e.e. [3.15]=3. Let {x}=x-[x].
Find a certain number x such that {x}+{1/x}=1
我有answer但係冇solution,而且諗左好耐都冇頭緒,
希望大家幫手解答...thz!!!
更新1:
第二行打錯字,係e.g.先岩,sor
fractional part function
2009-04-01 7:56 am
Define [x] as the largest integer not greater than x,
e.e. [3.15]=3. Let {x}=x-[x].
Find a certain number x such that {x}+{1/x}=1
我有answer但係冇solution,而且諗左好耐都冇頭緒,
希望大家幫手解答...thz!!!
e.e. [3.15]=3. Let {x}=x-[x].
Find a certain number x such that {x}+{1/x}=1
我有answer但係冇solution,而且諗左好耐都冇頭緒,
希望大家幫手解答...thz!!!
回答 (2)
2009-04-02 12:22 am
✔ 最佳答案
If x >= 1, let n-1 <= x < n for some integer n >= 2.{x} + {1/x} = 1
x - (n - 1) + 1/x = 1
x - n + 1/x = 0
x^2 - nx + 1 = 0
x = [n +/- √(n^2 - 4)] / 2
If x <= -1, let -n <= x < -n + 1 for some integer n >= 2.
{x} + {1/x} = 1
x - (-n) + 1/x + 1 = 1
x + n + 1/x = 0
x^2 + nx + 1 = 0
x = [-n +/- √(n^2 - 4)] / 2
For 0 < x < 1 and -1 < x < 0, we have similar results(you can try).
Solution for {x} + {1/x} = 1:
x = [+/-n +/- √(n^2 - 4)] / 2 for ineger n >= 2
(note: when n = 2, x = 1 or -1)
2009-04-01 16:23:11 補充：
'ineger' in last two line should be integer
2009-04-01 19:41:40 補充：
CORRECTION (x <> +/- 1)
When n = 2, x = 1 or -1 (rejected)
Solution for {x} + {1/x} = 1:
x = [+/-n +/- √(n^2 - 4)] / 2 for integer n >= 3
2009-04-01 10:48 am
Can I answer this in English? The answer I think is
((n+1) + ((n+1)^2 - 4) ^ (1/2)) / 2
((n+1) - ((n+1)^2 - 4) ^ (1/2)) / 2
(-(n+1) + ((n+1)^2 - 4) ^ (1/2)) / 2
(-(n+1) - ((n+1)^2 - 4) ^ (1/2)) / 2
for n = 2, 3, 4, ....
((n+1) + ((n+1)^2 - 4) ^ (1/2)) / 2
((n+1) - ((n+1)^2 - 4) ^ (1/2)) / 2
(-(n+1) + ((n+1)^2 - 4) ^ (1/2)) / 2
(-(n+1) - ((n+1)^2 - 4) ^ (1/2)) / 2
for n = 2, 3, 4, ....
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