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What about the measure-theoretic foundations of probability theory. The section on probability is very sorely lacking on this point!
Also, the link to measure should go to measure_(mathematics).
As a native French speaker who's had little exposure to English terminology, I wonder what the differences in meaning and use between "analysis" and "calculus" are. I know that "calculus" is more frequent than "analysis", while "analyse" is more frequent in French than "calcul" in the sense of "calculus", but I don't know much more. _R_ 00:23, 16 Oct 2004 (UTC)
Thanks! It seems then that "analysis" always translates to "analyse" and "calculus" (alone) also translates to "analyse" while "differential calculus" (resp "integral", etc.) translates to "calcul différentiel" (resp. "intégral", etc.) The translation problem is further complicated by the fact that "calcul" usually means "computation" or sometimes "counting". Interwikis to and from fr: may become quite a mess... _R_ 14:12, 16 Oct 2004 (UTC)
" Historically, analysis originated in the 17th century, with the invention of calculus by Newton and Leibniz. "
This is not only POV but false.
One reference is here: [1]
" ...We may consider Madhava to have been the founder of mathematical analysis. Some of his discoveries in this field show him to have possessed extraordinary intuition. [GJ, P 293] " [G Joseph]
This page gives the wrong impression that mathematical analysis started with Newton and Leibniz, while Madhava pre-dated them by 300 years. I am restoring the POV tag.
Just to be trendy, I wonder if there isn't one or two Muslim mathematicians who could be found to have priority in the discovery of Mathematical Analysis.66.82.9.58 14:17, 14 January 2007 (UTC)AubreyAubervilliers
This article is biased and inaccurate. There should be no mention of Dedekind, Weierstrass or anyone else besides Archimedes. Dedekind contributed nothing with his theory of cuts. He was considered an idiot by his fellow mathematicians. As for Weierstrass, his attempts to 'rigorize' limits using epsilon-delta proofs have resulted in more confusion rather than rigour. First, Weierstrass dismisses infinitesimals and then in the same breath introduces terminology such as "...as close as you like...". The common epsilon-delta definition of a limit is conceptually flawed: to say that epsilon and delta can be as small as you like but not exactly zero directly implies the existence of infinitesimals and is thus contradictory. How close is 'close'? The theory does not strictly conform to 'reality' for there is a number that follows 0 which is less than every positive number even though we can't find it. If this number does not exist, it implies a discontinuity in the real interval (0,1) which is of course untrue. Thus infinitesimals exist and Weierstrass's assumptions are false. To say that numbers can be as small as you like, as long as they are not exactly zero is neither rigorous nor precise. Archimedes did not believe in infinitesimals yet he used the idea of infinitesimals in his methods of exhaustion. Mathematics is no more rigorous today than it was in the time of Archimedes. --68.238.102.180 11:18, 29 October 2005 (UTC)
Yes, it's been in textbooks for well over 100 years but it's still wrong. "as close as you like" applies to numbers, not sets. Just look at how both of you responded: for one thing, you stated the definition incorrectly for it does not say for all epsilon there exists a delta. Read the definition carefully. Here is a common example where it fails (but there are others):
If and only if for each epsilon > 0 there exists a delta > 0 with the property that
|f(x) - L| < epsilon whenever 0 < |x - a| < delta
then
Lim f(x) = L x->0
This is the definition. Now consider the function f(x) = |x| and let's investigate L = 0.
Then |x - 0| < epsilon whenever 0 < |x - 0| < delta.
This implies that
Lim f(x) = 0 (Choose any epsilon you like and set delta equal to it) x->0
hence, by this non-sense Weierstrass definition, f(x) = |x| could essentially have any limit you like. The value of L in this case is quite irrelevant since we can always find a delta and epsilon greater than zero such that the above definition is true. Now it is taught that f(x) has no derivative at zero which is in total contradiction to the Weierstrass definition that is used! [In actual fact the derivative of this function is zero at x=0 but you cannot use Weierstrass' definition to prove it. You will require true rigourous mathematics to show this fact.]
Nonsense. You cannot have x=0. The above as I stated it is perfectly correct. By the definition you cannot say |0 - L| < epsilon. Seems you made an elementary mistake. -- unsigned by 68.238.102.180.
I still see a problem with this: If you assume L=1, then you cannot let epsilon = 0.5 for then ||x| - L | is not less than epsilon. You in Europe/Russia? Good night. -- unsigned by 68.238.102.180.
