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    (Original post by shamika)
    ...
    Suppose x_{0}^{3}-4x_{0}+7=0, x_{0}=\frac{p}{q}, \gcd(p,q)=1. So, \displaystyle \left(\frac{p}{q}\right)^{3}-4\frac{p}{q}+7=0 \Rightarrow p^{3}+7q^{3}=4pq^{2}. The last equation gives 7q^{3} \equiv 0 \pmod p, hence  p= \pm1 ,\pm7. Next, p^{3} \equiv 0 \pmod q, which implies q= \pm 1. We have to check that none of  \pm 1, \pm 7 is a solution.
    There is a more general result, which is easy to prove. If I am not mistaken, it is called rational root theorem.
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    (Original post by Felix Felicis)
    I think Boy Wonder's solution should get credited in the OP too :teehee:
    But that solution is so ugly...
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    (Original post by Mladenov)
    Suppose x_{0}^{3}-4x_{0}+7=0, x_{0}=\frac{p}{q}, \gcd(p,q)=1. So, \displaystyle \left(\frac{p}{q}\right)^{3}-4\frac{p}{q}+7=0 \Rightarrow p^{3}+7q^{3}=4pq^{2}. The last equation gives 7q^{3} \equiv 0 \pmod p, hence  p= \pm1 ,\pm7. Next, p^{3} \equiv 0 \pmod q, which implies q= \pm 1. We have to check that none of  \pm 1, \pm 7 is a solution.
    There is a more general result, which is easy to prove. If I am not mistaken, it is called rational root theorem.
    Yep! I think this is one of those instances where the question would've been easier if they'd ask you to prove the general result, because its easier to see what do to there.

    Also thanks, because I was too lazy to latex myself
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    Problem 66*

    For any fixed number q, let \displaystyle A_n=\frac{q^n}{n!}\int^{\pi}_{0} \left[ x\left( \pi-x \right) \right]^{n}sin{x}dx.

    Show that \displaystyle A_n=\left(4n-2\right)qA_{n-1}-\left(q\pi\right)^{2}A_{n-2} and use this result to deduce that A_n is an integer for all n if q and q\pi are integers.

    Show also that A_n \to 0 as n \to \infty. Deduce that \pi is irrational.
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    (Original post by und)
    ...
    Interestingly you seem to have changed problem 60 from * to **/*** :eyeball: I was just looking at that and thinking how am I going to do this with A-level knowledge
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    (Original post by Felix Felicis)
    Interestingly you seem to have changed problem 60 from * to **/*** :eyeball: I was just looking at that and thinking how am I going to do this with A-level knowledge
    On further examination I decided that I should be kind to people by not giving them false hope.
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    (Original post by und)
    Problem 66*

    For any fixed number q, let \displaystyle A_n=\frac{q^n}{n!}\int^{\pi}_{0} \left[ x\left( \pi-x \right) \right]^{n}sin{x}dx.

    Show that \displaystyle A_n=\left(4n-2\right)qA_{n-1}-\left(q\pi\right)^{2}A_{n-2} and use this result to deduce that A_n is an integer for all n if q and q\pi are integers.

    Show also that A_n \to 0 as n \to \infty. Deduce that \pi is irrational.
    Done some introductory example sheets have we?
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    (Original post by Indeterminate)
    See above - I have a problem with the 'and'.

    If q \in \mathbb{Z} then q\pi is not an integer.
    you're using prior knowledge about pi. how do you know pi can't be expressed in the form p/q ?
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    (Original post by ben-smith)
    you're using prior knowledge about pi. how do you know pi can't be expressed in the form p/q ?
    Yes, it was an oversight on my part
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    (Original post by Star-girl)
    I haven't seen very many inequality questions around, so I'll throw some in.

    Problem 62*

    Show that if p>m>0, then
    \dfrac{p-m}{p+m} \leq \dfrac{x^2-2mx+p^2}{x^2+2mx+p^2} \leq \dfrac{p+m}{p-m} ,\ \forall x \in \mathbb{R}
    Solution 62
    \dfrac{x^2-2mx+p^2}{x^2+2mx+p^2} = \dfrac{(x-m)^2-m^2+p^2}{(x+m)^2-m^2 + p^2}
    This implies that the denominator and numerator are greater than 0 where p>m>0

    Assume True.
    Expansion and simplification. Of left hand inequality:
    \ 2mx^2 + 2mp^2 - 4pmx \geq 0

\Rightarrow \ x^2 - 2px + p^2 \geq 0

\Rightarrow \ (x-p)^2 \geq 0
    No Contradiction. True for all real x.

