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    (Original post by Mladenov)
    As someone has said, it is (\ln(1-x))^{2}.



    How do you derive your first inequality? It is not true. And, just by the way, this inequality is quite sharp.
    Oh gosh please spare me the embarrassment
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    (Original post by metaltron)
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    Solution 393

    Squaring both sides:

     a+b+c + 2(\sqrt{ab} + \sqrt{bc} +\sqrt{ca}) = 9 + 2(\sqrt{ab} + \sqrt{bc} +\sqrt{ca})  \geq ab + bc +ca

     (ab - 2\sqrt{ab}) + (bc - 2\sqrt{bc}) + (ca-2\sqrt{ca}) \leq 9

     (\sqrt{ab} -1)^2 + (\sqrt{bc} -1)^2 + (\sqrt{ac} -1)^2 \leq 12

    Using the AM-GM we have:

     \displaystyle \frac{a+b+c}{3} = 3 \geq \sqrt{abc}

     \displaystyle \sqrt{ab} \leq \frac{3}{\sqrt{c}}

    Hence since a,b,c => 1:

     \displaystyle 1 \leq \sqrt{ab}, \ \  \sqrt{bc}, \ \  \sqrt{ac} \leq 3

    Hence:

     (\sqrt{ab} -1)^2 + (\sqrt{bc} -1)^2 + (\sqrt{ac} -1)^2 \leq 2^2 + 2^2 + 2^2 = 12

    and we are done.

    So either I have redeemed myself or dug an even bigger hole. I have my fingers crossed
    Nope, it is still incorrect. The inequality 3 \ge \sqrt{abc} does not hold.
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    (Original post by Mladenov)
    Nope, it is still incorrect. The inequality 3 \ge \sqrt{abc} does not hold.
    Then I apologise for wasting your time with ridiculous solutions, hopefully I have woken up with a clearer mind today
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    (Original post by Mladenov)

    Problem 394***

    Find \displaystyle I = \int_{0}^{1} \frac{\ln x \cdot \ln^{2}(1-x)}{x} dx.
    If you make the substitution, x=\sin^2(\theta), then

    \displaystyle I = 16 \int_{0}^{\frac{\pi}{2}} \ln (\sin(\theta)) \cdot \ln^{2}(\cos(\theta)) \cot(\theta) d\theta .

    This is simply equal to

    \frac{1}{2}\frac{ \partial^2}{ \partial y^2}\frac{ \partial}{ \partial x} \beta (x=0, y=1).

    However, this is not defined. Has something gone awry in the substitution?

    Excellent problems, by the way!
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    This is the better alternative for 384, I'm pretty sure this was not mentioned in the official IMO solutions.

    WLOG assume that \displaystyle a < b

    let \displaystyle n = \frac {a^2 + b^2}{ab + 1}

    We define the infinite integer sequence \displaystyle \cdots,x_{ - 1},x_0,x_1,x_2,\cdots :
    \displaystyle x_0 = a,\quad x_1 = b,\quad x_{i + 1} = nx_i - x_{i - 1}\mbox{ for }i = 1,2,\cdots,\mbox{n}
    By induction we can easily show
    \displaystyle x_i^2 + x_{i - 1}^2 = n(x_ix_{i - 1} + 1)\mbox{ for }k = \cdots, - 1,0,1,2,\cdots
    and
    \cdots < x_{ - 1} < x_0 < x_1 < x_2 < \cdots,\mbox{(i.e. we have a strictly increasing sequence)}
    We must now to prove that one of the term \displaystyle x_i must be \displaystyle 0

    To prove this, let \displaystyle x_m be the smallest positive term in the sequence such that \displaystyle x_m > 0, by definition that this term is smallest, we have \displaystyle x_{m - 1}\leq 0

    suppose that \displaystyle x_{m - 1} < 0, then in the following equation (*), LHS is greater than 0 while RHS is less than or equal to 0, which is a contradiction
    \displaystyle x_m^2 + x_{m - 1}^2 = n(x_mx_{m - 1} + 1)\qquad(*)
    therefore \displaystyle x_{m - 1} = 0 , substitute into equation (*) we have \displaystyle n = x_m^2 = \frac {a^2 + b^2}{ab + 1} Q.E.D.
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    (Original post by henpen)
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    If you make the substitution, x=\sin^2(\theta), then

    \displaystyle I = 16 \int_{0}^{\frac{\pi}{2}} \ln (\sin(\theta)) \cdot \ln^{2}(\cos(\theta)) \cot(\theta) d\theta .

