Discussion of Integral calculus

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Boris Lagutin

Re: Discussion of Integral calculus

Post by Boris Lagutin »

a.s.h.,

I know what the limit is equal to 2 in your case. I try to prove another position, however, now I don't know how to prove it so that nobody has any doubts. In my previous post, I have already written that it requires additional contemplation.
Right now I have the following. If we take some function, say, p(x) = (7*x) / x. So if x = 0, p(0) = (7*0) / 0 = 0/0 that is undefined. That means that we cannot find any value of the function p(x) at x = 0. However, if we take limit_{x->0} (7*x) / x = limit_{x->0} (7) = 7, but this limit does not make (7*x) / x defined at x = 0.
Similarly, a slope cannot be defined at h = 0 because it is a quotient by nature as well.

P.S. What I wrote in my previous post and this post requires an additional proof, as I understand now. I feel that my arguments that I gave in this thread are not enough.

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. »

Even though p(0) is not defined in your example, the limit as x -> 0 of p(x) is perfectly well-defined. Remember, a slope (derivative) at a particular point is a limit, so as long as the limit is well-defined, the derivative exists.
Boris Lagutin

Re: Discussion of Integral calculus

Post by Boris Lagutin »

a.s.h,

Yes, we need a slope at some x-value. When we say that a slope (derivative) at, say, x = 1 of some function, we mean a precise value of a slope at x = 1 (not x = 1.01 or less), but the quotient expression defining the slope at x = 1 is undefined at x = 1 in x =1! As I see, it is not persuasive and possibly unclear.

Have a nice day, a.s.h.
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,

Yes, we need a slope at some x-value. When we say that a slope (derivative) at, say, x = 1 of some function, we mean a precise value of a slope at x = 1 (not x = 1.01 or less), but the quotient expression defining the slope at x = 1 is undefined at x = 1 in x =1! As I see, it is not persuasive and possibly unclear.

Have a nice day, a.s.h.
The slope at 1 would be defined as the limit as h -> 0 of ((f(1+h) - f(1))/h). Clearly this limit can exist, and if it does, the slope at 1 is defined as this limit.
Boris Lagutin

Re: Discussion of Integral calculus

Post by Boris Lagutin »

a.s.h.,

I don't know how to explain. I see I am not understood. I need to contemplate more. Please, look at the example which I gave you: p(x) = (7*x) / x. It is clear that the function p(x) is undefined at x = 0, i.e. the function output is undefined. Here is my question:

Can one use limit_{x->0} p(x) to define a value of the function at x = 0, if a value (output) of the function at x = 0 does not exist?

I want to say that we define a limit but not a real value of the function at x = 0 because a real value of the function does not exist at x = 0.
In terms of the definition of the derivative, if we divide zero distance by zero distance, we logically don't get any slope at all. It is well illustrated by the function p(x) = (7*x) / x. So if we replace x with h and set the interval for h from zero to infinity, we get a function of a straight line which expresses the slope equal to 7 for any h, except zero, which is logical.

Thank you.
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Boris Lagutin

Re: Discussion of Integral calculus

Post by Boris Lagutin »

a.s.h.,

I have one more question. Assume some original function has a domain from negative infinity to x = 3. In other words, at x = 3 the function has its last output.

How do you think the original differentiable function has a derivative at the last point x = 3 ? If you think that it does not have, then why?

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. »

In such cases, we would talk about the one-sided derivative of the function. In the example you've given, we would need to exam whether the function is left differentiable at 3. See for instance this page for definitions and information on one-sided derivatives: https://en.wikipedia.org/wiki/Semi-differentiability .

(As that page shows, a one-sided derivative is basically the same limit of a difference quotient as a regular derivative, except that we only use a one-sided limit for it.)
Boris Lagutin

Re: Discussion of Integral calculus

Post by Boris Lagutin »

a.s.h.,

By the way, you did not answer my last-to-previous question about defining an output of the function by limit_{x->0}. In terms of my last question, I suppose you did not understand me. I ask about the last value x = 3 of some function, i.e. the interval (-infinity, 3]. I repeat my question:

How do you think some function has a derivative at its last value at x = 3? If it does not have, then why?

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. »

We can only talk about differentiability at an interior point of the domain of the function. So at the endpoint 3, we can't talk about differentiability, but we can talk about one-sided derivatives, as I said in the previous post.
Boris Lagutin

Re: Discussion of Integral calculus

Post by Boris Lagutin »

a.s.h.,

So we come to an interesting point, in my opinion. So let me ask you two questions:

1) How do you think which derivative (left-sided or right-sided) of some function is at x = 3 if the function domain is (-infinity, 3] ?

2) How do you think why we cannot talk about differentiability of some function at its last point at x = 3 (if there is no any derivative at x =3) ?

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. »

We can't take a regular derivative at the endpoint 3, because the function wouldn't be defined on the right of 3, so we can't take a limit from that side (remember, for a regular limit, h can approach 0 from both sides). But we can still consider a one-sided limit, since the function would be defined on one side of 3 (the left side).
Boris Lagutin

Re: Discussion of Integral calculus

Post by Boris Lagutin »

a.s.h.,

We reach some interesting point, in my view. How do you think why we must care of h approaching zero (for regular limit) from both sides if we calculate a derivative by using the limit which takes h as zero? Please, notice that we don't care of a value (output) of a quotient (in the limit) expressing a value of a slope because the value of the quotient at h = 0 does not exist (undefined).

P.S. One could simply put x = 3 in the formula of some function derivative found by using the definition of the derivative (limit) and calculate a value of the derivative function because one does not need to care of h after x = 3. Remind that x = 3 is a last point of some function.

Thank you.
Boris Lagutin

Re: Discussion of Integral calculus

Post by Boris Lagutin »

Hello,
So I do not see that a.s.h. has answered my two last questions, I have prepared an updated proof of my integral paradox with a foundation of the paradox (three first pages attached to this post). To be precise, it is incomplete yet, maybe later (I am not concerned with math too much).
In my opinion, there is an interesting situation between the definition of the derivative (limit) and the conception of the function differentiability which can be solved by the following approaches:
1. change the current conception of the function differentiability, however, it may in my view lead to changing Mean Value Theorem and the other fundamental calculus conceptions;
2. understand the definition of the derivative (limit) as a limit when h goes to zero but not equal to zero for all the resulting derivative formulas.

P.S. I am sorry if something is wrong - I am not a mathematician. Have a nice weekend.
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Boris Lagutin

Re: Discussion of Integral calculus

Post by Boris Lagutin »

Hello,

Here are the three last pages of my integral paradox proof.

Have a nice day.
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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:Hello,

Here are the three last pages of my integral paradox proof.

Have a nice day.
So in summary, what are you claiming the paradox to be? You wrote an expression for f(x) (the derivative of F(x) := (x^3)/3) but didn't take the limit as Δx → 0. When we do this, we see that f(x) = x^2, and the Riemann sum can be evaluated.
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