A percentage change problem for debate
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Since I could not find out what rule 9(A)i is in the complete site guidelines, may I please ask you for the precise statement or the source of it?We probably should close this thread: I think it is against rule 9(A)i for me to be right about anything.
JeffM is always wrong.
D
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"always" is a very l-o-o-o-o-ng time.
No. I was being a clown.I see. I was worried that I made any inappropriate comments unknowingly, which might cause this thread to be closed by the moderators.
Very interesting posts you have made.
Allow me to introduce myself, I hold a Maths degree and am currently doing research on Pure Mathematics. This question was originally posted on a Hong Kong forum called LIHKG (https://lihkg.com/thread/2242892/page/1) and a LIHKG member sent a direct message to my social platform account to ask for my help.
I have always been writing popular (university level) Maths on a column of a newspaper called Standnews, I have written a solution to this problem in Chinese weeks ago.
As written in my article, I set B = f(A) for (positive) real numbers A and B and attain the functional equation 1.25^n f(A) = f(1.35^n A) for all natural numbers n by induction.
Then, I wrote that we now have 2 routes to go. The 1st route is that we construct a function f that satisfies this functional equation. Begin with an arbitrary function g on [1, 1.35), then we can extend g to the whole (positive) real line by multiplying g with 1.25ᵐ on the intervel [1.35ᵐ, 1.35ᵐᐩ¹) for integers m. In this way, we obtain a function f from g, and f satisfies the functional equation. Since g is arbitrary, there are infinitely many solutions.
The 2nd route is to make the functional equation satisfied by all real numbers n. Since I have written an article on the extension of the exponential function from natural numbers to real numbers before, I explained that we can do similarly this time, assuming that f is continuous. Hence by extending the functional equation 1.25^n f(A) = f(1.35^n A) to all real numbers n, we obtain a unique solution, i.e. f(A) = kA^(log1.25/1.35).
HERE COMES THE QUESTION: WHY DOES THIS QUESTION APPEAR ON THIS FORUM?
Vincent1 (His name is Vincent Chu) saw my article on Standnews. He understood barely, but he kept leaving so many comments which are mathematically incorrect. I myself, and many other people including other researchers in Maths, have explained for A WHOLE WEEK and pointed out his mistakes, we even suggested Maths textbooks for him to read and learn. Unfortunately, he was not even communicating, he kept looping his horrible arguments and threatened to prove us wrong by putting the problem on a Maths forum, which is here. He said if we do not open an account and discuss here, we are cowards and we are all wrong. This is why I decide to come here.
You all may visit my article here , and read the comment section below. We left comments in English.
I am very sorry that I have created such a trouble to every one of you. Perhaps I should not have written this article, as Vincent Chu has been annoying and harassing many people.
Another unimportant fact is that, some people noticed Vincent's misbehavior and recognized him, they told me about him: He was a Maths teacher in Hong Kong but his Mathematical ability is extremely poor such that he was quickly fired and complained by many students. All of his students became very weak at Maths and got very poor results. But before him, these students were not bad at Maths at all.
Once again, I apologize for every trouble I have made.
I have always been writing popular (university level) Maths on a column of a newspaper called Standnews, I have written a solution to this problem in Chinese weeks ago.
As written in my article, I set B = f(A) for (positive) real numbers A and B and attain the functional equation 1.25^n f(A) = f(1.35^n A) for all natural numbers n by induction.
Then, I wrote that we now have 2 routes to go. The 1st route is that we construct a function f that satisfies this functional equation. Begin with an arbitrary function g on [1, 1.35), then we can extend g to the whole (positive) real line by multiplying g with 1.25ᵐ on the intervel [1.35ᵐ, 1.35ᵐᐩ¹) for integers m. In this way, we obtain a function f from g, and f satisfies the functional equation. Since g is arbitrary, there are infinitely many solutions.
The 2nd route is to make the functional equation satisfied by all real numbers n. Since I have written an article on the extension of the exponential function from natural numbers to real numbers before, I explained that we can do similarly this time, assuming that f is continuous. Hence by extending the functional equation 1.25^n f(A) = f(1.35^n A) to all real numbers n, we obtain a unique solution, i.e. f(A) = kA^(log1.25/1.35).
HERE COMES THE QUESTION: WHY DOES THIS QUESTION APPEAR ON THIS FORUM?
Vincent1 (His name is Vincent Chu) saw my article on Standnews. He understood barely, but he kept leaving so many comments which are mathematically incorrect. I myself, and many other people including other researchers in Maths, have explained for A WHOLE WEEK and pointed out his mistakes, we even suggested Maths textbooks for him to read and learn. Unfortunately, he was not even communicating, he kept looping his horrible arguments and threatened to prove us wrong by putting the problem on a Maths forum, which is here. He said if we do not open an account and discuss here, we are cowards and we are all wrong. This is why I decide to come here.
You all may visit my article here , and read the comment section below. We left comments in English.
I am very sorry that I have created such a trouble to every one of you. Perhaps I should not have written this article, as Vincent Chu has been annoying and harassing many people.
Another unimportant fact is that, some people noticed Vincent's misbehavior and recognized him, they told me about him: He was a Maths teacher in Hong Kong but his Mathematical ability is extremely poor such that he was quickly fired and complained by many students. All of his students became very weak at Maths and got very poor results. But before him, these students were not bad at Maths at all.
Once again, I apologize for every trouble I have made.
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