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OK, using the calculus of variations, I want to find a function $f$ that maximizes:

$$J = \int_0^n L(x,f(x)) \text{d}x$$.

But $L$ has multiple integrals in it (for example, $\displaystyle \int_0^n y f(y) \text{d}y$). Can I still use the Euler-Lagrange equation to find f?

Here's the full equation: I want to maximize

$$J = \frac{\int_0^n x f(x) \text{d}x}{\sqrt{\int_0^n \left( y - \int_0^n z f(z) \text{dz} \right)^2 f(y) \text{d}y}} = \int_0^n \frac{x f(x)}{\sqrt{\int_0^n \left( y - \int_0^n z f(z) \text{d}z \right)^2 f(y) dy}} \text{dx}$$

Thanks in advance!

Edit: Actually, I should say, I don't care whether I use the calculus of variations or not, I just want to find $f$. I assumed calculus of variations is the way to do that.

  • 0
    What are the constraints on $f$? If $f$ is unsigned, the denominator could possibly be 0...2011-06-24
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    f is a probability distribution, so it is between zero and one. (I'm not sure how to enforce that, however....)2011-06-24
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    chapter 7 of http://books.google.com/books?id=YkFLGQeGRw4C&printsec=frontcover&source=gbs_ge_summary_r&cad=0#v=onepage&q&f=false might help2011-06-25
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    Thanks. Actually I'm thinking now that the denominator can be zero. Does that mean I'm screwed? (I'd still like to be able to find function that produces the "max" of infinity if possible....)2011-06-25
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    If I understand correctly, you're just dividing the mean by the standard deviation. You can make the standard deviation arbitrarily small by making $f$ arbitrarily peaked; as $f$ tends towards a delta distribution around any value $x_0\neq0$, $J$ grows without limits.2011-06-25
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    BTW, a philosophical point: solutions to the Euler-Lagrange equation do not necessarily have to be extremizers of the action integral. You still need to first show that extremizers exist. If your action is unbounded, then considering ELE won't help at all.2011-06-25
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    To continue joriki's comment: In a perverse sense, you can argue that the delta distribution centered at $n$ is the maximizer: it maximizes the numerator (the mean) among probability distributions on $[0,n]$, while minimizes the denominator (the standard deviation). The question is how to rigorously convert this to a statement about $J$, that is not done in an *ad hoc* manner.2011-06-25
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    Thanks for the input. joriki is exactly right. You caught me -- this is a simplified version of the problem I'm working on, which is to find the probability distribution with the best mean/stddev under some constraints. So I'd still like to know how to analytically find the optimal function using calculus of variations -- it looks a lot like the one posted....2011-06-25
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    Hi, @usul, have you found the solution? Now I have a very similar problem.2013-12-21

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