Comments on: CE. Big Numbers http://mathfactor.uark.edu/2007/03/24/ce-big-numbers/ The Math Factor Podcast Site Wed, 20 Aug 2008 12:22:57 +0000 http://wordpress.org/?v=2.5.1 By: strauss http://mathfactor.uark.edu/2007/03/24/ce-big-numbers/#comment-40 strauss Mon, 02 Apr 2007 13:37:22 +0000 http://mathfactor.uark.edu/2007/03/24/ce-big-numbers/#comment-40 Hi there; I was counting these as the same error; to be more precise, each <i> particular moment of spacing out</i> was counted as one error! Hi there; I was counting these as the same error; to be more precise, each particular moment of spacing out was counted as one error!

]]> By: rmjarvis http://mathfactor.uark.edu/2007/03/24/ce-big-numbers/#comment-32 rmjarvis Mon, 26 Mar 2007 15:52:04 +0000 http://mathfactor.uark.edu/2007/03/24/ce-big-numbers/#comment-32 I only spotted those two. But (10^500)! is definitely larger than googol^googol. Take the natural log of each: ln(g^g) = g ln (g) = 10^100 ln(10^100) = 10^100 * 100 ln(10) For the second one, use Sterling's formula for factorials of very large numbers: ln(x!) ~= (x/e) ln(x) ln(10^500 !) = 10^500/e ln(10^500) = 10^500 * 500 ln(10)/e So basically, as long as there were more than around 100 9's on that page, then that entry beats the googol^googol entry. But it doesn't beat googol plex ^ googol plex. In fact, this kind of analysis makes this week's puzzle quite tenable. Think about taking n logs of the nth entry in each series. That is, compare the series: ln(1) ln(ln(2^2)) ln(ln(ln(3^3^3))) ... and ln(1) ln(ln(2!)) ln(ln(ln(3!!))) .... I won't go the next step and give the answer here, but suffice to say, one series gets way way bigger than the other. It's no contest. Peace, Mike I only spotted those two. But (10^500)! is definitely larger than googol^googol.

Take the natural log of each:

ln(g^g) = g ln (g) = 10^100 ln(10^100) = 10^100 * 100 ln(10)

For the second one, use Sterling’s formula for factorials of very large numbers:

ln(x!) ~= (x/e) ln(x)

ln(10^500 !) = 10^500/e ln(10^500) = 10^500 * 500 ln(10)/e

So basically, as long as there were more than around 100 9’s on that page, then that entry beats the googol^googol entry. But it doesn’t beat googol plex ^ googol plex.

In fact, this kind of analysis makes this week’s puzzle quite tenable. Think about taking n logs of the nth entry in each series.

That is, compare the series:

ln(1) ln(ln(2^2)) ln(ln(ln(3^3^3))) …

and

ln(1) ln(ln(2!)) ln(ln(ln(3!!))) ….

I won’t go the next step and give the answer here, but suffice to say, one series gets way way bigger than the other. It’s no contest.

Peace,
Mike

]]> By: Sye Heinlein http://mathfactor.uark.edu/2007/03/24/ce-big-numbers/#comment-31 Sye Heinlein Mon, 26 Mar 2007 11:57:32 +0000 http://mathfactor.uark.edu/2007/03/24/ce-big-numbers/#comment-31 would another error be that a googol ^ googol is larger then the 10^500. You stated that the later was larger? Didn't really think this response through but i guess it is. I could only find the mistake that was previously mentioned and the one I just pointed out. What is the other?? would another error be that a googol ^ googol is larger then the 10^500. You stated that the later was larger?

Didn’t really think this response through but i guess it is.

I could only find the mistake that was previously mentioned and the one I just pointed out. What is the other??

]]> By: stampy http://mathfactor.uark.edu/2007/03/24/ce-big-numbers/#comment-30 stampy Sun, 25 Mar 2007 03:03:30 +0000 http://mathfactor.uark.edu/2007/03/24/ce-big-numbers/#comment-30 Hi there, well, obviously a googol ^ googol is not a googol multiplied by itself 100 times!! And 10^500 raised to the 10^500th power is not 10^500 multiplied by itself 500 times!! What are the other two mistakes? Great podcast, thanks! Hi there, well, obviously a googol ^ googol is not a googol multiplied by itself 100 times!! And 10^500 raised to the 10^500th power is not 10^500 multiplied by itself 500 times!!

What are the other two mistakes? Great podcast, thanks!

]]>