01[20070607-03:36] <Kal_El> Ok...
01[20070607-03:36] <Kal_El> This time if it breaks we go to bed
01[20070607-03:36] <Kal_El> Agreed?
[20070607-03:36] <null-> ok
01[20070607-03:36] <Kal_El> Ok
[20070607-03:36] <null-> allez
[20070607-03:36] <null-> so
01[20070607-03:36] <Kal_El> f(x) is concave for x>=0 => convex for x<=0
[20070607-03:37] <null-> if f(x) is odd, yes
01[20070607-03:37] <Kal_El> So P(a) is below P(-b)P(0) [remember a<0]
01[20070607-03:37] <Kal_El> ...by convexity
01[20070607-03:37] <Kal_El> as |a|<|b|
01[20070607-03:38] <Kal_El> agreed?
[20070607-03:38] <null-> yes
01[20070607-03:39] <Kal_El> Therefore, P(a) is below P(0)P(b)
[20070607-03:39] <null-> yes
01[20070607-03:39] <Kal_El> So, P(a)P(b) is below P(0)P(b)
[20070607-03:40] <null-> yes
01[20070607-03:40] <Kal_El> Now, P((a+b)/2) is above P(0)P(b), so above P(a)P(b)
[20070607-03:41] <null-> yes
01[20070607-03:41] <Kal_El> So, f((a+b)/2) >= (f(a)+f(b))/2
01[20070607-03:41] <Kal_El> [take mid point through P(a), P(b)]
[20070607-03:42] <null-> seems fair
01[20070607-03:43] <Kal_El> ok so that was the lemma... then take f=sin
[20070607-03:43] <null-> so indeed if f() is odd, convex for x>0, ((f(a) + f(b))/2 <= f((a+b)/2) for a+b > 0
01[20070607-03:43] <Kal_El> sin(x)+sin(y)<=2sin((x+y)/2)
01[20070607-03:44] <Kal_El> for x,y in (-Pi/2,Pi/2), x+y>0.
01[20070607-03:44] <Kal_El> let a=Pi/2-x, b=Pi/2-y
01[20070607-03:44] <Kal_El> cos(a)+cos(b)<=cos((a+b)/2), for a,b>0, a+b<Pi
01[20070607-03:45] <Kal_El> fair?
[20070607-03:45] <null-> looks right
01[20070607-03:45] <Kal_El> we're almost done
01[20070607-03:46] <Kal_El> cos(a)+cos(b)+cos(c) >= 2cos((a+b)/2) + cos(c)
[20070607-03:46] <null-> <=
01[20070607-03:46] <Kal_El> <=
[20070607-03:46] <null-> yes
01[20070607-03:46] <Kal_El> What we have done, is to replace two of the angles by their average.
01[20070607-03:46] <Kal_El> and the new some is greater...
01[20070607-03:47] <Kal_El> What happens if we iteratively keep replacing all 3 pairs of angles by their means?
[20070607-03:48] <null-> what?
01[20070607-03:48] <Kal_El> We reach the limit: (Pi/3, Pi/3, Pi/3), and each time the sum is more, everything being continuous, we get the maximum sum to be the sum of the limit
[20070607-03:49] <null-> wouwou
[20070607-03:49] <null-> what?
01[20070607-03:49] <Kal_El> Which part? The limit?
[20070607-03:49] <null-> erh
[20070607-03:49] <null-> so
[20070607-03:49] <null-> cos(a)+cos(b)+cos(c) <= 2cos((a+b)/2) + cos(c)
[20070607-03:49] <null-> we have this
01[20070607-03:49] <Kal_El> Yes
01[20070607-03:49] <Kal_El> Ok
01[20070607-03:50] <Kal_El> From then on, choose 2 angles one <Pi/3, one >Pi/3
01[20070607-03:50] <Kal_El> and replace them with their means
[20070607-03:50] <null-> ok..
01[20070607-03:50] <Kal_El> Actually, do it iteratively... chose a,b, then b,c, then c,a
01[20070607-03:51] <Kal_El> After each iteration we get closer to (Pi/3, Pi/3, Pi/3)
01[20070607-03:51] <Kal_El> Now this requires proof again
01[20070607-03:51] <Kal_El> Which I'm sure neither of us is up for
[20070607-03:51] <null-> we get : S = cos(a) + cos(b) + cos(c); S <= 2cos((a+b)/2) + cos(c); S <= 2cos((b+c)/2) + cos(a) ; S <= 2cos((a+c)/2) + cos(b)
01[20070607-03:51] <Kal_El> oh no
01[20070607-03:52] <Kal_El> a,b,c ---> (a+b)/2, (a+b)/2, c [call them a',b',c']
[20070607-03:52] <null-> "Actually, do it iteratively... chose a,b, then b,c, then c,a"
[20070607-03:52] <null-> ah
01[20070607-03:52] <Kal_El> a',b',c' ---> a',(b'+c')/2,(b'+c')/2 [call them a'',b'',c'']
[20070607-03:52] <null-> yes got it now
01[20070607-03:53] <Kal_El> we reach limit (Pi,Pi,Pi)/3
[20070607-03:53] <null-> but the "getting closer to pi/3 pi/3 pi/3" part?
