- 30 Jun '11 15:50

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?*Originally posted by David113***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. - 30 Jun '11 16:04

Edit: BIG OOPS. Too early in the day.*Originally posted by JS357***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. - 01 Jul '11 17:23

OK I'll start over. Taks the first 9^9 compared to 9*Originally posted by David113***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. - 02 Jul '11 00:12 / 4 editslets 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 - 02 Jul '11 16:47

Oops again.*Originally posted by David113***NOT 387 420 489^9 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!**

9^387 420 489 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

(The exclamation marks are not factorials)

According to google calculator,

((2^2)^2)^2 = 256

and

2^(2^(2^2)) = 65 536 - 03 Jul '11 22:22

Now trying 9s*Originally posted by iamatiger***lets do it with tens**slower than the nested power[/b]

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

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