hobz Posted April 26, 2010 Share Posted April 26, 2010 The product rule: [math] y=u(x)\cdot v(x) [/math] [math] y = u\cdot v [/math] [math] y+dy=(u+du)\cdot(v+dv) [/math] [math] y+dy=uv+udv+vdu+dudv [/math] now [math]dudv[/math] is discarded on the grounds of being "too small". If I were to include it, later on (by subtracting y and dividing through by dx), it would become [math]\frac{dudv}{dx}[/math]. What does that mean? 1 Link to comment Share on other sites More sharing options...

D H Posted April 26, 2010 Share Posted April 26, 2010 Nothing. You really shouldn't be treating things the way you did unless you know either differential forms or non-standard analysis. Link to comment Share on other sites More sharing options...

triclino Posted April 26, 2010 Share Posted April 26, 2010 (edited) The product rule: [math] y=u(x)\cdot v(x) [/math] [math] y = u\cdot v [/math] [math] y+dy=(u+du)\cdot(v+dv) [/math] [math] y+dy=uv+udv+vdu+dudv [/math] now [math]dudv[/math] is discarded on the grounds of being "too small". If I were to include it, later on (by subtracting y and dividing through by dx), it would become [math]\frac{dudv}{dx}[/math]. What does that mean? Definition: if y = f(x) ,then dy= f'(x)Δx ,and dx = 1.Δx. Theorem : if y = u(x)v(x) ,then dy = duv + udv. hence y+dy = vdu +udv +uv. Thus [math]y+dy\neq(u+du)(v+dv)[/math]. However y+Δy = u(x+Δx)v(x+Δx) = (u+Δu)(v+Δv) Edited April 26, 2010 by triclino correction Link to comment Share on other sites More sharing options...

hobz Posted April 27, 2010 Author Share Posted April 27, 2010 What I write looks identical to both http://en.wikipedia.org/wiki/Product_rule#Discovery_by_Leibniz and http://en.wikipedia.org/wiki/Product_rule#Using_non-standard_analysis. In both cases the dudv is discarded. But what happens if I keep it? Link to comment Share on other sites More sharing options...

ajb Posted April 27, 2010 Share Posted April 27, 2010 You can keep such terms if you wish. In essence you will be keeping another term in an expansion. You end up considering "second order variations". The way to think about it is in terms of Taylor's theorem, which holds for any smooth function. Then consider the Taylor expansion about [math]x + \epsilon dx[/math], where [math]\epsilon[/math] is the "perturbation parameter" and [math]dx[/math] is considered a formal object. You think of this as a "morphism of the real line"* of the form [math] x \rightarrow x + \epsilon dx[/math]. We think of things quite formally and take the only possible real value of [math]\epsilon[/math] to be zero. (You also think the same of [math]dx[/math]). Then you end up with [math]U(x) \rightarrow \overline{U}(x,dx) = U(x) + \epsilon dx \frac{\partial U}{\partial x} + (\epsilon dx)^{2}\frac{1}{2!} \frac{\partial^{2} U}{\partial x^{2}}[/math], where we have assumed that [math]\epsilon^{3}=0[/math]. Or you could keep going. You should be able to build up what you need from there. However, don't think about division by formal objects. The better way of doing things is to pick terms in order of [math]\epsilon[/math]. For instance, the standard derivative is (in essence) the first order term. As an aside I have wondered about any possible role of a "second order Lie derivative" acting on geometric objects over a manifold. But that is a different, but similar story. * Really we are considering the "space" with local coordinates [math]\{x,dx\}[/math]. Link to comment Share on other sites More sharing options...

hobz Posted April 27, 2010 Author Share Posted April 27, 2010 But surely there are no terms in the Taylor expansion that are not there when taking the derivative. I mean, the Taylor expansion must yield zero for the second term, right? If the Taylor expansion is used on [math]y=u(x)\cdot v(x)[/math] would the term [math]\frac{dudv}{dx}[/math] also appear? On another note. Is the differential notation not perfectly valid and equivalent to the difference quotient notation? If [math] \frac{dy}{dx} = \lim_{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x} [/math] then what is [math]\frac{dy}{dx}[/math] in the differential notation? If: [math] y = x^2 [/math] Changing x by the amount dx gives the change dy in y: [math] y + dy = (x + dx)^2 [/math] [math] y + dy = x^2 + 2x\cdot dx + (dx)^2 [/math] [math] dy = 2x\cdot dx + (dx)^2 [/math] [math] \frac{dy}{dx} = 2x + dx [/math] Suggesting that the rate of change in y with respect to x also depends on dx. Now you have to argue that dx is negligible, which is handled by limits in the difference quotient way of writing this. So is the differential notation inferior? Link to comment Share on other sites More sharing options...

