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Why this is a reasonable approximation?

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    http://www.trin.cam.ac.uk/show.php?dowid=4
    Question 9.

    I don't know how to explain why logM is reasonable approximation to the one of the roots.

    The only thing I could come up with is lnMX=lne^x and so x=lnM + ln X and if M gets too large the ln X is very small comparing to x itself.
    But, I'm not sure if it's sensible at all.
    If substitute x=lnM you get M.lnM=M
    And so M=e?
    :confused:
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    When M=e, the approximation is perfect. I think one of the ways to do it is to consider the function e^x - Mx. Specifically, are there are any turning points and what is the gradient like around those turning points?
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    What was your explanation for the first part?

    A fancy way to approach this would be to say \displaystyle M(w) = \frac{e^w}{w}.
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    Clearly, M gets large when w \to \infty, and equivalently \log(M).

    If we consider the rate of change of \log(M) = w - \log(w) with respect to w, we find that \displaystyle \frac{d\log(M)}{dw} = 1 - \frac{1}{w}.

    This suggests that when w is very large, the rate of change of \log(M) to w becomes nearly a constant.

    Hence, \displaystyle w \to \infty \Rightarrow \frac{d\log(M)}{dw} = 1 and by integrating this we find \displaystyle \log(M) = w + c.
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    "rather hard"

    lolz
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    (Original post by electriic_ink)
    When M=e, the approximation is perfect. I think one of the ways to do it is to consider the function e^x - Mx. Specifically, are there are any turning points and what is the gradient like around those turning points?
    I don't want to cheat and use wolfram.


    ?

    EDIT: it should be 1 instead of e, and the bottom curve shouldn't touch the y-axis. And I can't follow why it's a perfect approximation? This approximation means that M can only be e?

    (Original post by gff)
    What was your explanation for the first part?
    I plotted them on the graph and said that, as M gets larger, one of the roots get closer and closer to 0. Then, consider mx=e^x at x=1 if M is large enough Mx>e^x, but as x gets bigger, no matter how big the M is, we'll get mx=e^x and then e^x>mx

    What's in your spoiler?
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    (Original post by Dog4444)
    What's in your spoiler?
    The graphical argument is good, and it is good that you have noted the other root.
    In the spoiler is what usually appears in spoilers.
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    So, how right is this?
    (Original post by Dog4444)
    The only thing I could come up with is lnMX=lne^x and so x=lnM + ln X and if M gets too large the ln X is very small comparing to x itself.
    And if lnM is a reasonable approximation of the root, we get M=e therefore only one possible value of M. But clearly, M can be any big number. What's wrong?
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    (Original post by Dog4444)
    So, how right is this?

    And if lnM is a reasonable approximation of the root, we get M=e therefore only one possible value of M. But clearly, M can be any big number. What's wrong?
    How did you get this, nobody said w = \log(M)? What has been suggested is that w \approx \log(M).
    You are also told that w > 1, which isn't the case when M = e. Otherwise, the equation will have one solution where the straight line is tangent to the curve.
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    (Original post by gff)
    How did you get this, nobody said w = \log(M)? What has been suggested is that w \approx \log(M).
    You are also told that w > 1, which isn't the case when M = e. Otherwise, the equation will have one solution where the straight line is tangent to the curve.


    If lnMX=lne^x and so x=lnM + ln X. And bigger M means bigger root. So as x get's larger difference between x and ln x get's bigger and you can ignore it, right?
    If so, when M is huge, it's pretty good approximation.
    The problem is, I tried M=10^35, 10^75. And lnM doesn't seem to be a good approximation for x. So, there is a problem somewhere?
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    (Original post by Dog4444)
    The problem is, I tried M=10^35, 10^75. And lnM doesn't seem to be a good approximation for x. So, there is a problem somewhere?
    For M = 10^{75} the root is at approximately x \approx 177, and \ln(10^{75}) \approx 172. What is your understanding of a reasonable approximation?

    The reasoning you have seems acceptable. Can you turn it in Maths?
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    (Original post by gff)
    For M = 10^{75} the root is at approximately x \approx 177, and \ln(10^{75}) \approx 172. What is your understanding of a reasonable approximation?

    The reasoning you have seems acceptable. Can you turn it in Maths?
    What is the y they are looking for?

