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Topic: Calculus of Variations problem (Read 789 times) |
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amangupta
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Calculus of Variations problem
« on: Feb 12th, 2008, 12:54pm » |
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I am a computer science researcher, and is facing the following problem:
Given a real number D > 0, minimize
Integrate[(f(x+1)/f(x)) dx] from 0 to f-1(D).
over all differentiable functions f: R+ union {0} --> R+, such that f-1(D) >= 1.
Note that the limits mean the range of f includes D.
Using f as the exponential function, I get this as O(log D). I want to know if this can be bettered.
Thanks in advance
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| « Last Edit: Feb 13th, 2008, 12:38am by amangupta » |
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Icarus
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Re: Calculus of Variations problem
« Reply #1 on: Feb 12th, 2008, 4:23pm » |
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The minimum value of the integral is 0, obtained by any function f for which f(0) = D, for example f(x) = D + x.
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amangupta
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Re: Calculus of Variations problem
« Reply #2 on: Feb 13th, 2008, 12:37am » |
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Oh...I didnt notice that. Can it be solved under the assumption that f-1(D) >= 1? My actual situation requires this; I am sorry I forgot to mention before.
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Eigenray
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Re: Calculus of Variations problem
« Reply #3 on: Feb 13th, 2008, 2:42am » |
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One can still make the integral as small as you want. Just pick f so that f(x) > D for x < 1, f(1)=D, f(x) < D for x>1, and f(x) <  for x > 1+. Then
 01 f(x+1)/f(x) dx < 0 1 dx + 1 /D dx = (1 + (1-)/D).
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amangupta
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Re: Calculus of Variations problem
« Reply #4 on: Feb 13th, 2008, 5:15am » |
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Looks like f increasing is also a condition 
Think of it as this way:if I use f(x) = xk, I get it to be O(D1/k), worse than O(log D)
if I use f(x) = 2x, I get it to be O(log D)
if I use f(x) = Gamma(x) (i.e. generalization of n factorial), then I get O(logk D), for some k such that 1 < k < 2. This is because 2x < x! < 2x^2.
So, as I am increasing the rate of change of f, I first decrease its value, then it starts increasing back again. So there is a minima lying somewhere in between - I want to find that.
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| « Last Edit: Feb 13th, 2008, 5:21am by amangupta » |
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Grimbal
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Re: Calculus of Variations problem
« Reply #5 on: Feb 13th, 2008, 6:57am » |
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It looks like on the large scale, the rate of change must be well distributed. So I guess an exponential would be a good choice.
The idea is that for a and c fixed, min(b/a+c/b) is at b=sqrt(ac), where a, b, c follow a geometric progression. You want exactly that condition on f if you take a=f(x), b=f(x+1), c=f(x+2). An exponential f(x)=ax satisfies that condition nicely.
Maybe eint(x) could be better on a small scale.
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| « Last Edit: Feb 13th, 2008, 6:57am by Grimbal » |
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Eigenray
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Re: Calculus of Variations problem
« Reply #6 on: Feb 13th, 2008, 9:34am » |
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I think I am still missing something. You could do f(x) = D + (x-1), and then
01 f(x+1)/f(x) dx = 1 + log[D/(D-)],
which can be made arbitrarily close to 1.
Should we also be assuming f(0) is fixed, say f(0) = 1?
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amangupta
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Re: Calculus of Variations problem
« Reply #7 on: Feb 13th, 2008, 9:47am » |
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f(0) = 1 was needed, but I assumed we can just scale the function by 1/f(0) for that, since f(0) was strictly positive. Though it seems how scaling changes f-1(D) is affecting.
So f(0) = 1 is another constraint.
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Eigenray
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Re: Calculus of Variations problem
« Reply #8 on: Feb 13th, 2008, 3:16pm » |
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We can still get arbitrarily close to 1. For 0 < p < 1, let f(x) = 1 + (D-1)*xp. Then for 0<x<1, f(x) > D*xp, and f(x+1) < D*2p. So
01 f(x+1)/f(x) dx < 01 D*2p/(D*xp) dx = 2p/(1-p).
But this can be made arbitrarily close to 1 by picking p small enough. Is there another condition you'd like to add? 
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