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    (Original post by Lord of the Flies)

    Problem 103* (there is a slick *** solution though)

    \displaystyle \int_0^{\infty} \arctan\left( \frac{x(e-1)}{ex^2+1}\right)\frac{dx}{x}
    Noticing the addition formula and regarding e as an arbitrary parameter:
    I(e)=\displaystyle \int_0^{\infty} \arctan\left( \frac{x(e-1)}{ex^2+1}\right)\frac{dx}{x}= \displaystyle \int_0^{\infty} \dfrac{arctan(ex)-arctanx}{x}dx

\Rightarrow I'(e)=\displaystyle \int_0^{\infty} \dfrac{1}{1+(ex)^2}dx=\frac{1}{e  } \frac{\pi}{2}

\Rightarrow I(e)=ln(e) \pi/2+C
    setting e=1 gives C=0 and letting e be eulers constant we get I(e)=pi/2.
    I've had a shocking day mathswise, am I close?
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    (Original post by ben-smith)
    I've had a shocking day mathswise, am I close?
    Can't be worse than me. I've been wrestling with "games" that are ironically not fun at all for the whole day :coffee:
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    (Original post by Lord of the Flies)
    I thought it had been solved! Here we go:

    Solution 98

    \dfrac{a_{n+1}-a_n}{\sqrt{n+1}-\sqrt{n}}=\dfrac{1}{a_{n}\big( \sqrt{n+1}-\sqrt{n}\big)}\to 0\;\;\Rightarrow \;\;\dfrac{a_n}{\sqrt{n}}\to 0 by Stolz-Cesaro.
    Hmmm...never heard of that test. It looks pretty cool. However; could you walk me through your justification as to why the original limit exists?

    Edit: The one with the a_n and roots on the denominator.
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    (Original post by ukdragon37)
    Can't be worse than me. I've been wrestling with "games" that are ironically not fun at all for the whole day :coffee:
    Part III sounds tough


    (Original post by Lord of the Flies)
    Yep - there is an arctan missing but that is just a typo.
    You have me intrigued as to what the slick way is. Complex numbers?
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    (Original post by ben-smith)
    Part III sounds tough
    Thankfully I only need a pass to move on to PhD. And if my exams go as I planned I should only need 65% in my dissertation to get a distinction (75%) overall (60% is the passmark).
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    (Original post by ukdragon37)
    Thankfully I only need a pass to move on to PhD. And if my exams go as I planned I should only need 65% in my dissertation to get a distinction (75%) overall (60% is the passmark).
    Are you going to minor in anything cool like, I don't know... maths?
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    (Original post by ben-smith)
    Are you going to minor in anything cool like, I don't know... maths?
    I was thinking of minor-ing in Law or Music actually Law was my first choice for applying to Cambridge before I changed my mind and went with my second choice, CompSci.
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    (Original post by SParm)
    Hmmm...never heard of that test. It looks pretty cool. However; could you walk me through your justification as to why the original limit exists?

    Edit: The one with the a_n and roots on the denominator.
    a_{n+1}-a_n>0 hence either a_n\to \ell or a_n\to \infty as n\to\infty. A quick contradiction shows that there is no such \ell

    (Original post by ben-smith)
    You have me intrigued as to what the slick way is. Complex numbers?
    You're going to be disappointed - essentially the same as what you did, but set up differently:

    \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}



    Another? You TSR folk are too good!

    Problem 105* (technically, if we are rigorous, this is a ***)

    \displaystyle \int_0^{\infty} \left\{x-\frac{x^3}{2}+\frac{x^5}{2\cdot 4}-\frac{x^7}{2\cdot 4\cdot 6}+\cdots\right\}\left\{1+\left( \frac{x}{2}\right)^2+\left(\frac  {x^2}{2 \cdot 4}\right)^2+\left(\frac{x^3}{2 \cdot 4 \cdot 6}\right)^2+ \cdots\right\}\,dx

    (the curly braces don't mean anything in particular - I just thought they looked nice)
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    (Original post by Lord of the Flies)
    a_{n+1}-a_n>0 hence either a_n\to \ell or a_n\to \infty as n\to\infty. A quick contradiction shows that there is no such \ell
    That bit is fine. I'm just wondering about what happens to the roots at n goes to infinity. Mind it's late and I'm tired but I certainly didn't get 0 for this (Well I didn't then I did by second guessing myself and then I didn't again once I realised I was being a tit.)
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    Problem 106

    Prove the formula  \displaystyle\sum_{k=0}^n \prod_{\mu=1}^a(k+\mu) = \dfrac{1}{a+1} \displaystyle\prod_{\mu=1}^{a+1}  (n+\mu)


    Problem 107

    Evaluate the series  \displaystyle\sum_{n=0}^{\infty}  \dfrac{(n+\mu)!}{2^n n!\mu!}
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    (Original post by SParm)
    ...
    Oh no, you're absolutely correct. I am the one being an idiot. I somehow omitted the roots in my working leaving (n+1) - n = 1. My sincere apologies.

    Edit: amended below.
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    (Original post by Lord of the Flies)
    Oh no, you're absolutely correct. I am the one being an idiot. I somehow omitted the roots in my working leaving (n+1) - n = 1. My sincere apologies.
    It's fine, I'm sure we've all had that one (or few dozen for me) shockers when writing maths down on paper. I'm reluctant to post my solution to this, even if I'm confident it works, simply because this thread is brilliant for pre-undergrad level peeps to have a go at some cool maths. My solving what I think is a lovely Analysis problem with a first year Analysis course under my belt kind of ruins that a bit.

    I'm sure Mladenov would be happy to give anyone who wants to do it some hints, as would I.

    Edit: Also my solution's laborious to the max. I'm sure one of you guys can come up with a lovely 4 or 5 liner.
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    A lot of big terrifying things flying around this thread
    Spoiler:
    Show
    :troll:
    so how about something simple:

    Problem 108*


    Find all pairs of integers (p,q) such that the the roots of the equation, \displaystyle(px-q)^2+(qx-p)^2=x, are integers.

    Spoiler:
    Show
    Mladenov isn't allowed to solve it and I'm sure he knows why


    -----------------

    Btw, can people make sure they put 'assumed knowledge' ratings on problems please.
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    Solution 98

    a_{n+1}^2=a_n^2+\dfrac{1}{a_n^2}  +2\Rightarrow a_{n+1}^2 =a_0^2+2(n+1)+\displaystyle\sum_  {k=0}^n \dfrac{1}{a_k^2}

    Square the limit:

    \displaystyle\left(\lim_{n\to \infty} \frac{a_n}{\sqrt{n}}\right)^2 =\lim_{n\to\infty} \frac{a_n^2}{n}=2+\frac{a_0^2}{n  }+\sum_{k=0}^{n-1}\frac{1}{a_k^2n}

    w_n=\displaystyle\sum_{k=0}^{n-1}\dfrac{1}{a_k^2} then \dfrac{w_{n+1}-w_n}{(n+1)-n}=\dfrac{1}{a_n^2}\to 0 hence \dfrac{w_n}{n}\to 0 by Stolz Cesaro.

    Hence \displaystyle\lim_{n\to\infty} \frac{a_n}{\sqrt{n}}=\sqrt{2}

    Solution 106

    \displaystyle\binom{\alpha}{ \beta}+\binom{\alpha}{\beta+1}= \binom{\alpha+1}{\beta+1} \Rightarrow\displaystyle\sum_{k=  0}^n \binom{k+a}{k}=\binom{n+a+1}{n} inductively.

