## Monday, February 14, 2011

in honor of st. valentine's day,

(from xkcd #55, but now available as a t-shirt)

on a related note, march 14th isn't just pi day but a day of reciprocation:
Custom dictates that women give chocolate to all of the important men in their lives — from fathers and schoolteachers, to office colleagues and of course boyfriends — so women will probably spend more this year to keep up appearances.

But in Japan, there's no such thing as free chocolate. The confectionery industry has deemed March 14 "White Day," when men are supposed to return the favor and give candy to women.

(the article continues @ NPR.)

on an unrelated note, my monday lectures are generally out of sorts, especially my proofs course.

today i couldn't seem to get my mind straight on what i wanted to prove. also, it bothered me that i couldn't find an easy proof of
$$\left(1+\frac{1}{n}\right)^n \;\leq\; 3$$ for large values of n .. or without limits, anyway. (to explain, i covered bernοulli's inequality and wanted to relate it to what you can say about the number e.)

maybe it's because i work on the weekends, which is a wonderfully uninterrupted time to do research, and it takes me a day to get back to being "responsible."

speaking of which, there's never enough time to get anything done .. [sighs]

Leonid said...

I may have seen a nice direct proof once, but if so, I don't remember it...
This could be used for an advanced lesson on induction: sometimes you have to prove more that you want. Namely, replace the exponent n with n+1. Since the expression evaluates to <3 when n=5, it remains to prove that it is decreasing. The latter boils down to the inequality
n^(2n+1) > (n+1)*(n^2-1)^n
Here, the left side can be estimated by Bernoilli (or binomial formula):
(n^2)^n = (n^2-1+1)^n > (n^2-1)^n + n*(n^2)^(n-1)
hence
n^(2n+1) > n*(n^2-1)^n + n^(2n) > (n+1)*(n^2-1)^n
as desired.

mmailliw said...

(1 + 1/n)^n: the first term is 1, the second term is n * 1^(n - 1) * 1/n = 1 too, and each term after that is less than half of the preceding term (multiply by something less than n, divide by something at least 2, and then multiply by 1/n for the extra 1/n) so we get - termwise - less than 1 + 1 + 1/2 + 1/4 + 1/8 + ... = 3.

Leonid said...

Ah, that's the proof that have forgotten. Nice one.