This is the first problem of the "Hundred-dollar, Hundred-digit Challenge problems":

https://en.wikipedia.org/wiki/Hundred-dollar,_Hundred-digit_Challenge_problems

The Maple solution was obtained by Robert Israel:

https://www.math.ubc.ca/~israel/challenge/challenge1.html

Edit: Mathematica seems to have better algorithms for computing oscillatory integrals.

Too many errors ...

neville:=proc(X::list,Y::list,x) # polynomial interpolation
local T,i,j,n;
n:=nops(Y)-1;
T:=Array(0..n, Y);
for j to n do
  for i from 0 to n-j do
  T[i] :=  normal( (T[i]*(X[j+i+1]-x) + T[i+1]*(x-X[i+1]) )/(X[j+i+1]-X[i+1]) )
od; od;
end:

Test:

f:=x->sin(3*x);
r:=neville([$1..10],[f(n)$n=1..10],x):
plot([f(x),r],x=1..10);

You were told that f is not a pdf unless d=0.

For d=0  f is a Negative binomial distribution which is implemented in Maple, so you can use it.

But you want d>0.

In this case, you probably want to mimic the moments defined for a pdf i.e.

M[k] = Sum( n^k * f(n), n=0..infinity);

But it can be shown that f(n) is equivalent (as n --> infinity) to K/n^2 where K is a nonzero constant.
Hence all the moments are infinite.

Actually, your with your procedure a sequence with both elements is produced. In Maple 2015, Eqplot(2,3) returns 2x+3 and the plot in a reduced size. 

If you use r := Eqplot(2,3), then r[1] equals 2*x+3 and r[2] equals a PLOT structure such that r[2] will display the plot when evaluated.

It would be better to use:

Eqplot:=proc(a,b)
local y,x;
y:=a*x+b;
print(plot(y,x));
y;
end proc:

This way, Eqplot(2,3) will return 2*x+3 ant the plot itself is produced as a side effect.

f:=2*x-2:
an:=2*int( f*cos(Pi*x*n),x=0..1) assuming n::integer;

a0:=2*int(f,x=0..1);

To compare graphically f with a partial Fourier sum:

s3:=a0/2 + add( eval(an,n=k) * cos(k*Pi*x), k=1..3):

plot([f, s3], x=0..1, view=[-0.2..1.2,-2.2..0.5]);

Install the package in a directory where you have write permission. After the .hdb conversion you can move it into Program Files if you want.

For d=0, f is a PDF  (actually a Negative binomial distribution).

It seems to me (after some investigations) that this is the only case when f is a PDF.
According to Maple, for [r=3, b=2, d=0.01411053479]  f seemed to be a distribution
just because  Sum_x f(x)   is very close to 1 for a small d.

Why don't you use 1D math? Later, you may convert to 2D if you want. All these problems will disappear.

I think that 1D for input and 2D for output is the best choice for now. 

se:=proc(f::procedure,a::realcons,b::realcons)
local x,s;
s:=(f(b)-f(a))/(b-a)*(x-a)+f(a);
print(plot([f(x),s], x=a..b));
s;
end proc:

f:=x -> x^2:

se(f,-1,2);

rad:=u -> `*`(op(map( z->op(1,z),factors(u)[2])));

g:=expand((x*sqrt(2)+y*sqrt(3)+z*sqrt(6)+x*y*sqrt(6)+x)^6*(x+2*y)^2):
rad(g);

If _C1 is constant just use _C1 instead of _C1(t).

If _C1(t) etc are already in an expression, use

subs( [_C1(t)=_C1, _C2(t)=_C2, _C3(t)=C3], expr);

You shoud not write in the Program Files directory.
Put your library in a directory where you have write permission. It will work.

L1toL2:=proc(L1,L2)
local f;
  f:=proc(x,L) local t; member(x,L,t); t end;
map(f,L2,L1);
end;

# it works for nondistinct elements, provided that {op(L2)} subset {op(L1)}  (of course)
L1:=[a,b,a,a]; L2:=[b,b,b,a];

{op(L2)} subset {op(L1)};
                              true
P:=L1toL2(L1,L2);
                          [2, 2, 2, 1]
L1[P],L2;
                   [b, b, b, a], [b, b, b, a]

It seems that you want to approach Riemann hypothesis using Maple. :-)

Mathematica seems to have a much better algorithm to approximate Zeta(z) for a large |z|.

Maple computes Zeta(0.6+I*1000000) very slowly (many seconds!). Other CAS-es even refuse to approximate.
It would be interesting to know the algorithm used by Mathematica and its robustness.

The probability of having 0 branch breakings after 40 steps is

1-(1-0.02)^40 = 0.5542995960 (> 1/2)

If you want a simulation to compute the number of branch breaking, you should run your code N times (in a loop) and compute the average number of breaks. (N = 10000 or something).