Why don't they teach Newton's calculus of 'What comes next?'
Why Mathematics Stack Exchange recommends it
Mathematics Stack Exchange answerers point people to it when a question involves finding closed forms for sums and products or fitting sequences and recurrences with Newton-Gregory interpolation.
Recommended by 0 people across 0 different months, 2021 to 2022.
What it is
A 47-minute Mathologer video on the discrete counterpart of calculus, the calculus of finite differences. It covers differences as derivatives, sums as integrals, difference equations, and the Newton-Gregory interpolation formula as a 'what comes next' oracle, with appearances by Fibonacci numbers, Pascal's triangle and Maclaurin series.
Cited in 2 Wikipedia articles, including Generating function transformation, Finite difference.
What people said
Cites Alexander Conrad · 2021-10 · Mathematics Stack Exchange · on “Methods and knowledge needed for the conversion of $\sum$ and $\prod$ into non-iterative expressions”
@user110391 I learned it using "A Synopsis of Elementary Results in Pure and Applied Mathematics" and then developed on it on my own deriving much of the identities now understood by commonplace math YouTubers such as Mathologer (who did a video on it recently: [this video] ) and Michael Penn (here: [LINK] ).
Cites IntegerUnderflow · 2022-08 · Mathematics Stack Exchange · on “How to discover a recursive relation in a data set?”
There are fairly simple and fast methods for finding polynomials that match data exactly (As long as the data set size doesn't exceed 10^6, it'll work) (Newton-Gregory Interpolation, and much more, see 1. for a video on that topic), but assuming you want a simpler solution than the nth-degree polynomial that Newton-Gregory spits out, then a first guess is to throw it into OEIS - usually you'll find a sequence which is well defined, but not recursive - converting into a recursive relation shouldn't be too hard. If the internet doesn't give you any answers, then you could look into making a neural net that guesses an answer - no promises. The training data should be randomly generated recursiv
Cites Collag3n · 2022-09 · Mathematics Stack Exchange · on “Question regarding 1-cycles in the Collatz conjecture”
Yes, you should have $2^m-3^n|2^{m-n}-1$ since $X_1+1=\frac{2^m-3^n+3^n-2^n}{2^m-3^n}=\frac{2^m-2^n}{2^m-3^n}$ and the denominator is odd so you can ignore the power of 2 in the numerator for the divisibility. Notice that since $m<2n$ you can see that $2^{m-n}-1<2^n$ , but we know (with Rhin's bound and for $n>5$ ) that $2^m-3^n>2^n$ [LINK] so in the end $2^m-3^n\nmid2^{m-n}-1$ Also, I didn't try it myself, but it made me think of this Mathlologer video. Perhaps something you would want to explore: [this video]