what u mean by shorter form

p:=plot((Spec(d)),d=-10..10,axes=boxed,title=tit,color=black,font=[2,3,18],thickness=2,tickmarks=[3,3],titlefont=[SYMBOL,14],font=[1,1,18],linestyle=1);

stell when I set the parameters theta:=Pi/3;phi:=Pi;

takes long time then i stop it. please need advise.

Thanks for your answer..

can you explane

A:= op([1,1], P):  #The data matrix from the plot
Smax1:= max(A[..,2]):
A[..,2]:= A[..,2]/Smax1:
P:= subsop([1,1]= A, P):
print(P);

where I put them before the plot??

for a simple example:

if f(x)=x^2+1

so its max value is:

Xmax:=max(seq(evalf(x^2+1),x=0..10));

then it is easy to plot

f(x)/Xmax

but in my problem it isnt work!!!!!

can we find the roots without substitute b and k

can I do it in two loops

one for t the other for n

I tried to find a root of polynomail of higer degree than that one, and i get the roots why this one not possible??

eq5:=u->-(2*(1+2*u+k*u^2))*(-2-3*k*u+k^2*u^3)*(-2*k^2*u^4-8*k*u^3-4*k*u^2+b*k*u^2-8*u^2-8*u-2-b);

sol:=(solve(eq5(u),u)): S:=array([],1..3): S[1]:=(eval(sol[1])):S[2]:=sol[2]:S[3]:=sol[3]:

r=S[1];
(1/2)
-1 + (-k + 1)
r = ------------------
k

please nedd help to find the roots of this polynomail of degree 6 withot setting values of k and b

eq5:=u->-3*b*k*u^2-3*b*k^2*u^4-k^3*u^6*b+k*u+5*k^2*u^3-b;

This's what i'm looking for...Many thanks 

This's what i'm looking for...Many thanks 

I tried:

eq1:=(alpha+(l+alpha)*u+alpha*k*u^2)*a=u*(alpha+(l+alpha)*u+alpha*k*u^2)*(1+l*alpha*b/((alpha+(l+alpha)*u+alpha*k*u^2))):
eq2:=a=(u*(1+l*v/(1+u+k*u^2))):
factor((rhs-lhs)(eq1)):
eq1:=collect(%,u): 

factor((rhs-lhs)(eq2)): 
eq2:=collect(%,u):

sp1:=unapply(eq1,u,a):
sp2:=unapply(eq2,u,a,v):

a:=array(0..N1,[]):u:=array(0..N1,[]):v:=array(0..N1,[]):

pt1:=array(0..N1,[]): #t0:=0.0:tN:=60.0: N1:=500;th:=evalf((tN-t0)/N1):
for i from 0 to N1 do u[i]:=evalf(th*i):

v[i]:=evalf(th*i):

evalf(sp1(u[i],a[i]));

evalf(sp2(u[i],a[i],v[i])));

pt1[i]:=[u[i],a[i]]:
od;

I tried:

eq1:=(alpha+(l+alpha)*u+alpha*k*u^2)*a=u*(alpha+(l+alpha)*u+alpha*k*u^2)*(1+l*alpha*b/((alpha+(l+alpha)*u+alpha*k*u^2))):
eq2:=a=(u*(1+l*v/(1+u+k*u^2))):
factor((rhs-lhs)(eq1)):
eq1:=collect(%,u): 

factor((rhs-lhs)(eq2)): 
eq2:=collect(%,u):

sp1:=unapply(eq1,u,a):
sp2:=unapply(eq2,u,a,v):

a:=array(0..N1,[]):u:=array(0..N1,[]):v:=array(0..N1,[]):

pt1:=array(0..N1,[]): #t0:=0.0:tN:=60.0: N1:=500;th:=evalf((tN-t0)/N1):
for i from 0 to N1 do u[i]:=evalf(th*i):

v[i]:=evalf(th*i):

evalf(sp1(u[i],a[i]));

evalf(sp2(u[i],a[i],v[i])));

pt1[i]:=[u[i],a[i]]:
od;

Thank you for your answer.

I was thinking in easiest way by plug these two equations in  for loop to get (a,u) from the first equation thenn (a,v) since v depend on (a,u).

Can you plz help on that.

Thank you for your answer.

I was thinking in easiest way by plug these two equations in  for loop to get (a,u) from the first equation thenn (a,v) since v depend on (a,u).

Can you plz help on that.

First I would like to thank u for ur quick reply.

I need some explanations plz..

First, is it possible to define an implicit eqs as:

eq1:=(alpha+(l+alpha)*u+alpha*k*u^2)*a=
u*(alpha+(l+alpha)*u+alpha*k*u^2)*(1+l*alpha*b/((alpha+(l+alpha)*u+alpha*k*u^2)));

and 

eq2:=a=(u*(1+l*v/(1+u+k*u^2)));

when you define

points:=seq([a,fsolve(eval(eq1,params),u)],a=0..20); was this means that points (a,u) becuase the roots of u are one real and two complex conjugates..and i want the real one.

Note eq1 is F(a,u) and eq2 is g(a,u,v)

my goal is to plot points (a,v)

Thanks in advanced.

First I would like to thank u for ur quick reply.

I need some explanations plz..

First, is it possible to define an implicit eqs as:

eq1:=(alpha+(l+alpha)*u+alpha*k*u^2)*a=
u*(alpha+(l+alpha)*u+alpha*k*u^2)*(1+l*alpha*b/((alpha+(l+alpha)*u+alpha*k*u^2)));

and 

eq2:=a=(u*(1+l*v/(1+u+k*u^2)));

when you define

points:=seq([a,fsolve(eval(eq1,params),u)],a=0..20); was this means that points (a,u) becuase the roots of u are one real and two complex conjugates..and i want the real one.

Note eq1 is F(a,u) and eq2 is g(a,u,v)

my goal is to plot points (a,v)

Thanks in advanced.