My Math Forum > Math 1 ,36,Z,38346,1246285..... find Z and Zn?

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 June 8th, 2019, 09:12 AM #1 Newbie   Joined: Jun 2019 From: Casablanca https://extrazlove.blogspot.com/ extrazlove https://extrazlove.blogspot.com Posts: 19 Thanks: 0 1 ,36,Z,38346,1246285..... find Z and Zn? Hello everyone, 1, 36, Z, 38346, 1246285 ..... find Z and Zn? If you want more values, I can supply you up to n? For info, I calculated Zn according to n with the help of a new axiom which I call axiom Z. Last edited by skipjack; June 9th, 2019 at 01:59 AM.
June 8th, 2019, 09:38 AM   #2
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Quote:
 Originally Posted by Zouha10 Hello everyone, 1, 36, Z, 38346, 1246285 ..... find Z and Zn? If you want more values, I can supply you up to n? For info, I calculated Zn according to n with the help of a new axiom which I call axiom Z.
There are too many possibilities for Z. (An infinite number of them, actually.) It would help if you could tell us where this series comes from? That would give us an idea of how to find the series.

-Dan

Last edited by skipjack; June 9th, 2019 at 02:00 AM.

 June 8th, 2019, 10:39 AM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 https://extrazlove.blogspot.com WHAT kinda site is THAT????? Thanks from topsquark
June 8th, 2019, 11:03 AM   #4
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According to the website he asked a similar question here:

https://forums.futura-sciences.com/m...trouver-z.html

And was told

Quote:
 Hello, This forum is reserved for mathematics, of which I see no trace in your message. albanxiii, for moderation.
Ouch.

 June 8th, 2019, 11:33 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,969 Thanks: 2219 This question may have come from an IQ test. Thanks from topsquark
June 8th, 2019, 03:00 PM   #6
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Quote:
 Originally Posted by topsquark There are too many possibilities for Z. (An infinite number of them, actually.) It would help if you could tell us where this series comes from? That would give us an idea of how to find the series. -Dan

No, my Z is unique, it comes from here by applying axiom Z to find a solution.
To this equation: ln (2 ^ t + t ^ 2-64) = t (1-log (5)):

Last edited by skipjack; June 9th, 2019 at 02:01 AM.

 June 8th, 2019, 03:26 PM #7 Member     Joined: Oct 2018 From: USA Posts: 90 Thanks: 61 Math Focus: Algebraic Geometry Is there any significance to the sequence or is it just some numbers?
June 8th, 2019, 03:59 PM   #8
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Quote:
 Originally Posted by Zouha10 No, my Z is unique, it comes from here by applying axiom Z to find a solution. To this equation: ln (2 ^ t + t ^ 2-64) = t (1-log (5)):
That's what I was trying to ask. Where does this series come from?

All right then. To speed up the converstation: What is axoim Z?

-Dan

Last edited by skipjack; June 9th, 2019 at 02:02 AM.

 June 8th, 2019, 04:02 PM #9 Newbie   Joined: Jun 2019 From: Casablanca https://extrazlove.blogspot.com/ extrazlove https://extrazlove.blogspot.com Posts: 19 Thanks: 0 J'ai Zn et les valeurs jusqu'à n tu veux le voir ? Ah l'axiome Z a donner un raclé a l'axiome ZF pour trouver Zn même avec ln(-n). Je peux aller jusqu'a Z70=2.3338556159279057e+104 avec cette calcultrice CALCULATRICE EN LIGNE Si ta un calcultrice qui peux calculer des grands nombres tu peux indiquer ou se trouver et je te donne la valeur de n=10^10000000000000000000 si tu veux.
June 8th, 2019, 04:25 PM   #10
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Quote:
 Originally Posted by skipjack This question may have come from an IQ test.
oui j'ai un QI plus 1000 pour trouver Zn lol
Voici Z70=2.3338556159279057e+104
Mais je peux pas aller loin si j'ai pas une calculatrice qui calculer les nombres grands.

Last edited by Zouha10; June 8th, 2019 at 04:28 PM.

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