ajb Posted April 27, 2010 Share Posted April 27, 2010 But surely there are no terms in the Taylor expansion that are not there when taking the derivative. I mean, the Taylor expansion must yield zero for the second term, right? One would usually think of [math]\epsilon[/math] as nilpotent, so [math]\epsilon^{2}=0[/math]. The only number that can do this is zero. So, to keep things non-trivial [math]\epsilon[/math] is formal, or we can think of it as taking "values" elsewhere. As you want to think about "higher order terms" it makes sense to consider [math]\epsilon[/math] either nilpotent at some higher order e.g. [math]\epsilon^{3}=0[/math] or not nilpotent at all. If the Taylor expansion is used on [math]y=u(x)\cdot v(x)[/math] would the term [math]\frac{dudv}{dx}[/math] also appear? So, as the product of smooth functions is a smooth function we can directly take the series. I get [math]\overline{y}(x,dx) = u \cdot v(x) + \epsilon dx \left( \frac{\partial u}{\partial x} v + u \frac{\partial v}{\partial x} \right) + \frac{1}{2!} (\epsilon dx)^{2}\left( \frac{\partial^{2} u}{\partial x^{2}} v + u \frac{\partial^{2} v}{\partial x^{2}} + 2 \frac{\partial u}{\partial x} \frac{\partial v}{\partial x}\right)[/math]. Again taking the cubic term to vanish. So we see that first order term is the derivative and the second order term contains term that measures the failure of a second order derivative [math]\frac{\partial^{2}}{\partial x^{2}}[/math] ("Laplacian") to be a derivation. It gets very interesting when you consider such terms, in a slightly different context. You end up with things related to the BV-antifield formalism in quantum field theory, but that is another story. On another note. Is the differential notation not perfectly valid and equivalent to the difference quotient notation? If [math] \frac{dy}{dx} = \lim_{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x} [/math] then what is [math]\frac{dy}{dx}[/math] in the differential notation? Partial derivatives of functions makes sense, this we can take in terms of limits as you have said or we could define it axiomatically. So under a true diffeomorphism of the real line [math]\gamma : x \rightarrow y[/math] with the usual abuse [math]\gamma^{*}y = y(x)[/math] (now a function in terms of the original coordinate) we have [math]dy = dx \frac{\partial y(x)}{\partial x}[/math]. Note we have not dived by [math]dx[/math]. The partial derivatives are defined in terms of limits, the differentials are formal. If: [math] y = x^2 [/math] Changing x by the amount dx gives the change dy in y: [math] y + dy = (x + dx)^2 [/math] [math] y + dy = x^2 + 2x\cdot dx + (dx)^2 [/math] [math] dy = 2x\cdot dx + (dx)^2 [/math] [math] \frac{dy}{dx} = 2x + dx [/math] Suggesting that the rate of change in y with respect to x also depends on dx. Now you have to argue that dx is negligible, which is handled by limits in the difference quotient way of writing this. You are almost right, but dividing by [math]dx[/math] is problematic. I mean in general we will not have a clear notion of dividing. What you get is [math]\overline{y}(x,dx) = y(x) + \epsilon dx 2x + (\epsilon dx)^{2}[/math] I then define the derivative to be the first order term, that is [math] \overline{y}(x,dx) - y(x) = \epsilon dx\frac{\partial y}{\partial x} + O(\epsilon^{2})[/math] then [math]dy = dx\frac{\partial y}{\partial x}[/math]. So is the differential notation inferior? Done properly the use of differentials is far better than just using partial derivatives. This becomes clear when dealing with differential geometry where the use of differentials allows coordinate free expressions. Look up differential forms for example. (I can say a lot more about them another time ) Link to comment Share on other sites More sharing options...

