Initialization

The cells below are initialization cells. These cells are automatically evaluated by Mathematica as soon as the notebook is opened.

darkblue=RGBColor[34/256,44/256,162/256];
lightblue=RGBColor[220/256,248/256,248/256];
darkgreen=RGBColor[0,0.5,0];
modulePlot[function_,{xmin_,xmax_},ticks_:None,plotrange_,imagesize_:600,aspectratio_:Automatic,{opts___}]:=Module[{x},Plot[function[x],{x,xmin,xmax},
PlotStyle->Red,AxesStyle->darkgreen, Background->lightblue,TicksStyle->darkblue,
Ticks->ticks,
PlotRange->plotrange,
ImageSize->imagesize,
AspectRatio->aspectratio,
opts]]

moduleParametricPlot3D[function_,αmin_,αmax_,tmin_,tmax_][mesh_,viewpoint_][imagesize_:400,colorfunction_:White,{opts___}]:=Module[{α,t},ParametricPlot3D[Evaluate[function[α,t]],{α,αmin,αmax},{t,tmin,tmax},
Axes->False,PlotRange->All,BoxRatios->{1,1,1},Boxed->False,
Mesh->mesh,
ViewPoint->viewpoint,
ImageSize->imagesize,
ColorFunction->colorfunction,
opts]]

Figure 1.1   Example 1.3

p1=Plot[(1/(Abs[Log[x]])),{x,0.00001`,0.5`},PlotStyle->Black,Ticks->{{0.5`},{1.4`}},PlotPoints->1800, MaxRecursion->15, BaseStyle->{FontFamily->"Times",FontSize->12.`}];
p2=Plot[(1/(x Abs[Log[x]])),{x,0.00001`,0.5`},PlotStyle->Black,Ticks->{{0.5`},{50}},PlotPoints->800,PlotRange->{-1,50},BaseStyle->{FontFamily->"Times",FontSize->12.`}];
Show[GraphicsRow[{p1,p2}],ImageSize->450]

Figure 1.2   Example 1.4 and Example 20.27

Simplify[Re[ComplexExpand[(1/(1- r E^(I x))),{x∈ℝ}]],{x∈ℝ}]

A[r_,x_]:=((1-r Cos[x])/(1+(r^2)-2 r Cos[x]))
Limit[A[r,x],r->1]
p[r_]:=Plot[Table[Evaluate[A[r+9 (∑_(j=2)^k)(10^(-j)),x]],{k,2,5}],{x,-0.105,0.105},PlotStyle->Black,PlotRange->{0,21},PlotPoints->20,MaxRecursion->15,Ticks->{Table[l 0.05,{l,-2,2}],{0,0.5,10,20}}, BaseStyle->{FontFamily->"Times",FontSize->14}]
q[r_]:=Plot[Table[A[r+9 (∑_(j=2)^k)(10^(-j)),x],{k,5,5}],{x,-3 π-1,3 π+1},PlotStyle->Black,PlotRange->{0,21},PlotPoints->20,MaxRecursion->15,Ticks->{Table[2 l π,{l,-1,1}],{0,(1/2),10,20}},BaseStyle->{FontFamily->"Times",FontSize->14},ImageSize->300]
GraphicsRow[{p[0.9],q[0.9]}]

The family of functions (Subscript[A, r]) has some remarkable properties.

prim0[r_,α_]:=Evaluate[FullSimplify[Integrate[A[r,α],α]]]
prim0[r,α]

FullSimplify[(∂_α)prim0[r,α]]

Plot[Table[prim0[k 0.1,α],{k,0,10}],{α,-π,π},Ticks->{{-π,-(π/2),(π/2),π},{-(π/2),(π/2)}},AspectRatio->Automatic,ImageSize->400]

prim1[r_,α_]:=Piecewise[{{prim0[r,α],α≤0},{prim0[r,α]+π,α>0}}]
Plot[{Table[Evaluate[prim1[k 0.1,α]],{k,0,9}],(α/2)},{α,-π,π},PlotStyle->Black,BaseStyle->{FontFamily->"Times",FontSize->13},Ticks->{{-π,-(π/2),(π/2),π},{-(π/2),(π/2),π,((3 π)/2)}},AspectRatio->Automatic,ImageSize->300]

prim[r_,α_]:=(α/2)-ArcTan[(((r+1) Tan[(α/2)])/(r-1))]
FullSimplify[D[prim[r,α],α]]

Figure 1.3

p1=Plot[Table[prim[k 0.1,α],{k,0,9}],{α,-π,π},Ticks->{{-π,π},{-π,-(π/2),(π/2),π}},PlotStyle->Black,BaseStyle->{FontFamily->"Times",FontSize->14},AspectRatio->Automatic,ImageSize->300];
p2=Graphics[Line[{{{-Pi,-Pi},{0,-Pi/2}},{{Pi,Pi},{0,Pi/2}}}]];
p3=Graphics[{Dashing[{0.02,0.02}],Line[{{Pi,0},{0,-Pi/2}}]}];
Show[p1,p2,p3]

Figure 1.4

Plot[Table[A[k 0.1,α],{k,0,10}],{α,-π-0.25,π+0.25},PlotStyle->Black,Ticks->{Table[l π,{l,-1,1}],{0.75,1,2}},PlotRange->{0.4,3.5},AxesOrigin->{0,0.5},BaseStyle->{FontFamily->"Times",FontSize->14},ImageSize->300]

(∫_(-π)^π)Table[A[0.1 k,α],{k,0,10}]\[DifferentialD]α==Join[Table[2 π,{k,0,9}],{π}]

Finding an antiderivative of (Subscript[A, r]). First apply Exercise 0.1 in [7] and use twice the duplication formula for the arctangent.

FullSimplify[(A[r,x,0]/.{Cos[x]->((1-t^2)/(1+t^2))})2/(1+t^2)]
Integrate[%,t]

FullSimplify[(((((1+r) t)/(1-r))-(((1+r) t)/(-1+r)))/(1+(((1+r) t)/(1-r))(((1+r) t)/(-1+r))))]

Expand[(1+r (-1+t)+t) (-1+r+t+r t)]

FullSimplify[(((((1+r) t)/(1-r))-(((1+r) t)/(-1+r)))/(1+((((1+r) t) ((1+r) t))/((1-r) (-1+r)))))==-((2((r+1)/(r-1))  t)/(1-(((((r+1)/(r-1)) t))^2)))]

The envelope of the graphs of the (Subscript[A, r]).

A[r_,x_,y_]:=((1-r Cos[x])/(1+(r^2)-2 r Cos[x]))-y

FullSimplify[D[A[r,x,y],r]]

FullSimplify[Solve[{D[A[r,x,y],r]==0,A[r,x,y]==0},{x,y},r]]

Plot[(1/2) ((Csc[x])^2) (1-((Cos[x])^2)+(Sqrt[((Sin[x])^2)])),{x,0,Pi}]

FullSimplify[(1/2) ((Csc[x])^2) (1-((Cos[x])^2)+(Sqrt[((Sin[x])^2)]))==(1/2) (1+(1/(Abs[Sin[x]]))),{-π<x<π}]

FullSimplify[A[r,x,0]/.{r->(1-Sin[x])/Cos[x]}]

Plot[(1/2) (1+1/Abs[Sin[x]]),{x,-1,1}]

Plot[{Table[A[k 0.1,α],{k,0,10}],(1/2) (1+1/Sin[α])},{α,-π-0.25,π+0.25},PlotStyle->Black,Ticks->{Table[l π,{l,-1,1}],{0.75,1,2}},PlotRange->{0.4,10},AxesOrigin->{0,0.5},BaseStyle->{FontFamily->"Times",FontSize->13},ImageSize->300]

FullSimplify[A[r,x,0]/.{r->(1-Sin[x])/Cos[x]}]

Figure 1.5

Clear["`*"]
Off[General::"spell1"]
α[x_]:=(E^(-1/x))
p0=Plot[α[x],{x,0,0.4`},PlotPoints->100,Ticks->None,AspectRatio->Automatic,PlotRange->All];
α[x_]:=(E^(-1/x))
p00=Plot[α[x],{x,0,10},PlotPoints->100,Ticks->None,AspectRatio->Automatic,PlotRange->All];
a1=-1;
b1=2;
β1[x_]:=0/;x≤a1
β1[x_]:=α[x-a1] α[b1-x]/;a1<x<b1
β1[x_]:=0/;b1≤x
Plot[β1[x],{x,a1,b1},Ticks->None,AspectRatio->1,PlotRange->All];
Int1=N[(∫_a1^b1)β1[x]\[DifferentialD]x];
γ1[x_]:=((β1[x])/Int1)
Plot[γ1[x],{x,-1,2},PlotStyle->{Black,Thickness[0.005]},Ticks->None,AspectRatio->(1/2),PlotRange->All,ImageSize->90]
N[(∫_Global`a1^Global`b1)Global`γ1[Global`x]\[DifferentialD]Global`x];
ϵ=(1/2);
Global`γ1ϵ[Global`x_]:=((γ1[(x/ϵ)])/ϵ)
Plot[γ1ϵ[x],{x,a1,b1},PlotStyle->{Black,Thickness[0.005]},Ticks->None,AspectRatio->1,PlotRange->All,ImageSize->90]
Show[GraphicsRow[{p1,p2}],Ticks->None,AspectRatio->Automatic,PlotRange->All,ImageSize->500];
a2=2;
b2=3;
β2[x_]:=0/;x≤a2
β2[x_]:=α[x-a2] α[b2-x]/;a2<x<b2
β2[x_]:=0/;b2≤x
Plot[β2[x],{x,a2,b2},Ticks->None,AspectRatio->1,PlotRange->All];
Int2=N[(∫_a2^b2)β2[x]\[DifferentialD]x];
γ2[x_]:=((β2[x])/Int2)
p3=Plot[γ2[x],{x,a2,b2},Ticks->None,AspectRatio->Automatic,PlotRange->All];
N[(∫_a2^b2)γ2[x]\[DifferentialD]x];

Figure 1.6   Problem 1.3

Clear["`*"]
p[x_]:=0/;x<-1
p[x_]:=(35/32) (((1-(x^2)))^3)/;-1≤x≤1
p[x_]:=0/;x>1
NIntegrate[p[x],{x,-1,1}]
Plot[p[x],{x,-1.1`,1.1`},PlotStyle->{Black,Thickness[0.002]},PlotPoints->100, MaxRecursion->15, Ticks->{{-1,1},{0,35/32}},PlotRange->All,BaseStyle->{FontFamily->"Times",FontSize->12.`},AspectRatio->Automatic,ImageSize->300]

Figure 1.7   Problem 1.11

Clear["`*"]
Off[NIntegrate::ploss]
p[x_]:=Piecewise[{{0,x<-1},{(35/32)(((1-(x^2)))^3),-1≤x≤1},{0,x>1}}]
p1=Plot[200p[100x],{x,-0.01,0.01},PlotStyle->Black,PlotPoints->100,Ticks->{{-0.01,0.01},{200}},BaseStyle->{FontFamily->"Times",FontSize->12.}];
p2=Plot[-1+2Integrate[1000p[1000t],{t,-0.001,x}],{x,-0.0015,0.0015},PlotStyle->Black,PlotPoints->100,PlotRange->All,Ticks->{{-0.0015,0.0015},{-1,1}},BaseStyle->{FontFamily->"Times",FontSize->12.}];
p3=Plot[-x+2Integrate[Evaluate[Integrate[1000p[1000t],{t,-0.001,ξ}]],{ξ,-0.001,x}],{x,-0.003,0.003},PlotStyle->Black,Ticks->{{-0.003,0.003},{0.003}},BaseStyle->{FontFamily->"Times",FontSize->12.},AspectRatio->Automatic];
Show[GraphicsArray[{p1,p2}],ImageSize->400]
Show[p3,ImageSize->400]

Figure 1.8   Problem 1.13

Plot[Composition[ArcCos,Cos][x],{x,-4 π-1,4 π+1}, MaxRecursion->15, PlotStyle->Black,Ticks->{Table[k π,{k,-4,4,1}],Table[k π,{k,-1,1,1}]},PlotRange->{-0.1,π+0.1},AspectRatio->Automatic,BaseStyle->{FontFamily->"Times",FontSize->11},ImageSize->300]
(D[ArcCos[t],t]/.t->Cos[x])D[Cos[x],x]

Notice that Mathematica defines cos ∘ arccos on a domain properly containing [ -1, 1 ] .

modulePlot[ArcCos,{-2,2},{Table[(k/2),{k,-3,3,1}],Table[((k π)/2),{k,0,4,1}]},{-0.1,π+0.1},400,{}]
modulePlot[Composition[Cos,ArcCos],{-2,2},{Table[(k/2),{k,-3,3,1}],Table[(k/2),{k,-3,3,1}]},{-(π/2)-0.1,(π/2)+0.1},400,{}]
modulePlot[Composition[ArcCos,Cos],{-4 π-1,4 π+1},{Table[k π,{k,-4,4,1}],Table[k π,{k,-1,1,1}]},{-0.1,π+0.1},300,{}]
MatrixForm[Table[{((k π)/2),ArcCos[Cos[((k π)/2)]]},{k,-4,4}]]
FullSimplify[(∂_x)Composition[ArcCos,Cos][x]==((Abs[Sin[x]])/(Sin[x]))]
modulePlot[((Abs[Sin[#1]])/(Sin[#1]))&,{-4 π,4 π},{Table[k π,{k,-4,4,1}],Table[k,{k,-1,1,1}]},{-1-0.1,1+0.1},400,{}]

We recognize arccos ∘ cos as a convolution product .

