Research in definite integration
Vladimir Bondarenko
http://mapleadvisor.com/cgi-bin/moin.cgi/Maple/Experts
Jacques Carette writes
http://www.mapleprimes.com/blog/alec/discontinuous-antiderivatives#comment-4353
"Research in definite integration seems to have stopped, yet
it is an area with lots and lots of interesting open problems."
Any comments?
Cheers,
Vladimir Bondarenko
Co-founder, CEO, Mathematical Director
http://www.cybertester.com/ Cyber Tester Ltd.
---------------------------------------------------------------
"We must understand that technologies
like these are the way of the future."
---------------------------------------------------------------
G. A. Edgar
<a6d4e577-702d-441e...@w24g2000prd.googlegroups.com>,
Vladimir Bondarenko <v...@cybertester.com> wrote:
> A Maple Expert,
>
> http://mapleadvisor.com/cgi-bin/moin.cgi/Maple/Experts
>
> Jacques Carette writes
>
>
> http://www.mapleprimes.com/blog/alec/discontinuous-antiderivatives#comment-435
> 3
>
> "Research in definite integration seems to have stopped, yet
> it is an area with lots and lots of interesting open problems."
>
> Any comments?
>
True. Except perhaps it should be amended to "an area with lots and
lots of open problems, of which a few are interesting" ... For
application, compared to a numerical value, there is little interest in
getting a five-line expression involving polylogs or hypergeometrics.
> Cheers,
>
> Vladimir Bondarenko
>
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
rjf
wrote:
>
> > "Research in definite integration seems to have stopped, yet
> > it is an area with lots and lots of interesting open problems."
>
> > Any comments?
>
> True. Except perhaps it should be amended to "an area with lots and
> lots of open problems, of which a few are interesting" ... For
> application, compared to a numerical value, there is little interest in
> getting a five-line expression involving polylogs or hypergeometrics.
The lack of progress on some interesting problems of algorithms for
symbolic definite integration is probably due to the fact that they
are nasty to study and solve. (And as GAE points out, numerical
results are much easier to get and often at least as useful.)
Topics like simplification of hypergeometric functions are sometimes
cleaner approaches to similar problems, but frankly, what amounts to
the theory of (many) complex variables is just hard.
Why do so many people study fundamentally uninteresting problems these
days?
(Why do I say this is happening? ....If you want to find areas with
lots and lots of uninteresting open symbolic math problems, just look
at some journals.)
Often a paper announces a problem, even though there is no external
or plausible internal (in the paper) motivation for it.
The paper then solves the "low hanging fruit" parts of the problem --
for example doing very little other than defining the problem
carefully, and giving it a name. It then announces open problems,
making it seem that the authors have identified an important kernel of
interest and one that has extensions of some sort, broadening the
horizon of research.
This allows a student or friend of the author(s)to refer to it,
reinforcing the idea that the problem is a worthy object of study,
when it is mostly an excuse for 2 friends to write about the same
uninteresting topic. Sometimes the author, once he gets a foothold of
one paper, can continue to write papers on the topic indefinitely, and
doesn't even need a friend (except perhaps for the editor who let the
first paper be published.)
The fact that the topic is essentially of no intrinsic interest is
irrelevant to all participants, who may not even be aware that there
is anything odd about all this study and paper-writing etc. It is
just the way things are done, and how one gets papers published,
achieves tenure, etc.
In my own thesis research (c. 1969), I met with John Tate, a very
distinguished Harvard mathematician, about results related to a
theorem of Chebotarev that I was hoping to use to simplify radicals in
a program in Macsyma. His incisive suggestion was to change the
problem to something that was different -- a problem that could indeed
be addressed far more effectively. The only difficulty was that the
altered problem was of no use whatsoever, since simplifying radicals
over a field of characteristic 1 was useless. The simplifications I
needed were, (incidentally, to close the loop on this message) ,
needed for constructing minimal algebraic extensions for symbolic
integration.
Happy new year, the first year of the post-cyber tester era.
RJF
PS I suppose I, or someone else, could now start writing papers on
radicals of ... noting that this topic is of interest having been
raised by Prof. Fateman (or far better, Prof. Tate!)
Vladimir Bondarenko
wrote:
> In article
> <a6d4e577-702d-441e-9294-6bba9f50b...@w24g2000prd.googlegroups.com>,
>
>
>
> Vladimir Bondarenko <v...@cybertester.com> wrote:
> > A Maple Expert,
>
> > http://mapleadvisor.com/cgi-bin/moin.cgi/Maple/Experts
>
> > Jacques Carette writes
>
> > 3
>
> > "Research in definite integration seems to have stopped, yet
> > it is an area with lots and lots of interesting open problems."
