Use * (asterisk) instead of . (dot) for multiplication:

while k<100000 do x:=f(x): k:=k+1 :
  if(x>0 and x<1) then a:=a+1
  elif x<2 then b:=b+1
  elif x<3 then c:=c+1
  elif x<4 then d:=d+1
  end if :
end do:

(edited)

To make an illustrative plot, give v_max and S_k arbitrary values, and use a tickmark option:

Download PiecewiseGraph.mw

Use a legend option with six empty legend texts:

A:=plot( [seq(seq(lambda(F,Nb,delta2), Nb=[0.1,0.2,0.3]), F=[0.1,0.2,0.5])],
          delta2=0.02..0.1, linestyle = [solid,longdash,dashdot],'
          thickness = 2',color=[red$3,blue$3,black$3],
          legend=["Nb=0.1","Nb=0.2","Nb=0.3",""$6]):

I suppose that "when x = -3..3 and y = -3..3" means the initial value y(-3.3)=-3.3. The differential equation happens to have an exact solution with discontinuities.

restart;
DE := diff(y(x),x) = x^2 - y(x)^2:
dsolve({DE,y(-3.3)=-3.3},y(x)): Y := subs(%,y(x)):
plot(Y,x=-4..4, discont);

So you have to plot the phaseportrait in a neighbourhood of x=-3.3.

DEtools[DEplot]( DE, y(x), x=-4..-3, {y(-3.3)=-3.3},linecolor=blue);

Download PhasePortrait.mw

Just solve for y:

Download SolveEq.mw

Simply use evalc:

Download ComplexSort.mw

In this case the remove command can be used:

Download RemoveListElements.mw

I suppose that you mean the norm in the meaning of the absolute value:
 

Download AbsError.mw

There are several difficulties:

• coeffs gives the coefficients of each power of q. In your example the coeff of q^2 is b-a; so we have to split it in "-a" and "b";
• subs doesn't want to subs -a=1, so we make it a=-1.

Download UnitRoots.mw

Use add instead of sum for non-symbolic summation:

restart;
L := [[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]:
add(L[i,1],i=1..5);
                               11

Perhaps you mean:

Download Torabi.mw

Which equation do you mean?

Download Parentheses.mw

The imaginary unit is capital I, not lowercase i