hobz Posted July 13, 2011 Share Posted July 13, 2011 What I think I know [math] \frac{\mathrm{d}y}{\mathrm{d}x} = \lim_{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x} [/math] [math] \lim\limits_{x \to p} (f(x)/g(x)) = {\lim\limits_{x \to p} f(x) / \lim\limits_{x \to p} g(x)} [/math] but NOT for [math] p = 0 [/math] So the [math]\frac{\mathrm{d}y}{\mathrm{d}x}[/math] can't really be separated, but in many cases it is treated as if it could (many textbooks do this). How is that justified? Link to comment Share on other sites More sharing options...

baric Posted July 13, 2011 Share Posted July 13, 2011 I'm not exactly sure I understand your reasoning of why they cannot be separated. We see [math]dx[/math] separated when we integrate. Link to comment Share on other sites More sharing options...

hobz Posted July 13, 2011 Author Share Posted July 13, 2011 They can't be separated because [math]\mathrm{d}y[/math] and [math]\mathrm{d}x[/math] are not defined by the difference equation. [math]\mathrm{d}x[/math] is not [math]lim_{\Delta x \rightarrow 0}\Delta x[/math] as it then would be zero. Link to comment Share on other sites More sharing options...

baric Posted July 13, 2011 Share Posted July 13, 2011 They can't be separated because [math]\mathrm{d}y[/math] and [math]\mathrm{d}x[/math] are not defined by the difference equation. [math]\mathrm{d}x[/math] is not [math]lim_{\Delta x \rightarrow 0}\Delta x[/math] as it then would be zero. As you saying that [math]\mathrm{d}y/{d}x[/math] is not a mathematical ratio between [math]\mathrm{d}y[/math] and [math]\mathrm{d}x[/math] , but just a single variable? Link to comment Share on other sites More sharing options...

hobz Posted July 13, 2011 Author Share Posted July 13, 2011 Yes, it is not a ratio. Link to comment Share on other sites More sharing options...

baric Posted July 13, 2011 Share Posted July 13, 2011 Yes, it is not a ratio. huh. I always assumed that dy/dx was a ratio of infinitesimals, which is why the inverse of dy/dx is dx/dy, and dx/dx = 1. Also, it explains why dx is a numerator in the symbol for integration (i.e. u substitution). It looks like a duck to me. Link to comment Share on other sites More sharing options...

DrRocket Posted July 13, 2011 Share Posted July 13, 2011 (edited) huh. I always assumed that dy/dx was a ratio of infinitesimals, which is why the inverse of dy/dx is dx/dy, and dx/dx = 1. Also, it explains why dx is a numerator in the symbol for integration (i.e. u substitution). It looks like a duck to me. Unfortunately, in the real numbers there is no such thing as an "infitesimal", so neither dy nor dx has any meaning that makes dy/dx a ratio. [math] \frac {dy}{dx}(x_0)= \displaystyle \lim_{h \to 0} \frac {y(x_0 +h) - y(x_0)}{h}[/math] There are ways to make sense of dy or dx (as the induced map on the tangent bundle) but they are a bit beyond introductory calculus. Similarly there are some unconventional ways to make sense of infitesimals (using nonstandard analysis, based on the nonstandard real numbers, which requires knowledge of ultrafilters) which is again well beyond calculus and has not gained much traction. Thinking of dy/dx as a ratio is a sometimes useful crutch, but it is not correct and can get you into trouble on occasion. When used properly it is just a shortcut for reasoning involving the chain rule. There are several ways to look at the dx in an integral, but at the level of calculus it is just a reminder that in a Riemann sum the multiplier for the value of a function at a point in a partition is just the length of the sub-interval. This will make more sense when you learn about measure theory or differential forms. Edited July 13, 2011 by DrRocket Link to comment Share on other sites More sharing options...

baric Posted July 14, 2011 Share Posted July 14, 2011 (edited) Unfortunately, in the real numbers there is no such thing as an "infitesimal", so neither dy nor dx has any meaning that makes dy/dx a ratio. Yes, but you can understand the confusion when dy/dx is defined thusly: [math] \frac{\mathrm{d}y}{\mathrm{d}x} = \lim_{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x} [/math] I mean, that is literally saying that "dy/dx is the ratio of delta-y over delta-x as delta-x approaches 0". The right side of the equation is a clear ratio While I understand that "infinitesimal" is an abstract concept (as is "infinity") which has no place in the real set, that does not seem to me to be reason enough to preclude calling dy/dx a ratio. After all, we can certainly discuss ratios of infinities without our heads exploding, why not infinitesimals? Also, I will be the first to admit that I am far from an expert in calculus (mostly self-taught ), so please don't take the above statements as a disagreement with you rather than an exposition of my level of understanding. If you could provide an example where treating dy/dx as a ratio could get you in trouble, it would be greatly appreciated! Edited July 14, 2011 by baric Link to comment Share on other sites More sharing options...

