Hello,
My paradox proof is based on the following. As I can see, the current conception of the function differentiability is based on the statement that h in the definition of the derivative (limit) goes to zero but not equal to zero, however, the current limit in the definition of the derivative uses h=0, i.e. the limit does not depend on h at all. Therefore, based on the current conception of the function differentiability, I take the limit in the definition of the derivative as h goes to zero but not equal to zero, and compare it with the current limit in the definition of the derivative as h=0. I will post an updated proof later.
Also, I offer to try distinguishing values of slopes of functions "tangent lines" from "existing or real" rates, i.e. "nonexisting" rates that are defined by the current limit from "existing" rates which are defined by the limit as h goes to zero but not equal to zero.
Have a nice day.
Discussion of Integral calculus
Moderator: msmod

 Posts: 178
 Joined: Thu Aug 28, 2014 9:17 pm
 Location: USA
Re: Discussion of Integral calculus
Boris Lagutin
student
student

 Posts: 178
 Joined: Thu Aug 28, 2014 9:17 pm
 Location: USA
Re: Discussion of Integral calculus
Hello Everybody,
I have added a summary to my foundation of the paradox (attachments, 2 pages). I think that the summary could matter for physics and engineering. I do not have free time to update the proof of the paradox.
Thank you.
I have added a summary to my foundation of the paradox (attachments, 2 pages). I think that the summary could matter for physics and engineering. I do not have free time to update the proof of the paradox.
Thank you.
 Attachments

 Updated_foundation_6.15.2017_page1.PNG (153.02 KiB) Viewed 1056 times

 Updated_foundation_6.15.2017_page2.PNG (212.16 KiB) Viewed 1056 times
Boris Lagutin
student
student

 Posts: 178
 Joined: Thu Aug 28, 2014 9:17 pm
 Location: USA
Re: Discussion of Integral calculus
Hello,
I have totally updated my mathematical work and called it as "Derivatives as NonExisting Rates in Physics". All of the work analysis led me to the terrific conclusion that the modern calculus seems to be paradoxical mathematics... In this post, I publish the updated foundation of the work which shows that derivatives express nonexisting values of the physical parameters defined by quotients in the definition of the derivative (limit). In the next two posts, I will publish the rest of the work.
Thank you,
Boris.
I have totally updated my mathematical work and called it as "Derivatives as NonExisting Rates in Physics". All of the work analysis led me to the terrific conclusion that the modern calculus seems to be paradoxical mathematics... In this post, I publish the updated foundation of the work which shows that derivatives express nonexisting values of the physical parameters defined by quotients in the definition of the derivative (limit). In the next two posts, I will publish the rest of the work.
Thank you,
Boris.
 Attachments

 Math_work(updated, December 2017)_by_B.Lagutin_1.PNG (162.8 KiB) Viewed 801 times

 Math_work(updated, December 2017)_by_B.Lagutin_2.PNG (244.68 KiB) Viewed 801 times
Last edited by Boris Lagutin on Thu Jan 18, 2018 2:57 am, edited 2 times in total.
Boris Lagutin
student
student

 Posts: 178
 Joined: Thu Aug 28, 2014 9:17 pm
 Location: USA
Re: Discussion of Integral calculus
In this second post, I publish the rest (final conclusion is in the next post) of my mathematical work in which it is shown that differentials in all integrals are equal to zero in according to the modern calculus (see attachments).
Thank you,
Boris.
Thank you,
Boris.
 Attachments

 Math_work(updated, December 2017)_by_B.Lagutin_3.PNG (144.58 KiB) Viewed 798 times

 Math_work(updated, December 2017)_by_B.Lagutin_4.PNG (127.02 KiB) Viewed 798 times

 Math_work(updated, December 2017)_by_B.Lagutin_5.PNG (116.57 KiB) Viewed 798 times
Last edited by Boris Lagutin on Thu Jan 18, 2018 2:48 am, edited 2 times in total.
Boris Lagutin
student
student

 Posts: 178
 Joined: Thu Aug 28, 2014 9:17 pm
 Location: USA
Re: Discussion of Integral calculus
In this third post, I publish the last page of my mathematical work with the final conclusion (see attachment). One can algebraically multiply any assmallasneeded value (except zero) by infinity. This algebraic operation is always resulting into infinity! Therefore, multiplying any assmallasneeded value (except zero) of one term of Riemann's sum limit (definite integral) by infinity results into infinity, but adding up infinitely many assmallasneeded terms gives some signed area within some interval in according to the conception of Riemann's sum limit (definite integral). That is paradoxical!
Thank you,
Boris Lagutin.
Thank you,
Boris Lagutin.
 Attachments

