> #sfb oct 9, 2001
> sinPi4:=sin(Pi/4);cosPi4:=cos(Pi/4);
> sinPi8:=sin(Pi/8);cosPi8:=cos(Pi/8);
> # sin^2 x = (1 - cos 2x)^/2;
> sinPi8:=solve(x^2=(1-cosPi4)/2);
> sinPi8:=solve(x^2=(1-cosPi4)/2)[1];
> op(4,eval(sin));#For advanced use only
> sin(Pi/8):=sinPi8;
> # two plots on the same line
> with(plots):
> F:=plot(sin(x),x=0..2*Pi,title="sin(x)",legend="sin(x)",linestyle=2):
> G:=plot(cos(x),x=0..2*Pi,title="cos(x)",legend="cos(x)",linestyle=3):
> display(array([F,G]));
> display([F,G]);
> F;
> G;
> #####
> #Sequence
> a:=1/2^n;b:=1-1/2^n;
> limit(a,n=infinity);limit(b,n=infinity);
> seq(a,n=1..10);
> sum(a,n=1..10);
> subs(n=10,b);
> with(student);
> Sum(a,n=1..10);
> f:=1/2^x;#f:=1/x;
> F:=leftbox(f,x=1..6,5):display(F,title="leftsum");
> G:=rightbox(f,x=1..6,5):display(G,title="rightsum");
> display(array([F,G]));
> restart;with(student):with(plots):
> ## The integral test: Given f(x)>=0 non-increasing defined x>=1
> ## Let a_n = f(n), then the fate of 
> Sum(f(n),n=1..infinity),Int(f(x),x=1..infinity);
> # are the same. Either both converge or both diverge.
> Sum(1/n^2,n=1..infinity)=sum(1/n^2,n=1..infinity);
> Int(1/x^2,x=1..infinity)=int(1/x^2,x=1..infinity);
> evalf(Pi^2/6);
> ## other tests
> Sum(a[n],n=1..infinity);
> #converges implies that
> Limit(a[n],n=infinity)=0;
> #comparison
> 0 <= a[n]; a[n] <= b[n];
> Sum(a[n],n=1..infinity), "diverges implies", Sum(b[n],n=1..infinity), "diverges also";
> Sum(b[n],n=1..infinity), "converges implies", Sum(a[n],n=1..infinity), "converges also";
> #RATIO test suppose
> Limit(abs(a[n+1])/abs(a[n]),n=infinity) = L;
> 0 <= L, L < 1," implies ",Sum(a[n],n=1..infinity),"converges";
> 1 < L," implies ",Sum(a[n],n=1..infinity),"diverges";
> L=1,"then the test fails";
> #
> a:=1/2^i;Limit(subs(i=n+1,a)/subs(i=n,a),n=infinity)=limit(subs(i=n+1,a)/subs(i=n,a),n=infinity);
> a:=1/i!;Limit(subs(i=n+1,a)/subs(i=n,a),n=infinity)=limit(subs(i=n+1,a)/subs(i=n,a),n=infinity);
> a:=1/i;Limit(subs(i=n+1,a)/subs(i=n,a),n=infinity)=limit(subs(i=n+1,a)/subs(i=n,a),n=infinity);
> a:=1/i^2;Limit(subs(i=n+1,a)/subs(i=n,a),n=infinity)=limit(subs(i=n+1,a)/subs(i=n,a),n=infinity);
> #
> #Power series sum c_n (x - a)^n or 
> a:='a';Sum(c[n]*(x-a)^n,n=1..infinity);
> #examples
> series(exp(x),x=0,10);series(ln(x),x=1,10);series(arctan(x),x=0,11);series(1/(1-x),x=0,10);#
> # Apply the ratio test
> Limit(abs(c[n+1]*(x-a)^(n+1))/abs(c[n]*(x-a)^n),n=infinity);#
> #
> f:=arctan(x);g:=array(1..5);
> for i from 1 to 5 do g[i]:=convert(series(f,x=0,2*i+1),polynom); od;
> n:=2;plot([f,g[1],g[2],g[3],g[4],g[5]],x=-n*Pi..n*Pi,y=-2..2);
> plot([f,convert(series(f,x=0,41),polynom)],x=-6*Pi..6*Pi,y=-2..2);
> plot([f,convert(series(f,x=0,41),polynom)],x=-6*Pi..6*Pi,y=-2..2,numpoints=2000);
> f:=sqrt(x);plot([f,convert(series(f,x=2,41),polynom)],x=0..20,y=-2..20,numpoints=2000);
> 
> # We are use to writing 1+2x+x^3 in the order x^3+2x+1
> # But from a power series viewpoint 1+2x+x^3 makes more sense
> # because we are thinking of `small' x so x^3 is smaller than x.
> #
> # intervals/radius of convergence graphically
> #
> # Geometric series radius 1
> f:=1/(1-x);deg:=20;g:=convert(series(f,x=0,deg+1),polynom);
> plot([f,g],x=-2..2,y=-2..6,thickness=[3,1],color=[red,blue]);
> #radius 4 -- sqrt at 4
> f:=sqrt(x);deg:=33;g:=convert(series(f,x=4,deg+1),polynom);
> plot([f,g],x=0..10,y=-2..4,thickness=[3,1],color=[red,blue]);
> #why here? -- nice function with radius 1
> f:=arctan(x);deg:=20;g:=convert(series(f,x=0,deg+1),polynom);
> plot([f,g],x=-5..5,y=-2..2,thickness=[3,1],color=[red,blue]);
>