mm-182
===
Subject: Re: topology and analysis/differential
geometry> Can anyone of you provide me with information about
the link between> topology and analysis / differential
geometry (maybe give me some> links or reference to good
books)?Well, a link between topology and analysis/Diff
Geometry is theAtiyah-Singer theorem. You could look at
Booss/Bleecker:Topology andAnalysis, 1985ish. If that interest
you start reading the CollectedWorks of Atiyah. He is a very
good expositor.-- Maarten Bergvelt
===
Subject: Re: penny on a
corner problemOriginator: tchow@markov.mit.edu.mit.edu
(Timothy Chow)>I think this is discussed in Littlewood's
Miscellany, but I'd have to check>to be sure.I checked and
it's not there. I think that what happened is that I
wasreading the Miscellany around the same time that I was
looking at someMathematical Tripos papers. I now believe that
the penny in the cornerproblem was a relatively recent Tripos
problem, but I just tried to 'ndit and failed. Sorry for the
confusion.-- Tim Chow tchow-at-alum-dot-mit-dot-eduThe range
of our projectiles---even ... the artillery---however great,
willnever exceed four of those miles of which as many thousand
separate us fromthe center of the earth. ---Galileo, Dialogues
Concerning Two New Sciences
===
Subject: integration is
harderEpigone-thread: zherdquuglamInirgration in general is
harder. For derivatives, there are manystraightforward rules
that can be used for complicated expressions,e.g. derivatives
of products, quotients, functions of functions, etc. For
inde'nite integrals, there are no such general
rules.
replied to my original 2/15/04 post. I was lessinterested in a
solution to this very nice problem than in trying to traceits
history and antecedents, and I am grateful to Bruce Reznick
for pointingout that this was a problem on the 1948 Putnam
Competition.--J. Wetzel
===
Subject: Full Modular Group
Maximal?Is the group SL(n,Z) maximal (as an discrete group) in
SL(n,R)? Theanswer is yes for n=2 and one would imagine it is
also for any n,however, I don't know how to prove
it.
the very nice post! Lots of good information there...> Well,
you can write the general solution of your example as > a(n) =
2 p_n(b) a(0)/n! + q_n(b) a(1)/n! where > p_n(b) = b p_{n-1}(b)
+ n (n-1) p_{n-2}(b) and similarly for q,> p_0(b) = 1/2, p_1(b)
= 0, q_0(b) = 0, q_1(b) = 1. For n >= 2,> p_n and q_n are monic
polynomials of degree n-2 and n-1 respectively,> with
nonnegative integer coef'cients;> p_n is an even function if 
n
is even and an odd function if n is odd,> q_n is an odd
function if n is even and an even function if n is odd.> Just
out of curiosity, how much of the work 'nding this
generalsolution was a cookbook receipe (that I'm evidently 
not
aware of!),and how much was good guessing, the weight of
experience, etc., onyour part? As I mentioned, there are other
sequences I am working onwith a similar form, but different
functions F(n), so I'd like to havea good method to solve
these generally.> This has singularities (in general, branch
points) at x = 1 and x = -1.> Now basically a singularity like
(x-1)^p, (x+1)^p, (x-1)^p ln(x-1) > or (x+1)^p ln(x+1) with p <
-1 will result in terms whose absolute> value goes to in'nity
as n -> in'nity; p > -1 will make terms go to 0; > (x-1)^(-1)
has terms that are constant, (x+1)^(-1) has terms that >
oscillate but are bounded. > Let me rephrase what you did with
the generating function, to see if Iunderstand this... Once you
get this function f(x), you look at itsanalytic behavior, to
see whether there are problems in any region inthe parameter
space of b, a(0) and a(1).The variable x in itself has no
singularity at x = +1, -1 exposes potentialproblems in the
choice of parameters. For example, as you showed,looking at x
= -1 examines the region b < 0; speci'cally, b = -2means you
have to choose a(0) and a(1) to satisfy the relation a(1) =[
(3 - 4 ln(2)) / (2 ln(2) - 1}] a(0) for a(n+1) - a(n) to be
boundedas n -> in'nity. The singularity at x = +1 highlights
the behaviorfor b > 2.By the way, does that mean there are no
worries in the interval 0 < b< 2? That a(n+1) - a(n) is
the help!-- Daniel
===
Subject: Re: Fibonacci-type
message>Just out of curiosity, how much of the work 'nding
this general>solution was a cookbook receipe (that I'm
evidently not aware of!),>and how much was good guessing, the
weight of experience, etc., on>your part? As I mentioned,
there are other sequences I am working on>with a similar form,
but different functions F(n), so I'd like to have>a good 
method
to solve these generally.There's a general theory of 
difference
equations. In particular thegenerating function is related to
the Z-transform (by z = 1/x). The techniques are pretty much
standard I think. There are also computeralgebra tools for
this, e.g. the Maple package LREtools has a procedureREtoDE
that will compute a differential equation for the generating
function.>By the way, does that mean there are no worries in
the interval 0 < b>< 2? That a(n+1) - a(n) is bounded for all
choices of b, a(0) and>a(1)?Yes, I think that is correct.
