Discussion of Integral calculus
Moderator: msmod
Re: Fundamental paradox of Integral calculus?
Thank you, Joe. If i don't mistake, infinity / infinity is an undetermined form. Moreover, I just say that zeros sum to zero. If each term of some series was zero, the limit of the series partial sums would be zero. In my opinion, h and delta_x are not zeros, and delta_x = (b-a)/n, where n goes to infinity.
Have a nice weekend
Have a nice weekend
Re: Fundamental paradox of Integral calculus?
That's why you have to be careful with limits and the language you use.
The limit of N*(1/N) as N goes to infinity does not mean infinity times zero. It is not adding zeros.
When we say "Limit of f(h) goes to F as h goes to zero", we define this to mean:
for any epsilon, however small, we can find a value of h small enough that |f(h)-F| < epsilon.
Note that this definition does not use infinities or infinitesimals.
Be careful only to quote and to judge mathematicians for what they say, not what you say they say.
The limit of N*(1/N) as N goes to infinity does not mean infinity times zero. It is not adding zeros.
When we say "Limit of f(h) goes to F as h goes to zero", we define this to mean:
for any epsilon, however small, we can find a value of h small enough that |f(h)-F| < epsilon.
Note that this definition does not use infinities or infinitesimals.
Be careful only to quote and to judge mathematicians for what they say, not what you say they say.
Re: Fundamental paradox of Integral calculus?
Thanks a lot, Joe.
What is F? In this forum, I write only my own understanding of the general conceptions of Calculus, Algebra, and Geometry. I agree that there are not infinities in the definition that you have just mentioned. But I consider the definition of the derivative (limit) from algebraic and geometrical perspectives, by the way, I have mentioned about it above. Therefore, the point about continuity of the original function and its derivative is open so far. You mentioned epsilon but, please, do not forget delta_x in this definition that cannot be arbitrary, otherwise, continuity of the derivative can be violated, in my opinion.joe wrote:When we say "Limit of f(h) goes to F as h goes to zero", we define this to mean:
for any epsilon, however small, we can find a value of h small enough that |f(h)-F| < epsilon.
Note that this definition does not use infinities or infinitesimals.
Re: Fundamental paradox of Integral calculus?
You ask
When we say "The derivative of f(x) is g(x) ", we mean
The limit of {f(x+h)-f(x)}/h goes to g(x) as h goes to zero.
And by that we mean (quoting from above):
for any epsilon, however small, we can find a value of h small enough that |{f(x+h)-f(x)}/h - g(x)| < epsilon.
Note that this definition does not use infinities or infinitesimals.
An exercise for you: make the analogous statement about integral calculus (without looking up a textbook on calculus).
When I wroteWhat is F?
F was the value of the limit of f(h) as h goes to zero.Limit of f(h) goes to F as h goes to zero
When we say "The derivative of f(x) is g(x) ", we mean
The limit of {f(x+h)-f(x)}/h goes to g(x) as h goes to zero.
And by that we mean (quoting from above):
for any epsilon, however small, we can find a value of h small enough that |{f(x+h)-f(x)}/h - g(x)| < epsilon.
Note that this definition does not use infinities or infinitesimals.
An exercise for you: make the analogous statement about integral calculus (without looking up a textbook on calculus).
Re: Fundamental paradox of Integral calculus?
Thank you, Joe. By the way, I specially try not to use additional resources so that I can follow my own line of reasoning. I try operating and analyzing the general conceptions of Algebra, Geometry, and Calculus which I know. Maybe I try making a new definition taking the limitation of delta_x with respect to the continuity of the original function and its derivative as your exercise. However, I don't know whether I will be able to do it. I seem I am not so smart.
Have a nice day.
Have a nice day.
Last edited by Boris Lagutin on Sat Oct 15, 2016 11:04 pm, edited 1 time in total.
Re: Fundamental paradox of Integral calculus?
Actually, based on your performance in the MOOC, I think that you are smart.
