baric Posted July 17, 2011 Share Posted July 17, 2011 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 What sense does it make to add 1 to an infinity? It's not a real number. Link to comment Share on other sites More sharing options...

DrRocket Posted July 18, 2011 Share Posted July 18, 2011 What sense does it make to add 1 to an infinity? It's not a real number. Oh, you can do it. The real question involves the context, and whether or not you would find a good reason to do such a thing,. In the context of this thread it is quite safe to ignore xcthulhu and all of nonstandard analysis. One can do lots of things in mathematics. Some of those things turn out to be very productive. Some don't. Non-standard analysis has not turned out to be particularly productive. But some of its adherents, particularly within the logic community, just enjoy flogging a dead horse. Non-standard analysis is just not appropriate for those trying to learn basic calculus. Those who understand calculus don't need non-standard analysis either. It has become a solution in search of a problem. Link to comment Share on other sites More sharing options...

hobz Posted July 22, 2011 Author Share Posted July 22, 2011 Take a look at http://en.wikipedia.org/wiki/Escape_velocity#Derivation_using_G_and_M Is this calculus, or pseudo-calculus? Link to comment Share on other sites More sharing options...

john_gabriel Posted January 6, 2012 Share Posted January 6, 2012 (edited) 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? Here is an interesting link: http://india-men.nin...l-quotient?page One of the attachments called meaning_of_dydx.pdf explains. Another interesting link is: http://thenewcalculus.weebly.com/uploads/5/6/7/4/5674177/lhopital.pdf @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). Non-standard analysis is absolute rot in my opinion. In fact the very concept of infinitesimal is nonsense. Rather than try to explain here, I suggest a visit to my New Calculus site: http://thenewcalculus.weebly.com Edited January 6, 2012 by john_gabriel Link to comment Share on other sites More sharing options...

D H Posted January 7, 2012 Share Posted January 7, 2012 (edited) Take a look at http://en.wikipedia.org/wiki/Escape_velocity#Derivation_using_G_and_M Is this calculus, or pseudo-calculus? I presume you're talking about those eight steps used to derive conservation of energy. I would call that "freshman physics calculus." There's a couple of steps in there that would make a mathematician cringe a bit. That said, those quicksteps are easily rectified. On the other hand, why not just start with conservation of energy? As far as the original question is concerned, you can often get away with interpreting Leibniz notation as a ratio. It works fine -- some of the time. Other times, not so fine. Consider a point on a sphere [math]x^2+y^2+z^2=a^2[/math]. What is [math]\frac{\partial y}{\partial x}\frac{\partial z}{\partial y}\frac{\partial x}{\partial z}[/math]? Treating these partials as ratios would suggest the answer is 1. It isn't. It's -1. Edited January 7, 2012 by D H Link to comment Share on other sites More sharing options...

DrRocket Posted January 7, 2012 Share Posted January 7, 2012 Non-standard analysis is absolute rot in my opinion. In fact the very concept of infinitesimal is nonsense. Rather than try to explain here, I suggest a visit to my New Calculus site: http://thenewcalculus.weebly.com That is great site for those with a strong stomach and a weak mind. As far as the original question is concerned, you can often get away with interpreting Leibniz notation as a ratio. It works fine -- some of the time. Other times, not so fine. In other words it is a somewhat useful mnemonic that one ought to follow up with rigorous reasoning since the mnemonic can get you into trouble on occasion. The derivative is not a ratio in the rigorous sense (outside of nonstandard analysis), but as a limit of ratios that reasoning is sometimes useful, so long as you don't get carried away. Consider a point on a sphere [math]x^2+y^2+z^2=a^2[/math]. What is [math]\frac{\partial y}{\partial x}\frac{\partial z}{\partial y}\frac{\partial x}{\partial z}[/math]? Treating these partials as ratios would suggest the answer is 1. It isn't. It's -1. The problem here is that [math]x^2+y^2+z^2=a^2[/math] does not define a function of x, y, and z but only gives one in terms of the other when you fix the third variable. Thus without further explanation [math]\frac{\partial y}{\partial x}\frac{\partial z}{\partial y}\frac{\partial x}{\partial z}[/math] is not really meaningful as they cannot all be defined under a single set of assumptions -- partial derivatives are defined for functions, not equations. Link to comment Share on other sites More sharing options...

