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    (Original post by DJMayes)
    I find it telling how the OP hasn't submitted a solution since about question 50, and we're now past question 120. This thread stopped being interesting (Or even accessible) for the vast majority of people quite a long time ago and now it seems very much to just be a battle of wits between Mladenov and Lord of the Flies.

    (If this comes across as jealous, then that's probably because it is - I'd love to be able to even attempt the questions that are being thrown around lately! :lol: Unfortunately, the questions that are being thrown out are aimed at a very small audience indeed and just looking at the list in the OP it's very easy to see. It reminds me very strongly of what the STEP thread was like before Christmas.)
    Well, the last question posted by Mladenov could be attemped by an A-level student.. (Though establishing the inequality between the 2nd and 3rd terms has got me in a bind at the moment! 1 < 2 and 1 < 3 were actually very easy (and im hoping the one im stuck on doesnt have an obvious solution im not missing though I don't think so, due to the three stars). Everyone could give the two sub-question inequalities i just outlined a go!
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    (Original post by shamika)
    LOL! At least you could stick my curve into Wolfram Alpha easily (which is what I was intending )

    What is it?
    :flute: Na na na na na na na na na na na na na na na na....
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    (Original post by FireGarden)
    Well, the last question posted by Mladenov could be attemped by an A-level student.. (Though establishing the inequality between the 2nd and 3rd terms has got me in a bind at the moment! 1 < 2 and 1 < 3 were actually very easy (and im hoping the one im stuck on doesnt have an obvious solution im not missing though I don't think so, due to the three stars). Everyone could give the two sub-question inequalities i just outlined a go!
    The triple stars means that I somewhat doubt it'll be simple either, although I do agree that 1 < 2 and 1 < 3 seem fair enough. However, the questions on here for the most part stopped being accessible a long time ago.
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    (Original post by ukdragon37)
    :flute: Na na na na na na na na na na na na na na na na....
    I'm really annoyed with myself I should've got that
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    Problem 126:
    I=\displaystyle \int^1_0 f(x)(x-1/2)dx=\displaystyle \int^{1/2}_0 f(x)(x-1/2)dx+\displaystyle \int^{1}_{1/2}f(x)(x-1/2)dx

\displaystyle \int^{1}_{1/2}f(x)(x-1/2)dx=\displaystyle \int^{1/2}_0f(x+1/2)xdx

\displaystyle \int^{1/2}_{0}f(x)(x-1/2)dx=-\displaystyle \int^{1/2}_{0}f(1/2-x)xdx

\therefore I=\displaystyle \int^{1/2}_{0}(f(x+1/2)-f(1/2-x))xdx \geq 0



\displaystyle \int^1_0 xf(x)dx=\displaystyle \int^{1/2}_0 xf(x)dx+\displaystyle \int^1_{1/2} xf(x)dx

=\displaystyle \int^1_{1/2}(x-1/2)f(x-1/2)+\displaystyle \int^1_{1/2} xf(x)dx
    So:
    \displaystyle \int^1_{1/2}f(x)dx-\displaystyle \int^1_0 xf(x)dx

=\int^1_{1/2} f(x)dx-\displaystyle \int^1_{1/2}(x-1/2)f(x-1/2)dx-\displaystyle \int^1_{1/2} xf(x)dx

=\displaystyle \int^1_{1/2} f(x)(1-x)dx-\displaystyle \int^1_{1/2}(x-1/2)f(x-1/2)dx

\displaystyle \int^1_{1/2} f(x)(1-x)dx=\displaystyle \int^1_{1/2}f(3/2-x)(x-1/2)dx
    So:
    \displaystyle \int^1_{1/2}f(x)dx-\displaystyle \int^1_0 xf(x)dx

=\displaystyle \int^1_{1/2}(x-1/2)(f(3/2-x)-f(x-1/2))dx \geq 0
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    Problem 127 *

    Two cards are chosen (randomly, without replacement) from a standard deck (52). Find the probability that

    i) both cards are aces given that at least one ace is chosen

    ii) both cards are aces given that the ace of spades is chosen

    and comment on your answers.
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    (Original post by Mladenov)
    Problem 44***

    Let P(n) be a polynomial with coefficients in \mathbb{Z}. Suppose that \deg(P)=p, where p is a prime number. Suppose also that P(n) is irreducible over \mathbb{Z}. Then there exists a prime number q such that q does not divide P(n) for any integer n.
    Assume for all primes q there exists n \in \mathbb{Z} such that P(n) \equiv 0\ (q). Hence P(x) is reducible in \mathbb{Z}/q\mathbb{Z}[X] for all primes q. By Chebotarev's density theorem (and a fair bit of kung fu), this implies that P(x) is reducible in \mathbb{Z}[X]; thus a contradiction.
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    (Original post by DJMayes)
    However, the questions on here for the most part stopped being accessible a long time ago.
    I'm thinking people miss the problems I was setting
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    Problem 127**/***

    Undergrads, you should be able to do this easily so give A-level people a go :P

    Prove analytically that the area under the curve E^-(x^2)=sqrt(pi)
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    Solution 127

    The probability that both cards are aces is \displaystyle \frac{\dbinom{4}{2}}{\dbinom{52}  {2}}. The probability that at least one card is an ace is \displaystyle 1-\frac{\dbinom{48}{2}}{\dbinom{52  }{2}}.
    Bayes gives \displaystyle \frac{ \frac{\dbinom{4}{2}}{\dbinom{52}  {2}}}{1-\frac{\dbinom{48}{2}}{\dbinom{52  }{2}}}.