Not suggesting that you remove it completely but rather that you state the facts correctly: Weierstrass's attempts were to rigorize calculus. I would say he failed and that his work is mostly in error. 68.238.102.180.
No. You cannot choose epsilon indiscretely. Why? Well, you have to start with an epsilon that makes the above statement true, i.e. | x - 0 | < e whenever | x - 0 | < d. Epsilon gives you some insight into the choice of delta. Your first epsilon must make the first part of the definition true for otherwise you have a false assumption and everything else that follows is also false. See? In the above example, all you need to do to make it all true, is set delta = epsilon. And then once you have done this, you will realize that you can essentially show it is *true* for any L which is of course *false*. You start with epsilon, not delta. It is a bit confusing that it states |f(x) -L| < e whenever |x-a| < d because you find a d using e and then turn the statement around so that it reads as it does above in the definition. There is most definitely a problem with this absolute value function when you try to use Weierstrass e-d proofs even though this tends to work correctly with most (but not all) other functions. 68.238.102.180.
No, you are wrong. The way I stated it is perfectly correct. You evidently have no idea what you are talking about. Let me repeat: It is true that | x - 0 | < e whenever | x - 0 | < d. This is exactly the same as: | f(x) - 0 | < e whenever | x - 0 | < d. You obviously did not think about it eh? In this case f(x) = |x| and | |x| - 0 | = | x - 0 |. Thus I have stated it perfectly correct and it is you who needs to pay attention! 68.238.102.180.
I passed my real analysis course a long time ago. Of course this does not mean I agreed with what was taught. The only one who is making mistakes is you! My posts here have one purpose: to *correct* you. I am not seeking your help. I know what I am talking about. I also know that I am completely correct. It is you who does not have the slightest idea of what you are talking about. You illustrated this by your first hasty response that showed you are still confused and I would bet that you passed your real analysis course a long time ago too. In fact, I would probably bet that you might even be teaching mathematics somewhere. Irony? Yes. The shame is that there are many others like you and they are too afraid to stand up against the establishment of academia who are as ignorant as they always were. 68.238.102.180
Never mind what you agree. What matters is what is correct. The purpose of Wikipedia is not to descibe prevailing opinions but to state subject matter *objectively*. It is both arrogant and fruitless to present topics such as this that have no absolute truth value. Again, I state with conviction that mathematics today is no more rigourous than in the time of Archimedes. 68.238.102.180
Oleg did not show this. I showed that Oleg's logic is incorrect. And yes, you can show that it is true for any L as I demonstrated. Pick any L you like, then you can always find an epsilon to make |f(x) - L| < epsilon true. In this case, for the absolute value function |x| you are *done*. All you need to say then is that |x-a| < epsilon. It's true because |f(x)-L| and |x-a| are *exactly* equal in this case. Weierstrass's definition fails miserably for this function and cannot be true. Please, before you respond again, think about this carefully. 68.238.102.180
You responded again without thinking. You do not know what you are saying. I will try again to explain this to you:
I said: Lim |x| = L (as x approaches 0)
I also said that L can be anything you want (greater than 0 of course since you have an absolute value). So, you begin with ||x|-L| < e
=> -e < |x| - L < e
<=> -e+L < |x| < e+L <=> -e+L < |x|-0 < e+L
The next line is incorrect. It would be true only if you had -(e+L) < |x|-0 < e+L. 192.67.48.22
<=> ||x|-0| < e+L
But we want ||x|-a| < d that is: ||x|-0| < d So all we need to do is set d = e, then
||x|-0| < d => ||x|-0| < e+L
<=> -(e+L) < |x|-0 < e+L
Same problem in the next line: If you subtract L, you have -(e+2L) < |x|-L < e which does not lead to ||x|-L|<e. So L cannot be any positive number you like but it does seem that 0 can be a limit if you set d=e. This would imply that f(x)=|x| is differentiable at x = 0. Hmmm? Seems the derivative must be zero at x=o then? What do the rest of you think? 192.67.48.22
<=> -e < |x|-L < e <=> ||x|-L| < e And we are done!
