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    (Original post by james22)
    You haven't actually given such an f that works yet, but you did the hard bit.
    f such that f(x) = 1 when x is rational and f(x) = 0 when x is irrational works right?
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    (Original post by 0x2a)
    f such that f(x) = 1 when x is rational and f(x) = 0 when x is irrational works right?
    No, but it does if you replace rational and irrational by the sets you defined.
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    (Original post by james22)
    No, but it does if you replace rational and irrational by the sets you defined.
    Oh woops, apparently I don't understand periodicity.

    I'll just set f(x) = 0 when x is algebraic, and create some bijection between T/\sim and (0,1], and if a,a' \in [\overline{a}], then f(a) = f(a').

    I guess this one doesn't even work, periodicity can't be defined for some x right?
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    (Original post by 0x2a)
    Oh woops, apparently I don't understand periodicity.

    I'll just set f(x) = 0 when x is algebraic, and create some bijection between T/\sim and (0,1], and if a,a' \in [\overline{a}], then f(a) = f(a').

    I guess this one doesn't even work, periodicity can't be defined for some x right?
    Looking over your proof again, it isn't obvious how it leads to a function with uncountably many periods. You haven't got any additive properties in there which are essential.
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    Problem 490**

     \displaystyle \int^{\pi}_{0} \dfrac{x \sin x}{1 + \cos^2 x} \ dx
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    (Original post by ThatPerson)
    Problem 490**

     \displaystyle \int^{\pi}_{0} \dfrac{x \sin x}{1 + \cos^2 x} \ dx
    \displaystyle \int_{0}^{\pi}\frac{x \sin x}{1 + \cos^2 x} = -\int_{0}^{\pi}\frac{\mathrm{d}}{  \mathrm{d}x}(\cos x)\frac{x}{1+\cos^2 x}

    By IBP, letting \displaystyle u = x, dv = \frac{(\cos x)'}{1+\cos^2x}, we arrive at:

    \displaystyle \bigg[x \arctan \cos x \bigg]_0^{\pi} + \int_{0}^{\pi}\arctan \cos x \, \mathrm{d}x

    But \displaystyle \int_{0}^{\pi}\arctan \cos x \, \mathrm{d}x = 0 since \arctan \cos x is symmetric x= \frac{\pi}{2}.

    Then the integral evaluates to \displaystyle \frac{\pi^2}{4}. Which is absolutely beautiful.
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    (Original post by Zacken)
    \displaystyle \int_{0}^{\pi}\frac{x \sin x}{1 + \cos^2 x} = -\int_{0}^{\pi}\frac{\mathrm{d}}{  \mathrm{d}x}(\cos x)\frac{x}{1+\cos^2 x}

    By IBP, letting \displaystyle u = x, dv = \frac{(\cos x)'}{1+\cos^2x}, we arrive at:

    \displaystyle \bigg[x \arctan \cos x \bigg]_0^{\pi} + \int_{0}^{\pi}\arctan \cos x \, \mathrm{d}x

    But \displaystyle \int_{0}^{\pi}\arctan \cos x \, \mathrm{d}x = 0 since \arctan \cos x is symmetric x= \frac{\pi}{2}.

    Then the integral evaluates to \displaystyle \frac{\pi^2}{4}. Which is absolutely beautiful.
    Nice method
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    (Original post by ThatPerson)
    Nice method
    Thanks!

    I think you can use the substitution x = \pi - y as well to evaluate the integral.

    Also, we could have used \displaystyle \int_0^\pi xf(\sin x )\mathrm{d}x=\frac \pi2\int_0^\pi f(\sin x )\mathrm{d}x to evaluate the integral. We were asked to prove this in a past STEP paper, if I remember correctly.
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    (Original post by Zacken)
    Thanks!

    I think you can use the substitution x = \pi - y as well to evaluate the integral.