FullSimplify[(∂_x)(Composition[ArcCos,Cos][x]-((π/2)-Composition[ArcSin,Sin][x+(π/2)])),x]
modulePlot[Evaluate[(∑_(k=-2)^2)Convolve[Boole[0<y<π],Boole[0<y<π],y,#1-2 k π]]&,{-4 π,4 π},{Table[k π,{k,-4,4,1}],Table[k π,{k,-4,4,1}]},{-0.1,π+0.1},600,{}]
modulePlot[(∑_(k=-2)^2)π UnitTriangle[((#1)/π)-2 k-1]&,{-4 π,4 π},{Table[k π,{k,-4,4,1}],Table[k π,{k,-4,4,1}]},{-0.1,π+0.1},600,{}]

modulePlot[Composition[ArcSin,Sin],{-4π,4 π},{Table[k π,{k,-4,4,1}],Table[k π,{k,-4,4,1}]},{-(π/2)-0.4,(π/2)+0.4},800,{}]
FullSimplify[(∂_x)Composition[ArcSin,Sin][x]]
modulePlot[(π/2)-Composition[ArcSin,Sin][#1+(π/2)]&,{-2 π,2 π},{Table[((k π)/2),{k,-4,4,1}],Table[((k π)/2),{k,-4,4,1}]},{-0.1,π+0.1},800,{}]

FullSimplify[Convolve[Boole[0<y<Pi],Boole[0<y<Pi],y,x]==Pi UnitTriangle[x/Pi-1]]

Figure 2.1   δ-neighborhood

Clear["`*"]
tubecurve[ϕ_][δ_][t_]:=ϕ[t]-((δ {{0,-1},{1,0}}.ϕ'[t])/(Sqrt[((ϕ'[t][[1]])^2)+((ϕ'[t][[2]])^2)]))
ϕ[t_]:=((3 {t,(t^2)})/(1+(t^3)))
d=Graphics[{GrayLevel[0.9`],Disk[ϕ[-0.5],0.2]}];
p=Table[ParametricPlot[{tubecurve[ϕ][δ][t],ϕ[t]},{t,-0.5,15},MaxRecursion->15,PlotStyle->{GrayLevel[0.9],GrayLevel[0]},Axes->None],{δ,-0.2,0.2,0.0005}];
Show[d,p,AspectRatio->Automatic]

Figure 2.2   Lemmas 2.8 and 2.9, and 2.9

Clear["`*"]
α[x_]:=(E^(-1/x))
p0=Plot[α[x],{x,0,0.52},PlotStyle->Black,PlotPoints->100,Ticks->{{0,1/2},{0,α[1/2]}},AspectRatio->Automatic,PlotRange->All,BaseStyle->{FontFamily->"Times",FontSize->14},ImageSize->300]
a1=-1;
b1=2;
β1[x_]:=Piecewise[{{0,x≤a1},{α[x-a1] α[b1-x],a1<x<b1},{0,b1≤x}}]
Int1=N[(∫_a1^b1)β1[x]\[DifferentialD]x]
γ1[x_]:=((β1[x])/Int1)
N[γ1[0.5],2]
p1=Plot[Evaluate[γ1[x]],{x,-1.1,2.1},Ticks->{{-1,1/2,2},{0.598}},PlotStyle->Black,AspectRatio->Automatic,PlotRange->All,BaseStyle->{FontFamily->"Times",FontSize->14}
,ImageSize->300]

Figure 2.3   Lemma 2.9

Plot3D[γ1[x1] γ2[x2],{x1,a1,b1},{x2,a2,b2},PlotStyle->Black,ViewPoint->2 {a1,b1,0.6`},Mesh->{50,50},Axes->None,Boxed->False,PlotRange->All,ImageSize->400 , ColorFunction->(White&)]

Figures 2.2 and  2.3.   Because of problems with rendering the graphics by Mathematica in the Versions from 6.0 on, the illustrations were produced using Version 5.2 with the code below.

Clear["`*"]
Off[General::spell1]
α[x_]:=(E^(-1/x))
p0=Plot[α[x],{x,0,0.4},PlotPoints->100,Ticks->None,AspectRatio->Automatic,PlotRange->All];
α[x_]:=(E^(-1/x))p00=Plot[α[x],{x,0,10},PlotPoints->100,Ticks->None,AspectRatio->Automatic,PlotRange->All];
a1=-1;
b1=2;
β1[x_]:=0/;x<=a1
β1[x_]:=α[x-a1]α[b1-x]/;a1<x<b1
β1[x_]:=0/;b1<=x
Plot[β1[x],{x,a1,b1},Ticks->None,AspectRatio->1,PlotRange->All]
Int1=N[(∫_a1^b1)β1[x]\[DifferentialD]x]γ1[x_]:=1/Int1 β1[x]p1=Plot[γ1[x],{x,-1,2},Ticks->None,AspectRatio->1/2,PlotRange->All];
N[(∫_a1^b1)γ1[x]\[DifferentialD]x]ϵ=1/2;γ1ϵ[x_]:=1/ϵ γ1[x/ϵ]p2=Plot[γ1ϵ[x],{x,a1,b1},Ticks->None,AspectRatio->1,PlotRange->All];Show[GraphicsArray[{p1,p2}],Ticks->None,AspectRatio->Automatic,PlotRange->All,ImageSize->500]
a2=2;b2=3;
β2[x_]:=0/;x<=a2
β2[x_]:=α[x-a2]α[b2-x]/;a2<x<b2
β2[x_]:=0/;b2<=x
Plot[β2[x],{x,a2,b2},Ticks->None,AspectRatio->1,PlotRange->All]
Int2=N[(∫_a2^b2)β2[x]\[DifferentialD]x]γ2[x_]:=1/Int2 β2[x]
p3=Plot[γ2[x],{x,a2,b2},Ticks->None,AspectRatio->Automatic,PlotRange->All];
N[(∫_a2^b2)γ2[x]\[DifferentialD]x]

Figure 2.4   Theorem 2.15

Clear["`*"]
s[0][x_]:=((x^3)/6);
s[1][x_]:=(1/6) ((((1+x))^3)-4 (x^3));
s[x_]:=0/;x<-2;
s[x_]:=s[0][x+2]/;-2<x<-1;
s[x_]:=s[1][x+1]/;-1<x<0;
s[x_]:=s[1][-x+1]/;0<x<1;
s[x_]:=s[0][-x+2]/;1<x<2;
s[x_]:=0/;2<x;
spl[x_]:=s[2 x];
spl0=Plot[Global`spl[x+1],{x,-2,0},PlotStyle->Black,PlotRange->{0,0.7`},PlotPoints->100,AspectRatio->Automatic,BaseStyle->{FontFamily->"Times",FontSize->12.`}];spl1=Plot[spl[x+(1/2)],{x,-1.5`,0.5`},PlotStyle->Black,PlotRange->{0,0.7`},PlotPoints->100,AspectRatio->Automatic,BaseStyle->{FontFamily->"Times",FontSize->12.`}];
spl2=Plot[spl[x],{x,-1,1},PlotStyle->Black,PlotRange->{0,0.7`},PlotPoints->100,AspectRatio->Automatic,BaseStyle->{FontFamily->"Times",FontSize->12.`}];
spl3=Plot[spl[x-(1/2)],{x,-0.5`,1.5`},PlotStyle->Black,PlotRange->{0,0.7`},PlotPoints->100,AspectRatio->Automatic,BaseStyle->{FontFamily->"Times",FontSize->12.`}];
spl4=Plot[spl[x-1],{x,0,2},PlotStyle->Black,PlotRange->{0,0.7`},PlotPoints->100,AspectRatio->Automatic,BaseStyle->{FontFamily->"Times",FontSize->12.`}];
ssa=Show[spl0,spl1,spl2,spl3,spl4,PlotRange->{0,0.7`},AspectRatio->Automatic,Ticks->{{-2,2},{0.6`}},ImageSize->500]
ssb=Plot[(∑_(j=-2)^2)spl[x+(j/2)],{x,-2,2},PlotStyle->Black,PlotRange->{0,1},AspectRatio->Automatic,PlotPoints->200,Ticks->{{-2,2},{1}},ImageSize->500,BaseStyle->{FontFamily->"Times",FontSize->12.`}]

Figure 2.5

Clear["`*"]
Plot[((E^(-(x^2)))/(Sqrt[π])),{x,-3,3},Ticks->None,PlotRange->All]
Plot3D[((E^(-(x1^2)-(x2^2)))/π),{x1,-3,3},{x2,-3,3},PlotPoints->50,Ticks->None,PlotRange->All,ViewPoint->2 {-3,-4,1},Boxed->False,Axes->None,ImageSize->550,ColorFunction->(White&)]
ParametricPlot3D[{r Cos[α],r Sin[α],((E^(-(r^2)))/π)},{r,0,4},{α,-π,π},PlotStyle->Black,Mesh->{30,60},PlotPoints->50,AspectRatio->0.7`,Ticks->None,ViewPoint->{-3,4,0.05`},Boxed->False,Axes->None,ImageSize->550,BoundaryStyle->Automatic,ColorFunction->(White&)]

Figure 2.6   Problem 2.2

Clear["`*"]
ϵ=(1/50);
γ[x_]:=(1/((Sqrt[π])ϵ))(E^(-((x^2)/(ϵ^2))))
Integrate[γ[x],{x,-∞,∞}]
Plot[γ[x],{x,-0.05,0.05},PlotStyle->Black,ImageSize->500,Ticks->{{-0.05,0.05},{25}},BaseStyle->{FontFamily->"Times",FontSize->15.}]
f[x_]:=Integrate[Abs[x-y]γ[y],{y,-∞,∞}]
Plot[f[x],{x,-0.5,0.5},PlotStyle->Black,AspectRatio->Automatic,ImageSize->500,Ticks->{{-0.5,0.5},{0.5}},BaseStyle->{FontFamily->"Times",FontSize->15.}]

Figure 5.1   Problem 5.3

p1=Plot[Table[(x/((x^2)+(ϵ^2))),{ϵ,0.025,0.25,0.025}],{x,-0.55,0.55},PlotRange->{-10.5,10.5},PlotStyle->Black,BaseStyle->{FontFamily->"Times",FontSize->12.},Ticks->{{-(1/2),(1/2)},{-10,10}}];
p2=Plot[Table[(ϵ/((x^2)+(ϵ^2))),{ϵ,0.025,0.25,0.025}],{x,-0.55,0.55},PlotRange->{-0.5,10.5},PlotStyle->Black,BaseStyle->{FontFamily->"Times",FontSize->12.},Ticks->{{-(1/2),(1/2)},{0,10}}];
GraphicsRow[{p1,p2},ImageSize->500]

Figure 5.2   Problem 5.5

Clear["`*"]
u[t_,x_]:=((E^(-((x^2)/(4 t))))/(Sqrt[4 π t]))
t=(1/10);
u[t,0]
p1=Plot[u[t,x],{x,-2,2},PlotStyle->Black,Ticks->{{-2,2},{0,u[t,0]}},AspectRatio->Automatic,BaseStyle->{FontFamily->"Times",FontSize->12.},ImageSize->300];
p11=Plot[u[t,x],{x,-2,2},PlotStyle->Black,AspectRatio->Automatic,BaseStyle->{FontFamily->"Times",FontSize->12.},Ticks->{{-2,2},{}}(*,ImageSize->200*)];
v[t_,x_]:=(∂_xx)u[t,xx]/.xx->x
sol=Solve[(∂_x)v[t,x]==0,x]/.Rule->Equal
p2=Plot[v[t,x],{x,-2,2},PlotStyle->Black,Ticks->{{-2,-(1/(Sqrt[5])),(1/(Sqrt[5])),2},{v[t,-(1/(Sqrt[5]))],v[t,(1/(Sqrt[5]))]}},AspectRatio->Automatic,BaseStyle->{FontFamily->"Times",FontSize->12.}(*,ImageSize->200*)];
p21=Plot[v[t,x],{x,-2,2},PlotStyle->Black,AspectRatio->Automatic,BaseStyle->{FontFamily->"Times",FontSize->12.},Ticks->{{-2,2},{}}(*,ImageSize->200*)];
w[t_,x_]:=(∂_xx)v[t,xx]/.xx->x
Solve[(∂_x)w[t,x]==0,x]
p3=Plot[w[t,x],{x,-2,2},PlotStyle->Black,Ticks->{{-2,-(Sqrt[(3/5)]),(Sqrt[(3/5)]),2},{(5/(E ))(Sqrt[(10/(E π))]),w[t,-(Sqrt[(3/5)])],w[t,(1/(Sqrt[5]))]}},AspectRatio->Automatic,BaseStyle->{FontFamily->"Times",FontSize->12.},ImageSize->200];
p31=Plot[w[t,x],{x,-2,2},PlotStyle->Black,Ticks->{{-2,2},{}},AspectRatio->Automatic,BaseStyle->{FontFamily->"Times",FontSize->12.}(*,ImageSize->200*)];
(∫_(-∞)^∞)u[t,x]\[DifferentialD]x
Show[GraphicsArray[{p1,p2,p3}],ImageSize->600]
Show[GraphicsArray[{p11,p21,p31}],ImageSize->600]
p11
p21
p31

Clear[t]
FullSimplify[(∂_t)u[t,x]-(∂_({x,2}))u[t,x]]

