@Markiyan Hirnyk

find it attached, it is based on a C routine due to Ooura, intdeo_hirnyk.mws

It is not a good shot here (depending too much on Digits)
and certainyl Pseudomodo's approach is far better (I often
forget to switch to the 'operator' form)

You mean to calculate it numerically ?

@Joe Riel 

Thank you for pointing to that (seems I quite often stumble into such) - is it ok using Vector?

Where I would prefer to use the 'polynomial' approach as in No. 2 anyway.

@Joe Riel 

Thank you for pointing to that (seems I quite often stumble into such) - is it ok using Vector?

Where I would prefer to use the 'polynomial' approach as in No. 2 anyway.

Sometimes I use a varaint of this by '%' = %:

'solve(exp(x)=3,x)':
'%'= %;
                     solve(exp(x) = 3, x) = ln(3)

for i from 1 to 2 do 'i^2': print('%'= %); end do :
                                 2
                                i  = 1
                                 2
                                i  = 4

Sometimes I use a varaint of this by '%' = %:

'solve(exp(x)=3,x)':
'%'= %;
                     solve(exp(x) = 3, x) = ln(3)

for i from 1 to 2 do 'i^2': print('%'= %); end do :
                                 2
                                i  = 1
                                 2
                                i  = 4

looking at it it seems data become linear in the large and thus doubt an polynomial would satisfy

without knowing more, that probably is not a way for a 'solution'

looking at it it seems data become linear in the large and thus doubt an polynomial would satisfy

without knowing more, that probably is not a way for a 'solution'

2^(z)-5: 
cos(x)=solve(%,z); 
solve(%,x); evalf(%);

                                     ln(5)
                            cos(x) = -----
                                     ln(2)

                                   ln(5)
                            arccos(-----)
                                   ln(2)

                            1.485570061 I

Now the rest comes in by periodics of log (solving for z)
and arccos (the above only takes 1 solution in each case)

2^(z)-5: 
cos(x)=solve(%,z); 
solve(%,x); evalf(%);

                                     ln(5)
                            cos(x) = -----
                                     ln(2)

                                   ln(5)
                            arccos(-----)
                                   ln(2)

                            1.485570061 I

Now the rest comes in by periodics of log (solving for z)
and arccos (the above only takes 1 solution in each case)

What is "Spielverderber" in Russian ?  :-)

Ok, look up "Little Picard" should proof it.

What is "Spielverderber" in Russian ?  :-)

Ok, look up "Little Picard" should proof it.

I only had a short look into the paper (and gave up), but for the Q "complex zeros as well?"
they seem to suggest using the integral(s) counting zeros (and poles), for which one would
want to know analiticity, of course.

But using some trials - Yes, there are complex zeros (for lamda towrads 0). And also some
real, but strictly positive (non-complex [to be verfied *beyond* Numerics]) zeros, see some
suggestions appended.

Kopie_von_KVasymptDa.mws

I only had a short look into the paper (and gave up), but for the Q "complex zeros as well?"
they seem to suggest using the integral(s) counting zeros (and poles), for which one would
want to know analiticity, of course.

But using some trials - Yes, there are complex zeros (for lamda towrads 0). And also some
real, but strictly positive (non-complex [to be verfied *beyond* Numerics]) zeros, see some
suggestions appended.

Kopie_von_KVasymptDa.mws

Will try to understand tonight, but I can not see at a glance, where the thing
is analytic (and that probably would be needed): complex R3 given lambda?