推 icenuclear:x1*logx1 < x2*logx2 iff 0<x1<x2 01/20 13:24
→ snaredrum:x*ln x 在(0,pi/4) 不是遞增吧... if x < 1/e f'<0 01/20 13:39
推 math1209:我沒有什麼好辦法,我用 Lagrange Multiplier Method. 01/20 14:10
按照大師的推文去做,
Let f(x,y)= x lnx - y lny constraint x^2+y^2=1 so (2x,2y)
\nobla f = (lnx +1 , lny +1)
Hessian f 對角線是 (1/x, 1/y) 0<x,y<1
所以函數在 x=y = sqr{2}/2 has min = 0
done.
am I right?
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推 math1209:yes~ 01/22 08:36