What this argument seems to be about is the definitions of limits, and some (understandable and common) misunderstandings about the use of 'to infinity' and 'to zero'.
What this argument seems to be about is the definitions of limits, and some (understandable and common) misunderstandings about the use of 'to infinity' and 'to zero'.
Re: Fundamental paradox of Integral calculus?
If I don't mistake, say, a limit of form: limit of n->0 (n/n) may be rewritten as limit of n -> infinity (1/n / 1/n).
Thank you.
Thank you.
Re: Fundamental paradox of Integral calculus?
Yes. In this case, both of those limits are just equal to 1, since the function we're taking the limit of is the constant function 1.Boris Lagutin wrote:If I don't mistake, say, a limit of form: limit of n->0 (n/n) may be rewritten as limit of n -> infinity (1/n / 1/n).
Thank you.
Re: Fundamental paradox of Integral calculus?
a.s.h.,
As far as I know, these limits are undetermined. To solve limits like this form when a numerator and a denominator go to zero or infinity, L'Hopital rule may be used.
In any case, I have never met any proof of lim_{h->0} (h) is NOT equal to lim_{n->infinity} (1/n). Maybe someone else knows such a proof?
Thank you.
a.s.h. wrote:Yes. In this case, both of those limits are just equal to 1, since the function we're taking the limit of is the constant function 1.
As far as I know, these limits are undetermined. To solve limits like this form when a numerator and a denominator go to zero or infinity, L'Hopital rule may be used.
In any case, I have never met any proof of lim_{h->0} (h) is NOT equal to lim_{n->infinity} (1/n). Maybe someone else knows such a proof?
Thank you.
Re: Fundamental paradox of Integral calculus?
No need for L'Hôpital here, the functions n/n and (1/n)/(1/n) are just 1 for all n =/= 0, so their limits approaching anything are 1.Boris Lagutin wrote:a.s.h.,
a.s.h. wrote:Yes. In this case, both of those limits are just equal to 1, since the function we're taking the limit of is the constant function 1.
As far as I know, these limits are undetermined. To solve limits like this form when a numerator and a denominator go to zero or infinity, L'Hopital rule may be used.
In any case, I have never met any proof of lim_{h->0} (h) is NOT equal to lim_{n->infinity} (1/n). Maybe someone else knows such a proof?
Thank you.
Also, lim_{h->0} (h) and lim_{n->∞} (1/n) are both equal to 0.
Re: Fundamental paradox of Integral calculus?
a.s.h.,
You earlier wrote in this thread: "Note that lim k-> oo (1/(10^k)) = 0. So if we say h equals this limit, we are saying h = 0. But this isn't the case, since then the difference quotient is just 0/0, which is indeterminate." I hope I quoted you right?
Don't you think that you contradict yourself in your last comment when you write that lim_{h->0} (h) is equal to zero (if I understand you right) ?
Thank you.
You earlier wrote in this thread: "Note that lim k-> oo (1/(10^k)) = 0. So if we say h equals this limit, we are saying h = 0. But this isn't the case, since then the difference quotient is just 0/0, which is indeterminate." I hope I quoted you right?
Don't you think that you contradict yourself in your last comment when you write that lim_{h->0} (h) is equal to zero (if I understand you right) ?
Thank you.
Re: Fundamental paradox of Integral calculus?
I didn't contradict myself (assuming I understood you correctly). If we were to say h = lim_{k-> oo} (1/(10^k)) , we would indeed be saying that h = 0, because lim_{k-> oo} (1/(10^k)) = 0. This doesn't contradict the fact that lim_{h->0} (h) = 0.Boris Lagutin wrote:a.s.h.,
You earlier wrote in this thread: "Note that lim k-> oo (1/(10^k)) = 0. So if we say h equals this limit, we are saying h = 0. But this isn't the case, since then the difference quotient is just 0/0, which is indeterminate." I hope I quoted you right?
Don't you think that you contradict yourself in your last comment when you write that lim_{h->0} (h) is equal to zero (if I understand you right) ?
Thank you.