Figure 5.3   Problem 5.9

Clear["`*"]
f[x_]:=(∑_(n=1)^10000)((Sin[(n^2)x])/(n^2))
Plot[Evaluate[f[x]],{x,0,π},PlotStyle->Black,PlotPoints->10000,ImageSize->500,Ticks->{{Pi/2,Pi},{1,1.2337}},BaseStyle->{FontFamily->"Times",FontSize->10.},AspectRatio->Automatic]

Clear["`*"]
f[x_]:=(∑_(n=1)^10000)((Sin[(n^2)x])/(n^2))
Plot[Evaluate[f[x]],{x,(π/2)-0.05,(π/2)+0.15},PlotStyle->Black,PlotPoints->10000,ImageSize->400,Axes->None,BaseStyle->{FontFamily->"Times",FontSize->10.},AspectRatio->Automatic]

f[x_]:=(∑_(n=1)^Infinity)((Sin[(n^2)x])/(n^2))
N[f[Pi/2],5]

Figure 9.1   See separateFile RCL.nb

Problem 9.14   A computation

DSolve[L J''[t]+R J'[t]+(1/C) J[t]==0,J,t]
Solve[L (x^2)+R x+(1/C)==0,x]

Ra=4;
L=2;
Ca=1;
Rb=2 (Sqrt[2]);
N[%]
Rc=1;
Rd=0;
(Ra^2)-((4 L)/Ca)
(Rb^2)-((4 L)/Ca)
-((Rc^2)-((4 L)/Ca))
((Sqrt[-((Rd^2)-((4 L)/Ca))])/(2 L))
(2/(Sqrt[-((Rd^2)-((4 L)/Ca))]))
Plot[{((2 HeavisideTheta[t] (E^(-((Ra t)/(2 L)))) Sinh[((t (Sqrt[(Ra^2)-((4 L)/Ca)]))/(2 L))])/(Sqrt[(Ra^2)-((4 L)/Ca)])),((HeavisideTheta[t] t (E^(-((Rb t)/(2 L)))))/L),((2 HeavisideTheta[t] (E^(-((Rc t)/(2 L)))) Sin[((t (Sqrt[-((Rc^2)-((4 L)/Ca))]))/(2 L))])/(Sqrt[-((Rc^2)-((4 L)/Ca))])),((2 HeavisideTheta[t] (E^(-((Rd t)/(2 L)))) Sin[((t (Sqrt[-((Rd^2)-((4 L)/Ca))]))/(2 L))])/(Sqrt[-((Rd^2)-((4 L)/Ca))]))},{t,0,17},PlotRange->{-0.8,0.8},PlotStyle->Black,BaseStyle->{FontFamily->"Times",FontSize->11.},Ticks->{{0,π(Sqrt[2]),2 π (Sqrt[2]),3 π (Sqrt[2])},{-(1/(Sqrt[2])),(1/(Sqrt[2]))}}]

Figure 10.1   Problem 5.5

Plot[(x^4)-(x^2),{x,-(Sqrt[2]),(Sqrt[2])},PlotStyle->Black,PlotRange->All,Ticks->{{-(Sqrt[2]),-1,1,(Sqrt[2])},{0,2}},BaseStyle->{FontFamily->"Times",FontSize->14},ImageSize->300]

Figure without caption   Corollary 11.7

Clear["`*"]
compact=Graphics[{GrayLevel[0.45],Polygon[Table[N[{-0.2+Cos[n π/5],0.3+Sin[n π/5]}],{n,8}]]}];
x1=Graphics[{GrayLevel[0.85],Disk[{0,1},2]}];
x2=Graphics[{GrayLevel[0.85],Disk[{0,-0.4},1.9]}];
n1=Graphics[{GrayLevel[0.75],Disk[{0,0.5},1.7]}];
n2=Graphics[{GrayLevel[0.65],Disk[{0,0.5},1.4]}]; 
t=Graphics[{Text["(Subscript[K, j])",{-0.4,-0.1}],Text["X",{0,-1.8}],Text["supp (Subscript[χ, j])",{0.9,0}],Text["(Subscript[d, j])-nbhd of supp (Subscript[χ, j])",{0.1,2.05}]}];
Show[x1,x2,n1,n2,compact,t,BaseStyle->{FontFamily->"Times",FontSize->14.},AspectRatio->Automatic]

Figure 12.1   Problem 12.2

Clear["`*"]
p=Table[PolarPlot[r(Sqrt[Abs[Cos[α]]]),{α,0,2π},PlotStyle->Black,Ticks->None,AspectRatio->Automatic],{r,1,7,1/2}];
l1=Graphics[{Thickness[Tiny],Line[{{-7.2,0},{7.2,0}}]}];
l2=Graphics[{Thickness[Tiny],Line[{{0,-4.8},{0,4.8}}]}];
p1=Show[p,l1,l2];
q=Table[ContourPlot[{(x1^2)+(x2^2)==c x1,(x1^2)+(x2^2)==-c x1},{x1,-7,7},{x2,-7,7},Frame->False,ContourStyle->Black,AspectRatio->Automatic],{c,1,7,1/2}];
l1=Graphics[{Thickness[Tiny],Line[{{-7.5,0},{7.5,0}}]}];
l3=Graphics[{Thickness[Small],Line[{{0,-4},{0,4}}]}];
q1=Show[q,l1,l3];
Show[GraphicsArray[{p1,q1}],ImageSize->500]

Figure 12.2   Problem 12.8

Clear["`*"]
r[a_]:={{Cos[a],Sin[a]},{-Sin[a],Cos[a]}}
arrow[x_,y_,h_,b_,a_]:=Polygon[{r[a].{-2 b,-4 h}+{x,y},{x,y},r[a].{2 b,-4 h}+{x,y},r[a].{0,-3 h}+{x,y}}]
l=ParametricPlot[{t,Abs[t]},{t,-2,2},PlotStyle->Black];
p=l/.(Line[pts_]:>{GrayLevel[0.88],Polygon[pts]});
a=Graphics[{Black,Line[{{0,0},{0,2.1}}],Line[{{-1.4,1.4},{1.4,1.4}}],Line[{{1.4,1.4},{1.4,2.2}}],Text["V",{0.5,1.1}],arrow[-0.8,0.8,0.06,0.03,3π/4],arrow[1,1,0.06,0.03,π/4]}];
Show[p,l,a,Axes->True,Ticks->None,AspectRatio->Automatic,AxesLabel->{"x","t"},BaseStyle->{FontFamily->"Times",FontSize->14.},ImageSize->300]

Clear["`*"]
r[a_]:={{Cos[a],Sin[a]},{-Sin[a],Cos[a]}}
arrow[x_,y_,h_,b_,a_]:=Polygon[{r[a].{-2 b,-4 h}+{x,y},{x,y},r[a].{2 b,-4 h}+{x,y},r[a].{0,-3 h}+{x,y}}]
l=ParametricPlot[{t,Abs[t]},{t,-2,2}]
p=l/.Line[pts_]:>{GrayLevel[0.92`],Polygon[pts]};
a=Graphics[Line[{{0,0},{0,2.1`}}]]
Show[p,l,a,Axes->True,Ticks->None,AspectRatio->Automatic,AxesLabel->{"x","t", FontSlant->Italic},BaseStyle->{FontFamily->"Times",FontSize->11.`}]

Problem 12.13   A computation

Clear["`*"]
n[x_List]:=(Sqrt[(∑_(j=1)^(Length[x]))x[[j]]^2])
n[{(Subscript[x, 1]),(Subscript[x, 2]),(Subscript[x, 3])}]
g[x1_,x2_,x3_]:=Cos[k n[{(Subscript[x, 1]),(Subscript[x, 2]),(Subscript[x, 3])}]]
h[x1_,x2_,x3_]:=((Sin[k n[{(Subscript[x, 1]),(Subscript[x, 2]),(Subscript[x, 3])}]])/(n[{(Subscript[x, 1]),(Subscript[x, 2]),(Subscript[x, 3])}]))

Print["Function g"]
Simplify[(∂_(Subscript[x, 1]))g[(Subscript[x, 1]),(Subscript[x, 2]),(Subscript[x, 3])]]/.(n[{(Subscript[x, 1]),(Subscript[x, 2]),(Subscript[x, 3])}]->"||x||")
Simplify[(∂_(Subscript[x, 1])) (∂_(Subscript[x, 1]))g[(Subscript[x, 1]),(Subscript[x, 2]),(Subscript[x, 3])]]/.(n[{(Subscript[x, 1]),(Subscript[x, 2]),(Subscript[x, 3])}]->"||x||")
Simplify[(∂_(Subscript[x, 1])) (∂_(Subscript[x, 1]))g[(Subscript[x, 1]),(Subscript[x, 2]),(Subscript[x, 3])]+(∂_(Subscript[x, 2])) (∂_(Subscript[x, 2])) g[(Subscript[x, 1]),(Subscript[x, 2]),(Subscript[x, 3])]+(∂_(Subscript[x, 3])) (∂_(Subscript[x, 3])) g[(Subscript[x, 1]),(Subscript[x, 2]),(Subscript[x, 3])]]

Print["Function h"]
Simplify[(∂_(Subscript[x, 1]))h[(Subscript[x, 1]),(Subscript[x, 2]),(Subscript[x, 3])]]/.(n[{(Subscript[x, 1]),(Subscript[x, 2]),(Subscript[x, 3])}]->"||x||")
Simplify[(∂_(Subscript[x, 1])) (∂_(Subscript[x, 1]))h[(Subscript[x, 1]),(Subscript[x, 2]),(Subscript[x, 3])]]/.(n[{(Subscript[x, 1]),(Subscript[x, 2]),(Subscript[x, 3])}]->"||x||")
Simplify[(∂_(Subscript[x, 1])) (∂_(Subscript[x, 1]))h[(Subscript[x, 1]),(Subscript[x, 2]),(Subscript[x, 3])]+(∂_(Subscript[x, 2])) (∂_(Subscript[x, 2])) h[(Subscript[x, 1]),(Subscript[x, 2]),(Subscript[x, 3])]+(∂_(Subscript[x, 3])) (∂_(Subscript[x, 3])) h[(Subscript[x, 1]),(Subscript[x, 2]),(Subscript[x, 3])]]

Figure 13.1   Nappes of cone

Clear["`*"]
Off[General::spell1]
(*transparant cone*)
x1[α_,t_]:=t Cos[α];
x2[α_,t_]:=t Sin[α];
x3[α_,t_]:=t;
x[α_,t_]:={x1[α,t],x2[α,t],x3[α,t]};

raster=0.04;
point=0.001;
list=Table[Point[x[α,t]],{α,-π,π,raster},{t,-3,3,raster}];
cone=Graphics3D[{PointSize[point],list},Axes->False,PlotRange->All,AspectRatio->Automatic];
cif=ParametricPlot3D[Evaluate[x[α,3]],{α,-π,π},PlotPoints->200,DisplayFunction->Identity];
cib=ParametricPlot3D[Evaluate[x[α,-3]],{α,-π,π},PlotPoints->200,DisplayFunction->Identity];
view=2{1,1,0.7};
Show[cone,cif,cib,PlotRange->All,Boxed->False,BoxRatios->{1,1,1.7},AspectRatio->Automatic,ViewPoint->view,DisplayFunction->$DisplayFunction]

Figure 13.2  Level surfaces of Lorentz metric

SetOptions[ParametricPlot3D, BoundaryStyle->Automatic,PlotPoints->70,Boxed->False,Ticks->None,Axes->False,ViewPoint-> {3,-2.8,1.4},ImageSize->400];g=Show[{Table[ParametricPlot3D[{{r Cos[α], r Sin[α],Sqrt[r^2-c]}, {r Cos[α], r Sin[α],-Sqrt[r^2-c]}},{r,0,3},{α, 0, 2 π}, MeshFunctions->{#3&, #5&},Mesh->{Range[-3,3,1/8],Range[0,2 π, 2 π/60]}, PlotRange->{All,All,{-3.01,3.01}},ColorFunction->(White&),MaxRecursion->5,BoxRatios->{1,1,1}],{c, {-1}}],
Table[ParametricPlot3D[{{r Cos[α], r Sin[α],Sqrt[r^2-c]}, {r Cos[α], r Sin[α],-Sqrt[r^2-c]}},{r,0,4},{α, 0, 3 π /2}, MeshFunctions->{#3&, #5&},Mesh->{Range[-3,3,1/8],Range[0,2 π, 2 π/60]}, PlotRange->{All,All,{-3.01,3.01}},ColorFunction->(White&),MaxRecursion->5,BoxRatios->{1,1,1}],{c, {0,1,4}}]
}]

Export["C:\Users\hp\Documents\Distributions\JPG\13_2.jpg",g,ImageResolution->800]

The code below is obsolete.

