A Philosophical Calculus Discussion

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Discus: SAT/ACT Tests and Test Preparation: March 2004 Archive: A Philosophical Calculus Discussion
 By Miscanon (Miscanon) on Friday, March 26, 2004 - 11:10 pm: Edit

Okay..all you math buffs..we all know it doesn't take a genius to score an 5 on the AP Calc exams.

Lets go beyond the intro course...and discuss something that I've been pondering the past few days...in the indefinite integral...what is the actual meaning of "dx" when we use separable differential equations..think about it..and don't respond with "it's telling us to integrate with respect to" [extend that]...because while that's true and accurate..i argue the "dx" is simply this -- notational bookeeping..and in which case...it is easy-to-teach but techincally 'misleading' to approach separable differential equations the way it is taught:

e.g.

1) separate dx's to one side, dy's to the other...this is okay with regards to meanings of the differentials
2) "integrate" both sides -- this is the convenience we have invented..but not physically meaningful [it's not a riemann sum, remember its the indefinite integral; we are not summing an infinite amount of infinitismally small components...so when we say "integrate both sides" -- while it is true that we are finding the function F(x[or y]) when differentiated will yield the integrand...it is wrong to interpret this in the same light as a definite integral, which has physical significance with these differentials]

...any thoughts?

 By Titanz05 (Titanz05) on Friday, March 26, 2004 - 11:35 pm: Edit

philosophically...

it doesnt matter.
it doesnt exist.
you dont exist.
nothing exist.
the end.

 By Ubercollegeman (Ubercollegeman) on Saturday, March 27, 2004 - 12:16 am: Edit

First, I'd like to say, wow, welcome to the club where people actually THINK about calculus rather than just look at formulas and apply them. You're a rarity .

I've actually done a lot of thinking about this independently, and I've come up with this explanation.

No, the dx is integral (haha, bad joke) to making calculus integration make sense. Do more calculus or see calculus applied to physics, and it is easy to see why..say.. (Integral)7 or (Integral)x don't make sense.

THIS IS BECAUSE THE DEFINITE INTEGRAL IS MEANT TO BE THE SUM OF AN INFINITE NUMBER OF THINGS THAT ARE INFINITELY SMALL.

The extra "dx" is actually multiplied to whatever is in front of it, making the thing INFINITELY small. Therefore, something like

dy = xdx

Both sides are infinitely small, but *how* infinitely small each is depends on whatever x-coordinate you are talking about.

dx = 7

That equation makes absolutely no sense whatsoever.

You cannot compare something that is inifintely small to a constant value. The truth is, the concept of a differential-something that is infinitely small-is a very strange one. It is not quite a variable, but it definitely isn't a constant. Think about this a little more and it will start to make sense. I cannot stand it when one of my friends asks me a calculus question and there is no dx or dy or whatever in the integral . Omitting that dx shows that you have minimal understanding of the theory behind integral calculus.

About your thoughts between indefinite vs definite integrals:

You must think of the indefinite integral as more of an anti-derivative. The definite integral is based upon the anti-derivative, and it is folly to just say that the "indefinite integral is just a definite integral without the bounds." That totally oversimplifies something very complicated. The definite integral is a method of summing an infinite number of infinitely small things, and taking anti-derivatives are simply parts of the process.

Hope that helps.

 By Anduin (Anduin) on Saturday, March 27, 2004 - 02:07 am: Edit

The following is a description of what it means to find the value of an integral from an analysis standpoint. I realize that a description of integration is far clearer when it includes pictures and mathematical symbols, but my tools are a bit limited here.

Definite integration describes the accumulation of a function over an interval. The accumulation of any integrable function can be approximated by the accumulation of similar functions that are either greater or less than the original function. These upper sums and lower sums are constructed by partitioning the interval, and taking the maximum value of the function across each subinterval for the value of the upper sum across the subinterval, and likewise with the minimum function value for the lower sum.

If you make these subintervals smaller and smaller, you can get the upper and lower sums closer and closer together, so that for any number that you pick, you can find upper and lower sums whose accumulations across the whole interval are less than that number. The least upper sum and the greatest lower sum both equal the integral of the function over the interval.

dx is the length of the subinterval that guarantees that you can get upper and lower sums whose values are as close together as you want.

Indefinite integration is the extension of the same principle to any and all intervals of the real line. An indefinite integral is just a function whose values equal the accumulation of another function from the origin to any other point on the real line. Again, dx is the length that guarantees that you can partition a general interval of the real line into subintervals that are so small that the upper and lower sums are as close together as you want. Thus, the accumulations of upper and lower sums over a given partition can be extended...indefinitely.

 By Miscanon (Miscanon) on Saturday, March 27, 2004 - 02:09 am: Edit

ubercollegeman..yes the Riemann Sum idea (delta(x) --> dx as n-->+infinity, or, more generally, delta(x)*delta(y)*delta(z)*.... --> dx*dy*dz*.... as number of n-dimensional 'boxes' --> infinity) is very clear to me..i am not arguing the logic of this...(i've done multiple calc-based physics courses where it is used quite a bit )and yes, i agree with your thoughts there.

regarding the indefinite vs. definite..it is most definitely clear that they are different based upon one being a reimann sum and the other being an "antiderivative" (we can think of the elongated S as an operator). My comment can perhaps be better captured by specifically looking at separable (simple) differential equations. When the two are separated, e.g.:

We have dy / dx = f'(x) , or in differential format, dy = f'(x) * dx

This is valid... the "propogation" in the differential dy is indeed (physically and intuitively) = f'(x) * dx , but HOWEVER,

when we "integrate" both sides...i argue this step is simply valid because of a convenience of notational "book-keeping"..and NOT physically significant (as it is with the Riemann Sum [definite integral]). In example:

dy / dx = f(x) / f(y) , f(y) * dy = f(x) * dx,

integral (f(y) * dy) = integral (f(x) * dx)
^
|
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this step: when we "integrate" the f(y) * dy we are NOT summing an infinite number of differential components (Riemann Sums) in an INdefinite integral but we are simply doing something that has conveniently turned out to prove true by mathematical logic, that is, it is simply finding the antiderivative but should not be interpreted that the (dy or dx) are physically significant -- they simply tell us which variable we must integrate with respect to.

 By Bahaa (Bahaa) on Saturday, March 27, 2004 - 12:05 pm: Edit

you can think of dx as the thickness of a straight line, technically, I believe (dx)^2 is approximated by 0. it's more complicated and I really need an actual paper because I hate math notation using ascii, so I guess , I can only recommend reading one of Serge Lang's books in calculus, he got all the details!

 By Miscanon (Miscanon) on Saturday, March 27, 2004 - 03:07 pm: Edit

right dx = the "thickness of a straight line" -- > it limits 0 ; and therefore lim((f(x))^2) = [lim(f(x)) ]^2 = 0 as well...etc.

but hmm...very pondersome topic, this is

 By Miscanon (Miscanon) on Saturday, March 27, 2004 - 08:28 pm: Edit

thanks for responses..any further thoughts, anyone?