Robert Israel israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Image of exp in
that exp(Mn(C)) =GLn(C). But, what is the image of exp in
Mn(R), i.e what is exp(Mn(R))?It is quite obvious that
exp(Mn(R)) is a subset of GLn+(R) = { A inMn(R), det(A)>0}.
But this is a *strict* subset if n=>2.Does anyone know
precisely what exp(Mn(R))is ? Is there a method toknow if a
G.
exponential of matrices. It is known that exp(Mn(C)) => GLn(C).
But, what is the image of exp in Mn(R), i.e what is exp(Mn(R))>
?> It is quite obvious that exp(Mn(R)) is a subset of GLn+(R) =
{ A in> Mn(R), det(A)>0}. But this is a *strict* subset if
n=>2.> Does anyone know precisely what exp(Mn(R))is ? Is there
a method to> know if a matrix is in the image of exp ?If the
eigenvalues of X are l_1,...,l_nthen the eigenvalues of A=e^X
are e^{l_1},...,e^{l_n}.It follows that negative real
eigenvalues of Amust occur with even multiplicity.I think this
is a necessary and suf'cient condition,as one can (probably)
see by expressing X = S + N,where S is semisimple, N is
nilpotent and SN=NS.Similarly A = TU where T is semisimple and
U is unipotent, and TU=UT.I think this reduces the problem to
the semisimple and unipotent cases,and I think one can
determine log U for unipotent matrices.I could be wrong.--
Timothy Murphy e-mail (<80k only): tim /at/
birdsnest.maths.tcd.ietel: +353-86-2336090,
+353-1-2842366s-mail: School of Mathematics, Trinity College,
Dublin 2, Ireland
===
Subject: Re: constructing Hadamard
about it a little> more before I read your response and am
happy that I came to the same> conclusion that you did that
gradient ascent methods would be a great> idea for
experiments. I am currently getting programs from Numerical>
Recipes in C together to try this. Lots of local extrema is
the only> barrier to this, but one can never know the extent
of the barrier> unless one tries it out.Craig,I'd be
interested to hear the outcome. For comparison, all of
theascent algorithms I'm aware of that work on the discrete
space seem toencounter insurmountable dif'culties somwhere
between order 20 andorder 24.Will
===
Subject: special
hyperplane sectionsHello --Is it known which degree d plane
curves can be hyperplane sections ofsmooth degree d
hypersurfaces in P3? There are some result about
higherdimensional varieties as hyperplane sections due to
Badescu and others,but they use Lefschetz theory for Picard
groups, which won't work in lowdimension. Does this mean the
lower dimensional case is harder, or is itsome classical
result which I just don't know?
