查看: 149|回复: 0

[标准] 斐波那契数列-Fibonacci numbers

[复制链接]

1534

主题

1564

帖子

3万

积分

论坛元老

Rank: 8Rank: 8

积分
30456
QQ
发表于 2017-5-18 21:51:29 | 显示全部楼层 |阅读模式 | 百度 
today_gg的评分为17--时间:2635-12-09 19:47:18.DillonWebb的评分为92--时间:2295-06-03 10:09:48,xlz888的评分为31--时间:2606-07-08 17:22:28,爱王源go的评分为52--时间:2159-02-14 12:16:28。安史之绿的评分为33--时间:2044-06-02 23:59:58.超声波清洗机90的评分为29--时间:2052-05-03 17:44:08,

马上注册硬科学论坛吧。上传说明:若上传附件过大,请上传至网盘或者优酷等视频网站,给出链接方式。下载说明:如果资源大于2G,请安装【百度网盘】或【360云盘】后下载!

您需要 登录 才可以下载或查看,没有帐号?立即注册

x
输入:
F_n
情节:
生成函数:
sum_(n=0)^∞ F_n t^n = 1/(1 - t - t^2)
序列属性:
F_n is a sequence with integer values for integer n.
价值观:
n | F_n0 | 01 | 12 | 13 | 24 | 35 | 56 | 87 | 138 | 219 | 3410 | 55
替代形式:
((1/2 (1 + sqrt(5)))^n - (2/(1 + sqrt(5)))^n cos(π n))/sqrt(5)
数值根:
n = 0
n ≈ -1.57077646820395...
n ≈ -2.47042683968583...
n ≈ -3.51085138466955...
n ≈ -4.49579533459999...
n ≈ -5.50159705685184...
n ≈ -6.49938868284049...
在n = 0级数展开:
-(n (log(4) - 2 log(1 + sqrt(5))))/sqrt(5) + (π^2 n^2)/(2 sqrt(5)) - (n^3 (log^3(4) - 2 log^3(1 + sqrt(5)) + log^2(2) log(64) - 6 log^2(2) log(1 + sqrt(5)) + log(64) log^2(1 + sqrt(5)) - log^2(4) (log(8) + 3 log(1 + sqrt(5))) - π^2 (log(8) - 3 log(1 + sqrt(5))) + log(4) log(64) log(1 + sqrt(5))))/(6 sqrt(5)) - (n^4 (π^4 - log^3(2) (log(256) - 8 log(1 + sqrt(5))) + log^2(2) (6 log^2(4) - log(16777216) log(1 + sqrt(5))) + π^2 (-6 log^2(2) - 6 log^2(4) + log(4) log(4096) + (log(4096) - 6 log(1 + sqrt(5))) log(1 + sqrt(5))) + log^2(4) (log^2(4) + log(4096) log(1 + sqrt(5)) - log(4) (log(16) + 4 log(1 + sqrt(5))))))/(24 sqrt(5)) + (n^5 (-10 π^2 log^3(1/2 (1 + sqrt(5))) - (log(4) - 2 log(1 + sqrt(5))) (5 log^4(2) + log^4(4) + log^4(1 + sqrt(5)) - 10 log^3(2) log(4) - 3 log^3(4) log(1 + sqrt(5)) - 2 log(4) log^3(1 + sqrt(5)) + 4 log^2(4) log^2(1 + sqrt(5)) + 10 log^2(2) (log^2(4) + log^2(1 + sqrt(5)) - log(4) log(1 + sqrt(5))) - 5 log(2) log(4) (log^2(4) + 2 log^2(1 + sqrt(5)) - 2 log(4) log(1 + sqrt(5)))) + 5 π^4 log(1/2 (1 + sqrt(5)))))/(120 sqrt(5)) + O(n^6)(Taylor series)
导数:
d/dn(F_n) = (?^(-n) (?^(2 n) log(?) + cos(π n) log(?) + π sin(π n)))/sqrt(5)
不定积分:
integral F_n dn = ((1/2 (1 + sqrt(5)))^n (π^2 + log^2(2/(1 + sqrt(5)))) - π (2/(1 + sqrt(5)))^n log(1/2 (1 + sqrt(5))) sin(π n) + (2/(1 + sqrt(5)))^n log^2(2/(1 + sqrt(5))) cos(π n))/(sqrt(5) (π^2 + log^2(2/(1 + sqrt(5)))) log(1/2 (1 + sqrt(5)))) + constant
替代表示:
F_n = 1/5 (L_(-1 + n) + L_(1 + n))
F_n = fibonacci(n, 1)
F_n = i^(n - 1) U_(n - 1)(-i/2) for n element Z
系列表示:
F_n = sum_(k=0)^(-1 + n) binomial(k, -1 - k + n) for (n element Z and n>=0)
F_n = sum_(k=0)^(-1 + n) binomial(-1 - k + n, k) for (n element Z and n>=0)
F_n = sum_(k=0)^(-1 + n/2) binomial(k + n/2, 1 + 2 k) for (n/2 element Z and n>=0)
积分表示:
F_n = 2^(-1 - n/2) 3^(-1 + n/2) n integral_0^π (1 + 1/3 sqrt(5) cos(t))^(-1 + n/2) sin(t) dt for n/2 element Z
Multiple-argument公式:
F_n = -F_(-4 + n) + 3 F_(-2 + n)
F_n = F_(n/2) L_(n/2) for n/2 element Z
F_n = F_(n/2 - p) F_(-1 + n/2 + p) + F_(1 + n/2 - p) F_(n/2 + p) for (n/2 element Z and n>=0 and p element Z and p>=0)


回复
百度搜狗360奇虎

使用道具 举报

您需要登录后才可以回帖 登录 | 立即注册

快速回复 返回顶部 返回列表