User:Evan.pacini/sandbox

= The Number Decomposing Formula or GOF(Getal Ontledingsformule) Made by Kasper van Maasdam =

$$f(x, y) = \left\lfloor \frac {|x| \cdot 10^y } {10^{(\lfloor\log(|x|\cdot10^y)\rfloor+1)-y}}\right\rfloor\mod10$$ $$y \in {\mathbb {Z}}$$ $$-1 \geq x \geq 1$$

The GOF gives the possibility, when used in a computer program, to have a computer decompose a value in a way that is comparable, just as we humans do.

The rest of this document explains how and why the GOF works. Applications are also given.

The GOF:

$$f(x, y) = \left\lfloor \frac {|x| \cdot 10^y } {10^{(\lfloor\log(|x|\cdot10^y)\rfloor+1)-y}}\right\rfloor\mod10$$

Also written as:

$$f(x, y) = \left\lfloor \frac {|x| \cdot 10^y } {10^{L-y}}\right\rfloor\mod10$$

Where:

$$L = \lfloor\log(|x|\cdot10^y)\rfloor+1$$

The GOF needs an $x$ ​​and a $y$. In this case, the $x$ is the value to be decomposed. The $y$ is in this case the variable that indicates which number of $x$  is the target from left to right.

Example
$$f(143, 2) = \left\lfloor \frac {|143| \cdot 10^2 } {10^{(\lfloor\log(|143|\cdot10^2)\rfloor+1)-2}}\right\rfloor\mod10$$

Because:

$$|143| \cdot 10^2 = 14300$$

And:

$$L = (\lfloor\log(14300)\rfloor + 1) = 5 $$

You can write:

$$f(143, 2) = \left\lfloor \frac {14300} {10^{5-2}}\right\rfloor\mod10$$

Because:

$$\frac {14300} {10^3} = 14.3$$

You can write that:

$$f(143, 2) = \lfloor 14.3\rfloor\mod10 $$

Which is equal to:

$$f(143, 2) = 14\mod10$$

Then you get:

$$f(143, 2) = 4$$

Explaination
In the example $x$ was equal to 143 and $y$  equal to 2. This meant that the second number from the left side of $x$ should be the outcome.

$L$ became 5 because the $x$  value was 5 numbers long: 14300 $x$ and $y$  may also be negative or higher numbers. $x$ may be negative, because only the absolute value of $x$  is taken and $y$  may be negative or higher, because the number sequence does not end. $y$ may not be a fraction or irrational number. This is because 1.5th number of a number sequence does not exist. The GOF only works if $x$ is lower or equal to -1, or if $x$  is higher than or equal to 1. If $x$ has a value such as 0.0453 and you want to know the third number (4) then a wrong value will come out. This is because:

$$0.0453 \cdot 10^3 = 45.3$$

$L$ should be 5 in this case, but is 2. If $L$ is two then this is in the denominator:

$$10^{-1} = \frac{1} {10}$$

Then it will eventually be:

$$\left\lfloor\frac {45.3} {\frac {1} {10}} \right\rfloor\mod10 = 453\mod10 = 3$$

And this is not the correct answer. Therefore, the following conditions are given to the GOF: $$y \in {\mathbb {Z}}$$ $$-1 \geq x \geq 1$$