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Posers and Puzzles

Posers and Puzzles

  1. 30 Jun '11 13:28 / 1 edit
    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)! ?
  2. 30 Jun '11 15:50
    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)! ?
    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.
  3. 30 Jun '11 16:04
    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.
    Edit: BIG OOPS. Too early in the day.
  4. 30 Jun '11 18:13
    Also, it is not g^9, but 9^g.
  5. 01 Jul '11 17:23
    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)! ?
    OK I'll start over. Taks the first 9^9 compared to 9

    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.
  6. Subscriber AThousandYoung
    It's about respect
    01 Jul '11 18:23
    So...the powers function beats the factorial function right?
  7. 01 Jul '11 19:09
    NOT 387 420 489^9 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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

    (The exclamation marks are not factorials)
  8. 02 Jul '11 00:12 / 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
  9. 02 Jul '11 16:47
    Originally posted by David113
    NOT 387 420 489^9 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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

    (The exclamation marks are not factorials)
    Oops again.

    According to google calculator,


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

    and


    2^(2^(2^2)) = 65 536
  10. 03 Jul '11 22:22
    Originally posted by iamatiger
    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]
    Now trying 9s

    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