    For ii) we have three possibilities such that one of the aces is spade. Hence the probability that two cards are aces and one is the ace of spades is \displaystyle \frac{3}{\dbinom{52}{2}}. The probability that one of the cards is the ace of spades and the oder is random is \displaystyle \frac{51}{\dbinom{52}{2}}.
    Again we use Bayes formula to obtain \displaystyle \frac{\displaystyle \frac{3}{\dbinom{52}{2}}}{ \displaystyle \frac{51}{\dbinom{52}{2}}}.


    (Original post by jack.hadamard)
    Assume for all primes q there exists n \in \mathbb{Z} such that P(n) \equiv 0\ (q). Hence P(x) is reducible in \mathbb{Z}/q\mathbb{Z}[X] for all primes q. By Chebotarev's density theorem (and a fair bit of kung fu), this implies that P(x) is reducible in \mathbb{Z}[X]; thus a contradiction.
    This is exactly what I had in mind.
    By the way, I found this result, when I was trying to prove a special case of it, and after annoying attempts to solve my problem using cyclotomic polynomials, I decided to employ Chebotarev's theorem.
    Edit: For the sake of completeness, I would like to add that we can use Chebotarev's density theorem to show that there are many primes q which satisfy the condition. In other words, we look for those primes q for which the degree of the splitting field of P over F_{q} is p.
    What do you think about the case when \deg (P) = p^{\alpha}, where \alpha &gt; 1.

    (Original post by ben-smith)
    Problem 126
    Yup, it is correct.
    Spoiler:
    Show
    The first inequality is a case of Chebishev's inequality for integrals.
    The second inequality can be done as follows:
    Spoiler:
    Show
    \displaystyle \int_{\frac{1}{2}}^{1} f(x)dx= \int_{\frac{1}{2}}^{1} ((1-x)f(x) +xf(x)) dx \ge \int_{\frac{1}{2}}^{1} ((1-x)f(1-x)+xf(x))dx= \int_{0}^{1} xf(x) dx
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    (Original post by natninja)
    Problem 127**/***

    Undergrads, you should be able to do this easily so give A-level people a go :P

    Prove analytically that the area under the curve E^-(x^2)=sqrt(pi)
    Solution 127

     \left(\int_{-\infty}^{\infty} e^{-x^2}dx \right)^2

     =\left(\int_{-\infty}^{\infty} e^{-x^2}dx \right)\left(\int_{-\infty}^{\infty} e^{-y^2}dy \right)

     =\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2+y^2)}dx dy

    Polar.

     =\int_{0}^{\infty}\int_{0}^{2\pi  } re^{-r^2}dr d\theta

     =\int_{0}^{\infty}2\pi re^{-r^2}dr

     = \pi and the result follows.
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    (Original post by bananarama2)
    Solution 127
    Very good, it's a rather neat trick imo :P

    Problem 129 **

    Show that:

    y1=(1/k)*integral [0,x] f(x')sinh{k(x-x')} dx'

    Is a particular solution to the second order differential equation:

    y''-(k^2)y=f(x)
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    (Original post by bananarama2)
     =\int_{0}^{\infty}\int_{0}^{2\pi  } re^{-r^2}dr d\theta

     =\int_{0}^{\infty}2\pi re^{-r^2}dr
    How did you do this step? Where did the theta go? And where do the new limits keep coming from?

    Nice proof btw, did you come up with it yourself? (As opposed to having seen it derived before)
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    (Original post by Jkn)
    How did you do this step? Where did the theta go? And where do the new limits keep coming from?

    Nice proof btw, did you come up with it yourself? (As opposed to having seen it derived before)
    I just integrated respects to theta. The new limits are because the range of r in polar coords is from 0 to infinity and the range of theta is 0 to 2pi. I just integrating over  \mathbb{R}^2 .

    Unfortunately not, well sort of. It was hinted at in a question I did once. (Kinda like a STEP questions I suppose)
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    (Original post by bananarama2)
    I just integrated respects to theta. The new limits are because the range of r in polar coords is from 0 to infinity and the range of theta is 0 to 2pi. I just integrating over  \mathbb{R}^2 .

    Unfortunately not, well sort of. It was hinted at in a question I did once. (Kinda like a STEP questions I suppose)
    Ohhhh I see I thought you had integrated with respect to the one on the left

    But why is it valid to say the total range of values in polar co-ordinates is exactly equivalent to the range in cartesian co-ordinates? You need some sort of formal derivation of these limits or a rigorous reason as to why this is a valid deduction.

    Haha fair enough Were you originally going to apply to do maths then?

