mm-155
Formulae 1 is the only advanced CAS applications
currently available for the P800. While there are several
different numerical based calculator programs for the P800,
there is nothing that comes close to the mathematical
capabilities of F1. F1 transforms your P800 into a
sophisticated math system that utilizes a unique user
interface that was custom designed to optimize the use of the
small screen area of the device. Most of the screen area is
available to view and manipulate math expressions, unlike
other calculators where most of the screen is populated by
buttons. For a complete list of features (with lots of screen
shots) and some more information go to:
http://www.poliplus.com/handheldproducts.htm Poliplus Software
is excited about this release as it demonstrates the
versatility and power of its F1 application. Carlos
Bazzarella, CEO of Poliplus Software commented Years ago, our
vision was aimed at the convergence of mobile computing. The
P800 is one of the devices leading the way in that direction,
and we are proud to offer our CAS for it as it further
validates our strength and innovative software written in
Java. Poliplus Software is dedicated to offering powerful
educational handheld applications. About Poliplus Software
Poliplus Software is an innovative developer of non-trivial,
feature rich applications for handheld devices. Poliplus'
comprehensive suite of applications in the 'elds of
Mathematics, Science and Engineering add tremendous value to
handheld devices. All of our products are developed in 100%
pure Java using various standard pro'les including MIDP,
PersonalJava, JavaSE and JavaEE. For more information please
visit www.poliplus.com. -30-Contact: Rafael Bazzarella
Poliplus Software rbazzaTAKE@OUTpoliplus.com
>Formulae 1
is the only advanced CAS applications currently available for
the P800. While there are several different numerical based
calculator programs for the P800, there is nothing that comes
close to the mathematical capabilities of F1. >F1 transforms
your P800 into a sophisticated math system that utilizes a
unique user interface that was custom designed to optimize the
use of the small screen area of the device. Most of the screen
area is available to view and manipulate math expressions,
unlike other calculators where most of the screen is populated
by buttons. For a complete list of features (with lots of
screen shots) and some more information go to: >And what about
customizing your post to 't average PC screen?...A.L.
=This
is the one by Mignotte and Stefanescu.The book covers good
stuff, but typos and worse force the reader to reconstruct the
reasoning in many of the proofs.I would like to compare notes
with another
reader. at http://www.TitanNews.com <<<<-=Every
Newsgroup - Anonymous, UNCENSORED, BROADBAND Downloads=-
=
Would anybody kindly give me any suggestion to symbolic math
software? I know mathematica from mathwork. However, it is a
commercial SW that I have to pay. Is there any conterpart of
GPL SW as good asmathematica? What's the major difference
between matlab and mathematica?--BR,Logia
 Would anybody
kindly give me any suggestion to symbolic math software?> I
know mathematica from mathwork. However, it is a commercial SW
that> I have to pay. Is there any conterpart of GPL SW as good
as> mathematica?> What's the major difference between matlab
and mathematica?> --> BR,> LogiaCheck out Maxima
(http://maxima.sourceforge.net/) and
YACAS(http://yacas.sourceforge.net/). They're free, and 
Maxima
is prettyimpressive. They are no match for Mathematica or
Maple, however.
 Would anybody kindly give me any
suggestion to symbolic math software?> - Maxima (again) -
Mupad light is free of charge (Maple like...) :
http://www.mupad.com/download.shtml - pari/gp (rather for
number-theory than symbolic integration butcomplementary.. and
may be used as a DLL)
http://www.gn-50uma.de/ftp/pari/00index.html For more products
you may try here :
it helped, Raymond
=I need to 'nd a solution f(n) of the
recurrence relation,f(n-1)+f(n+1)= C1*(2n+1)*
f(n),f(0)=C2,f(1)=C3,where C1=1/z, C2=1/z, C3=1/z^2The
relation is connected with a representation of the
sphericalBessel functions, and appears as formula (10.1.10) of
Abramowitz &Stegun.I have the bad feeling it has no solution.
Well, but it will be verynice if it has. I will be content to
obtain some approximation off(n) - series, etc. as well.Any
ideas?
=|>I need to 'nd a solution f(n) of the recurrence
relation,|>f(n-1)+f(n+1)= C1*(2n+1)*
f(n),|>f(0)=C2,|>f(1)=C3,|>where C1=1/z, C2=1/z, C3=1/z^2For
convenience let's write z = 1/c. I don't have 
a closed form,
butyou could try a series in c. It looks to me like f(2k) =
(-1)^k c + a_k c^3 + O(c^5)f(2k+1) = b_k c^2 + O(c^4)with a_0
= 0 and b_0 = 1.Substituting these into the recurrence
relation, I getb(k)+b(k-1) = (4k+1) (-1)^ka(k)+a(k+1) = (4k+3)
b(k)and then b(k) = (-1)^k (k+1)(2k+1) and a(k) = (-1)^k
(1-4k^2)(k+1)/2Clearly it should be possible to get more terms
along these lines.Robert Israel israel@math.ubc.caDepartment of
Mathematics http://www.math.ubc.ca/~israel University of
British Columbia Vancouver, BC, Canada V6T 1Z2
 I need to
'nd a solution f(n) of the recurrence relation,>
f(n-1)+f(n+1)= C1*(2n+1)* f(n),> f(0)=C2,> f(1)=C3,> where
C1=1/z, C2=1/z, C3=1/z^2> The relation is connected with a
representation of the spherical> Bessel functions, and appears
as formula (10.1.10) of Abramowitz &> Stegun.from (10.1.10) and
(10.1.8) of Abramowitz & Stegun one has, for n>=0,j_n(z) =
f_n(z) sin(z) + (-1)^(n+1) f_{-n-1}(z) cos(z) = [ P(n+1/2,z)
sin(z-Pi*n/2) + Q(n+1/2,z) cos(z-Pi*n/2)] / zwhere P and Q are
polynomials in (1/z). Apply the addition theoremsfor sin and
cos in the second line, then read off f_n and f_{n-1}according
to the 'rst line. Note that for any integer n, one
ofsin(Pi*n/2) and cos(Pi*n/2) will be zero and the other will
be 1 or-1, so it should be useful to look at the cases of even
and odd nseparately.Icke
=I'm not a mathematician, so please
forgive me if I am not 100% correct in my math terminology. I
am developing an equation solving program (AutoAbacus
http://singularsys.com/autoabacus) which should be able to
solve systems of equations by automatically recognizing the
type of equation system, and applying an appropriate solution
method. I hope you might be able to recommend an internal
canonical representation for polynomials.I need an appropriate
way of representing any polynomial of any number of variables
in a canonical fashion. I have come up with the following
template (here shown for two variables x1, x2): (A * x1^a1 *
x2^a2) + (B * x1^b1 * x2^b2) + (C * x1^c1 * x2^c2)...Within my
software I could ef'ciently represent a polynomial as a 
matrix
with the following structure: Coeff. A B C ... Exp1 a1 b1 c1
... Exp2 a2 b2 c2 ... ... ... ... ...And to clarify it even
more with an example:A polynomial such as 1 + 2 x2 + 4 x3
y2would be represented internally as Coeff. 1 2 4 Exp. x 0 2 3
Exp. y 0 0 2Here are my questions: - Am I correct with my
interpretation of polynomial? - Is this an acceptable
representation, or am I missing something? - Are there other
standards used in common CAS that would have bene'ts for
solving systems of polynomial equations (if I ever get around
to implementing that...)?Nathan
Maple can be used quite ef'ciently forobject-oriented
programming if you don't create too manyobjects, because
method invocation is quite ef'cientwhile object creation is
quite inef'cent in thisimplementation.The cost for the latter
correlates with the size of themodule code. It seems that the
implementation of modulecreation indeed uses a method similar
to the onedescribed on the help page module:the implementation
for----------------------------------------z5 := module()
export plus, times; plus := (a,b) -> a + b mod 5; times :=
(a,b) -> a * b mod 5;end
module;-----------------------------------------being similar
to:-----------------------------------------makezp := proc(
N::posint ) local plus, times; # will be exported plus :=
(a,b) -> a + b mod N; times := (a,b) -> a * b mod N; plus,
times # return exportsend
proc:------------------------------------------creating new
procedures (as escaped local variables) for eachnew object in
turn.In the maple session below I did some measurements of
thetime needed for object creation and invocation of a
methodplus compared to the procedure call of a
correspondingfunction. For a large module I got a ratio of
about 17 inthe execution times (the object-oriented version
being 17times slower than the procedural version), for
smallestmodules the ratio was 4.An alternative implementation
of object oriented programmingcould do the following (as is
done in many other object-orientedprogramming languages): add
the formal parameter this to themethod in front of the
parameter list for representation of thecurrent object the
method is called upon and (Maple-speci'c)if the variables 
args
and nargs are used in the method, thenadd a line correcting
them in the beginning of the method code:args := args[2..-1];
nargs := nargs-1;Changing the method code in this way could be
done using thebe performed only once for each class
de'nition.This way, there would only be one generated
procedure in memoryfor each method, not one per object as in
the above implementation.For method invocation and 'eld
access, one could use/overload the. (Dot) operator. One would
have to change the code of Dotcarefully in order to keep the
functionality of the LinearAlgebrapackage ef'cient.
Inheritance of methods, polymorphism and handlingof super
could be implemented using the usual Maple scheme
forspecialization of procedures on different types of
arguments:procedure `a` delegating to procedure
`a/className`.What do you think? Is it worth worth while to
implement ef'cientobject-oriented programming in Maple? Bruno
Daniel--------------------------------------------------------
-----------Institut fuer Angewandte PhysikUniversitaet
KarlsruheGermany
= restart;> timing := proc(expr::uneval)>
local startTime, result;> startTime := time();> result :=
eval(expr);> result, time() - startTime> end:>> newA :=
proc()> module()> export plus, plus1, plus2, plus3, plus4,
plus5, plus6, plus7,> plus8,plus9,plus10,plus11;>
plus:=(a,b)->a+b;> plus1:=(a,b)->a+b;> plus2:=(a,b)->a+b;>
plus3:=(a,b)->a+b;> plus4:=(a,b)->a+b;> plus5:=(a,b)->a+b;>
plus6:=(a,b)->a+b;> plus7:=(a,b)->a+b;> plus8:=(a,b)->a+b;>
plus9:=(a,b)->a+b;> plus10:=(a,b)->a+b;> plus11:=(a,b)->a+b;>
plusa6:=(a,b)->a+b;> plusa7:=(a,b)->a+b;> plusa8:=(a,b)->a+b;>
plusa9:=(a,b)->a+b;> plusa10:=(a,b)->a+b;>
plusa11:=(a,b)->a+b;> plusb6:=(a,b)->a+b;>
plusb7:=(a,b)->a+b;> plusb8:=(a,b)->a+b;> plusb9:=(a,b)->a+b;>
plusb10:=(a,b)->a+b;> plusb11:=(a,b)->a+b;>
plusc6:=(a,b)->a+b;> plusc7:=(a,b)->a+b;> plusc8:=(a,b)->a+b;>
plusc9:=(a,b)->a+b;> plusc10:=(a,b)->a+b;>
plusc11:=(a,b)->a+b;> plusd6:=(a,b)->a+b;>
plusd7:=(a,b)->a+b;> plusd8:=(a,b)->a+b;> plusd9:=(a,b)->a+b;>
plusd10:=(a,b)->a+b;> plusd11:=(a,b)->a+b;> end;> end;Warning,
`plusa6` is implicitly declared local to moduleWarning,
`plusa7` is implicitly declared local to module[...] newA :=
proc() module() local plusa6, plusa7, plusa8, plusa9, plusa10,
plusa11, plusb6, plusb7, plusb8, plusb9, plusb10, plusb11,
plusc6, plusc7, plusc8, plusc9, plusc10, plusc11, plusd6,
plusd7, plusd8, plusd9, plusd10, plusd11; export plus, plus1,
plus2, plus3, plus4, plus5, plus6, plus7, plus8, plus9,
plus10, plus11; plus := (a, b) -> a + b; plus1 := (a, b) -> a
+ b; plus2 := (a, b) -> a + b; plus3 := (a, b) -> a + b; plus4
:= (a, b) -> a + b; plus5 := (a, b) -> a + b; plus6 := (a, b)
-> a + b; plus7 := (a, b) -> a + b; plus8 := (a, b) -> a + b;
plus9 := (a, b) -> a + b; plus10 := (a, b) -> a + b; plus11 :=
(a, b) -> a + b; plusa6 := (a, b) -> a + b; plusa7 := (a, b) ->
a + b; plusa8 := (a, b) -> a + b; plusa9 := (a, b) -> a + b;
plusa10 := (a, b) -> a + b; plusa11 := (a, b) -> a + b; plusb6
:= (a, b) -> a + b; plusb7 := (a, b) -> a + b; plusb8 := (a, b)
-> a + b; plusb9 := (a, b) -> a + b; plusb10 := (a, b) -> a +
b; plusb11 := (a, b) -> a + b; plusc6 := (a, b) -> a + b;
plusc7 := (a, b) -> a + b; plusc8 := (a, b) -> a + b; plusc9
:= (a, b) -> a + b; plusc10 := (a, b) -> a + b; plusc11 := (a,
b) -> a + b; plusd6 := (a, b) -> a + b; plusd7 := (a, b) -> a +
b; plusd8 := (a, b) -> a + b; plusd9 := (a, b) -> a + b;
plusd10 := (a, b) -> a + b; plusd11 := (a, b) -> a + b end
moduleend proc> f1:= proc(n::integer)> local i, obj;> for i to
n do> obj := newA();> obj:-plus(3, i);> end> end; f1 :=
proc(n::integer)local i, obj; for i to n do obj := newA();
obj:-plus(3, i) end doend proc> timing(f1(100000)); 100003,
17.074> f2:= proc(n::integer)> local i;> for i to n do> 3 +
i;> end> end; f2 := proc(n::integer) local i; for i to n do 3
+ i end do end proc> timing(f2(100000)); 100003, 0.192>
f3:=(a,b)->a+b; f3 := (a, b) -> a + b> f4:= proc(n::integer)>
local i;> for i to n do> f3(3, i);> end> end; f4 :=
proc(n::integer)local i; for i to n do f3(3, i) end doend
proc> timing(f4(100000)); 100003, 1.073>
are concerned with speed, you should probably>not be writing
programs in an interpreted language.>>If you write in Lisp,
the object-oriented programs>can be compiled to machine
language.I like Lisp, but what if I want to have all
thefunctionality of the Maple libraries at my disposal?As far
as I know, Common Lisp Object System doesn'toffer this.Maple
is not that slow, after all, when dealing withgeneral symbolic
data.I just propose another way to implement the
object-orientedphilosphy into Maple. In the meantime, I had an
idea fordoing this in a way that seamlessly 'ts into
Maple'sphilosophy of delegating a function f to `f/class`.