Print["Cone"]
cone=#2{Cos[#1],Sin[#1],1}&;
cone[α,t]
composfun[f_]:={f[#1,#2],f[αmin,#2],f[αmax,#2], f[#1,tmax]}&[#1,#2]&
composfun[cone][α,t]

αmin=(π/6);
αmax=((9 π)/5);
tmin=0;
tmax=3;
mesh={41,25};
viewpoint={3,-0.7,1.4};
coneupp=moduleParametricPlot3D[composfun[cone],αmin,αmax,tmin,tmax][mesh,viewpoint][{}];

composfun[f_]:={f[#1,#2],f[αmin,#2],f[αmax,#2], f[#1,tmin]}&
αmin=-((5π)/6);
αmax=((4 π)/5);
tmin=-3;
tmax=0;
conedow=moduleParametricPlot3D[composfun[cone],αmin,αmax,tmin,tmax][mesh,viewpoint][{}];

Print["One-sheeted hyperboloids"]
Clear[composfun,αmin,αmax,tmin,tmax,mesh]
hyperb1=#1 {Cos[#2] Cosh[#3],Sin[#2] Cosh[#3],Sinh[#3]}&;
composfun[r_][f_]:={f[r,#1,#2],f[r,αmin,#2],f[r,αmax,#2], f[r,#1,tmin], f[r,#1,tmax]}&
r=(Sqrt[2]);
αmin=(π/6);
αmax=((9 π)/5);
tmin=ArcSinh[-(3/2)(Sqrt[2])];
tmax=-tmin;
mesh={50,50};
hyperb1large=moduleParametricPlot3D[composfun[r][hyperb1],αmin,αmax,tmin,tmax][mesh,viewpoint][{}];
r=1;
tmin=ArcSinh[-3];
tmax=-tmin;
hyperb1small=moduleParametricPlot3D[composfun[r][hyperb1],αmin,αmax,tmin,tmax][mesh,viewpoint][{}];

Print["Two-sheeted hyperboloid"]
Clear[composfun,αmin,αmax,tmin,tmax,mesh]
hyperb2up=#1 {Cos[#2] Sinh[#3],Sin[#2] Sinh[#3],Cosh[#3]}&;
hyperb2down=#1 {Cos[#2] Sinh[#3],Sin[#2] Sinh[#3],-Cosh[#3]}&;
composfunup[r_][f_]     :={f[r,#1,#2],f[r,#1,   tmax]}&
composfundown[r_][f_]:={f[r,#1,#2],f[r,#1,-tmax]}&
r=1;
αmin=-π;
αmax=π+(π/124);
tmin=0;
tmax=ArcCosh[3];
mesh={70,21};
hyperb2upp=moduleParametricPlot3D[composfunup[r][hyperb2up],αmin,αmax,tmin,tmax][mesh,viewpoint][{}];
hyperb2dow=moduleParametricPlot3D[composfundown[r][hyperb2down],αmin,αmax,tmin,tmax][mesh,viewpoint][{}];

Show[hyperb2upp,hyperb2dow,coneupp,conedow,hyperb1small,hyperb1large]

Figure 13.3  Graph of | Γ |

Clear["`*"]
Plot3D[Abs[Gamma[x + I y]], {x, -2.5, 3.2}, {y, -2, 2.2},PlotRange -> {0, 10}, PlotPoints -> {131, 129},ColorFunction->White, Mesh->80,BoxRatios -> Automatic, ViewPoint -> {-1, -2, 1}, Boxed -> False, Axes -> None]

Figure 13.4  Graph of restriction of  (1/Γ)       FOLLOWING INPUT IS CONFUSING

Clear["`*"]
c1={0,0};
c2={1,0};
e=0.2;
r[a_]:={{Cos[a],Sin[a]},{-Sin[a],Cos[a]}}
arrow[x_,y_,h_,b_,a_]:=Polygon[{r[a].{-2 b,-4 h}+{x,y},{x,y},r[a].{2 b,-4 h}+{x,y},r[a].{0,-3 h}+{x,y}}]
ci1[t_]:=0.4{Cos[t],Sin[t]}
ci2[t_]:={1,0}-0.4{Cos[t],Sin[t]}
p1=Graphics[{Text["0",c1+{0,0.08}],Text["1",c2+{0,0.08}],Text["ϵ",c1+{e+0.05,0.08}],Text["1-ϵ",c2+{-e-0.08,0.08}],Text["(1/2)+R",{1.5+0.13,0}],Text["(Subscript[γ, R])",{1.01,0.98}],Text["(Subscript[δ, ϵ])",{1.01,0.27}],{AbsolutePointSize[4],Point[c1],Point[c2],Point[c1+{e,0}],Point[c2-{e,0}],Point[{1.5,0}]},{Circle[{1/2,0},1]},Line[{c1+{e,0},c2-{e,0}}],Line[{ci1[Pi/4],ci2[7Pi/4]}],Line[{ci1[7Pi/4],ci2[Pi/4]}],Circle[c1,e],Circle[c2,e],arrow[-e,-0.04,0.02,0.01,π],arrow[0.3,0.04,0.02,0.01,-π/2],arrow[0.7,-0.04,0.02,0.01,π/2],arrow[0.5-0.02,0.285,0.02,0.01,-π/2],arrow[0.57,-0.285,0.02,0.01,π/2],arrow[1+e,0.04,0.02,0.01,0],arrow[0.5-0.02,1,0.02,0.01,-Pi/2]},AspectRatio->Automatic,PlotRange->All,DisplayFunction->Identity];
p2=ParametricPlot[ci1[t],{t,Pi/4,7Pi/4},AspectRatio->Automatic,PlotRange->All,DisplayFunction->Identity];
p3=ParametricPlot[ci2[t],{t,Pi/4,7Pi/4},AspectRatio->Automatic,PlotRange->All,DisplayFunction->Identity];
Show[p1,p2,p3,BaseStyle->{FontFamily->"Times",FontSize->15.},ImageSize->400,DisplayFunction->$DisplayFunction];

FindMaximum[(1/(Gamma[x])),{x,-3.5}]
Solve[D[(1/(Gamma[x])),x]==0]
Plot[PolyGamma[0,x],{x,-4,-3}]
FindRoot[PolyGamma[0,x]==0,{x,-3.5}]
1/Gamma[-3.635293366436902]
FindRoot[PolyGamma[0,x]==0,{x,1.5}]
1/Gamma[1.4616321449683622]

Plot[1/Gamma[x],{x,-4.1,4.1},PlotStyle->Black,PlotPoints->200,PlotRange->All,Ticks->{{-4,-3,-2,-1,0},{1.129,4.076}},AspectRatio->Automatic,BaseStyle->{FontFamily->"Times",FontSize->14.}]

Figure 13.5   Problem 13.11

Clear["`*"]
r[a_]:={{Cos[a],Sin[a]},{-Sin[a],Cos[a]}}
arrow[x_,y_,h_,b_,a_]:=Polygon[{r[a].{-2 b,-4 h}+{x,y},{x,y},r[a].{2 b,-4 h}+{x,y},r[a].{0,-3 h}+{x,y}}]
poly=Polygon[{{2.8,2.8},{0,0},{2.8,0}}];
g=Graphics[{GrayLevel[0.85],poly}];
l=ParametricPlot[{t,t},{t,0,2.8},PlotStyle->Black,DisplayFunction->Identity];
a=Graphics[{Line[{{1.4,0},{1.4,1.4}}],Line[{{1.4,1.4},{2.8,1.4}}],arrow[1.4,0.8,0.04,0.02,0],arrow[2.2,1.4,0.04,0.02,π/2]}];
Show[g,l,a,Axes->True,Ticks->None,AspectRatio->Automatic,BaseStyle->{FontFamily->"Times",FontSize->16.},AxesLabel->{"t","|| x ||"},LabelStyle->Directive[FontFamily->"Times",FontSize->14.],ImageSize->350]

Problem 13.12   A computation

A={{1,0,0,0},{0,1,0,0},{0,0,1,0},{-2y1,-2y2,-2y3,2y4}}
Inverse[A]
Transpose[Inverse[A]]
Det[A]
Det[Inverse[A]]

Simplify[(((yj Sinh[a]+yn Cosh[a]))^2)-(((yj Cosh[a]+yn Sinh[a]))^2)]

Figure 14.1   Graph of sinc

Clear["`*"]
Plot[((Sin[x])/x),{x,-21 π,21 π},PlotStyle->Black,AspectRatio->(1/4),PlotRange->All,Ticks->{{-20 π,20 π},{-0.217,1}},BaseStyle->{FontFamily->"Times",FontSize->12.`}]
FindMinimum[((Sin[x])/x),{x,1}]

Computation of a tempered distribution

Plot[Cos[(E^x)],{x,0,4.5},PlotPoints->10000,ImageSize->400, Ticks->None]

Problem 14.9   A computation

f[x_]:=(E^(-((x^2)/2)))
g[x_]:=x (E^(-((x^2)/2)))
Ff[ξ_]=FullSimplify[(∫_(-∞)^∞)(E^(-I x ξ)) f[x]\[DifferentialD]x]
Fg[ξ_]=FullSimplify[(∫_(-∞)^∞)(E^(-I x ξ)) g[x]\[DifferentialD]x]

FullSimplify[D[E^(-I x  xi+x^2/2),{x,5}]==I^5 E^(xi^2/2)D[E^((x-I xi)^2/2),{xi,5}]]

Problem 14.10   A computation

FullSimplify[(∫_(-∞)^∞)(∫_(-∞)^∞)(E^(-(1/2) ((x^2)+(y^2))-I (x ξ+y η))) ((x^2)-(y^2))\[DifferentialD]y\[DifferentialD]x]
FullSimplify[(∫_(-∞)^∞)(∫_(-∞)^∞)(E^(-(1/2) ((x^2)+(y^2))-I (x ξ+y η))) ((x^3)-3 x (y^2))\[DifferentialD]y\[DifferentialD]x]

Problem 14.16   Solution   A computation

Integrate[((Sin[a ξ]Cos[x ξ])/(ξ(1+ξ^2))),{ξ,0,Infinity},Assumptions->{a>0,Im[x]==0}]

f[x_]:=Piecewise[{{(1/2) (1-((Cosh[x])/E)),-1≤x≤1},{(1/2) Sinh[1] (E^(-Abs[x])),x>1}}]
Plot[Evaluate[f[x]],{x,0,10},PlotRange->All]
Plot[Evaluate[f'[x]],{x,0.8,1.1},PlotRange->All]
Plot[Evaluate[f''[x]],{x,0.8,1.1},PlotRange->All]

Problem 14.19   A computation

(∫_(-∞)^∞)(1/(((1+(x^2)))^2))\[DifferentialD]x
(∫_(-∞)^∞)((x^2)/(((1+(x^2)))^2))\[DifferentialD]x

Problem 14.20   A computation

(∫_0^∞)((Sin[a x] Sin[b x])/(x^2))\[DifferentialD]x
FullSimplify[Abs[a+b]-Abs[a-b]==2Min[a,b],{a>0,b>0}]

Problem 14.28   A computation

Simplify[Solve[L (((I ξ))^2)+R I ξ +(1/C)==0,ξ],C>0]
Simplify[Solve[L (((η))^2)+R η+(1/C)==0,η],C>0]
(*{(SubPlus[ξ]),(SubMinus[ξ])}={(Simplify[Solve[L (((I ξ))^2)-R I ξ +(1/C)==0,ξ],C>0] /.{Rule->Equal})[[1]],(Simplify[Solve[L (((I ξ))^2)-R I ξ +(1/C)==0,ξ],C>0] /.{Rule->Equal})[[2]]}
(SubPlus[ξ])[[1]]
Apart[(1/(L (((I ξ))^2)+R I ξ +(1/C))),ξ]*)
Q=(Sqrt[((4L)/C)-(R^2)])
FullSimplify[Together[(1/(ξ-(((I R-Q)/(2 L)))))-(1/(ξ-(((I R+Q)/(2 L)))))]]

(SuperMinus[c])[α]=FullSimplify[(FullSimplify[(∫_0^∞)((x^(α-1))/(Gamma[α]))Fg[x]\[DifferentialD]x,0<Re[α]<1]-FullSimplify[(∫_(-∞)^0)((((-x))^(α-1))/(Gamma[α]))Fg[x]\[DifferentialD]x,0<Re[α]<1])/(FullSimplify[(∫_0^∞)((x^(-α))/(Gamma[1-α]))g[x]\[DifferentialD]x,0<Re[α]<1]-FullSimplify[(∫_(-∞)^0)((((-x))^(-α))/(Gamma[1-α]))g[x]\[DifferentialD]x,0<Re[α]<1]),-0<Re[α]<1]

F[((SubPlus[χ])^α)]=(((SuperPlus[c])[α])/2)(((SubPlus[χ])^(1-α))+((SubMinus[χ])^(1-α)))+(((SuperMinus[c])[α])/2)(((SubPlus[χ])^(1-α))-((SubMinus[χ])^(1-α)))//FullSimplify
F[((SubMinus[χ])^α)]=(((SuperPlus[c])[α])/2)(((SubPlus[χ])^(1-α))+((SubMinus[χ])^(1-α)))-(((SuperMinus[c])[α])/2)(((SubPlus[χ])^(1-α))-((SubMinus[χ])^(1-α)))//FullSimplify

Problem 14.38   A computation

u[x_]:=((E^(-p Abs[x-xi]))/(2p))+a (E^(p x))+b (E^(-p x))
u0=u[x]/.x->0
u1=u[x]/.x->1
FullSimplify[u[x]/.Flatten[Solve[{u0==0,u1==0},{a,b}]]]
FullSimplify[%,Assumptions->{0<x<1&&x<xi&&0<xi<1&&Im[x]==0&&Im[xi]==0&&Im[p]==0}]
FullSimplify[%%,Assumptions->{0<x<1&&xi<x&&0<xi<1&&Im[x]==0&&Im[xi]==0&&Im[p]==0}]

