mm-279
Subject: algebraic numbersIs the following statement
true?Given an algebraic number a in C (the complex numbers),
there exists aunique monic polynomial P of smallest degree
with rationalcoefficients such that(1) a is a root of P(2) P
contains no repea roots?It seems to me that this should be
true, for if you found such apolynomial with a repea root, you
could always take out that linearfactor and re-normalize by
multiplying by the appropriate rationalnumber.
===
Subject: Re:
algebraic numbers> Is the following statement true?> Given an
algebraic number a in C (the complex numbers), there exists a>
unique monic polynomial P of smallest degree with rational>
coefficients such that> (1) a is a root of P> (2) P contains
no repea roots?> It seems to me that this should be true, for
if you found such a> polynomial with a repea root, you could
always take out that linear> factor and re-normalize by
multiplying by the appropriate rational> number. Actually, the
situation is simpler: if P is monic and ofsmallest degree among
polynomials having a as a root, then ithas no repea rots
anyway. Why? Suppose P has repea roots. By rational
operations, you candivide P by the (monic) greatest common
divisor of P and itsderivative P'. The quotient will have the
same roots as P, eachexactly once. ZVK(Slavek)
===
Subject:
Intelligence and Induction for proofHere is what seems to
determine a person's mathematical intelligence, somethingmore
than mere mathematical learning:A typical preCalc book may
have topics in the order of arithmetic sequences,geometric
sequences, and then mathematical induction. The first two
topicsseem quite understandable; just find patterns and use
formulas for boththeoretical exercises and applied-type
excersises. Then at the engagement withthe induction topic,
something changes. Proving that if a k term gives anassumed
result, then the k+1 term should/must also give a correct
match to theformula. I am finding the process difficult, not
being sure what sub-stepswill be needed in the actual proof
process. I understand the basic idea:(1) the n=1 statement
must be true;(2) assume the k statement is true, and prove
that the k+1 statement is alsotrue.In part number (2), I am
unsure how to continue. This is not for anyparticular problem,
but for most possible problems. Is this something thatrequires
Mathematical intelligence beyond what most of the rest of
thePreCalculus textbook requires? I notice that in three
different preCalculusbooks, there is only one mathematical
induction section in each book. Thisseems that the induction
is worth only a small emphasis for a PreCalculuscourse. Is
this far off from the truth? Does math induction truely
requiremore mathematical intelligence than most of the rest of
PreCalculus? G C
===
Subject: Re: Intelligence and Induction for
proofGC-> Here is what seems to determine a person's
mathematical intelligence,something> more than mere
mathematical learning:> A typical preCalc book may have topics
in the order of arithmeticsequences,> geometric sequences, and
then mathematical induction. The first twotopics> seem quite
understandable; just find patterns and use formulas for both>
theoretical exercises and applied-type excersises. Then at the
engagementwith> the induction topic, something changes. Proving
that if a k term gives an> assumed result, then the k+1 term
should/must also give a correct match tothe> formula. I am
finding the process difficult, not being sure whatsub-steps>
will be needed in the actual proof process.> I understand the
basic idea:> (1) the n=1 statement must be true;> (2) assume
the k statement is true, and prove that the k+1 statement
isalso> true.> In part number (2), I am unsure how to
continue. This is not for any> particular problem, but for
most possible problems. Is this somethingthat> requires
Mathematical intelligence beyond what most of the rest of the>
PreCalculus textbook requires? I notice that in three
differentpreCalculus> books, there is only one mathematical
induction section in each book.This> seems that the induction
is worth only a small emphasis for a PreCalculus> course. Is
this far off from the truth? Does math induction
truelyrequire> more mathematical intelligence than most of the
rest of PreCalculus?Most induction problems at the introduction
do not require more intelligenceper se, but they do often
demand techniques and ways of thinking about maththat you may
not yet have encountered. Whereas most precalculus
isdeductive, i.e. trying to simplify expressions, many
induction problemsrequire backward thinking (and
un-simplifying expressions).Consider the following
