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    (Original post by Jkn)
    Problem 110*

    Prove that \displaystyle (\sum_{k=1}^{\infty}{a_k}{b_k})^  2 \leq \sum_{k=1}^{\infty}{a_k}^2 \sum_{k=1}^{\infty}{a_k}^2 for all real numbers \displaystyle a_k, b_k with using Cauchy-Schwartz.
    Solution 110

    Consider the quadratic equation:

     \big{(}\displaystyle\sum_{k=1}^{  \infty}a_{k}^2\big{)}x^2 + 2\big{(}\displaystyle\sum_{k=1}^  {\infty}a_{k}b_{k}\big{)} x + \big{(}\displaystyle\sum_{k=1}^{  \infty}b_{k}^2\big{)} = 0

    which one can verify is the same as  \displaystyle\sum_{k=1}^{\infty}  (a_kx+b_k)^2 = 0, which is why the above quadratic is non-negative.

    Since the equation is non-negative, this equation can have at most 1 root in \mathBB{R} (hopefully obvious) and this implies  b^2 - 4ac \leq 0 (discriminant test for no. of roots), from which the result follows.
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    (Original post by FireGarden)
    Solution 110

    Consider the quadratic equation:

     \big{(}\dfrac{1}{4}\displaystyle  \sum_{k=1}^{\infty}a_{k}^2\big{)  }x^2 + \big{(}\displaystyle\sum_{k=1}^{  \infty}a_{k}b_{k}\big{)}^2 x + \big{(}\displaystyle\sum_{k=1}^{  \infty}b_{k}^2\big{)} = 0

    Since all terms are non-negative, this equation can have at most 1 root in \mathBB{R} The this implies  b^2 - 4ac \leq 0 , from which the result follows.
    Your result implies  \displaystyle (\sum_{k=1}^{\infty}{a_k}{b_k})^  4 \leq \sum_{k=1}^{\infty}{a_k}^2 \sum_{k=1}^{\infty}{a_k}^2 which is simply not true.

    Also, why is it that all the coefficients being non-negative implies b^2 \leq 4ac? Consider x^2+x+1=0 etc... (perhaps i have misunderstood you. If so I apologise.)
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    (Original post by Jkn)
    Your result implies  \displaystyle (\sum_{k=1}^{\infty}{a_k}{b_k})^  4 \leq \sum_{k=1}^{\infty}{a_k}^2 \sum_{k=1}^{\infty}{a_k}^2 which is simply not true.

    Also, why is it that all the coefficients being non-negative implies b^2 \leq 4ac? Consider x^2+x+1=0 etc... (perhaps i have misunderstood you. If so I apologise.)
    Wrote very badly/lazy.. to the point of making factual errors out of a lack of thinking.. but hopefully the corrections/additions should be sufficient now!
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    (Original post by FireGarden)
    Wrote very badly/lazy.. to the point of making factual errors out of a lack of thinking.. but hopefully the corrections/additions should be sufficient now!
    Well the infinite series I have presented are not necessarily bounded so are you able to justify the use of the discriminant in the quadratic equation and perhaps even the validity of forming a quadratic equation in the cases where the coefficients are infinite? (It may be correct, but you'll need to justify this assumption either way.)

    Note that a very straightforward solution exists whereby the considerations of infinity are trivial (if it turns out your method is false.)
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    Subbing!
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    (Original post by elixir)
    We have f(xy)=f(x)f(y) and f(x+y)=f(x)+f(y) for all x,y\in \mathbb{R^{+}}.

    How did you get these?
    Can someone tell me how he got that implication, it's driving me nuts!

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    (Original post by elixir)
    Can someone tell me how he got that implication, it's driving me nuts!

    f(z)^{f(xy)}=f(z^{xy})=f(z^{x})^  f(y)=f(z)^{f(x)f(y)}.
    f(z)^{f(x+y)}=f(z^{x+y})=f(z)^{f  (x)}f(z)^{f(y)}.

    Solution 111

    There are two possibilities:
    If, \lambda = p^{5}, we have \min(a+b) = p^{2}+p.
    If, \lambda = p_{1}p_{2}^{2}, then \min(a+b) = p_{1}p_{2}+p_{1}^{2}p_{2}.

    A quick question to Lord of the Flies:
    Are you sure about problem 104? For example, the right-handed side is not defined when x=0, whereas the left-handed side is.

    (Original post by Jkn)
    Hahahaa, did you figure out why? :rolleyes:
    Nope, I cannot think of anything that impels you to put such restrictions.
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    (Original post by Mladenov)
    f(z)^{f(xy)}=f(z^{xy})=f(z^{x})^  f(y)=f(z)^{f(x)f(y)}.
    f(z)^{f(x+y)}=f(z^{x+y})=f(z)^{f  (x)}f(z)^{f(y)}.

    Solution 111

    There are two possibilities:
    If, \lambda = p^{5}, we have \min(a+b) = p^{2}+p.
    If, \lambda = p_{1}p_{2}^{2}, then \min(a+b) = p_{1}p_{2}+p_{1}^{2}p_{2}.

    A quick question to Lord of the Flies:
    Are you sure about problem 104? For example, the right-handed side is not defined when x=0, whereas the left-handed side is.
    thanks
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    (Original post by Mladenov)
    Nope, I cannot think of anything that impels you to put such restrictions.
    I found it when I looked at the Bulgarian Olympiads hahaha!

