1-200r = (1+r)^(-60), r=?
*此行湊字數不用理會
*此行湊字數不用理會
*此行湊字數不用理會
*此行湊字數不用理會
*此行湊字數不用理會
*此行湊字數不用理會
*此行湊字數不用理會
*此行湊字數不用理會
*此行湊字數不用理會
超難數學題一條 [g.math]
2008-03-12 8:54 am
回答 (4)
2008-03-20 11:38 am
✔ 最佳答案
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參考資料：
my maths knowledge + MATLAB
2008-03-16 11:50 pm
這是個 61 次的多項式, 可能會有好多 root
XD
2008-03-16 16:04:05 補充：
用 Mathematica 計過, 有3 個實根的~
{r -> 0.}, {r -> -1.90566}, {r -> -0.0334108}
XD
2008-03-16 16:04:05 補充：
用 Mathematica 計過, 有3 個實根的~
{r -> 0.}, {r -> -1.90566}, {r -> -0.0334108}
2008-03-12 7:33 pm
其中之一個答案是 0
LHS = 1-200r = 1-200 (0) = 1
RHS = (1+r)^(-60) = (1+0)^(-60) = 1
LHS = 1-200r = 1-200 (0) = 1
RHS = (1+r)^(-60) = (1+0)^(-60) = 1
2008-03-12 9:35 am
obvious answer, r = 0
not so obvious answer, r = -1.905655...., can be found be numerical methods.
Observation is that when |1 ＋ r| ＞1, (1 ＋ r)^-60 ~= 0, therefore plotting from r = -2 to 0 is going to give you best chance finding the root. Will wrap up and write the solution nicely.
not so obvious answer, r = -1.905655...., can be found be numerical methods.
Observation is that when |1 ＋ r| ＞1, (1 ＋ r)^-60 ~= 0, therefore plotting from r = -2 to 0 is going to give you best chance finding the root. Will wrap up and write the solution nicely.
收錄日期: 2021-04-23 17:47:15
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080312000051KK00134