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## #26 2007-04-11 19:09:07

Stanley_Marsh
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Registered: 2006-12-13
Posts: 345

### Re: Janes exercises

To 10 :

I think  it converge to

Last edited by Stanley_Marsh (2007-04-11 19:22:40)

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## #27 2007-04-11 20:40:20

JaneFairfax
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Registered: 2007-02-23
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### Re: Janes exercises

The a[sub]n[/sub] have to be rational numbers. Are they?

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## #28 2007-04-12 02:44:16

Stanley_Marsh
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Posts: 345

### Re: Janes exercises

Awwwww , Let me come up with another one.

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## #29 2007-04-12 04:07:16

JaneFairfax
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Registered: 2007-02-23
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### Re: Janes exercises

Youd be surprised  its actually much simpler than you think.

And dont be afraid to try #9 and #11  they really arent 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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## #30 2007-04-12 09:16:53

Stanley_Marsh
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Posts: 345

### Re: Janes exercises

I happened to see the Riemann zeta-function today~lol

Last edited by Stanley_Marsh (2007-04-12 09:17:10)

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## #31 2007-04-12 09:18:57

Stanley_Marsh
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Registered: 2006-12-13
Posts: 345

### Re: Janes exercises

9 ,11 are hard for me , I haven't that much knowledge of math , I just learn math randomly by myself .haha

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## #32 2007-04-12 09:51:34

kylekatarn
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Registered: 2005-07-24
Posts: 445

### Re: Janes exercises

Zhylliolom wrote:

but a real decent calculator would notify you something like "Warning: 0^0 replaced by 1" .

... like a ti92+ or voyage 200 ; )

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## #33 2007-04-12 10:27:24

JaneFairfax
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Registered: 2007-02-23
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### Re: Janes exercises

Stanley_Marsh wrote:

I happened to see the Riemann zeta-function today~lol

Yes, thats fine. In fact, youve 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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## #34 2007-04-12 10:42:54

Zhylliolom
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Registered: 2005-09-05
Posts: 412

### Re: Janes exercises

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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## #35 2007-04-12 15:58:27

JaneFairfax
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Registered: 2007-02-23
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### Re: Janes exercises

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.

Im glad you spotted the trick.

Last edited by JaneFairfax (2007-04-12 16:05:41)

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## #36 2007-04-15 07:58:16

JaneFairfax
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## #37 2007-04-15 08:18:24

Stanley_Marsh
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### Re: Janes exercises

Last edited by Stanley_Marsh (2007-04-15 08:26:09)

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## #38 2007-04-15 08:25:44

Stanley_Marsh
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### Re: Janes exercises

Oh , forget it ,

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## #39 2007-04-15 08:39:54

JaneFairfax
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### Re: Janes exercises

But thats not the question.

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## #40 2007-04-15 08:50:26

JaneFairfax
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### Re: Janes exercises

Whats more,

is false. Try a = b = 0, k = 1⁄2.

And

You mean the CauchySchwarz inequality?

Last edited by JaneFairfax (2007-12-18 14:51:06)

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## #41 2007-04-15 12:51:25

Stanley_Marsh
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Registered: 2006-12-13
Posts: 345

### Re: Janes exercises

I think I got it

Last edited by Stanley_Marsh (2007-04-15 12:59:56)

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## #42 2007-04-15 21:30:39

JaneFairfax
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Registered: 2007-02-23
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### Re: Janes exercises

Stanley_Marsh wrote:

Sorry, I dont get you here.

In the second part, youre also dividing by a, b, and c. Shouldnt 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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## #43 2007-04-16 09:54:01

Stanley_Marsh
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Registered: 2006-12-13
Posts: 345

### Re: Janes exercises

actually , the first part  ( They are all positive)

The first part can be done by

U can also think that a,b,c are distinct ,

Last edited by Stanley_Marsh (2007-04-16 10:03:07)

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## #44 2007-04-16 10:04:11

JaneFairfax
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Registered: 2007-02-23
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### Re: Janes exercises

Okay for the first part, but Im still not happy with the division bit in the second part. If youre gonna divide, you should make sure youre 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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## #45 2007-04-16 10:10:58

Stanley_Marsh
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### Re: Janes exercises

Numbers are the essence of the Universe

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## #46 2007-04-16 10:19:34

JaneFairfax
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### Re: Janes exercises

Yes, youve got it!!

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## #47 2007-04-25 09:00:24

JaneFairfax
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Registered: 2007-02-23
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### Re: Janes exercises

#13

Last edited by JaneFairfax (2007-04-25 21:36:08)

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## #48 2007-04-25 12:15:42

Stanley_Marsh
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Registered: 2006-12-13
Posts: 345

### Re: Janes exercises

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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## #49 2007-04-25 22:07:06

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

### Re: Janes exercises

I forgot to state that the three vectors a,b,c must be distinct (Ive edited my post and added it now).

Anyway:

(i) You must show that at least one of x, y, z is not 0. I dont 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 dont forget also to prove the converse.

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## #50 2007-04-27 03:30:06

Stanley_Marsh
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Registered: 2006-12-13
Posts: 345

### Re: Janes exercises

I have to dicuss those situation separately . Hmmm.....

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