# Rigorous treament of differentials?

1. Nov 26, 2007

### pivoxa15

In applied subjects, the differential is often treated as i.e C'(x) approximately equals C(x+1)-C(x)

1 is used instead of h as h->0 because we are talking about discrete units such as items or people. They argue it works because x>>1. i.e considering lots of items, x. However what is rigorous treatment of this argument?

2. Nov 26, 2007

### Hurkyl

Staff Emeritus
C(x+a) = C(x) + a C'(x) + error term.

This is differential approximation, or if you prefer, a Taylor series.

Last edited: Nov 26, 2007
3. Nov 26, 2007

### Xevarion

If you want a rigorous treatment of differentials, try Abraham Robinson's Nonstandard Analysis. To some extent, that's what you're looking for. Maybe not exactly...

4. Nov 26, 2007

### Hurkyl

Staff Emeritus
1 is a finite number, so the ideas of NSA would only be relevant to his situation if he is working with a transfinite quantity of items.

5. Nov 26, 2007

### Xevarion

Right but my impression is that the OP was asking if there was a rigorous theory that is based on the sort of approximations often found in applied work. Obviously as it's stated there's not much to it; basically it's a particular case of "the error in an approximation is small enough that we can ignore it" which happens pretty much everywhere in analysis...

6. Nov 27, 2007

### HallsofIvy

Staff Emeritus
As far as I know, the differential is NEVER "treated as C'(x) approximately equals C(x+1)- C(x)". I'm sure you must have misunderstood. C(x+1)- C(x) would be the "difference quotient (f(x+h)-f(x))/h with h= 1. Since C'(x) is the limit of the difference quotient as h goes to 0. Since 1 is not anywhere near 0, C(x+1)- C(x) is, generally, nowhere near C'(x)!
(C(x+1)- C(x) is often used as the "difference quotient" in discrete functions. That has nothing at all to do with C'(x).)

Okay, I should have read further. You ARE talking about "discrete functions" and thus about "the calculus of finite differences" which is very different from the regular calculus. There is no "rigorous treatment" for using C(x+1)- C(x) as "C'(x)" because they are not at all the same. In fact, in the situations you are talking about, with "discrete units", C'(x) cannot even be defined. The "calculus of finite differences" has a lot of solid rigorous foundation but it is not "calculus"- it's a completely separate subject.

7. Nov 28, 2007

### WWGD

Back to the continuous case of differentials, I always wondered

what allowed us to go from:

[Tex]
\frac{dy}{dx}=y
[/Tex]

to:

[Tex]
\frac{dy}{y}=dx
[/Tex]

and then just integrate both sides.

It does not seem clear--even after doing a good amount of reading--
what the "algebra" of these differentials is. I can see why it does not
make sense to go from (as dx->0. Sorry, I am still learning Tex):

[Tex]
\frac {dy}{dx}=f'(x)
[/Tex]

to:

[Tex]
\frac{dx}{dy}=\frac{1}{f'(x)}
[/Tex]

But I don't see the justification for the first manipulation clearly.

Anyone know?.
Thanks.

What Would Gauss Do?

8. Nov 28, 2007

### WWGD

Sorry about that. I thought I had figured it out. I'll go back to the
practice pen.

9. Nov 28, 2007

### pivoxa15

I don't think that is it.

They are talking about instead of change in x ->0 in the limit as in calculus, we can only let change in x = 1 as we are dealing wth discrete units like money.

So d(C(x))/dx = lim dx->0 (C(x+dx)-C(x))/(dx)

But if C(x) is a cost function then dx=1 as a min

But given C(x)=300+3x

we can still differentiate C(x) like it's a function in calculus. However what is the argument for the legimaticy of it?

10. Nov 29, 2007

### WWGD

.

If I understand well, it is because C(x) is linear. C'(x) , if you treat it as being
discrete, gives you the slope of the tangent line at x ( I am being loose in
here), or the best local approximation to the change of the function, i.e
the differential. So, if C(x) is linear, the best local linear approximation to
the change of the function is the function itself.

Still, an additional assumption you are doing when finding C'(x), is that
C(x) is defined outside of the discrete points x1,x2,...,xn , and, that
near these values, C(x) is also 300+3x,i.e, as you said,in order to talk about
dx for dx<1 (or, actually, as dx->0), you need to (and seem to) be assuming
that C(x) acts infinitesimally in the same way as it does discretely.

11. Nov 29, 2007

### pivoxa15

You could have a C(x)=x^2+2 hence nonlinear.

12. Nov 30, 2007

### WWGD

Please give me some time, until after finals, and I will get back to it.