I think I've seen quite a few with those problems; Stirling approximation: log(n!)~ n log n (using log as the natural logarithm) (more like log(n!)=n log n - n + O(log(n))).... but we aren't going to be so careful) instead of comparing 2^(24!) and (2^24)!... compare their logs A=log (2^(24!))=24! log 2 B=log ((2^24)!)=2^24 log (2^24)=24(2^24)log 2 turns out 24! > 24(2^24)... so A = 24! log 2 > 24(2^24)log 2 = B (to do that you can do the same reasoning used on the video... 23! > 2^23)
Thumbs-down on your "very easy" comment. If it were "very easy," it would not be taking the host seven minutes, 40 seconds to explain his method in the video.
Thanks
This is really cool.
Very cool!
I think I've seen quite a few with those problems;
Stirling approximation: log(n!)~ n log n
(using log as the natural logarithm)
(more like log(n!)=n log n - n + O(log(n))).... but we aren't going to be so careful)
instead of comparing 2^(24!) and (2^24)!... compare their logs
A=log (2^(24!))=24! log 2
B=log ((2^24)!)=2^24 log (2^24)=24(2^24)log 2
turns out 24! > 24(2^24)... so A = 24! log 2 > 24(2^24)log 2 = B
(to do that you can do the same reasoning used on the video... 23! > 2^23)
2^24 = 16,777,216.
True color nerds be like
Nice and very easy😊
Thumbs-down on your "very easy" comment. If it were "very easy," it would not be taking the host seven minutes, 40 seconds to explain his method in the video.
@@robertveith6383 well,wehen i say very easy i mean it is very easy to me.i can explain it in 1 min: (2^24)!(2^24)!.
@@yoav613-- Your third and fourth lines are not clear in the writing. I think they are clear to you in your mind.
Is there a solution for 2^x! = (2^x)! ?
Yes indeed, the graphing site desmos shows us where 🙂
@@stephenlesliebrown5959 Thank you
x=1. no graphing needed for that
5:37 The exact value of 2^24 is 16,777,216.
Therefore (2^24)! is 16,777,216!
So the comparison can be written as 2^(24!) vs 16,777,216!.
1st comment