#$&* course MTH 173 8/7/11 026. `query 26
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Given Solution: 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qQuery 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&&&&&&&&&&&&&& **************** If you get the integral of a function from x = a to x = c, you get F(c) - F(a), integral of function from x = a to x = c: F(c) - F(a) integral of function from x = c to x = b: F(b) - F(c) so F(c) - F(a) + F(b) - F(c) = F(b) - F(a), which is the integral of the function from x = a to x =b. &&&&&&&&&&&&&&&& **************** confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** The lowest sum doesn't exceed the least upper bound. A lower bound is any quantity which is less than the value you are seeking. There is no limit to how low a lower bound can go. There's a limit to how high it can go--it can't exceed the sought value. There's also a limit to how low an upper bound can go (the 'least upper bound'), but no limit to how high it can go (no such thing as a 'greatest upper bound'). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!