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作者 Vulpix 在 PTT [ Math ] 看板的留言(推文), 共7171則
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18F→:著眼點要放在歐拉的哪裡...07/15 03:07
5F推:1.「最大下限」發生在某個「不是三角形」的情況:07/13 20:01
6F→:a:b:c=(sqrt(2)+1):(sqrt(2)-1):207/13 20:02
7F→:c/(a+b)+b/c的最大下限為sqrt(2)-0.507/13 20:02
9F→:嗯,我也想問「何來最小值?」這題根本找不到"min."07/14 10:59
12F推:嗯,我的意思就是題目既然這樣出,那只好把"最小值"07/15 13:02
13F→:理解成"最大下限"來算。畢竟,在能夠形成三角形的07/15 13:03
14F→:情況下,c/(a+b)+b/c想要多靠近sqrt(2)-0.5都可以,07/15 13:04
15F→:只是永遠碰不到sqrt(2)-0.5。07/15 13:04
16F→:再說...sqrt(2)不是最小值,因為a:b:c=7:2:6會讓07/15 13:05
17F→:c/(a+b)+b/c=1,比sqrt(2)更小。07/15 13:06
18F→:總之1.的作法是這樣:因為 |b-c| < a < b+c07/16 15:11
19F→:所以c/(a+b)+b/c > c/(2b+c)+b/c = 1/(2b/c+1)+b/c07/16 15:13
20F→:= 1/(2b/c+1)+(b/c+1/2)-1/2 >= sqrt(2)-1/207/16 15:14
21F→:最後還要確定sqrt(2)-1/2真的是"最大"下限才行07/16 15:14
1F→:10*(10-1)/2! 就是C(10,2)啦...07/13 12:03
18F推:意思是你本來要的是"某個函數恆正"結果你發現"這個07/12 11:25
19F→:函數在某個地方是負的"?那就不可能了啊~07/12 11:26
20F→:不過呢...你需要的到底是"處處負定的Hessian"還是07/12 11:27
21F→:"在特定臨界點的負定Hessian"?這兩個東西不太一樣07/12 11:28
22F→:例如:f(x)=x^3-x,我們來算算看Hessian吧。07/12 11:30
23F→:f"(x)=6x,不是負定。可是在x=-1/sqrt(3),f"是負定07/12 11:31
2F→:應該是題目漏了一個0,1要改成10。07/10 10:38
10F→:願意推你這篇文章的板友本來應該都願意幫你的,還是07/08 21:23
11F→:把題目放上來吧。這樣對幫你的人比較方便。07/08 21:23
21F→:照這情況看起來,只要能算出正的行列式值就好了。07/09 10:48
2F→:試試看 y = π/2 - x 吧。07/07 16:41
2F→:先假設f可微分吧。(a)否,微分為0是對的,但無法推論07/06 19:57
3F→:是極值與否。(b)"單調"不能有此推論。凹函數或凸函數07/06 19:58
4F→:可以保證極值唯一。或許還有其他好的簡單條件吧。07/06 19:59
7F→:那你還要多假定f二次可微07/07 09:49
10F→:喔,對了,應該強調"嚴格"凹(凸)函數XD07/07 09:54
3F→:Rudin那本沒有曾經唸過會很難在這麼短時間內全消吧07/06 18:01