wayne 发表于 2010-8-12 09:24:25

连分数趣题一则

1/{1+1/{2+1/{3+1/{4+1/{...}}}}}
的准确值是多少？

wayne 发表于 2010-8-12 09:26:17

还有，
1/{1+1/{3+1/{5+1/{7+1/{...}}}}}

hujunhua 发表于 2010-8-12 09:59:11

In:= N], 50]
Out= 1.4331274267223117583171834557759918204315127679060

搜1.4331274267223117捕空，搜1.43312742有一页，是台湾的国立电离层研究所的网页，不知道它是什么数据。
http://www.google.com.hk/search?client=aff-cs-360se&forid=1&ie=utf-8&oe=UTF-8&q=1%2E43312742

wayne 发表于 2010-8-12 10:00:42

3# hujunhua

:)N], 50]我可以瞬间给出前一百万位，这是前一千位
0.6977746579640079820067905925517525994866582629980212323686300828165308527646411129969656541826765687239828218773964133931131922961195325839482671540233685720770846879316532596768026096993447735279134807392866925472877889269341631325163541360922351694910777667127019798991789043551299822748847417815118582827474312800016883973575031589630558148456722812773785313893537964574949111443995739496545408641490244407439658462383405191698214657075454152356161978927702157019980844153256948432472055320438254601089536953956756141086175951613153820732931364439051157889913997941184531707255433214244317404753282387468232949778600917592531885601921774744917302475827510588530039799891961428677298872920269118479725541584489103832653246261506026959580395171325335518290539864261160122414588261244351825022559389263137501312747096674905266754096536040252450838548835894016411556364834073454971407636882729626651342464624332583445931037716279976454941621295291266608179784300381978775455761063199246771331988090510

hujunhua 发表于 2010-8-12 10:07:32

In:= N], 50]
Out= 0.69777465796400798200679059255175259948665826299802

搜到1页
http://www2b.abc.net.au/science/k2/stn/newposts/4544/topic4542551.shtm

wayne 发表于 2010-8-12 10:12:08

5# hujunhua

该链接的源头是：
http://www.research.att.com/~njas/sequences/A052119

wayne 发表于 2010-8-12 10:15:23

貌似 这个 差分方程不好解

a_{n+1}=n*a_n+a_{n-1}

wayne 发表于 2010-8-12 10:35:29

设a(0)=a,a(1)=b,将上述方程进行Z变换，得到微分方程：

x^2*y'(x)=(1-x^2)y(x)+x(a+b*x)

wayne 发表于 2010-8-12 11:04:08

BesselI=\sum _{n=0}^{\infty } {n}/{(n!)^2}
BesselI=\sum _{n=0}^{\infty } 1/{(n!)^2}
那个无穷连分数的值为： BesselI/BesselI =0.69777465796400798200679059255

第二个无穷连分数的值为 tanh(1)=0.76159415595576488811945828260479359

wayne 发表于 2010-8-12 11:10:08

BesselI 函数叫Modified Bessel Function of the First Kind，
http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html
是下面方程的解：
z^2y''(z)+zy'(z)-(z^2+n^2)y(z)=0
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