assertion:For all integers n >= 1, 1+3^(2n-1) is divisible by
4.Testing the first case (usually called the base case),
1+3^[2(1)-1] = 1+3 = 4, which is divisible by 4Now, let's
assume this is true for n = k. Thus, there is some integer
msuch that 1+3^(2k-1) = 4mLet's consider the n = k+1 case: our
expression is then 1+3^(2(k+1)-1) = 1+3^(2k+1) =
1+(3^2)*3^(2k-1) = 1+9*3^(2k-1)Here's the trick: we write to
rewrite our expression to include 1+3^(2k-1) somewhere
(because we know that expression to be divisible by4, from our
n=k case). Since we have 3^(2k-1) multiplied by 9, let's add
9and subtract 9 from the total expression (you'll see why in a
moment): = 1+9*3^(2k-1)+9-9Factoring a 9 out of the second and
third terms, = 1+9*[1+3^(2k-1)]-9 = 9*[1+3^(2k-1)]-8Now, we
know 1+3^(2k-1) = 4m; so, substituting: = 9*4m-8Writing 8 as
4*2, = 4*9m-4*2 = 4*(9m-2)But this is still our expression
(albeit in disguise) for our n = k+1 case,and this is clearly
divisible by 4 (m is an integer, so 9m-2 is an integer).Thus,
we have both our base case true, and our recursive (or
iterative)statement true as well; by induction, our claim that
1+3^(2n-1) is divisibleby 4 is true. QEDHope this
helps.Travis
===
Subject: Re: Intelligence and Induction for
proof>I understand the basic idea:>(1) the n=1 statement must
be true;>(2) assume the k statement is true, and prove that
the k+1 statement is also>true.>In part number (2), I am
unsure how to continue. This is not for any>particular
problem, but for most possible problems. Is this something
that>requires Mathematical intelligence beyond what most of
the rest of the>PreCalculus textbook requires? Not really, no.
It's just that induction is no longer taught in Algebra I as it
used to be, so it seems strange and frightening.Think of
Chairman Mao: A journey of a thousand miles begins with a
single step. That single step is the n=1 case. Then assume
true for n=k, and prove that n=k+1 must also be true as a
result says wherever I am, I can take one step forward. If you
can always take a step forward, and you have a starting point,
then you can reach any possible point; in other words, the
statement must be true for all n.You say>This is not for any
particular problem, but for most possible problems.But really
the devil is in the details: One particular induction proof
won;t look like another because each one depends on what
you're trying to prove.here's one example: Prove that the sum
of 1+2+3+...+n = n(n+1)/2.Step 1: Is it true for n=1? Yes,
because 1 = 1(1+1)/2.Step 2: assume that it's true for n=k.
That is, assume that 1+2+3+...+k = k(k+1)/2Now, is it true for
n=k+1? Well, on the left-hand side it would be
1+2+3+...+k+(k+1), meaning that you add k+1 to the left-hand
side. But you can always make a true equation by adding the
same thing to both sides of an existing true equation. Since
you assumed the equation was true for n=k, if you add k+1 to
both sides it will still be true: 1+2+3+...+k = k(k+1)/2
1+2+3+...+k+(k+1) = k(k+1)/2 + (k+1)Now what would you have on
the right-hand side if the conjecture were true for n=k+1?
You'd have n(n+1)/2 but with (k+1) in place of n. In other
words, you'd have (k+1)(k+2)/2. Now, can you make the
right-hand side that you do have, k(k+1)/2 + (k+1), look like
(k+1)(k+2)/2? Yes, you can: 1+2+3+...+k+(k+1) = k(k+1)/2 +
(k+1) 1+2+3+...+k+(k+1) = k(k+1)/2 + (k+1)(2/2)
1+2+3+...+k+(k+1) = (k+1)(k/2) + (k+1)(2/2) 1+2+3+...+k+(k+1)
= (k+1)[(k/2) + (2/2)] 1+2+3+...+k+(k+1) = (k+1)(k+2)/2What
you have done is to prove that _if_ the conjecture is true for
any particular n (call it k), _then_ it is also true for the
next number n (call it k+1). But you also know that it _is_
true for n=1. Therefore it is also true for n=2. But if it's
true for n=2 then it's also true for n=3. And so on, forever.
You have proved that it's true for all n >= 1.-- Stan Brown,
Oak Road Systems, Cortland County, New York, USA
http://OakRoadSystems.comFortunately, I live in the Uni States
of America, where we aregradually coming to understand that
nothing we do is ever ourfault, especially if it is really
stupid. --Dave Barry
===
Subject: Re: Strings: Need a
ProofContent-transfer-encoding: 8bit> Hello everybody,> Here
is my question: Suppose L is the set of all strings over the
set> {a, b}, including the null string, that can be construc
using the> following rules,> - if x is in L, then axb and bxa
are in L> - if x is in L and y is in L, then xy is in L> Show
that if x has an equal number of a's and b's, then x is in L.>
I figured this would involve some proof by contradiction, but
I've> found nothing of a contradiction in any of my attempts.
