#$&* course Mth 163 21 query*********************************************
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Given Solution:`a** You can have as many as 6 linear and 3 irreducible quadratic factors for a polynomial of degree 6. For a polynomial of degree 6: If you have no irreducible quadratic factors then to have degree 6 you will need 6 linear factors. If you have exactly one irreducible quadratic factor then this factor is of degree 2 and you will need 4 linear factors. If you have exactly two irreducible quadratic factors then the product of these factors is of degree 4 and you will need 2 linear factors. If you have three irreducible quadratic factors then the product of these factors is of degree 6 and you can have no linear factors. For a polynomial of degree 7:If you have no irreducible quadratic factors then to have degree 7 you will need 7 linear factors. If you have exactly one irreducible quadratic factor then this factor is of degree 2 and you will need 5 linear factors to give you degree 7. If you have exactly two irreducible quadratic factors then the product of these factors is of degree 4 and you will need 3 linear factors to give you degree 7. If you have three irreducible quadratic factors then the product of these factors is of degree 6 and you will need 1 linear factor to give you degree 7. ** ********************************************* Question: `qFor a degree 6 polynomial with one irreducible quadratic factor and four linear factors list the possible numbers of repeated and distinct zeros. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: there could be 1 root repeated 4 times, 2 roots with 1 repeated 3 times and the other distinct from it, 2 distinct roots each repeated twice, three distinct roots with one of them repeated twice, or four distinct roots. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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. ** ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your 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. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution:The graph will pass through the x axis 4 times. Near each of the 4 zeros the graph will appear to be a straight line, though of course as you get away from the x axis the graph in between the zeros will need to curve in order to return to the axis. The graph will get increasingly steep at far right and far left, with the slopes at left and right having opposite signs (i.e., if the slope at far right is positive, with the graph moving up and to the right, then at far left the slope must be negative, moving down as it moves to the right; similarly if the slope at far left is positive, the slope at far right will be negative). The graph might look like the figure below; it might also be 'upside down' compared to the graph in the figure. ********************************************* Question: `qIt doesn't matter if you don't have a graphing utility--you can answer these questions based on what you know about the shape of each power function. Why does a cubic polynomial, with is shape influenced by the y = x^3 power function, fit the first graph better than a quadratic or a linear polynomial? What can a cubic polynomial do with this data that a quadratic can't? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The curve of a cubic can change. Linear graphs don't curve, quadratic graphs can be concave either upward or downward but not both on the same graph. Cubics can change concavity from upward to downward confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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. ** ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When degrees are more higher and even the more flatten out their vertices are confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution:`a** higher even degrees flatten out more near their 'vertices' ** ********************************************* Question: `qOn problem 5, how do the shapes of 4th-degree polynomials and 6th-degree polynomials progressively differ from the shape of a 2d-degree polynomial in such a way as to permit a better and better fit to the graph? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: On a degree-2 polynomial there is only one change of direction confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution:`aSTUDENT RESPONSE: progressively more flexing, because more curves, and fit graph better the that of a lesserdegreeINSTRUCTOR 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. ********************************************* Question: `qWhat 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)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2.718 .217 less than the function confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution:`a** The degree 2 Taylor polynomial is f(t) = 1 + 2t + (2t)^2 / 2! = 1 + 2 t + 4 t^2 / 2 = 1 + 2 t + 2 t^2.Therefore we have T(.5) = 1 + 2 * .5 + 2 * .5^2 = 1 + 1 + 2 * .25 = 1 + 1 + .5 = 2.5. The actual value of e^(2t) at t = .5 is f(.5) = e^(2 * .5) = e^1 = e = 2.718, approx..The approximation is .218 less than the actual function. ** ********************************************* Question: `qBy how much does your accuracy improve when you make the same estimate using the degree 3 Taylor approximation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4 times closer confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution:`a** 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. ** ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: .000023354 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 1
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Given Solution:`a**The value of e^.5 to 10 significant figures is 1.648721270. Each approximation should be rounded to a number of significant figures great enough to give at least a 2-significant-figure difference.The Taylor approximation for e^x is 1 + x + x^2 / 2! + x^3 / 3! + ... + x^n / n! + ...Thus for degree 2 the approximation is 1 + .5 + .5^2/2 = 1.625 F0r degree 3 the approximation is 1 + .5 + .5^2/2 + .5^3/6 = 1.6458 For degree 4 the approximation is 1 + .5 + .5^2/2 + .5^3/6 + .5^4/24 = 1.64838 approx For degree 5 the approximation is 1 + .5 + .5^2/2 + .5^3/6 + .5^4/24 + .5^5/120 = 1.648697917 The errors, if calculations are done accurately enough to yield a difference with 3 significant figures, are as follows: ê^.5 - (1 + .5 + .5^2/2 ) = 0.0237 ê^.5 - (1 + .5 + .5^2/2 + .5^3/6 ) = 0.00288 ê^.5 - (1 + .5 + .5^2/2 + .5^3/6 + .5^4/24 ) = 0.000283 e^.5 - (1 + .5 + .5^2/2 + .5^3/6 + .5^4/24 + .5^5/120) = .000023354 ********************************************* Question: `qWhat is the function which gives the quadratic approximation to the natural log function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: P2(x) = (x-1) - (x-1)^2/2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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. ** ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There is no error at x=1 and as the move is farther and farther away from 1 it becomes less accurate. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution:`a** The respective errors are .03, .00314, .00232, .016472. There is no error at x = 1, since both the function and the approximation give us 0. As we move away from 1 the approximation becomes less and less accurate. ** ********************************************* Question: `qproblem 12. What does the 1/x graph do than no quadratic function can do? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The y = 1/x graph has vertical asymptotes at the y axis and horizontal asymptotes at the x axis. The parabolas we get from quadratic functions do have neither vertical nor horizontal asymptotes confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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. ** ********************************************* 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 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph of approximation error vs. x gets greater as we move away from x = 1 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a** The quadratic approximation to 1/x is the second-degree Taylor polynomial P2(x) = 1 - (x - 1) + (x - 1)^2. A table of values of 1/x, P2(x) and P2(x) - 1/x: x 1/x P2(x) P2(x) - 1/x .6 1.666... 1.56 -.1066... .8 1.25 1.24 -0.01 1.2 0.8333.... 0.84 0.006666 1.4 0.714285... 0.76 0.457 A graph of approximation error vs. x decreases at a decreasing rate to 0 at x = 1, then increases at an increasing rate for x> 1. This shows how the accuracy of the approximation decreases as we move away from x = 1. The graph of approximation error vs. x gets greater as we move away from x = 1.** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating:
#$&* course Mth 163 21 query*********************************************
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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. ** ********************************************* Question: `qFor a degree 6 polynomial with one irreducible quadratic factor and four linear factors list the possible numbers of repeated and distinct zeros. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 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. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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. ** ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your 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. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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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. ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The curve 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 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution:`a** the concavity (i.e., the direction of curvature) of a cubic can change. Linear graphs don't curve, quadratic graphs can be concave either upward or downward but not both on the same graph. Cubics can change concavity from upward to downward. ** ********************************************* Question: `qOn problem 5, how do the shapes of 4th-degree polynomials and 6th-degree polynomials progressively differ from the shape of a 2d-degree polynomial in such a way as to permit a better and better fit to the graph? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When degrees are more higher and even the more flatten out their vertices are confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution:`a** higher even degrees flatten out more near their 'vertices' ** ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: On a degree-2 polynomial there is only one change of direction confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution:`aSTUDENT RESPONSE: progressively more flexing, because more curves, and fit graph better the that of a lesserdegreeINSTRUCTOR COMMENT: On a degree-2 polynomial there is only one change of direction, which occurs at the vertex. For degrees 4 and 6, respectively, there can be as many as 3 and 5 changes of direction, respectively. For higher degrees the graph has more ability to 'wobble around' to follow the data points. ********************************************* Question: `qWhat is the degree 2 Taylor approximation for f(t) = e^(2t), and what is your approximation to f(.5)? How close is your approximation to the actual value of e^(2t)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2.718 .217 less than the function confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution:`a** 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. ** ********************************************* Question: `qBy how much does your accuracy improve when you make the same estimate using the degree 3 Taylor approximation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4 times closer confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution:`a** The degree 3 Taylor polynomial is f(t) = 1 + 2t + (2t)^2 / 2! + (2t)^3 / 3! = 1 + 2 t + 4 t^2 / 2 + 8 t^3 / 6 = 1 + 2 t + 2 t^2 + 4 t^3 / 3. Therefore we have T(.5) = 1 + 2 * .5 + 2 * .5^2 + 4 * .5^3 / 3 = 1 + 1 + 2 * .25 + 4 * .125 / 3 = 1 + 1 + .5 + .167 = 2.667. The actual value of e^(2t) at t = .5 is f(.5) = e^(2 * .5) = e^1 = e = 2.718, approx..The approximation is .051 less than the actual function. This is about 4 times closer than the approximation we obtained from the degree-2 polynomial. ** ********************************************* Question: `qDescribe your graph of the error vs. the degree of the approximation for degree 2, degree 3, degree 4 and degree 5 approximations to e^.5. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: .000023354 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 1
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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 ********************************************* Question: `qWhat is the function which gives the quadratic approximation to the natural log function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: P2(x) = (x-1) - (x-1)^2/2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution:`a** The function is P2(x) = (x-1) - (x-1)^2/2. A table of values of ln(x), P2(x) and P2(x) - ln(x): x ln(x) P2(x) P2(x) - ln(x) .6 -0.5108256237 -0.48 0.03082562376 .8 -0.2231435513 -0.22 0.003143551314 1.2 0.1823215567 0.18 -0.002321556793 1.4 0.3364722366 0.32 -0.01647223662 At x = 1 we have ln(x) = ln(1) = 0, and P2(x) = P2(1) = (1-1) - (1-1)^2 / 2 = 0. There is no difference in values at x =1. As we move away from x = 1 the approximation becomes less and less accurate. ** ********************************************* Question: `qWhat is the error in the degree 2 approximation to ln(x) for x = .6, .8, 1.2 and 1.4? Why does the approximation get better as x approaches 1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There is no error at x=1 and as the move is farther and farther away from 1 it becomes less accurate. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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. ** ********************************************* Question: `qproblem 12. What does the 1/x graph do than no quadratic function can do? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 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 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution:`a** The y = 1/x graph has vertical asymptotes at the y axis and horizontal asymptotes at the x axis. The parabolas we get from quadratic functions do have neither vertical nor horizontal asymptotes. ** ********************************************* Question: `qWhat are the errors in the quadratic approximation to 1/x at x = .6, .8, 1, 1.2, and 1.4? Describe a graph of the approximation error vs. x YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph of approximation error vs. x gets greater as we move away from x = 1 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a** 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 1.666... 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: #*&!