hobz Posted July 14, 2011 Author Share Posted July 14, 2011 (edited) Yes, but you can understand the confusion when dy/dx is defined thusly: [math] \frac{\mathrm{d}y}{\mathrm{d}x} = \lim_{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x} [/math] I mean, that is literally saying that "dy/dx is the ratio of delta-y over delta-x as delta-x approaches 0". The right side of the equation is a clear ratio While I understand that "infinitesimal" is an abstract concept (as is "infinity") which has no place in the real set, that does not seem to me to be reason enough to preclude calling dy/dx a ratio. After all, we can certainly discuss ratios of infinities without our heads exploding, why not infinitesimals? Also, I will be the first to admit that I am far from an expert in calculus (mostly self-taught ), so please don't take the above statements as a disagreement with you rather than an exposition of my level of understanding. If you could provide an example where treating dy/dx as a ratio could get you in trouble, it would be greatly appreciated! Let me first state that I too am no expert (which is why I am asking people in sf). As far as I know, the concept of "infinitisimal" has been abandoned since Weierstrass gave a rigorous definition of the concept of limits and used limits to "derive" calculus. The exception is, as DrRocket says, the non-standard calcuclus, which is, as far as I know, also a rigorous method worked out by a guy in the 60's. I think the problem is that dy/dx is really a notation and not a ratio. The notation is sometimes abused by (e.g.) multiplying by dx on both sides, which has no meaning from the dy/dx limit definition. My question is, how can you justify doing it anyways? P.S. If you like the infinitisimals, I can recommend "Calculus made easy" by Silvanus Thompson. It's very informal, only relying on algebra. Edited July 14, 2011 by hobz Link to comment Share on other sites More sharing options...

baric Posted July 14, 2011 Share Posted July 14, 2011 Let me first state that I too am no expert (which is why I am asking people in sf). As far as I know, the concept of "infinitisimal" has been abandoned since Weierstrass gave a rigorous definition of the concept of limits and used limits to "derive" calculus. The exception is, as DrRocket says, the non-standard calcuclus, which is, as far as I know, also a rigorous method worked out by a guy in the 60's. So I read up on what is essentially the history of establishing the foundation of calculus. Weierstrass dispensed with infinitesimals in favor of limits, and then Robinson defined the hyperreals in order to provide the necessary rigor to infinitesimals. ok, fine. So it sounds to me that dy/dx can be considered a ratio of infinitesimals, at least with regards to non-standard calculus. I don't understand Dr. Rocket's concern about it not "gaining traction". What does that mean? Is there dispute about the validity of Robinson's reasoning? Link to comment Share on other sites More sharing options...

ajb Posted July 14, 2011 Share Posted July 14, 2011 So it sounds to me that dy/dx can be considered a ratio of infinitesimals, at least with regards to non-standard calculus. Well as DrRocket has said, you can informally think of it as ratio and this can be useful in things like the chain rule and solving differential equations etc. The trouble is that infinitesimals are nilpotent, in the sense that squaring an infinitesimal is zero. This leads to difficulties in defining division. Link to comment Share on other sites More sharing options...