Theconsequence is that you get polymorphism not only
forobjects, but also for other kinds of expressions, just
byimplementing `f/+`, `f/*`, `f/=` (which could, for
example,use map on a recursive call of f on the operands).
This means,that these kinds of object-oriented methods should
be writtenin functional notation:f(obj, args)instead
ofobj:-f(args) or obj.f(arg) etc.Inheritance of methods, which
is missing in Maple modules, could also be implemented quite
easily this way. I did some performance measurements on Maple
modules, Maple Arrays,Maple lists and Maple Records and I
found out, that the latter are the most ef'cient candidates
for holding an object's state, concerning object creation as
well as setting/getting attributes.So I suppose that there are
not only performance advantages ofthis kind of
object-orientation, but also conceptual advantages.What do you
***
=ef'cient way of 'nding the counts of different 
sets
taking intoaccount the overlap between the different sets.
Also the counts of thedifferent setsshould take into account
the way the overlaps occur between the sets.i.e. the counts to
disperse the elements of the overlap between thetwo sets needs
to take into account the universe count of each set andthen
use a ratio of the universe counts to disperse the
overlappingelements.To better illustrate this, lets say there
are two sets, whoseuniverses are depicted by A and B
respectively:Scenario I (Two Sets A &
B):----------------------------------- Count (A) = 2000 //
Universe Count for Set A Count (B) = 3000 // Universe Count
for Set B Count (A & B) = 100 // Count for overlappingelements
b/w Sets A & B Count (A - B) = 1900 // Count for
elementsbelonging uniquely to Set A (and not B) Count (B - B)
= 2900 // Count for elementsbelonging uniquely to Set B (and
not A) Now the new counts for sets A & B will be computed as
follows: Real_Count_A = Count(A - B) + Count(A & B) *
[Count(A) / (Count(A) + Count(B))] Real_Count_B = Count(B - A)
+ Count(A & B) * [Count(B) / (Count(A) + Count(B))] i.e.
Real_Count_A = 1900 + 100 * [2000 / (2000 + 3000)]
Real_Count_A = 1900 + 40 = 1940 Real_Count_B = 2900 + 100 *
[2000 / (2000 + 3000)] Real_Count_B = 2900 + 60 = 2960 So the
count for Set A is made up of 1940 elements of which
1900elements belong exclusively to it plus the 40% of the 100
overlappingelements.Similarly the count for Set B is made up
of 2960 elements of which2900 elements belong exclusively to
it plus the 60% of the 100overlapping elements. [ This is
because the ratio to disperse the overlappingelements b/w A
and B was Count A / Count B = 2000 / 3000 = 40% / 60% ]Now if
you were to extend this scenario to 3 Sets, the logic
becomes:Scenario II (Three Sets A, B &
C):-----------------------------------Real_Count_A = Count(A -
B - C) + Count(A & B & ~C) * [Count(A) / (Count(A) + Count(B))]
+ Count(A & ~B & C) * [Count(A) / (Count(A) + Count(C))] +
Count(A & B & C) * [Count(A) / (Count(A) + Count(B) +
Count(C))] Real_Count_B = Count(B - A - C) + Count(B & A & ~C)
* [Count(B) / (Count(A) + Count(B))] + Count(B & ~A & C) *
[Count(B) / (Count(B) + Count(C))] + Count(A & B & C) *
[Count(B) / (Count(A) + Count(B) + Count(C))] Real_Count_C =
Count(C - A - B) + Count(C & B & ~A) * [Count(C) / (Count(B) +
Count(C))] + Count(C & ~B & A) * [Count(C) / (Count(A) +
Count(C))] + Count(A & B & C) * [Count(C) / (Count(A) +
Count(B) + Count(C))]Scenario N: (N Sets)---------------------
If we generalize this for N Sets, I think there will be N *
2^(N-1) Total Terms to compute the counts for the N
Sets.---------------------------------------------------------
-----------# of Sets = N Total Terms (N X 2 N-1
)-------------------------------------------------------------
------- 1 1 2 4 3 12 4 32 5 80 10 5120 15 491520 20
10485760------------------------------------------------------
-------------- If we use the above formula (using mathematical
induction), there willbe anexponential rise in the number of
terms to be computed as the numberof sets increases.What I
would like to know is if there is any ef'cient method 
of'nding
the same counts (at least approximate counts) for the sets
byusing a more simpler formulae or other heuristic methods /
tools?Leo
=Take a look at this book: Discrete and
Combinatorial Mathematics: AnApplied Introduction by Ralph P.
Grimaldi.He should have an answer to your question. Pay
special attention to section5.3 Stirling numbers of the second
kind, if I correctly understood yourproblem.Yours is more a
problem of counting than anything else.> ef'cient way of
'nding the counts of different sets taking into> account the
overlap between the different sets. Also the counts of the>
different sets> should take into account the way the overlaps
occur between the sets.> i.e. the counts to disperse the
elements of the overlap between the> two sets needs to take
into account the universe count of each set and> then use a
ratio of the universe counts to disperse the overlapping>
elements.>> To better illustrate this, lets say there are two
sets, whose> universes are depicted by A and B respectively:>>
Scenario I (Two Sets A & B):>
----------------------------------->> Count (A) = 2000 //
Universe Count for Set A>> Count (B) = 3000 // Universe Count
for Set B>> Count (A & B) = 100 // Count for overlapping>
elements b/w Sets A & B>> Count (A - B) = 1900 // Count for
elements> belonging uniquely to Set A (and not B)>> Count (B -
B) = 2900 // Count for elements> belonging uniquely to Set B
(and not A)> Now the new counts for sets A & B will be
computed as follows:>> Real_Count_A = Count(A - B) +> Count(A
& B) * [Count(A) / (Count(A) + Count(B))]>> Real_Count_B =
Count(B - A) +> Count(A & B) * [Count(B) / (Count(A) +
Count(B))]>> i.e.>> Real_Count_A = 1900 +> 100 * [2000 / (2000
+ 3000)]>> Real_Count_A = 1900 + 40 = 1940> Real_Count_B =
2900 +> 100 * [2000 / (2000 + 3000)]>> Real_Count_B = 2900 +
60 = 2960> So the count for Set A is made up of 1940
elements of which 1900> elements belong exclusively to it plus
the 40% of the 100 overlapping> elements.> Similarly the count
for Set B is made up of 2960 elements of which> 2900 elements
belong exclusively to it plus the 60% of the 100> overlapping
elements.> [ This is because the ratio to disperse the
overlapping> elements b/w A and B was Count A / Count B = 2000
/ 3000 = 40% / 60% ]> Now if you were to extend this scenario
to 3 Sets, the logic becomes:>> Scenario II (Three Sets A, B &
C):> ----------------------------------->> Real_Count_A =
Count(A - B - C)> + Count(A & B & ~C) * [Count(A) / (Count(A)
+ Count(B))]> + Count(A & ~B & C) * [Count(A) / (Count(A) +
Count(C))]> + Count(A & B & C) * [Count(A) / (Count(A) +
Count(B) + Count(C))]>> Real_Count_B = Count(B - A - C)> +
Count(B & A & ~C) * [Count(B) / (Count(A) + Count(B))]> +
Count(B & ~A & C) * [Count(B) / (Count(B) + Count(C))]> +
Count(A & B & C) * [Count(B) / (Count(A) + Count(B) +
Count(C))]>> Real_Count_C = Count(C - A - B)> + Count(C & B &
~A) * [Count(C) / (Count(B) + Count(C))]> + Count(C & ~B & A)
* [Count(C) / (Count(A) + Count(C))]> + Count(A & B & C) *
[Count(C) / (Count(A) + Count(B) + Count(C))]> Scenario N:
(N Sets)> --------------------->> If we generalize this for N
Sets, I think there will be N * 2^(N-1)> Total Terms to
compute the counts for the N Sets.>
--------------------------------------------------------------
------> # of Sets = N Total Terms (N X 2 N-1 )>
--------------------------------------------------------------
------> 1 1> 2 4> 3 12> 4 32> 5 80> 10 5120> 15 491520> 20
10485760>
--------------------------------------------------------------
------>> If we use the above formula (using mathematical
induction), there will> be an> exponential rise in the number
of terms to be computed as the number> of sets increases.>>
What I would like to know is if there is any ef'cient method
of> 'nding the same counts (at least approximate counts) for
the sets by> using a more simpler formulae or other heuristic
methods / tools?>> Leo>
 Well, sky = blue means sky is
equal to blue, not> sky is blue. If you wanted to write sky
is blue,> you'd have to de'ne blue as a 
property of sky,
not> as an equiv[i|e]lant thereof.>> Or equivalent,
even. :-)>> Actually, you'd have to de'ne 
colour as a
property> of sky, and set the value of that property to
blue.> I was simply using it as an example to get across my
point. Not as an> has at least attempted to convert standard
English sentences into> mathematical equations. Anyone know if
anyone has written such? If so,> title of book and/or URL
please.> Your point was indeed understood, and his point
remains mostpertinent. When it comes to the mathematics, we
can have no room forabiguity (notice how ambiguous that
statement is itself!). Yourexchanges illustrated precisely why
the extemely more precise languageof mathematics developed.The
closest thing I can think of -- not quite what you have in
mind,of course -- would be what was formerly referred to as
the logicalcalculus, viz., symbolic logic.
=Wrong newsgroup.
You want arti'cial intelligence / naturallanguage
understanding.Seehttp://www.cyc.com/products2.htmlfor an
example of what people have been doing for thelast couple of
decades.> ...>>has at least attempted to convert standard
English sentences into>>mathematical equations. Anyone know if
anyone has written such? If so,>>title of book and/or URL
please.> Your point was indeed understood, and his
point remains most> pertinent. When it comes to the
mathematics, we can have no room for> abiguity (notice how
ambiguous that statement is itself!). This is quite wrong.
Mathematics is quite ambiguous and needsa natural language
context to disambiguate. a(b+c) is eithera times (b+c) or
apply the function a to the quantity (b+c).> The closest thing
I can think of -- not quite what you have in mind,> of course
-- would be what was formerly referred to as the logical>
calculus, viz., symbolic logic.Getting warmer. See link
above.RJF
 Wrong newsgroup. You want arti'cial
intelligence / natural> language understanding.I've posted
this question before to comp.ai.philosophy with little
results.The only newsgroup on natural languages that my news
server provides iscomp.edu.languages.natural. Is this the one
you're referring to?> See> 
http://www.cyc.com/products2.html>
for an example of what people have been doing for the> last
couple of decades.Yes, I'm aware of what cyc has done. They
are pursuing something differentthan what I am.>>has at
least attempted to convert standard English>>sentences into
mathematical equations. Anyone>>know if anyone has written
such? If so, title of>>book and/or URL please.>> Your
point was indeed understood, and his point> remains most
pertinent. When it comes to the> mathematics, we can have no
room for abiguity> (notice how ambiguous that statement is
itself!).>> This is quite wrong. Mathematics is quite
ambiguous> and needs a natural language context to
disambiguate.> a(b+c) is either a times (b+c) or apply the
function a> to the quantity (b+c).In a conversation with a
friend that's into math, he suggested I pose thequestion
here.> The closest thing I can think of -- not quite what
you> have in mind, of course -- would be what was formerly referred to as the logical calculus, viz., symbolic
logic.What is it now referred to as?-- Got a business problem
or question? Like to start a business?You can explore those
problems, questions, and dreams inmisc.business.consulting,
misc.business.marketing.moderatedmisc.business.moderated, and
misc.entrepreneurs.moderated
=Newsgroup Poster
<ng_poster@yahoo.com> schrieb in im
> Well, sky = blue means sky is equal to blue, not sky is blue. If you wanted to write sky is blue,>
you'd have to de'ne blue as a property of sky, 
not> as
an equiv[i|e]lant thereof.>> Or equivalent, even. :-)> Actually, you'd have to de'ne colour as a 
property of sky, and set the value of that property to blue.>> I
was simply using it as an example to get across my point. Not
as anthat> has at least attempted to convert standard
English sentences into> mathematical equations. Anyone know
if anyone has written such? If so,> title of book and/or URL
please.>> Your point was indeed understood, and his point
remains most> pertinent. When it comes to the mathematics, we
can have no room for> abiguity (notice how ambiguous that
statement is itself!). Your> exchanges illustrated precisely
why the extemely more precise language> of mathematics
developed.>> The closest thing I can think of -- not quite
what you have in mind,> of course -- would be what was
formerly referred to as the logical> calculus, viz., symbolic
logic.It's not possible to reduce English, or any other
language, to mathematicalsymbols. What you can do is to parse
sentences and to assign various symbolsto different parts of
speech (e.g. a symbol that says, This is the subjectof the
sentence and is in the form of a noun phrase). Or you can
usemathematical symbols to explain the structure of a
sentence; e.g., thesentence I saw the dog, and he saw me has
the structure [S P O] + [S P O].But that's not what you 
mean,
is it?There is a theory -- and I can't now remember all the
details or all thefancy names for various things -- that,
underlying every language, there isa sort of a meta-language
made up of about a dozen concepts, such asword and negative.
