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    (Original post by TheMagicMan)
    I've posted this problem before (not on this thread)

    Problem 242 *
    This has been posted already
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    (Original post by jack.hadamard)
    Problem 241 * / **

    i) Find \displaystyle \sin \left(\frac{\pi}{5}\right).

    ii) Prove that \displaystyle \pi \phi > 5, where \phi is the golden ratio.


    I made this problem up and it is not straightforward for a single star.
    Challenge accepted

    Solution 241
    *

    First we note that \cos(x)=\sin(4x) \Rightarrow \cos(x)(2 \sin(x)-1)(4 \sin(x)+2 sin(x) -1)=0 \Rightarrow \cos(x)=0 , \ \sin(x)=\frac{1}{2}, \ \sin(x)=\frac{1}{4}(-1 \pm \sqrt{5})

    Solving for x, we get x=2 \pi n \pm (\frac{\pi}{2}-4x) \Rightarrow x= \frac{\pi}{10} , \ \frac{\pi}{6} for acute x.

    Comparing the acute angled routes we deduce that \begin{aligned} \sin(\frac{\pi}{10})=\frac{1}{4}  (\sqrt{5}-1) \Rightarrow \sin(\frac{\pi}{5})=2 \sin(\frac{\pi}{10}) \cos( \frac{\pi}{10}) \end{aligned}

= 2 \sin(\frac{\pi}{10}) \sqrt{1-\sin^2(\frac{\pi}{10})}=\sqrt{ \frac{5}{8}-{\sqrt{\frac{5}{8}}}} \ \square.

    Now, the arc of a regular n-gon of circumradius 1 (which is enclosed by a surrounding circle of radius 1) is clearly \frac{n}{2} \sin(\frac{2 \pi}{n}) \Rightarrow \frac{n}{2} \sin(\frac{2\pi}{n} < \pi \ \Rightarrow \pi > 10\sin(\frac{\pi}{10}) by considering the case n=20.

    \therefore \displaystyle \pi \phi >10 (\frac{1}{4} (\sqrt{5}-1))(\frac{1}{2} (\sqrt{5}+1))=\frac{5}{4}(5-1)=5 \ \square

    Edit: Btw, the problem would be improved upon by saying that values for \pi and \sqrt{5} are not quotable (forcing people to find a clever method or give a long tedious derivation of \pi to 2 decimal places whereby, if an infinite series is used, it must be derived). That or what TheMagicMan said in adding a "hence"
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    (Original post by Lord of the Flies)
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    If I have correctly understood what you mean, yes the limiting area is that of the other shape, but the limiting perimeter need not be.

    The calculus version asks whether over some interval (a,b) and for two functions sequences of continuous functions the following is true: \displaystyle d_n=\underset{a\leq x\leq b}{\text{max}}\big| |f_n(x)|-|g_n(x)|\big|\to 0\Rightarrow \displaystyle\int_a^b \big||f_n(x)|-|g_n(x)|\big|\,dx\to 0 as n\to\infty, which it clearly is:

    \displaystyle 0\leq \int_a^b \big||f_n(x)|-|g_n(x)|\big|\,dx\leq \int_a^b \underset{a\leq x\leq b}{\text{max}}\big||f_n(x)|-|g_n(x)|\big|\,dx=(b-a)d_n\to 0

    On the other hand, the arc-length of a function f is \displaystyle \int_a^b \Big(f'^2(x)+1\Big)^{\frac{1}{2}  }\,dx
    Just because \displaystyle \int_a^b |f_n| and \displaystyle \int_a^b |g_n| tend to the same thing doesn't imply that \displaystyle \int_a^b \Big(f'^2_n+1\Big)^{\frac{1}{2}} and \displaystyle \int_a^b \Big(g'^2_n+1\Big)^{\frac{1}{2}} behave the same.