Re: Fundamental paradox of Integral calculus?
Dear a.s.h.,
I am sorry, but I don't understand you. Let's take: limit_{h->0} (h) , limit_{n -> infinity} (1/n) , and h in the definition of the derivative (limit). Now let me analyze what you have written at your very last comment.
Do you suppose that h in limit_{h->0} (h) is different from h in the definition of the derivative (limit)? Do I understand you right? Here, I mean the definition of the derivative (limit) when h goes to zero.
Thank you.
I am sorry, but I don't understand you. Let's take: limit_{h->0} (h) , limit_{n -> infinity} (1/n) , and h in the definition of the derivative (limit). Now let me analyze what you have written at your very last comment.
a.s.h. wrote:If we were to say h = lim_{k-> oo} (1/(10^k)) , we would indeed be saying that h = 0, because lim_{k-> oo} (1/(10^k)) = 0. This doesn't contradict the fact that lim_{h->0} (h) = 0.
Do you suppose that h in limit_{h->0} (h) is different from h in the definition of the derivative (limit)? Do I understand you right? Here, I mean the definition of the derivative (limit) when h goes to zero.
Thank you.
Re: Fundamental paradox of Integral calculus?
In what sense do you mean they are (or are not) "different"? lim h->0 (h) and lim h -> 0 ((f(x+h)-f(x))/h) are just two limits, with h being the variable that is tending towards 0.Boris Lagutin wrote:Dear a.s.h.,
I am sorry, but I don't understand you. Let's take: limit_{h->0} (h) , limit_{n -> infinity} (1/n) , and h in the definition of the derivative (limit). Now let me analyze what you have written at your very last comment.
a.s.h. wrote:If we were to say h = lim_{k-> oo} (1/(10^k)) , we would indeed be saying that h = 0, because lim_{k-> oo} (1/(10^k)) = 0. This doesn't contradict the fact that lim_{h->0} (h) = 0.
Do you suppose that h in limit_{h->0} (h) is different from h in the definition of the derivative (limit)? Do I understand you right? Here, I mean the definition of the derivative (limit) when h goes to zero.
Thank you.
Re: Fundamental paradox of Integral calculus?
a.s.h.,
As I know, limit_{h->0} is used in the definition of the derivative (limit). We discuss h in limit, i.e. a value of h when h goes to 0. It means that a value of h is defined by limit_{h->0}. If I don't mistake, you wrote that h is not equal to 0, however, later you wrote that limit_{h->0} (h) = 0. For example, analyze g(x) = x^2 , then the definition of its derivative (limit) can be written like:
g'(x) = limit_{h->0} ( (x+h)^2-x^2 / h ) = limit_{h->0} (2x+h) = 2x + limit_{h->0} (h) = 2x
In this way, h that, you wrote, is not equal to 0 can't be 0 if reason algebraically. Although, limit_{h->0} (h) = 0 breaks this logic.
Please do not forget that h can't be arbitrary because an arbitrary value of h in the limit of the definition of the derivative can violate the continuity of the derivatives, as I proved it in my supplemented proof of the fundamental paradox above in this thread.
As I know, limit_{h->0} is used in the definition of the derivative (limit). We discuss h in limit, i.e. a value of h when h goes to 0. It means that a value of h is defined by limit_{h->0}. If I don't mistake, you wrote that h is not equal to 0, however, later you wrote that limit_{h->0} (h) = 0. For example, analyze g(x) = x^2 , then the definition of its derivative (limit) can be written like:
g'(x) = limit_{h->0} ( (x+h)^2-x^2 / h ) = limit_{h->0} (2x+h) = 2x + limit_{h->0} (h) = 2x
In this way, h that, you wrote, is not equal to 0 can't be 0 if reason algebraically. Although, limit_{h->0} (h) = 0 breaks this logic.
Please do not forget that h can't be arbitrary because an arbitrary value of h in the limit of the definition of the derivative can violate the continuity of the derivatives, as I proved it in my supplemented proof of the fundamental paradox above in this thread.