Comparing Very Large Numbers--Which Is Bigger?
Date: 12/01/2004 at 22:40:50 From: Adam Subject: Writing Down the Biggest Number Hello, I was curious about the following game. Students are given a sheet of paper and say 30 seconds to write down any number they wish. Whoever writes down the biggest wins. How can you be sure to win? Also, how do you decide which is bigger? How do you know whether 10^10!^10 is bigger than 9!^9!^9? Is there some clever set of numbers that would certainly be bigger than all standard attempts of things like 10!^10!^... or maybe (...(((10!)!)!)!....)!? Thank you.
Date: 12/02/2004 at 10:52:22 From: Doctor Vogler Subject: Re: Writing Down the Biggest Number Hi Adam, Thanks for writing to Dr. Math. Throwing on lots of factorials will certainly get you a really big number. Of course, I don't think there's any practical application to being able to write very large numbers with few symbols. But there is some value to being able to compare large numbers. The main trick is to use logarithms. You want to compare two large numbers. Take the logs of both and compare them. For factorials, it might also be useful to use Stirling's Approximation http://mathforum.org/library/drmath/view/55996.html For example, in your question you want to compare 9!^(9!^9) and 10^(10!^10). (I assume you meant the parentheses to go that way, since otherwise you would just have a product in the exponent.) So you take the log of each (9!^9) * log (9!) and (10!^10) * log 10 and these are still pretty big, so you take the log again 9 log (9!) + log log (9!) and 10 log (10!) + log log 10. Then you use Stirling's Approximation which says that log(n!) = n log n - n (approximately). So you use that and find that the log-log of the first number is about 9 (9 log 9 - 9) + log (9 log 9 - 9) and the log-log of the second number is about 10 (10 log 10 - 10) + log log 10, each of which you can punch into your calculator and get a number. Which is bigger? If the two numbers are *really* close, then you might want to look at upper bounds and lower bounds on Stirling's Approximation, but I think you'll find that big numbers tend to differ by such huge amounts that this won't be necessary except in very rare circumstances. If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum