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    (Original post by Zacken)
    And I'll have to make less mistakes.
    Me too!
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    Problem 573*:

    Compute the 100th derivative of \displaystyle \frac{x^2 + 1}{x^3 - x}.
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    (Original post by Zacken)
    Problem 573*:

    Compute the 100th derivative of \displaystyle \frac{x^2 + 1}{x^3 - x}.
    Spoiler:
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    \dfrac{x^2+1}{x^3-x}=\dfrac{x(x+1)+x(x-1)-(x-1)(x+1)}{x(x-1)(x+1)}

\implies \dfrac{d^{100}}{dx^{100}}\left( \dfrac{x^2+1}{x^3-x} \right) = 100!\left( \dfrac{1}{(x-1)^{101}}+ \dfrac{1}{(x+1)^{101}}-\dfrac{1}{x^{101}}\right)
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    (Original post by joostan)
    Spoiler:
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    \dfrac{x^2+1}{x^3-x}=\dfrac{x(x+1)+x(x-1)-(x-1)(x+1)}{x(x-1)(x+1)}

\implies \dfrac{d^{100}}{dx^{100}}\left( \dfrac{x^2+1}{x^3-x} \right) = 100!\left( \dfrac{1}{(x-1)^{101}}+ \dfrac{1}{(x+1)^{101}}-\dfrac{1}{x^{101}}\right)
    Yep!
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    (Original post by Lord of the Flies)
    Well the concept of winning/losing moves presupposes knowledge of the game's outcome, which some sense answers your question. It requires that the winning player be perfect (i.e. will not make mistakes beyond said move), and that one player plays only winning moves, the other losing moves. That should be quite clear.So it is more whether the idea of a winning move makes sense in any game where you can only win or lose, or equivalently, whether there is a "correct" outcome to any such game. The answer is essentially yes under appropriate, almost obvious conditions, and is given in full detail by Zermelo's theorem in game theory:For every (finite) game of perfect information (that is, one where all useable information is available to both players at all times; chess, go, or reversi are examples), not involving chance (so not backgammon for instance), between two players who take moves in alternation, there is either a winning strategy for one of the players, or both players can force a draw. If we disregard the possibility of draws (as in our painting game), then we have existence of winning & losing moves/strategies.
    I love stuff like this. I wonder, do you get to delve into some game theory as a Cambridge mathmo? I might consider economics if not!
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    Problem 574* :

    Define f^{n}(a) to be the n-th derivative of f evaluated at a. Find f^{2005}(0) where

    \displaystyle

\begin{equation*} f(x) = \frac{1}{1 + 2x + 3x^2 + \cdots + 2005x^{2004}}\end{equation*}
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    (Original post by Mathstatician)
    I love stuff like this. I wonder, do you get to delve into some game theory as a Cambridge mathmo? I might consider economics if not!
    Not at undergrad level no, but I am sure there is in part III!

    To answer the \omega-extension of that problem, for those interested: we know that player 1 wins n\times m for all n,m (excluding the trivial case). Player 2 actually wins 2\times \omega (and this implies player 1 wins n\times \omega for all n\neq 2). Explanation in spoiler.

    Spoiler:
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    For 2\times \omega, player 2 can always paint so that after their turn, the board looks like this:
    and hence they win.
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    (Original post by Zacken)
    Problem 573*:

    Compute the 100th derivative of \displaystyle \frac{x^2 + 1}{x^3 - x}.
    These are great questions for quickly learning a new rearranging technique/idea, how do you come up with them?
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    (Original post by EnglishMuon)
    These are great questions for quickly learning a new rearranging technique/idea, how do you come up with them?
    I've come across these, can't take the credit for coming up with them myself, unfortunately!
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    (Original post by Lord of the Flies)
    Not at undergrad level no, but I am sure there is in part III!
    Oh. I find that surprising! I'd expect the mathematicians approach to be more rigorous and so feature later than in economics but my understanding is Warwick maths teach game theory in years two and three. Shame!
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    (Original post by Zacken)
    Problem 574* :