 Math_work(updated, December 2017)_by_B.Lagutin_6.PNG (146.32 KiB) Viewed 798 times
Boris Lagutin
student
student
Re: Discussion of Integral calculus
You seem to be saying that since each term of a series goes to 0 as n > ∞, the entire sum goes to 0 as n > ∞. Unfortunately, this is not true because the number of terms in the sum is also going to infinity! That is, ∞*0 is an indeterminate form.
A simple example of a series where the terms goes to 0 but the series does not: consider the sum
(1/n) + (1/n) + … + (1/n),
where there are n terms in total. Then this sum equals n*(1/n) = 1, for every positive integer n. So as n > ∞, the sum has a limit of 1, even though each individual term goes to 0.
A simple example of a series where the terms goes to 0 but the series does not: consider the sum
(1/n) + (1/n) + … + (1/n),
where there are n terms in total. Then this sum equals n*(1/n) = 1, for every positive integer n. So as n > ∞, the sum has a limit of 1, even though each individual term goes to 0.

 Posts: 178
 Joined: Thu Aug 28, 2014 9:17 pm
 Location: USA
Re: Discussion of Integral calculus
Hello,
Your manipulation with the series is called algebraic trick. Such tricks have been misleading many people in my opinion. Let's reason physically. If follow the logic in your last post, then there are some assmallasneeded (except zero) amounts of matter which multiplied by infinity give limited amounts of matter. However, it is clear that any amount (except zero) of matter multiplied by infinity results in infinite amount of matter; otherwise, it would violate the conception of infinity. I hope this helps you to understand what infinity is:
infinity * infinity * infinity = infinity = (any number, except zero) * infinity
P.S. n = infinity (not goes to infinity) in the calculations for every term of the limit of Riemann's sum (definite integral). Dividing any number (except zero and infinity) by infinity is always resulting in zero (not goes to zero) as far as I know.
Thank you,
Boris Lagutin.
Your manipulation with the series is called algebraic trick. Such tricks have been misleading many people in my opinion. Let's reason physically. If follow the logic in your last post, then there are some assmallasneeded (except zero) amounts of matter which multiplied by infinity give limited amounts of matter. However, it is clear that any amount (except zero) of matter multiplied by infinity results in infinite amount of matter; otherwise, it would violate the conception of infinity. I hope this helps you to understand what infinity is:
infinity * infinity * infinity = infinity = (any number, except zero) * infinity
P.S. n = infinity (not goes to infinity) in the calculations for every term of the limit of Riemann's sum (definite integral). Dividing any number (except zero and infinity) by infinity is always resulting in zero (not goes to zero) as far as I know.
Thank you,
Boris Lagutin.
Boris Lagutin
student
student
Re: Discussion of Integral calculus
I don't think n ever "equals" infinity in these infinite sums. The infinite sums are defined as limits as n → ∞.
In the sum I wrote ((1/n) + (1/n) + … + (1/n), where there are n terms in total), when we take the limit as n → ∞, the answer is 1. Taking the limit does not mean we are multiplying a term by infinity. There is a precise mathematical definition of limits that is being used, and essentially forms the basis of Calculus, since the key operations of Calculus like integration and differentiation are defined in terms of limits. This is why elementary Calculus textbooks often begin with an introduction to limits before they teach any differentiation or integration.
(cont. in next post)
In the sum I wrote ((1/n) + (1/n) + … + (1/n), where there are n terms in total), when we take the limit as n → ∞, the answer is 1. Taking the limit does not mean we are multiplying a term by infinity. There is a precise mathematical definition of limits that is being used, and essentially forms the basis of Calculus, since the key operations of Calculus like integration and differentiation are defined in terms of limits. This is why elementary Calculus textbooks often begin with an introduction to limits before they teach any differentiation or integration.
(cont. in next post)
Re: Discussion of Integral calculus
Basically, the definition of saying "the limit of that sum as n → ∞ is 1" is that the sum can be made to stay arbitrarily close to the value 1 whenever n is sufficiently large. The value of n never has to "equal infinity" when using the definition of the limit. Precise formulations of the definition of the limit, as well as intuitive explanations, may be found in many places online. For example, here is one place that introduces the formal definition of a limit of a function while providing intuition as well: https://www.mathsisfun.com/calculus/limitsformal.html .