    This is simply equal to

    \frac{1}{2}\frac{ \partial^2}{ \partial y^2}\frac{ \partial}{ \partial x} \beta (x=0, y=1).

    However, this is not defined. Has something gone awry in the substitution?

    Excellent problems, by the way!
    The substitution is fine and nice.

    How did you obtain the mixed beta derivative?
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    (Original post by Felix Felicis)
    Solution 388

    \displaystyle \sum_{n\geq 0} \frac{4n}{3^{n+1}} = \sum_{n\geq 1} \frac{4n}{3^{n+1}}

    |x| < 1:

    \displaystyle\begin{aligned} \sum_{n\geq 1} x^n = \frac{x}{1-x} \implies \sum_{n\geq 1} n x^{n-1} & = \frac{d}{dx} \frac{x}{1-x} \\ & = \frac{1}{(x-1)^2} \\ \implies \sum_{n\geq 1} 4n x^{n+1} & = \frac{4x^2}{(x-1)^2} \\ \implies \sum_{n\geq 0} \frac{4n}{3^{n+1}} & = \frac{4x^2}{(x-1)^2} \bigg|_{x=\frac{1}{3}} = 1

    Solution 389
    It is very well known that \displaystyle\sum_{n\geq 1} \frac{1}{n^2} = \frac{\pi^2}{6} (unless that was the intention of the question).

    \displaystyle\begin{aligned} \sum_{n\geq 1} \frac{(-1)^n}{n^2} & = \sum_{n\geq 1} \frac{1}{(2n)^2} - \sum_{n\geq 1} \frac{1}{(2n-1)^2} \\ & = \frac{\pi^2}{24} - \left( \sum_{n\geq 1} \frac{1}{n^2} -  \sum_{n\geq 1} \frac{1}{(2n)^2} \right) \\ & = -\frac{\pi^2}{12}
    (Original post by Mladenov)
    Alternative:

    Solution 389

    Consider \displaystyle f(z)= \frac{1}{z^{2} \sin \pi z}. Choose some beautiful contour C_{n}; for example, one the most popular - square with vertices at \displaystyle \pm (n+\frac{1}{2}) \pm i(n+\frac{1}{2}) will do. We can easily (indeed, all we need is to bound \sin \pi z below) see that \displaystyle \int_{C_{n}} f \to 0 as n \to \infty.

    Our function has simple poles at z=n \in \mathbb{Z}, n \not= 0 and 3-pole at z=0.
    For n \not= 0, we have \displaystyle \text{Res}(f,n) = \frac{(-1)^{n}}{\pi n^{2}}, which is obvious.

    We turn to the case z=0. We note \displaystyle \sin \pi z = \sum_{0}^{\infty} (-1)^{k}\frac{(z \pi)^{2k+1}}{(2k+1)!} and \displaystyle f(z) = z^{-3}\frac{1}{\frac{\sin z \pi}{z}}. Hence, \displaystyle \text{Res}(f,0) = \frac{\pi}{6}.

    Whence, \displaystyle \frac{1}{\pi}\sum_{n  \in \mathbb{Z} | n \not= 0} \frac{(-1)^{n}}{n^{2}} + \frac{\pi}{6}= 0.
    Actually I had thought up the problem with the following solution in mind.

    Alternative solution II to problem 389:

    Consider the Fourier series

    \displaystyle x^2 = \dfrac{\pi^2}{3} + 4\sum_{n \geq 1} \dfrac{(-1)^n \cos(nx)}{n^2}

    Where setting x=0 leads to the required solution.
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    (Original post by Mladenov)
    The substitution is fine and nice.

    How did you obtain the mixed beta derivative?
    \displaystyle \frac{\partial}{\partial x} \beta(x,y) =2 \int_0^{\frac{\pi}{2}} \frac{\partial}{\partial x}  \cos(t)^{2y-1} \sin(t)^{2x-1}dt

    \displaystyle  =2 \int_0^{\frac{\pi}{2}} 2 \log(\sin(t))  \cos(t)^{2y-1} \sin(t)^{2x-1}dt

    \displaystyle \frac{\partial^2}{\partial y^2} \frac{\partial}{\partial x} \beta(x,y) =2 \int_0^{\frac{\pi}{2}}2 \log(\sin(t)) 4 \log^2(\cos(t)) \cos(t)^{2y-1} \sin(t)^{2x-1}dt

    Let x=0, y=1.
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    If you are having a boring Wednesday night, then play with n!^{(k)} + a = m^2 to find solutions (n,m) for k,a \in \mathbb{N}.