Any hints or> clues?> Thanks in advance,> BerndOK, I'll take
another stab at it.Again, I'll use induction on half string
length.Since the empty string is in L, ab and ba are in L so
any string oflhalf ength 1 (or 0) with the same number of a's
and b's is in L.Now suppose that every string of half length
less than k with the samenumber of a's as b's is in L.Let w be
a string with length 2k and k a's and k b's. Wlog, assume
wstarts with a. If w = abw_1 then since ab is in L and w_1 has
halflength k-1 and the same number of a's as b's, by
assumption, w_1 is inL and thus w is in L. Similarly, if w
ends in b or in ba, w is in L.So, suppose w = aaw_1aa. Let
A(n) be the number of a's in the initialsubstring of w of
length n and let B(n) be the number of b's in thatinitial
substring.We have A(2) = 2, B(2) = 0, A(2k-2) = k-2 and
B(2k-2) = k. So, theremust be a smallest t such that A(t) <=
B(t). But then A(t-1) > B(t-1).Consider the t-th entry of w.
It can't be a or else A(t) = A(t-1) + 1 > B(t-1) + 1 >= B(t).
So the t-th entry must be b.Thus, A(t-1) = A(t) <= B(t) =
B(t-1)+1. If A(t) < B(t) then A(t-1) < B(t-1)+1 which is
false.Consequently A(t) = B(t) so the initial substring of
length t has asmany a's as b's and thus is in L. For the same
reason, the remainingfinal substring is in L so w is in L.I
suspect there is a more elegant way to do this.That completes
the induction.-- Paul SperryColumbia, SC (USA)Subject: stupid
question about tensors
===
I am currently trying to familiarize
myself with tensor calculus. However Ihave come across a
question to which I cannot find an answer. Namely, in thebook
I am using (D.Kay Tensor Calculus p.37) it is sta that ( _
issubscript ):If T_i are the components of covariant vector T,
then S_ij = T_i T_ j - T_ jT_i are the components of a skew
symmetric covariant tensor S.As far as I can understand each
S_i_ j should equal 0 ( T_i T_ j = T_ jT_i). But how can a
tensor, having all of it's components equal to 0, beskew
symmetric? Or is it a tensor at all? But it is said to be a
tensor. Soshould I deduce that multiplication of tensor
components is not commutative?
===
Subject: Re: stupid question
about tensors> I am currently trying to familiarize myself
with tensor calculus. HoweverI> have come across a question to
which I cannot find an answer. Namely, inthe> book I am using
(D.Kay Tensor Calculus p.37) it is sta that ( _ is> subscript
):> If T_i are the components of covariant vector T, then S_ij
= T_i T_ j - T_j> T_i are the components of a skew symmetric
covariant tensor S.> As far as I can understand each S_i_ j
should equal 0 ( T_i T_ j = T_ j> T_i).Correct.> But how can a
tensor, having all of it's components equal to 0, be> skew
symmetric?Well, s_i_j = s_j_i = -s_j_i = 0, so it is both
symmetric and skewsymmetric, no problem.> Or is it a tensor at
all?Yes, it's a zero tensor, which is OK although rather dull!>
But it is said to be a tensor. So> should I deduce that
multiplication of tensor components is notcommutative?No, the
tensor components themselves are just numbers... I don't have
thebook you're reading, so I'm not sure what the author was
getting at. Thingsare more interesting with two vectors T and
U. Then S_ij = T_i U_j - T_jU_i is a skew symmetric tensor
(rela to the vector cross product of T&U -note that the cross
product of any vector with itself is also zero).Mike.(remove
dead stones to reply by email)
===
Subject: Re: stupid question
about tensorsI have the same Tensor book by D. Kay and looked
up the example. I thinkthe author is merely trying to
establish that given that T_i is covariantandgiven that S_ij =
T_iT_j - T_jT_ishow that S_ij is covariant.This part of the
question can be answered by substituting the definition forthe
covariant tensor of order one on page 27 (eq 3.10) everywhere
for T_iand T_j . and substituting the definition for a
covariant tensor of ordertwo on page 29 (eq. 3.12)everywhere
for S_ijThe skew symmetric part of the question, he says is
obvious.(I hate it whenthey do that!) He means he has defined
S in such a way that it equals theequavalent of AB - BA (to