john_gabriel Posted January 10, 2012 Share Posted January 10, 2012 (edited) "That is great site for those with a strong stomach and a weak mind." And I'd say you have a weak mind. "In other words it is a somewhat useful mnemonic that one ought to follow up with rigorous reasoning since the mnemonic can get you into trouble on occasion. " Nonsense. It is well-defined in my new calculus. There is no doubt as to its meaning. "The derivative is not a ratio in the rigorous sense (outside of nonstandard analysis), but as a limit of ratios that reasoning is sometimes useful, so long as you don't get carried away. " Again, not true. The derivative is exactly a ratio. Assuming that x, y and z are not functions of each other, then [math]\frac{\partial y}{\partial x}=2y\frac{\partial y}{\partial x}[/math] (1) [math]\frac{\partial z}{\partial y}=2z\frac{\partial z}{\partial y}[/math] (2) [math]\frac{\partial x}{\partial z}=2x\frac{\partial x}{\partial z}[/math] (3) How is this equal to -1? You are confusing yourself by failing to realize that the symbols are not numbers until numbers are assigned to one or more symbols. Even if x, y and z are functions of each other, your understanding of partial derivatives is still wrong. "The problem here is that [math]x^2+y^2+z^2=a^2[/math] does not define a function of x, y, and z but only gives one in terms of the other when you fix the third variable." Actually, it very much defines a function of x, y, and z. In fact, one can parameterize all of these in terms of a 4th variable. So what you say is false. "Thus without further explanation [math]\frac{\partial y}{\partial x}\frac{\partial z}{\partial y}\frac{\partial x}{\partial z}[/math] is not really meaningful as they cannot all be defined under a single set of assumptions -- partial derivatives are defined for functions, not equations." I am not certain what you mean by "under a single set of assumptions"? There are no assumptions. Partial derivatives are well-defined. The following links explain more: http://thenewcalculu...ing_of_dydx.pdf http://thenewcalculu...ean_exactly.pdf http://thenewcalculu...tes_example.pdf You have to assign a meaning to each partial differential. You can't say dy/dx x dz/dy x dx/dz where the dees are partial derivatives unless you know what these are. It's like saying turkey/dog x cat/turkey x dog/cat = 1. As for checking everything in mathematics, this goes without say. Edited January 10, 2012 by john_gabriel Link to comment Share on other sites More sharing options...

ajb Posted January 10, 2012 Share Posted January 10, 2012 Assuming that x, y and z are not functions of each other, then [math]\frac{\partial y}{\partial x}=2y\frac{\partial y}{\partial x}[/math] (1) [math]\frac{\partial z}{\partial y}=2z\frac{\partial z}{\partial y}[/math] (2) [math]\frac{\partial x}{\partial z}=2x\frac{\partial x}{\partial z}[/math] (3) If we take [math]y \neq y(x)[/math] then [math]\frac{\partial y}{\partial x} =0[/math]. And similar for the other partial derivatives. Is that what you mean? Link to comment Share on other sites More sharing options...

john_gabriel Posted January 10, 2012 Share Posted January 10, 2012 If we take [math]y \neq y(x)[/math] then [math]\frac{\partial y}{\partial x} =0[/math]. And similar for the other partial derivatives. Is that what you mean? Yes. Link to comment Share on other sites More sharing options...

ajb Posted January 10, 2012 Share Posted January 10, 2012 This is not the case DrRocket and DH are considering. It maybe more prudent to consider the equation [math]x^{2}+ y^{2} = 1[/math]. This does allow us to define [math]y(x)[/math]... Link to comment Share on other sites More sharing options...

