Alec, I admire your patients and how you still keep good manners in that case or however one should express that in English :-)

@hafizudin: so I understand it is convenient to fire questions and get replies you are going beyond some limit (for my taste). At least you may want to do some searching on your own and struggle through the given hints. You may also try Google, especially google.scholar as well as you may not hesitate to visite the library at the university.

Or in short: try to do some work on your own.

It is a long time ago that I read (parts of) Bloch who called your attitude "bürgerliche Jugendfreude" (w.r.t. 'the best' and the rest), but in short: it is very unkind, at least for me. Neither the world nor the www is going to answer questions in that style and a programmer would get the answer 'RTFM', while students would hear 'try to find out more using the library'. Just try that.

Sorry for that somewhat personal statement, but think it is in place (@admin: delete if you dislike ... no problem for me)

Cheers

You may wish to look up your notes and Maple's help before firing questions.
Try to answer the following:

1. How may asymptotic series for f(x) and series in 0 for f(1/x) related?
2. What does it mean for log?
3. For your task?

You may wish to look up your notes and Maple's help before firing questions.
Try to answer the following:

1. How may asymptotic series for f(x) and series in 0 for f(1/x) related?
2. What does it mean for log?
3. For your task?

:-)

My concern is, that (see "Gamma(z) vs Gamma(0,z)" with JacquesC)
it internally still 'simplifies' Gamma(0,z) to Gamma(z).

The original post was to point to a bug only. For converting to
a Sum I meanwhile pass through hypergeometrics:

  convert(Ei(1,x),hypergeom): convert(%,Sum, dummy=n);

                              /infinity                 \
                              | -----        n  (n + 1) |
                              |  \       (-1)  x        |
             -gamma - ln(x) + |   )     ----------------|
                              |  /      (n + 1)! (n + 1)|
                              | -----                   |
                              \ n = 0                   /

which is the same as given by Robert or Alejandro or yours.

I do not like to use Standard interface quite much (even if the
graphics are sometimes much nicer), but here is an example how
it should not reply:
 

As one sees the axes have "1, * 10^7" etc. Thus it seems to use
local language settings (mine is German, of course, Win NT SP2),
while 1.0*1e-7 is properly displayed after hitting enter.

To cross check just change country settings and the same sheet
will display it differently (without re-running).

And I would expect "1.0*1e-7" in the graphics (not a missing 0).
 

As a new version appeared with upgrades:

  restart: interface(version); Digits:=14:

    Classic Worksheet Interface, Maple 12.02, Windows, Dec 10 2008 Build ID 377066

  Ei(1,1/2*Pi*(1+2*k)):
  %=convert(%,Sum);
  subs(k=0,%);
  evalf(%);
 
                             Pi            Pi
                      Ei(1, ----) = GAMMA(----)
                             2             2


                 0.090082461803865 = 0.89056089038154

Not that funny.

"The Handbook of Essential Mathematics" at edoutreach.wpafb.af.mil/ed_outreach/pages/ed-resources/math_resources.html

"The Handbook of Essential Mathematics" at edoutreach.wpafb.af.mil/ed_outreach/pages/ed-resources/math_resources.html

Thank you, it is more to process data from pages related to one I gave (and usually did that with Excel, it is somewhat more adeaquate for organizing data).

Just wanted to get an impression how to do it in Maple.

Thus your application is fine as an exmaple as well (for grabbing a nice way is to have a depot at an online bank and access it, it was also possible through Excel in most cases, if the password could be sent as part of the URL).

Yes, the solutions are equal I think over the whole unit square (p,q) in (0,1)^2 
and the diagonal restriction is not needed.

For this you 'only' have to show the following (difference of both solutions) is
zero in that domain, since as a function in q it is regular off q=0 on the Reals
and the 4th derivative w.r.t. q vanishes:

-1/6*q^2*(p^3*(-ln(-q*p)+ln(-p*(2-p)^2*q/(-2+p)^2))+p*ln((2-p)^24/(-2+p)^24)+
p^2*ln((-2+p)^12/(2-p)^12)+ln(1/(2-p)^16*(-2+p)^16));

For a 'proof' just plot it, plot3d(%, p=0..1, q=0..1, axes=boxed);

Yes, the solutions are equal I think over the whole unit square (p,q) in (0,1)^2 
and the diagonal restriction is not needed.

For this you 'only' have to show the following (difference of both solutions) is
zero in that domain, since as a function in q it is regular off q=0 on the Reals
and the 4th derivative w.r.t. q vanishes:

-1/6*q^2*(p^3*(-ln(-q*p)+ln(-p*(2-p)^2*q/(-2+p)^2))+p*ln((2-p)^24/(-2+p)^24)+
p^2*ln((-2+p)^12/(2-p)^12)+ln(1/(2-p)^16*(-2+p)^16));

For a 'proof' just plot it, plot3d(%, p=0..1, q=0..1, axes=boxed);

Alec, thank you - seems I am a bit too formalistic in viewing at problems ...

BTW: I wish you a Good New Year! Axel

Alec, thank you - seems I am a bit too formalistic in viewing at problems ...

BTW: I wish you a Good New Year! Axel

Thank you - but how you see it?