 Posts: 178
 Joined: Thu Aug 28, 2014 9:17 pm
 Location: USA
Re: Discussion of Integral calculus
Dear a.s.h.,
I told you about physical reasoning in my previous post, but you tell me about calculus textbooks. I know what is written in calculus textbooks. I offer to critically analyze the content of textbooks. I am sorry but you even did not understand what I tried to tell you in my previous post.
I think you mix the formulas for limits with calculations defining values of limits. The formulas have an element showing that some variable goes to infinity, zero or some number, but some variable is replaced with infinity or zero, or some number in the calculations.
By the way, I tried to make the first versions of my mathematical work based on logic, but later I realized that the modern calculus is paradoxical. My some conclusions based on the work have been already estimated. If I have a possibility, I will improve my work, then send it for publication in a scientific magazine.
P.S. I am going to close my account here, so unfortunately I need to stop my studies for some reasons.
Thank you,
Boris Lagutin.
I told you about physical reasoning in my previous post, but you tell me about calculus textbooks. I know what is written in calculus textbooks. I offer to critically analyze the content of textbooks. I am sorry but you even did not understand what I tried to tell you in my previous post.
I think you mix the formulas for limits with calculations defining values of limits. The formulas have an element showing that some variable goes to infinity, zero or some number, but some variable is replaced with infinity or zero, or some number in the calculations.
By the way, I tried to make the first versions of my mathematical work based on logic, but later I realized that the modern calculus is paradoxical. My some conclusions based on the work have been already estimated. If I have a possibility, I will improve my work, then send it for publication in a scientific magazine.
P.S. I am going to close my account here, so unfortunately I need to stop my studies for some reasons.
Thank you,
Boris Lagutin.
Boris Lagutin
student
student
Re: Discussion of Integral calculus
Sometimes we can replace a variable by infinity or 0 and have something that makes sense. When we do this, we are using the extended reals, which is basically the real numbers together with +∞ and ∞. For example, for the limit
lim n → ∞ (3/n),
we can indeed say this is 3/∞ = 0 (i.e. replace n by ∞) by viewing this as a limit in the extended reals. Here we are using arithmetic in the extended reals, and there are various rules of arithmetic that do apply here. See e.g. here for a list of such rules and more info: https://en.wikipedia.org/wiki/Extended_ ... operations .
However, as is displayed at that link, even in the extended reals, we cannot define certain things, like 0*∞ or 0/0. These are known as indeterminate forms ( https://en.wikipedia.org/wiki/Indeterminate_form ), and we can't define them because different limits yield different results when naïvely replacing variables with 0 or ∞ would give us such an indeterminate form.
(cont. in next post)
lim n → ∞ (3/n),
we can indeed say this is 3/∞ = 0 (i.e. replace n by ∞) by viewing this as a limit in the extended reals. Here we are using arithmetic in the extended reals, and there are various rules of arithmetic that do apply here. See e.g. here for a list of such rules and more info: https://en.wikipedia.org/wiki/Extended_ ... operations .
However, as is displayed at that link, even in the extended reals, we cannot define certain things, like 0*∞ or 0/0. These are known as indeterminate forms ( https://en.wikipedia.org/wiki/Indeterminate_form ), and we can't define them because different limits yield different results when naïvely replacing variables with 0 or ∞ would give us such an indeterminate form.
(cont. in next post)
Last edited by a.s.h. on Tue Jan 30, 2018 1:22 pm, edited 1 time in total.
Re: Discussion of Integral calculus
(cont.)
So perhaps you were trying to convey your physical intuition in your posts, but from a mathematical standpoint, I don't see any paradoxes here.
(For example with your Riemann sums, you are basically saying that since ∆x → 0, our sum is 0 + 0 + ... 0 = 0. But this is not valid because there are infinitely many terms in this sum, so we have something of the form 0*∞, which is an indeterminate form, and mathematically speaking, we can't just conclude that this is equal to 0.)
So perhaps you were trying to convey your physical intuition in your posts, but from a mathematical standpoint, I don't see any paradoxes here.
(For example with your Riemann sums, you are basically saying that since ∆x → 0, our sum is 0 + 0 + ... 0 = 0. But this is not valid because there are infinitely many terms in this sum, so we have something of the form 0*∞, which is an indeterminate form, and mathematically speaking, we can't just conclude that this is equal to 0.)
Who is online
Users browsing this forum: No registered users and 1 guest