course Mth 173 Calculus I11-24-2007
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22:31:17 These are subtle concepts, but very necessary to our understanding of both the power and the limitations of the Calculus. These are the ideas at the heart of the Calclulus, and it is these ideas that give us confidence in the very foundations of the subject. After Newton Leibnitz invented the calculus it took well over a century for basic questions about the foundations of calculus to emerge and it took some very great minds to figure out how to put it all on a rigorous foundation.
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RESPONSE --> okay
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22:31:26 What you are seeing here is that foundation.
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RESPONSE --> okay
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22:45:05 Query class notes #32 completeness, nested interval thm. Suppose that the value of a polynomial is < 0 at x = 1.7 and > 0 at x = 1.8, and that the derivative is positive on this interval. Let S stand for the set of x values in this interval for which the polynomial is less than 0. Then what is an upper bound for S? What must be the value of the polynomial at the least upper bound of S? What makes us think that there is the least upper bound for this set?
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RESPONSE --> The upper value for S would be the number that is squeezed out between the values as the function nears 0. The least upperbound would be 1.7 as that is the given upperbound of S.
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22:46:46 ** 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 and 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. **
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RESPONSE --> okay
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22:52:16 What does the Nested Interval Theorem tell us?
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RESPONSE --> That if we are given an inteval y
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22:54:03 ** 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. **
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RESPONSE --> The intreval is like the walls of a never ending hall way, even if you walk a very long ways the walls still converge to a single point.
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22:59:10 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?
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RESPONSE --> The intermediate theorem tells us that if 1.7<0<1.8 then somewhere between those number there is a zero however it may be a never ending nonrepeating decimal, it's still there.
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22:59:38 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.
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RESPONSE --> okay
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23:06:06 Query 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?
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RESPONSE --> Beacause r is in a nested interval then no matter how small the interval becomes then r is still in the middle of the interval and it is the only number that is, therefore making it unique.
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23:08:31 STUDENT 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. **
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RESPONSE --> okay
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23:17:14 Query Theory 2 Problem 12 from Limits and Continuity (was problem 12 p 134) various limits at infinity How do you algebraically manipulate the expression (x+3) / (2-x) so that its limit as x -> infinity can be easily obtained, and what is the resulting limit?
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RESPONSE --> since a limit isn't what happens at the limit but what happens as we approach the limit, we can use increasingly large values to map out the function to see what the function does. y= (x+3) / (2-x) x y 100 .515 500 .503 1000 .5015 5000 .5003 we can see that with increasing values y becomes larger but the larger the x value the smaller 'dy will be.
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23:23:35 STUDENT 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 closed as we wish to 1 / -1 = -1. The limit is therefore -1. **
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RESPONSE --> my answer was in the same boat as the first student's, although I can see the problems with this, where as with the second one you are thinking about the equation and it's seperate parts to figure out what it will ultimatly be.
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23:31:21 STUDENT 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. **
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RESPONSE --> As x grows larger the function ultimatly nears 3/2.
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23:39:10 Query 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.
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RESPONSE --> y= -2x +3 for x={0, 3} The function is linear and therfore will be a straightline and continuos and the limits as x->0 is 3 and as x->3 is -3
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23:48:34 ** 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. **
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RESPONSE --> I misunderstood the problem and the terminology, so epsilon is essentially another ""delta"" variable? I understand your limit and how you found it by making the rectangle exit the sides of the graph.
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23:54:56 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> Overall I think that I understand the diffrent theorums about intrevals and why they are the fundimentals of calculus.
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00:01:10 This has been by far the most difficult assignment ever in this course. I have not had much experience at proving things that I know to be true, just because they are. I need lots of help on this assignment.
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RESPONSE --> I was thinking this in the back of my mind but didn't want to bother asking it because it is maybe a little philisophical, but why must things that are known truths be proven. Somethings are just assumptions and must be proven, but usually you can ""feel"" the differance between the two. For instance, mathematically if I stick my hand behind my back there is a chance it's not really there, but I don't feel as though I need prove to tell me that it is there. Is it just one of those ""because it can be done"" things?
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