Youtube comments of Samuel De Andrade (@samueldeandrade8535).
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@ethanrin9518 "Well, rigor is useful for two very different things".
I don't understand why you are saying that to me. But ok. I will only notice that your statement is not rigorous enough. The rigorous version would be
"rigor is useful for AT LEAST two reasons"
No need for "two very different things". When we talk about reasons they don't have multiplicity, meaning we don't count them more than one time. And we count equivalent reasons as the same.
"When a research paper is not rigorous, it happens that the proofs is contains are wrong"
Well, not exactly. Rigorous and right are not the same. You have to work with a very restricted definition of right or a very relaxed definition of rigor to make that statement.
"It doesn't mean the results are wrong ... results are as important as proofs"
Well, for me, they are the same thing. There are no results, by the way we usually interpret the word, without proofs. And every proof is associated with a result. So ...
"... when a paper is well written ..."
This has nothing to do with rigor. A paper can be rigorous, but badly written.
"... it's understandable even by non specialists"
Not really. That's ... That's a manifestation of Einstein's (attributed) quote "You don't really understand something unless you can explain it to your grandmother". This ignores the cases of intellectually challenged and/or not interested grandmothers.
"But clearly, many papers do not apply this philosophy"
Haha. What philosophy? You didn't defined any philosophy. You should write well your comments so any person can understand it. Hahahahaha. (Obvious punchline:) See what I did here?
"... making the reading painful and longer than it should be"
Man, I know that pretty well. You have no idea. I always had a sharp critical thinking. And, yes, some papers were a torture to read. There is one I always remember: Partial Representations and Partial Group Algebras, Dokuchaev, Exel, Piccione. Man, some parts of this paper could be made SOOOOOO much easier, it is insane they didn't. Something papers lack nowadays is time for the mathematicians to write them (at least on some places, I admit I don't know the time demands in all the world) and an artistic perspective. By that I mean mathematicians should write papers trying to find the best proofs possible. They don't do that. Not in papers. Not in classes. Not in Youtube videos. I will give you one recent example. Will find the video (actually 4 videos). For the people that can watch it, compare the proof the teacher presents with the one I gave in the comments. The difference is insane. Just for us to have a example of right proofs but with very different "readibilities". My proof, I would say, is more rigorous, in a sense. My sense. Because rigor to me involves this search for clean proofs. Just a second.
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If you already know that the ratio between consecutive terms converges, then the 1, x², x³, ... thing comes from the following:
suppose a, b, c, d are consecutive terms of the tribonnaci sequence, then
d/c = (a+b+c)/c
= a/c + b/c + 1
= (a/b)(b/c) + (b/c) + 1
If x is the limit of the ratio between a term and the previous term, then taking the limit of the expression before, we have
x = (1/x)(1/x) + 1/x + 1
= 1/x² + 1/x + 1
In other words,
x³ = x² + x + 1
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Or, if you prefer, if you start with a sequence
a_1, a_2, a_3, a_4, ...
with non zero terms, consider the ratios
R_1 = 1 and R_n = a_n/a_{n-1}
Notice that, for example,
a_5/a_2
= (a_5/a_4)*(a_4/a_3)*(a_3/a_2)
= R_5*R_4*R_3
Knowing this, suppose the original sequence is recursive, so
a_{n+k+1} = F(a_n, ... , a_{n+k})
with F a continuous function in k+1 variables such that
c*F(x_0, ... , x_k) = F(c*x_0, ... , c*x_k)
for any scalar c. Of course, for c different of zero,
F(x_0, ... , x_k)/c = F(x_0/c, ... , x_k/c)
Then
R_{n+k+1} R_{n+k} ... R_{n+1}
= a_{n+k+1}/a_n
= F(a_n, ... , a_{n+k})}/a_n
= F(a_n/a_n , ... , a_{n+k}/a_n)
= F(1, R_{n+1}, ... , R_{n+k}*R_{n+k-1}*...*R_{n+1})
So, if R_n converges to x, taking limit when n goes to infinity of the last expression, we obtain
x^{k+1} = F(1, x, ... , x^k)
Cheers! If F is the sum of the coordinates of a N-dimensional vector
F(x_1, ... , x_n) = x1 + ... x_n
we have the N-bonacci numbers. So, the N-bonacci constant satisfies the polynomial
x^n = 1+x+ ... + x^{n-1}
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