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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?

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lolz 
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Offline0ReputationRep:(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? 
Offline0ReputationRep:(Original post by Dog4444)
What's in your spoiler?
In the spoiler is what usually appears in spoilers. 
 Thread Starter
Offline0ReputationRep: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. 
Offline0ReputationRep:(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. 
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Offline0ReputationRep:(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? 
Offline0ReputationRep:(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? 
Offline3ReputationRep:(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 
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Offline0ReputationRep:(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. 
Offline0ReputationRep:(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. 
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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?

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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. 
Online2ReputationRep:(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. 
Offline2ReputationRep:(Original post by Slumpy)
This is right(with the correction of the two typos). 
Online2ReputationRep:(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). 
Offline3ReputationRep:(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. 
Offline0ReputationRep:(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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