rid of cumbersome notation). Note if S=AB-BAit is not
necessarily equal to zero. Let me give you an example.Suppose
A is a counterclockwise rotation of 90 degrees in an xy plane
thensuppose B is a reflection in the y axis. Draw a picture
and you will see ifyou reverse the operations, they do not
commute. However, S itself would beskew symmetric since it own
negative transpose is the neg. transpose of(AB-BA) So you get
S-S(transpose)=AB-BA-(-BA-AB) =0> I am currently trying to
familiarize myself with tensor calculus.However> I> have come
across a question to which I cannot find an answer. Namely,
in> the> book I am using (D.Kay Tensor Calculus p.37) it is
sta that ( _ is> subscript ):>>If T_i are the components of
covariant vector T, then S_ij = T_i T_ j -T_> j> T_i are the
components of a skew symmetric covariant tensor S.>>As far as
I can understand each S_i_ j should equal 0 ( T_i T_ j = T_ j>
T_i).> Correct.> But how can a tensor, having all of it's
components equal to 0, be> skew symmetric?> Well, s_i_j =
s_j_i = -s_j_i = 0, so it is both symmetric and skew>
symmetric, no problem.> Or is it a tensor at all?> Yes, it's a
zero tensor, which is OK although rather dull!> But it is said
to be a tensor. So> should I deduce that multiplication of
tensor components is not> commutative?> No, the tensor
components themselves are just numbers... I don't have the>
book you're reading, so I'm not sure what the author was
getting at.Things> are more interesting with two vectors T and
U. Then S_ij = T_i U_j - T_j> U_i is a skew symmetric tensor
(rela to the vector cross product ofT&U -> note that the cross
product of any vector with itself is also zero).> Mike.>
(remove dead stones to reply by email)
===
Subject: Re: The
mistical value of e > yes, but HOW if presen with this
question, could you narrow> the limit down to e? (heck, I'm
guessing it's impossible)Please don't top-post.There are many
ways to see that the limit as x goes to zeroof (1+x)^(1/x) is
e. But once you throw away the informationthat the base is 1+x
and the exponent is 1/x, and simplytry to understand it as
1^infty, then yes, it *is* impossible.It may have fallen out
of fashion, but older calculus booksused to talk about
indeterminate forms; once you reducean expression to an
indeterminate form, you know you've reducedit too far, and you
need to go back and use some of the informationyou've thrown
away, because an indeterminate form can a prioritake many
values. The classic indeterminate forms are 0/0 0^0
infty/infty infty-infty 1^infty
===
Subject: Re: The mistical
value of e> The classic indeterminate forms are> 0/0> 0^0>
infty/infty> infty-infty> 1^inftyOh, I left out an important
one: 0*inftyAlso, if you wind up with infty+infty, you needto
check carefully to make sure the two infinitiesare signed and
have the same sign. If you're usingunsigned infinity, then
infty+infty is also anindeterminate form.
===
Subject: Re: The
mistical value of e>Well, it's not that mystical, how it is
achieved I understand, what I don't>understand is how the
limit below could have its value:> lim(x->0) (1+x)^(1/x) =
e>heck I would simplify it as follows instead:> lim(x->0)
(1+0)^(infinitum) = 1>But that's wrong, or more correctly
perhaps....not precise enough, but why?That's not how a limit
works. 1^infinity is what we call an indeterminate form,
meaning that you have to go back to the original expression
and do some more work to evaluate the limit.Informally, try it
this way. Make a table of x and (1+x)^(1/x). This should be
easy to do with any calculator, or of course with Excel:x
(1+x)^(1/x)-------- -----------1 20.1 2.593742460.01
2.7048138290.001 2.7169239320.0001 2.7181459270.00001
2.7182682370.000001 2.718280469This is not a proof, of course,
but it should be enough to convince you that as x gets closer
and closer to 0, (1+x)^(1/x) gets closer and closer to the
particular decimal value that you know as e. And that's pretty
much what a limit means: if you can get the function as close
as you like to the supposed limit just by keeping x close
enough to the sta value, then the supposed limit IS the limit.
