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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

1904.0. "Concours General, France, 1992, problem 4" by EVTSG8::ESANU (Au temps pour moi) Thu Oct 13 1994 09:56

Let u(n) be a sequence of real numbers, defined by

	0 < u(0) < 1
	0 < u(1) < 1
		 (sqrt(u(n)) + sqrt(u(n-1))
	u(n+1) = --------------------------
			     2

a) Prove that the sequence u(n) is convergent.
b) Prove that, beginning with a certain term n0, the sequence (u(n))n>=n0
is strictly increasing.


(The "Concours General" is the French Mathematics Olympiad).


Mihai.
T.RTitleUserPersonal
Name
DateLines
1904.1A solutionBALZAC::QUENIVETMargins are often too small.Mon Dec 12 1994 11:4654
1904.2re .1: Retry...EVTSG8::ESANUAu temps pour moiMon Dec 12 1994 12:1410
>       f(1) = 1/2(1+sqrt(u_(n-2)) > 0,

Herve, you made a slight mistake:
f(1) =  (1/2)(-1 + sqrt(u_(n-2)) < 0

So...

I found no way of proving b) before a).

Mihai.
1904.3RTL::GILBERTMon Dec 12 1994 18:3312
>2. The sequence u_n is (strictly) increasing.

I'm suprised there's no simpler way of proving this (and I've looked).

>3. The real sequence u_n has an upper bound ( =1 ) and is increasing.
>Then it has a limit l.

This isn't clear.  For example, the sequence:
	v_1 = 0, v_2 = 1/4, v_n = v_<n-1> + 2^(-n)
has an upper bound of 1 and is increasing.  But that doesn't make its limit 1.

(FWIW, the v sequence's _least_ upper bound is 1/2, as is its limit).
1904.4Let's put it in other wordsBALZAC::QUENIVETMargins are often too small.Tue Dec 13 1994 06:5821
1904.5AUSSIE::GARSONachtentachtig kacheltjesTue Dec 13 1994 23:404
    re .4
    
    I too misread the l as a 1. I would humbly request that l not be used
    in proofs. Anyway .4 clears it up.
1904.6.0 has not been proved,EVTSG8::ESANUAu temps pour moiThu Dec 15 1994 11:311
and it is most enjoyable!