This is favourite of mine. A lot of you will know it, so please don't give the answer away if you've seen it before
Problem 43 **/***
Evaluate
(To do it rigorously you obviously need analysis, but don't bother if you've done no Analysis)

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 12042013 16:03

und
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 12042013 17:34
(Original post by shamika)
This is favourite of mine. A lot of you will know it, so please don't give the answer away if you've seen it before
Problem 43 **/***
Evaluate
(To do it rigorously you obviously need analysis, but don't bother if you've done no Analysis) 
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 12042013 17:38
(Original post by und)
My guess is that it's 1 if x is rational and 0 otherwise. 
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 12042013 17:42
(Original post by shamika)
Indeed! Can you explain why? 
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 12042013 17:42
Let r = 90:
Let r = 89:
As such, we are summing:
In this sum, all terms except for two will cancel out:
We know that:
Let r = 0:
So cos1 = 1  sin1, so our sum is , as required.Last edited by DJMayes; 12042013 at 17:44. 
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 12042013 17:46
(Original post by DJMayes)
Solution 42:
Let r = 90:
Let r = 89:
As such, we are summing:
In this sum, all terms except for two will cancel out:
We know that:
Let r = 0:
So cos1 = 1  sin1, so our sum is , as required.
cos(1)=0.9998476952
1sin(1)=0.9825475936.
So I got confused an gave up. I checked the first bit out numerically too and that was dodgy too. 
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 12042013 17:50
(Original post by bananarama2)
See that's what I did except.
cos(1)=0.9998476952
1sin(1)=0.9825475936.
So I got confused an gave up. I checked the first bit out numerically too and that was dodgy too. 
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 12042013 17:53
(Original post by DJMayes)
I know. I don't like it; and wouldn't have submitted the solution if it weren't for the partial fractions bit saying that it's apparently true. 
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 12042013 17:53
(Original post by und)
The only thing I'm not sure about is the order of the limits. It feels like they should be the other way round, although perhaps it doesn't matter? 
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 12042013 17:56
(Original post by bananarama2)
I think the problem is with the partial fractions bit. You're finding A and B only for the cases where r= 90 and 89. It isn't an identity. 
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 12042013 17:56
(Original post by und)
The only thing I'm not sure about is the order of the limits. It feels like they should be the other way round, although perhaps it doesn't matter?Spoiler:ShowWere you thinking the bit in the brackets tended to an integer multiple of pi? 
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 12042013 17:59
Solution 43
Lemma: If is not rational then is not an integer for any . We will prove this by contraposition. If is an integer then , hence is rational. Therefore if is not rational then is not an integer.
We consider the two cases: is rational; is not rational.
If is rational, then we can write for some with , and hence , which is clearly an integer when . Since in the limit, we can make arbitrarily large, so is an integer multiple of and hence for all .
If is irrational, then from the lemma is not an integer multiple of , and hence .Last edited by und; 12042013 at 20:43. 
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 12042013 18:01
Solution 41
For black color we write , for white  0.
By we denote the set of all the students and .
Let be the set of all maps (colorings) from to .
For each student from , we say if the set is finite. Since is an equivalence relation on , by AC there exists map , such that . Also leads to .
Define to be a guessing function for the student under the coloring .
Since each student knows , we define . Consequently, all but finitely many students will guess correctly.
Let me try something different.
 the set of all the students,  the set of colors, in our case . We consider again the set of all mappings . For each student , define the equivalence relation on by if and only if the colorings and are not distinguishable for the student . Let denote the equivalence class of the coloring under the relation .
Now, let be a wellordering of (AC). Define to be the guessing function such that is the least element of .
Let .
Here, I am left to define a ''good'' binary relation over . I mean, if the set has no infinite chains, then is "good".Last edited by Mladenov; 12042013 at 18:08. 
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 12042013 18:10
(Original post by ukdragon37)
Problem 41***
A (countably) infinite class of Cambridge students who believe in the Axiom of Choice are taking their finals for the Mathematical Tripos. At the start of the exam, the examiner places either a white or a black hat on each student at random. There is only one question in the exam: "What colour is your hat?" If only a finite number of students answers incorrectly then everyone becomes a Wrangler (very good), otherwise the Wooden Spoon is given to the whole class (very bad). Everyone can see the hats of everyone else besides their own, and since it's exam conditions they are not allowed to communicate. Students may not remove their hats or try to look at it in any way.
What strategy could the students devise together so to avoid failure and the impending doom of unemployment? Being clever clogs from Cambridge, memorising infinitely large amounts of data is no problem to the students.
Note: A student suggested that each person just guess at random. However this has already been ruled out by the cleverer students who realised that if they were really unlucky, infinite of them could guess wrong.
to the ttheta to the x has planck quantity root g minus m plus l minus 2q
plus t hbar to the m plus l minus q plus t minus xe to the minus m minus l minus 7q minus 3t plus 6xk to the minus 2x epsilon 0 to the q. 
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 12042013 18:12
(Original post by bananarama2)
I've got it. It you plot lateral binary graph associated with the symbolic mathematical solutions, you can reduced the problem down to simple SNF's. And by using Dalek's contraction algorithm the answer is: the physical quantity of dimensions m to the ml to the lq to the qt
to the ttheta to the x has planck quantity root g minus m plus l minus 2q
plus t hbar to the m plus l minus q plus t minus xe to the minus m minus l minus 7q minus 3t plus 6xk to the minus 2x epsilon 0 to the q.
http://youtu.be/lTDcYi_uG08?t=9s
Also, The proof is trivial! Just view the problem as a Noetherian complexity class whose elements are countable logistic systems.Last edited by und; 12042013 at 18:15. 
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 12042013 18:18
(Original post by und)
This is what I was thinking so I'm guessing it's wrong!
Spoiler:ShowLemma: If is not rational then is not an integer for any . We will prove this by contraposition. If is an integer then , hence is rational. Therefore if is not rational then is not an integer.
We consider the two cases: is rational; is not rational.
If is rational, then we can write for some with , and hence , which is clearly an integer when . So in the limit , is an integer multiple of and hence for all .
If is irrational, then from the lemma is not an integer multiple of , and hence . 
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 12042013 18:20
(Original post by shamika)
You're almost there. You don't need to consider the case where m < q. Why? This is subtle so don't worry if you're getting confused 
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 12042013 18:25
(Original post by shamika)
You're almost there. You don't need to consider the case where m < q. Why? This is subtle so don't worry if you're getting confusedLast edited by und; 12042013 at 18:40. 
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 12042013 18:40
(Original post by und)
This is why I said I found the order of the limits confusing. In the case m<q, the first limit is 0, but then in the second limit we have m>∞. I would then say that for m<q, the limit is 0, but you told me that it was 1. Hmm...
(Remember that in the first limit, m is arbitrary; the limit is with respect to n.
Also note that the limits have to be this way around, because otherwise it causes issues when you have irrational x)Last edited by shamika; 12042013 at 18:44. 
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 12042013 18:40
(Original post by DJMayes)
I dunno. We'll probably have to find out what Felix was after; as the solution isn't an agreeable one.
Felix's solution is correct, though I'm trying to think of a clever method to show that it is
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