The idea is that, if you can translate a sentenceinto this
meta-language, you can then translate it into any language
youwish. Each concept had its own symbol.Unfortunately, while
this was an interesting idea, it was also unhelpful. Itturns
out that even the simplest sentence, like I made a cake,
becomessomething along the lines of I am thinking of a world
in which a part ofthat world contains a world in which a cake
is not a part and is becoming aworld in which a cake is a
part, or something equally verbose and obscure.It's rather
like using quantum theory to explain how an athlete
runs.
> Wrong newsgroup. You want arti'cial intelligence /
natural>> language understanding.>>I've posted this question
before to comp.ai.philosophy with little results.>You posted
to wrong group. comp.ai.philosophy is about philosophy,not
about problem solving. Post to comp.ai and
comp.ai.nat-langA.L.
 Wrong newsgroup. You want arti'cial
intelligence / natural> language understanding.> See>
http://www.cyc.com/products2.html> for an example of what
people have been doing for the> last couple of decades. ...>>has at least attempted to convert standard English
sentences into>>mathematical equations. Anyone know if
anyone has written such? If so,>>title of book and/or URL
please.> Your point was indeed understood,
and his point remains most> pertinent. When it comes to the
mathematics, we can have no room for> abiguity (notice how
ambiguous that statement is itself!). > This is quite wrong.
Mathematics is quite ambiguous and needs> a natural language
context to disambiguate. a(b+c) is either> a times (b+c) or
apply the function a to the quantity (b+c).> I respectfully
disagree for two reasons. First, mathematics is not ambiguous,
mathematical notation is. Theambiguity occurs mostly because of
context dependencies. If thecontext information such as a, b, c
are elements of algebra A, ...or alternatively b and c are
element of algebra A, a is a function A->X, ...is added, there
is no ambiguity anymore. It is much easier to limitthe context
information when interacting with a human then
when'explaining' things to a computer. So, the 
real issue is
how to addthe necessary context information and still have
things done. I seethat as a proper topic for this
newsgroup.Second, Scott's posting is a great refresher after
Dar Kabanoff andsuch productions.:-)AK> The closest thing I
can think of -- not quite what you have in mind,> of course
-- would be what was formerly referred to as the logical>
calculus, viz., symbolic logic.> Getting warmer. See link
above.> RJF
=There was a book I read ages ago, it was put
out by MIT Press and was acollection of essayson Arti'cial
Intelligence. If I rerember correctly, this book was
fromaround 1967.One of the contributors, I think it was Marvin
Minsky (or was he just thecode on automating the solution of
algebraic word problems. It used sometranslationallisp code
which had to be obtained elsewhere but actually converted
englishalgebra word problemsinto a system of equations and
then solved them.I can't rerember the title of that book - 
in
fact about the only thing thatsticks in my mind about
itSemantic Memory by Quilliam,another early MIT AI
researcher.Sorry can't be of more help.Jimis> Well, sky
= blue means sky is equal to blue, not> sky is blue. If
you wanted to write sky is blue,> you'd have to 
de'ne
blue as a property of sky, not> as an equiv[i|e]lant
thereof.>> Or equivalent, even. :-)>>
Actually, you'd have to de'ne colour as a 
property> of
sky, and set the value of that property to blue.>> I was
simply using it as an example to get across my point. Not as
anthat> has at least attempted to convert standard English
sentences into> mathematical equations. Anyone know if
anyone has written such? If so,> title of book and/or URL
please.>> Your point was indeed understood, and his point
remains most> pertinent. When it comes to the mathematics, we
can have no room for> abiguity (notice how ambiguous that
statement is itself!). Your> exchanges illustrated precisely
why the extemely more precise language> of mathematics
developed.>> The closest thing I can think of -- not quite
what you have in mind,> of course -- would be what was
formerly referred to as the logical> calculus, viz., symbolic
logic.
 There was a book I read ages ago, it was put out
by MIT Press and was a> collection of essays> on Arti'cial
Intelligence. If I rerember correctly, this book was from>
around 1967.> One of the contributors, I think it was Marvin
Minsky (or was he just the> code on automating the solution of
algebraic word problems. It used some> translational> lisp code
which had to be obtained elsewhere but actually converted
english> algebra word problems> into a system of equations and
then solved them.You are thinking of the STUDENT program by
Bobrow & Minsky, which could solve algebraic word problems.
The program itself was, I believe, written by Daniel
Bobrow.Bobrow, D. G. Natural Language Input for a Computer
Problem-Solving System, in Minsky (ed.) Semantic Information
Processing, Cambridge: M.I.T. Press, 1968.A later approach,
using frames, is inMinsky, M. A Framework for Representing
Knowledge, in Winston (ed.) One predecessor was Newell, Simon,
& Shaw's Logic Theorist, later to become the General Problem
Solver.At the same time Winograd's SHRDLU program was 
handling
the block word: his thesis is one of the few by someone else
that I have a copy of.> I can't rerember the title of that
book - in fact about the only thing that> sticks in my mind
about it> Semantic Memory by Quilliam,> another early MIT AI
researcher.> Sorry can't be of more help.> Jim> is>Well, sky = blue means sky is equal to blue,
not>sky is blue. If you wanted to write sky is
blue,>you'd have to de'ne blue as a property of 
sky,
not>as an equiv[i|e]lant thereof.>>Or equivalent,
even. :-)>>Actually, you'd have to de'ne colour 
as a
property>>of sky, and set the value of that property to
blue.>>I was simply using it as an example to get across
my point. Not as an> that>has at least attempted to
convert standard English sentences into>mathematical
equations. Anyone know if anyone has written such? If
so,>title of book and/or URL please.>Your point was
indeed understood, and his point remains most>>pertinent. When
it comes to the mathematics, we can have no room for>>abiguity
(notice how ambiguous that statement is itself!).
Your>>exchanges illustrated precisely why the extemely more
precise language>>of mathematics developed.>>The closest
thing I can think of -- not quite what you have in mind,>>of
course -- would be what was formerly referred to as the
logical>>calculus, viz., symbolic logic.> 
is is posible to solve systems of equationslike equation in
Mupad described
assolve(sum(x(i)^3*y(i),i=1..n)=a*sum(x(i)^6,i=1..n),a)solve(
sum(x(i)^3*y(i),i)=a*sum(x(i)^6,i),a)Can you recomend me any
software doing such work,please?I would like hearing from
youYouirs faithfullyPeter
solve systems of equations> like equation in Mupad described
as> solve(sum(x(i)^3*y(i),i=1..n)=a*sum(x(i)^6,i=1..n),a)>
solve(sum(x(i)^3*y(i),i)=a*sum(x(i)^6,i),a)> Can you
recomend me any software doing such work,please?> I would
like hearing from you> Youirs faithfully> Peter> Here is
what Maple does... What solutions did you want to get?> T1 :=
sum(x(i)^3*y(i),i=1..n); n ----- 3 T1 := ) x(i) y(i) / ----- i
= 1> T2 := a*sum(x(i)^6,i=1..n); / n |----- | | 6| T2 := a | )
x(i) | | / | |----- | i = 1 /> solve(T1=T2,a); n ----- 3 )
x(i) y(i) / ----- i = 1 ---------------- n ----- 6 ) x(i) /
----- i = 1> T3 := sum(x(i)^3*y(i),i); ----- 3 T3 := ) x(i)
y(i) / ----- i> T4 := a*sum(x(i)^6,i); /----- | 6| T4 := a | )
x(i) | | / | |----- | i /> solve(T3=T4,a); ----- 3 ) x(i) y(i)
/ ----- i ---------------- ----- 6 ) x(i) / ----- i
like equation in Mupad described as>>
solve(sum(x(i)^3*y(i),i=1..n)=a*sum(x(i)^6,i=1..n),a)>
solve(sum(x(i)^3*y(i),i)=a*sum(x(i)^6,i),a) Can you
recomend me any software doing such work,please?>> I would
like hearing from you>> Youirs faithfully>> Peter>> Here is what Maple does... What solutions did you want to
get?> T1 := sum(x(i)^3*y(i),i=1..n);>> n> -----> 3> T1 :=
) x(i) y(i)> /> -----> i = 1>> T2 := a*sum(x(i)^6,i=1..n);>>
/ n > |----- |> | 6|> T2 := a | ) x(i) |> | / |> |----- |> i =
1 />> solve(T1=T2,a);>> n> -----> 3> ) x(i) y(i)> /> ----->
i = 1> ----------------> n> -----> 6> ) x(i)> /> -----> i = 1>
I am ableto get Maple.I was just asked if any software is able
to do so...Yors faithfullyPeter
Fodrek
=
Resume = I wanna program an extension to NTL which
mutivariate polynomialfactorization. What are the actual best
algorithm (in term of speed) ?Like this in Maple ? Have you
any reference ?THANKS you very much for your
ansersThx
==
deep thinkingbefore beginning the developpement.First before
entering into the real subject I wishtell you (if it has not
been before) that GiNaC+CLNcompile well on W2K under cygwin
with gcc(gcc version 3.2 20020927 (prerelease))All chech and
tests are ok! Great job.NOW THE REAL SUBJECT OF THE
MAIL--------------------------------For one of our project WE
DEFINITLY need a packagefor multivariate polynomial
factorization under Z.For this I will have rapidely the
man.power to beginthis great challenge (next week).My great
wish would be to work within open-sourceframework. As this
subject is one of the ToDo listof GiNac, and a very dif'cult
task.Before compiling GiNaC, I beleived that I coulduse it as
a native library un Win32. But it seemsthat not, unless
modi'cation. Perhaps someone hasdid it ? I have a constraint
that I can't modify :I MUST BE ABLE to use and link with a
classic nativewin32 application.So, unless someone tell me the
modi'cation to doto be able to compile cln + ginac under 
VC6++I
cannot use ginac as a basis for further
developpment.-------------------------------------------------
----------The solution for me today is the use of NTL (which
compilewell with VC6) as univariate
basis.--------------------------------------------------------
---BUT the developpement of this package under NTL couldserve
as a basis for GiNac also !So I need you help for this
developpement at a
informationlevel.---------------------------------------------
------------What is today the best method for multivariate
polynomialfactorization under Z
?---------------------------------------------------------I
have the old algorithm from Berlekamp & Hensel for
univariate(p-adique method) and only an extension for
multivariate,but the algorithm complexity seems to be very
high, and I guessthat new algorithms have been found far now
?Someone could help me to indicate what is today the
bestalgorithm for this task ? REMARK = The rapidity and
accuracy of the result of such task made by Maple5 indicates
that there should be a quit rapid algorithm !For 'nish this -
call to contribution - , I must be honestand indicate that the
polynom I have to factorize are littlespecial :- they are very
very very long in term of monomial : the number of monomial
could be 100, 1000 !!!- they could be a lot of variable (when
I say a lot, it could be 10 and more) !!!!- THEY ARE
homogeneous : the total degree of monomial are always the
same- The coef'cient of each monomial is : 1, 2 or not very
high- IT SEEMS, after lot experiment under Maple5 that they
are squarre free (BUT we have not demonstrate this
assertion)UNDER this condition I can tell you that the kernel
of Maple5(in Matlab), for a HUGE HUGE polynomial like I
describe,result is almost immediate !!!!! Very very speed.I
hope that it will interest you to collaborateand help me to
develop this piece of softwareunder open-source.HervePS = If
you the modi'cation to do to GiNaC to be able to be a native
win32 library is not to complex I promiss to collaborate to do
this task.
