It's 3 in the morning and this partial solution involves the axiom of choice, so it's probably wrong but I'll give it a go.(Original post by Lord of the Flies)
Problem 487***
satisfies . Let can take uncountably many values?
Take {1}, and extend it to a Hamel basis for . Remove 1 from this basis and normalise everything so we have an uncountable linearly independent subset of , call it .
For each define a sequence by . Clearly . Linear independence gives us that there sequences never intersect. Define by . Conditions are clearly fufilled for the points where f is defined, just need to deal with the other values of f, including what f does to X itself. This may not be possible, and I need to go to bed right now.
EDIT: I'm fairly sure this is possible, but it looks like it requires choice again.
EDIT2: Doesn't letting f(x)=x/2 on all other values work, with f being well defined due to linear independence?
EDIT3: I think I have a slightly cleaner solution which i will post below.

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 06012015 04:37
Last edited by james22; 06012015 at 15:32. 
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 06012015 15:38
(Original post by Lord of the Flies)
Problem 487***
satisfies . Let can take uncountably many values?
Let be a Hamel basis for
, this is obvious but fiddly to prove, and nothing in the proof of this is interesting so i won't post it. Note that we are mapping in (1/2,1) not (0,1), important later*.
Let
For each define a sequence by .
Clearly . Linear independence gives us that there sequences never intersect (or we would have a nontrivial linear dependence between elements of . Define by on these sequences, and otherwise. We have that f is well defined here because of *.
We now have that takes all values in , which is uncountable (it isn't trivial to prove that there is no countable basis, but I think I can assume it in this question).Last edited by james22; 06012015 at 15:48. 
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 07012015 17:34
Since our derivative is 0, we can conclude that is a constant w.r.t b. So then:
Last edited by Zacken; 07012015 at 17:35. 
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 07012015 17:45
(Original post by Zacken)
I've just discovered this neat little technique, so please point out any problems with it.
Since our derivative is 0, we can conclude that is a constant w.r.t b. So then:

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 07012015 18:03
(Original post by ThatPerson)
I don't see how you changed the limits to ? Your working looks fine to me. 
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 07012015 18:12
(Original post by Zacken)
The curve is symmetrical about the yaxis for all real values of b, so the area from to is the same as the area from to . The function also has period . So I integrated from 0 to 2pi instead. I saw the method being used on a similar integral with the same properties, so I tried it out here. Worked fine, still coming to grips with it though. 
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 09012015 19:02
?Last edited by ThatPerson; 09012015 at 19:05. 
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 09012015 19:44
Post away! 
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 09012015 20:22
(Original post by Zacken)
Pretty much, you just have an extra factor of 2 in your answer, it should be 2 arctan...
Post away!
Last edited by ThatPerson; 09012015 at 20:23. 
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 09012015 21:11
(Original post by Zacken)
Ooh, nice. Slightly different from how I did it, but I like it.
If you haven't heard of the formula before, it is:

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 09012015 23:46
(Original post by ThatPerson)
It's an interesting question, especially as the expression relates to Euler's reflection formula. Which admittedly was the first thing I thought of when I saw the question, but I couldn't see the gamma function approach. I'm sure that using hyperbolic functions is much simpler, but I'd be interested to see if there was a way to do this purely via the Gamma function.
If you haven't heard of the formula before, it is:
Ooh, that looks gorgeous! I've seen plenty of Gamma functions on this thread, never got around to learning about them though. I've picked up on the relation between them and the factorial function.
Where'd you learn about them, anywhere you could point me to specifically? Especially examples of using them. I'd like to learn a bit more 'bout it. How to use it, blah blah. Thanks! 
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 10012015 00:14
(Original post by Zacken)
Ooh, that looks gorgeous! I've seen plenty of Gamma functions on this thread, never got around to learning about them though. I've picked up on the relation between them and the factorial function.
Where'd you learn about them, anywhere you could point me to specifically? Especially examples of using them. I'd like to learn a bit more 'bout it. How to use it, blah blah. Thanks! 
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 10012015 00:17
(Original post by ThatPerson)
Nowhere specific, I just read a lot on Wikipedia, pdfs on Google, and Stackexchange. The latter is particularly good to find brilliant applications of special functions, though be aware that a lot of the material discussed is quite advanced, so you may not understand substantial portions of some solutions (I know I don't at least ). The Gamma function is also related to the Beta function and Riemann Zeta Function. 
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 10012015 00:35
Problem 489*/**/*** (more knowledge will give you a stronger answer)
We say that a function
is periodic with periods if
and and y is a period of f, for each period .
How large (cardinality wise) can you make . Can you prove that it cannot be any larger?
An example to see if it makes things clearer, f(x)=sin(x) has S={pi,pi}. You could not say that S={3pi,4pi} as these are multiples of a smaller period.
Please let me know if the above is not clear, this question is easy to visualise but not so easy to write down.Last edited by james22; 10012015 at 01:28. 
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 10012015 01:17
(Original post by james22)
Problem 489*/**/*** (more knowledge will give you a stronger answer)
We say that a function
is periodic with periods if
and and y is a period of f, for each period .
How large (cardinality wise) can you make . Can you prove that it cannot be any larger?
An example to see if it makes things clearer, f(x)=sin(x) has S={pi,pi}. You could not say that S={3pi,4pi} as these are multiples of a smaller period.
Please let me know if the above is not clear, this question is easy to visualise but not so easy to write down.
Does this denote the set of the naturals not including 1 and 1? I thought that was the positive integers anyway?
Also, does your definition of include 0?
I suspect that I'll think these questions are stupid when I wake up tomorrow.Last edited by ThatPerson; 10012015 at 01:24. 
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 10012015 01:26
(Original post by ThatPerson)
Does this denote the set of the naturals not including 1 and 1? I thought that was the positive integers anyway?
Also, does your definition of include 0?
I suspect that I'll think these questions are stupid when I wake up tomorrow. 
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 10012015 17:23
(Solution 489)
Consider the relation on the set of transcendental numbers where if and only if is a rational multiple of . Clearly such a relation is welldefined as the multiple of any transcendental number by an algebraic number is again a transcendental number.
It is clear that this relation is reflexive , and it is symmetric , and furthermore, it is transitive . Thus the relation is an equivalence relation and partitions the set of transcendental numbers into disjoint sets.
Now the cardinality of each partition is countable due to the countability of the rationals, thus the number of partitions must be uncountable so as to satisfy the requirement that the cardinality of the set of transcendental numbers is uncountable.
Now if you consider the elements of the set , none of them are rational multiples of one another, as if , then this means , which is a contradiction as the partitions are disjoint.
Thus since none of the partitions are rational multiples of one another, they are definitely not integer multiples of one another, and thus we have shown that can be the size of the continuum.
cannot be larger than the continuum, as that would imply that is larger than itself.Last edited by 0x2a; 10012015 at 18:22. 
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 10012015 17:28
(Original post by 0x2a)
Solution 489
Consider the relation on the set of transcendental numbers where if and only if is a rational multiple of . Clearly such a relation is welldefined as the multiple of any transcendental number by an algebraic number is again a transcendental number.
It is clear that this relation is reflexive , and it is symmetric , and furthermore, it is transitive . Thus the relation is an equivalence relation and partitions the set of transcendental numbers into disjoint sets.
Now the cardinality of each partition is countable due to the countability of the rationals, thus the number of partitions must be uncountable so as to satisfy the requirement that the cardinality of the set of transcendental numbers is uncountable.
Now if you consider the elements of the set , none of them are rational multiples of one another, as if , then this means , which is a contradiction as the partitions are disjoint.
Thus since none of the partitions are rational multiples of one another, they are definitely not integer multiples of one another, and thus we have shown that can be the size of the continuum.
cannot be larger than the continuum, as that would imply that is larger than itself.
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