One equation (the 2nd, a parabola) can be solved (here even analytically) and feeding reduces to dim=1

After pondering and trying a longer time: it is easy to show 'unimodal',
essentially one has to show that Psi(r) - Psi(m-r) is increasing in r,
for which one should not use Maple's definition.

  C:= (n,k)-> binomial(n,k)*2^k; #  binomial theorem 
  psi:= z -> Sum(1/_k1-1/(_k1+z-1),_k1 = 1 .. infinity)-gamma;

  'psi(z)': '%' = value(%); # better look at FunctionAdvisor

                           psi(z) = Psi(z)

  'diff(C(n,k), k)': '%' = simplify(%);

    d
    -- C(n, k) =
    dk
                         k
        -binomial(n, k) 2  (Psi(k + 1) - Psi(n - k + 1) - ln(2))

Now one wants to see that the very factor has at most 1 zero:


  select(has, %, Psi);
  eval(%, Psi=psi);
  combine(%);
  diff(%, k);
  #value(%);  #subs(n=8, %);  plot(%, k=0.1 .. 9);


                infinity
                 -----
                  \      /      1              1     \
                   )     |-------------- + ----------|
                  /      |             2            2|
                 -----   \(_k1 + n - k)    (_k1 + k) /
                _k1 = 1

That means that the very factor is strictly increasing and I think
it is easy to see +- infinity in the boundaries for Psi(0), giving
the needed sign change.

Ok. One can read the Jacobi Matrix as linear map on the tangent spaces
or differential forms and rank means to compute the linear dimension of
the image (though rank can be done more simple). I am not aware whether
Maple directly provides that linear algebra over (polynomial) rings.

The other thing may be be geometrics, looking for dimensions of the
fibres of f and try to argue by 'generic' cases. Have not thought about
Groebner - that would need to give your f in the dual notion between
the algebras (and not the spaces).

May be a dedicated system like Macaulay provides more.

Ok, but a parametrization is 'onto', else you would not catch exactly all points (ignoring structures and allowing multiple 'hits' [i.e. not injective])

So what is the precise setting?

gradient ? or Jacobi matrix? and what means "not zero" ? and if it is a parametrization of a variety then you know the dimension - that of the variety ...

May be you you rephrase, I still do not understand your setting

What do you mean by that - "locally" or "parmeters"? f: C^n -> C^p a morphism of algebraic varieties, i.e. polynomial?

the formal solution for the integral is a kind of gamma distribution, I think, and I get

870*(GAMMA(17/dd+1)-GAMMA(17/dd+1,70/dd)) * dd * 
  exp(-17/dd*ln(17/dd)+17/dd)

val := value of integral, denum(val) = (dd+17)*(2*dd+17)*(3*dd+17)
with solution -17, -17/2, -17/3, so -8.5003.... is too close to
a singularity for the real valued solver?

No, it is not Whittaker per se as they sum up to be real valued
(as the integral is real), but it may have spurious imaginary parts.

But it may be (1/dd)^something, which *is* imaginary in general and
may not cancel out numerically.

eval(V=val, dd=sol);

  57002400 = 57002399.9972916 - 0.895403686845066e-6*I

So there may be 2 reasons

V=Re(val):
fsolve(%, dd, complex);

             -8.50029502704987 + 0.0000389502640690873 I

A nice example, indeed.

Hacker, you may wish to post in this forum (and I do not answer via Google's gmail or such).

My results for 14800 and solving F(b,'C'(b)) are

  b0 = 136.810925806632, c0 = -136.810007282565

needing ~ 5 seconds.

Using that tmp3 I have

       136.810925806271,      -136.810007282204

and using that approx_eq2 additionally I get

       136.810925564333,      -136.810007040264

Digits = 15, no guess from a plot is needed.

More if the originator is willing to talk about his task,
where it comes from and for what he is using these results.

NB1: below 14xxx one can not use the 'simple' solution
NB2: a kind of (numerical) 'discontinuity' occurs for
     intensity ~ 14240

Hm ... I got my results from yesterday through it in some way.

Provide the initial guess b = 140 for the final fsolve and it is immediate.

Ok, will provide an improved version later

I already gave you another way, see the other threads

The essential part involves exp of b and c, their negatives and combinations, especially adding those terms and I think one has to prepare/handle it carefully (no, I do not want to do that ...) instead of just feeding

The solution is ~ b0 = 136.810925806288, c0 = -136.810007282221, using the way I sketched already in another thread on this task(s).

Where - of course - it is not clear, that there is one solution at most (i.e. none beyond towards infinity) and complex solutions are ignored anyway.

Is this the same question = system as in your other 2 threads, except a different value for "intensity"?

Difficult to answer beyond what I said, roughly you ask how the solution depends
on variation of your "intensity". For a rought guess you may try an implicite plot
like Markiyan did in the original thread for the task.

I think DirectSearch is quite reliable - have you feed its results into your task?