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    is it log log M?


    Do you think you get bonus marks for noting that if f(x)=log(x), then x= \displaystyle\sum_{i=1}^{ \infty} f^i(M) is the solution
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    (Original post by gff)
    For M = 10^{75} the root is at approximately x \approx 177, and \ln(10^{75}) \approx 172. What is your understanding of a reasonable approximation?

    The reasoning you have seems acceptable. Can you turn it in Maths?
    I don't think I can turn it into proper maths. Do you think, it's not acceptable to explain it in words?
    And thanks, and I got what I did wrong.
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    (Original post by TheMagicMan)
    What is the y they are looking for?
    EDIT: Is there a factorial missing in this?

    (Original post by Dog4444)
    Do you think, it's not acceptable to explain it in words?
    To be honest, I don't know what is acceptable and what is not.
    Similarly, I don't think they're looking for something too specific, as many people have done only C1-C4 (+ FP1) at the time of the interviews.
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    I'm intrigued, the answer to this
    There is a pile of 129 coins on a table, all unbiased except for one which has heads on both sides. Bobchooses a coin at random and tosses it eight times. The coin comes up heads every time. What is the
    probability that it will come up heads the ninth time as well?
    is this?
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    \frac{65}{129}
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    (Original post by Miss Mary)
    I'm intrigued, the answer to this


    is this?
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    \frac{65}{129}
    I don't think so.

    I think we'd say:

    P(coin is unbiased) = 128/129
    P(coin is biased) = 1/129
    P(unbiased coin is heads 8 times) = (1/2)^8 = 1/256
    P(biased coin is heads 8 times) = 1

    So the probability of an unbiased coin doing what has been done is 1/258

    And the probability of a biased coin doing what is done is 2/258

    So the probability we have a biased coin is 2/3 and the probability we have a unbiased coin is 1/3

    Then finally the probability it will be heads again is (2/3)(1) + (1/3)(1/2) = 5/6


    I don't really know if this is correct, and sorry I couldn't use proper notation - I've never really done probability except a tiny bit in S1 so I've just done this out of logic alone.
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    (Original post by hassi94)
    I don't think so.

    I think we'd say:

    P(coin is unbiased) = 128/129
    P(coin is biased) = 1/129
    P(unbiased coin is heads 8 times) = (1/2)^8 = 1/256
    P(biased coin is heads 8 times) = 1

    So the probability of an unbiased coin doing what has been done is 1/258

    And the probability of a biased coin doing what is done is 2/258

    So the probability we have a biased coin is 2/3 and the probability we have a unbiased coin is 1/3

    Then finally the probability it will be heads again is (2/3)(1) + (1/3)(1/2) = 5/6


    I don't really know if this is correct, and sorry I couldn't use proper notation - I've never really done probability except a tiny bit in S1 so I've just done this out of logic alone.
    This is right.
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    (Original post by Slumpy)
    This is right(with the correction of the two typos:p:).
    Thanks, but why is it 256 and not 258?
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    (Original post by hassi94)
    Thanks, but why is it 256 and not 258?
    Oops, didn't actually check what you were doing there, just assumed you were essentially restating the previous bit:p: What you wrote it fine I think.
    (For the record: P(A|B)=P(AnB)/P(B) is a handy formula for this kinda thing, to make it clearer what's going on).
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    (Original post by hassi94)
    I don't think so.

    I think we'd say:

    P(coin is unbiased) = 128/129
    P(coin is biased) = 1/129
    P(unbiased coin is heads 8 times) = (1/2)^8 = 1/256
    P(biased coin is heads 8 times) = 1

    So the probability of an unbiased coin doing what has been done is 1/258

    And the probability of a biased coin doing what is done is 2/258

    So the probability we have a biased coin is 2/3 and the probability we have a unbiased coin is 1/3

    Then finally the probability it will be heads again is (2/3)(1) + (1/3)(1/2) = 5/6


    I don't really know if this is correct, and sorry I couldn't use proper notation - I've never really done probability except a tiny bit in S1 so I've just done this out of logic alone.
    I'm pretty sure this is right
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    (Original post by TheMagicMan)
    ...
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    Can you elaborate on the convergence of this power series you've posted? I don't see how it works out.

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