=some data in an ASCII 'le. Everything seems ok,
but the importdoesn't seems to work. I just recently started
working with MuPAD andI don't know if maybe the import 
library
doesn't work on the Macversion. Could somebody help me or 
give
me some idea of how toproceed.The command looks like
this:Data:=import::readdata(r4); FAILThe name of my 'le is 
r4,
I have checked and it seems to be ok (3columns of numbers
separated by one space each). Any insigth will beCarlos
=:
some data in an ASCII 'le. Everything seems ok, but the
import: doesn't seems to work. I just recently started 
working
with MuPAD and: I don't know if maybe the import library
doesn't work on the Mac: version. Could somebody help me or
give me some idea of how to: proceed.: : The command looks
like this:: : Data:=import::readdata(r4);in 1.4.2 the function
was io::readdata, it was renamed to import::readdata with
version 2.0.0. And the argument isin both versions expected to
be a string, so try:io::readdata(r4);-- *---* MuPAD -- The Open
Computer Algebra System *---*| *---*
 The command looks
like this:>> Data:=import::readdata(r4);>> FAIL>> The name of
my 'le is r4, I have checked and it seems to be ok (3 Try
import::readdata(r4). If that doesn't work either, please
sendme the 'le. What you were telling MuPAD is to take the
value of thevariable r4 as a 'lename. Most likely, r4 does 
not
have a value,though. Note that more recent versions of MuPAD
will tell you so:>> import::readdata(r4);Error: Illegal
'lename [import::readdata]I'm sorry I 
can't tell you when
MuPAD 2.5 for Mac will be out of betatest. You might wish to
contact info@sciface.com on this subject.-- +--+ +--+| |+-|+
Christopher Creutzig (ccr@mupad.de)
 in 1.4.2 the function
was io::readdata, it was renamed to > import::readdata with
version 2.0.0. And the argument is> in both versions expected
to be a string, so try:> io::readdata(r4);manuals with
version 1.4.2. For some reason they don't open, each timeI 
try
day.Carlos H
=At http://functions.wolfram.com, Equation
07.27.03.0003.01 gives a(complicated) analytic expressionfor
3F2(a,b,c;d,e;1) with very general restrictions on
theparameters.I found this surprising, since I didn't know
that such an expressionexisted - I have never before seen any
mention of such a result in theliterature.But, given that it
exists:I need to evaluate A1:=3F2(b,1,3/2; b+1,3-s;1) as a
function of s andb, but only have access to Maple(6) which
seems never to have heard ofsuch a result;So, could someone
who has access to Mathematica please try to evaluateA1 and let
me know what you get?.I have tried all the special results I
can 'nd both on the Mathematicawebsite and in the literature,
but can't 'nd anything that will touchthis sum 
using pen and
paper.Mike
 [...]> I need to evaluate A1:=3F2(b,1,3/2;
b+1,3-s;1) as a function of s and> b, but only have access to
Maple(6) which seems never to have heard of> such a result;A
hypergeometric function in pFq notation with 3 upper and 2
lowerparameters. In Maple, this
isA1:=hypergeom([b,1,3/2],[b+1,3-s],1);but I know of no
approach to simplify this special case...-- Thomas
RichardMaple SupportScienti'c Computers
GmbHhttp://www.scienti'c.de
 [...]> I need to
evaluate A1:=3F2(b,1,3/2; b+1,3-s;1) as a function of s and>
b, but only have access to Maple(6) which seems never to have
heard of> such a result;> A hypergeometric function in pFq
notation with 3 upper and 2 lower> parameters. In Maple, this
is> A1:=hypergeom([b,1,3/2],[b+1,3-s],1);> but I know of no
approach to simplify this special case...> --> Thomas
Richard> Maple Support> Scienti'c Computers GmbH>
http://www.scienti'c.des = [1/2,-1/2,-3/2,-5/2], A1
=[-3/(2*b-3)*b*(2-gamma-2*ln(2))+3/(2*b-3)*b*Psi(b),-3/(2*b-3)
*b*(2-gamma-2*ln(2))+3/(2*b-3)*b*Psi(b),5/(2*b-3)*b-15/(15-16*
b+4*b^2)*b*Psi(b)+15/(15-16*b+4*b^2)*b*(8/3-gamma-2*ln(2)),-21
*b/(15-16*b+4*b^2)+7/2/(2*b-3)*b+105/(-105+142*b-60*b^2+8*b^3)
*b*Psi(b)-105/(-105+142*b-60*b^2+8*b^3)*b*(46/15-gamma-2*ln(2)
),-945*b/(2*b-5)/(2*b-7)/(2*b-9)/(2*b-3)*Psi(b)-3/2*(-16*b^3+
204*b^2-1040*b-105+1260*ln(2)+630*gamma)*b/(2*b-5)/(2*b-7)/(2*
b-9)/(2*b-3)];Chris
=STPT is an invaluable tool if you need
to import or extract data from anytext (ASCII) 'le. STPT 
makes
dealing with ASCII data easy and ef'cientthrough a patented
model and a graphical user interface where parameters canbe
de'ned and the extraction operation tested. The data which 
may
benumeric (tables, lists, numbers) or non-numeric (words,
sentences) can beimported as an array into languages such as
MATLAB, Visual Basic, C or C++where it can be used in
calculations. STPT can also be used to automaticallypopulate
tables in MS Word, worksheets in MS Excel or databases in
MSAccess.A free 10-day trial can be found at
http://www.stiwww.com/Parser.htm
=Heureka !!op(2,u);this is
or get right part of equation of type>> ax^3+bx^2+.... =
x^3*(p^t^3+.....)=-x^2*(...>> I am to get roots
such type of eqution where a,b... are solutions of> another
systm of equtions...>> Can anyone recomend me how to do
so,please?>> I am not to 'nd any solution nor simple
solution..>> Thak you for any help.>> Yours faithfully>>
Peter>>
 Heureka !!> op(2,u);> this is what I
want....Okay, but there are more elegant functions in
Maple:lhs(u) and rhs(u) return the left hand side and right
hand side of anequation u, respectively.-- Thomas RichardMaple
SupportScienti'c Computers 
GmbHhttp://www.scienti'c.de
>
Heureka !!>> op(2,u);>> this is what I want....>>
Okay, but there are more elegant functions in Maple:> lhs(u)
and rhs(u) return the left hand side and right hand side of
an> equation u, respectively.thank you I will try...
experts!Is there any way how to solve or get right part of
equation of typeax^3+bx^2+.... =
x^3*(p^t^3+.....)=-x^2*(...I am to get roots such
type of eqution where a,b... are solutions ofanother systm of
equtions...Can anyone recomend me how to do so,please?I am not
to 'nd any solution nor simple solution..Thak you for any
help.Yours faithfullyPeter
=If one considers that life is
based on Carbon 6, then one can also assumethat the hidden
messages that determine human destiny can also be exploredby
using some kind of decoding method using 6.If one assume that
all letters A to Z of English language are based onCarbon 6
code, so then for example A=6, B=12, C=18 etc and Z=156.If any
words that have historical meaning, and their numbers add up to
anyof the three fundamental 3 digit number 222, 444, 666, this
may help todecipher some hidden messages.For example: 666=
Computer = 18+ 90+ 78+ 96+ 126+120+30+108 ( the modern
civilizationbased on microchip? the ultimate limit of human
knowledge?)similarly 444= Cross, English, Jesus, Lucifer,
Messiah,and 222= India, ( some hidden energy source?
Enlightenment? )and also note the sum of all numbers for
letters A to Z is = 2025 (Spooky?)So, can anybody help us
decipher any more words, which may guide Humancivilization
towards the Holy Grail?R2 D2 & 4 Horsemen
=I have two
questions about Maclaurin series expansions of
goniometricfunctions (which I hope I should be asking here).
An exercise I can't'gure out is as follows:Use 
the series for
sin x and cos x to obtain the Maclaurin series fortan x as far
as the term in x^7. Deduce the series for ln cos x.Should I
start from the de'nition of tan x as sin x / cos x, and
thentry to divide the series for sin x by that for cos x
somehow? I trieddifferentiating tan x before going to series,
but that still seems toleaves me with a fraction of series, in
the second derivative already(2 * sin x / cos^3 x).As might be
expected, I don't see how the second part of the exerciseis
helped much by working out the 'rst part, unless I should
betransforming things to powers of e. I approximated the ln
cos x seriesby inserting z = cos(x) - 1 into ln(1 + z) = z -
z^2 / 2! + z^4 / 4! -... and ignoring higher powers (over 4)
of the cos x series (minus 1)to get an approximation of the
powers of z. I plotted the result inMatlab and it was indeed a
bad approximation. But the book wants anexact
answer.Thomas
I have two questions about Maclaurin series
expansions of goniometric>functions (which I hope I should be
asking here). An exercise I can't>'gure out is 
as follows:>Use
the series for sin x and cos x to obtain the Maclaurin series
for>tan x as far as the term in x^7. Deduce the series for ln
cos x.>Should I start from the de'nition of tan x as sin x /
cos x, and then>try to divide the series for sin x by that for
cos x somehow?Yes.>As might be expected, I don't see how the
second part of the exercise>is helped much by working out the
'rst part, unless I should be>transforming things to powers 
of
e.Robert Israel israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2
I have two questions
about Maclaurin series expansions of goniometric>functions
(which I hope I should be asking here). An exercise I can't'gure out is as follows:> I published may FOUR-BOX algorithm
athttp://www.wehner.org/eulerThis helps with the VISUALISATION
of McLaurin's series.Charles Douglas Wehner
Experts,Does anyone know how to correctMAPLE V 5.1
problemError, (in expand/bigprod) object too largeAfter
solving 1st and 2nd part of the 've or more parts of my
problem via:L1
:=sum(x(i)^6,i=1..n)*a+sum(x(i)^5,i=1..n)*b+sum(x(i)^4,i=1..n)
*c+sum(x(i)^3,i=1..n)*d;L2
:=sum(x(i)^5,i=1..n)*a+sum(x(i)^4,i=1..n)*b+sum(x(i)^3,i=1..n)
*c+sum(x(i)^2,i=1..n)*d;L3
:=sum(x(i)^4,i=1..n)*a+sum(x(i)^3,i=1..n)*b+sum(x(i)^2,i=1..n)
*c+sum(x(i),i=1..n)*d;L4:=sum(x(i)^3,i=1..n)*a+sum(x(i)^2,i=1.
.n)*b+sum(x(i),i=1..n)*c+d;R1:=sum(y(i)*x(i)^3,i=1..n);R2:=sum
(y(i)*x(i)^2,i=1..n);R3:=sum(y(i)*x(i),i=1..n);R4:=sum(y(i),i=
1..n);SOL:=solve({L1=R1,L2=R2,L3=R3,L4=R4},{a,b,c,d});EQV:=SOL
[1]*p^3+SOL[2]*p^2+SOL[3]*p+SOL[4];EQV2:=op(2,EQV);result:=
solve(EQV2=0,p);Can anyone send me any recomendation what to
do,please?thank you for any helpYours faithfullyPeter
Does anyone know how to correct MAPLE V 5.1 problem Error,
(in> expand/bigprod) object too largeSolve the cubic 'rst 
with
simple symbolic coef'ecints, then subs in thesolution of the
linear system.> After solving 1st and 2nd part of the 've or
more parts of my problem via:> L1 :
sum(x(i)^6,i=1..n)*a+sum(x(i)^5,i=1..n)*b+sum(x(i)^4,i=1..n)*c
+sum(x(i)^3,i 1..n)*d;> L2 :
sum(x(i)^5,i=1..n)*a+sum(x(i)^4,i=1..n)*b+sum(x(i)^3,i=1..n)*c
+sum(x(i)^2,i 1..n)*d;> L3 :
sum(x(i)^4,i=1..n)*a+sum(x(i)^3,i=1..n)*b+sum(x(i)^2,i=1..n)*c
+sum(x(i),i=1.> .n)*d;>
L4:=sum(x(i)^3,i=1..n)*a+sum(x(i)^2,i=1..n)*b+sum(x(i),i=1..n)
*c+d;>> R1:=sum(y(i)*x(i)^3,i=1..n);>
R2:=sum(y(i)*x(i)^2,i=1..n);> R3:=sum(y(i)*x(i),i=1..n);>
R4:=sum(y(i),i=1..n);>>
SOL:=solve({L1=R1,L2=R2,L3=R3,L4=R4},{a,b,c,d});>>
EQV:=SOL[1]*p^3+SOL[2]*p^2+SOL[3]*p+SOL[4];I assume that you
wnat SOL[1] to be the solution for a, SOL[2] thesolution for
b, etc. The above will not in general work because you 
can'tbe
sure of the order that a, b, c, d, appear in SOL. You need to
do this:EQV:= eval(a*p^3+b*p^2+c*p+d, SOL);and this will just
take the right sides of the equations, which is whatyou want
anayway.> result:=solve(EQV2=0,p);> Can anyone send me any
recomendation what to do,please?Take off the eval. Solve the
cubic 'rst, then use the eval:EQV:= a*p^3+b*p^2+c*p+d;SOL2:=
[solve(EQV, p)];eval(SOL2, SOL);Do not try to simplify this
expression.
> Does anyone know how to correct MAPLE V
5.1 problem Error, (in> expand/bigprod) object too large>>
Solve the cubic 'rst with simple symbolic 
coef'ecints, then
subs in the> solution of the linear system.It seems to be
elegant and functionable, but I was never used substitutionin
Maple, yet..