hobz Posted April 27, 2010 Author Share Posted April 27, 2010 Thanks for the reply. So we see that first order term is the derivative and the second order term contains term that measures the failure of a second order derivative [math]\frac{\partial^{2}}{\partial x^{2}}[/math] ("Laplacian") to be a derivation. I don't understand this. Failure of a derivative to be a derivation? So under a true diffeomorphism of the real line [math]\gamma : x \rightarrow y[/math] with the usual abuse [math]\gamma^{*}y = y(x)[/math] (now a function in terms of the original coordinate) we have [math]dy = dx \frac{\partial y(x)}{\partial x}[/math]. I am unfamiliar with this notation. Does it mean that [math]\gamma[/math] maps x to y? (Can't figure out [math]\gamma^{*}y=y(x)[/math].) Note we have not dived by [math]dx[/math]. The partial derivatives are defined in terms of limits, the differentials are formal. That is to say that [math]dx[/math] is a differential where [math]\partial x[/math] is not? I thought they were identical execept the partials denoted a function of several variables being differentiated with respect to one while treating the other as constant. You are almost right, but dividing by dx is problematic. I mean in general we will not have a clear notion of dividing. Does that mean that the derivative is not dependent on division to exist in a given context in order to work? (This would mean that the notation is sort of abused). I cases where division is defined (a*b=c <=> b=c/a i guess) is it then problematic to divide? [math] dy = dx\frac{\partial y}{\partial x} [/math] Is this different from [math] dy = dx\frac{dy}{dx} [/math] when y is only dependent on x? (Related to a previous encounter http://www.scienceforums.net/forum/showthread.php?t=44865) Link to comment Share on other sites More sharing options...

triclino Posted April 27, 2010 Share Posted April 27, 2010 But surely there are no terms in the Taylor expansion that are not there when taking the derivative. I mean, the Taylor expansion must yield zero for the second term, right? If the Taylor expansion is used on [math]y=u(x)\cdot v(x)[/math] would the term [math]\frac{dudv}{dx}[/math] also appear? On another note. Is the differential notation not perfectly valid and equivalent to the difference quotient notation? If [math] \frac{dy}{dx} = \lim_{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x} [/math] then what is [math]\frac{dy}{dx}[/math] in the differential notation? If: [math] y = x^2 [/math] Changing x by the amount dx gives the change dy in y: [math] y + dy = (x + dx)^2 [/math] [math] y + dy = x^2 + 2x\cdot dx + (dx)^2 [/math] [math] dy = 2x\cdot dx + (dx)^2 [/math] [math] \frac{dy}{dx} = 2x + dx [/math] Suggesting that the rate of change in y with respect to x also depends on dx. Now you have to argue that dx is negligible, which is handled by limits in the difference quotient way of writing this. So is the differential notation inferior? hobz ,how do you know that: [math] y + dy = (x + dx)^2 [/math] Link to comment Share on other sites More sharing options...

ajb Posted April 28, 2010 Share Posted April 28, 2010 I don't understand this. Failure of a derivative to be a derivation? Try it yourself. Take the operator [math]\frac{\partial^{2}}{\partial x \partial x}[/math] and apply it to a product of functions. I am unfamiliar with this notation. Does it mean that [math]\gamma[/math] maps x to y? (Can't figure out [math]\gamma^{*}y=y(x)[/math].) Yes, so we have [math]x[/math] and we kick out [math]y[/math]. In order to proceed we then have to pull-back [math]y[/math] to get a function of [math]x[/math]. So you think of the map as giving a function [math]y(x)[/math]. That is to say that [math]dx[/math] is a differential where [math]\partial x[/math] is not? I thought they were identical execept the partials denoted a function of several variables being differentiated with respect to one while treating the other as constant. You usually write [math]\frac{dy}{dx} = \frac{\partial y }{\partial x}[/math] in the situation you have described. But saying [math]dx = \partial x[/math] I do not understand. Does that mean that the derivative is not dependent on division to exist in a given context in order to work? (This would mean that the notation is sort of abused). I cases where division is defined (a*b=c <=> b=c/a i guess) is it then problematic to divide? You can define the notion derivatives or differential on vector spaces, algebras, Lie algebras etc... You don't need the notion of division to do this. Is this different from [math] dy = dx\frac{dy}{dx} [/math] when y is only dependent on x? (Related to a previous encounter http://www.scienceforums.net/forum/showthread.php?t=44865) You should still really say [math] dy = dx\frac{\partial y }{\partial x} [/math] and then you can "informally" think of cancelling the [math]dx[/math] and the [math]\partial x[/math]. But this is very informal, it will help you keep track of things. Link to comment Share on other sites More sharing options...

