course Mth173 assignment #026??????Ux?x????{+??Calculus I
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19:58:35 Query Theory 3 Problem 6 (was problem 6 page 186) integral from 1 to 4 of 1 / `sqrt(1+x^2). How many subintervals did you use, and how are you sure that the lower and upper sums differ by less than 0.1? What is the number of intervals required to ensure a discrepancy of less than .05?
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RESPONSE --> y= 1/('sqrt(1+1x^2)) Using the lf(b)-f(a)l * 'dt inequality method we get. x y 1 .71 4 .24 l(.71) - (.24)l *'dt < .1 l .47 l * 'dt <.1 'dt< .21 So as long as 'dt is less than .21 then the interval will have a descrepancy of .1 between the sums. It would make sense that since the first interval has a greater rate of change if it is below the maximum descrepancy then the other intervals would be below the descrpancy. x y 1 .71 1.2 .64 This is a maximum descrepancy of .07 a bit lower than .1 for an interval l(.47)l * 'dt < .05 'dt<.11 the interval would be under .05 x y 1 .71 1.1 .67 This gives us a maximum descrepancy of .04
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20:02:38 STUDENT RESPONSE: Sums given. I would estimate that a n-value of 20 would yield a value of .05 of the true value. My reasoning is that when n = 10 there is a difference between the left and right hand sums of .139. And, when n = 15 there is a .093 difference in the left and right sums: .139 - .093 = .046. So, I predict that with another increase of n, by a value of 5 we will gain another .05 (or so) of accuracy. Therfore, my estimate is n = 20 (Though I didn't calculate it) INSTRUCTOR CRITIQUE: Good thinking, but since you are dividing by the number of intervals the situation isn't that linear. The max error is | f(b) - f(a) | * `dx. Since f(1) = 1/`sqrt(2) = .707 and f(4) = 1 / `sqrt(17) = .245, approx., we have max error | .245 - .707 | * `dx Setting this equal to 0.1 we get |.245 - .707| * `dx < 0.1 so `dx < .1 / .462; `dx < = .2 would do. Since the interval has length 4-1 = 3, then `dx < = .2 implies n > = 3 / .2 = 15. This agrees with your result. To get the discrepancy lower than .05 we use the same means, this time obtaining equation |.245 - .707| * `dx < 0.05, satisfied for `dx < .05 / .462. We could get away with `dx = .11, but .10 is nicer and divides evenly into interval length. If `dx = .10 we get n = 3 / .10 = 30. ** Note that n = 20 won't assure us of a discrepancy of < .05. ** ALTERNATIVE SOLUTION: ** You need to get a good picture of this situation in your head; of course that should probably start on paper. Sketch a graph of this function from x=1 to x=4 and subdivide into 3 intervals. Sketch the lower rectangles of the 3-interval Riemann Sum, then the upper rectangles. Shade in the small rectangles that represent the differences between the upper and lower rectangles. Now imagine moving each of these rectangles straight to the right, until they're at the right-hand side of your figure, with one on top of the other. The three rectangles now stack into a single rectangle whose altitude is f(b) - f(a) and whose width is 1, the width of a rectangle. The area of this stack is the difference between the upper and lower sum, and its area is [ f(b) - f(a) ] * `dt, where `dt is the width of a single rectangle. Now if you made the `dt smaller, say, .1, this would make all the little rectangles thinner; and when you stacked them the stack would be just as high as before, but skinnier. Its area would be area = diff between upper and lower=(f(b)-f(a))'dt. Since `dt is smaller so is the area, and so therefore is the area. You found that if `dt = .22, you can reduce the difference to .1. If `dt is .22 or less, then, the difference will be no greater than .1. On an interval of length 4 - 1 = 3 this implies that n > 3 / .22 = 13.7 or so; any n >= 14 will do it. **
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RESPONSE --> yes the intervals where .2 and .1 with a range of 3 so that would give us a number of intervals: 'dt # of intervals .1 30 .2 15 this would make sense because the number of intrevals is doubled for a descrepancy that is twice and accurate.
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20:30:19 Query Theory 3 Problem 20 (was problem 20 p. 187) prove for continuous f, c in [a,b] that int from a to b = int from a to c + int from c to b Explain how you have proved the stated result.
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RESPONSE --> If we have an interval of a and b with a number c within this interval. Since the maximum and minimum of the limit is always going to be a and b (what I mean my this is if all of our rieman sums recangles are shifted on top of each other this will be the height of the greater rectangle.) And since the function is applied to all of the numbers the same way then they will be proportional to the numbers a,c,and b. So lets let the numbers equal 3, 7, and 17, since the function is applied to each of them the same and there values after being run through the function will be proportional to there orginal values we can just use the original values. a=3 c=7 b=17 b-a= 14 c-a= 4 b-c= 10 so (c-a) + (b-c) = 14
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20:36:51 ** To understand what this problem is talking about, you can think first about areas. The area under a curve from a to b is equal to its area from a to c, plus its area from c to b. That is, you can split the region under the curves at any x value between a and b and the total area will be the sum of the areas. The strategy shown in the text is to show that if you find a lower sum from a to c, and a lower sum from c to b, that when you add them you have a lower sum from a to b. Why should this be so? Then you show that for any lower sum on the interval [a, b], you can find lower sums on [a, c] and on [c, b] whose total is greater. This isn't hard to show because when you refine lower sums by splitting the interval up, the lower sums tend to get larger. So if a = x0 < x1 < ... < xn = b is a partition of [a,b], how can you refine this partition to get partitions of [a, c] and [c, b] for which the lower sum is no less than the lower sum you get over [a, b]? Note that it is not necessarily the case that c is equal to one of the numbers x0, x1, ..., xn, so you can't assume that it is; hence there is at least some work to do here. See if you can do it. What you show in the first two parts can be used to show that the least upper bound of lower sums on [a, b] is equal to the sum of the least upper bound of lower sums on [a, c] and on [c, b]. Since the least upper bound of lower sums of a continuous function is equal to the integral, this gives the desired result. See if you can put the proof together. **
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RESPONSE --> essentially we are saying the same thing but you are describing it graphically, where I used a more algebric approach, you go into more depth about what the parts represent on a graph. I do however think that my explanation is reasonable, do you agree, or would you have me to give more detail?
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20:40:19 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> Using the definate integral is similar to finding limits you are predicting what will happen as the interval becomes smaller and smaller.
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20:41:07 I am still having a lot of trouble on the 'Focus on Theory' parts of the text. This tells me that my weakness in math lies in proofs. I have had little experience with proofs, so I do need a lot more practice.
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RESPONSE --> I asked this question in assignment 24.
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