    For example, n!! + 2 = m^2.
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    (Original post by henpen)
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    \displaystyle \frac{\partial}{\partial x} \beta(x,y) =2 \int_0^{\frac{\pi}{2}} \frac{\partial}{\partial x}  \cos(t)^{2y-1} \sin(t)^{2x-1}dt

    \displaystyle  =2 \int_0^{\frac{\pi}{2}} 2 \log(\sin(t))  \cos(t)^{2y-1} \sin(t)^{2x-1}dt

    \displaystyle \frac{\partial^2}{\partial y^2} \frac{\partial}{\partial x} \beta(x,y) =2 \int_0^{\frac{\pi}{2}}2 \log(\sin(t)) 4 \log^2(\cos(t)) \cos(t)^{2y-1} \sin(t)^{2x-1}dt

    Let x=0, y=1.
    Well, we get the same thing without the substitution, so it does not give us any further information using it that way. I gave it some thought, and it seems that we can employ neutrices to evaluate this derivative (we get the correct result). If you are interested, I can provide some references.
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    (Original post by Mladenov)
    Well, we get the same thing without the substitution, so it does not give us any further information using it that way. I gave it some thought, and it seems that we can employ neutrices to evaluate this derivative (we get the correct result). If you are interested, I can provide some references.
    Very much so, could you quickly summarise what neutrices are? It seems counterintuitive that a function can have a defined derivative at a point where it is not defined.
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    Problem 395**


    If  I = \displaystyle \int_{-\infty}^{\infty} e^{-x^2} dx.

    Show that  I = \sqrt\pi
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    Solution 395 (Not exactly 3*s o.O) \displaystyle I=\int_{- \infty}^{\infty} e^{-x^2} dx

    \displaystyle I^2= (\int_{- \infty}^{\infty} e^{-x^2} dx)^2=(\int_{- \infty}^{\infty} e^{-x^2} dx)(\int_{- \infty}^{\infty} e^{-x^2} dx)

    \displaystyle I^2= (\int_{- \infty}^{\infty} e^{-x^2} dx)(\int_{- \infty}^{\infty} e^{-y^2} dy) (it doesn't really matter which variable you use...)

    \displaystyle I^2= \int_{- \infty}^{\infty}\int_{- \infty}^{\infty} e^{-(x^2+y^2)} dxdy Switching to polar coordinates r=x^2+y^2

    \displaystyle I^2= \int_{0}^{2 \pi} \int_{0}^{ \infty} re^{-r^2} dr \d \theta= -\dfrac{1}{2} \int_{0}^{2 \pi} \left[ re^{-r^2} \right]_0^{\infty} d \theta =\int_{0}^{2 \pi} -1 d \theta= \pi

    \therefore I^2= \pi \Rightarrow I=\sqrt{\pi}...
    I must leave now, if I've made any mistakes I'll correct on my return.
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    Hm, my method is similarish

    y=xv,\ dy=xdv

\displaystyle I=\int^{\infty}_{-\infty}e^{-x^2}dx=  2\int^{\infty}_{0}e^{-x^2}dx

\displaystyle I^2=4\int^{\infty}_{0} \int^{\infty}_{0}e^{-(x^2+y^2)}dxdy

\displaystyle I^2=4\int^{\infty}_{0} \int^{\infty}_{0}xe^{-x^2(1+v^2)}dxdv

\displaystyle\int xe^{-x^2}dx=-\dfrac{e^{-x^2}}{2}

\displaystyle\int^{\infty}_{0}xe  ^{-x^2(1+v^2)}dx=\dfrac{1}{2(1+v^2)  }

\displaystyle I^2=4\int^{\infty}_{0}\dfrac{1}{  2(1+v^2)}dv= 2[\arctan{v}]_0^\infty=\pi

\displaystyle I=\sqrt\pi

    I think most people have seen that question before, you guys should continue to post alternative methods
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    Nice. Glad i was able to do something...
    (Original post by Flauta)
    Hm, my method is similarish<br />
    <br />
    y=xv,\ dy=xdv&lt;br /&gt;