(We can dress that up with epsilons and deltas if you like.)--
Stan Brown, Oak Road Systems, Cortland County, New York, USA
http://OakRoadSystems.comFortunately, I live in the Uni States
of America, where we aregradually coming to understand that
nothing we do is ever ourfault, especially if it is really
stupid. --Dave Barry
===
Subject: Re: The mistical value of
eContent-transfer-encoding: 8bit> Well, it's not that
mystical, how it is achieved I understand, what I don't>
understand is how the limit below could have its value:>
lim(x->0) (1+x)^(1/x) = e> heck I would simplify it as follows
instead:> lim(x->0) (1+0)^(infinitum) = 1> But that's wrong, or
more correctly perhaps....not precise enough, but why?> If you
know L'Hopital's rule, now is the time for it.Failing that,
you may know lim((1 + (1/x))^x, x -> oo) = e. If so,observe
lim((1+x)^(1/x), x -> 0) = lim((1 + (1/x))^x, x -> oo) = e.--
Paul SperryColumbia, SC (USA)
===
Subject: Re: The mistical
value of e> anywho, but how would you go about simplifying
that statement> lim(x->0) (1+x)^(1/x)> to obtain e. Or more
correctly, is it even possible to somehow simplify> this as we
do> with the limit definition of a derivative? I mean, can we
do anything to> this equation> to somehow narrow it down to e
Note ln[(1+x)^(1/x)] = [ln(1+x)]/x = [ln(1+x) - ln(1)]/x ->
ln'(1) = 1 as x -> 0, by the definition of the derivative.
Therefore (1+x)^(1/x) -> e^1 = e as x -> 0.
===
Subject: Re:
rational arithmetic library?> I've been looking for a means to
implement a numerical linear algebra> algorithm, but without
the concomitant round-off error (I'm fortunate> enough that my
initial matrices are of integer type). > I needed a more
general library, though. I had the linear algebra> algorithm
all coded up, but I simply wan to exchange the arithmetic>
operators to ones which would understand & simplify both
rational> numbers and integers. Check out www.linalg.org. You
can download extensive C++ code therefor the exact solution of
matrix problems. Rational arithmetic isusually not the best
thing to use for integer matrix problems, execptperhaps that
it may be required in the final steps if the answersthemselves
are rational.
===
Subject: Re: rational arithmetic library?===>
Check out www.linalg.org. You can download extensive C++ code
there> for the exact solution of matrix problems. Rational
arithmetic is> usually not the best thing to use for integer
matrix problems, execpt> perhaps that it may be required in
the final steps if the answers> themselves are rational.Why
so? I need the answers to be exact, not approximate. My
matrixstarts with positive & negative integers between -2 and
2. In theworst-case, I can imagine a bignum implementation
chewing up a lot ofresources. But I imagine that the average
case will be acceptable (Ineed accuracy, not speed).--
()
===
Subject: Re: rational arithmetic library?You don't say
what you are doing, or how long you expectyour computation to
take. There are studies that suggestdoing some exact
operations can be done more efficientlyby using finite field
arithmetic (perhaps several times)and sticking the pieces
together with the Chinese RemainderAlgorithm. e.g. McClellan's
work on exact solution of linearequations using modular
arithmetic.You might also look at papers on Exact Linear
Algebraby Ashoff(try google).>>Check out www.linalg.org. You
can download extensive C++ code there>>for the exact solution
of matrix problems. Rational arithmetic is>>usually not the
best thing to use for integer matrix problems, execpt>>perhaps
that it may be required in the final steps if the
answers>>themselves are rational.> Why so? I need the answers
to be exact, not approximate. My matrix> starts with positive
& negative integers between -2 and 2. In the> worst-case, I
can imagine a bignum implementation chewing up a lot of>
resources. But I imagine that the average case will be
acceptable (I> need accuracy, not speed).
===
Subject: Re:
rational arithmetic library?