john_gabriel Posted January 10, 2012 Share Posted January 10, 2012 (edited) This is not the case DrRocket and DH are considering. It maybe more prudent to consider the equation [math]x^{2}+ y^{2} = 1[/math]. This does allow us to define [math]y(x)[/math]... I see no difference on first inspection. But I am not a mind-reader, so I don't know exactly what Dr Rocket is referring to. I do know that partial derivatives are well-defined, but like everything else, one must consider the context and use of the same. This does not mean that one should cease questioning or investigating this knowledge further. Had I accepted the wrong ideas of the infinitesimal concept, I would never have discovered the New Calculus. I have also shown that the base concept (Cauchy's derivative definition) from which the infinitesimal was born is indeed flawed. Edited January 10, 2012 by john_gabriel Link to comment Share on other sites More sharing options...

DrRocket Posted January 10, 2012 Share Posted January 10, 2012 "That is great site for those with a strong stomach and a weak mind." And I'd say you have a weak mind. You say a lot of things. Most of them are wrong. I see no difference on first inspection. That is evidence of nothing more than a failing on your part. One more time -- partial derivatives are defined for functions, not equations. If your equation does not at least implicitly define a function then there is no meaning to a partial derivative. Link to comment Share on other sites More sharing options...

D H Posted January 10, 2012 Share Posted January 10, 2012 The problem here is that [math]x^2+y^2+z^2=a^2[/math] does not define a function of x, y, and z but only gives one in terms of the other when you fix the third variable. Thus without further explanation [math]\frac{\partial y}{\partial x}\frac{\partial z}{\partial y}\frac{\partial x}{\partial z}[/math] is not really meaningful as they cannot all be defined under a single set of assumptions -- partial derivatives are defined for functions, not equations. You're right. I should have justified what I wrote. So, on a sphere we have [math]x^2+y^2+z^2=a^2[/math]. Thus [math] \begin{aligned} y^2 &= a^2-(z^2+x^2)\qquad(1) \\ z^2 &= a^2-(x^2+y^2)\qquad(2) \\ x^2 &= a^2-(y^2+z^2)\qquad(3) \end{aligned} [/math] Taking partial derivatives with of equation (1) respect to x, equation (2) with respect to y, and equation (3) with respect to z yields [math] \begin{aligned} \frac{\partial y}{\partial x} = -\frac x y \\ \frac{\partial z}{\partial y} = -\frac y z \\ \frac{\partial x}{\partial z} = -\frac z x \\ \end{aligned} [/math] Thus [math] \frac{\partial y}{\partial x}\frac{\partial z}{\partial y}\frac{\partial x}{\partial z} = \left(-\frac x y\right) \left(-\frac y z\right) \left(-\frac z x\right) = -1 [/math] Link to comment Share on other sites More sharing options...