=I'm using maxima and I'd like to know if
there's a way to make it evaluatesomething I enter at the
command line (I have the 5.9.0 version forWindows).I mean
something like that:c:maximasomethingbinary-gclsomething>
 I'm using maxima and I'd
like to know if there's a way to make it evaluate> something 
I
enter at the command line (I have the 5.9.0 version for>
Windows).> I mean something like that:>
c:maximasomethingbinary-gclsomething> maxima.exe -eval>
integrate(3x+2,x);> 2> x> --> 2Under the unix versions, you
could type maxima --batch-string=integrate(3*x+2,x);to do what
you want. Unfortunately, this functionality is broken
underWindows in 5.9.0. I plan to 'x the Windows version in
5.9.1.In the meantime, you could experiment by setting the
followingenvironment variables:
MAXIMA_INT_INPUT_STRING=integrate(3x+2,x);
MAXIMA_INT_BATCH_FLAG=:batchI haven't tested this 
workaround.
In any event, the next release of Maximashould solve the
problem.--Jim Amundson
=James Amundson
<amundson@users.sourceforge.net> ha scritto nel messaggio> I
haven't tested this workaround. In any event, the next 
release
of Maxima> should solve the problem.
=James Amundson
<amundson@users.sourceforge.net> ha scritto nel messaggio>
maxima --batch-string=integrate(3*x+2,x);OK. If I put the
command into a batch (.mac 'le), is there a way to have
itevaluated and printed to the screen?
=I have a question
which, I guess, should be very stupid, but I 
couldn't'gure it
out looking at the manual (btw, where should I have looked
to'nd it?). The problem is that when I expand a polynomial 
the
output isorder in some way that I can 'gure out why. For
example:>> a 2 2 (x - 1) (x + 1)>> expand(a) 2 3 2 x - 2 x + 2
x - 2Why? How to have the normal (deg) ordering for 1 variable
polynomials?Fabio
 2 3> 2 x - 2 x + 2 x - 2>> Why? How to
have the normal (deg) ordering for 1 variable polynomials? By
telling MuPAD you want a polynomial:>> 2*poly(x-1)*poly(x+1)^2
3 2 2 poly(x + x - x - 1, [x]) If you happen to have an idea
how to *reliably* and *fast* check*any* expression whether it
should be ordered as a polynomial and ifso, in which
variables, please tell me.-- +--+ +--+| |+-|+ Christopher
Creutzig (ccr@mupad.de)
 2*poly(x-1)*poly(x+1)^2>> 3 2>
2 poly(x + x - x - 1, [x]) Oh, sorry: I still think the 2
*should* be automatically multipliedinto the polynomial and
keep writing as though it would.>> poly(2*(x-1)*(x+1)^2) 3 2
poly(2 x + 2 x - 2 x - 2, [x])-- +--+ +--+| |+-|+ Christopher
Creutzig (ccr@mupad.de)
=Ever wonder about original and
independent mathematical research?Well I'm happy to have my
own mathematical research to talk about, andit includes what I
call my prime counting function.You can read an exposition on
how my prime counting function works
athttp://groups.msn.com/AmateurMath/
primecountingexplained.msnwand there are two things to
remember about my work:It's correct, and it 
doesn't look like
anybody else's.However the other thing to know is that 
I've
been in a lot ofarguments with people who try to downplay its
value.Well you can try to judge for yourself at the page
linked to above,and then maybe follow the links down the left
side of that page tomore, like a short C++
implementation.James Harris
=How much would it cost to pay
you to stop posting in math newsgroups?> Ever wonder about
original and independent mathematical research?
=Martin
Johansen Algebra Calculator is a new innovative algebra
calculator,which aims at working on the most basic concepts,
giving it dynamiccapabilities.For more information visit
<http://mjac.8bit.co.uk>
=Wolfram Research is celebrating 15
years of unprecedented advances insymbolic and numeric
computation with the release of Mathematica5--the new
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technicalcomputing software. Key new technologies enable
Mathematica 5 tooutperform dedicated numerical systems in raw
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Fortran libraries or specializedpackages. With the release of
Mathematica 5, this choice is no longernecessary. Numerical
computations in Mathematica 5 are vastly fasterthan in earlier
versions and often, for example for numerical linearalgebra,
surpass the speed of dedicated numerical systems like
MATLAB,MATRIXx, and O- Matrix.Mathematica 5 introduces
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algorithms that are exclusive to Mathematica 5.
Otheralgorithms provide functionality that was up to now only
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original research by in-housedevelopers.Mathematica 5 also
extends Wolfram Research's position as the leaderin 
providing
integration with other software and standards. Inaddition to
updated versions of J/Link and 
MathLink--Mathematica'sstandard
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toseamlessly integrate Mathematica into applications using
Microsoft's.NET framework.The advances in Mathematica 5
strengthen Mathematica's position of notonly having the
broadest scope of any technical computing system, butalso
providing the best performance. Mathematica accomplishes this
bycombining outstanding numeric, symbolic, and graphical
capabilitieswith a uniquely productive programming language
and interactivedocument system.Mathematica's overall
capabilities without compromising any of itsaccuracy or expert
nature. Mathematica 5's unique combination ofspeed, scope, 
and
scalability will take you from start to 'nish, 
fromprototyping
to 'nal computations, from research to presentation, inthe
shortest possible time.For more information about Mathematica
5, visit: www.wolfram.com/mathematica/newin5
=This web-site
is intended for scientists, experts, teachers andstudents in
natural sciences such as mathematics, physics,
mechanics,engineering, computer science, cybernetics, computer
algebra (Maple,Mathematica), etc. The web-site represents
books, monographies andsoftware in the above 'elds created
under direction of Prof. V.Z.Aladjev and published by
International Academy Noosphere (Tallinn,Estonia) and Vilnius
Technical University (Vilnius, Lithuania)
duringhttp://www.geocities.com/noosphere_academyMoreover, we
shall be extremely grateful, if you will 'nd anopportunity to
acquaint with the above information the colleagues withwhich
you support business relations.
=I have a polynomial that
has an a radical in it and I was wondering if therewas a way
to solve it without squaring out the radical?For example when
trying to solve for x:a*x^2 - b*x + c - (x^2 - d^2)^(1/2) =
0What I've been doing is solving via:a*x^2 - b*x + c = (x^2 
-
d^2)^(1/2)Squaring both sides(a*x^2 - b*x + c)^2 = (x^2 -
d^2)and then solving the equation... problem is that you 
can't
solve foranything higher than a quadratic this way since 
you're
doubling theexponents to remove the radical. Unfortunately 
I'm
trying to solve anequation that's formatted like this but is
4th order and when I do thesquaring it's 8th order... which 
I
don't know how to solve.Btw, all the 
coef'cients are rational
and I'm only interested in therational
solutions...-Tom
=whoops.. I meant to say the coef'cients
and the solutions are real numbersnot that they had to be
rational :)> I have a polynomial that has an a radical in it
and I was wondering ifthere> was a way to solve it without
squaring out the radical?>> For example when trying to solve
for x:>> a*x^2 - b*x + c - (x^2 - d^2)^(1/2) = 0>> What I've
been doing is solving via:>> a*x^2 - b*x + c = (x^2 -
d^2)^(1/2)>> Squaring both sides>> (a*x^2 - b*x + c)^2 = (x^2
- d^2)>> and then solving the equation... problem is that you
can't solve for> anything higher than a quadratic this way
since you're doubling the> exponents to remove the radical.
Unfortunately I'm trying to solve an> equation 
that's
formatted like this but is 4th order and when I do the>
squaring it's 8th order... which I don't know 
how to solve.>>
Btw, all the coef'cients are rational and I'm 
only interested
in the> rational solutions...> -Tom>
 Unfortunately
I'm trying to solve an> equation that's 
formatted like this
but is 4th order and when I do the> squaring it's 8th
order... which I don't know how to solve.If all terms are of
even degree after your squaring, it is still solvable as a
quartic in the square of x.
 Btw, all the coef'cients are
rational and I'm only interested in the> rational
solutions...Then you're in luck. Even degree 8 can be done
under these rules.Clear denominators to get integer
coef'cients. Say highest terma*x^n, constant term b. 
Determine
all factors of a and all factors ofb. Try rationals of the form
b1/a1, where b1 is a factor of b anda1 is a factor of a. See if
any of them solve the equation. Allrational zeros may be found
in this way.Example... 3*x^8-3*x^3-7*x^2+x^7-2*x^6+4 ;factors
of 3 are: 3, 1, -1, -3;factrs of 4 are: 4, 2, 1, -1, -2,
-4;try all quotients:
4/3,4,-4,-4/3,2/3,2,-2,-2/3,1/3,1,-1,-1/3.Rational zeros are:
-1 and 2/3. Done.
to 'nd parameters a,b,c,d of cubic polynomialax^3+bx^2+cx+d
via cubic regression and therfor I am to solve equation viain
Maximasolve([Sxi3yi=a*Sxi6+b*Sxi5+c*Sxi4+d*Sxi3,Sxi3yi=a*Sxi5+
b*Sxi4+c*Sxi3+d*Sxi2,Sxiyi=a*Sxi4+b*Sxi3+c*Sxi2+d*Sxi,Syi=a*
Sxi3+b*Sxi2+c*Sxi+d],[a,b,c,d]);or in symbolic toolbox of
MatlabSOL=solve(ïSxi3yi=a*Sxi6+b*Sxi5+c*Sxi4+d*Sxi3[CapitalOTi
lde],'Sxi3yi=a
*Sxi5+b*Sxi4+c*Sxi3+d*Sxi2','Sxiyi=a*Sxi4+b*Sx
i3+c*Sxi2+d*Sxi'
,'Syi=a*Sxi3+b*Sxi2+c*Sxi+d','a
,b,c,d');a=simple(simple(SOL.a)
);b=simple(simple(SOL.b));c=simple(simple(SOL.c));d=simple(
simple(SOL.d));disp ï ïdisp ï 
ïdisp ï ïdisp
ïa='pretty(a,800)disp ï 
ïdisp ï ïdisp ï 
ïdisp
ïb='pretty(b,800)disp ï 
ïdisp ï ïdisp ï 
ïdisp
ïc='pretty(c,800)disp ï 
ïdisp ï ïdisp ï 
ïdisp
ïd='pretty(d,800)And I am to 'nd 
pretty form of solution, but
I am not able to do so...Can you recomend me how to do
so,please?... Or can you recomend me anysoftware (GPL
preffered) which can do so,please?I would like haering form
youYours faithfullyPeter Fodrek, PhD candidate in Control
Engineering....Yours faithfullyP.S.:S is abriviation for Suma
form i=1 to n of input/outputrt i i-thsample...numers after xi
of yi is power of such input/output variable so xi3yi2
meansxi^3*yi^2
to 'nd parameters a,b,c,d of cubic polynomial> ax^3+bx^2+cx+d
via cubic regression and therfor I am to solve equation via>
in Maxima????Not sure whether I understand what you want.*The*
third degree polynomial that takes value yi in xi (i=1..4)
is(x-x2)(x-x3)(x-x4)/[(x1-x2)(x1-x3)(x1-x4)]
y1+(x-x1)(x-x3)(x-x4)/[(x2-x1)(x2-x3)(x2-x4)]
y2+(x-x1)(x-x2)(x-x4)/[(x3-x1)(x3-x2)(x3-x4)]
y3+(x-x1)(x-x2)(x-x3)/[(x4-x1)(x4-x2)(x4-x1)] y4Also known as
Lagrange interpolation polynomial.Hope that helps
??Best,Amities, Olivier
=----- Original Message ----->>
*The* third degree polynomial that takes value yi in xi
(i=1..4) is>> (x-x2)(x-x3)(x-x4)/[(x1-x2)(x1-x3)(x1-x4)] y1>
+> (x-x1)(x-x3)(x-x4)/[(x2-x1)(x2-x3)(x2-x4)] y2> +>
(x-x1)(x-x2)(x-x4)/[(x3-x1)(x3-x2)(x3-x4)] y3> +>
(x-x1)(x-x2)(x-x3)/[(x4-x1)(x4-x2)(x4-x1)] y4>This is known
form for me, but I am to compute third degree polynomial
frommore than 4 elemets of x(i) and y(i), to correct errors in
real timemeasurement such as noise of real variables , for
example when x(i) is timeand y(i) is speed of the AC motor at
time x(i)...I do not know exact english term for that, but it
may be called solution ofpolynomial parameters from
overdeterminated (overde'ned) system ofequations whic is
variation of Least Squares-Rank Method specialized forthird
degree polynomial coe'cient solution..But thank you for
answer...Yours faithfullyPeter Fodrek
=I am pleased to
announce the publication of the book:Gene Expression
Programming: Mathematical Modeling by an
Arti'cialIntelligenceby Candida Ferreira, 2002.272 pp. ISBN
972-95890-5-4In this 'rst book on gene expression programming
the author describesthoroughly the basic gene expression
algorithm and numerous modi'cations tothis new algorithm,
providing all the implementation details so that anyonewith
elementary programming skills will be able to implement it
themselves.As a powerful meta-language, gene expression
programming touches all the'elds of computer intelligence and
everyone who faces challenging problemsand cannot solve them
using either traditional mathematical approaches
orsophisticated machine learning techniques can bene't from
the practicalunderstanding of this new powerful technique.The
book is self-contained and can be used by people with little
knowledgeof calculus and no prerequisites associated with
knowledge of anyprogramming language are required.The book
provides an introduction to this new exciting 'eld of
computerintelligence, including a large body of previously
unpublished materialssuch as:o data miningo classi'er 
systemso
parameter optimizationo evolution of Kolmogorov-Gabor
polynomialso times series predictiono evolution of linking
functionso multicellular systemso automatically de'ned
functionso user de'ned functionso complete neural network
inductionThe book also discusses some important and
controversial evolutionary topicsthat might be refreshing to
both evolutionary computists and evolutionarybiologists.For
further information and to order online, please go
to:http://www.gene-expression-programming.com/gep/Books/
index.aspCandida
Ferreira------------------------------------------------------
-----------------------Candida Ferreira, Ph.D.Chief Scientist,
Gepsoft73 Elmtree DriveBristol BS13 8NA, UKph: +44 (0) 117 330
9272www.gepsoft.com/gepsoftwww.gene-expression-programming.com
/author.asp---------------------------------------------------
--------------------------
=Please, look at
thishttp://www.paolo.zavarise.name/mathematica/First line was
evaluated by mathematica 4.0Second line (In[1]) was evaluated
by mathematica 5.0Why the difference?What's the meaning of
SimplifyDump`x ?Paolo
=Good morning!I wish to have some news
about Riemann hypothesis.I read some news about a geometrical
proof and an advert for aprobabilistic book who disagree with
riemann's hypothesis.Where is the truth???