Problem 14.39   A computation

(SuperPlus[c])[α]=FullSimplify[(FullSimplify[(∫_0^∞)((x^(α-1))/(Gamma[α]))Ff[x]\[DifferentialD]x,0<Re[α]<1]+FullSimplify[(∫_(-∞)^0)((((-x))^(α-1))/(Gamma[α]))Ff[x]\[DifferentialD]x,0<Re[α]<1])/(FullSimplify[(∫_0^∞)((x^(-α))/(Gamma[1-α]))f[x]\[DifferentialD]x,0<Re[α]<1]+FullSimplify[(∫_(-∞)^0)((((-x))^(-α))/(Gamma[1-α]))f[x]\[DifferentialD]x,0<Re[α]<1]),0<Re[α]<1]

Problem 14.39   Solution   A computation

Integrate[x^(a-1)E^(-x^2/2),{x,0,Infinity}]

Integrate[(-x)^(a-1)E^(-x^2/2),{x,-Infinity,0}]

FullSimplify[Sqrt[Pi]2^a Gamma[a/2]Gamma[1-a]/(Gamma[a]Gamma[(1-a)/2]),Assumptions->{a>0}]

FullSimplify[ Gamma[a/2]/Gamma[a]==Sqrt[Pi]2^(1-a)/Gamma[(a+1)/2],Assumptions->{a>0}]

FullSimplify[ Gamma[1-a]/Gamma[(1-a)/2]==Gamma[1-a/2]/(Sqrt[Pi]2^a),Assumptions->{a>0}]

FullSimplify[Sqrt[Pi]2^(1-a) Gamma[1-a/2]/Gamma[(1+a)/2],Assumptions->{a>0}]

FullSimplify[Gamma[1-a] Gamma[a/2]Gamma[1-a/2]/(Gamma[a]Gamma[1-a])==Out[3],Assumptions->{a>0}]

Problem 14.41   Solution   A computation

(∫_(-1)^1)(((1-(s^2)))^((a/2)-1))(((I α s+β))^(-a))\[DifferentialD]s

Below the original output in Version 5.1 is reproduced

Problem 14.47   Solution   A computation

Integrate[((Sin[a x]Sin[b x])/x),{x,0,Infinity},Assumptions->{a>0,b>0}]

Figures 16.1 and 16.6   Problem 16.5

Clear["`*"]
ComplexExpand[FullSimplify[(∑_(n=-N)^N)(E^(I n x))]]
Plot[(∑_(n=-35)^35)(E^(I n x)),{x,-1-2π,2π+1},PlotPoints->400,PlotRange->All,Frame->False];
Limit[((Sin[(35+1/2)x])/(Sin[x/2])),x->0]
p=Plot[((Sin[(35+1/2)x])/(Sin[x/2])),{x,-1-2π,2π+1},PlotStyle->Black,PlotPoints->10001,PlotRange->All,Frame->False,Ticks->{{-π,π},{71}},BaseStyle->{FontFamily->"Times",FontSize->18.}]

plot1=Plot[(1/2) Cot[(x/2)],{x,-3 π-1,3 π+1},AspectRatio->Automatic,PlotStyle->Black,PlotRange->{-11,11},PlotPoints->20,MaxRecursion->10,Ticks->{Table[l π,{l,-2,2}],{-10,0,10}},BaseStyle->{FontFamily->"Times",FontSize->14},ImageSize->300];
P=(1/2) Log[1-Cos[#1]]&;
FullSimplify[(∂_x)P[x]==(1/2) Cot[(x/2)]]
plot2=Plot[P[x],{x,-3 π-1,3 π+1},AspectRatio->Automatic,PlotStyle->Black,PlotRange->{-7,1},PlotPoints->55,MaxRecursion->10,Ticks->{Table[l π,{l,-2,2}],{-5,0}},BaseStyle->{FontFamily->"Times",FontSize->18},ImageSize->300];
GraphicsRow[{plot1,plot2}]

Figure 16.2    Truncated graphs of (Subscript[D , 20]) and ...

p1=Plot[Sin[(25+1/2)x]/Sin[x/2],{x,-0.5-2π,2π+0.5},PlotStyle->Black,PlotPoints->1000,PlotRange->{-3.1,3.1},Frame->False,Ticks->False,AspectRatio->Automatic];
p2=Plot[{1/Sin[x/2],-1/Sin[x/2]},{x,-0.5-2π,2π+0.5},PlotStyle->Black,PlotPoints->100,PlotRange->{-3.1,3.1},Ticks->{{-2π,-π,π,2π},{-2,2}},BaseStyle->{FontFamily->"Times",FontSize->16.},Frame->False,Ticks->False,AspectRatio->Automatic];
Show[GraphicsArray[{{p1},{p2//.Line[{a___,b:{_,y1_},c:{_,y2_},d___}]:>Sequence[Line[{a,b}],Line[{c,d}]]/;Abs[y1-y2]>4}}],AspectRatio->Automatic,ImageSize->500]

Figure 16.3   Graph of (Subscript[F, 8])

F[n_,x_]:=(((Sin[((n x)/2)])^2)/(n ((Sin[(( x)/2)])^2)))
Plot[F[8,x],{x,-2π-0.4,2π+0.4},PlotStyle->Black,Ticks->{{-2π,-π,π,2π},{8}},BaseStyle->{FontFamily->"Times",FontSize->16.},AspectRatio->Automatic,PlotRange->All,ImageSize->500]

Figure 16.4   Example 16.24

BernoulliB[1,0]

BernoulliB[2,0]
BernoulliB[2,1/2]

FindMaximum[BernoulliB[3,x],{x,0.25}]
Solve[D[BernoulliB[3,x],x]==0,x]
FullSimplify[BernoulliB[3,(1/6) (3-(Sqrt[3]))]]
N[FullSimplify[BernoulliB[3,(1/6) (3-(Sqrt[3]))]]]

FindMaximum[BernoulliB[4,x],{x,0.5}]
FindMinimum[BernoulliB[4,x],{x,1}]
FullSimplify[BernoulliB[4,1/2]]
FullSimplify[BernoulliB[4,0]]

FindMaximum[BernoulliB[5,x],{x,0.25}]
Solve[D[BernoulliB[5,x],x]==0,x]
N[FullSimplify[BernoulliB[5,(1/30) (15-(Sqrt[15 (15-2 (Sqrt[30]))]))]]]

FindMaximum[BernoulliB[6,x],{x,0.5}]
FindMaximum[BernoulliB[6,x],{x,1}]
FullSimplify[BernoulliB[6,1/2]]
FullSimplify[BernoulliB[6,0]]

bpl[n_,ticks__]:=Plot[BernoulliB[n,x-⌊x⌋],{x,-1,1},PlotPoints->600,ImageSize->150,Ticks->{{-1,0,1},ticks},PlotStyle->Black,BaseStyle->{FontFamily->"Times",FontSize->11.}];
bpl[1]=Show[bpl[1,{-1/2,1/2}]//.Line[{a___,b:{_,y1_},c:{_,y2_},d___}]:>Sequence[Line[{a,b}],Line[{c,d}]]/;Abs[y1-y2]>0.5,Ticks->{{-1,0,1},{-1/2,1/2}},BaseStyle->{FontFamily->"Times",FontSize->11.}];
bpl[2]=Show[bpl[2,{-1/12,1/6}]];
bpl[3]=Show[bpl[3,{-0.0481,0.0481 }]];
bpl[4]=Show[bpl[4,{-1/30,7/240}]];
bpl[5]=Show[bpl[5,{-0.0244,0.0244}]];
bpl[6]=Show[bpl[6,{-31/1344,1/42}]];
t1=Table[bpl[n],{n,1,3}];
t2=Table[bpl[n],{n,4,6}];
Show[GraphicsArray[{t1,t2}],ImageSize->500]

Computation in Example 16.23

Series[Pi Cot[Pi z]-1/z,{z,0,5}]
Series[(Pi z Cot[Pi z]-1)/z^2,{z,0,5}]
Limit[(1-Pi z Cot[Pi z])/(2z^2),z->0]
Simplify[Cot[z]^2==-(1+Cot'[z])]

Computation in Example 16.24

Series[Cot[t/2],{t,0,3}]
Plot[Cot[t/2]-2/t,{t,-Pi,Pi}]

Plot[Evaluate[Integrate[Sin[t]/t,{t,0,x}]],{x,0,1000},PlotRange->All];

Figure 16.5    Graph of sawtooth

Clear["`*"]
p=Plot[Evaluate[(1/2)-(∑_(n=1)^100)((Sin[2n π x])/(n π))],{x,-2-0.2,2+0.2},PlotStyle->Black,Ticks->{{-2,-1,0,1,2},{0,2}},MaxRecursion->8,PlotPoints->500,PlotRange->Full];
ov=0.89490
l1=Graphics[{White,Thickness[0.01],Line[{{-2,ov},{-2,1-ov}}]}]
c1=Table[Graphics[{GrayLevel[1],Disk[{k,1/2},0.09]}],{k,1,2}];
c2=Table[Graphics[{GrayLevel[1],Disk[{k,1/2},0.09]}],{k,-2,-1}];
Show[(p//.Line[{a___,b:{_,y1_},c:{_,y2_},d___}]:>Sequence[Line[{a,b}],Line[{c,d}]]/;Abs[y1-y2]>0.8),(*c1,c2,*)AspectRatio->Automatic,BaseStyle->{FontFamily->"Times",FontSize->14.},ImageSize->450]

ov=0.0895;
l=Table[Graphics[{White,Thickness[0.01],Line[{{k,-1/2+ov},{k,1/2-ov}}]}],{k,-2,2,1}];
p=Plot[Evaluate[-(∑_(n=1)^100)((Sin[2n π x])/(n π))],{x,-2.3,2.3},PlotStyle->Black,Ticks->{{-2,-1,0,1,2},{-1/2,1/2}},MaxRecursion->8,PlotPoints->4000,PlotRange->Full,AspectRatio->Automatic,BaseStyle->{FontFamily->"Times",FontSize->11.},ImageSize->300];
Show[p,l]

l=Table[Graphics[{White,Thickness[0.01],Line[{{k,-1/2+ov},{k,1/2-ov}}]}],{k,-2,2,1}];
p=Plot[Evaluate[-(∑_(n=1)^100)((Sin[2n π x])/(n π))],{x,-2.3,2.3},PlotStyle->Black,Ticks->{{-2,-1,0,1,2},{-1/2-ov,1/2+ov}},MaxRecursion->8,PlotPoints->4000,PlotRange->Full,AspectRatio->Automatic,BaseStyle->{FontFamily->"Times",FontSize->6.},ImageSize->300];
Show[p,l]

Problem 16.2   A computation

FullSimplify[ComplexExpand[1/(1-r E^(I x))]==((1-r Cos[x]+I r Sin[x])/(1+r^2-2r Cos[x]))]

Simplify[Re[ComplexExpand[(1/((1- r E^(I x))^2)),{x∈ℝ}]],{x∈ℝ}]

Limit[((1-2 r Cos[x]+(r^2) Cos[2 x])/(((1+(r^2)-2 r Cos[x]))^2)),r->1]

FullSimplify[-(1/4) Cos[x] ((Csc[(x/2)])^2)==(1/2)-(1/(4 ((Sin[(x/2)])^2)))]

Simplify[Im[ComplexExpand[(1/((1- r E^(I x))^2)),{x∈ℝ}]],{x∈ℝ}]

Limit[-2((r (-1+r Cos[x]) Sin[x])/(((1+(r^2)-2 r Cos[x]))^2)),r->1]

Problem 16.3   A computation

We want to compute the Fourier series of the periodic function arccos ∘ cos. 
Mathematica does not recognize a pattern.

FullSimplify[Composition[ArcCos,Cos][x],-π<x<π]
modulePlot[Composition[ArcCos,Cos],{-4 π-1,4 π+1},{Table[k π,{k,-4,4,1}],Table[k π,{k,-1,1,1}]},{-0.1,π+0.1},800,{}]

Instead we compute the Fourier series of the absolute value function made 2π - periodic.

