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# Why this is a reasonable approximation?

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1. Why this is a reasonable approximation?
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?
2. Re: Why this is a reasonable approximation?
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?
Last edited by electriic_ink; 14-04-2012 at 03:43.
3. Re: Why this is a reasonable approximation?
What was your explanation for the first part?

A fancy way to approach this would be to say .
Spoiler:
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Clearly, gets large when , and equivalently .

If we consider the rate of change of with respect to , we find that .

This suggests that when is very large, the rate of change of to becomes nearly a constant.

Hence, and by integrating this we find .
4. Re: Why this is a reasonable approximation?
"rather hard"

lolz
5. Re: Why this is a reasonable approximation?
(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

Last edited by Dog4444; 14-04-2012 at 15:26.
6. Re: Why this is a reasonable approximation?
(Original post by Dog4444)
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.
7. Re: Why this is a reasonable approximation?
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?
8. Re: Why this is a reasonable approximation?
(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 ? 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.
9. Re: Why this is a reasonable approximation?
(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?
10. Re: Why this is a reasonable approximation?
(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 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?
11. Re: Why this is a reasonable approximation?
(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?
What is the y they are looking for?

Spoiler:
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is it log log M?

Do you think you get bonus marks for noting that if , then is the solution
Last edited by TheMagicMan; 14-04-2012 at 19:12.
12. Re: Why this is a reasonable approximation?
(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?
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.
13. Re: Why this is a reasonable approximation?
(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.
Last edited by gff; 14-04-2012 at 19:48.
14. Re: Why this is a reasonable approximation?
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?
Spoiler:
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15. Re: Why this is a reasonable approximation?
(Original post by Miss Mary)
I'm intrigued, the answer to this

is this?
Spoiler:
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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.
16. Re: Why this is a reasonable approximation?
(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.
Last edited by Slumpy; 14-04-2012 at 23:24.
17. Re: Why this is a reasonable approximation?
(Original post by Slumpy)
This is right(with the correction of the two typos).
Thanks, but why is it 256 and not 258?
18. Re: Why this is a reasonable approximation?
(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 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).
19. Re: Why this is a reasonable approximation?
(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
20. Re: Why this is a reasonable approximation?
(Original post by TheMagicMan)
...
Spoiler:
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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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Last updated: April 15, 2012
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