Daleth function (Martinez)

The daleth function, as described by Gilbert Martinez on 27 Sep 2018, allows one to compute the sum of any geometric series.

$\daleth (a, r) = a \sum_{n=0}^{\infty} r^{n} = \frac{at}{t-s} \iff |r| < 1 \and r = \frac{s}{t}$

Here, a is the first term of the series, and r is the fraction that is exponentially multiplied to each term.

$a \sum_{n=0}^{\infty} r^{n} = a + ar + ar^{2} + ar^{3} + ...$

Here are a few examples of the daleth function in action.

$\daleth (1,{\frac {1}{2}})=\sum _{n=0}^{\infty }({\frac {1}{2}})^{n}={\frac {1*2}{2-1}}={\frac {2}{1}}=2$

$\daleth ({\frac {5}{2}},{\frac {3}{8}})={\frac {5}{2}}\sum _{n=0}^{\infty }({\frac {3}{8}})^{n}={\frac {{\frac {5}{2}}*8}{8-3}}={\frac {\frac {40}{2}}{5}}={\frac {20}{5}}=4$

$\daleth ({\frac {10}{3}},{\frac {4}{9}})={\frac {10}{3}}\sum _{n=0}^{\infty }({\frac {4}{9}})^{n}={\frac {{\frac {10}{3}}*9}{9-4}}={\frac {\frac {90}{3}}{5}}={\frac {30}{5}}=6$

$\daleth ({\frac {1}{2}},{\frac {5}{7}})={\frac {1}{2}}\sum _{n=0}^{\infty }({\frac {5}{7}})^{n}={\frac {{\frac {1}{2}}*7}{7-5}}={\frac {\frac {7}{2}}{2}}={\frac {7}{4}}$

The set of all numbers generated by the daleth function is notated as Σ, where

$\mathbb {N} \in \Sigma \iff \daleth (a,r)=\mathbb {N}$

The daleth theorem, proven on 28 Sep 2018, states that any rational number can be generated with the daleth function. Thus, any rational number can be generated with the geometric series.

Observations

Martinez noted that, using only rational numbers, the daleth function could never produce an irrational number, since the numbers used would always be whole number ratios.

He also noted that any whole number can be trivially generated using the following:

$\daleth (1, \frac{x-1}{x}) = \frac{1 * x}{x - (x - 1)} = \frac{x}{x - x + 1} = \frac{x}{1} = x$

For his proof, however, he also wanted to find nontrivial ways to generate any whole number.

On 22 Nov 2019, he made the following observation.

$\daleth (1,{\frac {1}{x}})={\frac {1*x}{x-1}}={\frac {x}{x-1}}$

e.g.

$\daleth (1,{\frac {1}{7}})={\frac {7}{6}}$

The next day, he found that he could generate any square number.

Failed to parse (unknown function "\if"): {\displaystyle \if \daleth (1, \frac{x-1}{x}) = x \then \daleth (x, \frac{x-1}{x} = x^{2}}

Daleth proof

Martinez was able to make a key observation for the first part of his hypothesis on 27 Sep 2018. He proved his hypothesis on 28 Sep 2018.

Part I: Nontrivial generation of any integer

There are a few ways to generate any integer using the daleth function. The following is the trivial way.

$\daleth (1, \frac{x-1}{x}) = \frac{1 * x}{x - (x - 1)} = \frac{x}{x - x + 1} = \frac{x}{1} = x$

For a nontrivial way, we'll have to get more general.

$a={\frac {p}{q}},r={\frac {s}{t}},|r|<1$

$\therefore \daleth (a,r)=\daleth ({\frac {p}{q}},{\frac {s}{t}})={\frac {pt}{q(t-s)}}$

Let's suppose this equals a whole number n.

$\frac{pt}{q(t-s)} = n$

$\frac{t}{t-s} = \frac{nq}{p}$

$\therefore t = nq \and p = t-s = nq-s$

This will work for multiples of p as well.

Part II: Generation of any fraction

The approach for this part of the proof will need to be the same as the first part.

$a={\frac {p}{q}},r={\frac {s}{t}},|r|<1$

$\therefore \daleth (a,r)=\daleth ({\frac {p}{q}},{\frac {s}{t}})={\frac {pt}{q(t-s)}}$

Let's suppose this equals a fraction $\frac{x}{y}$, where y ≠ 0.

$\frac{pt}{q(t-s)} = \frac{x}{y}$

$\frac{t}{t-s} = \frac{xq}{yp}$

$\therefore t=xq\land yp=t-s=xq-s\therefore p={\frac {xq-s}{y}}\square$

Evidently, the first part of the proof applies for instances where y = 1. The second part of the proof applies for all values of y ≠ 0 and is thus the general case.

Additional Notes

• $1\leq |s|<t\because |r|<1\land r={\frac {s}{t}}$
• p, q, s, and t must be whole numbers.
• Originally, Martinez wanted to use the Hebrew letter shin (ש) for the function, since it could be read as "s" for summation. However, this letter was unable to be used for Wikipedia's LaTeX, and so daleth was used instead.