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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?
Why this is a reasonable approximation?
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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?

lolz 
(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?
?
EDIT: it should be 1 instead of e, and the bottom curve shouldn't touch the yaxis. 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?
What's in your spoiler? 
(Original post by Dog4444)
What's in your spoiler?
In the spoiler is what usually appears in spoilers. 
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. 
(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?
You are also told that , which isn't the case when . Otherwise, the equation will have one solution where the straight line is tangent to the curve. 
(Original post by gff)
How did you get this, nobody said ? What has been suggested is that .
You are also told that , which isn't the case when . 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? 
(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?
The reasoning you have seems acceptable. Can you turn it in Maths? 
(Original post by gff)
For the root is at approximately , and . What is your understanding of a reasonable approximation?
The reasoning you have seems acceptable. Can you turn it in Maths?
Spoiler:Showis it log log M?
Do you think you get bonus marks for noting that if , then is the solution 
(Original post by gff)
For the root is at approximately , and . What is your understanding of a reasonable approximation?
The reasoning you have seems acceptable. Can you turn it in Maths?
And thanks, and I got what I did wrong. 
(Original post by TheMagicMan)
What is the y they are looking for?
(Original post by Dog4444)
Do you think, it's not acceptable to explain it in words?
Similarly, I don't think they're looking for something too specific, as many people have done only C1C4 (+ FP1) at the time of the interviews. 
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?

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. 
(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. 
(Original post by Slumpy)
This is right(with the correction of the two typos). 
(Original post by hassi94)
Thanks, but why is it 256 and not 258?
(For the record: P(AB)=P(AnB)/P(B) is a handy formula for this kinda thing, to make it clearer what's going on). 
(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. 
(Original post by TheMagicMan)
...Spoiler:Show
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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