Joined: Tue 09-09-2003 10:52PM Posts: 1145 Location: High in the Rocky Mountains
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kevintb wrote:
infinity^infinity ftw
What higher, [infinity]^infinity or [infinity]! ?
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so n^n is always going to be bigger than n! (well, I suppose with the exception of 0 and 1) since you're multiplying smaller numbers with the factorial
Joined: Mon 09-22-2003 3:29PM Posts: 4317 Location: Find out on irc
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hotmail wrote:
Not to be left out of the storage bonanza, a Hotmail representative said that while they "can't offer unlimited storage, they can delete all your e-mail at random intervals in conjunction with their Live OneCare service, to make sure you never run out of space."
hahahha.. and this is why i use google
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so n^n is always going to be bigger than n! (well, I suppose with the exception of 0 and 1) since you're multiplying smaller numbers with the factorial
But infinity! is the same as (infinity * infinity-1 * infinity-2 * etc.) which is equal to (infinity * infinity * infinity * etc.) anyway...
so n^n is always going to be bigger than n! (well, I suppose with the exception of 0 and 1) since you're multiplying smaller numbers with the factorial
But infinity! is the same as (infinity * infinity-1 * infinity-2 * etc.) which is equal to (infinity * infinity * infinity * etc.) anyway...
Which is exactly why the whole exercise is pointless to begin with.
_________________ The solution of this problem is trivial and is left as an exercise for the reader.
so n^n is always going to be bigger than n! (well, I suppose with the exception of 0 and 1) since you're multiplying smaller numbers with the factorial
But infinity! is the same as (infinity * infinity-1 * infinity-2 * etc.) which is equal to (infinity * infinity * infinity * etc.) anyway...
but if infinity+1 > infinity, then infinity -1 < infinity
Joined: Mon 08-18-2003 2:33PM Posts: 1188 Location: Somewhere East Of Pittsburgh
Source: Beta Sigma Psi
kjk437 wrote:
devil wrote:
snocat17 wrote:
kevintb wrote:
infinity^infinity ftw
What higher, [infinity]^infinity or [infinity]! ?
infinity^infinity
5^5 is 5*5*5*5*5 5! is 5*4*3*2*1
so n^n is always going to be bigger than n! (well, I suppose with the exception of 0 and 1) since you're multiplying smaller numbers with the factorial
But infinity! is the same as (infinity * infinity-1 * infinity-2 * etc.) which is equal to (infinity * infinity * infinity * etc.) anyway...
It's only inifinity by rounding. Ask Vy Le.
([infinity+infinity]^[infinity+infinity])+(whatever the next guy says)
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