Simplifying this expression for large values of D

In summary, the problem at hand is to prove the statement p(r'+e) \cong p(r')*exp( -3e2 / 2s2 ) for large D, given that p(r) = rD-1 * exp( -r2 / 2s2 ) and r' = sqrt(D) * s. This arises from a D-dimensional Gaussian distribution and r' is the radius at which the probability is maximized. The approach taken is to expand around r' = sqrt(D-1) * s and make use of the Taylor expansion for ln. After simplifying, the resulting expression is p(r'+e) approx. = exp(-(r'+e)^2/(2s^2
  • #1
CuppoJava
24
0

Homework Statement


Show that:
p(r'+e) [tex]\cong[/tex] p(r')*exp( -3e2 / 2s2 ) for large D

given that:
p(r) = rD-1 * exp( -r2 / 2s2 )
r' = sqrt(D) * s

Homework Equations


This arises out of a D-dimensional gaussian distribution and r' is the radius at which the probability is maximized.

The Attempt at a Solution


I've gotten to the point:

p(r'+e) [tex]\propto[/tex] exp( -(r'+e)2 / 2s2 + (D-1)ln(r'+e) )

And the solution manual says that using the Taylor expansion for ln, will help everything reduce down, but all I get is a mess. What really baffles me is where the dependence on D went.

Thanks very much for your help
-Patrick
 
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  • #2
According to Wolfram Alpha, this is true for large s but not for large D.
 
  • #3
Thanks for helping foxjwill. I'm almost certain that the result is correct though. There's an probabilistic interpretation of it, and it requires that this holds true for large D not s.

Wolfram Alpha, though, seems like an extremely useful tool. I haven't heard of it before. May I ask exactly what you typed into the query box?

Thanks
-Patrick
 
  • #4
For large [itex]D[/itex], what can you say about the size of [itex]\frac{e}{r'}[/itex]? Use that to Taylor expand the logarithm:

[tex]\ln(r'+e)=\ln(r'(1+\frac{e}{r'}))=\ln(r')+\ln(1+\frac{e}{r'})[/tex]
 
  • #5
First, I typed in

Code:
(exp(-3epsilon^2/(2s^2)) *r^(D-1)*exp(-r^2/(2s^2)) /  (r+epsilon)^(D-1)*exp(-(r+epsilon)^2/(2s^2)) )

Then I copied the result given in the Alternate form box:
Code:
r^(D-1) (r+epsilon)^(1-D) e^(-(r^2+r epsilon+2 epsilon^2)/s^2)

Finally, I pasted that into the input box, replaced each instance of r with "(s*sqrt(D))", then prepended "limit" and appended "as D->infinity":
Code:
 limit (s*sqrt(D))^(D-1) ((s*sqrt(D))+epsilon)^(1-D) e^(-((s*sqrt(D))^2+(s*sqrt(D)) epsilon+2 epsilon^2)/s^2) as D->infinity

Wolfram Alpha says this equals 0.
 
  • #6
Something in this isn't quite right. If you work out the expansion around sqrt(D)*s you get a linear term in e. That's because the maximum value of your p(r) is at sqrt(D-1)*s NOT sqrt(D)*s, isn't it? Why don't you expand around the maximum? And I also don't get exp(-3*e^2/s^2) for the correction factor several different ways. I get just exp(-e^2/s^2).
 
Last edited:
  • #7
Thank you very much for your help.
I haven't gotten then answer actually. It's a textbook question that asks me to prove the statement. (Exercise 1.20 from Pattern Recognition and Machine Learning by Bishop).

You're right. The maximum is actually at sqrt(D-1)*s, which the solution manual says can be approximated to sqrt(D)*s for large D. I will try expanding around sqrt(D-1)*s and see if I can get it.

Thank you for helping me.
-Patrick
 
  • #8
Hi Dick,
Would you mind showing me how to arrived at exp(-e^2/s^2)?
I've tried expanding around r = sqrt(D-1)*s
and by noticing that (D-1)*s^2 = r^2

I simplified the equation to
p(r+e) approx. = exp( -(r+e)^2 + r^2*[ r+e-1 - ((r+e-1)^2)/2 ] )

By the way: is there a general rule for deciding when some approximations are valid and when some aren't?
eg. for large D: can I say e/r -> 0? or that r+e -> r?
 
