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[cos(x)]^infinity, not continuous? Infinity even or odd?for any value "a" less then 1 but greater then zero, a^infinity = 0, correct? Last edited by mikau (20060710 11:47:49) A logarithm is just a misspelled algorithm. #2 20060710 12:17:43
Re: [cos(x)]^infinity, not continuous? Infinity even or odd?Interesting! Play with it: Plot of cos(x)^1000 vs cos(x)^1001 "The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  Leon M. Lederman #3 20060710 12:43:09
Re: [cos(x)]^infinity, not continuous? Infinity even or odd?precisly! Just look at the difference (+ 1) can make! Now get out there and vote! A logarithm is just a misspelled algorithm. #4 20060710 14:45:10
Re: [cos(x)]^infinity, not continuous? Infinity even or odd?mikau, if you've come across this by just thinking about math (i.e. not reading it in a book or from a professor), then I am really impressed. Even more so since (I think) you've never taken a reals course.
Not to nitpick or anything, and I'm pretty sure you know what you mean when you say the above, but it should read:
Let's apply the same terminology: This is called squences of functions. It works just like a squence of numbers, just replace numbers with a function. And they are really weird. I mean really, really weird. And you just showed the weird thing about them:
And yes, it sounds like there must be. But there isn't. The graph is not continuous. Let me introduce some terminology to make talking about these easier: "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #5 20060710 15:46:44
Re: [cos(x)]^infinity, not continuous? Infinity even or odd?As I always claim, using the concept of Real Infinity is using guessing instead of logic, and selfdefeating. X'(yXβ)=0 #6 20060710 23:50:30
Re: [cos(x)]^infinity, not continuous? Infinity even or odd?Are you saying limits shouldn't be used, George? "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #7 20060711 08:14:51
Re: [cos(x)]^infinity, not continuous? Infinity even or odd?Thanks, Ricky! Yes I came up with this while eating dinner at a local pizza shop ( I said it before: I'm such a nerd) and I wanted to tear off my clothes and run down the street screaming "Eureka!" :D A logarithm is just a misspelled algorithm. #8 20060711 10:49:28
Re: [cos(x)]^infinity, not continuous? Infinity even or odd?I mean the Real Infinity concept can only be "used" after Approaching method. Without approaching, equating approaching with being can cause fault. X'(yXβ)=0 #9 20060711 11:01:54
Re: [cos(x)]^infinity, not continuous? Infinity even or odd?Simply put, to mikau, do not use an independent, nonvariable infinity or infinitesimal at first and that will save you from lots of faults. X'(yXβ)=0 #10 20060711 11:02:41
Re: [cos(x)]^infinity, not continuous? Infinity even or odd?Ok, I think I'm starting to understand what you are saying, Geroge. It was, "Without approaching, equating approaching with being can cause fault." that got me. I've never heard of the words "Real Infinity" (nor can I find them used on the internet), so I think the language barriar was the thing that got me. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #11 20060711 11:51:59
Re: [cos(x)]^infinity, not continuous? Infinity even or odd?heheh! A logarithm is just a misspelled algorithm. #12 20060711 12:55:39
Re: [cos(x)]^infinity, not continuous? Infinity even or odd?By putting reals in quotes, mikau, I take it you've never really heard of real analysis before. So let me take a brief minute to explain what it is. Newton started off calculus, an entirely new form (well, not really) of math. Then some more funky dudes with wigs on countinued studying it. But it wasn't really till the 1800's that they started noticing weird things. There were all sorts of crazy functions being defined, and they had really weird properties. They started seeing some contradictions in different problems, some theorems were'nt working as expected. So a few guys sat down and said, "All right, let's take this back to square 1, and start off from the beginning making sure we are careful." And so Real Analysis was born. It is called real analysis because it is the study of the real numbers and functions on the real numbers, trying to find out their properties and analyzing what they do. Now I will prove an earlier statement. I will do the proof, and then an explanation of it. I suggest you read the proof, then the explanation, then go back and reread the proof. Prove f_n(x) = x^n converges pointwise to 0 for all values 0 < x < 1 Proof: Let a_n be a squence such that a_n = {a, a^2, a^3, ...