Re: 是變分法嗎？

看板Math作者llrabel (不屈不撓)時間10年前 (2014/01/15 01:27)推噓4(4推 0噓 19→)

留言23則, 9人參與討論串2/3 (看更多)

※ 引述《physmd (smd)》之銘言： : 我變分法是很多年前學的，而且只懂皮毛...... : 底下這個問題不好意思我用英文輸入啦 :P : 有請版友指點，多謝～～ : Consider real valued functions in the closed domain [0, 1] : Find f(x) that minimizes: Integrate{ g(x)^2 / f(x) }, : with respect to a given g(x), integrate over x from 0 to 1, : subject to the constraints : (1) f > 0 for all x in [0, 1] : (2) |f| = 1, that is, Integrate{ f(x) } over x = 0 to 1 is unity. : Suppose everything you need to know about g(x) is completely specified. : (the derivative or anti-derivative of g(x) etc) : In general g(x) can be just some real valued integrable function (that doesn't : even have to be piece-wise continuous), but if needed we can focus on the solution : f(x) for some smooth g(x). : p.s. : Some of you might notice that f is a probability density. : In fact, this question is simplified from a question regarding minimizing : the variance of some stuff. 是變分沒錯, 不要太要求嚴謹, 就像下面這樣做. Let f(x) be a function which satisfies (1), (2) and minimizes Int{ g(x)^2 / f(x) }. Let h(x) be any smooth function satisfying Int{ h(x) }=0. Then f(x)+εh(x) also satisfies (1) and (2) when ε is very close to zero. Therefore, the function I(ε) := Int{ g(x)^2 / (f(x)+εh(x)) } has a local maximum at ε=0, which implies I'(0)=0. By direct computation, we find this means Int{ [g(x)^2/f(x)^2]．h(x) } = 0. Note that this is true for arbitrary smooth function h(x) satisfying Int{h(x)}=0. The remaining problem is to ask what f(x) makes this possible. By observation, we find it's true when g(x)^2/f(x)^2 is a constant. Hence f(x)=c|g(x)|, where c is the constant so that (2) holds. Precisely, c = 1/Int{|g(x)|}. -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 111.243.114.220

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wohtp

01/15 04:42, , 1^F

01/15 04:42, 1^F

已修正, 謝謝.

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wohtp

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yw1002

01/15 05:31, , 3^F