That's kind of the point  it's to demonstrate that the expression is meaningless. That's the context it was from originally(Original post by Jkn)
Spoiler:ShowNot to be a ****, but surely you can both see that all of this is utterly wrong?
If you are intending it as a 'joke problem' of the sort where the object is to spot the error (e.g. prove that 1=2), then please make sure that this is clear in the statement of the problem.

Smaug123
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 06082013 20:55

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 06082013 21:18
Problem 304*
Does there exist an integer, k, such that using only at most k coin flips, a number in the set {1,2,3} can be randomly picked with each being equally likely? 
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 06082013 22:21
(Original post by Jkn)
I hadn't actually tried this problem when I posted it (as it was a 'problem 6' from the Putnam exam and I was hunting around for things that were going to challenge people), but I've given it a go and just managed to solve it! Really enjoyable so I hope people take a crack at it! 
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 06082013 22:27
(Original post by james22)
Problem 304*
Does there exist an integer, k, such that using only at most k coin flips, a number in the set {1,2,3} can be randomly picked with each being equally likely? 
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 06082013 22:30
(Original post by Jkn)
Spoiler:ShowNot to be a ****, but surely you can both see that all of this is utterly wrong?
If you are intending it as a 'joke problem' of the sort where the object is to spot the error (e.g. prove that 1=2), then please make sure that this is clear in the statement of the problem.
It's as Smaug says. 
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 06082013 22:35
(Original post by Smaug123)
That's kind of the point  it's to demonstrate that the expression is meaningless. That's the context it was from originally
Also, showing that it is greater than 2 does not prove that it is indeterminate. We need to theorems that rely on it having been established that a limit exists and so all that needs to be done is to show that the utilization of such theorems can lead to more than one value for (a contradiction).
Also, they have not even proven that a solution greater than 2 exists as they are supposing that
which, even if the final limit exists, is invalid as does not exists and so the rule cannot be applied. 
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 06082013 22:47
(Original post by Smaug123)
Hmm  I've proved a much weaker result than what I need, but it's an interesting problem. I'll sleep on it
Yes it certainly is! I did half an hour yesterday, went round in circles, and then tried it again this morning and got it in just over half an hour!
My solution is really elegant (in my opinion) but in my searching online I have found that the most commonly found solution was very long and very (very!) insane. One guy had an similarl method to mine though but I am unable to find any officially solutions (yet)
Spoiler:ShowBut please don't do any more bad maths, I'm not sure I can take it!
Had a go at the awesome sequence question yet? I have the feeling it is the kind of problem you would demolish (so long as you have the proper toolbox of techniques at your disposal)! 
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 06082013 22:54
(Original post by Smaug123)
It's easy to get "arbitrarily likely to have made a random pick" if we get k high enough… after nine flips, I can have a 1/256 chance of not having made a decision, for instance, while after 27 flips, I can have a chance of not having made a decision. I'm still thinking about ways to improve this. 
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 06082013 23:34
(Original post by james22)
Problem 304*
Does there exist an integer, k, such that using only at most k coin flips, a number in the set {1,2,3} can be randomly picked with each being equally likely?
There exists no such integer k.
The problem is essentially equivalent to constructing a probability of out of any linear combination of probabilities of the form where n is an integer. Here we are essentially considering the construction of one such value of as this is a necessary condition (and disproof of any such condition is sufficient for contradiction).
We suppose that which is a necessary condition (though not sufficient as we choose not to place constraints on besides from the obvious one that it is a positive integer). This is also implied by the existence of a value of an integer k as .
The impossibility of this construction can be proven in a number of ways. The most simple way is to prove this is by noting that 2 and 3 are coprime and so no fraction on the right hand side could be constructed with a denominator of 3. To bring mathematical justification to this statement we multiply through by :
which is a contradiction as the left hand side is not an integer as 2 and 3 being coprime implies that and 3 are coprime and the right hand side is obviously as integer as n is finite.
This is implies that no such linear combination exists and, hence, that no integer k exists 
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 06082013 23:38
(Original post by Jkn)
What weaker result have you proven? The special case in which k=1?Spoiler:ShowFor all integer k, we have that the sequence is strictly decreasing; it is bounded below by 0, and hence converges. So the limit exists. (Argument: suppose it's not strictly decreasing for some n, then make the inequality and do a binomial expansion, using that is increasing, which is evident from the definition, being a sum of two positive things.)
Depressingly little progress, I know
ETA: Also, the starting value a_0 shouldn't matter, because a_n is unbounded so eventually we'll exceed wherever we started. A little thing, but it might help!Last edited by Smaug123; 07082013 at 00:16. Reason: ETA 
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 07082013 00:54
(Original post by Smaug123)
Spoiler:ShowFor all integer k, we have that the sequence is strictly decreasing; it is bounded below by 0, and hence converges. So the limit exists. (Argument: suppose it's not strictly decreasing for some n, then make the inequality and do a binomial expansion, using that is increasing, which is evident from the definition, being a sum of two positive things.)
Depressingly little progress, I know
ETA: Also, the starting value a_0 shouldn't matter, because a_n is unbounded so eventually we'll exceed wherever we started. A little thing, but it might help!
They need to be greater than zero in order to restrict to the set of real numbers.
Here's some problems I just shared on ASOM (though I would pop them on here too as, if people are going to be writing solutions without extensive discussion, this is certainly the place for it). The first 4 are from BMO2 and the 5th is from Putnam (though it is a 'problem 1').
