git remote add origin https://username@bitbucket.org/your.new.repo
2014年5月12日 星期一
2014年5月6日 星期二
2014年5月5日 星期一
2014年5月4日 星期日
[ProjectEuler] Notes on Problem #435
Lemma: F_n(x) = (f_{n+1} x^{n+1} + f_{n} x^{n+2} - x) / (x^2 + x - 1)
F_n(x)
= sum_{0 <= i <= n} f_i x^i
= sum_{1 <= i <= n} f_i x^i ...... (A)
x F_n(x)
= sum_{0 <= i <= n} f_{i} x^{i+1}
= sum_{1 <= i+1 <= n+1} f_{i+1 - 1} x^{i+1}
= sum_{1 <= j <= n+1} f_{j-1} x^{j}
= sum_{1 <= i <= n} f_{i-1} x^{i} + f_n x^{n+1} ...... (B)
(A) + (B) implies that (x+1) F_n(x) = sum_{1 <= i <= n} f_{i+1} x^i + f_n x^{n+1}
Multiply x on both sides:
x(x+1) F_n(x) = sum_{1 <= i <= n} f_{i+1} x^{i+1} + f_n x^{n+2}
= sum_{2 <= i+1 <= n+1} f_{i+1} x^{i+1} + f_n x^{n+2}
= sum_{2 <= i <= n+1} f_i x^i + f_n x^{n+2}
= sum_{1 <= i <= n} f_i x^i - f_1 x + f_{n+1} x^{n+1} + f_n x^{n+2}
= F_n(x) + f_{n+1} x^{n+1} + f_n x^{n+2} - x
(x^2 + x - 1) F_n(x) = f_{n+1} x^{n+1} + f_n x^{n+2} - x
So, F_n(x) = (f_{n+1} x^{n+1} + f_n x^{n+2} - x) / (x^2 + x - 1)
不過這離解題還有一大段的距離。
訂閱:
文章 (Atom)