wu :: forums
« wu :: forums - Chords of length 1/n »

Welcome, Guest. Please Login or Register.
Apr 27th, 2024, 11:19am

RIDDLES SITE WRITE MATH! Home Home Help Help Search Search Members Members Login Login Register Register
   wu :: forums
   riddles
   putnam exam (pure math)
(Moderators: towr, Eigenray, Grimbal, Icarus, SMQ, william wu)
   Chords of length 1/n
« Previous topic | Next topic »
Pages: 1  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print
   Author  Topic: Chords of length 1/n  (Read 1171 times)
Aryabhatta
Uberpuzzler
*****






   


Gender: male
Posts: 1321
Chords of length 1/n  
« on: Nov 22nd, 2004, 12:24pm »
Quote Quote Modify Modify

f:[0,1] -> [bbr] is a continuous function such that f(0) = f(1) -- ( 1 )
 
([bbr] = set of Reals)
 
Show that for all natural numbers n, the graph of f has a 'horizontal' chord of length 1/n. ( horizontal = parallel to the x-axis )
 
This leads to the open ended question:
 
Given any real [alpha] [in] [0,1] - {1,1/2,1/3,....} is there some f ( satisfying ( 1 ) ) for which there is no horizontal chord of length [alpha]?
« Last Edit: Nov 22nd, 2004, 12:25pm by Aryabhatta » IP Logged
Obob
Guest

Email

Re: Chords of length 1/n  
« Reply #1 on: Nov 22nd, 2004, 1:03pm »
Quote Quote Modify Modify Remove Remove

Define g:[0,(n-1)/n] [smiley=to.gif] [smiley=bbr.gif] by g(x)=f(x+1/n)-f(x), and suppose that g has no root.  Then either g(x)>0 for every x or g(x)<0 for every x, as g is a continuous function.  Then g(0)>0, so f(1/n)>f(0).  Since g(1/n)>0, we have f(2/n)>f(1/n).  Continuing in this manner, we get f(0)<f(1/n)<f(2/n)<[smiley=cdots.gif]<f(1)=f(0), a contradiction.  Therefore g has a root, and f has a horizontal chord of length 1/n.
 
For every other number [smiley=alpha.gif] there is a continuous function f with f(0)=f(1) for which there is no horizontal chord of length [smiley=alpha.gif].  Let n be a positive integer for which 1/(n+1)<[smiley=alpha.gif]<1/n.  Then n[smiley=alpha.gif]<1, so 0<1-n[smiley=alpha.gif]<[smiley=alpha.gif].  Now define f(k[smiley=alpha.gif])=k and f(1-k[smiley=alpha.gif])=-k (including when k=0 in both cases), and extend f to all of [0,1] by making it piecewise linear (we can do this since all the numbers 1-k[smiley=alpha.gif] and j[smiley=alpha.gif] are all distinct).  Now f(0)=f(1)=0, f is continuous, and it is easily verified that f(x+[smiley=alpha.gif])-f(x)=1[smiley=ne.gif]0 for every x.  Hence f has no chord of length [smiley=alpha.gif].
IP Logged
Aryabhatta
Uberpuzzler
*****






   


Gender: male
Posts: 1321
Re: Chords of length 1/n  
« Reply #2 on: Nov 22nd, 2004, 3:55pm »
Quote Quote Modify Modify

Nice job Obob.  
Your counterexample for [alpha] is really nice.
 
Doesn't it seem like a surprising result though?
 
This property is true only for 1,1/2,1/3,... !
not even for a single other rational like 2/5!
 
Seems like a really strange result to me, but I guess I will have to get used to it  Grin
 
IP Logged
Pages: 1  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print

« Previous topic | Next topic »

Powered by YaBB 1 Gold - SP 1.4!
Forum software copyright © 2000-2004 Yet another Bulletin Board