Understanding differentials in Calculus 1?

[lurflurf]

Do not confuse detail with rigor. An approximation can be rigorous. Often when we wish to establish some fact C and we observe that A->B->C. It may happen that A is difficult to establish and may be false, while B is easy to establish. It is not more rigorous to prove A, proving A or B proves C. In fact proving B first may make it easier to prove or disprove A, or maybe A is unimportant.

Mathematics in general and calculus in particular often involves approximations. For example
$$\lim_{x \rightarrow 0}\frac{\sin (x)}{x}=1$$
is a fancy way to say sin(x)/x is approximately 1.
If we accept dy and dx together in dy/dx it is silly to reject them when they appear apart.
Saying dy/dx is "more rigorous" than dx is like saying 3/4 is "more rigorous" than 4. In fact dy/dx is less general as it assumes division has been defined, is possible, dx is not 0, and our interest is in the ratio. Often these conditions are not met.
We can use dx and dy when we move to vectors while dy/dx requires strange trickery to work.

Books often include warning about avoiding extremely stupid yet common errors. Introduction to Calculus and Analysis, Volume 1 by Richard Courant and Fritz John warns "We emphasize that this [the differential] has nothing to do with the vague concept of 'infinitely small quantities.'"

 

[Mandelbroth]

Mathematics in general and calculus in particular often involves approximations. For example
$$\lim_{x \rightarrow 0}\frac{\sin (x)}{x}=1$$
is a fancy way to say sin(x)/x is approximately 1.

Your words. They burn my eyes.

Limits are NOT approximations. You've been in the wilderness of No-Rigor Land for far too long if you think that.

If we accept dy and dx together in dy/dx it is silly to reject them when they appear apart.
Saying dy/dx is "more rigorous" than dx is like saying 3/4 is "more rigorous" than 4. In fact dy/dx is less general as it assumes division has been defined, is possible, dx is not 0, and our interest is in the ratio. Often these conditions are not met.
We can use dx and dy when we move to vectors while dy/dx requires strange trickery to work.

I accept $dy$ and $dx$. They are the exterior derivatives of x and y, respectively. They are covectors. If you wish to define $\frac{dy}{dx}$ as a ratio, this is possible for a function $y:\mathbb{R}\rightarrow\mathbb{R}$ because then $dy$ and $dx$ are necessarily scalar multiples of each other. However, a few people (micromass, for example) do not like defining division of vectors and covectors that are scalar multiples of each other. It is rigorous, per say, but it is not necessarily preferred. Further, it does not generalize very well.

 

[Stephen Tashi]

Hmm...32 posts so far. This is what usually happens in discussions about dy and dx. Someone needs to lay down the law from one of those calculus books where the explanation is perfectly clear and logical.

 

[lurflurf]

Rigor should not be confused with pedantry.

Limits are NOT approximations.

So |f(x)-L|=0 does it?
|f(x)-L|<ε looks approximate to me.
I know you will say limit is a mapping and therefore exact.
That is technically true, but the mapping is defined to provide approximations.
I can define a function approx(x) such that
approx(pi)=22/7. That is an exact equation, but it provides an approximation.
In other words limits are both approximations and mappings, but we only care about the latter because of the former.
As far as exterior derivatives, as I mentioned above they are very handy, but the properties d^2x=0 and dx^0=0 are not desired for all applications. They also require some motivation.

For calculus I several reasonable definitions of differentials are possible.
(one)
(x+dx,y+dy) is a point on a line tangent at (x,y) to a given function.
(two)
if y=f(x)
dy=f'(x) dx
(three)
if y=f(x)
dy is the best linear approximation to Δy=f(x+dx)-f(x)
by this we mean that the ratio between |dy-Δy| and |dx| can by made as small as desired

 

[Mandelbroth]

Rigor should not be confused with pedantry.

But they aren't exactly mutually exclusive.

So |f(x)-L|=0 does it?
|f(x)-L|<ε looks approximate to me.
I know you will say limit is a mapping and therefore exact.
That is technically true, but the mapping is defined to provide approximations.
I can define a function approx(x) such that
approx(pi)=22/7. That is an exact equation, but it provides an approximation.
In other words limits are both approximations and mappings, but we only care about the latter because of the former.

No. Forget what you're thinking. Observe:
$$\lim_{x\rightarrow\alpha}f(x)=\mathfrak{L}\iff\forall \varepsilon>0 \, \exists \delta>0:0<|x-\alpha|<\delta \implies |f(x)-\mathfrak{L}|<\varepsilon.$$
This is the symbolic definition of the (real) limit. Isn't it pretty?

From this, we can make $f(x)$ arbitrarily close to $\mathfrak{L}$ (if the limit exists), but there is no guarantee that $f(\alpha)=\mathfrak{L}$. It is not an approximation.

 

[jasonRF]

Backing up a bit, it is a fact of life that the intro calculus sequence has multiple roles - one of which is to serve as a service course for engineers, scientists and others. Differentials are a very useful tool that those hundreds of engineers and scientists need to know, and the typical Calc I justification has done fine by me for many years now. I'm glad they don't skip them and wait until that-course-I-would-never-have-time-to-take in order to teach them with more rigor.

jason