Problem 305*
Prove that the sequence defined by , for consists only of integers.
[As a bonus, can you find any 'similar' sequences for which this is true?]
Problem 306*
Find all solutions in nonnegative integers a,b to
[As a bonus, why not generalise this in the obvious way? Giving all solutions a,b,c dependent on either c having a certain property or c simply being any integer]
Problem 307*
Find the minimum possible value of where x,y are real numbers such that and .
Problem 308*
Find all pairs of integers (x,y) such that
Problem 309*/**/***
Find the volume of the region of points (x,y,z) such that
Time to see who has been paying attention to this thread (if you have been you will do this in minutes)! This one is a 'problem 4' from Putnam:
Problem 310**
Evaluate
Some easier ones. Please do not post a solution if they do not interest/challenge you  though whilst they are mostly easier than those above, they are still by no means 'trivial' or 'easy' and would still be considered difficult by the standards of numerous institutions/exams. Also, do not be discouraged if you cannot do them as problemsolving takes practice and is certainly something that has to be learnt in almost all cases:
Problem 311*/**/*** (the one* solution does exist and is not hard to find  please don't moan on about it!)
Let , and for .
Show that the sequence convergences as and determine its limit.
Problem 312*
Evaluate
Problem 313*
Find all real values of x,y and z such that
[As a bonus, can you solve the problem with a more general set of constraints?]
Problem 314*
Find all positive integers n such that divides and divides .
This is the kind of problem you could find really easy or rather difficult:
Problem 315**/***
Determine, with proof, whether the series converges.Last edited by Jkn; 07082013 at 10:17. 
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 07082013 12:02
(Original post by Jkn)
Ah good!
Spoiler:ShowBut please don't do any more bad maths, I'm not sure I can take it!
Had a go at the awesome sequence question yet? I have the feeling it is the kind of problem you would demolish (so long as you have the proper toolbox of techniques at your disposal)!
Is that problem 161? Looks like a mess, but I might take a punt, right now I'm having a go at some of the others that you just posted
(Original post by Jkn)
Problem 306*
Find all solutions in nonnegative integers a,b to
[As a bonus, why not generalise this in the obvious way? Giving all solutions a,b,c dependent on either c having a certain property or c simply being any integer]
Note that:
The obvious solutions are therefore:
To demonstrate that these are the only solutions with the given restrictions, consider:
Solving for a:
Observing that:
For this to be an integer, observe that must hold.
Therefore: for some integer n
Now we can solve similarly for b and state that:
for some integer m
So we find that: are of the form
Whereby the previously found solutions are the only solutions.
Consider the general case:
If for some integers and
Using the same method as above we can deduce that:
Such that:
If is not of the above form then for integer solutions either orLast edited by joostan; 07082013 at 13:45. Reason: General Case 
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 07082013 12:25
(Original post by joostan)
OK, I won't.
Is that problem 161? Looks like a mess, but I might take a punt, right now I'm having a go at some of the others that you just posted
Solution 306: 
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 07082013 12:44
Posted from TSR Mobile 
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 07082013 12:50
(Original post by Smaug123)
Damn, you got there before me  it generalises in the obvious way, of course, if you just replace c with p^2 q, but I can't be bothered to look at p^2 q r, etc :P
Yeah, you can do the generalisation, I don't really know what he's after.
Unless he wanted:
In which case:
which I suspect can be shown after a bit of algebra. 
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 07082013 12:57
(Original post by Jkn)
Yeah but I get that horrible sick feeling in the pit of my stomach whenever I see incorrect mathematical reasoning  it reminds me of every Physics lesson I've ever been in!
Also, showing that it is greater than 2 does not prove that it is indeterminate. We need to theorems that rely on it having been established that a limit exists and so all that needs to be done is to show that the utilization of such theorems can lead to more than one value for (a contradiction).
Also, they have not even proven that a solution greater than 2 exists as they are supposing that
which, even if the final limit exists, is invalid as does not exists and so the rule cannot be applied.
So is an indeterminate form. I don't know why you think you need to prove the limit exists; that's contradictory to it being an indeterminate form in the first place. You do need to show that the limits exist of the functions that you use to make a contradiction about the value, which we shan't bother doing since one's trivial (constant function) and the others a standard result. We did not suppose your second paragraph, either .Last edited by FireGarden; 07082013 at 12:59. 
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 07082013 13:37
(Original post by joostan)
Ninjaed
Yeah, you can do the generalisation, I don't really know what he's after.
Unless he wanted:
In which case:
which I suspect can be shown after a bit of algebra. 
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 07082013 13:49
(Original post by Smaug123)
Yup, it's the same algebra, too 
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 07082013 15:21
(Original post by Jkn)
What weaker result have you proven? The special case in which k=1?
k=1 case: I used Mathematica to get a couple of hundred iterations, I just couldn't intuit the limit… then it was easy enough to show by induction that always, after getting a recurrence relation for . Also shown previously that is strictly decreasing. Now I just need to show that 2 is indeed the limit. 
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 07082013 16:38
(Original post by Jkn)
g
Problem 307*
Find the minimum possible value of where x,y are real numbers such that and .
[B]+y^2)
Problem 310**
Evaluate
Let . Then, .
Now, we have to minimize , essentially.
Solution 310
Now, note that:
Last edited by MW24595; 07082013 at 16:41.
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