    I've been trying to come up with a solution of my own. I got as far as using repeated applications of integration by parts to equate it to e^{-x^2} multiplied by an infinite polynomial with coefficients in a well-defined (but complicated!) pattern. Now it gets a bit trickier. The Maclaurin's expansion doesn't seem to help. The only thing that seems promising is to consider the possible convergence of terms in the form \frac{e^{x^2}}{x^n} which is pretty tricky!

    Any ideas?
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    (Original post by Lord of the Flies)
    



You're going to be disappointed - essentially the same as what you did, but set up differently:



[tex]\displaystyle\begin{aligned} \int_0^{\infty} \frac{\arctan ex-\arctan x}{x}\,dx &= \int_0^{\infty}\int_x^{ex}\frac{  1}{x(1+t^2)}\,dt \,dx\\&=\int_0^{\infty}\int_{t/e}^{t}\frac{1}{x(1+t^2)}\,dx \, dt=\int_0^{\infty}\frac{dt}{1+t^  2}=\frac{\pi}{2}\end{aligned}
    Is this the (***) solution? If not, I don't have any knowledge of double integrals at A Level! :lol:
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    (Original post by natninja)
    Very good, it's a rather neat trick imo :P

    Problem 129 **

    Show that:

    y1=(1/k)*integral [0,x] f(x')sinh{k(x-x')} dx'

    Is a particular solution to the second order differential equation:

    y''-(k^2)y=f(x)
    Solution 129

     \frac{d}{dx} \frac{1}{k} \int_0^{x} f(x')\sinh (k(x-x')) dx'

    = \frac{1}{k} f(x) \sinh (k(x-x)) - f(0) \frac{1}{k}\sinh(kx) \times 0 + \frac{1}{k} \int_0^x k f(x') \cosh (k(x-x')) dx'

    = \int_0^x f(x') \cosh (k(x-x')) dx'

     \frac{d}{dx} \int_0^x f(x') \cosh (k(x-x')) dx'

     = f(x)\cosh (k(x-x)) - 0 + \int_0^x kf(x')\shin (k(x-x')) dx'

     = y_1''

    SO

     y_1'' - k^2y

     = f(x)\cosh (k(x-x)) + \int_0^x kf(x')\shin (k(x-x')) dx' - \int_0^x kf(x')\shin (k(x-x')) dx'

     = f(x)
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    (Original post by Jkn)
    Ohhhh I see I thought you had integrated with respect to the one on the left

    But why is it valid to say the total range of values in polar co-ordinates is exactly equivalent to the range in cartesian co-ordinates? You need some sort of formal derivation of these limits or a rigorous reason as to why this is a valid deduction.
    Rigorous? I'm a Natsci :rofl: Well I don't see the need for it here. I'm just evaluating the integral over all possible coords in the x-y plane. I don't think I understand you problem?

    Haha fair enough Were you originally going to apply to do maths then?

    I've been trying to come up with a solution of my own. I got as far as using repeated applications of integration by parts to equate it to e^{-x^2} multiplied by an infinite polynomial with coefficients in a well-defined (but complicated!) pattern. Now it gets a bit trickier. The Maclaurin's expansion doesn't seem to help. The only thing that seems promising is to consider the possible convergence of terms in the form \frac{e^{x^2}}{x^n} which is pretty tricky!

    Any ideas?
    No, I've never wanted to apply to do maths

    I don't at the minute, give me a few minutes.
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    (Original post by bananarama2)
    Rigorous? I'm a Natsci :rofl: Well I don't see the need for it here. I'm just evaluating the integral over all possible coords in the x-y plane. I don't think I understand you problem?
    For ****s sake :lol: classic physicist :lol:

    Well why is it that the cartesian domain (-\infty,\infty) translates exactly to [0,2\pi) and [0,\infty) in each parameter of the polar plane respectively? You must derive them using the standard techniques for variable-changing substitutions. You've plucked the limits out of nowhere without thought as to the size of each numerical set. For example: when substituting r^2=x^2+y^2, in order to ensure you don't have r \in (\infty,\infty), which is nonsense, you would have to split the limit exactly into two halves. This means you now have a "2" and the fact this cancels out is merely fortunate as per the elegance of the final result, rather than being an immediate deduction (though there might be an obvious way to deduce it that I'm missing ...)
    I don't at the minute, give me a few minutes.
    Hahaha :lol:

    Edit: wait.... "integrating over R squared"?! How/why have you learnt this already? ;o Why have you been doing STEP then? And how good are you at the questions?
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    As you want another proof for problem 128, try to prove the following result:

    Problem 130*

    \begin{aligned} \displaystyle \frac{n}{2n+1}\frac{((2n)!!)^{2}  }{((2n-1)!!)^{2}(2n+1)} &lt; \left(\int_{0}^{\infty} e^{-x^{2}}dx \right)^{2} &lt; \frac{n}{2n-1}\frac{(2n-3)!!)^{2}(2n-1)}{((2n-2)!!)^{2}}\left(\frac{\pi}{2} \right)^{2} \end{aligned}.
    Then note that e^{-x^{2}} is even.
 
 
 
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