Mathematical obscuration?

[Boris Lagutin]

Thanks a lot, Joe. Differential in Integral comes from the conception of Limit of Riemann's sum when delta(something) goes to zero. The limit of Riemann's sum is based on the general conception of Limit when delta(something) goes to zero. Does this limit conception contain some tool to increase or decrease a size of delta(something)? I am sorry but I don't remember such a tool.

Have a nice day.

[Boris Lagutin]

Based on my previous posts and my current knowledge of Calculus (possibly I don't know something special about infinitely small values applied in Limits), my suggestion is that Jacobian cannot be applied to differentials, however, it can be applied to a whole area of a figure (Integral). Maybe I mistake, of course. I could try to prove it mathematically but don't have a lot of time in order to do it. I study Calculus, Physics and Programming at the same time which is a good burden for me.

Have a nice day.

[Boris Lagutin]

In my opinion, there is an interesting collision between Algebra and Calculus. Let's take a definition of a derivative: a limit (as h goes to 0) of [f(x+h) - f(x)]/h and take f(x)=x^2. We have: limit_{h -> 0} of [(x+h)^2 - x^2]/h. As a ending result, we get: limit_{h -> 0} of (2x+h). Here is the point where we remove h because h -> 0 based on the conception of limit, however, (h -> 0) does not mean that h is zero. Therefore, the following equation should be FALSE algebraically if h does not equal zero: "a limit as h goes to zero of (2x + h) is equal to (2x)" . Do I mistake?
By the way, h (delta) is not considered to be zero in the limit of Riemann's sum.

P.S. Algebraically we cannot state that a sum of some number and something, that is not equal to zero, is equal to just some number!

Thank you.

[a.s.h.]

In my opinion, there is an interesting collision between Algebra and Calculus. Let's take a definition of a derivative: a limit (as h goes to 0) of [f(x+h) - f(x)]/h and take f(x)=x^2. We have: limit_{h -> 0} of [(x+h)^2 - x^2]/h. As a ending result, we get: limit_{h -> 0} of (2x+h). Here is the point where we remove h because h -> 0 based on the conception of limit, however, (h -> 0) does not mean that h is zero. Therefore, the following equation should be FALSE algebraically if h does not equal zero: "a limit as h goes to zero of (2x + h) is equal to (2x)" . Do I mistake?
By the way, h (delta) is not considered to be zero in the limit of Riemann's sum.

P.S. Algebraically we cannot state that a sum of some number and something, that is not equal to zero, is equal to just some number!

Thank you.

It is correct to say that lim h->0 (2x+h) = 2x.

[Boris Lagutin]

Ok, then, please, prove mathematically and/or geometrically that it is correct to say so.

Thank you.

[joe]

ASH is correct to say "It is correct to say that lim h->0 (2x+h) = 2x."

Boris then writes "Ok, then, please, prove mathematically and/or geometrically that it is correct to say so. "

It follows directly from the definition of a limit. See for example
https://en.wikipedia.org/wiki/Limit_(mathematics)
and there are quite a few other introductions to limits on the web.

[Boris Lagutin]

Ok, Joe. Then, how to explain the difference between the limit of Riemann's sum where delta x goes to zero but delta x is not treated as zero and the limit as the definition of the derivative where h (delta x) goes to zero but h is treated as zero? Conceptually, delta x and h is infinitely small values (no difference between them). Moreover, they come from the same limit conception. However, the same conceptions of delta x and h are used differently in these two limits. How so?

Thank you.

[joe]

"delta x goes to zero but delta x is not treated as zero"

In the definition of limit, delta x is a small quantity.

[Boris Lagutin]

I agree that delta x is an infinitely small value as well as h is an infinitely small value. Both h and delta x are not zero. However, they are used differently despite they are the same values conceptually. If we put the definition of the derivative in the limit of Riemann's sum then we get some kind of paradoxical situation (please, see attachment).

Thank you.

[joe]

h and delta x are both very small quantities, as required in the definition of limits.

[a.s.h.]

I agree that delta x is an infinitely small value as well as h is an infinitely small value. Both h and delta x are not zero. However, they are used differently despite they are the same values conceptually. If we put the definition of the derivative in the limit of Riemann's sum then we get some kind of paradoxical situation (please, see attachment).

In that sum, you have an iterated limit, where first the h is sent to 0 (independently of Delta x), and then after that, Delta x is sent to 0 (equivalently, n is sent to infinity).

Since as Delta x goes to 0 the number of terms we are summing up (n) goes to infinity, we can't just say the whole thing becomes 0. Yes, Delta x is getting very small (so each term is getting very small), but the number of terms we are adding up is getting very large. These two competing behaviours lead to the value of the integral in the end, and make it plausible that the result need not be 0 (and in fact isn't 0 here).

[Boris Lagutin]

Yes, a.s.h. I thought of the sum of infinitely many infinitesimal quantities.
Note that we could claim that delat x = h (infinitely small values defined by the limit conception). Moreover, you cannot quantitatively discern h from delta x based on the limit conception. Ok, you remove the terms 3*x*h and h^2, but then you must multiply x^2 by delta x (or h). If follow the conception of the limit for the derivative, we should calculate like this: x^2 * (delta x) = 0. Ok, we know that there are infinitely many such multiplications.
a.s.h., you state that delta x is not treated like zero because the infinitely many infinitesimal quantities add up to something bigger if I understand you right. In other words, you want to say that if the limit for derivative was hypothetically used infinitely many times to add up to some bigger number, then we could NOT treat (ignore) the terms 3*x*h and h^2 as zeros?! Do I understand you right?

Thank you.

[Boris Lagutin]

To sum up, I ask you, a.s.h., two questions - one from my previous post and a new one:

1) a.s.h., you state that delta x is not treated like zero because the infinitely many infinitesimal quantities add up to something bigger if I understand you right. In other words, you want to say that if the limit for derivative was hypothetically used infinitely many times to add up to some bigger number, then we could NOT treat (ignore) the terms 3*x*h and h^2 as zeros?! Do I understand you right?

2) What is a difference between two limits on the picture?

Thank you.

[Boris Lagutin]

Hello,

I have just corrected my calculations on the picture in my previous post. Also I am preparing a complete proof with some more calculations of this paradox found by me.

Have a nice day

[joe]

If you are preparing a formal argument, please make sure that you use the formal definition of limits. See for example
https://en.wikipedia.org/wiki/Limit_(mathematics)
and there are quite a few other introductions to limits on the web.
Please do not use the terms 'infinitesimal' or 'infinite', because these are only shorthand for longer formal statements.
For instance, '1/x goes to infinity as x becomes infinitesimally close to zero' is shorthand for
'1/x can be larger than any given large number if we choose a sufficiently small x'.

Some of the posts above demonstrate that using the words 'infinitesimal' or 'infinite' without defining them rather than using the simple, formal mathematical statements can lead to confusion.