>
> > Any comments?
>
> True. Except perhaps it should be amended to "an area with lots and
> lots of open problems, of which a few are interesting" ... For
> application, compared to a numerical value, there is little interest in
> getting a five-line expression involving polylogs or hypergeometrics.
>
> > Cheers,
>
> > Vladimir Bondarenko
>
> --
> G. A. Edgar http://www.math.ohio-state.edu/~edgar/
G. A. Edgar writes
GAE> For application, compared to a numerical value, there
GAE> is little interest in getting a five-line expression
GAE> involving polylogs or hypergeometrics.
Up to an esthetic value, I almost agree.
I hope you do not mean here the cases when the integrand
has parameter(s), for in these cases to have a relatively
compact formula could be better as compared to a library
of volumes of the numerical values.
Yet another point. The other day, for the integral
int(arctan(z)*arctan(2*z)/(1+z)^2, z= 0..infinity);
an answer involving 7 polylogs was reported
http://groups.google.com/group/sci.math.symbolic/msg/a0c376a876b2c57f
1/160 (-64 Catalan + (25 - (10 I)/3) Pi^2 - (20 + 20 I) *
ArcTan[4/3]^2 + 4 Pi (5 I ArcTan[4/3] +
ArcTan[3116/237] - 12 Log[2] +
18 Log[3] - 5 Log[5]) + 40 I PolyLog[2, -(3/5) + (4 I)/5] +
10 PolyLog[2, 1/81] - 48 PolyLog[2, 1/9] +
40 PolyLog[2, 1/5 - (4 I)/15] + 40 PolyLog[2, 1/5 + (4 I)/15] -
48 PolyLog[2, 1/3] + 40 I PolyLog[2, 3/5 + (4 I)/5])
and soon, an answer involving only 2 polylogs was reported
http://groups.google.com/group/sci.math.symbolic/msg/9514186439b41102
(1/40)*(-16*Catalan + 6*Pi^2 - 6*Log[2]^2 + 10*ArcTan[2]*
Log[5] + Log[8]*Log[9] - 2*Pi*Log[200000/19683] -
10*I*(PolyLog[2, 1/5 - (2*I)/5] - PolyLog[2, 1/5 + (2*I)/5]) -
3*PolyLog[2, 1/4])
From my experience, by and large, it is not totally unusual
for a "reasonably looking" 1-D definite integral to find a
reasonably compact exact answer.
Also, from time to time, somehow a thing with no real value in
eyes of engineers or physicists becomes a thing of practical
importance.
Say, a good physicist can recognize easily an experimental error.
"3 is a prime, 5 is a prime, 7 is a prime, 9 is an experimental
error, 11 is prime, 13 is prime..."
Yet number theory brought to the world RSA encryption.
http://mathworld.wolfram.com/RSAEncryption.html
A vague, badly formulated question, How do you feel,
How typical is that the answer to a non-parametric 1-D
definite integral with a "nice looking" integrand can
be expressed in terms of sum of many ( >=2 ) polylogs,
and not via only 1 or 2 polylogs?
rjf
>
> I hope you do not mean here the cases when the integrand
> has parameter(s), for in these cases to have a relatively
> compact formula could be better as compared to a library
> of volumes of the numerical values.
>
A plot (not a table of numerical values) of a numerical integral
with respect to one or more parameters may be quite useful. Also,
a fast numerical evaluation routine that will find an
approximation for any desired values of parameters would be useful.
Indeed, for
a complicated formula it is likely that either of these would provide
an important step in understanding it, where the routine can be used
for
interactive numerical exploration, and the plot gives some graphical
insight.
The formula per se, especially if it is 7 or 9 or ... polylogs etc,
will present difficulties.
When is a formula interesting? A very small subset of what you seem
to
think of as interesting:
1. if the question comes from some actual need, and is not just
something pasted together by random
selection by some computer heuristic.
2. the answer is small, and surprisingly so.
Regarding any answer involving polylogs, whether 7 or 9 or 2...
Can you understand that this is not very interesting?
Even if it is interesting to 5 people, it is not very interesting.
Herman Rubin
G. A. Edgar <ed...@math.ohio-state.edu.invalid> wrote:
><a6d4e577-702d-441e...@w24g2000prd.googlegroups.com>,
>Vladimir Bondarenko <v...@cybertester.com> wrote:
>> A Maple Expert,
>> http://mapleadvisor.com/cgi-bin/moin.cgi/Maple/Experts
>> Jacques Carette writes
>> http://www.mapleprimes.com/blog/alec/discontinuous-antiderivatives#comment-435
>> 3
>> "Research in definite integration seems to have stopped, yet
>> it is an area with lots and lots of interesting open problems."
