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June 10th, 2019, 12:55 AM   #1
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a Zn suite that begins with an anomaly and can not be made with the classical math

Hello


log (2 ^ t + t ^ 2-64) = t (1-log (5))

I put i = t with i ^ 2 = -1

So log (2 ^ i + i ^ 2-64) = i (1-log (5)) so log (2 ^ i-65) = i (1-log (5))

ln (2 ^ i-65) + i * ln (5) -i = ln (2 ^ i-65) + ln (5-i) = -i 0

ln (((2 ^ i-65) * 5 ^ i) = i

therefore (2 ^ i-65) * 5 ^ i = exp (i) = -10-325 * i = - (10 + 325 * i)

therefore exp (i) * exp (pi) = - (10 + 325 * i) * exp (pi) = exp (i * pi)

And since the t does not like to be a complex i it wants to be a Z so it will go back to the equation in Z form and by R because exp (i * pi) = exp (i * pi * t) or t belongs to Z not a R

So exp (i * pi * t) = - (10 + 325 * i) * exp (pi) so exp (i * t) = - (10 + 325 * i)

So i * t = log (-10 + 325 * i)

So t = log (-10 + 325 * i) / i = -i * log (-10 + 325 * i) = log- (10 + 325 * i) ^ i =

So I have my t = i * log- (10 + 325 * i)

And since the axiom Z says that an imaginary number can reduce its complexity if by transforming it into a counter 0 1 2 3 ....

So

Here are the solutions for i = 0 t = log (-10) ^ 0

for i = 1 t = log- (10 + 325 * 1) ^ 1

i = 2 t = log- (10 + 325 * 2) ^ 2

...

i = n t = log- (10 + 325 * n) ^ n so this is my V (t) possible solution

the solutions are t = log- (10 + 325 * n) ^ n = -n * log- (10 + 325 * n) =

verification:

** ln (2 ^ t + t ^ 2-64) = t (1-ln (5))? **

log (2 ^ log (n) + log (n) * log (n) -64) = log (n) (1-ln (5))

so log (-n + log (-n) * log (-n)) = log (-n) (1-log (5)) because 2 ^ log (-n) = - n

log (-n + log (-n) * log (-n)) = log (-n) -log-n * log (5) = log (-n) -log-n

Vn Zn = (- exp (Vn) / (3 * 10 * 10 ^ n) * (n ^ n))



***************************

i = 0 t = log (-10) 0/0

for i = 1 t = log- (10 + 325 * 1) ^ 1 (10 + 325 * 1) ^ 1 / (3 * 10 * 10 ^ 1 * 1 ^ 1) = 1.1166666666666667 = (

i = 2 t = log- (10 + 325 * 2) ^ 2 (10 + 325 * 2) ^ 2 / (3 * 10 * 10 ^ 2 * 2 ^ 2) = 36.3

************************************************** ***************(10 + 325 * 3) ^ 3 / (3 * 10 * 10 ^ 3 * 3 ^ 3) = 1179.8415123456790123

************************************************** ***************(10 + 325 * 4) ^ 4 / (3 * 10 * 10 ^ 4 * 4 ^ 4) = 38346.3438802083333333

************************************************** ***************(10 + 325 * 5) ^ 5 / (3 * 10 * 10 ^ 5 * 5 ^ 5) = 1246285.403469

************************************************** **************(10 + 325 * 6) ^ 6 / (3 * 10 * 10 ^ 6 * 6 ^ 6) = 40504909.9617741197988112

************************************************** *********

************************************************** **************...

i = n t = log- (10 + 325 * n) n



and the use of this sequence changes the infinitsimal calculus after a certain time the Z1 = 1.1133333333333333 and before was 1.1166666666666667 probably because of the anomaly 0/0 that there change ca?
Zn represents a bijection towards Z without turning to the left nor to the right and a simple forge
Zn = (- exp (Vn) / (3 * 10 * 10 ^ n) * (n ^ n))
So my Zn is math or not because with the math of axiom ZF can not find it

Zn=(10 + 325 * n) ^ n / (3 * 10 * 10 ^ n * n ^ n) starts 0/0 and can not build with classical math if not find Z10 here or Z19 without this formula with math even if I give you a lot of numbers

Last edited by Zouha10; June 10th, 2019 at 01:24 AM.
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June 10th, 2019, 01:50 AM   #2
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Zn=(10 + 325 * n) ^ n / (3 * 10 * 10 ^ n * n^ n) et vous avez pas réussi a trouver Z car z0=log(-10)=anomalie et cette suite en peux pas le calculer avec du math classique et a chaque fois que vous calculer ses valeur apres un certain temps vous refaite le callcul vous trouver autre et le cause l'anomalie de depart ln-10.

Last edited by skipjack; June 10th, 2019 at 04:31 AM.
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June 10th, 2019, 04:34 AM   #3
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Let's give this subject a rest.
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