Difference between revisions of "User:Objectz/Math"

(Created page with "This is my math subpage. These contain functions that I have created.")
 
Line 1: Line 1:
 
This is my math subpage. These contain functions that I have created.
 
This is my math subpage. These contain functions that I have created.
 +
 +
<math>x \downarrow y = x^{-(x \uparrow y)}</math> where <math>x \uparrow y = x^{y}</math> so <math>3 \downarrow 3 = 3^{-(3 \uparrow 3)} = 3^{-27}</math>.
 +
 +
<math> x \downarrow \downarrow y = (x \uparrow y)^{-(x \uparrow y)}</math> so <math>3 \downarrow \downarrow 3 = (3 \uparrow 3)^{-(3 \uparrow 3)} = 27^{-27} = 3^{-81}</math>.
 +
 +
<math> x \downarrow \downarrow \downarrow y = (x \uparrow \uparrow y)^{-(x \uparrow y)}</math>. So, <math>3 \downarrow \downarrow \downarrow 3 = (3 \uparrow \uparrow 3)^{-(3 \uparrow 3)} = 7625597484987^{-27} = 3^{-729}</math>.
 +
 +
<math> x \downarrow \downarrow \downarrow \downarrow y = (x \uparrow \uparrow \uparrow y)^{-(x \uparrow y)}</math>. So, <math>3 \downarrow \downarrow \downarrow \downarrow 3 = (3 \uparrow \uparrow \uparrow 3)^{-(3 \uparrow 3)} = (3^{7625597484987})^{-27} = 3^{-205891132094649}</math>.
 +
 +
<math>10 \underbrace{\downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \cdots \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow}_{10 \downarrow \downarrow \downarrow \downarrow 10} 10 = ObjectZ_{1}</math>.
 +
 +
<math>10 \underbrace{\downarrow \downarrow \downarrow\downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \cdots \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow}_{10 \underbrace{\downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \cdots \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow}_{10 \downarrow \downarrow \downarrow \downarrow 10} 10} 10 = ObjectZ_{2}.</math>
 +
 +
<math>ObjectZ_{513298}</math> is ObjectZ's Number.

Revision as of 11:41, 9 November 2020

This is my math subpage. These contain functions that I have created.

$x \downarrow y = x^{-(x \uparrow y)}$ where $x \uparrow y = x^{y}$ so $3 \downarrow 3 = 3^{-(3 \uparrow 3)} = 3^{-27}$.

$x \downarrow \downarrow y = (x \uparrow y)^{-(x \uparrow y)}$ so $3 \downarrow \downarrow 3 = (3 \uparrow 3)^{-(3 \uparrow 3)} = 27^{-27} = 3^{-81}$.

$x \downarrow \downarrow \downarrow y = (x \uparrow \uparrow y)^{-(x \uparrow y)}$. So, $3 \downarrow \downarrow \downarrow 3 = (3 \uparrow \uparrow 3)^{-(3 \uparrow 3)} = 7625597484987^{-27} = 3^{-729}$.

$x \downarrow \downarrow \downarrow \downarrow y = (x \uparrow \uparrow \uparrow y)^{-(x \uparrow y)}$. So, $3 \downarrow \downarrow \downarrow \downarrow 3 = (3 \uparrow \uparrow \uparrow 3)^{-(3 \uparrow 3)} = (3^{7625597484987})^{-27} = 3^{-205891132094649}$.

$10 \underbrace{\downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \cdots \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow}_{10 \downarrow \downarrow \downarrow \downarrow 10} 10 = ObjectZ_{1}$.

$10 \underbrace{\downarrow \downarrow \downarrow\downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \cdots \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow}_{10 \underbrace{\downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \cdots \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow}_{10 \downarrow \downarrow \downarrow \downarrow 10} 10} 10 = ObjectZ_{2}.$

$ObjectZ_{513298}$ is ObjectZ's Number.