    Expansion and simplification. Of Right hand inequality:
    \ 2mx^2 + 2mp^2 + 4pmx \geq 0

\Rightarrow \ x^2 + 2px + p^2 \geq 0

\Rightarrow \ (x+p)^2 \geq 0
    No contradiction. True for all real x.
    Steps are reversible. Inelegant but functional
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    Problem 67*

    A lense maker wishes to measure the radius of a hemispherical lense by placing a 1 meter ruler on top of the lense and causing it oscillate with a small amplitude. Assuming the ruler is much larger than the radius of the lense, show, using appropriate approximations that:

     \displaystyle r= \frac{1}{3g} \left(\frac{\pi}{T} \right) ^2

     T is the time period.

     r radius.
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    (Original post by joostan)
    Solution 62
    \dfrac{x^2-2mx+p^2}{x^2+2mx+p^2} = \dfrac{(x-m)^2-m^2+p^2}{(x+m)^2-m^2 + p^2}
    This implies that the denominator and numerator are greater than 0 where p>m>0

    Assume True.
    Expansion and simplification. Of left hand inequality:
    \ 2mx^2 + 2mp^2 - 4pmx \geq 0

\Rightarrow \ x^2 - 2px + p^2 \geq 0

\Rightarrow \ (x-p)^2 \geq 0
    No Conradiction. True for all real x.

    Expansion and simplification. Of Right hand inequality:
    \ 2mx^2 + 2mp^2 + 4pmx \geq 0

\Rightarrow \ x^2 + 2px + p^2 \geq 0

\Rightarrow \ (x+p)^2 \geq 0
    No contradiction. True for all real x.
    Inelegant but functional
    I see where you're coming from, but what you have actually shown is that if the result is true, then it implies another true result... but this is not actually a proof that the result is true itself because the implication here is the wrong way round.

    Hint:
    Spoiler:
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    Consider y= \dfrac{x^2-2mx+p^2}{x^2+2mx+p^2} and then rearrange to get a quadratic in  x .
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    Solution 60

    \displaystyle\int_{-\infty}^{\infty} \sin \left(\pi^4x^2+\frac{1}{x^2} \right)\,dx

    \displaystyle = \cos 2\pi^2\underbrace{\int_{-\infty}^{\infty} \sin \left(\pi^2x-\frac{1}{x}\right)^2\,dx}_{ \alpha}+\sin 2\pi^2\underbrace{\int_{-\infty}^{\infty} \cos \left(\pi^2x-\frac{1}{x}\right)^2\,dx}_{\beta  }

    In general, \gamma >0:

    \begin{aligned}\displaystyle \int_{0}^{\infty}f \left(\gamma^2 x-\frac{1}{x}\right)^2\,dx &= \frac{1}{\gamma}\int_{0}^{\infty  }f \left(\gamma x- \frac{\gamma}{x}\right)^2\,dx \quad(\textstyle x\to  \frac{x}{\gamma} )\\ &=\frac{1}{\gamma} \left[ \int_{0}^{1} f \left(\gamma x-\frac{\gamma}{x} \right)^2\,dx+ \int_{1}^{\infty} f \left(\gamma x-\frac{\gamma}{x}\right)^2\,dx \right]\\&= \frac{1}{\gamma}\left[\int_{1}^{\infty} \frac{1}{x^2}f \left(\gamma x-\frac{\gamma}{x} \right)^2\,dx+\int_{1}^{\infty} f\left(\gamma x-\frac{\gamma}{x}\right)^2\,dx \right]\quad( x\to\textstyle \frac{1}{x})\\&=\frac{1}{\gamma^  2} \int_{0}^{\infty} f (t^2 )\,dt \quad \big(t=\gamma x-\textstyle\frac{\gamma}{x}\big) \end{aligned}

    Hence we have:

    \displaystyle\alpha =\frac{1}{\pi^2} \int_{-\infty}^{\infty} \sin t^2 dt and \displaystyle\beta =\frac{1}{\pi^2} \int_{-\infty}^{\infty} \cos t^2 dt

    Consider:

    \begin{aligned}\displaystyle\int  _{-\infty}^{\infty} \cos t^2 dt-i\int_{-\infty}^{\infty} \sin t^2 dt &=\int_{-\infty}^{\infty} \cos(- t^2) dt+i\int_{-\infty}^{\infty} \sin(- t^2) dt\\&=\int_{-\infty}^{\infty} e^{-it^2}dt\\&=\int_{-\infty}^{\infty} e^{-\left[\left(\frac{\sqrt{2}}{2}+i\frac{ \sqrt{2}}{2}\right)t\right]^2}dt\\&=\sqrt{\pi}\left( \frac{ \sqrt{2}}{2}+i\frac{ \sqrt{2}}{2}\right)^{-1}\\&=\frac{\sqrt{2\pi}}{2}-i\frac{\sqrt{2\pi}}{2}\end{align  ed}

