What is the concept of dx in calculus?

[spaghetti3451]

What is dx?

Wikipedia defines it as an infinitesimal change in x.
Is Wikipedia correct?
Assuming it is, what is not to say that if we halve a piece of string an infinite number of times, we shouldn't end up with zero? In that sense, d has to be zero!?

More importantly, why is pdx = d(px) and (1/a)(d/dx) = 1/d(ax)?

 

[micromass]

From physics point-of-view, wikipedia is certainly correct that dx is an infinitesimal change of x.
However, as a mathematician, I strongly dislike this point-of-view since infinitesimals have no meaning in (standard) analysis. So always see dx as a notation with no meaning.

Of course, there are mathematical ways to give meaning to dx. But it's always been easier for me to just see dx without any meaning.

 

[spaghetti3451]

Then if dx without meaning, the whole edifice of physical sciences rests on a very shaky foundation!?

 

[micromass]

Let me first say, if I say that I consider dx to be without meaning, that I'm expressing my personal opinion. I'm sure a lot of people will disagree with me. So what I'm saying is only my point-of-view.

But the fact that dx has no meaning in mathematics, doesn't mean that the concepts of integrals and derivatives have no meaning. They are well defined. And it are these concepts that the physical sciences uses, not the tricky dx.

There is however a way to give a mathematical meaning to dx, most notably nonstandard analysis tries to do. So if you really want to use with infinitesimals, you should adopt a nonstandard approach.

 

[Hurkyl]

I would have argued the language of differential geometry is more notable -- differential forms specifically.

 

[micromass]

Yes, of course, how could I forget that... Maybe it's a good suggestion to the OP: try to read "calculus on manifolds" by Spivak. There they try to give a meaning to the term dx

 

[gomunkul51]

dx is a ghost of a long departed quantity, my friend.

 

[lavinia]

From physics point-of-view, wikipedia is certainly correct that dx is an infinitesimal change of x.
However, as a mathematician, I strongly dislike this point-of-view since infinitesimals have no meaning in (standard) analysis. So always see dx as a notation with no meaning.

Of course, there are mathematical ways to give meaning to dx. But it's always been easier for me to just see dx without any meaning.

From reading physics books I get the impression that dx is a small change in a physical variable - how small depends on the situation but it is always meant to be small enough. The infinitesimal is always implied in this way of thinking and sits behind the curtain of observable quantities but dx is always a finite small increment, an observable quantity.

The relation between this way of thinking and the mathematical way seems to come from the approximation of dy by y'(x)dx keeping in mind that this approximation is good when dx is "small enough".

 

[lurflurf]

What is dx?

More importantly, why is pdx = d(px) and (1/a)(d/dx) = 1/d(ax)?

That should be
(1/a)(d/dx) = d/d(ax)

I think the best way to think of differentials is as a function that maps functions to linearizations.

df=f'(x) dx
dx need not be small, though we may note that when it is it may be close to
f(x+dx)-f(x)

 

[HallsofIvy]

Then if dx without meaning, the whole edifice of physical sciences rests on a very shaky foundation!?

No, micromass did not say that dx was "without meaning". He just said that he found it easier to think of it that way. perhaps because the definition is very subtle.

 

[Studiot]

The traditional way out of this question is not to regard dx as change in x, small or otherwise, but to regard it as the difference between two points, say x₁ and x₂, where x₂ is in the neighbourhood of x₁

Then dx = x₂ - x₁ and dy = f(x₂) - f(x₁)

The distinction is very subtle but consider this:

If x₂ is a neighbourhood point for x₁ then this guarantees the conditions for differentiabilty. It allows us to use a limiting process/argument as there are always more neighberhood points between x₁ and x₂, no matter how close we get. That is continuity is guaranteed. Alternatively, if x₁ has no neighbourhood points in a certain direction that is the condition for the derivative not to exist.

I also think it is better to reserve the Greek lower case delta (\delta) for incremental (small) chages in something.

Edit

One advantage of the two point approach is that x₁ and x₂ can belong to the same set of points, X : x_n is in X.

However there is nothing to say that the difference dx has to belong to this set.

 

[AlephZero]

It may be "traditional", but that doesn't mean it is valid.

The idea that you can "guarantee the conditions for differentiability" by definiing something is nonsense. Either the function is differentiable at a point or it isn't.

You can define anything you like in math, but before you can use it, you have to be sure it exists.

You can prove anything you like by defining something that doesn't exist - for example "let X to be an integer greater than 0 and less than 1".

 

[Studiot]

is nonsense

If you think it is nonsense can you prove it?

You can define anything you like in math,

Are you suggesting that in mathematics I can take three self consistent axioms and define a fourth, contrary to each of the other three?

Either the function is differentiable at a point or it isn't.

Such a statement says nothing about the conditions under which a function is differentiable.

Why another personal attack?

If you genuinely don't understand what I have said ask for clarification.
If you can prove it wrong, show your working for all to see and I can accept it.