    Take f_n(x)=\frac{1}{n}(\cos n x+1) and g_n(x)=0 over (0,\pi) for instance. Here d_n=2n^{-1}\to 0, the areas tend to  0, yet the length of the first curve f_n is >\pi even at the limit (I have chosen the function so that it mimics the "circle approximation paradox" you posted: the length of f_n is in fact constant).
    Cheers dude! Fully satisfied! I've always looked at the difference is perimeter being because of fractal dimension but perhaps it's not? Could it perhaps be true that one of f and g are required to be discontinuous at n number of points with n \to \infty?

    I really need to start using that arc length formula more! 'Tis certainly a beast
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    (Original post by Jkn)
    Now, the arc of a polygon of circumradius 1 (which is enclosed by a surrounding circle of radius 1)...
    Area of a regular n-gon with unit circumradius. The two-star method I intended for the question was to first find \sin\left(\frac{\pi}{5}\right). It is a root of the equation x(1 + 4(1 - x^2)(1 - 4x^2)) = 0. Hence, get \sin\left(\frac{\pi}{10}\right) and use it in

    \displaystyle \int_{0}^{\frac{1}{2}\phi^{-1}} 1 - \frac{1}{\sqrt{1 - x^2}}\ dx

    to obtain the inequality.
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    (Original post by jack.hadamard)
    Area of a regular n-gon with unit circumradius. The two-star method I intended for the question was to first find \sin\left(\frac{\pi}{5}\right). It is a root of the equation x(1 + 4(1 - x^2)(1 - 4x^2)) = 0. Hence, get \sin\left(\frac{\pi}{10}\right) and use it in

    \displaystyle \int_{0}^{\frac{1}{2}\phi^{-1}} 1 - \frac{1}{\sqrt{1 - x^2}}\ dx

    to obtain the inequality.
    Wow, that seems rather elaborate! Did you intend that the * solution be precisely what I did?

    So how are you actually obtaining the inequality in that method? What specific ** technique is being used?
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    (Original post by Jkn)
    What specific ** technique is being used?
    The 'technique' is simply if f(x) < 0 for a < x < b, then \int_{a}^{b} f(x)\ dx < 0.
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    (Original post by joostan)
    By parts is reverse product rule which is C3 :lol:
    Don't discourage people from posting questions. If you don't like a question, then don't do it.
    • PS Helper
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    (Original post by jack.hadamard)
    Don't discourage people from posting questions. If you don't like a question, then don't do it.
    To be fair it was an awful question :laugh:

    We need some more * questions though, I just don't know any which would be of interest.
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    (Original post by Mladenov)
    ...
    What universities are you considering (or have applied to, or have been accepted in)?
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    (Original post by jack.hadamard)
    Don't discourage people from posting questions. If you don't like a question, then don't do it.
    That's not what I meant by any means, I hope I didn't discourage anyone
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    (Original post by jack.hadamard)
    Yes, I did, but still thought the answer was kind of irritating.

    Spoiler:
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    With some luck, it should be \displaystyle \frac{1}{2} - \frac{3}{4(2k - 1)}. I did the limiting probability two different ways (counted with replacement too) and got the same limit (the other one approaching from above).
    Yes that's right! Why did you think the answer is irritating?

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    You have no idea how long it took me to simplify your expression into a single fraction to verify its the same as mine. Urgh, I don't belong on this thread!
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    (Original post by shamika)
    Yes that's right! Why did you think the answer is irritating?
    It's just because I find it not-quite-so intuitive. Having the same probability of the three numbers (picked from all positive integers) being the sides of a triangle as they being not the sides of a triangle is puzzling. If we try to formalise such statements, then considering the probability space (\Omega, \mathcal{F}, \text{P}) with

    i) \Omega := \{\ 1, 2, 3, ...\ \}, \mathcal{F} = \mathcal{P}(\Omega),

    ii) if A \in \mathcal{F}, then \displaystyle \text{P}(A) = \sum_{a \in A} \text{P}(\{a\}) = \pi |A| where \pi is the (equal) probability of any given positive integer.

    makes no sense, since \text{P}(\Omega) = \begin{cases} 0\quad\ \text{when}\ \pi = 0 \\ \infty\quad \text{when}\ \pi > 0 \end{cases}.

