# Does dx have mulitple personalites?

1. Dec 13, 2003

### Johnny

Can someone, who really knows and understands, tell me what dx (or whatever variables given) means behind the integral sign? I have seen more disagreement in Calculus books concerning this. Some authors say it's there just to "indicate the variable" your integrating with respect to, i.e., its not formally required. I had a mathematics professor tell me otherwise, namely that dx is a "real" variable and is required, multiplying at every point with f(x), defining a virtual infinite number of Reimannian rectangles as in S f(x)dx, (S means sum) and performing the normal integration.

Also, (and this ties in) if I have for example, a simple separable ODE such as:

m dv/dt = mg -kv

(where g is accelleration due to gravity, m is mass, v is velocity and k is frictional constant), and if we assume dv/dt is the standard Leibniz operator notation for a derivative, then how can one simply multiply dt through, when it's actually part of an operator? Now I have read that by "appropriately" defining dt, then defining dv as dv = v' dt, we get of course v' = dv/dt. Now, having defined dv/dt as a ratio of differentials, using the same Leibniz notation, we should NOW be able to use basic algebraic techniques with little worry. But wait, if we now integrate both sides:

S (1/(g - cv))dv = S dt where S means sum, and c = k/m,

I still don't know how to interpret S dt? Or now S (1/g-cv)dv for that matter? Do I sound confused? I am! Any help on this is greatly appreciated. I need to understand this.

Last edited: Dec 14, 2003
2. Dec 14, 2003

### Hurkyl

Staff Emeritus
Yes

$dx$ does indeed have "multiple personalities".

The intuition is that $dx$ is "supposed" to be an "infinitessimally small quantity". The notation for many concepts in analysis was specifically chosen so that it looked like you really were manipulating infinitessimally small quantities.

Am I presuming correctly that when you write

You mean

$$\frac{1}{g - cv}$$

? If so, you wrote it wrong; you need to wrap the denominator in parentheses, such as: (1 / (g - cv))

Anyways:

$$\int \frac{dv}{g - cv}$$

is certainly in the form you mentioned earlier. If we write $f(z) = 1 / (g - cz)$, then we have:

$$\int \frac{dv}{g - cv} = \int f(v) \, dv$$

so you can use your favorite intuitive interpretation of $\int f(x)\,dx$ to interpret this integral.

3. Dec 14, 2003

### chroot

Staff Emeritus
The entity $dx$ is really something called a "1-form." You can feed a vector field $v$ to a 1-form $\omega$, and it spits out a real-valued function $\omega(v)$.

In more precise terms, a 1-form on a manifold M is a map from $\text{Vect}(M)$ to $C^\infty(M)$ that is linear over $C^\infty(M)$. $\text{Vect}(M)$ is the set of all vector fields that can be defined on M, and $C^\infty(M)$ is the set of all smooth (real-valued) functions on M.

So, really, the differential $dx$ has only one real personality -- that of a 1-form. Much of this formality is not needed in basic calculus classes, however, and this 1-form can be somewhat abused so that it appears to take on other roles, particularly in flat spaces.

- Warren

4. Dec 14, 2003

### lethe

in addition to being a 1-form, i was going to mention that dx is also a measure, which is a totally different kind of gadget.

i would say, yeah, it definitely has multiple personalities.

5. Dec 14, 2003

### chroot

Staff Emeritus
Yeah, I forgot about that.. hehe... suppose you're right, it's an entity that can function as any one of a handful of mathematical devices.

- Warren

6. Dec 14, 2003

### Johnny

Re: Yes

Sorry about the parentheses; I corrected it. Actually, I want to know what dx represents behind the integral sign.

Why is it needed, other than to indicate the variable your integrating with respect to? In my calculus book, the first time dx shows up behind the integral, is in the definition of an integral as the limit of a Reimann sum. But my author indicates that dx in that context is not to be confused with dx as it applies to a derivative. I find that hard to believe. But because I'm not a mathematician, for now, I'm going to assume I don't understand. He mentions it's there only for convenience, and useful for such purposes as u du substitution. Once I understand this, I think manipulation of dx in a DE will make sense. In the ODE example, dx is part of a derivative one second, then (by simple multiplcation) behind the integral in the next. Does dx mean the same thing in both cases? I know it's required for a derivative, but what about the integral? What do you think?

7. Dec 14, 2003

### Hurkyl

Staff Emeritus
Does this manipulation make it easier to digest?

$$m v'(t) = mg - kv(t)$$
$$\frac{1}{g - c v(t)} v'(t) = 1$$
$$\int \frac{1}{g - c v(t)} v'(t) \, dt = \int 1 \, dt$$

By substitution, we know

$$\int \frac{1}{g - c u} \, du = \int \frac{1}{g - c v(t)} v'(t) \, dt$$

Which let's us finish off the problem.

Last edited: Dec 14, 2003