You have such a hard head!!! Where did you learn anything? I don't mean to be nasty but you are a very ignorant individual. Everything I have written is 100% correct and logical. Please, think again before you post rubbish!!! And YES, |f(x)-L| and |x-a| are in this particular case *equal*(for L=0). This is one of the reasons why Weierstrass theory is unsound and does not cover every case. It fails because it rejects infinitesimals and then uses the same in its formulation of the definition. Again, how small is *small* and how *close* is *close* ? And, pray tell, what do *really small numbers* mean in mathematics Professor parlance?? 68.238.102.180
What are you saying guy?! You read it wrong again!! It's not -(e+L) < |x|-0 , it is: -(e+L) < |x|-0 < e+L. It would help if you could read too I suppose... 68.238.102.180
You might want to seriously reexamine your knowledge. There is *nothing* wrong with what I have written. Your 'chained/inequality' link is irrelevant. If you have something you can disprove, then do it, otherwise you are just taking up space and showing everyone exactly what a klutz you are. 68.238.102.180
Once again, what matters isn't what is "correct" according to someone who thinks standard math books are wrong (I doubt if he'll believe us any more than he believes Weierstrass.) What matters is Wikipedia policy, including Wikipedia:No original research. Please read it. Art LaPella 04:43, 31 October 2005 (UTC)
It's not what is *correct* but what is presented *objectively*. I have proved to you very clearly and without error that in this instance, e-d proofs do not work. In fact, they introduce a serious contradiction since most academia agree that no derivative exists at x=0 for this function (if there is doubt about the lim |x|, there will be doubt about the derivative). If you can find fault with what I write, prove it. A lot of words are just useless wind. This article should include this example I stated because any one else who might read this may think it is 'devine law' when they get confused. The problem is probably not with their intelligence but rather with the establishment. Wikipedia's policy on original research is a contradiction of itself - all human knowledge and theory starts out as *original research*. To publish 'knowledge' that has the backing of the masses (ignorant in most cases) is fruitless. All human knowledge should always be subject to continuous correction, reproof and scrutiny. There is no such thing as knowledge that is set in stone. Real Analysis has failed miserably to do anything else but contradict, confuse and mislead. It is high time to search for something better. 68.238.102.180
You are correct. Like all other encyclopedic content, I shall expect no more from Wikipedia. What was I thinking? What makes Wikipedia different from any of the other sources? Nothing. Generally that which is mainstream is controlled by those in authority. One of Wiki's claims is that it is a peoples' encyclopedia - evidently this is untrue. It is controlled by the few editors, sysops and their puppets. In the end it will be no better than World Book, Encyclopedia Brittanica or the lecture room of some ignorant mathematics professor. It's a real shame. 68.238.102.180
Hey there, 68.238.102.180, welcome to wikipedia. It's always nice to have people with unconventional views around. I haven't been following too closely. Could you please clarify? It sounds like you are claiming that according to Weierstrass's so-called "definition" of limits, the limit of |x| as x approaches 0 can actually be any (non-negative) number I like. Is my understanding of your position correct? Dmharvey File:User dmharvey sig.png Talk 02:39, 1 November 2005 (UTC)
I first stated that there is a limit as x approaches 0 and this limit is 0. However, as someone pointed out above, I made a mistake (twice) in thinking that the limit can be any positive real number. 68.238.102.180
Seems like 68.238.102.180 made an arithmetical booboo. However, the e-d proof in this case seems to imply a derivative exists at x=0? See also his postings on proof of 0.999... = 1 page. 192.67.48.22
has a limit for h->0. That is, if and only if exists.
Yes, 68.238.102.180 was confusing limits with derivatives. However, the example he mentioned is interesting (f(x)=|x|) because even though the rh limit and lh limit are different, f'(x) is equal to 0 if f'(x) is defined as follows: f'(x) = Lim (h->0) [f(x+h)-f(x-h)]/2h This is every bit as valid as the classical definition. IMO it is more representative and meaningful to use this as the definition rather than Lim (h->0) [f(x+h)-f(x)]/h So, I still think that f(x) = |x| has a derivative at x=0:
[|0+h| - |0-h|]/2h => [|h|-|-h|]/2h => 0/2h => 0.
192.67.48.22
Poor Rasmus - you made a mistake again: |-h| = h *always*. You are probably confusing yourself (again) with the result that states |h| = -h iff h < 0. Your arithmetic in the second formula is wrong!! |0-h|-|0--h| = 0 and not 2h!! Just stay with the 0.999... topic. You can hardly cope with this and now you are trying to multitask? Unlike you Rasmus, I am in my late forties and I don't have a beautiful wife. Guess this would make me a lesbian if I did. Ha, ha. But I do have a lot of hot young guys. :-) Some like to call me a 'bitch from hell' but I just pretend they do it affectionately. 192.67.48.22
You would be funny if you were not so pathetic. By the way, how is |-1|-|1|=2 ?? Okay, let's do this slowly:
Step A:
|-1| = 1
Step B:
|1| = 1
Step C:
|-1|-|1| = 1 - 1 = ? DoHHHHH!