    Following the above, I also find other "intuitive" statements irritating. Like picking an even integer among all the positive integers, which has a 'probability' of exactly one half. It's maybe because I am broken; I don't know.
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    (Original post by jack.hadamard)
    It's just because I find it not-quite-so intuitive. Having the same probability of the three numbers (picked from all positive integers) being the sides of a triangle as they being not the sides of a triangle is puzzling. If we try to formalise such statements, then considering the probability space (\Omega, \mathcal{F}, \text{P}) with

    i) \Omega := \{\ 1, 2, 3, ...\ \}, \mathcal{F} = \mathcal{P}(\Omega),

    ii) if A \in \mathcal{F}, then \displaystyle \text{P}(A) = \sum_{a \in A} \text{P}(\{a\}) = \pi |A| where \pi is the (equal) probability of any given positive integer.

    makes no sense, since \text{P}(\Omega) = \begin{cases} 0\quad\ \text{when}\ \pi = 0 \\ \infty\quad \text{when}\ \pi > 0 \end{cases}.

    Following the above, I also find other "intuitive" statements irritating. Like picking an even integer among all the positive integers, which has a 'probability' of exactly one half. It's maybe because I am broken; I don't know.
    Oh, I see what you mean. This is precisely why I posted the question because I also thought it was a really strange result. But isn't this why we study maths (i.e. to see how to think about these things properly?)
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    Nobody does my problems anymore .

    Oh well, simple, boring things.. simply 'cause I'm interested in what proofs might emerge..

    Problem 243*

    Find and prove formulae for the sum of the first n odd, and first n even numbers.
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    (Original post by FireGarden)
    Nobody does my problems anymore .

    Oh well, simple, boring things.. simply 'cause I'm interested in what proofs might emerge..

    Problem 243*

    Find and prove formulae for the sum of the first n odd, and first n even numbers.
    Solution 243*

    Odd: \displaystyle \sum^n_{r=1}(2r-1) = 2 \sum^n_{r=1}r - n = n(n+1) - n = n^2

    Even: \displaystyle \sum^n_{r=1} (2r) = 2\sum^n_{r=1} r = n(n+1)

    These problems are far too easy for this thread.
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    (Original post by Ateo)
    Solution 243*

    Odd: \displaystyle \sum^n_{r=1}(2r-1) = 2 \sum^n_{r=1}r - n = n(n+1) - n = n^2

    Even: \displaystyle \sum^n_{r=1} (2r) = 2\sum^n_{r=1} r = n(n+1)

    These problems are far too easy for this thread.
    Okay, solve it without using Mathematics.
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    (Original post by Zakee)
    Okay, solve it without using Mathematics.
    Solve it using bananas. (Say this in Rowan Atkinsons' voice)
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    (Original post by jack.hadamard)
    What universities are you considering (or have applied to, or have been accepted in)?
    Apologies for the late response, I was unable to reply earlier.

    Most probably, I will take a gap year...

    Apropos, as solution to your problem has already been posted, I just wanted to say that I generalized it a bit to 2^{2n}-3=m^{3} and then to y^{2}-3=m^{3}; the latter being a special case of Mordell's equation.

    (Original post by FireGarden)
    Nobody does my problems anymore .
    It is quite difficult to trace all the problems which are posted, for the OP has not been updated recently.

    Solution 231

    We can define f : A \to C with \displaystyle f(x,y) = (\frac{x}{\sqrt{x^{2}+y^{2}}}, \frac{y}{\sqrt{x^{2}+y^{2}}}, \sqrt{x^{2}+y^{2}}-1) and let g : C \to A be g(x,y,z) = (x(1+z),y(1+z)).
    It is sufficient to check that f \circ g = Id_{C} and  g \circ f = Id_{A}.
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    (Original post by bananarama2)
    Solve it using bananas. (Say this in Rowan Atkinsons' voice)

    Oh yes, he pronounces his b's rather distinctively, doesn't he?
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    (Original post by Mladenov)
    Most probably, I will take a gap year...
    You're not just doing that for another shot at the IMO, are you? :/ I can picture you getting into a rather strange situation if you started university in the first year... perhaps you could take 2/3 years of exams all at once though hmm...
 
 
 
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