DrRocket Posted July 14, 2011 Share Posted July 14, 2011 Yes, but you can understand the confusion when dy/dx is defined thusly: [math] \frac{\mathrm{d}y}{\mathrm{d}x} = \lim_{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x} [/math] I mean, that is literally saying that "dy/dx is the ratio of delta-y over delta-x as delta-x approaches 0". The right side of the equation is a clear ratio No, it is not a ratio. There is that "limit" thing. While I understand that "infinitesimal" is an abstract concept (as is "infinity") which has no place in the real set, that does not seem to me to be reason enough to preclude calling dy/dx a ratio. After all, we can certainly discuss ratios of infinities without our heads exploding, why not infinitesimals? A "ratio of infinities" can be literally anything. So can a ratio of "infinitesimals. Also, I will be the first to admit that I am far from an expert in calculus (mostly self-taught ), so please don't take the above statements as a disagreement with you rather than an exposition of my level of understanding. If you could provide an example where treating dy/dx as a ratio could get you in trouble, it would be greatly appreciated! Try treating dy/dx as a ratio of two things that you can't even define and then proving 1) that a differentiable function is necessarily continuous or 2) the mean value theorem. The mean value theorem is at the heart of calculus and allows you to prove the "fundamental theorem of calculus". So I read up on what is essentially the history of establishing the foundation of calculus. Weierstrass dispensed with infinitesimals in favor of limits, and then Robinson defined the hyperreals in order to provide the necessary rigor to infinitesimals. ok, fine. So it sounds to me that dy/dx can be considered a ratio of infinitesimals, at least with regards to non-standard calculus. I don't understand Dr. Rocket's concern about it not "gaining traction". What does that mean? Is there dispute about the validity of Robinson's reasoning? Abraham Robinson's reasoning is just fine, but you have no idea what that means. The whole construction of the non-standard real numbers relies on the axiom of choice and the construction of topological entities called ultrafilters. Basically you need both the ordinary real numbers plus a lot of machinery before you can construct the non-standard reals. By "no traction" I mean that non-standard analysis has not received much acceptance as a useful method in the mathematical community. For a short time there was a theorem in operator theory proved by non-standard techniques that took a bit of work by Paul Halmos to find a standard proof. Non-standard analysis is simply mostly ignored. This is not unusual. Quite often new ideas and techniques pique some early interest but then fade into oblivion when they don't live up to early promise. Yeah, you can find examples of people on the fringe who use non-standard analysis, and even have written elementary calculus texts based on the non-standard real numbers, but they do the student a great disservice since they are not prepared to follow the mainstream texts using standard techniques. If you are going to pursue non-standard analysis you should first attain a solid grasp of standard analysis. It is not a replacement. Link to comment Share on other sites More sharing options...

baric Posted July 14, 2011 Share Posted July 14, 2011 (edited) No, it is not a ratio. There is that "limit" thing. Yes! A "ratio of infinities" can be literally anything. So can a ratio of "infinitesimals. No one is suggesting to actually resolve a ratio (or product) of infinities into some other value. However, maintaining a distinction between different infinities is possible. If dx and dy are both infinitesimals, for example, this obviously does not mean they are the same. Try treating dy/dx as a ratio of two things that you can't even define and then proving 1) that a differentiable function is necessarily continuous or 2) the mean value theorem. The mean value theorem is at the heart of calculus and allows you to prove the "fundamental theorem of calculus". All I'm saying is that dy/dx seems to function as a ratio. Everyone with more knowledge of calculus is saying that it isn't, and I'm not disputing that. But I would be interested in some example where treating it as a ratio would not work. By "no traction" I mean that non-standard analysis has not received much acceptance as a useful method in the mathematical community. OK, gotcha. There are multiple ways to interpret your comment about traction, so I appreciate the clarification. It's not invalid, but not particularly useful. Edited July 14, 2011 by baric Link to comment Share on other sites More sharing options...

DrRocket Posted July 14, 2011 Share Posted July 14, 2011 But I would be interested in some example where treating it as a ratio would not work. How about you show me one instance in which all terms are clearly defined in which it does work. Link to comment Share on other sites More sharing options...

baric Posted July 15, 2011 Share Posted July 15, 2011 (edited) How about you show me one instance in which all terms are clearly defined in which it does work. I'll take this as a 'no'. I am clearly the calculus novice in this discussion so I think your request is a bit more onerous. I did not realize that asking for some sort of example would provoke such a combative response. Edited July 15, 2011 by baric Link to comment Share on other sites More sharing options...