I guess you are quite angry with me now and you are responding as a typical male would: like the proverbial bull in a China shop? 192.67.48.22
What my dear Rasmus, have I shaken up your world? You have been beaten into submission by an older woman? 192.67.48.22
I agree that if one defines the derivative as
then it follows that has limit at 0. But this definition is only a marginal improvement, for instance, the function
would still not be differentiable. And the above definition of the derivative is not useful, it would follow that even a discontinuous function is differentiable, like for and Oleg Alexandrov (talk) 19:41, 22 November 2005 (UTC)
http://www.geocities.com/john_gabriel/avsum.jpg - I think this article describes fairly well what is meant by average sum. 192.67.48.22
I think neither definition is that good and the 'symm' defn (actually it is better to call this a central difference and not a symm defn as Rasmus calls it) has some advantage in that one can calculate derivatives without an 'infinitesimal' factor. example: f(x) = x^2
[(x+h)^2 - (x-h)^2 ]/2h = [x^2 + 2xh + h^2 -x^2 + 2xh -h^2]/2h = 4xh/2h = 2x
Notice that it is not even required to take a limit here because there are no terms with h in them. Thus in this respect it is more precise (in fact it is *exact*). The classic form is only an approximation.
Differentiability is based on the classic form, i.e. if the classic limit exists. Had we started off with the central difference defn, we might have had a different definition of differentiability, i.e. provided the central difference limit exists, then the fn is differentiable. Furthermore, because the difference is central, there is no need to consider limits from 'both sides' but only one limit. Gabriel illustrates this by defining the limit in terms of w/n partitions and considering what happens when n approaches infinity. Continuity is *not* well defined. Using the current definition one can state that a function is continuous at *one* point and discontinuous everywhere else. For continuity, lh limit = rh limit and f(a) = L for some x. Example: f(x) = 1 for x = 3 and f(x) = 0 for all other x. Is f(x) continuous at x = 3? Yes. Why? lh limit = rh = limit and f(3) is defined, i.e. f(3) = 1. The function is continuous at that point (i.e. at that instant) even though f(3) is not equal to 0. In other words what is continuity at a point exactly? It is based on the notion of the lh limit = rh limit and f(x) = lh limit = rh limit which in turn is based on the understanding of the classical definition. One more example: Going back to f(x) = |x|. Is f(x) continuous at x = 0? Using classical definition, lh limit is not equal to rh limit but f(0) = 0. Do you then conclude that f(x) is not continuous at x = 0 ? This is what 68.238.102.180 demonstrated with his example of epsilon-delta proofs even though he was confusing it with differentiability. 192.67.48.22
Yes, it's true for even functions only. No, I am not Gabriel. I don't agree however that the classical defn is better because it proves continuity. I believe the central difference version demonstrates continuity better. --unsigned by anon
Interesting discussion. Anon may have a point here: I don't think he is trying to say the central difference (c.d) definition proves continuity (are you Anon?). Perhaps he is saying that the definition of continuity and differentiability would be different if the c.d definition was used. Well, the example of f(x) = |x| is puzzling - It is continuous at x=0 but not differentiable according to classical/traditional definitions/methods. Using traditional methods, you can show that it is not continuous, i.e. LH limit is not equal to RH limit is not equal to f(0). Oleg: how do you explain this? By definition of continuity, f(x) = |x| should not be continuous at x=0 ...—The preceding unsigned comment was added by 71.248.130.218 (talk • contribs) .
You are wrong. How do you reach this conclusion? The LH limit is not equal to the RH limit. Is this not a requirement for continuity in the classical definition?
Sorry, you are correct. I got muddled up reading the stuff about the derivative. The derivative LH limit is not equal to the RH limit. He was using the derivative limit incorrectly to find the limit of f(x) = |x| at x = 0. Guess I din't study this page long enough. —The preceding unsigned comment was added by 71.248.136.206 (talk • contribs) .
There is nothing in the guidelines that says to sign comments with tildes so I was unaware of this. I just started using Wikipedia. Will try to remember if I make more posts. Don't know if I want to make an account. Prefer to post anonymously. 71.248.136.206 00:44, 26 November 2005 (UTC)