January 5th, 2017, 05:44 AM  #1 
Newbie Joined: Jan 2017 From: Tehran, Iran Posts: 5 Thanks: 0  Rigorous definition of "Differential"
First of all I want to clarify that I posted this question on many forums and Q&A websites so the chances of getting an answer will be increased. So don't be surprised if you saw my post somewhere else. Now let's get started: __________________________________________________ _______________ When it comes to definitions, I will be very strict. Most textbooks tend to define differential of a function/variable in a way like this:  Let $\displaystyle f(x)$ be a differentiable function. By assuming that changes in $\displaystyle x$ are small, with a good approximation we can say:$\displaystyle \Delta f(x)\approx {f}'(x)\Delta x$ Where $\displaystyle \Delta f(x)$ is the changes in the value of function. Now if we consider that changes in $\displaystyle f(x)$ are small enough then we define differential of $\displaystyle f(x)$ as follows:$\displaystyle \mathrm{d}f(x):= {f}'(x)\mathrm{d} x$ Where $\displaystyle \mathrm{d} f(x)$ is the differential of $\displaystyle f(x)$ and $\displaystyle \mathrm{d} x$ is the differential of $\displaystyle x$. What bothers me is this definition is completely circular. I mean we are defining differential by differential itself. Although some say that here $\displaystyle \mathrm{d} x$ is another object independent of the meaning of differential but as we proceed it seems that's not the case: First of all we define differential as $\displaystyle \mathrm{d} f(x)=f'(x)\mathrm{d} x$ then we deceive ourselves that $\displaystyle \mathrm{d} x$ is nothing but another representation of $\displaystyle \Delta x$ and then without clarifying the reason, we indeed treat $\displaystyle \mathrm{d} x$ as the differential of the variable $\displaystyle x$ and then we write the derivative of $\displaystyle f(x)$ as the ratio of $\displaystyle \mathrm{d} f(x)$ to $\displaystyle \mathrm{d} x$. So we literally (and also by stealthily screwing ourselves) defined "Differential" by another differential and it is circular. Secondly (at least I think) it could be possible to define differential without having any knowledge of the notion of derivative. So we can define "Derivative" and "Differential" independently and then deduce that the relation $\displaystyle f'{(x)}=\frac{\mathrm{d} f(x)}{\mathrm{d} x}$ is just a natural result of their definitions (using possibly the notion of limits) and is not related to the definition itself. Though I know many don't accept the concept of differential quotient($\displaystyle \frac{\mathrm{d} f(x)}{\mathrm{d} x}$) and treat this notation merely as a derivative operator($\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}$) acting on the function($\displaystyle f(x)$) but I think that it should be true that a "Derivative" could be represented as a "Differential quotient" for many reasons. For example think of how we represent derivatives with the ratio of differentials to show how chain rule works by cancelling out identical differentials. Or how we broke a differential into another differential in the $\displaystyle u$substitution method to solve integrals. And it's especially obvious when we want to solve differential equations where we freely take $\displaystyle \mathrm{d} x$ and $\displaystyle \mathrm{d} y$ from any side of a differential equation and move it to any other side to make a term in the form of $\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}$, then we call that term "Derivative of $\displaystyle y$". It seems we are actually treating differentials as something like algebraic expressions. I know the relation $\displaystyle \mathrm{d} f(x)=f'(x)\mathrm{d} x$ always works and it will always give us a way to calculate differentials. But I (as an strictly axiomaticist person) couldn't accept it as a definition of Differential. So my question is: Can we define "Differential" more precisely and rigorously? Thank you in advance. P.S. I prefer the answer to be in the context of "Calculus" or "Analysis" rather than the "Theory of Differential forms". And again I don't want a circular definition. I think it is possible to define "Differential" with the use of "Limits" in some way(though it's just a feeling). 
January 5th, 2017, 05:54 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,210 Thanks: 555 
Yes, "differential" can be defined rigorously but such a definition really has to wait until "differential geometry". See http://people.math.gatech.edu/~ghomi...ureNotes8U.pdf for "differential map". until then, it is better to stay with the rough concept which is, yes, circular. Essentially, the concept of "differential" in basic Calculus is an attempt to make use of a general observation although the derivative, dy/dx, is NOT a fraction, it can be used AS if it WERE a fraction because it is a limit of a fraction. The "differentials", dy and dx are introduced, in a fairly handwaving way, in order to be able to treat dy/dx as if it were a fraction. 