DrRocket Posted July 15, 2011 Share Posted July 15, 2011 (edited) I'll take this as a 'no'. I am clearly the calculus novice in this discussion so I think your request is a bit more onerous. I did not realize that asking for some sort of example would provoke such a combative response. I gave you a couple of examples earlier, but you seem to be looking for some really stupid blunder based on loooking at derivativesvas a ratio. I don't make such blunders and so don't have a ready example. The real problem is the failure to recognize what a derivative really is. The whole point of bdifferential calculus is to use simple functions, linear functions, to study a much larger class of functions and to reach some deep conclusions about them. In that sense the derivative at a point is not really a number, but rather a linear operator. This notion generalizes easily to functions of several variables and to functions of complex variable as well. In one dimension this is hidden by the fact that linear functions are just multiplication by a number. The more general, and useful, idea is as follows: The basic idea is that you are trying to approximate a somewhat arbitrary function. near some given point, with a linear function. This is what you are doing in the case of a function of one variable, with the simplification that a linear function of one variable is just multiplication by a number -- the derivative of the function at the point in question. The generalization to several variables goes as follows: Let [math]f: \mathbb R^n \to \mathbb R^m[/math] and [math]x_0 \in R^n[/math] then [math]f[/math] is said to be differentiable at [math]x_0[/math] if there exists a linear function [math]D:\mathbb R^n \to \mathbb R^m[/math] such that [math]f(x_0 + x) = f(x_0) + D(x) + o(x)[/math] where [math]\displaystyle \lim_{\|x \|\to 0} \frac {\| o(x) \|}{\| x \|} =0[/math] Here if [math]x \in \mathbb R^k[/math], [math]x=(x_1,...,x_k)[/math] , then [math] \|x \|= \sqrt {x_1^2+...+x_k^2}[/math] If such a linear function [math]D[/math] exists it is called the derivative of [math]f[/math] at [math]x_0[/math] [math]\|x \|[/math] is called the norm of [math]x[/math] and is a generalization of absolute value, or length. The idea is that [math]D[/math] is the best linear approximation to [math]f[/math] near [math]x_0[/math] and the error [math]o(x)[/math] goes to 0 faster than linearly as [math]x[/math] goes to zero as a vector. This generalizes easily to functions on complex vector spaces and to infinite dimensional Banach spaces as well, which is useful in calculus of variations and partial differential equations. It is also central to the development of differential forms, which is the setting in which expressions like [math] dx[/math] and more generally [math]df[/math] are really formulated in a useful way. This has essentially nothing to do with non-standard analysis or "infinitesimals, but is crucial to differential geometry and modern mathematical physics. Derivatives are approximating linear functions, not ratios. Edited July 15, 2011 by DrRocket Link to comment Share on other sites More sharing options...

ajb Posted July 15, 2011 Share Posted July 15, 2011 Derivatives are approximating linear functions, not ratios. This is the way to think of derivatives in slightly wider sense. For instance the "geometric variation" aka Lie derivative. Though I will say I tend to formulate this using infinitesimals It is also central to the development of differential forms, which is the setting in which expressions like [math] dx[/math] and more generally [math]df[/math] are really formulated in a useful way. This has essentially nothing to do with non-standard analysis or "infinitesimals, but is crucial to differential geometry and modern mathematical physics. The modern way of treating differential forms (differential calculi )is quite formal. The object [math]dx[/math] is understood as formal variable. In fact on a manifold one can identify [math]dx[/math] with a coordinate on the total space of a supermanifold. In particular it is a Grassmann odd coordinate. You can now ask about the values [math]dx[/math] can take? Well, as it is Grassmann in nature the only real value it can take is zero (or it can take values in some other Grassmann algebra). The point is, as a real (or complex) number it vanishes and so thinking of ratios is impossible. One can do this even more generally and in fact one can construct differential calculi for any associative, but not necessarily commutative algebra. But this is a story for another time. Link to comment Share on other sites More sharing options...

baric Posted July 15, 2011 Share Posted July 15, 2011 I gave you a couple of examples earlier, but you seem to be looking for some really stupid blunder based on loooking at derivativesvas a ratio. I don't make such blunders and so don't have a ready example. Just trying to gain a better understanding. The more general, and useful, idea is as follows: The basic idea is that you are trying to approximate a somewhat arbitrary function. near some given point, with a linear function. This is what you are doing in the case of a function of one variable, with the simplification that a linear function of one variable is just multiplication by a number -- the derivative of the function at the point in question. The generalization to several variables goes as follows: Let [math]f: \mathbb R^n \to \mathbb R^m[/math] and [math]x_0 \in R^n[/math] then [math]f[/math] is said to be differentiable at [math]x_0[/math] if there exists a linear function [math]D:\mathbb R^n \to \mathbb R^m[/math] such that [math]f(x_0 + x) = f(x_0) + D(x) + o(x)[/math] where [math]\displaystyle \lim_{\|x \|\to 0} \frac {\| o(x) \|}{\| x \|} =0[/math] Here if [math]x \in \mathbb R^k[/math], [math]x=(x_1,...,x_k)[/math] , then [math] \|x \|= \sqrt {x_1^2+...+x_k^2}[/math] If such a linear function [math]D[/math] exists it is called the derivative of [math]f[/math] at [math]x_0[/math] [math]\|x \|[/math] is called the norm of [math]x[/math] and is a generalization of absolute value, or length. The idea is that [math]D[/math] is the best linear approximation to [math]f[/math] near [math]x_0[/math] and the error [math]o(x)[/math] goes to 0 faster than linearly as [math]x[/math] goes to zero as a vector. ok, that makes a lot more sense. Link to comment Share on other sites More sharing options...

