Loading [MathJax]/jax/output/HTML-CSS/fonts/TeX/fontdata.js

Monday, June 14, 2021

Continued fractions

coth

Continued fraction expansion for coth(z)

We will show the proof for this continued fraction posted by @infseriesbot. e+1e1=coth(12)=2+16+110+114+ Update We add the proof of this result, which is bascially the same function evaluated at other point: 0=coth(iπ2)=2π26π210π214
Recall the continued fraction representation for ratios of confluent hypergeometric functions of the type 0F1(c,z): Let (an) be a sequence of complex numbers defined by an=1(c+n1)(c+n) where c is a complex constant such that cZ{0}. Then 1+Kn=1(anz1)=0F1(c,z)0F1(c+1,z)
Now, for this particular case, we will use the hypergemetric representation of sin(z) and cos(z) sin(z)=z0F1(32,z24) sin(z)=0F1(12,z24) Then tan(x)=sin(x)cos(x)=xx33!+x55!x77!+...1+x22!+x44!x66!+...=x(1x23!+x45!x67!+...)1+x22!+x44!x66!+...=x0F1(32,x24)0F1(12,x24)=x1x23x25x27 If we evaluate at x=iz2 tan(iz2)=iz22z26z210z214=iz2+z26+z210+z214coth(z2)z=cot(iz2)iz=2+z26+z210+z214+ If we evaluate at z=1 e+1e1=coth(12)=2+16+110+114+ Since the result is valid for al the complex plane, if we evaluate at z=iπ then coth(iπ2)=0 and 0=coth(iπ2)=2π26π210π214

No comments:

Post a Comment

Series of the day

Series involving the digamma and the zeta functions The sum 1(n+1)pnq ...