john_gabriel Posted January 10, 2012 Share Posted January 10, 2012 (edited) You say a lot of things. Most of them are wrong. This is your opinion. That is evidence of nothing more than a failing on your part. Again, your opinion. One more time -- partial derivatives are defined for functions, not equations. If your equation does not at least implicitly define a function then there is no meaning to a partial derivative. Wrong. See http://mathworld.wol...Derivative.html (don't feel like typing it out, sorry) This is an equation, so I don't know what you are talking about... Besides, what is the difference between a function and an equation? Please define both. As far as I am concerned, the difference is superficial. You're right. I should have justified what I wrote. So, on a sphere we have [math]x^2+y^2+z^2=a^2[/math]. Thus [math]\begin{aligned}y^2 &= a^2-(z^2+x^2)\qquad(1) \\z^2 &= a^2-(x^2+y^2)\qquad(2) \\x^2 &= a^2-(y^2+z^2)\qquad(3)\end{aligned}[/math] Taking partial derivatives with of equation (1) respect to x, equation (2) with respect to y, and equation (3) with respect to z yields [math]\begin{aligned}\frac{\partial y}{\partial x} = -\frac x y \\\frac{\partial z}{\partial y} = -\frac y z \\\frac{\partial x}{\partial z} = -\frac z x \\\end{aligned}[/math] Thus [math]\frac{\partial y}{\partial x}\frac{\partial z}{\partial y}\frac{\partial x}{\partial z} = \left(-\frac x y\right) \left(-\frac y z\right) \left(-\frac z x\right) = -1[/math] I see what you mean here but this is due to the fact that you are interpreting the product out of context. The ratios in the product are symbolic so you cannot simply assume that the suggested product is 1 when it is in fact -1. The symbolic ratios become actual ratios when replaced by numbers - this is a part of your inability to understand this difference. A symbolic ratio does not carry a sign in it. I am surprised you missed this! The sign is understood once the ratio transitions from symbol to number. Observe that none of this changes the fact that partial derivatives are proper ratios. They are not infinitesimal (whatever this rot means) nor are limits required. One more thing: If you are going to say I am wrong, please have the decency to tell me where you think I have gone wrong. Just stating your opinion is valueless to me. Who knows, you may yet correct me... Edited January 10, 2012 by john_gabriel Link to comment Share on other sites More sharing options...

mississippichem Posted January 10, 2012 Share Posted January 10, 2012 Wrong. See http://mathworld.wol...Derivative.html (don't feel like typing it out, sorry) This is an equation, so I don't know what you are talking about... Besides, what is the difference between a function and an equation? Please define both. As far as I am concerned, the difference is superficial. An equation equates two algebraic expressions. A function assigns one output to every input. The equation of a circle is not a function. There is more than one value for f(x) for some values of x. A function has one [or zero] values for f(x) for every x as far as I know. Observe that none of this changes the fact that partial derivatives are proper ratios. They are not infinitesimal (whatever this rot means) nor are limits required. Incorrect, the limit definition of the partial derivative disagrees with you. The partial derivative and derivatives in general are limits of the difference quotient. The word infinitesimal is hardly "rot", it might be one of the most important concepts in analysis and calculus which are without doubt some of the most important concepts in mathematics. Link to comment Share on other sites More sharing options...

john_gabriel Posted January 11, 2012 Share Posted January 11, 2012 (edited) An equation equates two algebraic expressions. A function assigns one output to every input. The equation of a circle is not a function. There is more than one value for f(x) for some values of x. A function has one [or zero] values for f(x) for every x as far as I know. Yeah, but I don't think this matters in our discussion. Incorrect, the limit definition of the partial derivative disagrees with you. The partial derivative and derivatives in general are limits of the difference quotient. Okay, and so how does that have anything to do with infinitesimals? Also, I have proved that the limit definition is flawed. The word infinitesimal is hardly "rot", it might be one of the most important concepts in analysis and calculus which are without doubt some of the most important concepts in mathematics. Actually, it is completely ill-defined and arose from Cauchy's Kludge which is in error. http://thenewcalculu...auchykludge.pdf Edited January 11, 2012 by john_gabriel Link to comment Share on other sites More sharing options...