01[20070607-03:53] <Kal_El> yes
[20070607-03:53] <null-> how do you prove it
01[20070607-03:53] <Kal_El> Lemma 2: If A+B+C=0, and on A,B,C we iterate the process above, then in limit we get (0,0,0)
[20070607-03:53] <null-> I mean I suppose it's true but
01[20070607-03:54] <Kal_El> Proof I haven
01[20070607-03:54] <Kal_El> 't thought of to be frank
01[20070607-03:54] <Kal_El> Can we leave that for tomorrow?
[20070607-03:54] <null-> I suppose the invariant, their sum, is a good thing to keep in mind
[20070607-03:54] <null-> yes
01[20070607-03:54] <Kal_El> ok
01[20070607-03:54] <Kal_El> thanks for your patience
[20070607-03:54] <null-> thanks for explaining
[20070607-03:55] <null-> glad I finally understood
[20070607-03:55] <null-> heh
01[20070607-03:55] <Kal_El> i'm glad too that i was not chasing a wild goose after all
01[20070607-03:55] <Kal_El> that i was right :)
01[20070607-03:55] <Kal_El> i guess goodnight then
[20070607-03:55] <null-> I seriously hope you are indeed right and we're not both too tired to spot a mistake
[20070607-03:56] <null-> good night
01[20070607-03:56] <Kal_El> i hope so too
03[20070607-03:57] * fwp is now known as smartalec
01[20070607-03:58] <Kal_El> Oh hey
01[20070607-03:58] <Kal_El> The proof is easy
01[20070607-03:58] <Kal_El> U asleep?
[20070607-03:59] <null-> nope
01[20070607-03:59] <Kal_El> Cool
01[20070607-03:59] <Kal_El> Proof of lemma 2
[20070607-03:59] <null-> allez
01[20070607-03:59] <Kal_El> Oh let me modify to make it even easier
01[20070607-03:59] <Kal_El> I'll not exactly prove lemma 2
01[20070607-03:59] <Kal_El> I'll modify it
03[20070607-04:00] * Mulder_ (~mulder@124-197-53-75.callplus.net.nz) has joined #math
01[20070607-04:00] <Kal_El> Lemma 2': If A+B+C=0, then we can choose a pair (wlg A,B) st A'=(A+B)/2, B'=(A+B)/2, C'=C, and max(|A'|,|B'|,|C')<=max(|A|,|B|,|C|)/2
01[20070607-04:01] <Kal_El> *|C'|
[20070607-04:01] <null-> yes
[20070607-04:01] <null-> yes
[20070607-04:02] <null-> it works
01[20070607-04:02] <Kal_El> Proof: Let wlg, |A| be more than |B|, |C|
01[20070607-04:02] <Kal_El> then wlg A>0
01[20070607-04:02] <Kal_El> so B,C<0 [by maximality of |A|]
01[20070607-04:02] <Kal_El> cool
01[20070607-04:02] <Kal_El> so that's it then i guess
01[20070607-04:02] <Kal_El> phew!
[20070607-04:02] <null-> pretty much
06[20070607-04:02] * Kal_El wipes forehead
[20070607-04:02] <null-> that's was quite a proof
[20070607-04:03] <null-> -that
01[20070607-04:03] <Kal_El> i'd like to hear about the other proof if u're here tomorrow
01[20070607-04:03] <Kal_El> and also
01[20070607-04:03] <Kal_El> whle i was thinking about it, i realized that i don't know how to proove (or disprove):
01[20070607-04:03] <Kal_El> a/(b+c)+b/(a+c)+c/(a+b)>=3/2
01[20070607-04:04] <Kal_El> In general, is it true that:
01[20070607-04:04] <Kal_El> Sum{i1/(i2+...+in),...,in/(i1+...+i(n-1))} >= n/(n-1) ?
[20070607-04:05] <null-> what are a, b, c ?
01[20070607-04:05] <Kal_El> Because I remember that I've seen the first one... or at least something very similar
01[20070607-04:05] <Kal_El> oh all are reals, >0
[20070607-04:06] <null-> I guess one way would be to multiply the whole thing with (a+b)(a+c)(b+c) and see if it gets you anywhere
[20070607-04:06] <null-> I mean it should get you somewhere if the inequality is true
01[20070607-04:06] <Kal_El> yeah...
01[20070607-04:06] <Kal_El> another way probably would be to use calculus and try to maximize
01[20070607-04:07] <Kal_El> may be we can discuss tomorrow :)
[20070607-04:07] <null-> perhaps arithmetic/geometric/whatever averages
[20070607-04:07] <null-> we will
[20070607-04:07] <null-> :)
01[20070607-04:07] <Kal_El> 4:10AM doesn't seem like a good time (after alcohol)
[20070607-04:07] <null-> now I'm really off to bed
[20070607-04:07] <null-> heh
01[20070607-04:07] <Kal_El> yeah... thanks for tolerating me
01[20070607-04:07] <Kal_El> goodnight!
Session Close: Thu Jun 07 04:07:43 2007