Discussion of Integral calculus

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Boris Lagutin

Re: Discussion of Integral calculus

Post by Boris Lagutin »

Dear a.s.h.,
a.s.h. wrote:If the difference between the x-values was 0, we'd get 0/0 for the difference quotient, which is indeterminate. So we instead look at a limit as the difference between the x-values (h) tends to 0.


Agree with your last post.

Thank you.
Boris Lagutin

Discussion of Integral calculus

Post by Boris Lagutin »

Dear a.s.h. and Joe,

I think that some wrong impression that I do not understand your understanding of the limits could appear. Of course, I know all these conceptions more or less well. Here, I offer somehow another view and try to prove it.

It would be interesting to know which exactly understanding of the limits the originators of Calculus had many many years ago and how it was developing historically.

Thank you.
a.s.h.
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Re: Discussion of Integral calculus?

Post by a.s.h. »

Boris Lagutin wrote:Dear a.s.h. and Joe,

I think that some wrong impression that I do not understand your understanding of the limits could appear. Of course, I know all these conceptions more or less well. Here, I offer somehow another view and try to prove it.

It would be interesting to know which exactly understanding of the limits the originators of Calculus had many many years ago and how it was developing historically.

Thank you.
There is probably a lot of information available online or in books about the history of Calculus. Here's the relevant Wikipedia page if you want to get started: https://en.wikipedia.org/wiki/History_of_calculus .

The standard definition of limits is the 'epsilon-delta' definition, the Wikipedia page for which is linked in a previous page of this thread.
Boris Lagutin

Re: Discussion of Integral calculus

Post by Boris Lagutin »

Thank you very much, a.s.h. Some good book(s) about the history of Calculus would be excellent. However, books are different. Choosing a book requires some effort. I like when details are described - details that might be unseen at the first look. History is full of details some of which are sometimes omitted by accident or because they are considered to be not meaningful, in my opinion. When I get some time and opportunity, I will probably search such a book or books.

Have a nice day.
Boris Lagutin

Re: Discussion of Integral calculus

Post by Boris Lagutin »

Hello,

I guess this website administrator has changed my original title of this thread. Ok. I don't object. Possibly, it is a right move.

Have a nice weekend.
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joe
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Re: Discussion of Integral calculus

Post by joe »

Yes, I changed the title of the thread. I don't think we have uncovered a paradox.
Joe
Last edited by joe on Sun Oct 16, 2016 6:23 am, edited 1 time in total.
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Boris Lagutin

Re: Discussion of Integral calculus

Post by Boris Lagutin »

Hello Joe,

I don't leave efforts yet. Possibly, the arguments and conclusion will be somehow changed. By now, there are two key points left - the violation of the continuity of the derivative functions and the undetermined limit (0/0) contradicting the algebraic and geometrical conception of the function slope (quotient and tangent), in my opinion. Although, I agree with your decision to change the title. I suppose, my title was pompous :oops: Your title is better.

Thank you.
Boris Lagutin

Re: Discussion of Integral calculus

Post by Boris Lagutin »

Hello,

Let me continue with my proof. If I understand all the participants of this discussion right, all agree that h (distance between two x-values of the original function) in the definition of the derivative (limit) cannot be zero. Then, I have the following question:

How many values of x do you think there could be within the interval, say, from x = 0 to (x + h) = h at x-axis of x,y-coordinate system? I just remind that x-axis is a continuous straight line.

P.S. I think that this question is possibly not easy. In terms of me, I would use "undefined amount of x-values" if h is not the most minimal value for my proof.

Thank you.
a.s.h.
Posts: 110
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Re: Discussion of Integral calculus

Post by a.s.h. »

Boris Lagutin wrote:How many values of x do you think there could be within the interval, say, from x = 0 to (x + h) = h at x-axis of x,y-coordinate system?
There are infinitely many such x-values, and the type of infinity is called an uncountable infinity. This is in a sense bigger than say the total number of whole numbers there are, which is called a countable infinity. You can read a bit more about countable vs. uncountable infinity here: http://mathforum.org/library/drmath/view/74268.html .
Boris Lagutin

Re: Discussion of Integral calculus

Post by Boris Lagutin »

Thank you, a.s.h.