Good morning!>I
wish to have some news about Riemann hypothesis.>I read some
news about a geometrical proof and an advert for
a>probabilistic book who disagree with riemann's
hypothesis.>Where is the truth???>>The truth?Either the
Riemann hypothesis is false or it is just barely true. Onthe
otherhand, the 'rst 450 billion or so zeros happen to be on
thecritical line, so Riemann hypothesis is `true' for a 
large
range.The Riemann hypothesis is now part of pop culture. Three
differentpopular books on RH appeared in April alone. I think
it safe to assumeany rumors of a proof will appear in the
press, in a timely manner.--
http://www.math.fsu.edu/~bellenotbellenot <At/> math.fsu.edu
+1.850.644.7189 (4053fax)
 Good morning!> I wish to have
some news about Riemann hypothesis.> I read some news about a
geometrical proof and an advert for a> probabilistic book who
disagree with riemann's hypothesis.> Where is the truth???No
proof or disproof having gone through the usual
mathematicalchain of veri'cation has yet appeared or even is
in the processor being read as far as I know. The
probabislitic disproofof Berliocchi is hanging around for some
years now: when itwas supposed to be a proof, someone pointed
out to the authorthat some exchange os summations was not
justi'ed. After thatthe proof became a disproof: since a
series can have two values,there is a contradiction. The catch
is that it is still the very sameseries with the very same
exchange. It's been published by somepublisher in economy 
and
paid by the author. As for the geometricalone, it's been
hanging around for some years now, several mistakesfound,
corrected, found, corrected, found, corrected, so no one
isparticularly keen on keeping an eye on it anymore (though
this onehas been written by a serious candidate).Best,Amities,
Olivier
=1) sum_{m_0, m_1, ..., m_Q} M! / (m_0! m_1! ...
m_Q!) sum_{k=0}^Q m_k (sin(k pi /(Q+1)))^22) sum_{m_0, m_1,
..., m_Q} M! / (m_0! m_1! ... m_Q!) (sum_{k=0}^Q m_k (sin(k pi
/(Q+1)))^2)^23) sum_{m_0, m_1, ..., m_Q} M! / (m_0! m_1! ...
m_Q!) (sum_{k=0}^Q m_k sin(k pi /(Q+1)) cos(k pi/(Q+1))
)^2where M! / (m_0! m_1! ... m_Q!) is actually the multinomial
coef'cient withsum_{k=0}^Q m_k = M and the outer summation is
over all possible combinationsof m_k values, Q is a positive
integer such that Q >= 2.I only could derive the closed-form
expression when Q = 1. For Q > 1, I don'ta lot.Wenyan
For interested people let's reproduce here part of my 
earlier
answersto Wenyan (M is considered 'xed) :> 1) sum_{m_0, m_1,
..., m_Q} M! / (m_0! m_1! ... m_Q!) sum_{k=0}^Q m_k (sin(k pi>
/(Q+1)))^2> = M/2*(Q+1)^M> 2) sum_{m_0, m_1, ..., m_Q} M! /
(m_0! m_1! ... m_Q!) (sum_{k=0}^Q m_k (sin(k pi>
/(Q+1)))^2)^2> = M*(2M+1)/8*(Q+1)^M> 3) sum_{m_0, m_1, ...,
m_Q} M! / (m_0! m_1! ... m_Q!) (sum_{k=0}^Q m_k sin(k pi>
/(Q+1)) cos(k pi/(Q+1)) )^2> = M/8*(Q+1)^M> where M! / (m_0!
m_1! ... m_Q!) is actually the multinomial coef'cient with>
sum_{k=0}^Q m_k = M and the outer summation is over all
possible combinations> of m_k values, Q is a positive integer
such that Q >= 2.> I only could derive the closed-form
expression when Q = 1. For Q > 1, I don't> a lot.> Wenyan A
'ne way to solve this is to use the equation (M positive 
'xed)
:(*) f(X1,X2,..,XQ)= (X0+X1+..+XQ)^M = sum_{m0+m1+..mQ=M}
M!/(m0!*m1!*...*mQ!) X0^m0*X1^m1*...*XQ^mQ (proof : apply the
binomial expansion to ((X0)+(X1+..+XQ))^M then to
((X1)+(X2+..+XQ))^M and so on...) some formal results : -
partial derivative relatively to Xk : df/dXk=
M*(X0+X1+..+XQ)^{M-1} (doesn't depend of k...) - Xk*df/dXk=
sum_{m0+m1+..mQ=M}
M!/(m0!*m1!*...*mQ!)mk*X0^m0*X1^m1*...*XQ^mQ Let's solve 3)
(and others) nicely using an operator approach ( 2)may be
computed the same way, for 1) the operator has to be applied
onlyonce) : - Start with the f equation (*) : - search the
operator that must be applied to f to get the additionalfactor
sum_{k=0}^Q m_k sin(k pi/(Q+1)) cos(k pi/(Q+1)) at the right or
the shorter sum_{k=0}^Q m_k sin(2k pi/(Q+1))/2 it is (d/dXk :
partial derivative relatively to Xk) Op= sum_{k=0}^Q sin(2k
pi/(Q+1))/2 Xk*d/dXk - apply this operator a second time
(since we want [sum_{k=0}^Q m_ksin(2k pi/(Q+1))/2] squared )
to get (at the left) :Op^2 f= sum_{n=0}^Q sin(2n pi/(Q+1))/2
Xn*d/dXn [sum_{k=0}^Q sin(2kpi/(Q+1))/2 Xk*d/dXk]
f(X0,X1,...,XQ) - ïexpand' the derivatives ; 
note that : d/dXn
[sum_{k=0}^Q h(k) Xk*d/dXk] f = h(n) df/dXn + sum_{k=0}^Q h(k)
Xk*d^2f/(dXn dXk) - set X0=X1=...=XQ=1 - expand the
trigonometric forms in exponential form - simplify using
sum_{k=0}^Q e^(2k pi/(Q+1)) = 0 : consequence thesimple sin(2k
pi/(Q+1)), cos(2k pi/(Q+1))... disappear entirely : thesin(2k
pi/(Q+1))^2 contribute only with their constant 1/2 part
(oncewritten in exponential form). Finally your sum 3) will be
identical to the sum 1) with twoadditional 1/2 factors and
that's it! Raymond
=I'm trying to, given an algebraic
expression, extract individual terms oneat a
time.Example:6a^3b^2 + a^2b + 4a + 1I would like to be able to
extract 6a^3b^2, a^2b, 4a and 1 using a simple,built-in
procedure.Also, I would like one that, given an algebraic
expression, returns thenumber of terms in the expression.I
original thought that nops and op would do the job, but they
don't doquite what I was looking for. For example,nops(
6a^3b^2 ) would return 3, with 6 being op 1, a^3 being op2 and
b^2being op3, where as I would have liked nopsto return 1.Any
suggestions? If my phrasing is to vague, or you'd like some
more info,just let me know. - Chris
 I'm trying to,
given an algebraic expression, extract individual terms one>
at a time.> Example:> 6a^3b^2 + a^2b + 4a + 1> I would
like to be able to extract 6a^3b^2, a^2b, 4a and 1 using a
simple,> built-in procedure.> Also, I would like one that,
given an algebraic expression, returns the> number of terms in
the expression.> I original thought that nops and op would
do the job, but they don't do> quite what I was looking for.
For example,> nops( 6a^3b^2 ) would return 3, with 6 being op
1, a^3 being op2 and b^2> being op3, where as I would have
liked nops> to return 1.> Any suggestions? If my phrasing is
to vague, or you'd like some more info,> just let me know.>
First check if the type is `+`; if not there is only one
term...> u := 6*a^3*b^2 + a^2*b + 4*a + 1; 3 2 2 u := 6 a b +
a b + 4 a + 1> type(u,`+`); true> nops(u); 4> op(3,u); 4 a>--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
=Ah, I
tried something similar to that before, but I must have done
it wrong. - Chris I'm trying to, given an algebraic
expression, extract individual termsone> at a time.>>
Example:>> 6a^3b^2 + a^2b + 4a + 1>> I would like to
be able to extract 6a^3b^2, a^2b, 4a and 1 using asimple,>
built-in procedure.> Also, I would like one that, given an
algebraic expression, returns the> number of terms in the
expression. I original thought that nops and op would
do the job, but they don't do> quite what I was looking for.
For example,> nops( 6a^3b^2 ) would return 3, with 6 being
op 1, a^3 being op2 and b^2> being op3, where as I would
have liked nops> to return 1.>> Any suggestions? If my
phrasing is to vague, or you'd like some moreinfo,> just let
me know.> First check if the type is `+`; if not there is
only one term...>> u := 6*a^3*b^2 + a^2*b + 4*a + 1;>> 3 2
2> u := 6 a b + a b + 4 a + 1>> type(u,`+`);>> true>>
nops(u);>> 4>> op(3,u);>> 4 a>> -- > G. A.
Edgarhttp://www.math.ohio-state.edu/~edgar/
=What is
practical signi'cance of direction 'eld plot of
DifferentialEquation ? One such plot in Maple is like
thiswith(DEtools):ode := diff(y(x), x) = x*y(x);inits := {[0,
1], [0, 2]};DEplot(ode, y(x), x = -3..3, y = 0..3, inits,
scaling = constrained);What do dashed lines and the curves
represent in the plot? Do theygive an idea of general solution
of the equation or something else?
=I'm trying to 'nd a
computer algebra system that allows for thesymbolic
manipulation and simpli'cation of matrix equations. In
otherwords, a system that can reduce the following(inv(A) .
B)' A to B'where A is a m-by-m symmetric, 
positive de'nite
matrix, B is anarbitrary m-by-n matrix, . is the matrix dot
product, inv() is thematrix inverse operator, and ï is the
matrix transpose operator.I want to do this SYMBOLICALLY
without having to explicitly de'ne thematrices in terms of
their component scalars.Does such a package exist
?Rudolph
=Hallo,I'm desperately looking for any support of
the Mathematica Signals&Systems package that goes further than
the Online-Documentation. Maybe someone is out there who has
considerable experience with this package.Oliver
Friedrich
=How are questions normally posted to this group?
I'll use LaTeX-ishnotation in this post. I'm 
not a
mathematician, just an engineer witha question.I understand
that for a vector r_i, that dr_i/dr_j = delta_ij.I also think
that for a tensor t_ij, that dt_ij/dt_kl = delta_ik
delta_lj.Is this correct?What about when t_ij = t_ji?
dt_ij/dt_kl = dt_ji/dt_kl = dt_ij/dt_lk, right?So then
does:dt_ij/dt_kl = delta_ik delta_lj + delta_il delta_kj -
delta_iklj?Kevin Van Workum, PhDX-Replace-Address:
yesreply-to: vanw a_t nist d_o_t gov
really like to know what does compact support means.I'm 
trying
to become more familiar with wavelets, and i sometimes see this
expression.brAnton.
to know what does compact support means. I'm trying> to 
become
more familiar with wavelets, and i sometimes see this>
expression.Wrong group, by the way, but anyway:The support of
a function f(x) is the closure of the set of points wheref(x)
is not 0. (Closure means you include boundary points). A
functionhas compact support if that set, the support of f(x),
is bounded. Theword compact, in this context, means closed and
bounded.-- David L. Johnson __o | When you are up to your ass
in alligators, it's hard to remember _`(,_ | that your 
initial
objective was to drain the swamp. -- LBJ (_)/ (_) | 
compact support means. I'm trying> to become more familiar
with wavelets, and i sometimes see this> expression.>
Wrong group, by the way, but anyway:> The support of a
function f(x) is the closure of the set of points where> f(x)
is not 0. (Closure means you include boundary points). A
function> has compact support if that set, the support of
f(x), is bounded. ^^^^^^^ strictly speaking, a compact
setGenerally a compact set S is one for which, from every open
covering onecan select a 'nite subcovering that is also a
covering.> The word compact, in this context, means closed and
bounded. True only for sets in E_n .-- Julian V. NobleProfessor
^^^^^^^^^^^^^^^^^^http://galileo.phys.virginia.edu/~jvn/
Science knows only one commandment: contribute to science. --
Bertolt Brecht, Galileo.