hobz Posted April 28, 2010 Author Share Posted April 28, 2010 hobz ,how do you know that:[math] y + dy = (x + dx)^2 [/math] I assume that by changing x I change y since y=f(x). So by changing x by dx, I change y by dy. My goal is then to find out how much the change in y (dy) is in response to a change in x (dx). Merged post follows: Consecutive posts mergedTry it yourself. Take the operator [math]\frac{\partial^{2}}{\partial x \partial x}[/math] and apply it to a product of functions. I will attempt this a quiet evening. You usually write [math]\frac{dy}{dx} = \frac{\partial y }{\partial x}[/math] in the situation you have described. But saying [math]dx = \partial x[/math] I do not understand. You can define the notion derivatives or differential on vector spaces, algebras, Lie algebras etc... You don't need the notion of division to do this. If I understand your notation correctly, then [math]\frac{\partial y}{\partial x}[/math] means the derivative. This a notation, and not a division; the notation cannot be separated into a [math]\partial y[/math] and [math]\partial x[/math]. It serves as an operator. If so, then this operator relates the change in x [math]dx[/math] to the change in y [math]dy[/math]. Am I off? It makes sense in systems where division is not defined that [math]\frac{\partial y}{\partial x}[/math] is NOT a division. Is the notation [math]\frac{\partial y}{\partial x}[/math] still related to Taylor's theorem? Link to comment Share on other sites More sharing options...

ajb Posted April 28, 2010 Share Posted April 28, 2010 You are right, however thinking of [math]dx[/math] as "a small change in" is problematic, that is why in part differential forms etc were developed. This is also why division by them is ill-defined. The only real value they can take is zero. If you want to make any sense of it, the best you will be doing is dividing by zero! In part this is why your [math]dudv/ dx[/math] has no obvious meaning. This is different to taking limits tending to zero. They may still be well defined. Really, this maybe a "chicken and egg" situation (I am no fan of doing analysis very carefully!) but as any smooth function has a Taylor expansion I think of the derivatives formally as terms in this expansion. Not that this helps you calculate them! However, it can set you up ready for more general things. In a more general setting you can define differentials etc more formally via the algebraic rules you want them to have. This quickly starts to not feel very much like your intuitive notion of differentiation as "rates of change". I hope I have not confused you further. Link to comment Share on other sites More sharing options...

triclino Posted April 28, 2010 Share Posted April 28, 2010 I assume that by changing x I change y since y=f(x).So by changing x by dx, I change y by dy. My goal is then to find out how much the change in y (dy) is in response to a change in x (dx). In mathematics every statement equation e.t.c is justified by an axiom ,theorem,definition,or a law of logic ,otherwise mathematics would not be mathematics , but a composition of ideas resulting in a chaotic situation. Yes a change in x will respectively produce a change in y since y =y(x). Now since by by definition ,dy = ([math]x^2[/math])'dx = 2xdx ,then y + dy = [math]x^2 + 2xdx[/math] and not equal to [math](x+dx)^2[/math]. Unless you have a different definition for differentials . Further more there is a difference between infinitesimals and differentials Link to comment Share on other sites More sharing options...

ajb Posted April 28, 2010 Share Posted April 28, 2010 Now since by by definition ,dy = ([math]x^2[/math])'dx = 2xdx ,then y + dy = [math]x^2 + 2xdx[/math] and not equal to [math](x+dx)^2[/math]. So here we take differentials to be nilpotent. Thus [math](x+dx)^{2} = x^{2} + xdx + dx x + dx^{2} = x^{2}+ 2 x dx[/math]. Link to comment Share on other sites More sharing options...

triclino Posted April 28, 2010 Share Posted April 28, 2010 So here we take differentials to be nilpotent. Thus [math](x+dx)^{2} = x^{2} + xdx + dx x + dx^{2} = x^{2}+ 2 x dx[/math]. What is the exact definition in mathematics of nilpotent ?? Link to comment Share on other sites More sharing options...

ajb Posted April 28, 2010 Share Posted April 28, 2010 Have a look at the Wikipedia article Nilpotent. If the order is not specified then we usually mean order two, as is usually meant in the physics literature. Link to comment Share on other sites More sharing options...

triclino Posted April 28, 2010 Share Posted April 28, 2010 Have a look at the Wikipedia article Nilpotent. If the order is not specified then we usually mean order two, as is usually meant in the physics literature. That definition does justify the equation: dy = (x+dx)^2 ,but only the equation : [math](x+dx)^2 = x^2+2xdx[/math] Link to comment Share on other sites More sharing options...

hobz Posted April 28, 2010 Author Share Posted April 28, 2010 So how would you go about defining differentiation where the is no division? Link to comment Share on other sites More sharing options...

triclino Posted April 28, 2010 Share Posted April 28, 2010 So how would you go about defining differentiation where the is no division? I don't know what do you mean when you say: "there is no division" ,but since dy = f'(x)dx ,if you divide by dx ,then dy/dx = f'(x) = [math]\frac{dy}{dx}[/math]. Hence the ratio of dy to dx is equal to the derivative of y w.r.t x Link to comment Share on other sites More sharing options...

