推 hwanger : Let g(z)= z^2 - 2z log(z) -1 08/13 08:19
→ hwanger : It's easy to see g is strictly increasing for 08/13 08:20
→ hwanger : z in [1,infty) and g(1)=0 08/13 08:20
→ hwanger : Let h(y) = (2log(1+y))/y - 1/(1+y) - 1 08/13 08:22
→ hwanger : Note that h goes to 0 as y goes to 0 08/13 08:22
→ hwanger : Since for all positive y, g(1+y)/(y(y+1)) > 0 08/13 08:24
→ hwanger : h(y) < or = 0 for y in [0,infty) 08/13 08:25
→ hwanger : Given non-negative t, consider the function 08/13 08:26
→ hwanger : f(x) = (1+t/x)^(x^2)exp(-tx) defined on [1,inf) 08/13 08:27
→ hwanger : Then f'(x) = t f(x) h(t/x), which is 08/13 08:29
→ hwanger : non-positive 08/13 08:30
→ hwanger : Note that f_n(t) = f(√n) 08/13 08:32
推 hwanger : the idea is stupid but works 08/13 08:35
推 hwanger : it may be better if i use the notation F_t(x) in 08/13 10:14
→ hwanger : stead of f(x) 08/13 10:14
→ hau : 厲害！完成了，感謝~~ 08/13 16:33
推 hwanger : 可參照"A Problem Seminar, Donald J Newman"第94題 08/13 20:02
→ hwanger : 有想法一樣但作法不太一樣的證明 08/13 20:02
推 Vulpix : 這題不是直接對 n 微分就好了嗎？ 08/13 21:35
→ cmrafsts : 你直接展開不就有 ( 1 + t/√n )^n < exp(√nt ) 08/13 22:09
推 hwanger : V大說的沒錯 "A Problem Seminar"就是對n微分 重點 08/13 22:20
→ hwanger : 在微分後如何說明 導函數小於等於0 08/13 22:21
推 hwanger : 至於c大的不等式可以說明 f_n(t) < 1 但我不太明白 08/13 22:24
→ hwanger : 接下來怎麼做 08/13 22:24
推 cmrafsts : 喔我瞭解了，不好意思剛剛看錯 08/13 23:38
推 cmrafsts : 如果你只要可積分的上屆，趨近0用常數壓，趨近infty 08/13 23:41
→ cmrafsts : 抱歉我又錯了不要理我 08/13 23:42
推 cmrafsts : log(f_1/f_n)的微分是(√n-1)t^2/(√n+t)(1+t)>0 08/14 00:11
推 hwanger : c大漂亮地證明不等式的本身 08/14 00:36
推 cmrafsts : 抱歉，我的意思是你要證明exp((\sqrt{n}-1)t)>= 08/14 00:43
→ cmrafsts : (1+t/\sqrt{n})^n/(1+t). t=0成立，然後兩邊取log 08/14 00:44
→ cmrafsts : 微分 08/14 00:44
推 hwanger : ??? 你那行不是證明了對於所有t log(f_1/f_n)非負 08/14 00:52
→ hwanger : 所以f_1>= f_n 08/14 00:52
推 hwanger : 我不太懂你後來再講交叉相乘的用意? 08/14 00:55
→ cmrafsts : 半夜腦袋不清楚0.0 08/14 02:17
→ hwanger : XDD 直接對 f_n/f_1 微分 就會得到 08/14 08:38
→ hwanger : -[(√n-1)t^2*f_n]/[(t+1)(t+√n)f_1] < or = 0 08/14 08:40