DrRocket Posted January 11, 2012 Share Posted January 11, 2012 You're right. I should have justified what I wrote. So, on a sphere we have [math]x^2+y^2+z^2=a^2[/math]. Thus [math]\begin{aligned}y^2 &= a^2-(z^2+x^2)\qquad(1) \\z^2 &= a^2-(x^2+y^2)\qquad(2) \\x^2 &= a^2-(y^2+z^2)\qquad(3)\end{aligned}[/math] Taking partial derivatives with of equation (1) respect to x, equation (2) with respect to y, and equation (3) with respect to z yields [math]\begin{aligned}\frac{\partial y}{\partial x} = -\frac x y \\\frac{\partial z}{\partial y} = -\frac y z \\\frac{\partial x}{\partial z} = -\frac z x \\\end{aligned}[/math] Thus [math]\frac{\partial y}{\partial x}\frac{\partial z}{\partial y}\frac{\partial x}{\partial z} = \left(-\frac x y\right) \left(-\frac y z\right) \left(-\frac z x\right) = -1[/math] You are missing the point. For the moment let's not worry about the sign ambiguity that arises because of the squares. With that understanding we restrict attention to the case where x,y,z are all non-negative. So the equation [math]x^2+y^2+z^2=a^2[/math] defines a 2-sphere of radius a in 3-space. We restrict attention to the portion of that sphere in the octant where all of the variables are non-negative. In that octant you can think of the equation as defining y implicitly as function of x and z, or defining x implicitly as a function of y and z or defining z implicitly as a function of x and y -- BUT not all three at the same time. Thus you cannot sensibly talk about [math]\frac{\partial y}{\partial x}\frac{\partial z}{\partial y}\frac{\partial x}{\partial z}[/math] becuse the objects that are purportedly being multiplied are functions that do not have a common domain. You can juxtapose the symbols, but the implied multiplication does not make sense, and that is why the "chain rule" does not appear to work here. It all comes back to the simple fact that derivatives and partial derivatives are defined in terms of functions, not equations. You can take derivatives of functions related by an equation, but you do need to have functions, not simply variables, to start with,and those functions must have a common domain. In this case you don't have that situation and the statement that [math]\frac{\partial y}{\partial x}\frac{\partial z}{\partial y}\frac{\partial x}{\partial z} = \left(-\frac x y\right) \left(-\frac y z\right) \left(-\frac z x\right) = -1[/math] is not meaningful. It is just a bunch of meaningless symbols being pushed around. I have seen similar things being done with symbol pushing sans meaning in physics texts before. Sometiime there are compensating errors or alternate, but correct, logic chains that can result in the final answer being valid -- and sometimes not. Wrong. See http://mathworld.wol...Derivative.html (don't feel like typing it out, sorry) This is an equation, so I don't know what you are talking about... Clearly you don't know what I am talking about. But that is not surprising since you don't know what you are talking about either Besides, what is the difference between a function and an equation? Please define both. As far as I am concerned, the difference is superficial. This is rather fundamental. A function from a set A to a set B is a subset of the cartesion product AxB such that if (a,x) and (a,y) both belong to that subset then x=y. If (a,x) is a pair in the function then we commonly write x=f(a). An equation is simply a statement that two thing are in fact the same. Your opinion on the superficiality of the difference is incorrect and evidence of abject ingnorance concerning the most important definition in all of mathematics, that of a function. Link to comment Share on other sites More sharing options...

ajb Posted January 11, 2012 Share Posted January 11, 2012 Thus you cannot sensibly talk about [math]\frac{\partial y}{\partial x}\frac{\partial z}{\partial y}\frac{\partial x}{\partial z}[/math] becuse the objects that are purportedly being multiplied are functions that do not have a common domain. Naively it seems to work, but then you get into a mess thinking what is a function of what? It does not make much sense to me, though you put it more elegantly that I would. I have seen similar things being done with symbol pushing sans meaning in physics texts before. Sometiime there are compensating errors or alternate, but correct, logic chains that can result in the final answer being valid -- and sometimes not. As a mathematical physicist I must concur. Link to comment Share on other sites More sharing options...