\displaystyle I=\int^{\infty}_{-\infty}e^{-x^2}dx=  2\int^{\infty}_{0}e^{-x^2}dx&lt;br /&gt;

\displaystyle I^2=4\int^{\infty}_{0} \int^{\infty}_{0}e^{-(x^2+y^2)}dxdy&lt;br /&gt;

\displaystyle I^2=4\int^{\infty}_{0} \int^{\infty}_{0}xe^{-x^2(1+v^2)}dxdv&lt;br /&gt;

\displaystyle\int xe^{-x^2}dx=-\dfrac{e^{-x^2}}{2}&lt;br /&gt;

\displaystyle\int^{\infty}_{0}xe  ^{-x^2(1+v^2)}dx=\dfrac{1}{2(1+v^2)  }&lt;br /&gt;

\displaystyle I^2=4\int^{\infty}_{0}\dfrac{1}{  2(1+v^2)}dv= 2[\arctan{v}]_0^\infty=\pi&lt;br /&gt;

\displaystyle I=\sqrt\pi&lt;br /&gt;

<br />
    <br />
    I think most people have seen that question before, you guys should continue to post alternative methods <img src="images/smilies/tongue.png" border="0" alt="" title="" smilieid="5" class="inlineimg" />
    <br />
    <br />
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    (Original post by keromedic)
    Nice. Glad i was able to do something...
    Me too, most of the questions on here are too hard for me to even attempt

    (I tried to rep your method but it won't let me so I'll just try again later)
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    (Original post by Flauta)
    Me too, most of the questions on here are too hard for me to even attempt

    (I tried to rep your method but it won't let me so I'll just try again later)
    I didn't realise that the question was fairly routine. I only post questions which I myself can do. Oh well

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    (Original post by Arieisit)
    I didn't realise that the question was fairly routine. I only post questions which I myself can do. Oh well

    Posted from TSR Mobile
    It's still a really good question, there're some quite complex in-depth ways of doing it which I don't understand at all, would love to see some people post them. What was your method?
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    (Original post by Flauta)
    Hm, my method is similarish

    y=xv,\ dy=xdv

\displaystyle I=\int^{\infty}_{-\infty}e^{-x^2}dx=  2\int^{\infty}_{0}e^{-x^2}dx

\displaystyle I^2=4\int^{\infty}_{0} \int^{\infty}_{0}e^{-(x^2+y^2)}dxdy

\displaystyle I^2=4\int^{\infty}_{0} \int^{\infty}_{0}xe^{-x^2(1+v^2)}dxdv

\displaystyle\int xe^{-x^2}dx=-\dfrac{e^{-x^2}}{2}

\displaystyle\int^{\infty}_{0}xe  ^{-x^2(1+v^2)}dx=\dfrac{1}{2(1+v^2)  }

\displaystyle I^2=4\int^{\infty}_{0}\dfrac{1}{  2(1+v^2)}dv= 2[\arctan{v}]_0^\infty=\pi

\displaystyle I=\sqrt\pi

    I think most people have seen that question before, you guys should continue to post alternative methods
    Is there a method of solving this integral using just elementary methods (i.e. no double integrals or change of coordinate systems).
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    Another solution to 295:

    \displaystyle I = \int_{-\infty}^{\infty} e^{-x^2} \ dx = 2 \int_0^{\infty} e^{-x^2} \ dx \overset{x^2 \mapsto x}= \int_0^{\infty} e^{-x} x^{-\frac{1}{2}} \ dx = \Gamma \left(\tfrac{1}{2}\right)

    Now, as \Gamma (z) = \int_0^{\infty} x^{z-1} e^{-x} \ dx and \text{B} (x,y) = \dfrac{\Gamma (x) \Gamma (y)}{\Gamma (x+y)}, we have:

    \text{B}\left(\frac{1}{2}, \frac{1}{2}\right) = \dfrac{\Gamma^2 \left(\frac{1}{2}\right)}{\Gamma (1)} = \Gamma^2 \left(\frac{1}{2}\right) = I^2

    \displaystyle \text{B}\left(\tfrac{1}{2}, \tfrac{1}{2}\right) = \int_0^1 x^{-\frac{1}{2}} (1-x)^{-\frac{1}{2}} \ dx \overset{x^{\frac{1}{2}}\mapsto x}= 2\arcsin{x}\bigg|_0^1 = \pi

    Therefore I^2 = \pi \implies I = \sqrt{\pi}
 
 
 
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