FourierCoefficient[Abs[x],x,n]
FourierSeries[Abs[x],x,10]
(π/2)-(2/π) (∑_(k=-4)^5)((E^(I (2 k-1) x))/(((2 k-1))^2))
FullSimplify[%==%%]
FullSimplify[FourierSeries[Abs[x],x,10]]

modulePlot[Evaluate[FourierSeries[Abs[#],#,30]]&,{-4π-1,4π+1},{Table[k π,{k,-4,4,1}],Table[k π,{k,-4,4,1}]},{-0.1,π+0.1},800,{}]

(∑_(k=1)^∞)(1/(((2 k-1))^2))
(((∫_(-π)^π)((Composition[ArcCos,Cos][x])^2)\[DifferentialD]x)/(2 π))
((2(∫_0^π)(x^2)\[DifferentialD]x)/(2 π))
((((π/2)))^2)+(∑_(n=-∞)^∞)((((2/(π (((2 n-1))^2)))))^2)
FullSimplify[(((∫_(-π)^π)((Composition[ArcCos,Cos][x])^2)\[DifferentialD]x)/(2 π))==((((π/2)))^2)+(∑_(n=-∞)^∞)((((2/(π (((2 n-1))^2)))))^2)]
(∑_(n=1)^∞)(1/(((2 n-1))^4))
(∑_(n=1)^∞)(1/( (n^4)))

FullSimplify[(π^2) (E^(-I π ξ)) ((Sinc[(π/2) ξ])^2)==(((I(( (E^(-I π ξ))-1)/ξ)))^2),ξ∈ℝ]

f=Composition[ArcSin,Sin][#1]&;
Plot[f[x],{x,-3 π-1,3 π+1},PlotStyle->Black,Ticks->{Table[k π,{k,-5,5}],{-(π/2),(π/2)}},AspectRatio->Automatic,PlotRange->{-0.2-(π/2),(π/2)+0.2},ImageSize->500]
FullSimplify[f[x]==Composition[ArcCos,Cos][x+(π/2)]-(π/2)]
FullSimplify[Cos[(2 k-1) (x+(π/2))]==-(((-1))^k) Sin[(2 k-1) x],k∈ℤ]
(∫_(-π)^π)Evaluate[f[x]] Sin[(2 k-1) x]\[DifferentialD]x
Plot[(4/π)  (∑_(n=1)^30)(((((-1))^(n-1)) Sin[(2 n-1) x])/(((2 n-1))^2)),{x,-3 π-1,3 π+1},PlotStyle->Black,BaseStyle->{FontFamily->"Times",FontSize->12.},Ticks->{Table[k π,{k,-5,5}],{-(π/2),(π/2)}},AspectRatio->Automatic,PlotRange->{-0.2-(π/2),(π/2)+0.2},ImageSize->400]
Plot[(4/π)  (∑_(n=1)^100)(((((-1))^(n-1)) Cos[(2 n-1) x])/(2 n-1)),{x,-3 π-1,3 π+1},PlotStyle->Black,BaseStyle->{FontFamily->"Times",FontSize->12.},Ticks->{Table[k π,{k,-5,5}],{-1,1}},AspectRatio->Automatic,PlotRange->{-0.2-1,1+0.2},ImageSize->400]
(∑_(n=1)^∞)(((((-1))^(n-1)) )/(2 n-1))

modulePlot[(4/π)  (∑_(k=1)^50)(((((-1))^(k-1)) Sin[(2 k-1) #])/(((2 k-1))^2))&,{-4 π-1,4 π+1},{Table[k π,{k,-4,4,1}],Table[k π/2,{k,-1,1,1}]},{-0.1-Pi/2,π/2+0.1},300,{}]

modulePlot[(4/π)  (∑_(k=1)^100)(((((-1))^(k-1)) Cos[(2 k-1) #])/(2 k-1))&,{-4 π-1,4 π+1},{Table[k π,{k,-4,4,1}],Table[k,{k,-1,1,1}]},{-0.1-Pi/2,π/2+0.1},300,{}]

(∂_x)((4 (∑_(k=1)^4)(((((-1))^(k-1)) Cos[(2 k-1) x])/(2 k-1)))/π)
f=-1+2 HeavisideTheta[-#1+(π/2)] HeavisideTheta[#1+(π/2)]&;
modulePlot[Evaluate[-1+2 HeavisideTheta[-#1+(π/2)] HeavisideTheta[#1+(π/2)]&],{-4 π-1,4 π+1},{Table[k π,{k,-4,4,1}],Table[k,{k,-1,1,1}]},{-0.1-(π/2),(π/2)+0.1},300,{}]
(∂_x)f[x]
g[x_]:=(E^(-((x^2)/2)))
FullSimplify[a (∑_(k=-∞)^∞)g[h+k a]==(∑_(n=-∞)^∞)(E^(-((I n h 2 π)/a))) (∫_(-∞)^∞)Evaluate[g[x] (E^((I n h 2 π x)/a))]\[DifferentialD]x,a∈ℝ]
Simplify[N[Table[Abs[(a/h)] EllipticTheta[3,((I a)/2),(E^(-((a^2)/2)))]-(Sqrt[2 E π]) EllipticTheta[3,((h π)/a),(E^(-((2 (π^2))/(a^2))))] Sign[a],{a,1,4,1},{h,1,2,0.5}]]]

Problem 16.6   A computation

FullSimplify[(∂_x)((Cos[(x/2)])/(Sinc[(x/2)]))]
FullSimplify[%==-((x-Sin[x])/(4 ((Sin[(x/2)])^2)))]

Limit[((Sin[x]-x)/(4 ((Sin[(x/2)])^2))),x->0]

Table[n Cos[n π],{n,1,10}]
Table[(((-1))^n) n,{n,1,10}]
s[n_]:=(∑_(k=1)^n)(((-1))^k) k
Table[s[n],{n,1,10}]
N[Table[Evaluate[(((∑_(k=1)^(n+1))s[k])/n)],{n,1,200}]];
ListPlot[%,Ticks->{Table[k 20,{k,1,10}],{-0.5}}]

s[n_]:=(∑_(k=1)^n)a[k]
c[n_]:=(((∑_(k=1)^n)s[k])/n)
(∑_(n=1)^∞)a[n] (r^n)
q=FullSimplify[(∑_(n=1)^∞)(r^n) a[n]==(((1-r))^2) (∑_(n=1)^∞)n c[n] (r^n)]

a[n_]:=(1/(n^2))
r=(1/2);
c[n]
(∑_(n=1)^∞)(r^n) a[n]
(((1-r))^2) (∑_(n=1)^∞)n c[n] (r^n)
q
Clear[a]

a[n_]:=(((-1)^n)/n) 
(∑_(n=1)^∞)(r^n) a[n]
q
Clear[a]

a[n_]:=(-1)^n n^2
(∑_(n=1)^∞)(r^n) a[n]
q
Clear[a]

a[n_]:=n^3 
(∑_(n=1)^∞)(r^n) a[n]
q
Clear[a]

Problem 16.8   A computation

ParametricPlot3D[{r Cos[α],r Sin[α],((1-(r^2))/(1-2r Cos[α]+(r^2)))},{r,0,1},{α,-π,π},PlotRange->{0,5},PlotPoints->{21,100},Shading->False,Boxed->False,Axes->False,Ticks->None,ViewPoint->{-2,-2,1},ImageSize->300]
Plot[Evaluate[((1-(r^2))/(1-2r Cos[α]+(r^2)))/.r->0.9],{α,-0.1,2π+0.8},PlotRange->All,PlotPoints->70,Ticks->{{Pi,Pi}}]

Problem 16.16   See Solutions

Mathematica is unable to simplify the integral at the end.

Clear["`*"]
s[x_]:=(1/2)(π((Cosh[x+π])/(Sinh[π]))-1)/;-2π<x<0
s[x_]:=(1/2)(π((Cosh[x-π])/(Sinh[π]))-1)/;0≤x≤2π
s[x_]:=(1/2)(π((Cosh[x-3π])/(Sinh[π]))-1)/;2π<x≤4π
s[x_]:=(1/2)(π((Cosh[x-5π])/(Sinh[π]))-1)/;4π<x≤6π
Plot[s[x],{x,-1/2,4π+1/2},Ticks->{{0,2π,4π},{0,1}},PlotPoints->2000,BaseStyle->{FontFamily->"Times",FontSize->8.},AspectRatio->1/(4Pi)]
Plot[s[x],{x,-1/2,4π+1/2},Ticks->{{0,2π,4π},{0,1}},PlotPoints->2000,BaseStyle->{FontFamily->"Times",FontSize->8.},AspectRatio->Automatic,ImageSize->500]
Plot[s[x],{x,-1/2,4π+1/2},Ticks->{{0,2π,4π},{0,1}},PlotPoints->2000,BaseStyle->{FontFamily->"Times",FontSize->10.},AspectRatio->Automatic]

N[s[0],16]
t[x_]:=(1/2)(π((Cosh[x-π])/(Sinh[π]))-1)
Simplify[t[x]-t''[x]]
Limit[t'[x],x->2π,Direction->-1]-Limit[t'[x],x->0,Direction->1]
2N[180ArcTan[Pi/2]/Pi]
(* Integrate[(E^(i x ξ))((E^(-2πi k ξ))/(1+(ξ^2))),{ξ,-∞,∞},Assumptions->{Im[x]==0,Element[k,Integers],k≥0}] *)
2π I Residue[((E^(I(x-2π k)ξ))/(1+(ξ^2))),{ξ,I}]
-2π I Residue[((E^(I(x-2π k)ξ))/(1+(ξ^2))),{ξ,-I}]

Plot[s[x],{x,-1/2,4π+1/2},Ticks->{{0,2π,4π},{0,1}},PlotPoints->2000,BaseStyle->{FontFamily->"Times",FontSize->10.},AspectRatio->Automatic,ImageSize->500]

Problem 16.19. (i)   A computation

Sin[Pi/4]
FullSimplify[2+4Sum[(-1)^(n-1)/(16n^2-1),{n,1,Infinity}]==Pi/Sqrt[2]]

Problem 16.20. (v)   A computation

FullSimplify[1/(2z^2)+Sum[Cos[n x]/(z^2-n^2),{n,1,Infinity}],{Im[z]==0,Im[x]==0}]

Problem 16.22   A computation

f=Composition[ArcTan,Tan][((#1)/2)]&
f[x]
Plot[f[x],{x,-3 π-1,3 π+1},PlotStyle->Black,Ticks->{Table[k π,{k,-5,5}],{-(π/2),(π/2)}},AspectRatio->Automatic,PlotRange->{-(π/2)-0.2,(π/2)+0.2},ImageSize->500]
FullSimplify[(∂_x)f[x]]

Plot[(x/2)-π Floor[((x+π)/(2 π))],{x,-3 π,3 π},Ticks->{Table[k π,{k,-5,5}],{-(π/2),(π/2)}},AspectRatio->Automatic,ImageSize->500]

f=(∑_(n=1)^∞)(((((-1))^(n+1)) Sin[n #1])/n)&
FullSimplify[Evaluate[f[x]],x∈ℝ]
Plot[%,{x,-3 π,3 π},Ticks->{Table[k π,{k,-5,5}],{-(π/2),(π/2)}},AspectRatio->Automatic,ImageSize->500]
(∂_x)f[x]
FullSimplify[(∂_x)f[x]]
Plot[((Log[(E^(I x))])/(2 I)),{x,-3 π,3 π},Ticks->{Table[k π,{k,-5,5}],{-(π/2),(π/2)}},AspectRatio->Automatic,ImageSize->500]
FullSimplify[f[Pi/4]]

(∑_(n=1)^∞)(1/(4 (n^2)-1))
Simplify[(∑_(n=1)^∞)((((-1))^(n-1))/(4 (n^2)-1))]
(∑_(n=1)^∞)(1/( (((4n-2))^2)-1))
Plot[Evaluate[(2/π)+(4/π) (∑_(n=1)^∞)(((((-1))^(n-1)) Cos[2 n x])/(4 (n^2)-1))],{x,-3 π,3 π},Ticks->{Table[k π,{k,-3,3}],{1}},AspectRatio->Automatic,ImageSize->600]
Plot[Evaluate[(4/π)-1+(8/π) (∑_(n=1)^∞)(((((-1))^(n-1))((Cos[n x])^2))/(4 (n^2)-1))],{x,-3 π,3 π},Ticks->{Table[k π,{k,-3,3}],{1}},AspectRatio->Automatic,ImageSize->600]
Plot[Evaluate[(2/π)-(4/π) (∑_(n=1)^∞)((Cos[2 n x])/(4 (n^2)-1))],{x,-3 π,3 π},Ticks->{Table[k π,{k,-3,3}],{1}},AspectRatio->Automatic,ImageSize->600]
Plot[Evaluate[(8/π) (∑_(n=1)^∞)(((Sin[n x])^2)/(4 (n^2)-1))],{x,-2π-0.4,2 π+0.4},PlotStyle->Black,BaseStyle->{FontFamily->"Times",FontSize->12.},Ticks->{Table[k π,{k,-3,3}],{1}},AspectRatio->Automatic,ImageSize->600]
Plot[Evaluate[(8/π) (∑_(n=1)^∞)((n Sin[2n x])/(4 (n^2)-1))],{x,-2 π-0.4,2 π+0.4},PlotStyle->Black,BaseStyle->{FontFamily->"Times",FontSize->12.},Ticks->{Table[k π,{k,-3,3}],{-1,1}},AspectRatio->Automatic,ImageSize->600]
Simplify[(1/(4 (n^2)-1))==(1/2) ((1/(2 n-1))-(1/(2 n+1)))]
FourierCoefficient[Abs[Sin[x]],x,n]

Mathematica is not very illuminating.

