The definition of the partial derivative of [math]f[/math] with respect to [math]x[/math] is

 

[math]\frac{\partial f}{\partial x} = \frac{f(x+\mathrm{d}x,y,z)-f(x,y,z)}{\mathrm{d}x}[/math]

 

right?

 

While [math]\mathrm{d}x[/math] has a meaning, an infinitesimal change in [math]x[/math], what does [math]\partial x[/math] mean?

In the formal definition you need to take the limit as [math]dx \rightarrow 0[/math].

 

Thinking of [math]dx[/math] as an infinitesimal change is ok informally, or you could start thinking of non-standard analysis in which [math]dx[/math] is an infinitesimal. (I quite like that idea and it would get you a good grip on how to do calculus very quickly).

 

I think you could try to interpret the partial in the same way. Personally, I would think of it all much more formally.

The definition of the partial derivative of [math]f[/math] with respect to [math]x[/math] is

 

[math]\frac{\partial f}{\partial x} = \frac{f(x+\mathrm{d}x,y,z)-f(x,y,z)}{\mathrm{d}x}[/math]

 

right?

 

While [math]\mathrm{d}x[/math] has a meaning, an infinitesimal change in [math]x[/math], what does [math]\partial x[/math] mean?

 

You can think of it the same way, except slightly different. dx and [math]\partial x[/math] both represent infintesimal change with respect to x, however, the difference lies in how they relate to the other variables. dx treats them like other functions(see implicit differentiation) whereas [math]\partial x[/math] treats them like constants.

You can think of it the same way, except slightly different. dx and [math]\partial x[/math] both represent infintesimal change with respect to x, however, the difference lies in how they relate to the other variables. dx treats them like other functions(see implicit differentiation) whereas [math]\partial x[/math] treats them like constants.

 

A good explanation!

 

So, [math]\frac{\mathrm{d} y}{\mathrm{d}x}[/math] might produce something that relates [math]y[/math] and [math]x[/math], while [math]\frac{\mathrm{d} y}{\partial x} = y[/math]?

[math]\frac{\mathrm{d} y}{\mathrm{d}x}[/math] means "How does the output of y (a function y(x)) change when x (its only parameter) changes?"

 

[math]\frac{\partial y}{\partial x}[/math] means "How does the output of y (a function y(x,....)) change when x (one of its parameters) changes and all of its other parameters stay the same?"

 

As far as I know, [math]\frac{\mathrm{d} y}{\partial x}[/math] means nothing.

Have a look at the notion of a total derivative, that may help.

 

What you can write is

 

[math]x \mapsto y[/math]

which you can think of as [math]y(x)[/math]

then

[math]dy(x) = dx \frac{\partial y(x)}{\partial x}[/math]

 

which is the closest thing I can think of to your expression. I informally think of as the [math]dx[/math] and [math]\partial x[/math] cancelling.

Edited October 16, 200916 yr by ajb

I was going to put this in another topic, but since you brought it up I'll ask here. I know, or have heard, that multiplying equations by infinitesimals is frowned upon. Can you actually do what you did above? If so, why is it frowned upon? If you can't, why?

I was going to put this in another topic, but since you brought it up I'll ask here. I know, or have heard, that multiplying equations by infinitesimals is frowned upon. Can you actually do what you did above? If so, why is it frowned upon? If you can't, why?

 

Because it's not a general rule, you can only treat [math]\frac{dy}{dx}[/math], as a fraction in particular cases, I think mainly restricted to ODE's.

 

I didn't think you could do it with partial differential equations, as ajb has shown, but I'm sure he knows better than I do.

I am thinking of [math]dx[/math] as being very formal. Which I think maybe the best way to view it.

 

I think that multiplying equations by infinitesimals is, as you put it "frowned upon" is because infinitesimals are not numbers.

 

You can extend the real line to include them.

 

The way to do this is to extend the real line to include a formal object [math]dx[/math]. The rules are simple.

 

1)[math]dx dx=0[/math]

2) under changes of coordinate [math]x \mapsto y[/math] we have [math]dy(x) = dx \frac{\partial y(x)}{\partial x}[/math].

 

The point being that I attach no meaning to the [math]dx[/math], it is only formal.

 

The resulting "space" is the superspace [math]R^{1|1}[/math]. Differential forms are functions on this space.

As far as I know, [math]\frac{\mathrm{d} y}{\partial x}[/math] means nothing.

I sure does. Given [math]f(x,y(x),z(x))[/math], the total derivative of f with respect to x is

 

[math]\frac {df}{dx} =

\frac{\partial f}{\partial x} +

\frac{\partial f}{\partial y}\frac {dy}{dx} +

\frac{\partial f}{\partial z}\frac {dz}{dx}[/math]

I sure does. Given [math]f(x,y(x),z(x))[/math], the total derivative of f with respect to x is

 

[math]\frac {df}{dx} =

\frac{\partial f}{\partial x} +

\frac{\partial f}{\partial y}\frac {dy}{dx} +

\frac{\partial f}{\partial z}\frac {dz}{dx}[/math]

 

 

But by itself (ie. not in the context of total derivatives) does it mean anything?

 

ajb, I'm sorry, but I'm not on a level high enough to make much sense of what you wrote. However I'm eager to know the answer. Do you think you could PM me (I don't want to lead this thread off topic) a dumbed down version of what you wrote, or the same thing with quick explanations/references to pretty much everything above high school level?

 

ajb, I'm sorry, but I'm not on a level high enough to make much sense of what you wrote. However I'm eager to know the answer. Do you think you could PM me (I don't want to lead this thread off topic) a dumbed down version of what you wrote, or the same thing with quick explanations/references to pretty much everything above high school level?

 

Ok, all I am really saying is that if you think hard about it you will end up with problems trying to think of [math]dx[/math] as an infinitesimal change. It can be much more useful to treat it formally, without direct reference to [math]x[/math].

 

However, intuitively you would like to think of [math]dx[/math] as a small change.

Formal as in treat it as I would a variable (ie.: x, y, a, b...)? I'm unfamiliar with the meaning of "formal" in this context.

Perhaps [math]\partial x = \mathrm{d}x[/math] and the soft 'd' is only to emphasize that we are talking about a partial derivative?

Yup, I'm almost certain that's correct. But if it was meant as an answer to my question, that's not what I asked

Yup, I'm almost certain that's correct. But if it was meant as an answer to my question, that's not what I asked

 

It was not meant as an answer. I am not sure what formal entails. My guess would be; the formal way of saying 'rigid'

Edited October 19, 200916 yr by hobz

Formal (in this context) just means "without meaning or interpretation". You know how to handle them, but you don't worry too much what it means.

I see. Thanks ajb