I would really like to know what does compact support means.
I'm>> trying to become more familiar with wavelets, and i
sometimes see this>> expression. Wrong group, by the
way, but anyway: The support of a function f(x) is the
closure of the set of points>> where f(x) is not 0. (Closure
means you include boundary points). A>> function has compact
support if that set, the support of f(x), is>> bounded.>
^^^^^^^> strictly speaking, a compact set> Generally a
compact set S is one for which, from every open covering one>
can select a 'nite subcovering that is also a covering.>>
The word compact, in this context, means closed and bounded. True only for sets in E_n .Or complete metric spaces, but
that was the context anyway. This issomeone who is asking
about terminology used in wavelets papers, and Ithought the
simplest explanation would be the best.-- David L. Johnson __o
| Business! cried the Ghost. Mankind was my business. The _`(,_
| common welfare was my business; charity, mercy, forbearance,
(_)/ (_) | and benevolence, were, all, my business. The
dealings of my trade were but a drop of water in the
comprehensive ocean of my business! --Dickens, A Christmas
ALL!> I would really like to know what does compact
support means. I'm>> trying to become more familiar with
wavelets, and i sometimes see this>> expression.
Wrong group, by the way, but anyway: The support of a
function f(x) is the closure of the set of points>> where
f(x) is not 0. (Closure means you include boundary points). A> function has compact support if that set, the support of
f(x), is>> bounded.> ^^^^^^^> strictly speaking, a
compact set> Generally a compact set S is one for which,
from every open covering one> can select a 'nite subcovering
that is also a covering.>> The word compact, in this
context, means closed and bounded.> True only for sets
in E_n .> Or complete metric spaces, but that was the
context anyway. This is> someone who is asking about
terminology used in wavelets papers, and I> thought the
simplest explanation would be the best.complete, so your
condition is clearly insuf'cient. Wavelet theoryand (closed
and bounded) is important.Simon-- The good Christian should
beware of mathematicians, and all those whomake empty
prophecies. The danger already exists that themathematicians
have made a covenant with the devil to darken thespirit and to
con'ne man in the bonds of Hell. -- St. Augustin
=This note
explains the story of the so-called
Bailey-Borwein-Plouffealgorithmand formula. The story began
many years ago in 1974 when I wanted to 'nda formula for the
n'th digit of Pi. I was studying rational
andirrationalnumbers. With my calculator I was computing
inverses of primes andcouldeasily 'nd a way to compute those
inverses in base 10 to many digitsusingcongruences and rapid
exponentiation. Since it appeared impossible todothe same for
Pi, I wanted then to 'nd a simple formula f(n) thatcould
computethe n'th digit of Pi. I had that idea for 20
years.Since the computation of Pi looksmore complicated than
the number E , i.e. exp(1), I studied a way tocomputethat
number instead. At that time (around 1983), I had a simple
Basicprogramthat used a spigot algorithm to compute E, as
expected that algorithmworked butof course but was taking an
increasing amount of memory. My questionwas :why can't we do
it for E or Pi or any irrational numbers like sqrt(2).It was
during the year 1994 that I began to compute arctan series
butI did notrealized that this meant a lot. I was able to use
an algorithm tocompute arctanof 1/5 with fast exponentiation
without realizing that it couldcompute arctan(1/5)in base 5
very fast since the rapid exponentiation was natural in
thatbase.Later in 1995, around august 7 of that year I
suddenly realized thatlog(2) wasfast computable in base 2.
Since I had a bit of experience with spigotalgorithmsand also
my little Basic program to compute arctan, it was notdif'cult
to adaptthe algorithm to log(2).In the next few days I made my
'rst program : A program to computelog(9/10)in base 10 using 
a
very small amount of memory and very fast. Theprogram had432
characters long.That discovery was a shock to me. I realized
that I had found it yesbut it was notnew to me since I could
do arctan(1/5) easily too but it took me 2years to realize
it.This is where I began to use Pari-Gp, that program could
'nd aninteger relationamong real numbers (up to a certain
precision), very fast.During my stay at Bordeaux University in
1992-1993 I perfected thatprogram I hadthat could interface
Pari-Gp and Maple. That little Unix script had
anenormousadvantage of §exibility because I could set up a
series of realnumbers to test among 1unknown. At that time I
was beginning to 'nd new results, theprograms were able 
to'nd
identities.That program was the one that found the formula for
Pi in hexadecimal(or binary).I also used another one : PSLQ. It
was a good program but a bitcumbursome to usesince it is
written in fortran. Nevertheless I made an interface toMaple
too.Pari-Gp was by far easier to use and faster for small
cases (up to 10real numbers atthe time with 100 digits
precision was enough for those kind ofproblems).This is where
I made the biggest mistake in my life : To accept
thecollaborationof Peter Borwein and David H. Bailey as
co-founders of that algorithmand formulawhen they have found
nothing at all. David Bailey was not even closeto me whenI
found the formula. He was added to the group 2 months after
thediscovery.I was naively thinking that I could negociate a
job as professor atSimon FraserUniversity, which failed. I am
very poor at negociations.in October1995. I was at Jon
borwein's house and he had a copy of the newspaperin 
hand.This
is where I asked him to become a professor at SFU. He
simplyreplied rightaway < don't even think about it >. I
thought, this is the best chanceI will ever haveto become a
professor there, since it failed, I decided that I had toleave
that place.I was very frustrated at that time, in late 1995
after the discovery.I realized thatmany small details where
terribly wrong. They were getting a lot ofcredit for
thediscovery and I had the impression of not getting anything
in return.My strategyasked Peter Borwein? Your brotherhas
nothing to do with this!. He simply replied that the
PublicRelations at theUniversity made a mistake.Later that
year, I was invited to a ceremony in Vancouver for the
CUFA(faculty ofthe year Award).This is a prize with plaque and
mention that those 2 brothers receivedfor thediscovery of the
formula. They simply mentioned my name at theceremony andI
received nothing at all. They made a toast to the queen of
England,I did notstand up.In late 1995, there was that
Canadian Math Soc. congress in Vancouver,I was notinvited to
talk about the discovery. There was even a guy (Stan
Wagon)that saidto me, I don't know if you have anything to 
do
with this but in allcase, this isgood for you isn't ?Then in
1996, I realized that if I get up at night to hate them it isa
very bad sign,it means that I have to leave that place (Simon
Fraser university).I was convinced I had no future at all with
those 2 guys around.I was making serious plans to leave.The
story of the formula (my formula), was not the only one. The
samething happenedwith the ISC (the Inverse Symbolic
Calculator). The story is even moreridiculous.I opened the
site with my constants in July 1995 and it was animmediate
success.The 2 Borweins had nothing to do with that thing, I
had made thetables and allof the Unix programs to run it. The
precious help I had was from AdamVan Tuyl, agraduate student,
he made most of the code behind the web pages, laterPaul
Irvinemade some additional code.At that time the local
administrator of the lab. tried to convince meto stay even to
pay mefor maintaining the ISC, I refused. I wanted to leave
with what I had: my tables ofreal numbers and sequences I
worked for years (since 1986). This iswhy I opened thePlouffe
Inverter with my name in 1998, to keep what was mine.When I
realized that I was about to loose the paternity of the ISC,
Ileft in march 1997.I went to Champaign Illinois to work for
Wolfram and Mathematica.(this time it took me less time), that
one was worst than the 2brothers combined. I simplyleft as soon
as I could, 5 months later.Peter Borwein wanted very much that
I do a Ph. D. on the ISC but hewanted also tothesis. Again
thesame story was going on, these 2 guys are so greedy I 
can't
believeit. The behavior theyhad with me was not exclusive,
especially Peter Borwein he was thesame with most ofhis
students, especially the good ones, sucking the maximum. Jon
isthe same but hehas more talent in politics (more money too).
He is good but has atendency to sitehimself a lot. He thinks
that if he had the idea of the sum of 2numbers at one pointin
his life then all formulas in mathematics are his own
discovery.About David H. Bailey. He came after the discovery
of the formula andmy small basicprogram , I had also a fortran
version. This is where Peter Borweinsuggested to addhim as a
collaborator to the discovery since he contributed to it (ashe
said), this is mywouldn't ?!David H. Bailey (and Ferguson) 
are
the authors of the PSLQ program.That program isthe <american>
version of the Pari-Gp program. I used it a little itis true,
but what madethe discovery was pari-Gp and Maple interface
program I had. Soactually, that personhas nothing to do with
the discovery of that algorithm and very littleto do with
the'nding of the formula. The mistake was mine. Saying that
Bailey foundthe formulais like saying that the formula was
found by the Maple and Basicprogram.I tried very hard to
correct the situation avoiding the subject of theactual
discoverybase 10. I thoughtnaively again that this would
re-establish the situation, it did not.I almost acceptedto do
a 'lm at one point in 1999 when a certain guy from England
thatwanted to make amovie on Pi and the discovery of the
formula. he asked me if I wouldaccept to talkabout my
<differents> with the Borweins. I did not wanted to go inthat
direction,I should had. There was that book of Jean-Paul
Delahaye (le fascinantnombre pi)that mentioned the Plouffe
algorithm and formula because I told himpart of thestory. In
some way I was afraid of revealing that enormous story.Why was
I so naive ?I had a previous collaboration with Neil Sloane and
the Encyclopediaof Integer Sequencesand the web site, this was
really a big success and Neil is the personI respect the
mostin mathematics so this is why I thought (wrongly ) that
mycollaboration with theBorweins had to go well, a big
mistake.Why do I write this ?To tell the truth and also the
arrogance of those people makes mesick.Will I gain something
from this ? I don't care, I have nothing toloose.Simon
Plouffe
=wow. and you said,your 5 months working for Sir
David andthe Wolframites was worse !?!> This note explains the
story of the so-called Bailey-Borwein-Plouffe> algorithm> and
formula.--UN HYDROGEN (sic; Methanex (TM) reformanteurs)
ECONOMIE?...La Troi Phases d'Exploitation de la Protocols 
des
Grises de Kyoto:(FOSSILISATION [McCainanites?]
(TM/sic))/BORE/GUSH/NADIR @
http://www.tarpley.net/aobook.htm.Http://www.tarpley.net/
bushb.htm (content partiale, below): 17 -- L'ATTEMPTER de 
COUP
D'ETAT, 3/30/81 23 -- Le FIN d'HISTOIRE 24 -- 
L'ORDEUR du MONDE
NOUVEAU 25 -- THYROID STORK !?!
=it may have been your
neglect of making the toast. what ever you do,don't let 
Betty
Dos raise her Excalibur over your bare neck,sir! > I went to
Champaign Illinois to work for Wolfram and Mathematica.> (this
time it took me less time), that one was worst than the 2>
brothers combined. I simply> left as soon as I could, 5 months
later.--Dec.2000 ïWAND' Chairman Paul 
O'Neill, reelectedto
Board.
Newsish?http://www.rand.org/publications/randreview/issues/rr
.12.00/http://members.tripod.com/~american_almanac
=A
fascinating story.Here is a re-post of Simon Plouffe's post,
which seems to be poorlyformatted. I have improved the layout
and corrected some minor spellingerrors. I have not corrected
grammatical errors. I have no connection withSimon
Plouffe.Clive
Tooth#########################################################
###############################################This note
explains the story of the so-called
Bailey-Borwein-Plouffealgorithm and formula.The story began
many years ago in 1974 when I wanted to 'nd a formula forthe
n'th digit of Pi. I was studying rational and irrational
numbers. Withmy calculator I was computing inverses of primes
and could easily 'nd a wayto compute those inverses in base 
10
to many digits using congruences andrapid exponentiation. Since
it appeared impossible to do the same for Pi, Iwanted then to
'nd a simple formula f(n) that could compute the 
n'th digitof
Pi. I had that idea for 20 years.Since the computation of Pi
looks more complicated than the number E , i.e.exp(1), I
studied a way to compute that number instead. At that time
(around1983), I had a simple Basic program that used a spigot
algorithm to computeE, as expected that algorithm worked but
of course but was taking anincreasing amount of memory. My
question was : why can't we do it for E orPi or any 
irrational
numbers like sqrt(2).It was during the year 1994 that I began
to compute arctan series but I didnot realized that this meant
a lot. I was able to use an algorithm tocompute arctan of 1/5
with fast exponentiation without realizing that itcould
compute arctan(1/5) in base 5 very fast since the rapid
exponentiationwas natural in that base.Later in 1995, around
august 7 of that year I suddenly realized that log(2)was fast
computable in base 2. Since I had a bit of experience with
spigotalgorithms and also my little Basic program to compute
arctan, it was notdif'cult to adapt the algorithm to log(2).
