CE. Big Numbers
We’re well on our way towards describing the two largest numbers that have ever been used! Unfortunately, there are at least three errors in this segment of the Math Factor–can you spot them all?
March 24, 2007 · errata, math puzzles, numbers, The Mathcast · Permalink
«« Q&A: Why is 0! = 1?· · · CF. Mind Boggling! »»
We’re well on our way towards describing the two largest numbers that have ever been used! Unfortunately, there are at least three errors in this segment of the Math Factor–can you spot them all?
RSS feed for comments on this post · TrackBack URL
You must be logged in to post a comment.
Download a great math factor poster to print and share!
Got an idea? Want to do a guest post? Tell us about it!
Heya! Do us a favor and link here from your site!
The Math Factor Podcast is brought to you by:C Goodman-Strauss·· KUAF 91.3 FM·· Math Dept·· Univ. Ark·· XHTML ·· CSS
stampy said,
March 24, 2007 at 9:03 pm
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!
Sye Heinlein said,
March 26, 2007 at 5:57 am
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??
rmjarvis said,
March 26, 2007 at 9:52 am
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
strauss said,
April 2, 2007 at 7:37 am
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!