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    (Original post by Lord of the Flies)
    Problem 487***

    f:\;(0,1)\to (0,1) satisfies f(x)<x. Let \hat{f}(x) =\displaystyle \lim_{n\to\infty}\underbrace{f(f  (\cdots f}_n(x)\cdots); can \hat{f} take uncountably many values?
    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.

    Take {1}, and extend it to a Hamel basis for \mathbb{R} / \mathbb{Q}. Remove 1 from this basis and normalise everything so we have an uncountable linearly independent subset of (0,1), call it X.

    For each x \in X define a sequence (x_n) by x_n = \frac{x_{n-1} + x}{2}, x_0=1. Clearly x_n-->x. Linear independence gives us that there sequences never intersect. Define f by f(x_n)=x_{n+1}. 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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    (Original post by Lord of the Flies)
    Problem 487***

    f:\;(0,1)\to (0,1) satisfies f(x)<x. Let \hat{f}(x) =\displaystyle \lim_{n\to\infty}\underbrace{f(f  (\cdots f}_n(x)\cdots); can \hat{f} take uncountably many values?
    Solution 487***

    Let B be a Hamel basis for \mathbb{R} / \mathbb{Q}

    \forall x \in B \exists q_x \in \mathbb{Q}:xq_x \in (1/2,1), 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 X=\{xq_x : x \in B\}

    For each x \in X define a sequence (x_n) by x_n = \frac{x_{n-1} + x}{2}, x_0=1.

    Clearly \lim_{x \to \infty} x_n=x. Linear independence gives us that there sequences never intersect (or we would have a non-trivial linear dependence between elements of X. Define f by f(x_n)=x_{n+1} on these sequences, and f(x)=\frac{x}{2} otherwise. We have that f is well defined here because of *.

    We now have that \hat{f} takes all values in X, 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).
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    (Original post by ThatPerson)
    Problem 483***
     \displaystyle \int^{\pi}_{-\pi} \dfrac{e^{\sin x + \cos x}\cos (\sin x )}{e^{\sin x} + e^{x}} \ dx
    (Original post by james22)
    For 483 I have reduced it to \displaystyle\int^\pi_0 \cos(\sin(x)) e^{\cos(x)}

    but I'm not sure how to solve this one.
    I've just discovered this neat little technique, so please point out any problems with it.

    \begin{aligned} \displaystyle I(b) = \int_0^{\pi}e^{b \cos (x)} \cos(b \sin(x)) \, \mathrm{d}x  &=\frac{1}{2}\int_{-\pi}^{\pi}e^{b \cos (x)} \cos(b \sin (x)) \, \mathrm{d}x  \\ & =\frac{1}{2}\int_{0}^{2\pi}e^{b \cos (x)} \cos(b \sin (x)) \, \mathrm{d}x \\ & =\Re\bigg[\frac{1}{2} \int_0^{2\pi} e^{{be}^{ix}}\bigg] \end{aligned}

    

\displaystyle \begin{aligned} I'(b) = \frac{1}{2} \frac{\mathrm{d} }{\mathrm{d} x} \int_{0}^{2\pi} e^{{be}^{ix}} &= \frac{1}{2} \int_{0}^{2\pi} \frac{\partial }{\partial b} \bigg[e^{{be}^{ix}}\bigg] \mathrm{d}x \\ & =\frac{1}{2} \int_{0}^{2\pi} ibe^{{be}^{ix}} e^{ix} \mathrm{d}x \\ & = \frac{1}{2} \bigg[e^{{be}^{ix}}\bigg]_{0}^{2\pi} = 0

\end{aligned}

    Since our derivative I'(b) is 0, we can conclude that I(b) is a constant w.r.t b. So then:

    I(b) = I(0) = \int_0^{2\pi} \mathrm{d}x = \pi
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    (Original post by Zacken)
    I've just discovered this neat little technique, so please point out any problems with it.

    \begin{aligned} \displaystyle I(b) = \int_0^{\pi}e^{b \cos (x)} \cos(b \sin(x)) \, \mathrm{d}x  &=\frac{1}{2}\int_{-\pi}^{\pi}e^{b \cos (x)} \cos(b \sin (x)) \, \mathrm{d}x  \\ & =\frac{1}{2}\int_{0}^{2\pi}e^{b \cos (x)} \cos(b \sin (x)) \, \mathrm{d}x \\ & =\Re\bigg[\frac{1}{2} \int_0^{2\pi} e^{{be}^{ix}}\bigg] \end{aligned}

    

