8x^3-6x-1=0
救命耶~~~~~~
更新1:
詳盡d呀- -5
數學解方程
回答 (4)
2009-05-20 8:22 am
✔ 最佳答案
With reference to the information in:http://en.wikipedia.org/wiki/Cubic_equation#Monic_formula_of_roots
We can set up the cubic stuff as follows first:
圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/May09/Crazyeqn3.jpg
Then the roots should be expressed in trigonometric form which are:
圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/May09/Crazyeqn4.jpg
2009-05-20 22:04:14 補充：
感謝兩位高手的高見, 原本想以 Newton's approximation 求根, 惟一時醒不起 cos 3x = 4 cos^3 x - 3 cos x 的公式.
參考: http://en.wikipedia.org/wiki/Cubic_equation#Monic_formula_of_roots + My Maths knowledge
2009-05-21 3:50 am
設 x = cosθ 無疑更簡潔，但必須先證明 IxI <= 1。
可如下：
設 f(x) = 8x^3 - 6x - 1
f(-1) = -3
f(-1/2) = 1
f(0) = -1
f(1) = 1
f(x) 連續且最多只有三根，所以 若 8x^3 - 6x - 1 = 0，IxI <= 1。
當然，亦可取巧地將所得三個答案分別代入方程驗證，從而解方程。
可如下：
設 f(x) = 8x^3 - 6x - 1
f(-1) = -3
f(-1/2) = 1
f(0) = -1
f(1) = 1
f(x) 連續且最多只有三根，所以 若 8x^3 - 6x - 1 = 0，IxI <= 1。
當然，亦可取巧地將所得三個答案分別代入方程驗證，從而解方程。
2009-05-20 9:12 pm
以 cos(3t)= 4(cost)^3 - 3(cost), cos(3t) = 1/2解之, 對本題比較容易些!
3t=π/3, 5π/3, 7π/3 => x= cost= cos(π/9), cos(5π/9), cos(7π/9)
3t=π/3, 5π/3, 7π/3 => x= cost= cos(π/9), cos(5π/9), cos(7π/9)
2009-05-20 6:17 am
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090519000051KK01805