 

[galoisjr]

dx and dy for that matter have very specific meanings. It is the differential change in x and y. And it is defined as such because of the definition of the derivative. You see the from the definition you cannot actually take the derivative at a point because once you are only considering that one point you throw away the rest of the line and then how is it possible for a limit to exist? It is the same thing with endpoints of segments of a function. If a function is defined in a closed interval then the derivative is defined in the open interval. This is because a limit does not exist at the endpoints. The derivative acts on infinitesimal segments of a function which gives rise to the need for infinitesimal values of x and y. They do in effect have no meaning without their corresponding variable. Just like a variable x and y are place holders for any value of their respective set the differentials are placeholders for continuous infinitesimal changes for the values in those sets.

 

[galoisjr]

To answer your question about the string you may want to look at zeno's paradox. The differential will never equal zero but under the limit you approach zero. And as for your final question, these properties come from the fact that the d/dx is a linear operator and the derivative is a linear transformation.

 

[Studiot]

More importantly, why is pdx = d(px) and (1/a)(d/dx) = 1/d(ax)?

No entirely sure what you are thinking of here, perhaps if you labelled your terms?

If you were thinking of some formula such as posted by lavinia, have a care

The relation between this way of thinking and the mathematical way seems to come from the approximation of dy by y'(x)dx keeping in mind that this approximation is good when dx is "small enough".

eg use of

y = f(x₀) + f'(x₀) dx

to obtain approximations to y, crossing x=3 eg starting from x₀ = 2.9 and dx = 0.2
for the functions

y = {2x, x < 3}; {3x-3, x \geq 3}

or

y = {2x, x < 3}; {2x+3, x \geq 3}

will get you the wrong answers

Since
The first equation is continuous at x =3, but has no derivative there.
The second equation has a derivative at x = 3, but is discontinuous there.

 

[TylerH]

dx isn't without meaning in standard analysis. It can be defined as the limit x-h as h->x. So, it's 0, unless you put it in ratio with another differential.

 

[Hurkyl]

dx isn't without meaning in standard analysis. It can be defined as the limit x-h as h->x. So, it's 0, unless you put it in ratio with another differential.

If you define dx to be that limit, and similarly for dy, then dy/dx is undefined, because it would be the quotient of zero by zero.

"dx" defined in this fashion would have nothing to do with calculus at all.

 

[TylerH]

If you define dx to be that limit, and similarly for dy, then dy/dx is undefined, because it would be the quotient of zero by zero.

"dx" defined in this fashion would have nothing to do with calculus at all.

You can do cool crap when you're using limits.
\frac{df}{dx}=\frac{lim_{g \to f}f-g}{lim_{h \to x}x-h}=lim_{h \to x}\frac{f(x)-f(h)}{x-h}

 

[Hurkyl]

You can do cool crap when you're using limits.
\frac{df}{dx}=\frac{lim_{g \to f}f-g}{lim_{h \to x}x-h}=lim_{h \to x}\frac{f(x)-f(h)}{x-h}

You can't do what you tried to do there. The middle term is clearly undefined, being 0/0.

Check your limit laws --
\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}
requires the denominator on the right hand side to be nonzero.

Honestly, this should be clear -- the right hand side is nonsense if the denominator is zero. You have the divide then take limits -- you can't take limits then divide in such a case.

 

[homology]

Hi failexam:

differential forms is definitely a good way to think about differentials. You might also find nonstandard analysis interesting in that it constructs infinitesimals (and infinite numbers) rigorously and has some poetic language in it (halos, shadows, etc)

But for a great deal of basic physics treating dx as a 'little change in x' doesn't get you into trouble. You'll see it in many derivations and you'll generally notice that you could redo the derivation with \Delta x instead and take limits.

You'll usually get a split of folks on this matter.

Physically, I certainly think it has meaning as it can contain. Consider

<br /> W=\int F\cdot dx<br />

for the work done by a force F. dx has units of length and it certainly looks, feels and smells like Fdx is a 'small' amount of work and we're adding it up via an integral. When I teach calculus this is how I derive these integrals (and volumes etc) as it appeals to the intuition. Can you get in trouble, sure, but you'll be fine most of the time.

I think your interpretation will depend on what you have to do with them and what level of rigor you're comfortable with.

 

[Rasalhague]

No entirely sure what you are thinking of here, perhaps if you labelled your terms?

If you were thinking of some formula such as posted by lavinia, have a care



eg use of

y = f(x₀) + f'(x₀) dx

to obtain approximations to y, crossing x=3 eg starting from x₀ = 2.9 and dx = 0.2
for the functions

y = {2x, x < 3}; {3x-3, x \geq 3}

or

y = {2x, x < 3}; {2x+3, x \geq 3}

will get you the wrong answers

Since
The first equation is continuous at x =3, but has no derivative there.
The second equation has a derivative at x = 3, but is discontinuous there.