FullSimplify[(8/π) (∑_(n=1)^∞)(((Sin[n x])^2)/(4 (n^2)-1))]
Plot[%,{x,-π,π}]
FullSimplify[(∂_x)((2 I (ArcTanh[(E^(-I x))]-ArcTanh[(E^(I x))]))/π)]
Plot[((2 I (ArcTanh[(E^(-I x))]-ArcTanh[(E^(I x))]))/π),{x,-π,π}]
FullSimplify[(2/π)+(4/π) (∑_(n=1)^∞)(((((-1))^(n-1)) Cos[n x])/(4 (n^2)-1))]

Problem 16.23  A computation

f[α_,λ_][x]:=(((λ^α) (E^(-λ x)) (x^(α-1)))/(Gamma[α]))UnitStep[x]
(∫_0^∞)f[5,7][x]\[DifferentialD]x
(∫_0^∞)x f[5,7][x]\[DifferentialD]x
(∫_0^∞)(x^2) f[5,7][x]\[DifferentialD]x
PDF[GammaDistribution[α,(1/λ)],x]
Mean[GammaDistribution[α,(1/λ)]]
Variance[GammaDistribution[α,(1/λ)]]
Integrate[(x^5) PDF[GammaDistribution[α,(1/λ)],x],{x,0,Infinity},Assumptions->α>0&&λ>0]

17.1 Appendix   A computation

Simplify[(∫_0^∞)((r^(n/2))/(((1+(r^2)))^l))BesselJ[(n/2)-1,x r]\[DifferentialD]r]
(*,Assumptions->{Element[n,Integers],n>0,Element[l,Integers],n>0 *)

Series[BesselJ[l,x],{x,0,2}]
Series[BesselJ[n+(1/2),x],{x,0,1}]

BesselK[0,0]

coef[x_]:=(x^(1-k))
a[r_]:=((r^(n/2))/(((1+(r^2)))^k))
bes[r_]:=BesselJ[(n/2)-1, r]
Jnt[x_]:=FullSimplify[Integrate[ a[r] bes[x r],{r,0,∞} ],Assumptions->{n>0,k>(n/2),x>0}]
Int[x_]:=FullSimplify[coef[x]Integrate[ a[r] bes[x r],{r,0,∞} ],Assumptions->{n>0,k>(n/2),x>0}]
Int[x]

Jnt[x]

FullSimplify[D[Int[x],x]==(1-k)(x^(-k)) Jnt[x]+(x^(1-k)) D[Jnt[x],x],Assumptions->{n>0,k>(n/2),x>0}]

D[ bes[y],y]/.{y->x r}

FullSimplify[(x^2)D[Int[x],{x,2}]+x D[Int[x],x]==2(1-k)(x^(1-k)) Integrate[ a[r](x r)D[ bes[y],y]/.{y->x r},{r,0,∞} ]+((((k-1))^2)+((((n/2)-1))^2))Int[x]-(x^(1-k)) Integrate[ a[r](((x r))^2)bes[x r],{r,0,∞} ],Assumptions->{n>0,k>(n/2),x>0}]

FullSimplify[D[r int[r],r]-(((n/2)+1-2k) (r^(n/2)) ((((r^2)+1))^(-k))+2k((r^(n/2))/(((1+(r^2)))^(k+1))))]
Expand[Simplify[2(1-k)((n/2)+1-2k)]]

FullSimplify[-2(1-k)((n/2)+1-2k)coef[x]Integrate[ a[r] bes[x r],{r,0,∞} ]-2(1-k)2k coef[x]Integrate[ ((a[r])/(1+(r^2))) bes[x r],{r,0,∞} ]-coef[x]Integrate[ a[r](((x r))^2) bes[x r],{r,0,∞} ]==((x^2)-2(k^2)+(n+2)k-((n^2)/2)+n-2)coef[x]Integrate[ a[r] bes[x r],{r,0,∞} ],Assumptions->{Element[k,Integers],Element[n,Integers],k>(n/2)}]

FullSimplify[(x^2)D[BesselK[n,x],{x,2}]+x D[BesselK[n,x],x]-((n^2)+(x^2))BesselK[n,x]==0]
BesselK[(1/2),x]
BesselK[n,1]

Together[(∂_ξ)((ξ/(((1+(ξ^2)))^κ)))]
i1=Expand[Numerator[%](1+(ξ^2)),ξ]
Simplify[i1==(1-2κ)(ξ^4)+(2-2κ) (ξ^2)+1]
Together[(∂_ξ) (ξ(∂_ξ)((ξ/(((1+(ξ^2)))^κ))))]
Simplify[%==((1+2(1-4κ)(ξ^2)+(((2κ-1))^2)(ξ^4))/(((1+(ξ^2)))^(κ+2)))]
Together[(∂_(ξ,ξ))((1/(((1+(ξ^2)))^κ)))]
i2=Expand[(m^2)(((1+(ξ^2)))^2)-2m(1+(1-2κ)(ξ^2))(1+(ξ^2))+1+2(1-4κ)(ξ^2)+(((2κ-1))^2)(ξ^4),ξ]
eq=Simplify[Expand[i2+a i1+b+c 2 κ((2κ+1)(ξ^2)-1),ξ]]
eq1=Coefficient[eq,(ξ^4)]
eq2=Coefficient[eq,(ξ^2)]
eq3=(m^2)+a+b-2 m-2 c κ+1
Solve[{eq1==0,eq2==0,eq3==0},{a,b,c}]

a[ξ_]:=(1/(((1+(ξ^2)))^κ))
eq=Expand[Numerator[Simplify[(∂_(ξ,ξ))(((ξ^2)-r) a[ξ])-p (∂_ξ)(ξ a[ξ])+q a[ξ] ]],ξ]
eq1=Coefficient[eq,(ξ^4)]
eq2=Coefficient[eq,(ξ^2)]
eq3=FullSimplify[eq-eq1 (ξ^4)-eq2 (ξ^2)]
Solve[{eq1==0,eq2==0,eq3==0},{p,q,r}]

a[ξ_]:=(1/(((1+(ξ^2)))^κ))
FullSimplify[(∫_0^∞)Cos[x ξ] a[ξ]\[DifferentialD]ξ,Assumptions->{Element[κ,Integers],κ>0}]

κ=2
FullSimplify[(∫_0^∞)Cos[x ξ] a[ξ]\[DifferentialD]ξ]

κ=2
FullSimplify[(∫_0^∞)Cos[x ξ] a[ξ]\[DifferentialD]ξ]

κ=3
FullSimplify[(∫_0^∞)Cos[x ξ] a[ξ]\[DifferentialD]ξ]

κ=(1/2)
FullSimplify[(∫_0^∞)Cos[x ξ] a[ξ]\[DifferentialD]ξ]

k[ξ_]:=(∫_x^∞)(E^(-t))((((t^2)-(ξ^2)))^(k-(1/2)))\[DifferentialD]t
FullSimplify[(ξ(∂_(ξ,ξ))-(2k-1) (∂_ξ)-ξ)k[ξ]==0]

l[ξ_]:=(∫_(-ξ)^ξ)(E^(-t))((((ξ^2)-(t^2)))^(k-(1/2)))\[DifferentialD]t
FullSimplify[(ξ(∂_(ξ,ξ))-(2k-1) (∂_ξ)-ξ)l[ξ]==0]

Solve[{(2k-1)p+q+4k^2-6k+2==0,(k-1)p+q-k(2k+1)r+2-5k==0,-p+q+2k r+2==0},{p,q,r}]
Simplify[(k-1)(r+3-2k)+(1-2k)r-2k+1-k(2k+1)r+2-5k==0]
Det[{{2k-1,1,0},{k-1,1,-k(2k+1)},{-1,1,2k}}]

FullSimplify[x^(-m)(x^2D[x^m u[x],{x,2}]+x D[x^m u[x],x]-m^2 x^m u[x])]
D[x^m u[x],{x,2}]

Limit[(x/2)^(3.5) BesselK[3.5,x]/Gamma[3.5],x->0]

Integrate[1/(1+x^2)^k,{x,0,Infinity}]

FullSimplify[Integrate[E^(-xi)(xi^2-x^2)^(k-1),{xi,x,Infinity}]==(k-1)! E^(-x)(∑_(j=0)^(k-1))(((2k-1-j)!(2x)^j)/(j!(k-1-j)!)),Assumptions->{Element[k,Integers],k>0}]

FullSimplify[((Sqrt[Pi] Gamma[k-1/2])/(Gamma[k]Gamma[2k-1]))]

Integrate[2(1-Cos[x xi])/Abs[x]^(1+s),{x,-∞,∞},Assumptions->{0<s<1}]

BesselK[3/2,x]

(∫_0^∞)((Cos[u])/(((1+(u^2)))^3))\[DifferentialD]u

((π (E^(-1)))/((2^5)2!))(∑_(m=0)^2)(((2^m)(6-m-2)!)/(m!(3-m-1)!))

(∫_0^∞)(((x^(ν+1))BesselJ[ν,a x])/((((x^2)+(k^2)))^(μ+1)))\[DifferentialD]x

(∫_0^∞)((1-Cos[t])/(t^(1+2s)))\[DifferentialD]t
Plot[((1-Cos[t])/(t^2)),{t,0,5}]
Plot[- Cos[π s]Gamma[-2 s],{s,0.01,0.99}]
N[- Cos[π s]Gamma[-2 s]/.s->1/2+1/10000]
Limit[- Cos[π s]Gamma[-2 s],{s->1/2}]

(∫_(-π)^π)((Sin[s Sin[θ]])^2)((Cos[θ])^(n-2))\[DifferentialD]θ

FullSimplify[Solve[m x(1+(1-m)^2x^2)==(1+m^2x^2)(m-1)x,m],x>0]
FullSimplify[Solve[m^2-m-1/x^2==0,m],x>0]

Problem 17.8

Integrate[((Abs[x])^m) BesselK[m,Abs[x]],{x,-∞,∞},Assumptions->{m>0}]
(1/(Gamma[k])) Integrate[ (E^(-t(1+(ξ^2)))) (t^(k-1)),{t,0,∞},Assumptions->{ξ>0&&k>0}]
Integrate[(E^(-I x ξ-t (x^2))),{x,-∞,∞},Assumptions->{ξ∈ℝ&&t>0}]
Integrate[(E^(-t-((p^2)/(4 t)))) (t^(m-1)),{t,0,∞},Assumptions->{p>0&&m∈ℝ}]
Integrate[(E^(-t-((p^2)/(4 t)))) (t^(-(1/2))),{t,0,∞},Assumptions->{p>0}]
%%/.m->(1/2)
FullSimplify[BesselK[m,p]==BesselK[-m,p],Assumptions->{m∈ℝ&&p>0}]

FullSimplify[((Integrate[((E^(-I x ξ))/(((1+(ξ^2)))^k)),{ξ,-∞,∞},Assumptions->{x∈ℝ&&k>(1/2)}])/(2 π))]
f[x_]:=(1/(2(Sqrt[π])Gamma[k])) Integrate[(E^(-((x^2)/(4 t)))) (E^(-t)) (1/(Sqrt[t])) (t^(k-1)),{t,0,∞},Assumptions->{x>0&&k>(1/2)}]
f[x]
g[x_]:=(1/((2^(k-(1/2)))(Sqrt[π])Gamma[k]))((Abs[x])^(k-1/2))BesselK[k-(1/2),Abs[x]]
g[x]
FullSimplify[f[x]==g[x],Assumptions->k>(1/2)&&x>0]

FindMaximum[(((r^(4-(1/2))) BesselK[4-(1/2),r])/((2^((1/2)+4)) (Sqrt[π]) Gamma[4])),{r,0}]

Limit[(((r^(4-(1/2))) BesselK[4-(1/2),r])/((2^((1/2)+4)) (Sqrt[π]) Gamma[4])),r->0]
N[%,5]

p1=Plot[Join[Table[Evaluate[(((r^(k-(1/2))) BesselK[k-(1/2),r])/((2^((1/2)+k)) (Sqrt[π]) Gamma[k]))],{k,0.2,1.4,0.2}],Table[Evaluate[(((((-r))^(k-(1/2))) BesselK[k-(1/2),-r])/((2^((1/2)+k)) (Sqrt[π]) Gamma[k]))],{k,0.2,1.4,0.2}]],{r,-1.1,1.1},PlotRange->{-0.2,2.3},PlotStyle->Black,BaseStyle->{FontFamily->"Times",FontSize->12.},Ticks->{Table[k,{k,-1,1}],{0,2}},AspectRatio->Automatic];
p2=Plot[Evaluate[(((r^(4-(1/2))) BesselK[4-(1/2),r])/((2^((1/2)+4)) (Sqrt[π]) Gamma[4]))],{r,-1,10.4},PlotStyle->Black,BaseStyle->{FontFamily->"Times",FontSize->12.},PlotRange->{-0.01,0.083},Ticks->{Table[10k,{k,1,2}],{5/64}}];
Show[GraphicsRow[{p1,p2}],ImageSize->650]

Chap. 18.   Computation for subordination principle

Clear["`*"]
q[x_,t_]:=(1/(Sqrt[π t]))(E^(-((x^2)/(4t))))
Simplify[(∂_t) q[x,t]==(∂_(x,x))q[x,t]]
Integrate[q[x,t],{x,0,∞},Assumptions->{t>0}]

Problem 18.11  A computation

Sum[Sinc[k Pi/2]^2,{k,-Infinity,Infinity,1}]
Sum[Sinc[k Pi]^2,{k,-Infinity,Infinity,1}]
Sum[Sinc[k Pi]^2,{k,-10000,10000,1}]
N[2-Sum[Sinc[k Pi/2]^2,{k,-1000,1000,1}]]
N[2-Sum[Sinc[k Pi/2]^2,{k,-100000,100000,1}]]

(* ρ=2
ω=(π/ρ) *)
Simplify[(1/(2π))(∫_(-ρ)^ρ)Sign[ξ](E^(I k ω ξ))(E^(I x ξ))\[DifferentialD]ξ,ρ>0]
FullSimplify[Simplify[(1/(2ρ))(∫_(-ρ)^ρ)Sign[ξ](E^(I n ξ))(E^(I x ξ))\[DifferentialD]ξ,ρ>0]==I (((Sin[ρ((x+n)/2)])^2)/(ρ((x+n)/2)))]
Simplify[(1/(2ρ))(∫_(-ρ)^ρ)(E^(I k ω ξ))(E^(- I l ω ξ))\[DifferentialD]ξ,ρ>0]
t[k_][x_]:=(1/(Sqrt[ω]))(((Sin[((ρ x+k π)/2)])^2)/((ρ x+k π)/2))
(* Integrate[t[1][x]t[0][x],{x,-∞,∞}]
Integrate[t[1][x]t[1][x],{x,-∞,∞}]
Integrate[t[2][x]t[2][x],{x,-∞,∞}] *)