    Define f^{n}(a) to be the n-th derivative of f evaluated at a. Find f^{2005}(0) where

    \displaystyle

\begin{equation*} f(x) = \frac{1}{1 + 2x + 3x^2 + \cdots + 2005x^{2004}}\end{equation*}
    Spoiler:
    Show
    Note that: f(x)=\left(\displaystyle\sum_{n=  1}^{2005} nx^{n-1} \right)^{-1}.
    Note that g(x)=\displaystyle\sum_{n=0}^{20  05} x^n = \dfrac{x^{2006}-1}{x-1} so f(x)=\dfrac{1}{g'(x)}.

    g'(x)=\dfrac{2006x^{2005}(x-1)-(x^{2006}-1)}{(1-x)^2} \Rightarrow f(x)=\dfrac{(1-x)^2}{1-2006x^{2005}+2005x^{2006}}.

    \implies f(x)=(1-x)^2(1+2006x^{2005}+\mathcal{O}(  x^{2006})=(1-x)^2+2006x^{2005}+\mathcal{O}(x^  {2006}).
    Thus f^{(2005)}(x)=2006!+\mathcal{O}(  x) which implies f^{(2005)}(0)=2006!.
    Pilfered from Putnam, to replace the one I did:
    Problem 575:*/**
    Prove that the limit:
    \displaystyle \lim_{n \to \infty} \dfrac{1}{n^2}\prod_{k=1}^n(n^2+  k^2)^{\frac{1}{n}} exists and find its value.
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    (Original post by joostan)
    Spoiler:
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    Note that: f(x)=\left(\displaystyle\sum_{n=  1}^{2005} nx^{n-1} \right)^{-1}.
    Note that g(x)=\displaystyle\sum_{n=0}^{20  05} x^n = \dfrac{x^{2006}-1}{x-1} so f(x)=\dfrac{1}{g'(x)}.

    g'(x)=\dfrac{2006x^{2005}(x-1)-(x^{2006}-1)}{(1-x)^2} \Rightarrow f(x)=\dfrac{(1-x)^2}{1-2006x^{2005}+2005x^{2006}}.

    \implies f(x)=(1-x)^2(1+2006x^{2005}+\mathcal{O}(  x^{2006})=(1-x)^2+2006x^{2005}+\mathcal{O}(x^  {2006}).
    Thus f^{(2005)}(x)=2006!+\mathcal{O}(  x) which implies f^{(2005)}(0)=2006!.

    Daaamn, nice!
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    (Original post by joostan)
    Spoiler:
    Show
    Note that: f(x)=\left(\displaystyle\sum_{n=  1}^{2005} nx^{n-1} \right)^{-1}.
    Note that g(x)=\displaystyle\sum_{n=0}^{20  05} x^n = \dfrac{x^{2006}-1}{x-1} so f(x)=\dfrac{1}{g'(x)}.

    g'(x)=\dfrac{2006x^{2005}(x-1)-(x^{2006}-1)}{(1-x)^2} \Rightarrow f(x)=\dfrac{(1-x)^2}{1-2006x^{2005}+2005x^{2006}}.

    \implies f(x)=(1-x)^2(1+2006x^{2005}+\mathcal{O}(  x^{2006})=(1-x)^2+2006x^{2005}+\mathcal{O}(x^  {2006}).
    Thus f^{(2005)}(x)=2006!+\mathcal{O}(  x) which implies f^{(2005)}(0)=2006!.
    Pilfered from Putnam, to replace the one I did:
    Problem 575:*/**
    Prove that the limit:
    \displaystyle \lim_{n \to \infty} \dfrac{1}{n^2}\prod_{k=1}^n(n^2+  k^2)^{\frac{1}{n}} exists and find its value.
    Forgive my ignorance but are we allowed to assume things like the arithmetic geometric mean inequality? (not saying I have found a solution using it yet :P)
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    (Original post by EnglishMuon)
    Forgive my ignorance but are we allowed to assume things like the arithmetic geometric mean inequality? (not saying I have found a solution using it yet :P)
    I don't see why not, but I'm not convinced it's going to help. . .
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    (Original post by EnglishMuon)
    Forgive my ignorance but are we allowed to assume things like the arithmetic geometric mean inequality? (not saying I have found a solution using it yet :P)
    It is Putnam so AM-GM is trivial. Results that you can assume in the IMO can be assumed in Putnam for sure (Have not put pen to paper so I don't know if it works)