If there are many values of x within the interval equal to h, can one guarantee that the derivative function can be made of values of the original function slopes that could get the derivative function continuous if h is not a most minimal value and constant?

My position is that it cannot be guaranteed.

Outputs (slopes of the original function) of the derivative function must build a continuous curve or straight line if the original function is differentiable, except piece-wise functions (I don't consider piece-wise functions here). In other words, the derivative function cannot consist of discrete values in this case.
As we know, a value of a slope is equal to a difference of the original function two y-values divided by a difference of the original function two x-values.

Thank you.
a.s.h.
Posts: 110
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Re: Discussion of Integral calculus

Post by a.s.h. »

I think I misinterpreted what you were asking. I thought you meant how many numbers are there between 0 and h, for some given h > 0. Now it seems like you're asking how many values in that interval are equal to h. Well the answer to this would be just one: h itself.
Boris Lagutin

Re: Discussion of Integral calculus

Post by Boris Lagutin »

a.s.h.,

I think you understood me right. To be clearer, let's consider a slope of some function at x = 0. Now look at h -> 0 in the definition of the derivative (limit). So if h is not a most minimal value close to zero, then, if I understood you right, you agree that there are many values of x at x-axis between x = 0 and x = h. Therefore, we can take, say, h / 6 or h / 8 and use them in the quotient - a difference of two y-values / a difference of two x-values. In other words, we may get different values of slopes for different values of x which may make the derivative function discontinuous because h is always not a most minimal value close to zero. We cannot guarantee the continuity of the derivative function in this case.
Outputs (slopes of an original function) of the derivative function are defined by the definition of the derivative (limit).

Thank you.
a.s.h.
Posts: 110
Joined: Tue Jan 28, 2014 3:50 pm

Re: Discussion of Integral calculus

Post by a.s.h. »

While we can use any arbitrarily small h to get difference quotients, the point is that these difference quotients will approach something, namely, the slope of the tangent (if the function is differentiable).

E.g. consider finding the slope at x = 1 for f(x) = x^2. With h = 0.1, the difference quotient is ((1+0.1)^2 -1^2)/0.1 = (0.21)/0.1 = 2.1.

Using h = 0.01, it's ((1.01)^2 - 1^2)/0.01 = (0.0201)/0.01 = 2.01.

As you can see if you carry on this process, the slopes approach the number 2, which is the value of the derivative at 1.
Boris Lagutin

Re: Discussion of Integral calculus

Post by Boris Lagutin »

a.s.h.,

That's is right, but h cannot be zero. The value of the slope in the situation that you described in your last post can be equal exactly to 2 IF AND ONLY IF h = 0! As far as I understand, we agreed that h (a difference of two x-values) cannot be zero for defining a slope. If h = 0, then it already is not a slope - it is something else, in my opinion. Applying algebraic tricks (operations) we can get 2 but this value appears only if h = 0.

P.S. Now I understand that I should have begun my proof from this point. However, it requires additional contemplation.

Have a nice day.
a.s.h.
Posts: 110
Joined: Tue Jan 28, 2014 3:50 pm

Re: Discussion of Integral calculus

Post by a.s.h. »

Boris Lagutin wrote:a.s.h.,

That's is right, but h cannot be zero. The value of the slope in the situation that you described in your last post can be equal exactly to 2 IF AND ONLY IF h = 0! As far as I understand, we agreed that h (a difference of two x-values) cannot be zero for defining a slope. If h = 0, then it already is not a slope - it is something else, in my opinion. Applying algebraic tricks (operations) we can get 2 but this value appears only if h = 0.

P.S. Now I understand that I should have begun my proof from this point. However, it requires additional contemplation.

Have a nice day.
The limit as h -> 0 is still 2, and the definition of the derivative is this limit.
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