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Last edited by Stanley_Marsh (2007-04-11 19:22:40)
Numbers are the essence of the Universe
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The a[sub]n[/sub] have to be rational numbers. Are they?
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Awwwww , Let me come up with another one.
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Youd be surprised its actually much simpler than you think.
And dont be afraid to try #9 and #11 they really arent as hard as they look. I reckon #9 can be done in 4 or 5 lines of proof, and #11 in 3 or 4 lines of proof.
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What about Riemann zeta-function
Last edited by Stanley_Marsh (2007-04-12 09:17:10)
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9 ,11 are hard for me , I haven't that much knowledge of math , I just learn math randomly by myself .haha
Numbers are the essence of the Universe
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What about Riemann zeta-function
I happened to see the Riemann zeta-function today~lol
Yes, thats fine. In fact, youve found a required example.
A simpler one would be
The sequence is bounded above (e.g. 2 is an upper bound), it is increasing, and the a[sub]n[/sub] are rational but
is not rational.
Last edited by JaneFairfax (2007-04-12 16:12:50)
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My TI-89 does it... I know a guy with a 92, the thing is huge.
9. Let X = {x[sub]n[/sub]} be a bounded monotone sequence. Since X is bounded, there exists some real M such that x[sub]n[/sub] ≤ M for all n. By the completeness of R, A = sup{x[sub]n[/sub]: n ∈ N} exists and is real. Given ε > 0, A - ε is not an upper bound for X, so we have some x[sub]k[/sub] such that A - ε < x[sub]k[/sub]. Since X is an increasing sequence, x[sub]k[/sub] ≤ x[sub]n[/sub] if k ≤ n. Then A - ε < x[sub]k[/sub] ≤ x[sub]n[/sub] ≤ A < A + ε, so |x[sub]n[/sub] - A| < ε, and hence lim X = A = sup{x[sub]n[/sub]: n ∈ N}.
10. This is almost like your other thread, where we said that (Q, d), where d is the Euclidean metric, is not complete. Anyway, define {a[sub]n[/sub]} recursively as a[sub]1[/sub] = 1 and a[sub]n+1[/sub] = a[sub]n[/sub]/2 + 1/a[sub]n[/sub]. This is monotone increasing but converges to √2.
11. Trivial. Problem 9 shows that every increasing monotone sequence converges. But every convergent sequence is a Cauchy sequence (this is well-known from analysis, I can prove it if you really want though), so the result follows immediately.
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Brilliant!
Yes, #11 was supposed to contain some sort of hidden trick. I put rational numbers there to try and catch the unwary ones off guard. The trick is that all rational numbers are real numbers, so instead of treating (a[sub]n[/sub]) as just a rational sequence, treat it as a real sequence. Then, though (by #10) it may not converge in
, it certainly will in , by #9. Convergence implies Cauchy; thus (a[sub]n[/sub]) is a Cauchy sequence.Im glad you spotted the trick.
Last edited by JaneFairfax (2007-04-12 16:05:41)
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Last edited by Stanley_Marsh (2007-04-15 08:26:09)
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Oh , forget it ,
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But thats not the question.
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Whats more,
is false. Try a = b = 0, k = 1⁄2.
And
You mean the CauchySchwarz inequality?
Last edited by JaneFairfax (2007-12-18 14:51:06)
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I think I got it
Last edited by Stanley_Marsh (2007-04-15 12:59:56)
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Sorry, I dont get you here.
In the second part, youre also dividing by a, b, and c. Shouldnt you perhaps also consider separate cases where each of them is not 0?
Last edited by JaneFairfax (2007-04-15 21:31:13)
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actually , the first part ( They are all positive)
The first part can be done by
Last edited by Stanley_Marsh (2007-04-16 10:03:07)
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Okay for the first part, but Im still not happy with the division bit in the second part. If youre gonna divide, you should make sure youre not dividing by zero.
As a matter of fact, there is a way to do the second part without doing any kind of division. Try it.
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What about
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Yes, youve got it!!
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#13
Last edited by JaneFairfax (2007-04-25 21:36:08)
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Don't know it will work tho.
If a ,b,c lie in the same straight line , We have a-b=zc , b-c=xa ,c-a=by , Add them together , get xa+yb+zc=0 ,
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I forgot to state that the three vectors a,b,c must be distinct (Ive edited my post and added it now).
Anyway:
(i) You must show that at least one of x, y, z is not 0. I dont think you did that. Did you?
(ii) m(x+y+z) = 0 does not imply (x+y+z) = 0. What if m = 0?
(iii) You must also consider the case where all three points lie in the same vertical line. The equation y = kx+m does not cover vertical lines.
(iv) And dont forget also to prove the converse.
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I have to dicuss those situation separately . Hmmm.....
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#14
Prove that if n is an odd positive integer, there exists a sequence of n consecutive integers whose sum is equal to n[sup]2[/sup].
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