    Multiply by a!:

    \displaystyle\sum_{k=0}^n \frac{(k+a)!}{k!}=\frac{1}{a+1} \frac{(n+a+1)!}{n!} \iff \sum_{k=0}^n \prod_{\mu=1}^a (k+\mu)=\frac{1}{a+1}\prod_{\mu=  1}^{a+1}(n+\mu)

    Solution 107

    \displaystyle \frac{1}{(1-x)^{\mu+1}}=\sum_{k=0}^{\infty} \binom{n+\mu}{n} x^n\Rightarrow 2^{\mu +1}=\sum_{k=0}^{\infty}\frac{1}{  2^n}\binom{n+\mu}{n}
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    (Original post by FireGarden)
    Problem 106
    Prove the formula  \displaystyle\sum_{k=0}^n \prod_{\mu=1}^a(k+\mu) = \dfrac{1}{a+1} \displaystyle\prod_{\mu=1}^{a+1}  (n+\mu)
    ...for all non-negative integers n.

    Solution 106

    When n=0, LHS\displaystyle =\prod_{\mu=1}^a(\mu). RHS\displaystyle =\frac{1}{a+1} \prod_{\mu=1}^{a+1}(\mu) = \prod_{\mu=1}^a(\mu)=LHS. So true for n=0.

    Assume true for n=m. This gives...

    \displaystyle\sum_{k=0}^{m+1} \prod_{\mu=1}^a(k+\mu) =\frac{1}{a+1}\prod_{\mu=1}^{a+1  }(m+\mu) + \prod_{\mu=1}^a(m+1+\mu)
    \displaystyle =\frac{1}{a+1} \prod_{\mu=0}^a(m+1+\mu) + \prod_{\mu=1}^a(m+1+\mu)
    \displaystyle =(\frac{m+1}{a+1}+1)\prod_{\mu=1  }^{a}(m+1+\mu)
     \displaystyle =\frac{1}{m+a+2}(\frac{m+1}{a+1}  +1)\prod_{\mu=1}^{a+1}(m+1+\mu)
    \displaystyle =\frac{1}{a+1}\prod_{\mu=1}^{a+1  }((m+1)+\mu) .

    Therefore, if true for n=m, true too for n=m+1. As true for n=0, true for all non-negative integers n by mathematical induction. QED.
    Spoiler:
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    My latex masterpiece
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    (Original post by Lord of the Flies)
    Solution 98
    \lim_{n\to\infty} \frac{a_n^2}{n}=2+\frac{a_0}{n}+ \sum_{k=0}^{n-1}\frac{1}{a_kn}[/tex]
    Awesome solution, Stolz Cesaro seems pretty useful. Where did the squares go though? (even if they were squares the proof would still work so I'm just wondering)
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    (Original post by Jkn)
    Awesome solution, Stolz Cesaro seems pretty useful. Where did the squares go though? (even if they were squares the proof would still work so I'm just wondering)
    I forgot to type them in! :lol: All fixed now, thanks for noticing!
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    (Original post by Lord of the Flies)
    I forgot to type them in! :lol: All fixed now, thanks for noticing!
    Dude, you're way too tired :lol: I noticed your little \sqrt{n+1}-\sqrt{n} blunder earlier btw! I assumed it was another one of those things that are only obvious to you and mladenov so spent a few minutes trying to convince myself that a_n(\sqrt{n+1}-\sqrt{n}) was divergent! As you can imagine, I was quite puzzled! :lol: :lol:
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    (Original post by Lord of the Flies)
    Solution 106
    Oi! You only just edited that in! :lol: I better see a "(2)" pretty soon or there's gonna be hell to pay (nice and elegant btw, I clearly missed the point!)
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    (Original post by Jkn)
    Dude, you're way too tired :lol:
    Ha! I think you're right, time for bed!

    (Original post by Jkn)
    Oi! You only just edited that in! :lol: I better see a "(2)" pretty soon or there's gonna be hell to pay (nice and elegant btw, I clearly missed the point!)
    I had already solved it but was eager to post a solution to 98, after that blunder. Also, it doesn't matter what order the solutions come in - this thread is not a competition, it's for fun!
 
 
 
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