Hyperbolic Cotangent

Author

Eric W. Weisstein
August 24, 2008

This notebook downloaded from http://mathworld.wolfram.com/notebooks/SpecialFunctions/HyperbolicCotangent.nb.

For more information, see Eric's MathWorld entry http://mathworld.wolfram.com/HyperbolicCotangent.html.

©2008 Wolfram Research, Inc. except for portions noted otherwise

Plots

<<Utilities`Plot`
<<Utilities`Typesetting`

Real Axis

PlotDiscontinuous[Coth[x],{x,-5,0,5},PlotRange->{Automatic,{-5,5}},
PlotStyle->Red,
HorizontalAsymptotes->{-1,1},
AxesLabel->TraditionalForm/@{x,Coth[x]}]

Complex plane

ComplexPlot3D[Coth[z],{z,-5,5},PlotPoints->70,ClipFill->None,Mesh->False]//Timing

ComplexContourPlot[Coth[z],{z,-5,5}]//Timing

Limits

Limit[Coth[x],x->0,Direction->#]&/@{-1,1}

Limit[Coth[x],x-># Infinity]&/@{-1,1}

Series

??

RealSeries2[f_,{x_,x0_,n_}]:=Module[{g=TrigToExp[f]},
Series[g /.{Exp[x]->y,E^(nn_Integer*x):>y^nn},{y,x0,n}]/.y->Exp[x]
]

FullSimplify@RealSeries2[Coth[x],{x,Infinity,15}]

About 0

Series[Coth[x],{x,0,15}]

V5

<<DiscreteMath`RSolve`

FullSimplify[SeriesTerm[Coth[z],{z,0,n}]]

Table[(((2^n) (BernoulliB[1+n]+BernoulliB[1+n,1]))/(Gamma[2+n])),{n,-1,15}]

V7

SeriesCoefficient[Coth[z],{z,0,n}]//InputForm

TimeConstrained[GeneratingFunction[%,n,z],60]//Timing

TimeConstrained[GeneratingFunction[If[Mod[n,2]==1,(((2^(1+n)) BernoulliB[1+n])/((1+n)!)),0],n,x],60]//Timing

Sum[% z^n,{n,-2,5}]

About ∞

Series[Coth[x],{x,∞,5}]

Derivative and Integral

Derivative

D[Coth[z],z]

Integral

Integrate[Coth[z],z]

Sums

(∑_(n=-1)^∞)(((z^n) ((2^n) ((Subscript[B, n+1])+(Subscript[B, n+1])(1))))/(Γ(n+2)))==coth(z)

TimeConstrained[Sum[(((2^n) (BernoulliB[1+n]+BernoulliB[1+n,1]))/(Gamma[2+n]))(z^n),{n,-1,∞}],60]//Timing

(∑_(n=-1)^∞)(z^n) Piecewise[(({{1, n==-1}, {(((2^(n+1)) (Subscript[B, n+1]))/((n+1)!)), n mod 2==1∧n≥0}, {0, "PiecewiseDefault"[True]}}))]==coth(x)

SeriesCoefficient[Coth[z],{z,0,n}]//InputForm

Sum[z^n(Piecewise[{{1, n==-1}, {(((2^(1+n)) BernoulliB[1+n])/((1+n)!)), Mod[n,2]==1&&n≥0}, {0, True}}]),{n,-1,∞}]//InputForm

(∑_(n=1)^∞)(((z^(2 n-1)) ((2^(2 n)) (Subscript[B, 2 n])))/((2 n)!))+(1/z)==coth(z)

(1/z)+Sum[(((2^(2n))BernoulliB[2n])/((2n)!))(z^(2n-1)),{n,1,∞}]

Simplify[%]