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    (Original post by physicsmaths)
    It is Putnam so AM-GM is trivial. Results that you can assume in the IMO can be assumed in Putnam for sure (Have not put pen to paper so I don't know if it works)


    Posted from TSR Mobile
    Thanks I knew Putnam is supposed to be tough but I wasn't sure if it was one of those prove everything you use type exams.
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    (Original post by joostan)
    Pilfered from Putnam, to replace the one I did:
    Problem 575:*/**
    Prove that the limit:
    \displaystyle \lim_{n \to \infty} \dfrac{1}{n^2}\prod_{k=1}^n(n^2+  k^2)^{\frac{1}{n}} exists and find its value.
    Spoiler:
    Show

    It might be easier to evaluate
    \displaystyle \lim_{n \to \infty} \prod_{k=1}^n \left(1+\left( \frac{k}{n} \right)^2 \right)^{\frac{1}{n}}

    Take logs. The resulting expression approximates the integral \int_0^1 \log(1+x^2) dx
    To evaluate this, use parts: \int_0^1 \log(1+x^2) dx = [x \log(1+x^2)]_0^1 - \int_0^1 \frac{2x^2}{1+x^2}dx.

    This is \log(2) - 2 \int_0^1 1-\frac{1}{1+x^2} dx = \log(2) - 2 + 2 \tan^{-1}(1) = \frac{\pi}{2} - 2 + \log(2)

    Finally, we want to exponentiate this to get the answer: 2 e^{-2 + \pi/2}
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    I like this because it looks wrong.

    Problem 576

    Show that there is a constant a>1 such that for any sequence (\alpha_n) of positive reals:

    \displaystyle\limsup_{n\to\infty  }\left(\frac{1+\alpha_{n+1}}{ \alpha_{n}}\right)^n\geqslant a

    What is the best possible value for a?
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    (Original post by Smaug123)
    Spoiler:
    Show

    It might be easier to evaluate
    \displaystyle \lim_{n \to \infty} \prod_{k=1}^n \left(1+\left( \frac{k}{n} \right)^2 \right)^{\frac{1}{n}}

    Take logs. The resulting expression approximates the integral \int_0^1 \log(1+x^2) dx
    To evaluate this, use parts: \int_0^1 \log(1+x^2) dx = [x \log(1+x^2)]_0^1 - \int_0^1 \frac{2x^2}{1+x^2}dx.

    This is \log(2) - 2 \int_0^1 1-\frac{1}{1+x^2} dx = \log(2) - 2 + 2 \tan^{-1}(1) = \frac{\pi}{2} - 2 + \log(2)

    Finally, we want to exponentiate this to get the answer: 2 e^{-2 + \pi/2}
    This is nice but annoying - I was trying to do it this way last night but was too tired to get it to work. However, have you actually proved that the limit exists? I'm not sure what they expect in Putnam problems. I assumed that they wanted to see some convergence test applied.
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    (Original post by atsruser)
    This is nice but annoying - I was trying to do it this way last night but was too tired to get it to work. However, have you actually proved that the limit exists? I'm not sure what they expect in Putnam problems. I assumed that they wanted to see some convergence test applied.
    The point is that the Riemann sum converges to the integral in the limit as n \to \infty, so the limit exists iff the integral exists.
    Granted a little exposition might be necessary to justify, but for the purposes of this thread I imagine this is sufficient.
 
 
 
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