john_gabriel Posted January 11, 2012 Share Posted January 11, 2012 (edited) "You are missing the point. " I believe you have missed the point and you know it. So you are trying to back out gracefully. "In that octant you can think of the equation as defining y implicitly as function of x and z, or defining x implicitly as a function of y and z or defining z implicitly as a function of x and y -- BUT not all three at the same time. Thus you cannot sensibly talk about [math]\frac{\partial y}{\partial x}\frac{\partial z}{\partial y}\frac{\partial x}{\partial z}[/math] becuse the objects that are purportedly being multiplied are functions that do not have a common domain. " That they have a common domain or nor matters none. In this case they do have a common domain, so you are once again wrong. The domain is (-a,a) for all the partial derivatives. Trying to BS me? I am a mathematician and you do not fool me. This previous sentence is irrelevant. "You can juxtapose the symbols, but the implied multiplication does not make sense, and that is why the "chain rule" does not appear to work here." No, your ideas don't work, because you don't know what you are talking about. Not implied multiplication (no such thing), but exact multiplication takes place. "It all comes back to the simple fact that derivatives and partial derivatives are defined in terms of functions, not equations. " Again, BS! There is only a superficial difference in this case and it makes no difference to our discussion whatsoever. "Clearly you don't know what I am talking about. But that is not surprising since you don't know what you are talking about either" Too funny. I corrected you and I don't know what I am talking about... I am a mathematician and you are evidently not. "A function from a set A to a set B is a subset of the cartesion product AxB such that if (a,x) and (a,y) both belong to that subset then x=y. If (a,x) is a pair in the function then we commonly write x=f(a). An equation is simply a statement that two thing are in fact the same. " Really? You don't say?! I am not inferior to you. Do you think I am unaware of basic facts? Your response is devoid of logic and you are dancing around like a chicken without a head because you know you are wrong. Rather than try to cover up your stupidity, just admit that you are wrong. You claim an "equation is simply a statement that two things are in fact the same". Gee, let me see: f(x,y,z)=3x^5-2z and y=z^2+x^3. Hmm, your statement applies to both the function and the equation. It is clearly irrelevant and you have no idea with whom you are dealing with! Please, no more BS, okay? "Your opinion on the superficiality of the difference is incorrect and evidence of abject ingnorance concerning the most important definition in all of mathematics, that of a function." Actually what shows is your embarrassment at having being corrected. The partial derivative cannot just be meaningless symbols as you claim. For if this were the case, we could not talk about a product at all. Now, I am correcting your wrong comments on this post - you are clearly clueless as to what these partial derivatives or their physical interpretation means. The previous statement is so devoid of truth and evidently false, that it's hard to even know where to start pointing out your lack of understanding or maybe just your refusal to admit you are wrong? Be careful how you respond as my next response to you may include abusive verbal terms regarding your lack of intelligence. I am not inferior to you. In fact I consider myself inferior to no one. The first step toward understanding is to admit you are wrong when you are wrong. And you are wrong here in case you think otherwise. If you are a sincere academic, then what you do is say that you are wrong and move on. Now, if you have any further questions, I will be glad to help you but watch the arrogant and disdainful tone of your communication! The derivative (or partial derivative) is not infinitesimal, has nothing to do with very small values and does not require any knowledge of limits. Edited January 11, 2012 by john_gabriel Link to comment Share on other sites More sharing options...

ajb Posted January 11, 2012 Share Posted January 11, 2012 I am a mathematician and you are evidently not. Not that I can fully vouch for DrRocket, but based on his other posts and conversations with him he is a mathematician of good calibre. Anyway, we should all avoid personal attacks. Let the quality and clarity of the posts "do the talking". Link to comment Share on other sites More sharing options...

john_gabriel Posted January 11, 2012 Share Posted January 11, 2012 (edited) Naively it seems to work, but then you get into a mess thinking what is a function of what? It does not make much sense to me, though you put it more elegantly that I would. As a mathematical physicist I must concur. Nonsense. It works every time. But just like everything else needs to be carefully considered, this also has to be evaluated within context. Please do not feed Dr. Rocket's huge ego. Not that I can fully vouch for DrRocket, but based on his other posts and conversations with him he is a mathematician of good calibre. Anyway, we should all avoid personal attacks. Let the quality and clarity of the posts "do the talking". You are correct. However, his communication to me has been very disdainful, arrogant and provocative. If he were correct, it would probably not be so bad, but he is clearly wrong. All he has to do is admit he is wrong and move on. And I think he knows he is wrong. Edited January 11, 2012 by john_gabriel Link to comment Share on other sites More sharing options...

