2/2 = 1+2t+2t^2 = 1+2(.5)+2(.5)^2 = 1+1+.5 = 2.5 f(t)=e^(2*.5) = 2.71828 difference = approx. .21828
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20:59:20 ** 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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21:05:00 By how much does your accuracy improve when you make the same estimate using the degree 3 Taylor approximation?
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RESPONSE --> degree 3: 1+2*.5+2(.5)^2+4(.5)^3/6 = 2.583333 f(t)=e^(2*.5)=2.71828 difference = approx. .134947
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21:08:21 ** 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 --> wouldn't 4 t^3 / 3 be 4t^3/3! so it would be equal to 4 t^3 / 6?
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21:18:12 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 --> errors: degree 2 = .023721 degree 3 = .002888 degree 4 =.000283 degree 5 =.000023 the graph decreases as the degree values get larger
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21:19:37 ** 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 --> the degree errors are different from what i got. this is probably again because i used things like 3! as equallying 3*2*1 This was in the notes, when do you use this?
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21:27:17 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 --> e^.2 = 1.2214, actual 1.221403, error, .000003 e^.4 = 1.491733, actual 1.491825, error .000092 e^.6 = 1.8214, actual 1.8221188, error .0007188 e^.8 = 2.2224, actual 2.225541, error .003141 e^1 = 2.708333, actual 2.718282, error .009949 the graph increases exponentially in a .2 spacing.
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21:27:26 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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21:28:57 What is the function which gives the quadratic approximation to the natural log function?
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RESPONSE --> ln(x) = (x-1)-(x-1)^2/2
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21:29:09 ** 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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21:46:03 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 --> .6 error = -.084159 .8 error = -.016477 1 error = 0 1.2 error = -.011011 1.4 error = -.036861 approx is better as x approaces 1 because ln (1) = 0
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21:46:47 ** 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 used the for (x-1)-(x-1)^2/2+2(x-1)^2/6 to come to my answers, however, they differ. What method is supposed to be used?
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21:48:02 problem 12. What does the 1/x graph do than no quadratic function can do?
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RESPONSE --> the graph of 1/x has vertical and horizontal asymptotes that the quadratic function cannot do.
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21:48:09 ** 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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21:58:43 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 --> Using 1-(x-1)+2(x-1)^2/2 .6 = 1.56, actual 1.666667, error .106667 .8=1.24, actual 1.25, error .01 1=1, actual 1, error 0 1.2=.84, actual .833333, error -.006667 1.4=.76, actual .714286, error -.045714 the graph appears to be that of a cubic function, it decreases at a decreasing rate to x=1 where y is 0 and then begins to increase at an increasing rate for x values above 1.
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21:58:52 ** 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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