超難Maths問題

JOHNLAM17

very 勁wor

抹茶可樂

Originally posted by ericfong at 2005-1-13 09:57 PM:
0.207879576

[ Last edited by ericfong on 2005-1-13 at 10:01 PM ]

要step.

oscar

我禁cal-tor都話Math Error...

snoopy11hk

By definition, i^i = e^(i log i)
By definition again, log i = log 1 + i * arg(i) ( here the arg is multi-valued)
so log i = 0 + (2n(pi)+(pi)/2)*i , n is an integer
so i log i = -2n(pi)-(pi)/2 , n is an integer
so i^i = e^(-2n(pi)-(pi)/2) , n is an integer

Maybe it's too difficult for a secondary school student to know these...-_-!

by Master

貝貝☆

Originally posted by oscar at 2005-1-14 10:32 AM:
我禁cal-tor都話Math Error...

咁 0.207879576岩唔岩?

oscar

Originally posted by ericfong at 2005-1-18 19:14:

咁 0.207879576岩唔岩?

錯...
答案並非一個數值來的~

抹茶可樂

但根據某網的計算器，是那個值，而且標明是「Real」的

Royal

Originally posted by oscar at 2005-1-18 07:19 PM:

錯...
答案並非一個數值來的~

i^i = 0.207879576
用google計數機計出來的…""

順帶一提(要睡覺了)，根據「數學論壇」對這條數的研究，
i^i = e^(-pi/2)
其實同S生提供的那個答案很像已相近了(我還未學過這類數)，
（S生提供的答案為 i^i = e^(-2n(pi)-(pi)/2) , n is an integer）
而再用google計數機，e^(-pi/2) = 0.207879576…""
不過我無法在google計e^(-2n(pi)-(pi)/2)

詳情數式可參考：
http://mathforum.org/epigone/sci ... @posting.google.com

賣廣告：google真係好好用(個系統唔好再話我爆粗！）

貝貝☆

我阿哥唔係用google ga-.-
step:
e^(ix)=cis x
put x =pi/2+2n pi

禁下cal機
finished

[ Last edited by ericfong on 2005-1-19 at 12:18 AM ]

貝貝☆

Originally posted by ericfong at 2005-1-19 12:10 AM:
我阿哥唔係用google ga-.-
step:
e^(ix)=cis x
put x =pi/2+2n pi

禁下cal機
finished

[ Last edited by ericfong on 2005-1-19 at 12:18 AM ]

夠竟個答覆係唔係咁姐T_T我要同阿哥講番ga