Math Is Fun Forum / Jane’s exercises

Stanley_Marsh · 2007-04-12 17:09:07

To 10 :

I think it converge to

Last edited by Stanley_Marsh (2007-04-12 17:22:40)

JaneFairfax · 2007-04-12 18:40:20

The a_n have to be rational numbers. Are they?

Stanley_Marsh · 2007-04-13 00:44:16

Awwwww , Let me come up with another one.

JaneFairfax · 2007-04-13 02:07:16

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.

Stanley_Marsh · 2007-04-13 07:16:53

What about Riemann zeta-function

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

Last edited by Stanley_Marsh (2007-04-13 07:17:10)

Stanley_Marsh · 2007-04-13 07:18:57

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

kylekatarn · 2007-04-13 07:51:34

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

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

JaneFairfax · 2007-04-13 08:27:24

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.

A simpler one would be

The sequence is bounded above (e.g. 2 is an upper bound), it is increasing, and the a_n are rational – but

is not rational.

Last edited by JaneFairfax (2007-04-13 14:12:50)

Zhylliolom · 2007-04-13 08:42:54

My TI-89 does it... I know a guy with a 92, the thing is huge.

9. Let X = {x_n} be a bounded monotone sequence. Since X is bounded, there exists some real M such that x_n ≤ M for all n. By the completeness of R, A = sup{x_n: n ∈ N} exists and is real. Given ε > 0, A - ε is not an upper bound for X, so we have some x_k such that A - ε < x_k. Since X is an increasing sequence, x_k ≤ x_n if k ≤ n. Then A - ε < x_k ≤ x_n ≤ A < A + ε, so |x_n - A| < ε, and hence lim X = A = sup{x_n: 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_n} recursively as a₁ = 1 and a_n+1 = a_n/2 + 1/a_n. 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.

JaneFairfax · 2007-04-13 13:58:27

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_n) 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_n) is a Cauchy sequence.

I’m glad you spotted the trick.

Last edited by JaneFairfax (2007-04-13 14:05:41)

JaneFairfax · 2007-04-16 05:58:16

Stanley_Marsh · 2007-04-16 06:18:24

Last edited by Stanley_Marsh (2007-04-16 06:26:09)

Stanley_Marsh · 2007-04-16 06:25:44

Oh , forget it ,

JaneFairfax · 2007-04-16 06:39:54

But that’s not the question.

JaneFairfax · 2007-04-16 06:50:26

What’s more,

is false. Try a = b = 0, k = 1⁄2.

And

You mean the Cauchy–Schwarz inequality?

Last edited by JaneFairfax (2007-12-19 13:51:06)

Stanley_Marsh · 2007-04-16 10:51:25

I think I got it

Last edited by Stanley_Marsh (2007-04-16 10:59:56)

JaneFairfax · 2007-04-16 19:30:39

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 (2007-04-16 19:31:13)

Stanley_Marsh · 2007-04-17 07:54:01

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 (2007-04-17 08:03:07)

JaneFairfax · 2007-04-17 08:04:11

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.

Stanley_Marsh · 2007-04-17 08:10:58

What about

JaneFairfax · 2007-04-17 08:19:34

Yes, you’ve got it!!

JaneFairfax · 2007-04-26 07:00:24

#13

Last edited by JaneFairfax (2007-04-26 19:36:08)

Stanley_Marsh · 2007-04-26 10:15:42

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 ,

JaneFairfax · 2007-04-26 20:07:06

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.

Stanley_Marsh · 2007-04-28 01:30:06

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

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#26 2007-04-12 17:09:07

Re: Jane’s exercises

#27 2007-04-12 18:40:20

Re: Jane’s exercises

#28 2007-04-13 00:44:16

Re: Jane’s exercises

#29 2007-04-13 02:07:16

Re: Jane’s exercises

#30 2007-04-13 07:16:53

Re: Jane’s exercises

#31 2007-04-13 07:18:57

Re: Jane’s exercises

#32 2007-04-13 07:51:34

Re: Jane’s exercises

Zhylliolom wrote:

#33 2007-04-13 08:27:24

Re: Jane’s exercises

Stanley_Marsh wrote:

#34 2007-04-13 08:42:54

Re: Jane’s exercises

#35 2007-04-13 13:58:27

Re: Jane’s exercises

#36 2007-04-16 05:58:16

Re: Jane’s exercises

#37 2007-04-16 06:18:24

Re: Jane’s exercises

#38 2007-04-16 06:25:44

Re: Jane’s exercises

#39 2007-04-16 06:39:54

Re: Jane’s exercises

#40 2007-04-16 06:50:26

Re: Jane’s exercises

#41 2007-04-16 10:51:25

Re: Jane’s exercises

#42 2007-04-16 19:30:39

Re: Jane’s exercises

Stanley_Marsh wrote:

#43 2007-04-17 07:54:01

Re: Jane’s exercises

#44 2007-04-17 08:04:11

Re: Jane’s exercises

#45 2007-04-17 08:10:58

Re: Jane’s exercises

#46 2007-04-17 08:19:34

Re: Jane’s exercises

#47 2007-04-26 07:00:24

Re: Jane’s exercises

#48 2007-04-26 10:15:42

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#49 2007-04-26 20:07:06

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#50 2007-04-28 01:30:06

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