hobz Posted April 29, 2010 Author Share Posted April 29, 2010 I don't know what do you mean when you say: "there is no division" ,but since dy = f'(x)dx ,if you divide by dx ,then dy/dx = f'(x) = [math]\frac{dy}{dx}[/math]. Hence the ratio of dy to dx is equal to the derivative of y w.r.t x I mean, when the notion of division is not defined. For instance in a vector space as ajb suggested. I think we arrived at the special notation [math]\frac{dy}{dx}[/math] is not a ratio, but defined in terms of limits of a ratio (the normal derivative definition). However, I have seen plenty of derivations of formulas in physics, where dy is treated as a "thing" that easily can be divided with dx. Or multiplying [math]\frac{dy}{dx} = f(x)[/math] by dx and integrate both sides is also frequently used. Link to comment Share on other sites More sharing options...

ajb Posted April 29, 2010 Share Posted April 29, 2010 Jumping in at the deep end look at the wikipedia article on differential graded category! Probably more useful is the notion of a differential algebra, for a differential vector space use the "forgetful functor" to forget the multiplication in an algebra. (Often more useful objects arise when we give them a grading.) Link to comment Share on other sites More sharing options...

hobz Posted May 1, 2010 Author Share Posted May 1, 2010 The notation looks advanced. I will have to look it up somewhere. Is it fair to say, that in Leibniz' days, the [math]\frac{dy}{dx}[/math] was thought of as a ratio, until the more rigid approach with limits gave a definition where nilpotents and arguments as to [math](dx)^2[/math] could be discarded while [math]dx[/math] could not, were not needed? The notation [math]\frac{dy}{dx}[/math] is a legacy that reminds us of the origin (a ratio) but really isn't. If division is defined, then [math]dy[/math] and [math]dx[/math] can be split up into each of their own limits and be multiplied on both sides etc. rendering algebraic operations valid in algebra that has +,-,*,/ defined (as I suppose was the case around Leibniz days). If division is undefined, the notion of [math]\frac{dy}{dx}[/math] doesn't change; it is still the derivative or how much [math]dy[/math] changes when [math]dx[/math] changes, but it is no longer a ratio, and the derivative has to be defined in another way. I mean, when introduced in high school, the classic secant to tangent method is introduced, and the limit is called the derivative. While I always thought that this presented the idea of the derivative in any space, it really presents a way of calculating the derivative in that particular space where +,-,*,/ are defined. If we move into a vector field, the idea of measuring how the vector field changes is still perfectly valid. But to calculate the change is no longer a ratio issue, but the term [math]\frac{dy}{dx}[/math] lingers on, and becomes an operator in its entirety. [math]dx[/math] looses it meaning, the only thing that matters is the operator: [math]\frac{dy}{dx}[/math]. Can you verify or completely obliterate my thoughts the past days? Link to comment Share on other sites More sharing options...

ajb Posted May 1, 2010 Share Posted May 1, 2010 (edited) The notation [math]\frac{dy}{dx}[/math] is a legacy that reminds us of the origin (a ratio) but really isn't. It is a limit of a ratio. I think that Leibniz was using what we would call today infinitesimals for [math]dx[/math]. He was thinking of "very small changes" (very informally). If we move into a vector field, the idea of measuring how the vector field changes is still perfectly valid. This is the Lie derivative. ...[math]\frac{dy}{dx}[/math]. You should think of [math]\frac{\partial}{\partial x}[/math] as an operator. Generally I would not worry too much about what [math]dx[/math] means. Depending on the exact context I know how to deal with it and that is usually enough. I tend to think of [math]dx[/math] in terms of Grassmann algebra, this takes us to the notion of a differential form. Now, [math]dx[/math] is a Grassmann number, it can be treated formally or we can think of it taking values "somewhere". The only real value it can take is zero. Other people think of them as infinitesimals. Edited May 1, 2010 by ajb Link to comment Share on other sites More sharing options...

hobz Posted May 2, 2010 Author Share Posted May 2, 2010 So is [math]\frac{\partial}{\partial x}[/math] more "correct" than [math]\frac{d}{dx}[/math] which is more of a special case? Link to comment Share on other sites More sharing options...

ajb Posted May 2, 2010 Share Posted May 2, 2010 (edited) Really it is all about notation. Both the things you have written make sense, the partial is "more useful" when we have several variables and want to vary one of them. See Partial derivative. You can then use these to build the total derivative: Let us consider a function of two variables [math]x [/math] and [math]y[/math] [math]df = dx \frac{\partial f}{\partial x} + dy \frac{\partial f}{\partial y}[/math]. Edited May 2, 2010 by ajb Link to comment Share on other sites More sharing options...

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