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

Stanley_Marsh
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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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Re: Jane’s exercises

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

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

Stanley_Marsh
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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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Re: Jane’s exercises

You’d be surprised – it’s actually much simpler than you think. smile

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. smile

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#30 2007-04-12 09:16:53

Stanley_Marsh
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Re: Jane’s exercises

What about Riemann zeta-function 

 
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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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 10:27:24

JaneFairfax
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Re: Jane’s exercises

Stanley_Marsh wrote:

What about Riemann zeta-function 

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

Yes, that’s fine. In fact, you’ve found a required example. smile

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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#33 2007-04-12 10:42:54

Zhylliolom
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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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#34 2007-04-12 15:58:27

JaneFairfax
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Re: Jane’s exercises

Brilliant! big_smile

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. tongue

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

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#35 2007-04-15 07:58:16

JaneFairfax
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Re: Jane’s exercises

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#36 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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#37 2007-04-15 08:25:44

Stanley_Marsh
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Re: Jane’s exercises

Oh , forget it ,


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#38 2007-04-15 08:39:54

JaneFairfax
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Re: Jane’s exercises

But that’s not the question. wink

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#39 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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#40 2007-04-15 12:51:25

Stanley_Marsh
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Re: Jane’s exercises

I think I got it

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


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#41 2007-04-15 21:30:39

JaneFairfax
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Re: Jane’s exercises

Stanley_Marsh wrote:

Sorry, I don’t get you here. dunno

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? smile

Last edited by JaneFairfax (2007-04-15 21:31:13)

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#42 2007-04-16 09:54:01

Stanley_Marsh
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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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#43 2007-04-16 10:04:11

JaneFairfax
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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. smile

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#44 2007-04-16 10:10:58

Stanley_Marsh
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Re: Jane’s exercises

What about


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#45 2007-04-16 10:19:34

JaneFairfax
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Re: Jane’s exercises

Yes, you’ve got it!!

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#46 2007-04-25 09:00:24

JaneFairfax
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Re: Jane’s exercises

#13

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

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#47 2007-04-25 12:15:42

Stanley_Marsh
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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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#48 2007-04-25 22:07:06

JaneFairfax
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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. tongue

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#49 2007-04-27 03:30:06

Stanley_Marsh
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Re: Jane’s exercises

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


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#50 2007-04-27 19:25:57

JaneFairfax
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Re: Jane’s exercises

#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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