  • #9
CuppoJava said:
Hi Dick,
Would you mind showing me how to arrived at exp(-e^2/s^2)?
I've tried expanding around r = sqrt(D-1)*s
and by noticing that (D-1)*s^2 = r^2

I simplified the equation to
p(r+e) approx. = exp( -(r+e)^2 + r^2*[ r+e-1 - ((r+e-1)^2)/2 ] )

By the way: is there a general rule for deciding when some approximations are valid and when some aren't?
eg. for large D: can I say e/r -> 0? or that r+e -> r?

Where did the '(r+e-1)' stuff come from? Start with your expression in the first post exp(-(r'+e)^2/(2s^2)+(D-1)ln(r'+e)). Expand the (r'+e)^2, expand ln(r'+e) using fabbagabbahey's suggestion. Keep the linear and quadratic terms in the ln series. Collect like powers of e. What do you get?
 
  • #10
Hi Dick,
Thank you very much for your help and patience.

Here's my work so far, I have an extra term at the end that I cannot get rid of

numerator:
[tex]-(r+e)^2+2(D-1)s^2(ln(r)+ln(1+e/r))[/tex]
[tex]=-(r+e)^2 + 2r^2(ln(r)+ln(1+e/r))[/tex]
[tex]=-r^2-2re-e^2+2r^2(r-1-(r-1)^2/2+e/r-(e/r)^2/2)[/tex]
[tex]=-r^2-2re-e^2+2r^2(r-1)(1-(r-1)/2)+2re-e^2[/tex]
[tex]=-r^2-2e^2+r^2(r-1)(1-r)[/tex]

Thanks
-Patrick
 
  • #11
You do not replace ln(r) with (r-1)-(r-1)^2/2. You can only series expand small quantities. r (or rather you should be writing r') is not necessarily small. Just leave that term as ln(r'). Remember you want f(r')*exp(...). The ln(r') is part of the f(r').
 
  • #12
Of course I can't expand around a large number. I should know that. Thanks again Dick.

I left ln(r') as it is, and finally I'm left with:

[tex]-r'^2+2r'^2ln(r') - 2e^2[/tex] in the numerator.

The first two terms are part of f(r'), and the last term is the correction term. I get a different coefficient (-2) in front of e^2 though. It's different from both your answer and the textbook answer. I'll keep on looking for what I did wrong.

Thank you very much for your help.
-Patrick
 
  • #13
You multiplied the quantity you are expanding by 2*s^2. You should divide it back out and get f(r')exp(-e^2/s^2).
 
  • #14
Right. I forgot I did that.
Thanks again Dick.
 

1. What does "simplifying this expression for large values of D" mean?

Simplifying an expression for large values of D means finding a more concise and manageable form of the expression that is easier to work with when D becomes a very large number.

2. Why is it important to simplify expressions for large values of D?

It is important to simplify expressions for large values of D because it makes calculations and analysis more efficient and accurate. It also helps to better understand the behavior of the expression as D approaches infinity.

3. How do you simplify an expression for large values of D?

To simplify an expression for large values of D, you can use algebraic techniques such as factoring, expanding, and cancelling out common terms. You can also use mathematical properties and rules, such as the limit rules, to evaluate the expression for large D values.

4. Can simplifying an expression for large values of D change its meaning?

Yes, simplifying an expression for large values of D can change its meaning. It may result in a different form of the expression that may have different properties and behaviors, especially as D approaches infinity. Therefore, it is important to check the validity and accuracy of the simplified expression in the context of the problem.

5. Is there a limit to how much an expression can be simplified for large values of D?

There is no specific limit to how much an expression can be simplified for large values of D. It depends on the complexity of the expression and the techniques used. In some cases, an expression cannot be simplified further, while in others, it may be possible to simplify it even more by using advanced mathematical concepts and methods.

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