發表回覆

主題 通關密語 訪客發文, 請參考 這裡 輸入通關密語.

顯示表情符號

站內上傳圖檔     Upload.cc免費圖片上傳

數學塗鴉工具     常用數學符號表    

用Latex打數學方程式

 


 

+ / -檢視主題

[數學]求f(x)

發表 飛向自由 於 星期日 四月 08, 2007 2:07 pm

其實我並不太清楚,但其實這題的關鍵就是你要先找出此多項式的次數(deg)

發表 gkw0824usa 於 星期日 四月 08, 2007 12:10 pm

#ed_op#DIV#ed_cl#我發現一個很奇怪的現象,不知道是不是我算錯了:#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#FONT style="BACKGROUND-COLOR: #40ffff"#ed_cl#1. Let f(x)=x+1,#ed_op#/FONT#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#FONT style="BACKGROUND-COLOR: #40ffff"#ed_cl#initial multinomial#ed_op#/FONT#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#FONT style="BACKGROUND-COLOR: #40ffff"#ed_cl#→  f(x+1) =x#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#+1-2(x+1)-1#ed_op#/FONT#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#FONT style="BACKGROUND-COLOR: #40ffff"#ed_cl#Let x=-1,#ed_op#/FONT#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#FONT style="BACKGROUND-COLOR: #40ffff"#ed_cl#f(0)=1  #ed_op#/FONT#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#FONT style="BACKGROUND-COLOR: #40ffff"#ed_cl#→  f(x)=x+1  holds .......(1)#ed_op#/FONT#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#FONT style="BACKGROUND-COLOR: #ffff80"#ed_cl#comparism:#ed_op#/FONT#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#FONT style="BACKGROUND-COLOR: #ffff80"#ed_cl##ed_op#/FONT#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#FONT style="BACKGROUND-COLOR: #ffff80"#ed_cl#Let f(x)=f(0),#ed_op#/FONT#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#FONT style="BACKGROUND-COLOR: #ffff80"#ed_cl##ed_op#/FONT#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#FONT style="BACKGROUND-COLOR: #ffff80"#ed_cl#initial multinomial#ed_op#/FONT#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#FONT style="BACKGROUND-COLOR: #ffff80"#ed_cl#→  f(f(0))=f(0#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#)-2f(0)-1#ed_op#/FONT#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#FONT style="BACKGROUND-COLOR: #ffff80"#ed_cl#→  f(1)=1-2-1#ed_op#/FONT#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#FONT style="BACKGROUND-COLOR: #ffff80"#ed_cl#→  f(1)=-2........(2)#ed_op#/FONT#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#paradox:#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#let x=1,#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#In (1): f(1) = 2#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#In (2): f(1)=-2     #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#there is a contradiction between (1) and (2)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#How come?     #ed_op#/DIV#ed_cl#

[數學]求f(x)

發表 飛向自由 於 星期六 四月 07, 2007 10:38 pm

sorry,打錯了,已更正

Re: [數學]求f(x)

發表 gkw0824usa 於 星期六 四月 07, 2007 10:32 pm

飛向自由 寫到:設f(x) 為實係數n次多項式,<math><mrow><mi>n</mi><mo>∈</mo><mi>N</mi></mrow></math>#ed_op#BR#ed_cl#,f(f(x))= f(<math><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></math>) - 2f(x)-1 #ed_op#FONT style="BACKGROUND-COLOR: #ff0000"#ed_cl#=0#ed_op#/FONT#ed_cl# ,且f(0)=1 ,求 f(x) = ?
#ed_op#DIV#ed_cl#打錯了吧?#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#若去掉 "=0"#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#f(x)=x+1.......先把答案湊出來了#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#慢慢想解法 #ed_op#/DIV#ed_cl#

[數學]求f(x)

發表 飛向自由 於 星期六 四月 07, 2007 10:17 pm

設f(x) 為實係數n次多項式,
,f(f(x))= f() - 2f(x)-1,且f(0)=1 ,求 f(x) = ?