Originally posted by David113Question: Taking the general case where "100" is replaced by m, wouldn't the same inequality relationship hold for m=100 as for m=1 and m=2? Or is there a "crossover point" where the inequality reverses? Why would there ever be a crossover point?
9^(9^(9^(9^...))), with 100 exponentiations, or
(...(((9!)!)!)!...), with 100 factorials?
Putting it another way, what is bigger - a(100) where a(0)=9 and a(n+1)=9^a(n), or b(100) where b(0)=9 and b(n+1)=b(n)! ?
Obviously with m = 1, 9^9 >> 9!. So, calling 9^9 "g", and 9! "h", then just as obviously, g^9 >>> h!. And so on.
Or so it seems at first glance.
Originally posted by JS357Edit: BIG OOPS. Too early in the day.
Question: Taking the general case where "100" is replaced by m, wouldn't the same inequality relationship hold for m=100 as for m=1 and m=2? Or is there a "crossover point" where the inequality reverses? Why would there ever be a crossover point?
Obviously with m = 1, 9^9 >> 9!. So, calling 9^9 "g", and 9! "h", then just as obviously, g^9 >>> h!. And so on.
Or so it seems at first glance.
Originally posted by David113OK I'll start over. Taks the first 9^9 compared to 9
9^(9^(9^(9^...))), with 100 exponentiations, or
(...(((9!)!)!)!...), with 100 factorials?
Putting it another way, what is bigger - a(100) where a(0)=9 and a(n+1)=9^a(n), or b(100) where b(0)=9 and b(n+1)=b(n)! ?
9*9*9*9*9*9*9*9*9 = 387 420 489
9*8*7*6*5*4*3*2*1 = 362 880
(The above can be done using Google)
Now compare 387 420 489^9 to 362 880!
That's 1.9662705 e77 which is BIG, and ...
See http://en.wikipedia.org/wiki/Stirling%27s_approximation
... a number somewhat larger than (n/e)^n where n = 362,880 and e ~2.71828183
which is about 1 e4^n
which tacks on "00000" to "1" n+1 times.
Meaning the scientific notation is about 1 e(10000*362,880)
which is HUGE.
And it keeps going that way.
But maybe there's another OOPS coming.
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4 edits
lets do it with tens
10^10 = 1e10
10! = 3 628 800
10^(10^10) = 1 followed by 10000000000 zeroes
at 64 characters per line and 200 lines per page that number would take 781250 pages to write down.
from stirling's apporoximation
n! ~ sqrt(2*pi*n)(n/e)^n
so 10!! ~ 4775*1334960^3628000
log10(10!!) ~ 3.6 + 3628000*6 ~ 21768003.6
so 10!! ~ 1 followed by 21768003 zeros
At 64 characters per line and 200 lines per page that number would be only 1700 pages.
So it looks like the nested factorial grows a lot slower than the nested power
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Originally posted by iamatigerNow trying 9s
lets do it with tens
10^10 = 1e10
10! = 3 628 800
10^(10^10) = 1 followed by 10000000000 zeroes
at 64 characters per line and 200 lines per page that number would take 781250 pages to write down.
from stirling's apporoximation
n! ~ sqrt(2*pi*n)(n/e)^n
so 10!! ~ 4775*1334960^3628000
log10(10!!) ~ 3.6 + 3628000*6 ~ 21768003.6
so 10!! ...[text shortened]... ges.
So it looks like the nested factorial grows a [b] lot slower than the nested power[/b]
9^9 = 387420489
9! = 362880
9^(9^9) = 9^387420489
log10(9^(9^9)) = 387420489*log10(9)
log10(9^(9^9)) = 346275.522
so 9^(9^9) ~ 1 followed by 346275 zeroes
9!! ~ sqrt(2*pi*362880)(362880/e)^9 {stirling's approximation}
9!! ~ 1510*133496^9
log10(9!!) ~ log10(1510) + 9*log10(133496)
log10(9!!) ~ 49
so 9!! ~ 1 followed by 49 zeroes
Nested powers win hands down