>> Any comments?
>True. Except perhaps it should be amended to "an area with lots and
>lots of open problems, of which a few are interesting" ... For
>application, compared to a numerical value, there is little interest in
>getting a five-line expression involving polylogs or hypergeometrics.
A five-line expression involving polylogs or hypergeometrics
is likely to be far easier to evaluate to good accuracy than
it would be to carry out a numerical integration; this is for
one dimension, In high dimensions, nothing is easy, but
evaluating functions instead of approximate integration is
likely to be even better.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
rjf
..
>
> A five-line expression involving polylogs or hypergeometrics
> is likely to be far easier to evaluate to good accuracy than
> it would be to carry out a numerical integration; this is for
> one dimension, In high dimensions, nothing is easy, but
> evaluating functions instead of approximate integration is
> likely to be even better.
> --
Maybe in this case but not always.
consider the integral of 1/(1+z^64). The explicit antiderivative in
Mathematica is a sum of 64 terms.
A typical term (this is the 32nd one) looks like
1/64 cos((15 Pi)/64) log(z^2+2 cos((15 Pi)/64) z+1)
So evaluating this formula would require computing over a hundred
constants involving sines and cosines,
and then for each evaluation point, 64 quadratics in z, a log, and 64
multiplies...
whereas evaluating the integrand takes about 5 multiplies.
There is no reason to think that evaluating the formula would be
either fast or accurate.
And you could easily construct an integrand with a number larger than
64 :)
RJF
Herman Rubin
rjf <fat...@gmail.com> wrote:
>> A five-line expression involving polylogs or hypergeometrics
>> is likely to be far easier to evaluate to good accuracy than
>> it would be to carry out a numerical integration; this is for
>> one dimension, =A0In high dimensions, nothing is easy, but
>> evaluating functions instead of approximate integration is
>> likely to be even better.
--
>Maybe in this case but not always.
>consider the integral of 1/(1+z^64). The explicit antiderivative in
>Mathematica is a sum of 64 terms.
>A typical term (this is the 32nd one) looks like
>1/64 cos((15 Pi)/64) log(z^2+2 cos((15 Pi)/64) z+1)
>So evaluating this formula would require computing over a hundred
>constants involving sines and cosines,
>and then for each evaluation point, 64 quadratics in z, a log, and 64
>multiplies...
The number of constants needed may be that high, but if one
uses the relations between them, the number of computations
is not that great.
>whereas evaluating the integrand takes about 5 multiplies.
I see no way to get it in less than 6 multiplies plus one
divide, and divisions are slow.
However, unless one needs to evaluate the integral near 1,
a few terms in the power series expansion will do an excellent
job. And if one needs it there, it might still be necessary to
compute 100 terms to handle numerical integration.
>There is no reason to think that evaluating the formula would be
>either fast or accurate.
>And you could easily construct an integrand with a number larger than
>64 :)
This particular function is like 1 for z a little less than 1,
and like 1/z^64 for z a little greater than 1, with the above
modifications. Numerical integration is going to have to handle
a lot of terms near z=1 to resolve the mess. I am not sure if
a continued fraction will not do better.
One can construct problems with any type of mess; numerical
analysis is an art, and doing art requires an artist.
>RJF
Herman Rubin
Herman Rubin <hru...@odds.stat.purdue.edu> wrote:
>In article <9f851a12-766d-46a0...@v18g2000pro.googlegroups.com>,
>rjf <fat...@gmail.com> wrote:
>>On Jan 18, 2:03=A0pm, hru...@odds.stat.purdue.edu (Herman Rubin) wrote:
>>..
>>> A five-line expression involving polylogs or hypergeometrics
>>> is likely to be far easier to evaluate to good accuracy than
>>> it would be to carry out a numerical integration; this is for
>>> one dimension, =A0In high dimensions, nothing is easy, but
>>> evaluating functions instead of approximate integration is
>>> likely to be even better.
> --
>>Maybe in this case but not always.
>>consider the integral of 1/(1+z^64). The explicit antiderivative in
>>Mathematica is a sum of 64 terms.
The title of this thread is "Research in definite integration."
The emphasis in Fateman's paper, and also in my remark, is
on indefinite integration. When we look at this function
from the standpoint of definite integration, the only
somewhat reasonable endpoints are 0, +-1, and +-infinity.
The function is even, and the integral from 0 to infinity
requires one sine function. From 0 to 1, or 1 to infinity,
it can be done as a difference of two psi functions.