}, for some number 0 < a < 1. Let ∈ > 0. Then what we must show is that there exists N in the natural numbers such that a_n < ∈ for every n ≥ N. If we show that a_n is constantly decreasing and bounded below by 0 (i.e. does not contain a value less than 0), then this will be sufficent to show that such a number exists. So first, we must show that a_n is decreasing. We do this by induction. Since a < 1, a*a < a (multiply both sides by a). So that is the base case. Assume that a_n > a_n+1. Now we must show that a_n+1 > a_n+2. Since 0 < a_n, and 0 < a, 0 < a_n*a, or rather, 0 < a_n+1. But since a_n > a_n+1, a_n*a > a_n+1*a since 0 < a, and so a_n+1 > a_n+2. Thus, 0 < a_n+2 < a_n+1, and so a_n is always decreasing. Now we must show a is bounded below by 0. Already did that with the 0 < a_n+2. So it holds that for every ∈ > 0, there exists some N in the natural numbers such that for every n ≥ N, a_n < ∈ And now I claim that we're done. But what the heck did we just do? Well, we showed that if you out far enough to the right (or rather, a higher value for n), for any real number you pick, there will be a point such that every single a_n will be less than that real number. So this means that if I pick ∈ = 0.00000000001, I can still find an N such that all n's past it, a^n < ∈. And the same thing will happen if I pick ∈ = 0.00000000000000000000000000000000000001. And the same for ∈ = 10^1000000. So this proves that the farther I go out to the right, the closer my sequence will get to 0, and thus, it is the limit of the sequence. Now all that's left to show that 1^n = 1 as n goes to infinity, which is easy enough, so we showed that the function f(x) (what f_n(x) approaches) is discontinuous, even though ever f_n(x) itself is continuous, as all polynomials are. I'm sure you must have some questions, so I'll stop here for now. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #13 20060711 13:56:21
Re: [cos(x)]^infinity, not continuous? Infinity even or odd?I probably would have questions if I understood it a little better. I'm just struggling to keep up. I've never really dealt with induction before. A logarithm is just a misspelled algorithm. #14 20060712 12:36:27
Re: [cos(x)]^infinity, not continuous? Infinity even or odd?Ok, so let me take a few giant steps back. That first post was really to see how much you understand, and I think I got a pretty good idea. This simple set looks like {1, 2, 3, 4, 5, ...} So a_0 = 1, a_1 = 2, and so on. What we wish to prove is that a_n < a_n+1 for all n in the natural numbers. It seems pretty obvious, no? But how do we prove it? First, note that 1 < 2. That is, a_0 < a_1. This is the base case. Now we make our inductive assumption. a_n < a_n+1. It seems like we are assuming the conclusion, but we aren't. I'll come back to this. Since a_n < a_n+1, it must be that (a_n) + 1 < (a_n+1) + 1. But we also know that (a_n)+1 = a_n+1 and that (a_n+1) + 1 = a_n+2. So it holds that a_n+1 < a_n+2. Now I can claim that a_n < a_n+1 for all n in the naturals. And here's why. We first showed that a_0 < a_1. We also showed that if a_n < a_n+1, then it must be true that a_n+1 < a_n+2. So since a_0 < a_1, it must be that a_1 < a_2. And since a_1 < a_2, it must be that a_2 < a_3. And since a_2 < a_3.... and so on. Since this pattern must continue for all natural n, we have showed it true for all natural numbers. Induction is extremely important in real analysis, as we always deal with sets where the elements have some sort of relationship to one another, just like the set {1, 2, 3, 4...}. Here is a formal induction proof: Prove that for all n∈N Proof: This proof is by induction on n. Note that . Thus, the base case holds. Assume that . Now it must be shown that , as we are just taking off the last number in the summation and writting it seperately. (remember the inductive assumption And so . Therefore, by PMI, for all n∈N. (PMI stands for Principal of Mathimatical Induction) Any questions about induction? If not, try a few good induction proofs. Prove that Prove that Prove that for any number 0 < a < 1, a^n > a^(n+1). "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #15 20060713 23:29:43
Re: [cos(x)]^infinity, not continuous? Infinity even or odd?Probably it's not the language barrier, Ricky. X'(yXβ)=0 #16 20060714 11:52:06
Re: [cos(x)]^infinity, not continuous? Infinity even or odd?I think I understood most of that, Ricky. I mean it makes sense it would be bounded above by 1 and below by 0, and I have little interest in proving such an obvious fact while most mathematicians do. (no offense or anything, your math knowledge is clearly quite extensive!) A logarithm is just a misspelled algorithm. #17 20060714 12:10:14
Re: [cos(x)]^infinity, not continuous? Infinity even or odd?
