assignment 21

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00:12:05 What are the possible number of linear and irreducible quadratic factors for a polynomnial of degree 6?

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RESPONSE --> 3...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

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00:12:33 ** 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 was halfway correct, i think

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00:16:48 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 --> i'm not sure how to solve the problem. i understand more for a degree 4 polynomial with one irreducible quadratic factor and 3 linear factors.

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00:17:38 ** 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 understand a little more

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00:27:01 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, 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. 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. 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. 2 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. 4 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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00:28:36 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 --> i was close i think...

You were. You're doing well so far.

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01:18:07 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 --> a cubic polynomial fits the first graph better than a quadratic or linear polynomial because our expectation is to have a true model with a smooth curve and this fits that graph the best.

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01:18:39 ** 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 --> yes, i understand the concept of concavity

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01:21:29 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 --> because of the concavity. the shape and width of the parabola wil cause it to have a more or less better fit. and they're not all linear.

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01:22:00 ** higher even degrees flatten out more near their 'vertices' **

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RESPONSE --> ok

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01:22:18 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 --> more flattened

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01:23:39 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 --> i was a little correct

You correctly noted the relative shapes of the graphs of the y = x^4 and y = x^6 functions.

When combined with terms involving x, x^2, x^3 and x^5 to get polynomials, the resulting polynomials can 'wobble around' as indicated.

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01:31:26 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 approximation to f(.5) is f(t) = e^1 or e. ?? the approx. is fairly close to the actual value of e^(2t).

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01:32:04 ** 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 --> i wasn't exactly correct, but i understand it a little more and how to solve it

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01:34:29 By how much does your accuracy improve when you make the same estimate using the degree 3 Taylor approximation?

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RESPONSE --> The degree 2 Taylor series for e^x is 1 + x + x^2 / 2!. but The degree 3 series for this function is 1 + x + x^2 / 2! + x^3 / 3!, therefore it is more accurate than that of the 2 Taylor but less than that of the 4 Taylor approx. ??

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01:36:06 ** 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 --> i was correct i believe...

You gave a good answer.

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01:38:39 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 --> as the degree increases, the more accurate the approximation will be, so from left to right, as the degree increases, the error decreases.

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01:39:29 ** 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 was correct, but i didn't put the numerical values for the degree errors.

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01:47:48 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 --> i don't exactly fully understand how to solve for these

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01:48:18 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 --> ok...i think i'm understanding slowly

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01:50:58 What is the function which gives the quadratic approximation to the natural log function?

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RESPONSE --> y = ln(x) ???

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01:51:53 ** 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 --> i had the wrong function...i'm not sure if i understand this one all the way

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01:55:52 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 --> x vs ln(x) .6 = -.5108 .8 = -.2231 1.2 = .1823 1.4 = .3364

.6 is not equal to -.5108. Don't use = for anything but equality.

Your intent is that x = .6 gives you ln(x) = -.5108.

If you plug x = .6 into the degree-2 polynomial, as seen in the preceding solution, you get -.48, which differs from the accurate value -.5108 by .0308. So the error is .0308.

The errors for x = .8, x = 1.2 and x = 1.4 are also noted above.

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01:57:08 ** 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 can't understand what i did wrong and why it wouldn't be the same as plugging in x for ln(x) on your calculator. ??

See my note above.

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01:59:11 problem 12. What does the 1/x graph do than no quadratic function can do?

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RESPONSE --> the y = 1/x graph cannot be duplicated by any quadratic polynomial ??

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01:59:34 ** 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 understand

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02:02:23 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 --> x vs 1/x .6 = 1.66 .8 = 1.25 1 = 1 1.2 = .833 1.4 = .714

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02:02:47 ** 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 was halfway correct

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I believe you understand most of this. Feel free to ask specific questions on anything you aren't sure about.