Michael A. Van OpstallPadelford
C-113opstall@math.washington.eduhttp://www.math.washington.edu
/~opstall/
===
Subject: Classi'cation of Banach spacesHi all,Are
there any results on the classi'cation of Banach spaces
withinisometric isomorphism? For 'nite dimensional spaces
('xing thedimension) the l^p spaces, for pin [1, infty] are
pairwisenon-isometric isomorphic. Are there any results on the
isometricclassi'cation for 'nite dimensional 
Banach spaces?With
===
Subject: Re: Classi'cation of
Banach spaces>Are there any results on the classi'cation of
Banach spaces within>isometric isomorphism? For 'nite
dimensional spaces ('xing the>dimension) the l^p spaces, for
pin [1, infty] are pairwise>non-isometric isomorphic. Are
there any results on the isometric>classi'cation for 
'nite
dimensional Banach spaces?If U is any bounded convex balanced
(i.e. -U = U) open set, the Minkowskigauge p_U(x) = inf{s: x
in sU} is a norm, and conversely any norm isthe Minkowski
gauge for its unit ball. So the isometric classi'cationfor
'nite-dimensional Banach spaces is equivalent to the
classi'cationof bounded convex balanced open sets under 
linear
transformations.I think there is too much freedom here to have
any really useful parametrization. Robert Israel
israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Paper published by
Algebraic and Geometric TopologyThe following paper has been
paper no. 6, pages
81--93URL:http://www.maths.warwick.ac.uk/agt/AGTVol4/agt-4-6.
abs.htmlTitle:A functorial approach to differential
charactersAuthor(s):Paul TurnerAbstract:We describe
Cheeger-Simons differential characters in terms of avariant of
Turaev's homotopy quantum 'eld theories based 
on chains ina
57R56Secondary: 53C05, 81T15, 58A10Keywords:Differential
character, Homotopy quantum 'eld theoryReceived: 6 February
address(es):School of Mathematical and Computer Sciences
Heriot-Watt University Edinburgh EH14 4AS ScotlandEmail:
paul@ma.hw.ac.uk
===
Subject: alternative analysis generated by
DEEpigone-thread: snumglaizhalFew people expressed some
interest in the topic. To give betterpicture, I copied my
message from the physics forum. Here it is.Since I am not a
physisist but work in abstract differentialequations, I would
like to address this message to knowledgeablepeople working in
or related to theoretical physics, namely, 'eld
andspace-timetheories. I was told that the following could
most probably be o'nterest to theoretical physisists and I am
trying to look into thisin more detail. I need your advice on
what physical problem should Ilookinto or read about. I am
trying to 'nd connections of interest.It was shown that under
mild conditions a differential equation (ODEas well as
quasi-linear PDE) admits in'nitely many binary
laws(superpositions) of addition of its solutions. These and
relatedpublished results can be found
/www.math.u-szeged.hu/ejqtde/2001/200106.htmlThe equation does
not have to be linear. Further, these laws ofaddition are used
to re-de'ne 'eld axioms of complex numbers 
incertain
alternative form (roughly speaking, different from
standardaddition, multiplication etc...) which allows to 
de'ne
analternative derivative and all other objects of
calculus.Basically, an alternative calculus can be
constructed. Theinteresting fact is that if written in its own
(alternative) calculus,the differential equation has a linear
form, even if it is non-linearin the standard calculus. Also,
standard and alternative calculi arealgebraically related to
each other, so for the sake of completenessboth calculi have
to somehow be taken into consideration. This has its toll, the
construction of alternative calculus involves'ber spaces of a
certain covering map, so that the alternativecalculus is
really a calculus on 'bers.In summary, every DE has its own
alternative calculus in which itappears to be linear.