\displaystyle \begin{aligned} I'(b) = \frac{1}{2} \frac{\mathrm{d} }{\mathrm{d} x} \int_{0}^{2\pi} e^{{be}^{ix}} &= \frac{1}{2} \int_{0}^{2\pi} \frac{\partial }{\partial b} \bigg[e^{{be}^{ix}}\bigg] \mathrm{d}x \\ & =\frac{1}{2} \int_{0}^{2\pi} ibe^{{be}^{ix}} e^{ix} \mathrm{d}x \\ & = \frac{1}{2} \bigg[e^{{be}^{ix}}\bigg]_{0}^{2\pi} = 0

\end{aligned}


    Since our derivative I'(b) is 0, we can conclude that I(b) is a constant w.r.t b. So then:

    I(b) = I(0) = \int_0^{2\pi} \mathrm{d}x = \pi
    I don't see how you changed the limits to [0,2{\pi}]? Your working looks fine to me.
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    (Original post by ThatPerson)
    I don't see how you changed the limits to [0,2{\pi}]? Your working looks fine to me.
    The curve is symmetrical about the y-axis for all real values of b, so the area from -\pi to is the same as the area from to \pi. The function also has period 2\pi. 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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    (Original post by Zacken)
    The curve is symmetrical about the y-axis for all real values of b, so the area from -\pi to is the same as the area from to \pi. The function also has period 2\pi. 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.
    Ah, I suspected it was something to do with the period. That's quite useful
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    Problem 488**


    

\displaystyle \int - \left( \frac{\pi \csc\left(\pi (\frac{-1-ix}{2})\right)}{4}+\frac{\pi \csc\left(\pi (\frac{-1+ix}{2})\right)}{4}\right)\, dx
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    (Original post by Zacken)
    Problem 488**


    

\displaystyle \int - \left( \frac{\pi \csc\left(\pi (\frac{-1-ix}{2})\right)}{4}+\frac{\pi \csc\left(\pi (\frac{-1+ix}{2})\right)}{4}\right)\, dx
    Before I type out the answer in TeX, is the answer

    Spoiler:
    Show
     4\arctan\left(tanh\left(\dfrac{{  \pi}x}{4}\right)\right) + C
    ?
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    (Original post by ThatPerson)
    Before I type out the answer in TeX, is the answer

    Spoiler:
    Show
     4\arctan\left(tanh\left(\dfrac{{  \pi}x}{4}\right)\right) + C
    ?
    Pretty much, you just have an extra factor of 2 in your answer, it should be 2 arctan...

    Post away!
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    (Original post by Zacken)
    Pretty much, you just have an extra factor of 2 in your answer, it should be 2 arctan...

    Post away!
    I managed to misplace a factor of a half in the integrating process

    Spoiler:
    Show

    Firstly, to simplify the expression let
     \displaystyle P = {\pi}\left(\dfrac{-1-ix}{2}\right), Q = {\pi}\left(\dfrac{-1+ix}{2}\right)

    

 \displaystyle -\dfrac{\pi}{4} \int \dfrac{1}{\sin P} + \dfrac{1}{\sin Q} \ dx  = -\dfrac{\pi}{4} \int \dfrac{\sin P + \sin Q}{\sin P \sinQ} \ dx

    =  \displaystyle -\dfrac{\pi}{4} \int \dfrac{-2 \cos\left(-\frac{\pi i x}{2}\right)}{\sin P \sin Q} \ dx

     \displaystyle = \dfrac{\pi}{2} \int \dfrac{ \cos\left(\frac{\pi i x}{2}\right)}{\cosh^{2} \left(\dfrac{\pi x}{2}\right)} \ dx

     \displaystyle = \dfrac{\pi}{2} \int {sech\left(\dfrac{\pi x}{2}\right)} \ dx =  2\arctan\left(\tanh\left(\dfrac{  {\pi}x}{4}\right)\right) + C
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    (Original post by ThatPerson)
    I managed to misplace a factor of a half in the integrating process

    Spoiler:
    Show

    Firstly, to simplify the expression let
     \displaystyle P = {\pi}\left(\dfrac{-1-ix}{2}\right), Q = {\pi}\left(\dfrac{-1+ix}{2}\right)

    

 \displaystyle -\dfrac{\pi}{4} \int \dfrac{1}{\sin P} + \dfrac{1}{\sin Q} \ dx  = -\dfrac{\pi}{4} \int \dfrac{\sin P + \sin Q}{\sin P \sinQ} \ dx

    =  \displaystyle -\dfrac{\pi}{4} \int \dfrac{-2 \cos\left(-\frac{\pi i x}{2}\right)}{\sin P \sin Q} \ dx

