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Hello to the members of Vengeance - from Vindictive!

Tryden

Tryden

Administrator
Staff member
So as everyone knows, we're merging into one guild with the members of Vengeance, a fellow guild in AoC. With the merger, Vengeance takes on the name Vindictive (same difference
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), and we combine our officers into one leadership team. From Vindictive, Ivy, Mikey and myself will be officers. Our leader will be Aten, who led Vengeance.

To welcome all of Vengeance's members who are going to join this site over the next few days, I thought I'd create this thread. I met Aten over vent last night and he's an awesome person - which seems to represent everyone over at Vengeance.

So I'll be the first to say it. Welcome members of Vengeance!
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Feel free to introduce yourselves here.
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RE: Hello to the members of Vengeance - from Vindicti...

I have spoken with the Oracles and they have foretold that this merger of power will result in this:

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Welcome aboard guys
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RE: Hello to the members of Vengeance - from Vindicti...

Well I will start the ball rolling.  I play as Aten on AoC.  My name is Ryan and live in Grand Rapids Michigan...  Which makes me a EST player however I have horrible sleeping habits. I have played and beta tested MMORPG's for about 5 years now.  The first MMORPG I played was SWG...it was a good game and I loved the PvP but well they ruined it. I played that for about two years and it is where I met most of the core players from Vengeance. I took a break after that because I got a girlfriend and well she hated video games
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and I did not mind so much as they ruined the game.   So after the whole girlfriend came and went I tried all sort of betas and other MMORPGS... I am a fanatic about NWN.... That game is classic and I still play.  I like to role play as much as I like to play the game.  Outiside of games I am a student and work full time.  I will most likely have less time in the fall to play as I have a heavy load at school.  I work in health care and I hate it.... the general public sucks.  Pretty much I am a glorified baby sitter.  As I have been told before I am a little random and may have some bi-polarish moments but thats me.  Other than that I am happy to help any one in game and please feel free to ask me any questions anyone might have.  Glad to have you all and be a part of what you have.
Thanks
Aten
 
RE: Hello to the members of Vengeance - from Vindicti...

Now how random would that be huh?....

So rather than do that lets start out with a nice math problem that if you figure out I will reward you... Sounds fun huh?

Can you find three nonzero integers x, y, and z, such that (x+y+z)^3=xyz?


Well give it a go huh.... See what you can come up with.
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RE: Hello to the members of Vengeance - from Vindicti...

<font face=Calibri color=#0000ff size=3>Well since Aten got the ball rolling I guess it’s my turn. My main character in AOC is Syphion. My name is Rob and I live in South Carolina. My gaming goes all the way back to EQ and then onto SWG where I recruited all the guys in vengeance. Yeah that’s right I was the guild leader. I stopped playing and gave leadership to Aten, Bower and Xaul. So we all go back a ways.&nbsp;Well i dont talk as much as Ryan so he can fill in the missing parts.
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</font>
 
RE: Hello to the members of Vengeance - from Vindicti...

Welcome Grefox and Aten&nbsp;
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RE: Hello to the members of Vengeance - from Vindicti...

3ntr0py said:
Now how random would that be huh?....

So rather than do that lets start out with a nice math problem that if you figure out I will reward you... Sounds fun huh?

Can you find three nonzero integers x, y, and z, such that (x+y+z)^3=xyz?


Well give it a go huh.... See what you can come up with.
face_wink.gif
Click to expand...


Oh goodie! I have a bachelors in mathematics to help solve this problem.

Well to first see if there is an actual solution to the problem you're giving we must first consider the nature of the equation, a cubed sum equaling their respective product.

Well, now we must consider the fact that integers are numbers on series {...-2,-1, 0, 1, 2...}. The set of integers also form a ring denoted as Z. A given integer may be negative, nonnegative, or zero, or positive. The ring Z of integers can be shown to have cardinality of aleph0, with a function equal for the nonnegative integers to:

f(x) = x/(1-x)^2 = x + 2x^2 + 3x^3 + 4x^4 ...

Although with this definition of integers and the nature of cubics versus products, you're creating a rough factorial with (xyz) versus the cubic value of (x + y + z)^3. I'm going to assume that there is probably no real solution to this problem at any point in space.

After some searching: http://www.orbifold.net/default/?p=534 it is indeed an unsolved problem.

Edit: I said fractal instead of factorial XD
 
RE: Hello to the members of Vengeance - from Vindicti...

Well great.... I had a wonderful story problem to put up next with Unicorns, the Easter Bunny,&nbsp; and Santa...&nbsp; but I won't now...
 