Similarly, alternative matrix algebras,metric tensors,
geometries can be constructed. Alternative andstandard calculi
are in homomorphic (sometimes isomorphic) relations.The
question is, What is the use of the alternative calculus
in
===
Subject: Re: why is integration harder than
differentiation> someone asked this question, and i have been
thinking about it for a > while. why is integration harder
than differentiation?> what we really want to know is why,
given a function built out of > certain elementary functions,
it is easy to construct the derivative > in terms of those
elementary functions, but hard (and sometimes > impossible) to
construct the antiderivative in terms of those elemntary >
functions.> from a practical standpoint, the reason is
clear: there are rules for > the derivatives of the two
constructions you can do to elementary > functions, namely the
product and the composition. if there were a rule > for the
integral of the composition of two functions and for the
product > of two functions, then from that, we could write any
integrals of > elementary functions in terms of elementary
functions.> but we cannot. so why not? what is different
about integration that > makes it not have these rules? The
rules for differentiation write a derivative in terms of
derivativesof simpler pieces (usign fewer letters to represent
them), hence thedifferentiation of any expression is a 'nite
process.Already integration by part (the product rule for
integration)does not have this property.> i wanted to mutter
something about differential Galois theory, but i > think that
would just have been a cover for the more honest i don't
know.The theory of differential 'elds gives a fairly precise
answer ofwhich classes of expressions can be integrated by
elementary functions(and, if desired, a 'nite number of new
functions de'ned as solutionsof differential equations -
generalizing the de'nitions of exp, sin, etc.)Thus it
delineated the limits of symbolic integration
packages.Numerically, integration is simpler than
differentiation in one dimension,but in higher dimension,
integration suffers from the curse of dimensionalitywhile
differntiation doesn't. In particular, it is very hard to
getaccurate integrals in dimensions >100, say.This has to do
with the global aspects of integration and the
in'nitesimalaspect of differentiation.Arnold
Neumaier
===
Subject: Re: why is integration harder than
>>someone asked this question, and i have been thinking about
it for a >>while. why is integration harder than
differentiation?>>'rst of all, is it true? integration is
de'ned on a larger class of>>functions than differentiation,
so in some sense, it is easier to show>>the existence of the
integral than the derivative.> If you are dealing with a
'nite precision machine, then integration is > actually 
easier
than differentiation. Integration is a stable process > while
differentiation is ill-posed.> I'll leave it to you to 
'nd
out why (hint: think about the de'nitions).> - Tim> i am not
sure right away what the answer to your question is; something
about needing more signi'cant digits to get an accurate
difference between two close numbers, than you do to take the
sum of a bunch of numbers?thinking about this, it struck me as
surprising that derivatives are de'ned as in terms of
subtraction and division, which are inverse operations in
algebra, whereas (Riemann) integration is de'ned in terms of
multiplication and summation, which are 
``primitive''
operations algebraically, so to speak. and yet differentiation
turns out to be the primitive operation on functions, and
integration the inverse operation.
===
Subject: Re: why is
integration harder than differentiationZig
this question, and i have been thinking about it for a
>while. why is integration harder than
differentiation?>>'rst of all, is it true? integration is
de'ned on a larger class>of functions than differentiation,
so in some sense, it is easier to>show the existence of the
integral than the derivative.>>> If you are dealing
with a 'nite precision machine, then integration>> is 
actually
easier than differentiation. Integration is a stable>> process
while differentiation is ill-posed.>>> I'll leave it to you
to 'nd out why (hint: think about the>> 
de'nitions). >>> -
Tim>> i am not sure right away what the answer to your
question is; > something about needing more signi'cant digits
to get an accurate> difference between two close numbers, than
you do to take the sum of a> bunch of numbers?> thinking
about this, it struck me as surprising that derivatives are >
de'ned as in terms of subtraction and division, which are
inverse > operations in algebra, whereas (Riemann) integration
is de'ned in> terms of multiplication and summation, which 
are
``primitive''> operations algebraically, so to 
speak. and yet
differentiation turns> out to be the primitive operation on
functions, and integration the> inverse operation. > It is
exactly these reasons that on a computer, differentiation is
harder than integration. If you look at the de'nition of the
derivative,lim(dx->0) [f(x+dx)-f(x)]/dx, if dx really is
small, almost zero, you are subtracting two almost equal
values f(x+dx) and f(x). This result is then also close to
zero so you lose many, many degrees of accuracy. Then, you are
dividing a number that is almost zero by another number that is
almost zero. This again leads to a lose of many degrees of
accuracy. In fact, depending on the function f(x), it could
blow up towards positive in'nty or negative 
in'nity, with only
a small change in how you put it into your
computer.Integration, on the other hand, is just addition,
which for the most part, does not lose any accuracy (sure
there are some cases) and multiplication which, depending, can
increase accuracy (and occasionally decrease
it).Differentiation is an example of an ill-posed or unstable
problem. Maybe a more concrete example, think of tan(3^x) For
x=1.411, it is positive, nearly 4000. For x=1.412 it is
negative, about -200. Just a small difference in the x and
there is a huge difference in the answer. Differentiation, for
a computer, is alot like this.Here's something else to think
about, when you learned about the exponential and natural
logarithm, which did you learn 'rst and how was the other one
presented to you? Most cases you learn exp 'rst then ln as 
the
inverse function of exp. But, this is misleading. In fact, exp
is the inverse function of ln. Why? ln exists as a derivative
of 1/x whether exp exists or not.But, don't think about it 
too
much. - Tim---Timothy M. BrauchGraduate StudentDepartment of
MathematicsWake Forest University
===
Subject: Re: why is
integration harder than differentiation> Zig
with a 'nite precision machine, then integration>is actually
easier than differentiation. Integration is a stable>process
while differentiation is ill-posed. > If you look at the
de'nition of the derivative,> lim(dx->0) [f(x+dx)-f(x)]/dx, 
if
dx really is small, almost zero, you are > subtracting two
almost equal values f(x+dx) and f(x). This result is then >
also close to zero so you lose many, many degrees of accuracy.