Figure 19.1   Example 19.3

Clear["`*"]
Off[General::spell1]
α[x_]:=(E^(-1/x))
p0=Plot[α[x],{x,0,0.4},PlotPoints->100,Ticks->None,AspectRatio->Automatic,PlotRange->All];
α[x_]:=(E^(-1/x))
p00=Plot[α[x],{x,0,10},PlotPoints->100,Ticks->None,AspectRatio->Automatic,PlotRange->All];
a1=-1;
b1=2;
β1[x_]:=0/;x≤a1
β1[x_]:=α[x-a1]α[b1-x]/;a1<x<b1
β1[x_]:=0/;b1≤x
Plot[β1[x],{x,a1,b1},Ticks->None,AspectRatio->1,PlotRange->All];
Int1=N[(∫_a1^b1)β1[x]\[DifferentialD]x]
γ1[x_]:=1/Int1 β1[x]
p1=Plot[γ1[x],{x,-1,2},Ticks->None,AspectRatio->1/2,PlotRange->All];
N[(∫_a1^b1)γ1[x]\[DifferentialD]x]
ϵ=1/5;
γ1ϵ[x_]:=1/ϵ γ1[x/ϵ]
p2=Plot[γ1ϵ[x],{x,a1,b1},Ticks->None,AspectRatio->1,PlotRange->All];
Show[GraphicsArray[{p1,p2}],Ticks->None,AspectRatio->Automatic,PlotRange->All,ImageSize->500];
a2=2;
b2=3;
β2[x_]:=0/;x≤a2
β2[x_]:=α[x-a2]α[b2-x]/;a2<x<b2
β2[x_]:=0/;b2≤x
Plot[β2[x],{x,a2,b2},Ticks->None,AspectRatio->1,PlotRange->All];
Int2=N[(∫_a2^b2)β2[x]\[DifferentialD]x]
γ2[x_]:=1/Int2 β2[x]
p3=Plot[γ2[x],{x,a2,b2},Ticks->None,AspectRatio->Automatic,PlotRange->All];
N[(∫_a2^b2)γ2[x]\[DifferentialD]x]

ψ[z_]:=(∫_(-∞)^z)(γ1ϵ[ζ+0.5]-γ1ϵ[ζ-1])\[DifferentialD]ζ
Plot[ψ[z],{z,a1,b1},Ticks->{{-1,-0.5,1,1.5,2},{1}},PlotStyle->Black,PlotPoints->400,ImageSize->400,BaseStyle->{FontFamily->"Times",FontSize->14.},AspectRatio->Automatic]

Remark 19.11   A computation

f[x1_,x2_]:=((x1^2)/(((x1^2)+(x2^2)) Log[(x1^2)+(x2^2)]))
FullSimplify[(∂_({x1,2}))f[x1,x2]]
(∂_({x1,2}))f[x1,x2]

Figure 19.2 and 19.3

Clear["`*"]
Off[General::spell1]
w=(1/(Sqrt[2]));
Apart[(ξ/((1+(ξ^2))(1+(ξ^4))))]
Print["Function f"]
f[x_]:=2Sign[x](E^(-Abs[x]))
Plot[f[x],{x,-6,6},BaseStyle->{FontFamily->"Times",FontSize->12.},PlotRange->All,Ticks->{{-6,-3,3,6},{-2,2}},PlotLabel->"f",AspectRatio->Automatic,ImageSize->500];
Print["Function h"]
k[ξ_]:=((ξ (E^(I x ξ)))/(1+(ξ^4)))
res1[x]=ComplexExpand[2π I(Residue[k[ξ],{ξ,w(-1+I)}]+Residue[k[ξ],{ξ,w(1+I)}])]
res11[x]=ComplexExpand[2π I(Limit[(ξ-w(-1+I))k[ξ],ξ->w(-1+I)])]
res12[x]=ComplexExpand[2π I(Limit[(ξ-w(1+I))k[ξ],ξ->w(1+I)])]
res11[x]+res12[x]==res1[x]
res2[x]=ComplexExpand[-2π I(Residue[k[ξ],{ξ,w(-1-I)}]+Residue[k[ξ],{ξ,w(1-I)}])]
h[x_]:=Integrate[I((ξ Sin[x ξ])/(1+(ξ^4))),{ξ,-∞,∞},Assumptions->{Im[x]==0}]
h[x]
Plot[(1/I)h[x],{x,-0.001,0.001}]
Print["Function –(∂^2)h"]
hpos[x_]:=π I (E^(-w  x))Sin[w x]
FullSimplify[-D[hpos[x],{x,2}]]
hneg[x_]:=π I (E^(w  x))Sin[w x]
FullSimplify[-D[hneg[x],{x,2}]]
Print["First identity in (✶✶)"]
Integrate[I(((ξ^3) Sin[x ξ])/(1+(ξ^4))),{ξ,-∞,∞},Assumptions->{Im[x]==0}]

Clear["`*"]
w=(1/2)(Sqrt[2]);
f[x_]:=(E^(-x))+(E^(-w x))(Sin[w x]-Cos[w  x]);
g[x_]:=-(E^x)+(E^(w x))(Sin[w x]+Cos[w x]);
f[x]
g[x]
f[0]
f1[x_]:=Simplify[f'[x]]
g1[x_]:=Simplify[g'[x]]
f1[x]
g1[x]
f2[x_]:=Simplify[f''[x]]
g2[x_]:=Simplify[g''[x]]
f2[x]
g2[x]
f2[0]
g2[0]
f3[x_]:=Simplify[f'''[x]]
g3[x_]:=Simplify[g'''[x]]
f3[x]
g3[x]
f3[0]
g3[0]
f4[x_]:=Simplify[f''''[x]]
g4[x_]:=Simplify[g''''[x]]
f4[x]
g4[x]
f4[0]
g4[0]

m00=FindMaximum[f[x],{x,2}][[1]];
m0=0.36;
F0=Plot[f[x],{x,0,12},PlotStyle->Black,BaseStyle->{FontFamily->"Times",FontSize->11.}];
G0=Plot[g[x],{x,-12,0},PlotStyle->Black,BaseStyle->{FontFamily->"Times",FontSize->11.}];
T0=Graphics[{Black,Text[u,{4.2,0.2}]}];
H0=Show[F0,G0,T0,PlotRange->{-0.4,0.4},FormatType->TraditionalForm,BaseStyle->{FontFamily->"Times",FontSize->14.},(*PlotLabel->"u",*)Ticks->{{-10,10},{-m0,m0}},ImageSize->300]

m10=f1[0.1];
f1[0]
m1=0.41;
F1=Plot[f1[x],{x,0,0.1},PlotStyle->Black];
G1=Plot[g1[x],{x,-0.1,0},PlotStyle->Black];
T1=Graphics[{Black,Text["∂u",{0.07,0.4128}]}];
H1=Show[F1,G1,T1,PlotRange->{0.409,0.415},BaseStyle->{FontFamily->"Times",FontSize->14.},FormatType->TraditionalForm,Ticks->{{-0.1,0.1},{0.4,-1+Sqrt[2]}},ImageSize->300]

f2[0.1];
F2=Plot[f2[x],{x,0,0.1},PlotStyle->Black];
G2=Plot[g2[x],{x,-0.1,0},PlotStyle->Black];
T2=Graphics[{Black,Text["(∂^2)u",{0.07,-0.045}]}];
H2=Show[F2,G2,T2,PlotRange->{-0.1,0.1},BaseStyle->{FontFamily->"Times",FontSize->14.},FormatType->TraditionalForm,Ticks->{{-0.1,0.1},{-0.09,0.09}},ImageSize->300]

f3[0.1];
F3=Plot[f3[x],{x,0,0.1},PlotStyle->Black];
G3=Plot[g3[x],{x,-0.1,0},PlotStyle->Black];
T3=Graphics[{Black,Text["(∂^3)u",{0.07,-0.9}]}];
H3=Show[F3,G3,T3,PlotRange->{-1,-0.8},BaseStyle->{FontFamily->"Times",FontSize->14.},FormatType->TraditionalForm,Ticks->{{-0.1,0.1},{-1,-0.81}},ImageSize->300]

F4=Plot[f4[x],{x,0,0.1},PlotStyle->Black];
G4=Plot[g4[x],{x,-0.1,0},PlotStyle->Black];
T4=Graphics[Text["(∂^4)u",{0.065,1.4}]];
H4=Show[F4,G4,T4,PlotRange->{-2,2},BaseStyle->{FontFamily->"Times",FontSize->14.},FormatType->TraditionalForm,Ticks->{{-0.1,0.1},{-2,2}},ImageSize->300,AxesOrigin->{0,0}]

Show[GraphicsArray[{{H0},{H1},{H2},{H3},{H4}}],ImageSize->450]

Show[GraphicsArray[{{H0,H1},{H2,H3},{H4}}],ImageSize->450];

F00=Plot[f[x],{x,103,110}];

a1[x_]:=(E^(-(x/(Sqrt[2]))))Sin[(x/(Sqrt[2]))]
a2[x_]:=(E^(x/(Sqrt[2])))Sin[(x/(Sqrt[2]))]
a1[x]
a2[x]
Simplify[a1''[x]]
Simplify[a2''[x]]

Example 20.27   see    Figure 1.2

Miscellenea

Sin[Pi/4]
FullSimplify[2+4Sum[(-1)^(n-1)/(16n^2-1),{n,1,Infinity}]==Pi/Sqrt[2]]

Series[Pi Cot[Pi z]-1/z,{z,0,5}]
Series[(Pi z Cot[Pi z]-1)/z^2,{z,0,5}]
Limit[(1-Pi z Cot[Pi z])/(2z^2),z->0]

Simplify[Cot[z]^2==-(1+Cot'[z])]

ParametricPlot3D[{r Cos[α],r Sin[α],((1-(r^2))/(1-2r Cos[α]+(r^2)))},{r,0,1},{α,-π,π},PlotRange->{0,5},PlotPoints->{21,100},Shading->False,Boxed->False,Axes->False,Ticks->None,ViewPoint->{-2,-2,1},ImageSize->300];

Plot[Evaluate[((1-(r^2))/(1-2r Cos[α]+(r^2)))/.r->0.9],{α,-0.1,2π+0.8},PlotRange->All,PlotPoints->70,Ticks->{{Pi<PI}}];

FullSimplify[1/(2z^2)+Sum[Cos[n x]/(z^2-n^2),{n,1,Infinity}],{Im[z]==0,Im[x]==0}]

TrigExpand[Cos[z x+n x]/(z+n)+Cos[z x-n x]/(z-n)]

Simplify[Limit[(Pi z Cos[z(Pi-x)]/Sin[Pi z]-1)/(2z^2),z->0]]

F[n_,x_]:=(((Sin[((n x)/2)])^2)/(n ((Sin[(( x)/2)])^2)))
Plot[F[8,x],{x,-2π-0.4,2π+0.4},Ticks->{{-2π,-π,π,2π},{8}},BaseStyle->{FontFamily->"Times",FontSize->16.},AspectRatio->Automatic,PlotRange->All,ImageSize->500];

p1=Plot[Sin[(25+1/2)x]/Sin[x/2],{x,-0.5-2π,2π+0.5},PlotPoints->1000,PlotRange->{-3.1,3.1},Frame->False,Ticks->False,AspectRatio->Automatic,DisplayFunction->Identity];
p2=Plot[{1/Sin[x/2],-1/Sin[x/2]},{x,-0.5-2π,2π+0.5},PlotPoints->100,PlotRange->{-3.1,3.1},Ticks->{{-2π,-π,π,2π},{-2,2}},BaseStyle->{FontFamily->"Times",FontSize->16.},Frame->False,Ticks->False,AspectRatio->Automatic,DisplayFunction->Identity];
Show[GraphicsArray[{{p1},{p2//.Line[{a___,b:{_,y1_},c:{_,y2_},d___}]:>Sequence[Line[{a,b}],Line[{c,d}]]/;Abs[y1-y2]>4}}],AspectRatio->Automatic,ImageSize->500,DisplayFunction->$DisplayFunction];

Integrate[E^(- I x ξ)Log[1+ξ^2],{ξ,-Infinity,Infinity},Assumptions->{Im[x]==0}]

Factor[x^4+1, GaussianIntegers->True,Extension->{Sqrt[2]}]
Apart[1/(x^4+1)]

(∫_(-∞)^∞)(1/(((1+(x^2)))^2))\[DifferentialD]x
(∫_(-∞)^∞)((x^2)/(((1+(x^2)))^2))\[DifferentialD]x

V[s_,t_]:=(E^(a x+b y))U[x,y]
x=Log[s]
y=((σ^2)/2)(T-t)
a=(1/2)-(r/(σ^2))
b=-((((1/2)+(r/(σ^2))))^2)
FullSimplify[(∂_t) V[s,t]+r s (∂_s) V[s,t]+(((σ^2)(s^2))/2)(∂_(s,s))V[s,t]-r V[s,t]]/.{Log[s]->x,((σ^2)/2)(T-t)->y}

ϵ=1/50;
Plot[ArcTan[x/ϵ]/ArcTan[1/ϵ],{x,-1,1},PlotStyle->Black,AspectRatio->Automatic,Ticks->{{-1,1},{-1,1}},ImageSize->200]