RE: Hello to the members of Vengeance - from Vindicti...

3ntr0py said:
Well great.... I had a wonderful story problem to put up next with Unicorns, the Easter Bunny,&nbsp; and Santa...&nbsp; but I won't now...
Click to expand...

Where's Chris Hansen though, man?
 
RE: Hello to the members of Vengeance - from Vindicti...

Aemon Algiz said:
3ntr0py said:
Well great.... I had a wonderful story problem to put up next with Unicorns, the Easter Bunny,&nbsp; and Santa...&nbsp; but I won't now...
Click to expand...

Where's Chris Hansen though, man?
Click to expand...

At my house. And it looks like he's going to be here for a while. Damnit, I knew it was a trap. . .
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RE: Hello to the members of Vengeance - from Vindicti...

Lord Tryden said:
Aemon Algiz said:
3ntr0py said:
Well great.... I had a wonderful story problem to put up next with Unicorns, the Easter Bunny,&nbsp; and Santa...&nbsp; but I won't now...
Click to expand...

Where's Chris Hansen though, man?
Click to expand...

At my house. And it looks like he's going to be here for a while. Damnit, I knew it was a trap. . .
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Click to expand...

My bad man, I shouldn't have sent you those pictures ):
 
RE: Hello to the members of Vengeance - from Vindicti...

You can get in trouble for having pictures of Unicorns and the Eater Bunny?... * quickly deletes all his pictures*

I can totally understand Santa but the other two.
 
RE: Hello to the members of Vengeance - from Vindicti...

3ntr0py said:
You can get in trouble for having pictures of Unicorns and the Eater Bunny?... * quickly deletes all his pictures*

I can totally understand Santa but the other two.
Click to expand...

IT CAN BE FURRY TIME NAO TRYDEN?
 
RE: Hello to the members of Vengeance - from Vindicti...

Aemon Algiz said:
3ntr0py said:
Now how random would that be huh?....

So rather than do that lets start out with a nice math problem that if you figure out I will reward you... Sounds fun huh?

Can you find three nonzero integers x, y, and z, such that (x+y+z)^3=xyz?


Well give it a go huh.... See what you can come up with.
face_wink.gif
Click to expand...


Oh goodie! I have a bachelors in mathematics to help solve this problem.

Well to first see if there is an actual solution to the problem you're giving we must first consider the nature of the equation, a cubed sum equaling their respective product.

Well, now we must consider the fact that integers are numbers on series {...-2,-1, 0, 1, 2...}. The set of integers also form a ring denoted as Z. A given integer may be negative, nonnegative, or zero, or positive. The ring Z of integers can be shown to have cardinality of aleph0, with a function equal for the nonnegative integers to:

f(x) = x/(1-x)^2 = x + 2x^2 + 3x^3 + 4x^4 ...

Although with this definition of integers and the nature of cubics versus products, you're creating a rough factorial with (xyz) versus the cubic value of (x + y + z)^3. I'm going to assume that there is probably no real solution to this problem at any point in space.

After some searching: http://www.orbifold.net/default/?p=534 it is indeed an unsolved problem.

Edit: I said fractal instead of factorial XD
Click to expand...

Like i said. No.
 
RE: Hello to the members of Vengeance - from Vindicti...

LanksCOH said:
Aemon Algiz said:
3ntr0py said:
Now how random would that be huh?....

So rather than do that lets start out with a nice math problem that if you figure out I will reward you... Sounds fun huh?

Can you find three nonzero integers x, y, and z, such that (x+y+z)^3=xyz?


Well give it a go huh.... See what you can come up with.
face_wink.gif
Click to expand...


Oh goodie! I have a bachelors in mathematics to help solve this problem.

Well to first see if there is an actual solution to the problem you're giving we must first consider the nature of the equation, a cubed sum equaling their respective product.

Well, now we must consider the fact that integers are numbers on series {...-2,-1, 0, 1, 2...}. The set of integers also form a ring denoted as Z. A given integer may be negative, nonnegative, or zero, or positive. The ring Z of integers can be shown to have cardinality of aleph0, with a function equal for the nonnegative integers to:

f(x) = x/(1-x)^2 = x + 2x^2 + 3x^3 + 4x^4 ...

Although with this definition of integers and the nature of cubics versus products, you're creating a rough factorial with (xyz) versus the cubic value of (x + y + z)^3. I'm going to assume that there is probably no real solution to this problem at any point in space.

After some searching: http://www.orbifold.net/default/?p=534 it is indeed an unsolved problem.

Edit: I said fractal instead of factorial XD
Click to expand...

Like i said. No.
Click to expand...

Haha!
 
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