    ----------

    Everyone who doing/is interested in STEP, check out the new thread I made: "STEP Mathematics Problem Solving Society"

    I'm basically hoping to create something just like this but for STEP questions! :bhangra:
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    (Original post by Mladenov)
    ...
    Typo, it is [-1,1]\to\mathbb{R}
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    Problem 112*** difficult..

    Evaluate \displaystyle \int_{0}^{\frac{\pi}{2}} x \ln(\tan x)dx.

    Problem 113*** I cannot resist posting number theory!

    Let p be a prime number and m,n - positive integers. We have \displaystyle \sum_{i \equiv 0 \pmod p} (-1)^{i} \dbinom{n}{i} i^{m} \equiv 0 \pmod {p^{\left[\frac{n-m-1}{p-1}\right]}}.

    (Original post by Jkn)
    I found it when I looked at the Bulgarian Olympiads hahaha!
    It is given on the third round, right?
    It is too easy for the fourth round; I have solved negligible number of questions of our third round.

    (Original post by Lord of the Flies)
    Typo, it is [-1,1]\to\mathbb{R}
    Well. I am gonna ace it now.
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    (Original post by Mladenov)
    It is given on the third round, right?
    It is too easy for the fourth round; I have solved negligible number of questions of our third round.
    That's right! Hahaaha! We can't all be Bulgarian geniuses :rolleyes:. Come do some STEP on my thread! It's really empty
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    (Original post by Jkn)
    We can't all be Bulgarian geniuses :rolleyes:.
    I know right :rolleyes:
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    I was feeling smug about having solved someone's quantum mechanics problem. Then I returned here....
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    (Original post by bananarama2)
    I know right :rolleyes:
    Fortunately though, we can be English geniuses :rolleyes:
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    (Original post by Jkn)
    Fortunately though, we can be English geniuses :rolleyes:
    You can! I can be hopeful.
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    (Original post by bananarama2)
    You can! I can be hopeful.
    We will work on it next year! Big up Emmanuel 2K13!
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    (Original post by Jkn)
    We will work on it next year! Big up Emmanuel 2K13!
    I think you and all the other freshers will think I'm some kind of social retard. :lol:
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    (Original post by bananarama2)
    I think you and all the other freshers will think I'm some kind of social retard. :lol:
    Hahahahahaha, **** it uni's a time to reinvent yourself

    I will teach you some drinking games :pierre:
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    Solution 112

    \begin{aligned}\displaystyle \int_0^{\frac{\pi}{2}}x\ln \tan x\,dx &=\int_0^{\frac{\pi}{2}}x\ln \sin x\,dx-\int_0^{\frac{\pi}{2}}x\ln \cos x\,dx \\&=\int_0^{\frac{\pi}{2}}2x\ln \sin x\,dx-\frac{\pi}{2}\int_0^{\frac{\pi}{  2}}\ln \sin x\,dx \\&=\int_0^{\frac{\pi}{2}}2x\ln \sin x\,dx+\frac{\pi^2}{4}\ln 2\end{aligned}

    Using standard formulae:

    \displaystyle \sum_{k=1}^n \sin 2kz =\frac{1}{2\sin z} \Big( \cos z-\cos (2n+1)z\Big) which upon integration with 0<x<\frac{\pi}{2} gives:

    \displaystyle \int_x^{\frac{\pi}{2}} \frac{\cos (2n+1) z}{\sin z}\,dz=\sum_{k=1}^n \frac{\cos 2k x}{k}+\sum_{k=1}^n \frac{(-1)^{k+1}}{k}+\ln \sin x

    Note that \displaystyle \int_x^{\frac{\pi}{2}} \frac{\cos (2n+1) z}{\sin z}\,dz=\frac{(-1)^{n+1}}{2n+1}+\frac{\sin (2n+1) x}{(2n+1)\sin x}- \int_x^{\frac{\pi}{2}} \frac{\sin (2n+1) z\cos z}{(2n+1)\sin^2 z}\,dz

    Which \to 0 as n\to\infty. Hence we obtain the relation: \displaystyle \ln \sin x= -\sum_{k=1}^{\infty} \frac{\cos 2k x}{k}-\ln 2

    \displaystyle \begin{aligned}\int_0^{\frac{\pi  }{2}}2x\ln \sin x\,dx &=-\int_0^{\frac{\pi}{2}}\sum_{k=1}  ^{\infty} \frac{2x\cos 2k x}{k}+2x\ln 2\,dx \\&=\sum_{k=1}^{\infty}\int_0^{ \frac{\pi}{2}}\frac{\sin 2kx}{k^2}\,dx-\frac{\pi^2}{4}\ln 2\\&=\sum_{k=1}^{\infty} \frac{1}{2k}\left[\frac{1-\cos k\pi}{k^2}\right]-\frac{\pi^2}{4}\ln 2\\&=\sum_{m=0}^{\infty}\frac{1}  {(2m+1)^3}-\frac{\pi^2}{4}\ln 2

    Hence:

    \begin{aligned}\displaystyle \int_0^{\frac{\pi}{2}}x\ln \tan x\,dx &= \sum_{m=0}^{\infty}\frac{1}{(2m+  1)^3}\\&=\sum_{q=1}^{\infty} \frac{1}{q^3}-\sum_{q=1}^{\infty}\frac{1}{(2q)  ^3}\\&= \frac{7}{8} \sum_{q=1}^{\infty}\frac{1}{q^3}  \\&= \frac{7}{8}\zeta (3)
 
 
 
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