By making "n" (which is f(x) in your equation) in terms of x, you are completely changing the problem. You have to set n in stone, so to speak, and then show that as you approach infinity , (1n)^x = 0.5. n can't vary with x. And again, I state that no such real number exists. I'll formulate a very clear and careful proof possibly tomorrow.
Language barrier, again, I think. By writing an infinite series, you are already implying that you are talking about limits, George. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #18 20060714 12:21:15
Re: [cos(x)]^infinity, not continuous? Infinity even or odd?Thanks I am gradually learning from you guys, but this equation has one mistake. The summation grows one step ahead of the n up 2. Maybe this is better? igloo myrtilles fourmis #19 20060714 12:38:00
Re: [cos(x)]^infinity, not continuous? Infinity even or odd?One more thing. Are the following two things equivalent with and without parenthesis?? igloo myrtilles fourmis #20 20060714 14:12:26
Re: [cos(x)]^infinity, not continuous? Infinity even or odd?I think you make a word misguidance, Ricky. Bounded below zero seems to imply that the bound is some number negative. Should it be bounded above zero? X'(yXβ)=0 #21 20060714 14:17:55
Re: [cos(x)]^infinity, not continuous? Infinity even or odd?for x, x>0 and x<1, x^n converges to 0 X'(yXβ)=0 #22 20060715 02:46:29
Re: [cos(x)]^infinity, not continuous? Infinity even or odd?Lets say the limit of f(x)^x as x approaches infinity is 0.5 and 0 <= f(x) <= 1 for all x and f(x) is continuous.... Last edited by mikau (20060715 02:52:57) A logarithm is just a misspelled algorithm. #23 20060715 08:09:17
Re: [cos(x)]^infinity, not continuous? Infinity even or odd?Consider this: we would get 0 for every value of x except for x such that cos x = ±1, since ±1^{∞} is indeterminate. Thus, we would have a line tracing the xaxis with point discontinuities at all x = nπ for n ∈ Z. #24 20060715 10:51:09
Re: [cos(x)]^infinity, not continuous? Infinity even or odd?
Right you are. Or rather would be, if there wasn't a trick, which I get to below. QED Edit: And furthermore, something I just saw, is that this proof makes complete sense if a = 1, because then the limit doens't have to be 0! Hah, don't you just love it when everything comes together so well? "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #25 20060716 00:58:40
Re: [cos(x)]^infinity, not continuous? Infinity even or odd?Great proof in Latex! Your proof of is ideal because it requires virtually little prerequisite. Good effort,Ricky! Considering seri monotunely decreasing and bounded below has a limit, if your proof does not require a property of real numbers I would like to see it. To Mikau, that is not a fair game, even when you choose 0.999999999 as a, n can be large enough to lower a^n down to 0.0000000001 the defination of limit is powerful enough to beat you. Last edited by George,Y (20060716 01:03:50) X'(yXβ)=0 