※ 引述《covari (魚昆)》之銘言：
: ※ 引述《yyc2008 (YYC丫逼)》之銘言：
: : 大家好
: : 我想請問一題決定函數形式的題目
: : 是來自於物理上性質良好的函數
: : Show that if one assumes the functional form
: : T = Σ[x_i f(x_i)]
: : i
: : Σx_i = 1
: : i
: : where f(x) is some function of x, then the requirement that
: : T is extensive implies f(x) = ?
: : 感謝回答
感謝你的回答
這個定理真的是關鍵
: T is extensive means T(x_1, x_2,... a x_i, x_i+1, ...) = a T(x_1,....)
: By Euler's homogeneous function theorem,
: http://en.wikipedia.org/wiki/Euler%27s_homogeneous_function_theorem#Properties
: x . grad T = T
^^^^^^^^^^^^^^^
想請問一下 怎麼知道T是homogeneous of degree 1的函數?
: where x = (x_1, x_2, ...)
: x . grad T = Σ_k x_k @T/@x_k
: = Σ_k x_k @/@x_k ( Σ_i x_i f(x_i) )
: = Σ_k x_k Σ_i ( @x_i/@x_k f(x_i) + x_i Σ_j @f/@x_j @x_j/@x_k )
: = Σ_k x_k Σ_i ( del_ik f(x_i) + x_i Σ_j @f/@x_j del_jk )
@f(x_i)/@x_k 不會等於 @f(x_k)/@x_k for i=== k
0 for i=/= k ??
f應該不太像是多變數函數
還是我理解錯了?
= Σ_k x_k Σ_i (δ_ik f(x_i) + x_i Σ_j @f/@x_jδ_ij δ_jk )
= Σ_k x_k Σ_i ( δ_ik f(x_i) + x_i @f/@x_k δ_ik)
= Σ_k x_k ( f(x_k) + x_k @f/@x_k )
: = Σ_k x_k ( f(x_k) + @f/@x_k)
: Also T = Σ_k x_k f(x_k) by definition,
: so the requirement that T is extensive is that Σ_k @f/@x_k = 0
^^^^^^^^^^^^^^^^^
如果從你導出來= Σ_k x_k ( f(x_k) + @f/@x_k) 這怎麼得到的?
是不是應該是 Σ_k x_k @f/@x_k = 0?
還是可以繼續導到Σ_k @f/@x_k = 0?
x . grad T = p T assume T a homogeneous function of degree p
p T = Σ_k p x_k f(x_k)
x . grad T = Σ_k x_k ( f(x_k) + x_k @f/@x_k)
=> Σ_k x_k [ (1-p)f(x_k) + x_k @f/@x_k] = 0
=> xf' = (p-1)f
=> f = c x^(p-1) for a const c
可是答案卻是f(x) = c ln x的形式
不知道我上面的推導哪個地方發生問題了?
感謝回答
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