Re: [OT] math fulguration
by salva (Canon) on Apr 06, 2021 at 14:27 UTC

Does some mathematician knows if the above fulguration I had is a known fact?
Yeah, fortunately! That succession of powers is the base of the positional notation used to represent numbers!
So, what do you get when you subtract 1 to a power or 10? a string of nines. What do you get when you do the same to a power of 16 in hexadecimal? a string of F's. What do you get when you subtract one to a power of 2 in binary? a string of ones.
For instance:
9000 = 9 * 1000 = (10  1) * 10 ** 3
900 = 9 * 100 = (10  1) * 10 ** 2
90 = 9 * 10 = (10  1) * 10 ** 1
+ 9 = 9 * 1 = (10  1) * 10 ** 0

9999 = 9 * 1111 = (10  1) * 1111
9999
+ 1

10000 = 10 ** 4
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Re: [OT] math fulguration
by choroba (Archbishop) on Apr 06, 2021 at 13:56 UTC

#! /usr/bin/perl
use warnings;
use strict;
use feature qw{ say };
use List::Util qw{ sum };
for my $A (2..20) {
for my $n (1..10) {
die "$A, $n\n"
unless $A ** $n == 1 + ($A  1) * sum(map $A ** $_, 0 .. $
+n  1)
}
}
The proof can be compiled in TeX:
\documentclass{article}
\title{Summation Fulguration}
\newtheorem{thm}{Theorem}
\newtheorem{prf}{Proof}[thm]
\begin{document}
\begin{thm}
$$a^n = 1 + (a  1) \sum_{i=0}^{n1} a^i$$
\end{thm}
\begin{prf}
$$1 + (a  1) \sum_{i=0}^{n1} a^i$$
$$= 1 + \sum_{i=0}^{n1}(a1)a^i$$
$$= 1 + \sum_{i=0}^{n1} a^{i+1}  a^i$$
$$= 1 + \sum_{i=0}^{n1}a^{i+1}  \sum_{i=0}^{n1}a^i$$
$$= 1 + (\sum_{i=1}^{n1}a^i) + a^n  (a^0 + \sum_{i=1}^{n1}a^i)$$
$$= 1 + (\sum_{i=1}^{n1}a^i) + a^n  a^0  \sum_{i=1}^{n1}a^i$$
$$= 1 + a^n  a^0$$
$$= 1 + a^n  1$$
$$= a^n$$
Q.E.D.
\end{prf}
\end{document}
The result can be checked here.
I have no idea how broadly it is known.
map{substr$_>[0],$_>[1]0,1}[\*{},3],[[]],[ref qr1,,1],[{}],[sub{}^*ARGV,3]
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> I have no idea how broadly it is known.
Pretty much, what Salva pointed out is discussed at university in CS in the context of numeral systems to basis n, before introducing binary system.
By intuition I'd say it's also connected to some school math, like the theorem that every periodic fraction can be expressed in the form of
n+(m/9...9) with n,m in N
like
DB<3> p 3+7834/9999
3.78347834783478
This works in any number system.
What I learned was the word "fulguration". ;)
Cheers Rolf
_{(addicted to the Perl Programming Language :)
Wikisyntax for the Monastery
}
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Re: [OT] math fulguration
by jo37 (Hermit) on Apr 06, 2021 at 14:35 UTC

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Re: [OT] math fulguration
by duelafn (Parson) on Apr 07, 2021 at 12:22 UTC

Yes, as jo37 said, it is known, though they meant geometric series – the binomial formula would have binomial coefficients.
Note: The wikipedia formula is a more general form and written a different way. Divide out the "a" then note that (1xⁿ)/(1x) = (xⁿ1)/(x1). Then solve for xⁿ.
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Re: [OT] math fulguration
by talexb (Chancellor) on Apr 06, 2021 at 13:56 UTC

This rule doesn't seem to work for 3^3 which following this equation would be 3^0 + 3^1 + 3^2 + 1 => 14; the correct answer is 27. Yes, it does work for powers of 2.
Update: Oof  clearly, my reading comprehension sucks. I never did well on exams. :/
Alex / talexb / Toronto
Thanks PJ. We owe you so much. Groklaw  RIP  2003 to 2013.
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I guess the English formula should be fixed with s/ multiplied\K for n1/ by a1/, i.e.
3^3 = 1 + (3  1) * (3^0 + 3^1 + 3^2) = 1 + 2 * (1 + 3 + 9) = 1 + 2 *
+13 = 1 + 26 = 27
map{substr$_>[0],$_>[1]0,1}[\*{},3],[[]],[ref qr1,,1],[{}],[sub{}^*ARGV,3]
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(3^0 + 3^1 + 3^2) * (3  1) + 1 = 27
HTH.
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