Many sources state that differentiability implies continuity, e.g.

http://www.zweigmedia.com/RealWorld/calctopic1/contanddiffb.html

But your second example seems to contradict this. The limit of difference quotients, by which the derivative is defined, is the same from both sides at x=3, yet the differentiable function is discontinuous there, having a limit of 6 from one side, and 9 from the other. I think the "proof", at the above site, that differentiability implies continuity fails in this case because it assumes that, if one-sided limits for the differentiable function exist, they will be the same from both sides. But I've also seen this theorem (differentiablility implies continuity) stated in respectable textbooks, so I'm a bit confused...

 

[Studiot]

Well spotted, Rasalhague.

That is what happens when one does things in a hurry. I tried to quickly construct examples to show the problems with the approximation formula.

Yes most authorities require continuity of the base functions f(x) for differentiability, though some allow an infinite result and some do not.
However these same authorities also state that if both the limits for the differential fraction from the left and from the right exist and are equal at some point then the function is differentiable at that point.

As an applied mathematician, I follow Oliver Heaviside's approach if it works and give the right answer then use it.

If you differentiate my function at x=3 and substitute into the formula to find f(4) you will obtain the correct answer.

I would certainly welcome more detailed discussion as there is more to this than meets the casual eye.

 

[Rasalhague]

I'm still puzzled as to how people who claim (i.e. work in a logical system where things are defined so that) "differentiability implies continuity" resolve this apparent contradiction.

On the one hand, we know that

f&#039;(x_0) = \lim_{x \rightarrow x_0} \frac{f(x)-f(x_0)}{x-x_0}

and

0 = \lim_{x \rightarrow x_0} x-x_0.

So, by the http://planetmath.org/encyclopedia/ProofOfLimitRuleOfProduct.html , we have

\lim_{x \rightarrow x_0} \left ( \frac{f(x)-f(x_0)}{x-x_0} \cdot (x-x_0) \right ) = f&#039;(x_0) \cdot 0

=\lim_{x \rightarrow x_0} \left ( f(x)-f(x_0) \right ).

But on the other hand, if f:R-->R is defined as in your second example, "if x < 3 then f(x) = 2x" and "if x >= 3 then f(x) = 2x +3", then

\lim_{h \rightarrow 0^-} \frac{2 \cdot (3+h)-2 \cdot 3)}{h} = \lim_{h \rightarrow 0^+} \frac{2 \cdot (3+h)+3 - (2 \cdot 3 + 3)}{h}

=\lim_{h \rightarrow 0} \frac{f(3+h)-f(3)}{h} = f&#039;(3) = 2,

even though f is not continuous at 3.

 

[qbert]

The example is flawed. Differentiability implies continuity, always.

you evaluated the limits for the difference quotient wrong.
f(3) = 2*3 + 3 = 9

take x=3+h
for h<0 we have
f(x) - f(3) = f(3+h) - 9 = 2*(3+h) - 9 = 2h-3
so
(f(x)-f(3))/(x-3) = (2h-3)/h which is undefined as h->0.

 

[Rasalhague]

Thanks, qbert! I understand now. I was forgetting that f(3) = 9 in as h --> 0 from below too (as well as when h --> 0 from above). So, the function in question is not, in fact, differentiable at 3.

 

[Studiot]

Thank you for the discussion and clarification, folks.

 

[lavinia]

Well spotted, Rasalhague.Yes most authorities require continuity of the base functions f(x) for differentiability, though some allow an infinite result and some do not.

A function that is differentiable must be continuous. However a continuous function may not be differentiable at a point or even anywhere in some extreme examples.

The idea that approximations work for small enough changes in an observable quantity implies continuity but even more, it implies the existence of an infinitesimal that we call the derivative.

If one assumes that the function is only continuous then small enough changes may never exist. The approximation dy = adx for some fixed constant may fail and arbitrarily small dx's may correspond to arbitrarily large dy's.

 

[vector22]

please please please you guys freak me out

There is no mystery to an infitesimal.! . First consider an increment of x say x = .001. Of course someone else could say that x = .0001 and so there would be an endless battle of who's value was the smallest. Mathematics introduced the notion of an infitesimal which put an end to the guessing game of how small a value could be. In essense the value of an infitesimal is undefined that is to say the value is not known but likes to hide out in the region:

0 > dx < 1

the value of an increment is always known such as (delta) x (sorry my latex is bad)
but the value of an infitesimal is never known.

 

[Hurkyl]

Infinitessimal has a precise definition -- usually it's something like:

x is infinitessimal if the size of x is smaller than 1/n for every positive integer n​

The description you have doesn't really fit with how mathematicians use the term "infinitessimal". Among other things, in the real number system, there is only one infintiessimal, and it's called "zero".

Other number systems have infinitessimal objects that really aren't all that mysterious. Other algebraic structures capture the notion too -- such as the notions of "differential form" and "tangent vector".