xcthulhu Posted July 15, 2011 Share Posted July 15, 2011 (edited) Abraham Robinson's reasoning is just fine, but you have no idea what that means. The whole construction of the non-standard real numbers relies on the axiom of choice and the construction of topological entities called ultrafilters. Basically you need both the ordinary real numbers plus a lot of machinery before you can construct the non-standard reals. Hi DrRocket, You appear to be an expert. I have some really pedantic caveats with what you say here. (1) Specifically, Abraham Robinson uses a non-principle ultrafilter to obtain an elementary extension of the real numbers called an ultra-product. While the existence of non-principle ultrafilters is indeed a consequence of the axiom of choice, it's not logically equivalent. I don't mean to get all reverse mathematics up in here but you can obtain them in models of set theory without choice, namely they are guaranteed to exist on the real numbers as a consequence of the axiom of determinacy. I forget if you can obtain them from various forms of Konig's lemma... Reading Abraham Robinson will turn your mind into mush, btw. If you can just accept the existence of a non-principle ultra-filter as an article of faith then the non-standard reals aren't any harder to obtain than the reals via Cauchy sequences. I like Goldblatt's Lectures on the Hyper-reals which takes this approach. Applying the transfer theorems is as difficult as you say, however. (2) There are other approaches to the construction of infinitesimals. One approach is smooth infinitesimal analysis (here's Bell's introduction to the subject). Most mathematicians despise this approach completely because it's purely intuitionistic, which is hard if you aren't a disciple of Brouwer. By "no traction" I mean that non-standard analysis has not received much acceptance as a useful method in the mathematical community. For a short time there was a theorem in operator theory proved by non-standard techniques that took a bit of work by Paul Halmos to find a standard proof. Non-standard analysis is simply mostly ignored. This is not unusual. Quite often new ideas and techniques pique some early interest but then fade into oblivion when they don't live up to early promise. Yeah, you can find examples of people on the fringe who use non-standard analysis, and even have written elementary calculus texts based on the non-standard real numbers, but they do the student a great disservice since they are not prepared to follow the mainstream texts using standard techniques. If you are going to pursue non-standard analysis you should first attain a solid grasp of standard analysis. It is not a replacement. Well... do I really need to mention that the mathematical community is vast? A modern application of nonstandard analysis is in the semi-automated proof assistant Isabelle/HOL. Typically a automated theorem prover like VAMPIRE (invoked by Isabelle) will stumble on quantifier manipulation, requiring more interaction from the mathematician/programmer. Isabelle/HOL has a mechanism "proof by transfer theorem" that turns ordinary analysis problems into non-standard analysis problems, performing quantifier elmination in the process. It's considered the preferred way to tackle analysis once you get advanced enough. And I don't mean to straw man you, but I suspect you are thinking something to the extent "But computer proof isn't real mathematics, it's just nerds with computers." I would suggest reading Donald MacKenzie's Computers and the Sociology of Mathematical Proof. TL;DR: Apparently the British government is of the opinion that computer assisted proof constitutes mathematical proof. Edited July 15, 2011 by xcthulhu Link to comment Share on other sites More sharing options...

