求一函數,在其自變數為非負整數的情況下,其應變數皆為非負整數,且最大為26,最小為0?

2015-08-18 12:49 pm
更新1:

抱歉,應該是在自變數為自然數的情況下,應變數皆為非負整數

回答 (2)

2015-08-20 12:39 am
✔ 最佳答案
前一個問題的回答用在這裡仍是可行的, 例如
f(x) = [27*(x/27-[x/27])]

f(0) = 0, f(1) = 1, f(2) = 2, . . . . f(26) = 26,
f(27) = 0, f(28) = 1, f(29) = 2, ... , f(52) = 26,


{0,1,...,26} 的任一排列函數與上列 f(x) 搭配可得
不同形狀的週期及數.


同樣, 可做出不同週期的週期函數, 例如
g(x) = [27*(x/54-[x/54])]


也可與其他非負整數值函數合成, 構造出非週期函數,
例如 h(x) = [27*(x^2/27-[x^2/27])]

h(0) = 0, h(1) = 1, h(2) = 4, h(3) = 9, h(4) = 16, h(5) = 25,
h(6) = 9, h(7) = 22, h(8) = 10, . . .
2015-08-18 1:08 pm
我任意給幾個例子,不保證絕對正確:
1.
f:N→N∪{0}
f(x)=26, when 2│x
f(x)=0, otherwise
2.
f:R→R
f(x)=-(x-1)^2+26, when --(x-1)^2+26≧0
f(x)=0, otherwise


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原文連結 [永久失效]:
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