Then, you > are dividing a number that is almost zero by
another number that is almost > zero. This again leads to a
lose of many degrees of accuracy. More precisely: The
subtraction is done without error but it turns thesmall
relative error in the function values into a small absolute
but largerelative error. The division has a small relative
error, and has littleeffect on the relative error; but it
turns the small absolute errorinto a large absolute error. The
result is useless.Nevertheless, one can get full accuracy
numerical derivatives usingextrapolation procedures, as
described in my numerical analysis book.One just needs
care.Arnold Neumaier
===
Subject: Re: why is integration harder
than differentiation> Concerning the dif'culty> I guess we
have to differentiate [no pun intended] between the de'nite>
integral which is always with respect to a given set in the
function's> domain and the process of inde'nite 
integration
which is just the reverse> of differentiation. In an algebraic
sense I don't think that inverse> differentiation is all 
that
much more dif'cult than forward> differentiation. We have a
set of rules to apply to elementary functions> and algebraic
(and some transcendental) combinations of them to>
differentiate a given function. If none of our usual rules
apply, then> differentiation is -- at least symbolically --
impossible and we're stuck> with a numerical 
approximation.but
this is exactly the issue; differentiation never gets stuck in
this way. given any 'nite composition of elementary 
functions,
i can use the chain rule and product rule to algorithmically
reduce this completely. i can differentiate any such function
with impunity.not so with integration (or inverse
differentiation, if you would like to make the distinction).>
Ditto the process of inverse> differentiation. This is
generally thought to be dif'cult because most> people simply
don't recognize the inverse rules as readily as the forward>
ones. There are tables to use in both cases.> Concerning the
inverse relationship> Assuming that we de'ne the 
inde'nite
integral in terms of the de'nite> integral, 
it's relatively
easy to see that integration doesn't really care> if a
function has, for example, discrete jump discontinuities, the
function> is still integrable. But that same function fails to
be differentiable at> the points of discontinuity so in that
case I'd have to say that the process> of integration is
easier than the process of differentiation.> Norm> yes, i
mentioned something about this in the original post. in some
sense, integration is easier, because more functions are
integrable than are differentiable. a lot more, i think. it is
easier for a function to have an integral than a derivative. if
we chose a function at random, it would be more likely to be
integrable than differentiable (although i suspect actually
both probabilities would be zero, eh?)anyway, maybe this is
also exactly the reason why the process of 'nding that
elementary antiderivative is harder. simply because the
antiderivative has to exist for a broader class of functions?
===
Subject: Re:
why is integration harder than differentiation> someone asked
this question, and i have been thinking about it for a >
while. why is integration harder than differentiation?> If you
have a sequence a(n), it is generally much easier to compute
its difference:a(n+1)-a(n)than the partial
sums:a(1)+a(2)+...+a(n).
===
Subject: Re: why is integration
harder than differentiation>>someone asked this question,
and i have been thinking about it for a >>while. why is
integration harder than differentiation?> If you have
a sequence a(n), it is generally much easier to compute its
difference:> a(n+1)-a(n)> than the partial sums:>
a(1)+a(2)+...+a(n).> this seems related to Norm Dresner's
suggestion that basically integration is nonlocal; you have to
know a lot more about the function. i wonder if this connection
could be made more explicit?
