Use the option gridlines = true (in Standard GUI: does not work in Classic).

The option filledregions was introduced, I think, in Maple 11.  In earlier versions you could use
filled=true. 

To get red for > 423e-6 and green for < 423e-6, try

> implicitplot(1/2*L*(14.4/(1.5+R))^2 = 423e-6, L=757.4e-6..1135.1e-6, R = 13.85..15.95,
   filled = true, coloring = [green, red]);

The option filledregions was introduced, I think, in Maple 11.  In earlier versions you could use
filled=true. 

To get red for > 423e-6 and green for < 423e-6, try

> implicitplot(1/2*L*(14.4/(1.5+R))^2 = 423e-6, L=757.4e-6..1135.1e-6, R = 13.85..15.95,
   filled = true, coloring = [green, red]);

Of course you have to give values to the parameters, and as Dave mentioned earlier in this thread you need quotes.  So:

> with(Statistics):
    X := RandomVariable('Beta(3,5)');
    f := PDF(X,u);
    plot(f, u = 0 .. 1);

It really would help if you showed us an example of what you'd like as output.

Yes, it will, but you need to give it a list (in [] brackets) rather than a set (in {}).

Try

pointplot3d([[p1],[p2],[p3],[p4],[p1],[p6],[p5],[p2],[p5],[p8],
[p3],[p8],[p7],[p4],[p7],[p6]],style=line, colour=black);

Yes, it will, but you need to give it a list (in [] brackets) rather than a set (in {}).

Try

pointplot3d([[p1],[p2],[p3],[p4],[p1],[p6],[p5],[p2],[p5],[p8],
[p3],[p8],[p7],[p4],[p7],[p6]],style=line, colour=black);

Great choice!  Congratulations, Doug.

> F:= c*ln((exp(a))/(1+exp(a)))+d*ln(1-(exp(a))/(1+exp(a)))+e*ln((exp(a+b))/(1+exp(a+b)))+f*ln(1-(exp(a+b))/(1+exp(a+b)));

(avoiding the problems Joe referred to)

> focs:= simplify([diff(F,a), diff(F,b)]); 

(I assume this is what you meant, not diff(F,a) for both)

> simplify(eval(focs, solve({p1=exp(a)/(1+exp(a)), p2 = exp(a+b)/(1+exp(a+b))},{a,b})));

[-c*p1+c-d*p1-e*p2+e-f*p2, -e*p2+e-f*p2]

The maple tag is acting up again: that result should be [-c*p1+c-d*p1-e*p2+e-f*p2, -e*p2+e-f*p2]

> F:= c*ln((exp(a))/(1+exp(a)))+d*ln(1-(exp(a))/(1+exp(a)))+e*ln((exp(a+b))/(1+exp(a+b)))+f*ln(1-(exp(a+b))/(1+exp(a+b)));

(avoiding the problems Joe referred to)

> focs:= simplify([diff(F,a), diff(F,b)]); 

(I assume this is what you meant, not diff(F,a) for both)

> simplify(eval(focs, solve({p1=exp(a)/(1+exp(a)), p2 = exp(a+b)/(1+exp(a+b))},{a,b})));

[-c*p1+c-d*p1-e*p2+e-f*p2, -e*p2+e-f*p2]

The maple tag is acting up again: that result should be [-c*p1+c-d*p1-e*p2+e-f*p2, -e*p2+e-f*p2]

My first try was using unevaluation quotes:

> plot( 2*cos(x/2-Pi/6), x=-2*Pi..2*Pi,
      tickmarks=[spacing(Pi/2,0),3],
      title=typeset("A plot of one period of \n", y='2*cos(x/2-Pi/6)') );
 

That is still not perfect: it gives you y=2*cos(1/2*x - 1/6*Pi).   So I tried this, which worked:

> plot( 2*cos(x/2-Pi/6), x=-2*Pi..2*Pi,
      tickmarks=[spacing(Pi/2,0),3],
      title=typeset("A plot of one period of \n", y=Typesetting[Typeset](2*cos(x/2-Pi/6)) ));
 

My first try was using unevaluation quotes:

> plot( 2*cos(x/2-Pi/6), x=-2*Pi..2*Pi,
      tickmarks=[spacing(Pi/2,0),3],
      title=typeset("A plot of one period of \n", y='2*cos(x/2-Pi/6)') );
 

That is still not perfect: it gives you y=2*cos(1/2*x - 1/6*Pi).   So I tried this, which worked:

> plot( 2*cos(x/2-Pi/6), x=-2*Pi..2*Pi,
      tickmarks=[spacing(Pi/2,0),3],
      title=typeset("A plot of one period of \n", y=Typesetting[Typeset](2*cos(x/2-Pi/6)) ));
 

Your question does not make sense as stated, but the answer to the question I think you meant to ask is no. 
Look at the symbolic solution I got, and ask yourself what happens as i -> infinity.

Your question does not make sense as stated, but the answer to the question I think you meant to ask is no. 
Look at the symbolic solution I got, and ask yourself what happens as i -> infinity.

Then you'd better use that.  For example:

> n := 6:
   solve( eval( {seq(x[i-1]+x[i]+x[i+1]=i,i=1..n)}, {x[0]=0, x[n+1]=0}));

{x[1] = -2, x[2] = 3, x[3] = 1, x[4] = -1, x[5] = 4, x[6] = 2}

However, note that there's no solution if n+1 is divisible by 3.

You could try for a symbolic solution.

> rsolve({x(0)=0, x(i-1)+x(i)+x(i+1)=1}, x(i));

(1/3*I)*sqrt(3)*x(1)*(-1/2-(1/2*I)*sqrt(3))^i-(1/3*I)*x(1)*sqrt(3)*(-1/2+(1/2*I)*sqrt(3))^i-(2/3)*(-2/(1-I*sqrt(3)))^i/(1-I*sqrt(3))-(2/3)*(-2/(1+I*sqrt(3)))^i/(1+I*sqrt(3))+1/3

> simplify(evalc(%));

(2/3)*sqrt(3)*x(1)*sin((2/3)*i*Pi)-(1/3)*cos((2/3)*i*Pi)-(1/3)*sin((2/3)*i*Pi)*sqrt(3)+1/3

Note that I didn't use the end condition x(n+1) = 0.  That can be used to determine x(1) (if sin((2/3)*(n+1)*Pi) is not 0).