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    (Original post by und)
    :hi:

    The original method of cataloguing the problems and solutions was very time consuming. Someone at some point wrote a script to do this automatically but it needed improvement and I am not sure what happened with that.

    Even if we were to collectively update the OP, which in theory shouldn't be too time consuming, bear in mind there is a strict character limit for posts, which does not play well with all those links. The first post is already out of space.

    Anyway I hope you are having a nice summer!
    :hi:

    I don't know why I didn't see this quote... :confused:

    Aaaaah that makes sense.

    Thanks I had an OK summer. I hope you've been having a nice summer too. :cute:
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    Anyone know how to do this??


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    (Original post by eternaforest)
    Anyone know how to do this??


    Posted from TSR Mobile
    Jhust call it unit length, bung in some angles, and work out the length of each side.

    Spoiler:
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    The ratio is 2
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    (Original post by Slumpy)
    Jhust call it unit length, bung in some angles, and work out the length of each side.
    Spoiler:
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    The ratio is 2
    Spoiler:
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    I get rt3
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    (Original post by Renzhi10122)
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    I get rt3
    My work was pretty scrappy (literally on the back of an envelope), so I'm not 100% it was right (now in the bin). May look again in a bit...
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    Man, this thread sure has deteriorated from what it once was.
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    (Original post by Zacken)
    Man, this thread sure has deteriorated from what it once was.
    I would've loved it back then
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    (Original post by Renzhi10122)
    I would've loved it back then
    I would've loved it too, probably wouldn't have been able to do any of the questions though. We should make an effort to revive the thread! You've got all your tricky number theory questions.
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    (Original post by Renzhi10122)
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    I get rt3
    Ive done it as well now (I used 'r' for the length of the squares) and got that the length is 3r+(rt3)r and the width is (rt3)r + r so the ratio is rt3:1
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    Problem 502

    The independent normal variates X and Y have standard deviations a and b respectively. The variate Z=X+Y has standard deviation c. Prove that, if a, b and c are positive integers, it is impossible for all three variates to have odd variances.

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    Problem 503

    Given integers a and b satisfying \dfrac{a}{b}  < 1, find integers c and d such that

    \displaystyle \frac{a}{b} + \frac{c}{d} + \frac{ac}{bd} = 1
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    Problem 504

    Evaluate \displaystyle \int \frac{\mathrm{d}x}{\sqrt{e^{2x} + 1}}
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    Problem 505

    (a) Use \displaystyle \int_0^1 \frac{x^4(1-x)^4}{1+x^2} \, \mathrm{d}x to show that \frac{22}{7} > \pi.

    (b) By replacing the denominator of the above integral with two suitable integers, obtain a lower and upper bound for \pi.

    Massive hint for (b):
    Spoiler:
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    \displaystyle \int_0^1 \int \frac{x^4(1-x)^4}{2} \mathrm{ and } \int_0^1 \frac{x^4(1-x)^4}{1}
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    (Original post by Zacken)
    Problem 503

    Given integers a and b satisfying \dfrac{a}{b}  < 1, find integers c and d such that

    \displaystyle \frac{a}{b} + \frac{c}{d} = \frac{ac}{bd} = 1
    This is definitely wrong atm.
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    (Original post by Renzhi10122)
    This is definitely wrong atm.
    Finger slip = should be a +, fixed it, thanks for pointing it out!
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    Solution 504 (I think haha)

    Let  u = \sqrt{e^{2x} + 1}
    Therefore  u^2 = e^{2x} + 1 \implies x = \frac{1}{2}\ln(u^2 - 1) \implies \frac{dx}{du} = \frac{2u}{u^2 - 1} \implies dx = \frac{u}{u^2 - 1} \ du

    Substituting into our integral:

    \displaystyle \int \frac{dx}{\sqrt{e^{2x} + 1}} =  \displaystyle \int \frac{1}{u} \frac{u}{u^2 - 1} \ du = \displaystyle \int \frac{du}{u^2 - 1} = - \displaystyle \int \frac{du}{1 - u^2} = - artanh (u) + C = - artanh (\sqrt{e^{2x} + 1}) + C

    P.S. Does anybody know the LaTeX for hyperbolic and inverse hyperbolic functions?
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    (Original post by 16Characters....)
    Solution 504
    Precisely the solution I wanted, most people don't think of replacing the whole integrand with a u-sub.

    Quick note, you could have tidied your solution a slight bit by using implicit differentiation instead of re-arranging:

    \displaystyle u^2 = e^{2x} - 1 \implies 2u \frac{\mathrm{d}u}{\mathrm{d}x} = 2e^{2x} \implies \mathrm{d}x = \frac{u}{u^2 - 1} \mathrm{d}u

    Edit: I don't think Latex has hyperbolic trig, you could always use \mathrm{artanh } \, x though.
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    Can someone here tell me how to do those javascript equations? I really wanna know
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    (Original post by Zacken)
    Precisely the solution I wanted, most people don't think of replacing the whole integrand with a u-sub.

    Quick note, you could have tidied your solution a slight bit by using implicit differentiation instead of re-arranging:

    \displaystyle u^2 = e^{2x} - 1 \implies 2u \frac{\mathrm{d}u}{\mathrm{d}x} = 2e^{2x} \implies \mathrm{d}x = \frac{u}{u^2 - 1} \mathrm{d}u

    Edit: I don't think Latex has hyperbolic trig, you could always use \mathrm{artanh } \, x though.
    Thanks for the LaTeX info.

    I'll have to get thinking of some nice questions to post.
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    (Original post by Alexion)
    Can someone here tell me how to do those javascript equations? I really wanna know
    It's called Latex, and here's how you use it on TSR.
 
 
 
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