#$&* course Mth 173 7/16 3:38 PM 024. `query 24
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Given Solution: `a** There is a set S of numbers for which f(x) < 0, and since for the increasing function f(x) we have f(1.7) < 0 < f(1.8), as you correctly point out, it should be obvious that that set S contains 1.7 and does not contain 1.8 or any number > 1.8. Therefore 1.8 is an upper bound for that set. The set S is nonempty (it contains 1.7) and it has an upper bound. Therefore by the Completeness Axiom it has a least upper bound--of all the numbers, 1.8 included, that bound the set from above, there is a smallest number among all the upper bounds. Since S is the set of all values for which f(1.7) < 0, then if r is the least upper bound of the set, we must have f(r) = 0. If f(r) > 0 then by the continuity and increasing nature of f(x) there would be some smaller x, i.e., x < r, for which f(x) > 0, in which case r couldn't be the least upper bound. So f(r) <=0. But f(r) can't be less than 0 because then there would exist an x > r for which f(r) < 0 and r wouldn't be an upper bound at all. So were stuck: if r is the least upper bound then f(r) = 0. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ????? I do not understand the last part of this at all? Why could 0 just not be the least upper bound?
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Given Solution: `a** It tells us that an infinite sequence of nested closed intervals contains at least one common point. These intervals, which contain their endpoints (i.e., are closed), can't just to put away to nothing. We cannot construct, even if we try, and infinite sequence of closed nested intervals for which there is not a common point. The sequence can't just dwindle away to nothing. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: What does the Intermediate Value Theorem tell us, and how does this help assure us that the polynomial of the first question has a zero between x = 1.7 and x = 1.8? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If a function y = f(x) is continuous on an interval (a, b) of the x axis, then it must take every y value between f(a) and f(b) Since the polynomial is less than 0 for x=1.7 and greater than 0 for x=1.8, the intermediate value theorem says that between this interval the polynomial must take every y value. This includes an infinite number of values between 1.7 and 1.8 which would include a 0 in the polynomial function values confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If f(x) is continuous between two points a and b, then f(x) takes every value between f(a) and f(b). Since the polynomial is negative and one of the given x values and positive at the others, and since a polynomial is continuous at every value of x, it follows from the Intermediate Value Theorem that the polynomial must take the value 0 somewhere between these x values. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery Theory 1, Problem 2 (was Problem 2 p. 85) r is contained in a sequence of nested intervals whose width approaches 0 as n -> infinity; prove r is unique. How do you prove that r is unique? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: r is unique because as the interval goes to 0 the n values approach infinity. This means that there is an infinite number of r values as the interval goes to 0. There will not be a the same value of r in this interval because the n value will always be increasing confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSTUDENT SOLUTION: This is all that I have: lim as n app infinity of b of n - a of n r is greater than or equal to a of n for all n r is less than or equal to b of n for all n when r app zero, there is a unique number r INSTRUCTOR CRITIQUE: r doesn't approach zero, the width of the intervals approaches zero. Suppose there were two values, r1 and r2, and that |r2 - r1| wasn't zero. Since the sequence of nested intervals has width approaching 0, eventually the width will be less than any number we might choose. In particular the width will eventually be < | r2 - r1 |, meaning that for some N, if n > N the width of (an, bn) must be < |r2 - r1|. The two values r2 and r1 therefore cannot both exist in any of these intervals, which contradicts the assumption that there could be two different numbers both contained in all the intervals. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ????I think what I said is similar to the given solution but I didn’t use two different values of r to explain it? ------------------------------------------------ Self-critique Rating: 1
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Given Solution: `aSTUDENT RESPONSE: Simply by plugging a large number into the equation for x. Then repeating with even larger numbers to verify. resulting limit -1. INSTRUCTOR CRITIQUE: That's not an algebraic manipulation. You can't rigorously prove anything by plugging in numbers--the behavior might be different if you used different numbers. ANOTHER STUDENT RESPONSE: I did this by using p. 129 to simplify and take quantities off to themselves to get an answer I found that this equation ultimately comes down to (inf+3)/(2-inf) The top portion will be positive by adding an infinitely large number, while the bottom portion becomes negative by adding an arbitrarily large number, whoch gives +/- or -inf INSTRUCTOR CRITIQUE AND SOLUTION: You have the right idea, but inf (meaning infinity) is not a number so you can't formally deal with a quantity like (inf + 3) / (2 - inf). If you divide both numerator and denominator of (x+3) / (2-x) by x you get ( 1 + 3/x) / ( 2/x - 1). Then as x -> infinity, meaning as x gets as large as you might wish, 3 / x and 2 / x both approach 0 as a limit, meaning that each of these quantities can be made to be as close to zero as we might wish. As a result ( 1 + 3/x) / ( 2/x - 1) can be made as close as we wish to 1 / -1 = -1. The limit is therefore -1. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ????I’m very confused on this… I understand your calculations but I know I would not have been able to come to this conclusion on my own. I thought I would need to use the properties of limits to find this ------------------------------------------------ Self-critique Rating:
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Given Solution: `aSTUDENT RESPONSE: As the x value grows arbitrarily larger, both the top and bottom portions will go to infinity, which yields positve infinity as the result INSTRUCTOR CRITIQUE AND SOLUTION The same could be said of x / x^2, but the limit isn't infinite. If we divide both numerator and denominator by e^x we get (3 + 2 / e^x) / (2 + 3 / e^x). The terms 2/e^x and 3 / e^x approach zero so the fraction approaches 3/2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understand this better after looking at your solution to the previous question and comparing it to this question ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery Theory 2 Problem 18 from Limits and Continuity (was problem 18 p. 135) find positive delta such that the graph leaves the 2*epsilon by 2*delta window by the sides; fn -2x+3, a = 0, b = 3. Give your solution. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: |f(x) - L| < epsilon |x-c| < delta a - delta < x < a + delta b - epsilon < y < b + epsilon epsilon = .2, .1, .02, .01, .002, .001 I really have no idea how to find this… I know the epsilon values correspond with the y values. But I’m not sure how to relate them to the x values so that the graph leaves the window by the sides? Here’s a guess: Since b = 3 and b goes with the epsilon values: would the interval be from 3- .2 < y < 3+ .2 and 3-.001 < y < 3 + .001 Then set these values equal to -2x + 3 to find the x value? confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The graph of this function is a straight line passing through (a, b) = (0, 3) with a slope of -2. The window we are talking about lies within a `delta-interval about x = a--i.e., with the interval (a - `delta, a + `delta). This is an interval that extends to distance `delta on either side of x = a, and is therefore centered at a. You should sketch a picture of such an interval. Since a = 0, the x interval is (-`delta, +`delta). You should sketch a picture of this specific interval, thinking of `delta as a small number which can be made as small as we wish. On the plane, this interval defines a vertical 'strip' centered at the y axis and extending distance `delta on either side. You should sketch a picture of this 'strip'. The number `epsilon defines base similar interval on y values--an `epsilon-interval about b, extending from b-`epsilon to b+`epsilon. You should sketch a picture of such an interval. For the specific value b = 3, the `epsilon-interval is (3 - `epsilon, 3 + `epsilon). You should sketch this interval. This interval defines a horizontal 'strip' centered at the line y = 3 and extending to distance `epsilon on both sides. You should sketch a picture of such a strip. The two strips, one vertical and one horizontal, meet to form a rectangle at their intersection. This rectangle is a 'window' whose width is 2 `delta and whose height is 2 * `epsilon. Sketch the graph of the function in your window. If your `delta is too large for your `epsilon, the graph will exit the window at top and bottom, not at the sides. We regard `epsilon as given, and we adjust `delta so that the graph exits the window at the sides. It should be clear that by making `delta small enough this is possible. It should also be clear that this will be possible no matter how small `epsilon is chosen. Thus FOR EVERY `epsilon, no matter how small, THERE EXISTS a `delta such that WHENEVER | x - a | < `delta, IT FOLLOWS THAT | f(x) - b | < `epsilon. This is what defines and proves the statement that limit {x -> a} (f(x)) = b. ** INSTRUCTOR SUGGESTIONS FOR THE PERPLEXED: If you don't understand this problem, try to following. Actually attempt to do the steps, sketch your window and the graph, before you read the explanation: First verify that the graph of the function y = -2x + 3 passes through the point (a, b) = (0, 3) of the xy plane. Now sketch the 'window' for epsilon = .1 and delta = .2: The 'window' is a rectangle which extends epsilon units above and below the point (a, b), and delta units to the right and left of this point. The point (a, b) will be in the middle of the window. The y coordinate therefore extends .1 unit above and below y = 3, from y = 2.9 to y = 3.1. The x coordinate extends .2 units to the left and right of x = 0, from x = -.2 to x = .2. So the 'window' consists of the rectangle 2.9 < y < 3.1, -.2 < x < .2. Sketch the graph of the function within this window: For x = -.2 we get y = 3.4, and for x = +.2 we get y = 2.6. The point (-.2, 3.4) lies above the window, and the point (.2, 2.6) lies below, so neither of these points lies on or within the boundary of the window. For y = 2.9, we find that x = .05. So the point (.05, 2.9) lies on the lower boundary of the window. For y = 3.1, we find that x = -.05. So the point (-.05, 3.1) lies on the upper boundary of the window. Thus our graph within the window runs from the point (-.05, 3.1) to the point (.05, 2.9). Does the graph leave the window by the sides, or by the top and/or bottom of the window? The graph leaves the window by the top and bottom. The x = -.2 and x = +.2 points are too far above and below y = 3 to exit by the sides of the window. So this window doesn't fulfill the condition of the problem. We conclude that epsilon = .1 and delta = .2 doesn't fulfill the specified condition. Now try it again for epsilon = .1 and delta = .01. Your window will be the rectangle -.01 < x < .01, 2.9 < y < 3.1. When x = -.01, y = 3.03. The corresponding point (-.01, 3.03) lies on the left-hand boundary of the window. When x = .01, y = 2.97. The corresponding point (.01, 2.97) lies on the right-hand boundary of the window. The graph now leaves the window through its sides. You might want to try a few more combinations of epsilon and delta to get a better feel for this situation. Then read the given solution once more, and feel free to submit additional questions. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): After reading your suggestions this question makes better sense. I wasn’t looking at the question as the equation for a straight line. That also helped me get a better understanding of the question. I didn’t draw the graphs as I was trying to complete this problem so that definitely would have helped. ??? I have one question though, where are you getting the values of epsilon and delta to use on the graph? Were those just made up or should I see where they’re coming from based on the problem? ------------------------------------------------ Self-critique Rating: 0