# Meaning of the Differential

1. Nov 16, 2004

### e(ho0n3

I started reading this ODE book and it first starts by introducing the concept of the differential of a function of one independent variable. Here is the definition:

Let y = f(x) define y as a function of x on an interval I. The differential of y, written as dy (or df) is defined by

$$(dy)(x,\Delta x) = f'(x)\Delta x$$

Then it goes on to say:

To distinguish between the function defined by y = x and the variable x, we place the symbol ^ over the x so that $y = \hat{x}$. If $y = \hat{x}$ then

$$(dy)(x,\Delta x) = (d\hat{x})(x,\Delta x) = \Delta x$$

since f'(x) = 1. The text generalizes further by restating the first equation as

$$(dy)(x,\Delta x) = f'(x) (d\hat{x})(x,\Delta x)$$

I'm still not clear why this substitution is made. It then goes on and states: The relation [the equation above] is the correct one, but in the course of time, it became customary to write [the equation above] in the more familiar form dy = f'(x) dx. So, as I understand it, $\Delta x = dx$!? I'm I missing something.

2. Nov 17, 2004

### HallsofIvy

In effect, yes. However, note that it really should be $\Delta x= d\hat{x}$ where $\hat{x}$ is the identity function. Since the derivative of the identity function is 1, it is certainly true that $d\hat{x}(x,\Delta x)= \Delta x$ and so we can write dx instead of $\Delta x$.

Notice the "abuse of terminology": we have replaced the function $d\hat{x}(x, \Delta x)$ with the symbol dx. dx is NOT "$\Delta x$" but it is a function such that $d\hat{x}(x, \Delta x)= \Delta x$. In practice we treat dx as if it were the "denominator" of the "fraction" dy/dx. Of course, dy/dx is NOT a fraction but since we can always "treat it like one" (to prove any property of a fraction, go back before the limit to the difference quotient which IS a fraction, use the property and take the limit again), it is useful to have a notation that lets us do that.

3. Nov 17, 2004

### e(ho0n3

So whenever I see dx, I should interpret it as $\Delta x$ or $d\hat{x}(x, \Delta x)$? Now you're telling me that dx is a function!? Why all this abuse and nonsense?

4. Nov 17, 2004

### chroot

Staff Emeritus
The reason for all the abuse is that dx is really something called a "one-form," (which is related to a vector) and is not really a number or variable or function at all. In many cases, however, it suffices to make some simplifying assumptions about dx and treat it as a number, or variable. Unfortunately, it will be a while before anyone at school will teach you the full meaning of it.

- Warren

5. Nov 17, 2004

### e(ho0n3

Well for one thing, I'm not in school (not yet at least). I just want to know if I can forget about all that crap I showed in my first post and just blindly assume that dy = f'(x) dx. And what is the difference then between the differential and a one-form or are they the same thing? Does the definition of the differential I gave in my first post satisfy the requisites needed to be a one-form? Maybe I should just skip all this ODE business and jump strainght into differential geometry. What do you think?

6. Nov 19, 2004

### HallsofIvy

For practical use, just think of dx as meaning "a very slight change in x" and dy as "a very slight change in y". That's what engineers and physicists do. It's just mathematicians who have to be very precise and distinguish between "numbers", and "functions", and "functionals".