john_gabriel Posted January 11, 2012 Share Posted January 11, 2012 (edited) To see how asinine Dr. Rocket's argument is, let's consider p=-a/b q=-b/c r = c/(-a) in ordinary arithmetic (no calculus) Now, according to his argument, he believes that pqr suggests the product is 1 when it is in fact -1. But, p, q and r are symbols. Before we know their values, we can tell the product might be 1 or -1. However, when we substitute the numeric values for the symbols, we know exactly that it is -1. In my New Calculus, the symbolic differentials (whether ordinary or partial) are exactly equal to the antecedent and consequent parts of the finite difference ratio - not Newton's finite difference ratio (because Newton was WRONG), but my finite difference ratio. When the symbols transition to numbers in my finite difference ratio, the differentials are exactly equal or proportional to the antecedent and consequent parts of my finite difference ratio. Dr. Rocket: Until you admit you are wrong, you are no different from most academics I've met and you certainly don't deserve my respect. Edited January 11, 2012 by john_gabriel Link to comment Share on other sites More sharing options...

DrRocket Posted January 11, 2012 Share Posted January 11, 2012 Dr. Rocket: Until you admit you are wrong, you are no different from most academics I've met and you certainly don't deserve my respect. Everythiing that I have said is correct, and mathematically rigorous. I would be most concerned if I had your respect, as you have repeatedly made statements that have no basis whatever and have demonstrated lack of understanding of mathematics and inability with logic. Your assertion that you have had trouble with "academics" speaks volumes, though I have not been in academia for some time. Nevertheless you should probably listen to what those academics have told you. Go learn some mathematics and stop making absurd statements. Link to comment Share on other sites More sharing options...

john_gabriel Posted January 11, 2012 Share Posted January 11, 2012 (edited) Everythiing that I have said is correct, and mathematically rigorous. Chuckle. Chuckle. Almost everything you have said is nonsense and irrelevant. I would be most concerned if I had your respect, as you have repeatedly made statements that have no basis whatever and have demonstrated lack of understanding of mathematics and inability with logic. Your assertion that you have had trouble with "academics" speaks volumes, though I have not been in academia for some time. Nevertheless you should probably listen to what those academics have told you. Go learn some mathematics and stop making absurd statements. You don't have my respect because you are a fool. "Your assertion that you have had trouble with "academics" speaks volumes, though I have not been in academia for some time." Where did you read that you moron? Are you delusional as well? "Go learn some mathematics and stop making absurd statements." If only you knew how much this applies to you! Once again, claiming that I make baseless statements is your opinion. People are not stupid. They will realize you are clueless once they read the entire discussion. State where I have made a baseless statement. I don't care for your dishonest opinion. Do tell what the baseless and absurd statements are... Oh wait, there aren't any. You are a lying blowhard. Tsk. Tsk. So far the only baseless and irrelevant statements have been made by you: "Thus you cannot sensibly talk about because the objects that are purportedly being multiplied are functions that do not have a common domain." - Dr. rocket Chuckle, chuckle. The previous statement is wrong, absurd, baseless and irrelevant. ajb: I know what your response will be. Hold it. This is my last post. Yes, yes, I know I have insulted our friend Dr. Rocket. I could not help it. I apologize for my insults but not the mathematics or logic because I am correct. I was pushing Rocket to see what kind of character he has. Evidently he does not have the gumption to admit he is wrong. This is fine with me. I don't really care. Edited January 11, 2012 by john_gabriel Link to comment Share on other sites More sharing options...

Cap'n Refsmmat Posted January 11, 2012 Share Posted January 11, 2012 You are correct. However, his communication to me has been very disdainful, arrogant and provocative. If he were correct, it would probably not be so bad, but he is clearly wrong. All he has to do is admit he is wrong and move on. And I think he knows he is wrong. That is not an excuse for insults. I can see this discussion is going nowhere, so I'm closing it before we drop all semblance of civility whatsoever. I hope you'll familiarize yourself with SFN rule 1. Link to comment Share on other sites More sharing options...

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