     \displaystyle = \dfrac{\pi}{2} \int \dfrac{ \cos\left(\frac{\pi i x}{2}\right)}{\cosh^{2} \left(\dfrac{\pi x}{2}\right)} \ dx

     \displaystyle = \dfrac{\pi}{2} \int {sech\left(\dfrac{\pi x}{2}\right)} \ dx =  2\arctan\left(\tanh\left(\dfrac{  {\pi}x}{4}\right)\right) + C
    Ooh, nice. Slightly different from how I did it, but I like it.
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    (Original post by Zacken)
    Ooh, nice. Slightly different from how I did it, but I like it.
    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:

     \Gamma(z)\Gamma(1-z) = \dfrac{\pi}{\sin (\pi z)}
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    (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:

     \Gamma(z)\Gamma(1-z) = \dfrac{\pi}{\sin (\pi z)}

    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.

    \displaystyle F(n) = (n-1)!

    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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    (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.

    \displaystyle F(n) = (n-1)!

    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!
    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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    (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.
    Thank you! I see. I shall investigate further after a good night of sleep. It's 3 in the morning and I'm too groggy now.
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    Problem 489*/**/*** (more knowledge will give you a stronger answer)

    We say that a function

    f:\mathbb{R} \rightarrow \mathbb{R}

    is periodic with periods S if

    \forall x \in \mathbb{R}, a \in S, f(x+a)=f(x) and \not\exists n \in \mathbb{N}_{>1}, y \in \mathbb{R}:ny=a and y is a period of f, for each period a.

    How large (cardinality wise) can you make S. 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.
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    (Original post by james22)
    Problem 489*/**/*** (more knowledge will give you a stronger answer)

    We say that a function

    f:\mathbb{R} \rightarrow \mathbb{R}

    is periodic with periods S if

    \forall x \in \mathbb{R}, a \in S, f(x+a)=f(x) and \not\exists n \in \mathbb{N}-\{1,-1\}, y \in \mathbb{R}:ny=a and y is a period of f, for each period a.

    How large (cardinality wise) can you make S. 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.
    \mathbb{N}-\{1,-1\}

    Does this denote the set of the naturals not including 1 and -1? I thought that \mathbb{N} was the positive integers anyway?

    Also, does your definition of \mathbb{N} include 0?

    I suspect that I'll think these questions are stupid when I wake up tomorrow.
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    (Original post by ThatPerson)
    \mathbb{N}-\{1,-1\}

    Does this denote the set of the naturals not including 1 and -1? I thought that \mathbb{N} was the positive integers anyway?

    Also, does your definition of \mathbb{N} include 0?

    I suspect that I'll think these questions are stupid when I wake up tomorrow.
    They are good questions. N does not contain 0.
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    \neg(Solution 489)

    Consider the relation on the set of transcendental numbers where a \sim b if and only if a is a rational multiple of b. Clearly such a relation is well-defined as the multiple of any transcendental number by an algebraic number is again a transcendental number.

    It is clear that this relation is reflexive a = 1 \times a, and it is symmetric a = \dfrac{m}{n}b \Rightarrow b = \dfrac{n}{m}a, and furthermore, it is transitive a = \dfrac{m}{n}b, b = \dfrac{p}{q}c \Rightarrow a = \dfrac{mp}{nq}c. Thus the relation \sim 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 T/\sim, none of them are rational multiples of one another, as if \overline{a} = \dfrac{m}{n}\overline{b}, then this means \overline{a} \sim \overline{b}, 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 S can be the size of the continuum.

    S cannot be larger than the continuum, as that would imply that \mathbb{R} is larger than itself.
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    (Original post by 0x2a)
    Solution 489

    Consider the relation on the set of transcendental numbers where a \sim b if and only if a is a rational multiple of b. Clearly such a relation is well-defined as the multiple of any transcendental number by an algebraic number is again a transcendental number.

    It is clear that this relation is reflexive a = 1 \times a, and it is symmetric a = \dfrac{m}{n}b \Rightarrow b = \dfrac{n}{m}a, and furthermore, it is transitive a = \dfrac{m}{n}b, b = \dfrac{p}{q}c \Rightarrow a = \dfrac{mp}{nq}c. Thus the relation \sim 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 T/\sim, none of them are rational multiples of one another, as if \overline{a} = \dfrac{m}{n}\overline{b}, then this means \overline{a} \sim \overline{b}, 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 S can be the size of the continuum.

    S cannot be larger than the continuum, as that would imply that \mathbb{R} is larger than itself.
    You haven't actually given such an f that works yet, but you did the hard bit.
 
 
 
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