DrRocket Posted July 16, 2011 Share Posted July 16, 2011 (2) There are other approaches to the construction of infinitesimals. One approach is smooth infinitesimal analysis (here's Bell's introduction to the subject). Most mathematicians despise this approach completely because it's purely intuitionistic, which is hard if you aren't a disciple of Brouwer. Brouwer did some good work. Then he turned to intuitionism. Well... do I really need to mention that the mathematical community is vast? Not that vast. I know quite a few mathematicians, having taught at three major universities, and associated with people from several more. I knew one guy who specialized in non-standard analysis. He was let go. A modern application of nonstandard analysis is in the semi-automated proof assistant Isabelle/HOL. Typically a automated theorem prover like VAMPIRE (invoked by Isabelle) will stumble on quantifier manipulation, requiring more interaction from the mathematician/programmer. Isabelle/HOL has a mechanism "proof by transfer theorem" that turns ordinary analysis problems into non-standard analysis problems, performing quantifier elmination in the process. It's considered the preferred way to tackle analysis once you get advanced enough. And I don't mean to straw man you, but I suspect you are thinking something to the extent "But computer proof isn't real mathematics, it's just nerds with computers." I would suggest reading Donald MacKenzie's Computers and the Sociology of Mathematical Proof. TL;DR: Apparently the British government is of the opinion that computer assisted proof constitutes mathematical proof. Words are simply not available to describe how little I care what the British government, or any other government, regards as mathematical proof. That said, Appel and Haken's proof of the four color theorem by reduction to a finite numberof cases that were checked by computer has my admiration. However, the admirable aspect is the reduction to a finite number of cases, not the computer evaluation of the cases. Link to comment Share on other sites More sharing options...

xcthulhu Posted July 16, 2011 Share Posted July 16, 2011 Words are simply not available to describe how little I care what the British government, or any other government, regards as mathematical proof. You and Plato. Not that vast. I know quite a few mathematicians, having taught at three major universities, and associated with people from several more. I knew one guy who specialized in non-standard analysis. He was let go. How much time have you spent in industry? Anyway, I think we can both agree that nonstandard analysis is practically useless and of limited real world utility. Link to comment Share on other sites More sharing options...

hobz Posted July 17, 2011 Author Share Posted July 17, 2011 Anyway, I think we can both agree that nonstandard analysis is practically useless and of limited real world utility. If nonstandard analysis ties the intuitive infinitesimal calculus of Leibniz, et al, to a rigorous foundation of the standard calculus (using limits), then I would have to completely disagree with that statement. In many engineering and physics classes, informal treatment of dy/dx very often leads to treating it as a ratio, even though it strictly is meaningless or undefined. Non the less, all the physics derived from this informal basis, can be rigorously based on the nonstandard analysis, which then is everything else but useless. Link to comment Share on other sites More sharing options...

xcthulhu Posted July 17, 2011 Share Posted July 17, 2011 (edited) @hobz: In all politeness I suspect you don't really know nonstandard analysis. You don't have to study it for very long to understand that it's nothing like physics at all. Just think about infinity in this formalism. For instance, if you have an infinitesimal [math]dx[/math] then [math]\frac{1}{dx}[/math] is an "unlimited" or infinitary number. Let's call [math]\frac{1}{dx}=\omega[/math]. Evidently [math]\omega + 1[/math] is infinitary, too (and unlike cardinal arithmetic and ordinal arithmetic, [math]\omega + 1 = 1 + \omega \neq \omega[/math]). In fact all of the points [math]\{\omega + r\ |\ r \in \mathbb{R}\}[/math] are going to be infinitary, too. Nonstandard analysts like to call all of the points a finite distance away from some particular infinitary number a galaxies. A tedious question arises: how many galaxies are there? With difficulty, we can prove that [math]\omega / 2[/math] is less than every number in the galaxy around [math]\omega[/math]. Also, apparently [math]e^\omega[/math] is in a separate galaxy. So... with a little work you can show there must be at least uncountably many galaxies. But how many really are there? Where's physics based intuition to help us like in physics problems? (btw, the answer to this is undecidable) In nonstandard analysis there are "integers" that are greater than every finitary integer. One can prove bizarro "infinitary" prime factorization theorems. Just like for the real unlimited numbers, there is no smallest galaxy around an unlimited integer. Another tedious question: Can you always "round" an unlimited number to an infinitary integer? Again, physics, analysis, really nothing helps you answer this. (my answer: Yes in some models but I'm not sure about all of them) In physics, you don't bother with actual nonstandard analysis, because it's rediculous. You wave your hands and pretend, or you use functionals on the category of Hilbert spaces or manifolds or whatever (if you want to make proper use of your PhD). Edited July 17, 2011 by xcthulhu Link to comment Share on other sites More sharing options...

DrRocket Posted July 17, 2011 Share Posted July 17, 2011 You and Plato. How much time have you spent in industry? A couple of decades +. Link to comment Share on other sites More sharing options...

hobz Posted July 17, 2011 Author Share Posted July 17, 2011 (edited) If you look at http://en.wikipedia.org/wiki/Capacitance (just the stuff before the TOC) there is an example of what I have been talking about. Is this not abusing the notation? Edited July 17, 2011 by hobz Link to comment Share on other sites More sharing options...

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