===
Subject: Re: why is integration
harder than differentiation> Zig <jhannon@wisc.REMOVEME.edu>
been thinking about it for a>>while. why is integration
harder than differentiation?> One major
difference is that differentiation is a local operation while integration 
involves 'nite intervals -- okay, measurable
sets, but> intuitively it is de'ned over an extended region
of the domain while> differentiation isn't.> Norm differentiation is local 
and integration is nonlocal. hmm.
i like that > idea, but i am not convinced. actually, didn't 
i
once read that one of > the amazing things about deRham
cohomology is that the differential > forms on a manifold,
which are all locally de'ned objects, can encode > global
information about the manifold? but i am not sure if that has
> any relevance here.> anyway, just thinking outloud here,
but the difference between > integration and differentiation
that makes the former hard and the > latter easy (in the
elementary calculus sense) is that differentiation > is a sort
of forward operation, and integration is an inverse >
operation, in a way that is not symmetric.> i think the
analogy with algebraic operations is good. anyone can > square
a small number in her head, but who can take the square root? >
solving the quadratic equation is hard, and solving certain
quintics is > impossible.> but why does differentiation have
to be the forward operation and > integration have to be its
inverse? could we make it go the other way? > in other words,
is there some fundamental property of integration that > rules
out the possibility of writing the integral of a product in
terms > of the integrals of the multiplicands? if we had a
rule like that, and > one for composition of functions, then
integration would be as > algorithmic as differentiation, and
all elementary functions would be known.> since i know that
not all elementary functions have elementary > integrals, then
i must conclude that such a product rule cannot exist. > but
this is a very indirect way to see this, and sort of relies on
the > property i am trying to understand to explain this
property. it is a > bit circular. is there an obvious reason
why there can be no > integration rule for products? (the
familiar product rule is not good > enough here)A few thoughts
... First, there is a rule of sorts for integratingthe product
of two functions, namely integration by parts. I'm notsure
whether this is something important in this context
butintegration doesn't give a unique answer, 
it's only unique
up to anadditive constant. Differentiation, when it can be
done at all givesa unique answer. Finally, if we restrict
ourselves to a suitableclass of functions then integration and
differentiation are equallydif'cult, each being a termwise
operation on power series.
===
Subject: Re: why is integration
harder than differentiation> A few thoughts ... First,
there is a rule of sorts for integrating> the product of two
functions, namely integration by parts.sure, but this is not
good enough. for example, consider the integralint
1/sqrt(b^2-x^2)*1/sqrt(a^2-x^2) dxthis is the product of two
functions, each of whose integrals i know in terms of
elementary functions (arcsine). and yet, the integral of their
product is not an elementary function (elliptic function).so i
know that integration by parts will never help me solve this
integral.compare this with differentiation, where i could
immediately write down the total derivative, knowing the
individual derivatives. this difference between integration
and differentiation makes differentiation an algorithm, and
integration an art form.but why the difference?> I'm not> 
sure
whether this is something important in this context but>
integration doesn't give a unique answer, it's 
only unique up
to an> additive constant. Differentiation, when it can be done
at all gives> a unique answer.yes. perhaps that has something
to do with it. it is a good suggestion.> Finally, if we
restrict ourselves to a suitable> class of functions then
integration and differentiation are equally> dif'cult, each
being a termwise operation on power series.> yes, of course,
if we view functions in terms of their power series, then
neither one is easier or harder. so i guess i am only talking
about 'nite compositions of elementary functions, not power
series.perhaps there is something unnatural about restricting
yourself to only talking about closed functions like that? but
differentiation doesn't care, why should
integration?