    Also, we could have used \displaystyle \int_0^\pi xf(\sin x )\mathrm{d}x=\frac \pi2\int_0^\pi f(\sin x )\mathrm{d}x to evaluate the integral. We were asked to prove this in a past STEP paper, if I remember correctly.
    Yeah that's the approach I took.
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    Problem 491:**

    The enthusiasts amongst you may recognize this:
    Let: (x_n) be a sequence of positive real numbers such that:
    \displaystyle\sum_{n=1}^{\infty} n^2x_n^2 converges.
    Show that:
    \displaystyle\sum_{n=1}^{\infty} x_n is convergent. And the converse?
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    (Original post by joostan)
    Problem 491:**

    The enthusiasts amongst you may recognize this:
    Let: (x_n) be a sequence of positive real numbers such that:
    \displaystyle\sum_{n=1}^{\infty} n^2x_n^2 converges.
    Show that:
    \displaystyle\sum_{n=1}^{\infty} x_n is convergent. And the converse?
    Can you use the ratio test to show that the ratio must be less than or equal to 1 for the first part?
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    (Original post by 0x2a)
    Can you use the ratio test to show that the ratio must be less than or equal to 1 for the first part?
    What if the ratio test is inconclusive?
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    (Original post by james22)
    What if the ratio test is inconclusive?
    Can that be covered by the = 1 part? My logic was that if the ratio was bigger than 1 then it must diverge, so it must be less than or equal to 1?

    Ah of course, it can fail to exist right?
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    (Original post by 0x2a)
    Can that be covered by the = 1 part? My logic was that if the ratio was bigger than 1 then it must diverge, so it must be less than or equal to 1?

    Ah of course, it can fail to exist right?
    It can fail to exist, but also if the limit of \frac{(n+1)x_{n+1}^2}{n^2 x_n^2} is 1, we can only say that the limit of x_(n+1)/x_n is 1 (if it exists), and if the ratio test gives an answer of 1 you cannot say anything about convergence or divergence.
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    (Original post by 0x2a)
    Can you use the ratio test to show that the ratio must be less than or equal to 1 for the first part?
    Perhaps, it is however not absolutely necessary.
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    (Original post by james22)
    It can fail to exist, but also if the limit of \frac{(n+1)x_{n+1}^2}{n^2 x_n^2} is 1, we can only say that the limit of x_(n+1)/x_n is 1 (if it exists), and if the ratio test gives an answer of 1 you cannot say anything about convergence or divergence.
    So something like \dfrac{n(x_{n + 1})^2}{n(x_n)^2} < \dfrac{(n + 1)(x_{n + 1})^2}{n(x_n)^2} \leq 1 and so \dfrac{x_{n + 1}}{x_n} < 1 doesn't work since the limits of \dfrac{n(x_{n + 1})^2}{n(x_n)^2} and  \dfrac{(n + 1)(x_{n + 1})^2}{n(x_n)^2} are the same?
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    (Original post by 0x2a)
    So something like \dfrac{n(x_{n + 1})^2}{n(x_n)^2} < \dfrac{(n + 1)(x_{n + 1})^2}{n(x_n)^2} \leq 1 and so \dfrac{x_{n + 1}}{x_n} < 1 doesn't work since the limits of \dfrac{n(x_{n + 1})^2}{n(x_n)^2} and  \dfrac{(n + 1)(x_{n + 1})^2}{n(x_n)^2} are the same?
    Limits don't have to respect strong inequalities, so those < can become \geq after taking the limit.
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    From IMO 2013 Q1.

    Problem 492**

    Prove that for any pair of positive integers k and  n, there exist  k positive integers  m_1,m_2,...,m_k (not necessarily different) such that

     1 + \dfrac{2^k - 1}{n} = \left(1 + \dfrac{1}{m_1}\right)\left(1 + \dfrac{1}{m_2}\right)...\left(1 + \dfrac{1}{m_k}\right)
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    Problem 493***

    Let \zeta(s) denote the Riemann-Zeta function in the usual way, for complex s=\sigma + it

    Prove that, for \sigma > 1,

    |\zeta(\sigma)|^3 |\zeta(\sigma + it)|^4 |\zeta(\sigma + 2it)| \geq 1
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    Problem 494 ***

    Evaluate:
    \displaystyle \int_{0}^{\infty }\frac{cos(ax)}{1+x^2}dx for a\geq 0.

    Sorry if it's been asked before.
 
 
 
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