    It follows that \alpha =\beta = \dfrac{1}{\pi^2}\cdot \dfrac{\sqrt{2\pi}}{2}= \dfrac{1}{\pi \sqrt{2\pi}} and hence:

    \displaystyle\int_{-\infty}^{\infty} \sin \left(\pi^4x^2+\frac{1}{x^2} \right)\,dx=\frac{\cos 2\pi^2+\sin 2\pi^2}{\pi\sqrt{2 \pi}}


    \star

    In fact, using similar tricks one can show that p,q>0:

    \displaystyle\int_{-\infty}^{\infty} \sin \left(px^2+\frac{q}{x^2} \right)\,dx=\sqrt{\pi}\left( \frac{\cos 2\sqrt{pq}+\sin 2\sqrt{pq}}{\sqrt{2 p}}\right)

    So for instance, we have the surprisingly clean result of:

    \displaystyle\int_{-\infty}^{\infty} \sin \left(\pi x^2+\frac{\pi}{x^2} \right)\,dx=\frac{1}{\sqrt{2}}
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    Solution 23

    Let us denote \displaystyle S_{n}= \sum_{i+j+k=n} ijk. Consider, \displaystyle F(x)= \sum_{n=0}^{\infty} S_{n}x^{n}= \left(\sum_{n=0}^{\infty} nx^{n}\right)^{3}.
    From \displaystyle \sum_{n=0}^{\infty} x^{n}=\frac{1}{1-x}, we have \displaystyle \sum_{n=0}^{\infty} nx^{n-1}=\frac{1}{(1-x)^{2}}.
    Therefore \displaystyle F(x)=  \left(\frac{x}{(1-x)^{2}}\right)^{3}= \frac{x^{3}}{(1-x)^{6}}= x^{3}\left(\sum_{n=0}^{\infty} \dbinom{n+5}{5}x^{n}\right)= \sum_{n=0}^{\infty} \dbinom{n+2}{5}x^{n}.
    Hence S_{n}= \dbinom{n+2}{5}.
    Double counting, I believe, also works; notwithstanding, I am in a quandary as to what to count to obtain the sum.

    Solution 62

    \displaystyle P(x)= \frac{x^{2}-2mx+p^{2}}{x^{2}+2mx+p^{2}}= 1-\frac{4mx}{x^{2}+2mx+p^{2}}. Further, \displaystyle P'(x)= -\frac{4m(p^2-x^2)}{(x^{2}+2mx+p^{2})^{2}}, \displaystyle P''(x)= -\frac{8mx^{3}-24mp^{2}x-16m^{2}p^{2}}{(x^{2}+2mx+p^{2})^  {3}}. Hence, \displaystyle \max_{x\in \mathbb{R}}P(x)=P(-p)= \frac{p+m}{p-m} \ge P(x) \ge \displaystyle \min_{x\in \mathbb{R}}P(x)=P(p)= \frac{p-m}{p+m}.
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    (Original post by Mladenov)
    Solution 23

    Let us denote \displaystyle S_{n}= \sum_{i+j+k=n} ijk. Consider, \displaystyle F(x)= \sum_{n=0}^{\infty} S_{n}x^{n}= \left(\sum_{n=0}^{\infty} nx^{n}\right)^{3}.
    From \displaystyle \sum_{n=0}^{\infty} x^{n}=\frac{1}{1-x}, we have \displaystyle \sum_{n=0}^{\infty} nx^{n-1}=\frac{1}{(1-x)^{2}}.
    Therefore \displaystyle F(x)=  \left(\frac{x}{(1-x)^{2}}\right)^{3}= \frac{x^{3}}{(1-x)^{6}}= x^{3}\left(\sum_{n=0}^{\infty} \dbinom{n+5}{5}x^{n}\right)= \sum_{n=0}^{\infty} \dbinom{n+2}{5}x^{n}.
    Hence S_{n}= \dbinom{n+2}{5}.
    Double counting, I believe, also works; notwithstanding, I am in a quandary as to what to count to obtain the sum.