<4pkad382oet.fsf@thar.lbl.gov>
===
> I took a second look at
R5and I realized my confusion; I guess my> current
implementation-of-choice, Guile, doesn't support 'numerator',>
'denominator', or 'rationalize'! So I tried some others:>
*DrScheme: yes> *Guile: no> *Chicken: no> *Bigloo: yes>
*Scheme48/Scsh: yes> *SCM: noI made a mistake: Bigloo does not
support rationals:1:=> rationalize*** ERROR:bigloo:eval:Unbound
variable -- rationalize#unspecified1:=> numerator***
ERROR:bigloo:eval:Unbound variable --
numerator#unspecified1:=> denominator***
ERROR:bigloo:eval:Unbound variable --
denominator#unspecified1:=> (/ 10 3)3.33333333333331:=> (/
1111111111111111111 999999999999999999999999)***
ERROR:bigloo:/:not a number --
#l1111111111111111111#unspecified---~T
===
Subject: Re: rational
arithmetic library?>>I took a second look at R5and I realized
my confusion; I guess my>>current implementation-of-choice,
Guile, doesn't support 'numerator',>>'denominator', or
'rationalize'! So I tried some others:>>*DrScheme:
yes>>*Guile: no>>*Chicken: no>>*Bigloo: yes>>*Scheme48/Scsh:
yes>>*SCM: no> I made a mistake: Bigloo does not support
rationals:But any CL implementation does. A piece of advice:
bite the bullet! Switch to CL :)Cheers--Marco
===
Subject: Re:
Riccati challenge to Maxima THIS group, sci.math.symbolic,is
presumably open to discussion of all matters rela to symbolic>
computation, as the name indicates.> Unfortunately it is also
open to other stuff as well :)>When I was a kid each movie
feature was accompanied by a cartoon.Although you don't see
Mickey Mouse, The Road Runner, Porky Pig, et al,at the movies
anymore, you CAN come to sci.math.symbolic for
amusement..
===
Subject: Maxima vs MuPAD by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i1QDt7605010; by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) with ESMTP id i1QCnOi30576 by
legacy.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.10 $, legacy) id i1QCnO508671
===
,I made a comparison between
Maxima 5.9.0 and MuPAD 2.5.0 using M.Wester's 131 test
problems; seehttp://homepage.mac.com/peso1[/index.html] .
Comments are welcome -especially on problems I or Maxima or
MuPAD could not solve.Pekka Sorjonen
===
Subject: Re: Maxima vs
MuPADThis is an interesting comparison of a program (Maxima)
that is22 years old and pre-dates Wester's test by many
years,and MuPAD program whose authors could look at Wester's
testsand fix them. Wester originally used MuPAD 1.2.1a and
this testis of 2.5.0.I looked through the examples, andit
seems to me that the tensor calculations in Maximarequire
reading in some additional programs, but that theycould be
done. (65,66).Your conclusion that MuPAD scored better seems
to be basedthe performance on a number of problems all of
which turn onhandling of assumptions or declarations of
domains.I think there are just a bunch of bugs in Maxima (that
werefixed during the subsequent 22 yeain Macsyma) that needto
be repaired. On the other hand it may be that MuPAD hasdone
this right from the start, and the lack of bugs isa natural
consequence of the design. Do you think that is thecase, or is
it just the same hodge-podge, but debugged more?I also think
that adding statistical analysis programs, ornumerical matrix
programs is just a matter of writing some more routine
programs. Anyone doing serious statistical studies should
beusing a different program, I suspect. Left out of the
comparison is the speed. I would expectthat MuPAD is slower.
Can you give any data on this? ForMaxima, typing showtime:all$
gives data on each command. Also left out is the fact that
Maxima is (now) open-sourceand I gather that MuPAD is not,
although it is sometimesfree. The quality of the documentation
is a serious issue formany users. Maxima documentation is not
so good, but theavailable documentation for Macsyma is
substantial. And the fundamental question to meis whether
MuPAD or Maxima is foundationally a better systemon which to
build programs which depend on mathematical knowledge.Although
you conclude that MuPAD wins, I suspect that yourconclusions
would have to be revised if someone spent a monthfixing bugs
in Maxima, targetting Wester's tests.My recommendation to
people who ask which is better.... isto find out what they are
trying to do, and if they have friendsdoing similar things with
a CAS. They should use the same programas their friends,
usually.Thanks for doing the tests!!> ,> I made a comparison
between Maxima 5.9.0 and MuPAD 2.5.0 using M.> Wester's 131
test problems; see> http://homepage.mac.com/peso1[/index.html]
. Comments are welcome -> especially on problems I or Maxima or
MuPAD could not solve.> Pekka Sorjonen