January 5th, 2017, 07:21 AM  #3 
Senior Member Joined: May 2016 From: USA Posts: 421 Thanks: 169 
I may be wrong, but I am missing the circularity in what, admittedly, is very obscurely written. Let's try to clean it up. $Given\ y = f(x).$ $\Delta y \equiv f(x + \Delta x)  f(x) \equiv \text{change in y if x changed by } \Delta x.$ No mention of a "differential." What is defined is quite clearly a difference. $\displaystyle f'(x) \equiv \lim_{\Delta x \rightarrow 0} \dfrac{f(x + \Delta x)  f(x)}{\Delta x}.$ $dy \equiv f'(x)\ * \Delta x \equiv \text{differential of y.}$ The differential is a product, not a difference. Once the definitions are clarified, I am missing any circularity. Finally we have a proposition to be proved: $Given\ arbitrary\ \epsilon > 0,\ \exists\ \lambda\ such\ that$ $0 <  x  \Delta x  < \lambda \implies  dy  \Delta y  < \epsilon.$ In English, that proposition says that if the change in x is sufficiently small, the differential of y approximates the change in y. That as yet unproved proposition may or may not be true, and it may or may not be provable, but it is not circular. What am I missing? Last edited by JeffM1; January 5th, 2017 at 07:25 AM. 
January 5th, 2017, 07:30 AM  #4  
Newbie Joined: Jan 2017 From: Tehran, Iran Posts: 5 Thanks: 0  Quote:
 
January 5th, 2017, 07:55 AM  #5 
Newbie Joined: Jan 2017 From: Tehran, Iran Posts: 5 Thanks: 0  Once again the main problem is when we reach this point we easily take $\Delta x$ as $\mathrm{d} x$ without any reason and then treat the derivative of $y$ as the ratio of two differentials: $\frac{\mathrm{d} y}{\mathrm{d} x}$. Was $\mathrm{d} x$ as the same as $\Delta x$ then we wouldn't need to introduce a new concept such as differential. Or maybe I'm missing something? 
January 5th, 2017, 08:56 AM  #6  
Senior Member Joined: May 2016 From: USA Posts: 421 Thanks: 169  Quote:
I simply did not address the issues of what is the most intuitive notation for introducing calculus or what is the best way to let students learn how to apply calculus. I am dubious that any analysis should be included in introductory calculus, and if it must, I might lean in favor of some form of simplified nonstandard analysis. To put it differently, I think understanding should precede rigor. Understanding the utility of calculus justifies rigor; without that understanding, rigor is without motivation.  
January 5th, 2017, 09:25 AM  #7  
Newbie Joined: Jan 2017 From: Tehran, Iran Posts: 5 Thanks: 0  Quote:
I think at least in mathematics circularity is very frowned upon. Anyway now I get what you meant. Quote:
This seems important too. I wish there were a way too keep rigor and intuition at the same time.  
January 5th, 2017, 01:01 PM  #8  
Senior Member Joined: Aug 2012 Posts: 824 Thanks: 140  Quote:
Quote:
* Nobody else has bothered to write a similar text; and * Hyperrealbased teaching of calculus has not caught on. Studies have been done showing that students are equally confused no matter which approach is taken. And if you teach calculus using nonstandard terminology and concepts ("nonstandard" having two meanings here, as in nonstandard analysis and also not the standard way of teaching calculus) you would leave students totally unprepared to study standard physics, math, biology, statistics, or any other subject. The only thing I can imagine that would change this state of affairs is if tomorrow morning professor soandso from Helsinki proves P $\neq$ NP using hyperreals. If that happened, everyone would suddently develop tremendous interest in nonstandard analysis. Absent a dramatic development along those lines, nobody is going to change the way calculus is taught. And as I say, I completely agree with you that the treatment of differentials in calculus texts is very confusing to students; and the more thoughtful and mathematically inclined the student, the more confused they are on this point. Last edited by Maschke; January 5th, 2017 at 01:15 PM.  
January 6th, 2017, 07:59 AM  #9  
Newbie Joined: Jan 2017 From: Tehran, Iran Posts: 5 Thanks: 0  Quote:
I remember someone pointed out that the main problem with defining Differential in "Real calculus" is that the mathematical objects $\displaystyle df(x)$ and $\displaystyle dx$ aren't even "Real numbers" or in some sense aren't "Real valued functions". But I also remember there were a section in our "Calculus" course where they discussed "Infinite limits" and "Infinite derivatives" which were Limits and Derivatives that have a value of "Infinity". Since infinity isn't a real number sure they were treated as a kind of "Nonexistent limit/derivative" but as a special kind of them. I mean this type of nonexistence of limit/derivative had some significant properties which made them important. Now why don't introduce "Infinitesimals" as some kind of nonexistent limits and then relate the idea to the concept of "Differentials". By doing so we neither entered into nonstandard analysis nor put down the differentials and even somehow made them more rigorous. Do you think it's possible?  
January 6th, 2017, 08:25 AM  #10  
Senior Member Joined: Dec 2012 Posts: 890 Thanks: 20  Quote:
But I hope we are on the way to let think more and more clear. New more simple and clear math is knocking on the door, it's just question of how many years it takes, to cancel out the old math generations...  

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