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 |
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.R | Title | User | Personal Name | Date | Lines |
---|---|---|---|---|---|
1904.1 | A solution | BALZAC::QUENIVET | Margins are often too small. | Mon Dec 12 1994 11:46 | 54 |
1904.2 | re .1: Retry... | EVTSG8::ESANU | Au temps pour moi | Mon Dec 12 1994 12:14 | 10 |
> 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. | |||||
1904.3 | RTL::GILBERT | Mon Dec 12 1994 18:33 | 12 | ||
>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). | |||||
1904.4 | Let's put it in other words | BALZAC::QUENIVET | Margins are often too small. | Tue Dec 13 1994 06:58 | 21 |
1904.5 | AUSSIE::GARSON | achtentachtig kacheltjes | Tue Dec 13 1994 23:40 | 4 | |
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::ESANU | Au temps pour moi | Thu Dec 15 1994 11:31 | 1 |
and it is most enjoyable! |