One thing that is wrong is that you have e^(4*t) instead of exp(4*t).

The other thing that is peculiar is your range t= t..0

This should be something like t=T..0 instead.

If you enter this:

int(-(2/3)*exp(4*t)/(exp(t)+t), t );

the integral is not evaluted. This means that the antiderivative cannot be written down in any conventional closed form. Maybe you meant this instead?

int(-(2/3)*exp(4*t)/(exp(t)+1), t );

Notice the slightly different denominator.

You might try this:

>  assume(m,integer);assume(n,integer);additionally((m+n)/2,integer);

 > int(cos(m*x)*sin(n*x),x=0..Pi) ;

                                   0
 

>  assume(m,integer);assume(n,integer);additionally((m+n+1)/2,integer);

>  int(cos(m*x)*sin(n*x),x=0..Pi) ;

                                 2*n/(n^2-m^2)

To differentiate under the integral, use Int instead of int...

diff(Int(A(B(q,r)),r=a..b),q);

gives what you want.

But

diff( int(A(B(q,r)),r=a..b),q);

evaluates to 0.

Use parametric equations:

x=3, y=t

plot( [3, t, t=-4..4] );

The general format is

plot( [x(t),y(t),t=a..b] );

Try this:

F:=(i,j)->piecewise(j>i+1,a,i>j+1,c,b);

Matrix(5,5,(i,j)->F(i,j));

Look at the case a=2/5. Here you see that the solution depends on the roots of a seventh degree polynomial. Thus in general, you cannot expect analytic solutions.

If a is not rational, it looks even worse, because you do not end up with a polynomial.

Perhaps you have numerical values in mind for the parameters, in which case you can of course find numerical solutions.

Do some experimenting with particular values of a.

For example,

a:=2;

g:=t->long mess;

solve(g(t)=0,t);

returns explicit solutions.

If you use a:=3, then you see that the solutions depend on the roots of a fourth degree poly. If a:=5, the solutions depend on the roots of a sixth degree poly. It looks like if a is a positive integer, then the solutions depend the roots of a poly. of degree a+1. Of course in general, there is no "analytic" way to represent the roots of a polynomial. So there is no analytic solution for t, unless the polynomials happen to be very special, which I doubt.

If a=-1, you get a nice simple solution.

What can you assume about a? This seems to be the key issue.

Try creating your 2D XYPlot as a 3D object as follows:

XYPlot:= seq([rhs(dsn(i)[2]),rhs(dsn(i)[3]), 0 ], i=0..100,0.01):

Then display it with pointplot3d.

Finally convert to a 2D structure by using plottools[transform]...

Something like

proj:=plottools[transform]((x,y,z)->(x,y)):

proj(pointplot3d({XYPlot});

Split up your expression into real and imaginary plots, and treat as a parametric curve:

ex:=(-1/2)^q+(-1/2)^(1-q) ;

X:=Re(ex);
Y:=Im(ex);

plot( [X,Y,q=0.25..0.50]);

So you want

$A=maple("[seq(round(RandomTools[Generate](distribution(Normal(40,5)),1)), i=1..20)]");

$B=maple("[seq(($A[i+1]+$A[i+2])/2, i=1..18)]");

Notice how MapleTA variables have a dollar sign at the beginning of their name.

  I bet you get it now!

I think you want

$A=maple("[seq(round(RandomTools[Generate](distribution(Normal(40,5)),1)), i=1..20)]");

You are correct. Consider this code:

q2:= (x)-> x*exp(x^2)*(1-erf(x)):
    Digits := 50:
   infolevel[plot] := 2: 
   plot(q2(x), x = 0 .. 10);

In Maple 8, Maple 9, Maple 10 and Maple 11 we get good plots.

BUT in Maple 9.5, we get the ringing trash.

Interesting...in Classic 11

solve(a*x^2 + b*x + c, x);

 returns the familiar form of a quotient. So the denominator is 2a.

In Standard 11 worksheet, we get a product. So we get 1/2*stuff.

In the best of all possible worlds, we could simply use LaTeX syntax

V_{BE} instead of

`#msub(mi("V"),mi("BE"))`

But Maple developers seem dedicated to ignoring LaTeX.

So then the problem is to minimize Int( abs(f-g),0..1).
 

I would focus on this:

Int(c^2*x-x^3,x=0..c)+Int(c^3+(1-c^3)/(1-c)*(x-c)-x^3,x=c..1);

This would be the area between the graphs if the  graph of g is a segment from [0,0] to [c,c^3] followed by a segment from [c,c^3] to [1,1].