    Solution 62

    \displaystyle P(x)= \frac{x^{2}-2mx+p^{2}}{x^{2}+2mx+p^{2}}= 1-\frac{4mx}{x^{2}+2mx+p^{2}}. Further, \displaystyle P'(x)= -\frac{4m(p^2-x^2)}{(x^{2}+2mx+p^{2})^{2}}, \displaystyle P''(x)= -\frac{8mx^{3}-24mp^{2}x-16m^{2}p^{2}}{(x^{2}+2mx+p^{2})^  {3}}. Hence, \displaystyle \max_{x\in \mathbb{R}}P(x)=P(-p)= \frac{p+m}{p-m} \ge P(x) \ge \displaystyle \min_{x\in \mathbb{R}}P(x)=P(p)= \frac{p-m}{p+m}.
    Finally someone does it!

    This is the way I would do it - there is a simple (but ingenious) way of doing it too,
    Spoiler:
    Show
    by considering ways of removing five objects from n+2 objects
    .

    Note that once you think of using a generating function, the problem becomes rather straightforward
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    (Original post by shamika)
    ...
    Recently, I have started studying analytic number theory, and it occurred to me that I can use generating functions to solve this problem.
    Spoiler:
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    Who on earth could come up with the idea of removing 5 objects from a set of n+2 objects? There must be a good motivation.


    Another interesting application of generating functions is the following problem:
    Determine whether there exist an infinite set S, S\subset \mathbb{Z^{+}} and a positive integer N \in \mathbb{Z^{+}} such that all positive integers which are greater than N have the same number of representations in the form a+b where a,b \in S, a \ge b.
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    Solution 66

    Why, in the world, is [.] used here? I spent some time to convince myself that the statement is not true if [.] is the well-known arithmetic function.

    \begin{aligned}\displaystyle \int_{0}^{\pi} \left(x(\pi-x)\right)^{n} \sin x dx&= \int_{0}^{\pi} n\left(x(\pi-x)\right)^{n-1}(\pi-2x)\cos x dx\\&=-\int _{0}^{\pi} n(n-1)(\pi-2x)^{2}\left(x(\pi-x)\right)^{n-2}\sin x dx + \int_{0}^{\pi} 2n\left(x(\pi-x)\right)^{n-1}\sin x dx\\& =-n(n-1)\pi^{2}\int_{0}^{\pi} \left(x(\pi-x)\right)^{n-2}\sin x dx + (4n^{2}-2n)\int_{0}^{\pi} \left(x(\pi-x)\right)^{n-1}\sin x dx \end{aligned}.

    Hence, \displaystyle A_{n}= (4n-2)qA_{n-1}-q^{2}\pi^{2}A_{n-2} .

    Next, suppose that q and q\pi are positive integers(WLOG). I assume n \ge 0. We have A_{0}=2 and A_{1}=4q. Therefore, it follows, by induction, that A_{n} is an integer. Notice that A_{n} >0.

    |(x(\pi-x))^{n} \sin x| \le \frac{\pi^{2n}}{4^{n}}.
    So, A_{n} \le \pi \frac{\left(\frac{q\pi^{2}}{4} \right)^{n}}{n!}= B_{n}.
    For n sufficiently large, B_{n} > B_{n+1}, and \displaystyle \lim_{n\to\infty} B_{n}= 0. Hence \displaystyle \lim_{n \to \infty} A_{n} =0.

    Consequently, for any \epsilon there exists m such that for n >m, |A_{n}| <\epsilon. This, together with the inequality A_{n} >0 and the fact that A_{n} is an integer, leads to a contradiction.

    We conclude that there in no integer q such that q\pi \in \mathbb{Z}. Therefore \pi is irrational.
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    (Original post by Mladenov)
    For n sufficiently large, A_{n} > A_{n+1}. Hence \displaystyle \lim_{n \to \infty} A_{n} =0.
    I don't like this bit, the same could be said about 1+(1/n) but that converges to 1 not 0.
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    (Original post by james22)
    ....
    It should have been B_{n}>B_{n+1}.

    As we know, a bounded, monotonically decreasing sequence has limit. Call it B. In our case, B=\displaystyle \lim_{n \to \infty} B_{n-1} \frac{\frac{q\pi^{2}}{4}}{n}= B  \lim_{n \to \infty} \frac{\frac{q\pi^{2}}{4}}{n}= 0, where B_{n}= \pi \frac{\left(\frac{q\pi^{2}}{4} \right)^{n}}{n!}\right).
    Now, 0< A_{n} \le B_{n}. Since \displaystyle \lim_{n\to \infty} B_{n} = 0, we have \displaystyle \lim_{n\to \infty} A_{n}= 0.
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    (Original post by Mladenov)
    Solution 66

    Why, in the world, is [.] used here? I spent some time to convince myself that the statement is not true if [.] is the well-known arithmetic function.
    Sorry, I don't understand. Are you complaining about the square brackets?

    I might post my solution at some point I did it slightly differently.
 
 
 
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