===
Subject: Re: why is integration harder than
differentiation <c0sqmp$720$1@news.doit.wisc.edu>
sha1:Msd5D8rCdx2seQXyJYsi2CJnOQw=> Zig == gowan
<gowan4@hotmail.com> told me: Zig> A few thoughts ... First,
there is a rule of sorts for integrating Zig> the product of
two functions, namely integration by parts. I'm not Zig> 
sure
whether this is something important in this context but Zig>
integration doesn't give a unique answer, it's 
only unique up
to an Zig> additive constant. Differentiation, when it can be
done at all gives Zig> a unique answer. Finally, if we
restrict ourselves to a suitable Zig> class of functions then
integration and differentiation are equally Zig> dif'cult,
each being a termwise operation on power series.A few thoughts
echoing some of the opinions/ideas already posted here.
Theclass of functions that we are willing to accept as giving
us *closed form*solutions limits our ability to consider
integrals as easy ordif'cult. The simplest examples being the
error function and certain logintegrals...That reminds me,
long ago somebody on this group had pointed out sometheory
that enables one to decide whether a particular integral
isevaluable (evaluatable?) in ``closed form,'' 
or not -- could
somebodyplease resend some of that information -- It might help
to shed light onthe dif'culty of integration!
Gr.9fe,-suvrit.
===
Subject: Re: why is integration harder than
differentiationOriginator: tchow@lagrange.mit.edu.mit.edu
(Timothy Chow)>That reminds me, long ago somebody on this
group had pointed out some>theory that enables one to decide
whether a particular integral is>evaluable (evaluatable?) in
``closed form,'' or not -- could 
somebody>please resend some
of that information -- It might help to shed light on>the
dif'culty of integration! A good reference is Symbolic
Integration I by Manuel Bronstein.I wonder if the question
being asked in this thread is a non-question.It seems to
presuppose that the class of elementary functions is theright
set of functions to consider. That is, we pick a class
offunctions that behaves well under differentiation, and
puzzle overwhy it doesn't behave well under
antidifferentiation. But given thatwe can pick other classes
of functions for which differentiationand antidifferentiation
are equally easy, why puzzle? Presumablyone can also pick
another class of functions that behaves well
underantidifferentiation and not under differentiation
(derivatives ofC^1 functions, say). So some classes have an
af'nity for one processand others have an 
af'nity for the
other process. Unless there'ssome reason to think that
elementary functions are canonical in somesense, rather than
an arbitrary artifact of notation, why would weexpect an
answer beyond what has already been stated?-- Tim Chow
tchow-at-alum-dot-mit-dot-eduThe range of our
projectiles---even ... the artillery---however great,
willnever exceed four of those miles of which as many thousand
separate us fromthe center of the earth. ---Galileo, Dialogues
Concerning Two New Sciences
===
Subject: Re: why is integration
harder than differentiation> That reminds me, long ago
somebody on this group had pointed out some> theory that
enables one to decide whether a particular integral is>
evaluable (evaluatable?) in ``closed form,'' 
or not -- could
somebody> please resend some of that information -- It might
help to shed light on> the dif'culty of integration! >
Gr.9fe,> -suvrit.> I think the math that tells you when some
integrals can be express in terms of elementary functions is
called differential Galois theory. this theory tells you, for
example, that the error function is not a 'nite composition 
of
elementary functions. it uses the same types of ideas as used
in showing that the quintic cannot be solved in terms of
simple root extractions.i don't know much about it beyond 
what
i have said here, so it will be cool if someone who knows a lot
about it weighs in.
===
Subject: Re: why is integration harder
>> That reminds me, long ago somebody on this group had
pointed out some>> theory that enables one to decide whether a
particular integral is>> evaluable (evaluatable?) in ``closed
form,'' or not -- could somebody>> please 
resend some of that
information -- It might help to shed light on>> the dif'culty
of integration! >>I think the math that tells you when some
integrals can be express in >terms of elementary functions is
called differential Galois theory. >this theory tells you, for
example, that the error function is not a >'nite composition 
of
elementary functions. it uses the same types of >ideas as used
in showing that the quintic cannot be solved in terms of
>simple root extractions.>i don't know much about it beyond
what i have said here, so it will be >cool if someone who
knows a lot about it weighs in.At this moment I have no
library at hand so I cannot be veryspeci'c. But I remember
something like the Rish method which isused in Maple and which
(here I have to rely on memory, so I may bewrong) enables one
(Maple?) to say whether of a function build fromelementary
functions is integrable or not.