In the next few days I made my'rst program : A program to
compute log(9/10) in base 10 using a very smallamount of
memory and very fast. The program had 432 characters long.That
discovery was a shock to me. I realized that I had found it yes
but itwas not new to me since I could do arctan(1/5) easily too
but it took me 2years to realize it.This is where I began to
use Pari-Gp, that program could 'nd an integerrelation among
real numbers (up to a certain precision), very fast.During my
stay at Bordeaux University in 1992-1993 I perfected that
programI had that could interface Pari-Gp and Maple. That
little Unix script had anenormous advantage of §exibility
because I could set up a series of realnumbers to test among 1
unknown. At that time I was beginning to 'nd newresults, the
programs were able to 'nd identities.That program was the one
that found the formula for Pi in hexadecimal (orbinary). I
also used another one : PSLQ. It was a good program but a
bitcumbersome to use since it is written in Fortran.
Nevertheless I made aninterface to Maple too. Pari-Gp was by
far easier to use and faster forsmall cases (up to 10 real
numbers at the time with 100 digits precision wasenough for
those kind of problems).This is where I made the biggest
mistake in my life : To accept thecollaboration of Peter
Borwein and David H. Bailey as co-founders of thatalgorithm
and formula when they have found nothing at all. David Bailey
wasnot even close to me when I found the formula. He was added
to the group 2months after the discovery.I was naively thinking
that I could negotiate a job as professor at SimonFraser
University, which failed. I am very poor at negotiations. I
rememberwas at Jon Borwein's house and he had a copy of the
newspaper in hand. Thisis where I asked him to become a
professor at SFU. He simply replied rightaway < don't even
think about it >. I thought, this is the best chance Iwill
ever have to become a professor there, since it failed, I
decided thatI had to leave that place.I was very frustrated at
that time, in late 1995 after the discovery. Irealized that
many small details where terribly wrong. They were getting
alot of credit for the discovery and I had the impression of
not gettingof the Globe and Mail, I asked Peter Borwein : why
did they putted the photothis!. He simply replied that the
Public Relations at the University made amistake. Later that
year, I was invited to a ceremony in Vancouver for theCUFA
(faculty of the year Award). This is a prize with plaque and
mentionthat those 2 brothers received for the discovery of the
formula. They simplymentioned my name at the ceremony and I
received nothing at all. They made atoast to the queen of
England, I did not stand up.In late 1995, there was that
Canadian Math Soc. congress in Vancouver, I wasnot invited to
talk about the discovery. There was even a guy (Stan
Wagon)that said to me, I don't know if you have anything to 
do
with this but inall case, this is good for you isn't ?Then 
in
1996, I realized that if I get up at night to hate them it is
a verybad sign, it means that I have to leave that place
(Simon Fraseruniversity). I was convinced I had no future at
all with those 2 guysaround. I was making serious plans to
leave.The story of the formula (my formula), was not the only
one. The same thinghappened with the ISC (the Inverse Symbolic
Calculator). The story is evenmore ridiculous. I opened the
site with my constants in July 1995 and it wasan immediate
success. The 2 Borweins had nothing to do with that thing,
Ihad made the tables and all of the Unix programs to run it.
The precioushelp I had was from Adam Van Tuyl, a graduate
student, he made most of thecode behind the web pages, later
Paul Irvine made some additional code.At that time the local
administrator of the lab. tried to convince me tostay even to
pay me for maintaining the ISC, I refused. I wanted to
leavewith what I had : my tables of real numbers and sequences
I worked for years(since 1986). This is why I opened the
Plouffe Inverter with my name in1998, to keep what was mine.
When I realized that I was about to loose thepaternity of the
ISC, I left in march 1997. I went to Champaign Illinois towork
for Wolfram and Mathematica. (this time it took me less time),
that onewas worst than the 2 brothers combined. I simply left
as soon as I could, 5months later.Peter Borwein wanted very
much that I do a Ph. D. on the ISC but he wantedthesis. Again
the same story was going on, these 2 guys are so greedy 
Ican't
believe it. The behavior they had with me was not
exclusive,especially Peter Borwein he was the same with most
of his students,especially the good ones, sucking the maximum.
Jon is the same but he hasmore talent in politics (more money
too). He is good but has a tendency tosite himself a lot. He
thinks that if he had the idea of the sum of 2numbers at one
point in his life then all formulas in mathematics are hisown
discovery.About David H. Bailey. He came after the discovery
of the formula and mysmall basic program , I had also a
Fortran version. This is where PeterBorwein suggested to add
him as a collaborator to the discovery since hecontributed to
it (as he said), this is my second big mistake. Of course
heFerguson) are the authors of the PSLQ program. That program
is the<American> version of the Pari-Gp program. I used it a
little it is true,but what made the discovery was Pari-Gp and
Maple interface program I had.So actually, that person has
nothing to do with the discovery of thatalgorithm and very
little to do with the 'nding of the formula. The mistakewas
mine. Saying that Bailey found the formula is like saying that
theformula was found by the Maple and Basic program.I tried
very hard to correct the situation avoiding the subject of
thefor the base 10. I thought naively again that this would
re-establish thesituation, it did not. I almost accepted to do
a 'lm at one point in 1999when a certain guy from England 
that
wanted to make a movie on Pi and thediscovery of the formula.
he asked me if I would accept to talk about my<differents>
with the Borweins. I did not wanted to go in that direction,
Ishould had. There was that book of Jean-Paul Delahaye (le
fascinant nombrepi) that mentioned the Plouffe algorithm and
formula because I told him partof the story. In some way I was
afraid of revealing that enormous story.Why was I so naive ? I
had a previous collaboration with Neil Sloane and
theEncyclopedia of Integer Sequences and the web site, this
was really a bigsuccess and Neil is the person I respect the
most in mathematics so this iswhy I thought (wrongly ) that my
collaboration with the Borweins had to gowell, a big
mistake.Why do I write this ? To tell the truth and also the
arrogance of thosepeople makes me sick.Will I gain something
from this ? I don't care, I have nothing to
loose.####################################################
=
> Will I gain something from this ? I don't care, > I have
nothing to loose.Who did build Versailles castle?Was it really
Louis XIV? None of the paintingsshows him with dirty
architects
Hardouin-Mansart and Le Notreyou won't have the correct
answer, do you?It is a sad story you told. But at least you
canbe very proud about your 'ndings.I am strongly reminded of
Eric Weisstein's troublesome time ago. Maths and Cleverness
are very verydifferent as it seems. But it is possible to
beweak in maths and unclever too. I know that :-)Rainer
=[Note
follow-ups]> This note explains the story of the so-called
Bailey-Borwein-Plouffe> algorithm> and formula.[SNIP - a tale
of misattribution of a truly miraculous discovery.](Note - a
version in French can be found in fr.sci.maths) The Miraculous
Bailey-Borwein-Plouffe Pi Algorithm =Lies infuriate fellow
author much... Oily rag? Pipe bomb?Note for anagramatists who
wish to 'nd better ïgrams - 
ïSFU', for Simon-Frasier
University, is entirely relevant, and in the above
letters.'Gram by Lardy Girl, as I couldn't get 
beyond
horrifying mathematical liePhil
=This note explains the
story of the so-called Bailey-Borwein-Plouffe algorithm and
formula. The story began many years ago in 1974 when I wanted
to 'nda formula for the n'th digit of Pi. I was 
studying
rational and irrationalnumbers. With my calculator I was
computing inverses of primes and couldeasily 'nd a way to
compute those inverses in base 10 to many digits
usingcongruences and rapid exponentiation. Since it appeared
impossible to dothe same for Pi, I wanted then to 'nd a 
simple
formula f(n) that could computethe n'th digit of Pi. I had 
that
idea for 20 years.Since the computation of Pi looksmore
complicated than the number E , i.e. exp(1), I studied a way
to computethat number instead. At that time (around 1983), I
had a simple Basic programthat used a spigot algorithm to
compute E, as expected that algorithm worked butof course but
was taking an increasing amount of memory. My question was
:why can't we do it for E or Pi or any irrational numbers 
like
sqrt(2).It was during the year 1994 that I began to compute
arctan series but I did notrealized that this meant a lot. I
was able to use an algorithm to compute arctanof 1/5 with fast
exponentiation without realizing that it could compute
arctan(1/5)in base 5 very fast since the rapid exponentiation
was natural in that base.Later in 1995, around august 7 of
that year I suddenly realized that log(2) wasfast computable
in base 2. Since I had a bit of experience with spigot
algorithmsand also my little Basic program to compute arctan,
it was not dif'cult to adaptthe algorithm to log(2).In the
next few days I made my 'rst program : A program to compute
log(9/10)in base 10 using a very small amount of memory and
very fast. The program had432 characters long.That discovery
was a shock to me. I realized that I had found it yes but it
was notnew to me since I could do arctan(1/5) easily too but
it took me 2 years to realize it.This is where I began to use
Pari-Gp, that program could 'nd an integer relationamong real
numbers (up to a certain precision), very fast.During my stay
at Bordeaux University in 1992-1993 I perfected that program I
hadthat could interface Pari-Gp and Maple. That little Unix
script had an enormousadvantage of §exibility because I could
set up a series of real numbers to test among 1unknown. At
that time I was beginning to 'nd new results, the programs
were able to'nd identities.That program was the one that 
found
the formula for Pi in hexadecimal (or binary).I also used
another one : PSLQ. It was a good program but a bit cumbursome
to usesince it is written in fortran. Nevertheless I made an
interface to Maple too.Pari-Gp was by far easier to use and
faster for small cases (up to 10 real numbers atthe time with
100 digits precision was enough for those kind of
problems).This is where I made the biggest mistake in my life
: To accept the collaborationof Peter Borwein and David H.
Bailey as co-founders of that algorithm and formulawhen they
have found nothing at all. David Bailey was not even close to
me whenI found the formula. He was added to the group 2 months
after the discovery.I was naively thinking that I could
negociate a job as professor at Simon FraserUniversity, which
failed. I am very poor at negociations.October1995. I was at
Jon borwein's house and he had a copy of the newspaper in
hand.This is where I asked him to become a professor at SFU.
He simply replied rightaway < don't even think about it >. I
thought, this is the best chance I will ever haveto become a
professor there, since it failed, I decided that I had to
leave that place.I was very frustrated at that time, in late
1995 after the discovery. I realized thatmany small details
where terribly wrong. They were getting a lot of credit for
thediscovery and I had the impression of not getting anything
in return. My strategyasked Peter BorweinYour brotherhas
nothing to do with this!. He simply replied that the Public
Relations at theUniversity made a mistake.Later that year, I
was invited to a ceremony in Vancouver for the CUFA (faculty
ofthe year Award).This is a prize with plaque and mention that
those 2 brothers received for thediscovery of the formula. They
simply mentioned my name at the ceremony andI received nothing
at all. They made a toast to the queen of England, I did
notstand up.In late 1995, there was that Canadian Math Soc.
congress in Vancouver, I was notinvited to talk about the
discovery. There was even a guy (Stan Wagon) that saidto me, I
don't know if you have anything to do with this but in all
case, this isgood for you isn't ?Then in 1996, I realized 
that
if I get up at night to hate them it is a very bad sign,it
means that I have to leave that place (Simon Fraser
university).I was convinced I had no future at all with those
2 guys around.I was making serious plans to leave.The story of
the formula (my formula), was not the only one. The same thing
happenedwith the ISC (the Inverse Symbolic Calculator). The
story is even more ridiculous.I opened the site with my
constants in July 1995 and it was an immediate success.The 2
Borweins had nothing to do with that thing, I had made the
tables and allof the Unix programs to run it. The precious
help I had was from Adam Van Tuyl, agraduate student, he made
most of the code behind the web pages, later Paul Irvinemade
some additional code.At that time the local administrator of
the lab. tried to convince me to stay even to pay mefor
maintaining the ISC, I refused. I wanted to leave with what I
had : my tables ofreal numbers and sequences I worked for
years (since 1986). This is why I opened thePlouffe Inverter
with my name in 1998, to keep what was mine.When I realized
that I was about to loose the paternity of the ISC, I left in
march 1997.I went to Champaign Illinois to work for Wolfram
and Mathematica.(this time it took me less time), that one was
worst than the 2 brothers combined. I simplyleft as soon as I
could, 5 months later.Peter Borwein wanted very much that I do
a Ph. D. on the ISC but he wanted also tothesis. Again thesame
story was going on, these 2 guys are so greedy I can't 
believe
it. The behavior theyhad with me was not exclusive, especially
Peter Borwein he was the same with most ofhis students,
especially the good ones, sucking the maximum. Jon is the same
but hehas more talent in politics (more money too). He is good
but has a tendency to sitehimself a lot. He thinks that if he
had the idea of the sum of 2 numbers at one pointin his life
then all formulas in mathematics are his own discovery.About
David H. Bailey. He came after the discovery of the formula
and my small basicprogram , I had also a fortran version. This
is where Peter Borwein suggested to addhim as a collaborator to
the discovery since he contributed to it (as he said), this is
mywouldn't ?!David H. Bailey (and Ferguson) are the authors 
of
the PSLQ program. That program isthe <american> version of the
Pari-Gp program. I used it a little it is true, but what
madethe discovery was pari-Gp and Maple interface program I
had. So actually, that personhas nothing to do with the
discovery of that algorithm and very little to do with
the'nding of the formula. The mistake was mine. Saying that
Bailey found the formulais like saying that the formula was
found by the Maple and Basic program.I tried very hard to
correct the situation avoiding the subject of the actual
discovery10. I thoughtnaively again that this would
re-establish the situation, it did not. I almost acceptedto do
a 'lm at one point in 1999 when a certain guy from England 
that
wanted to make amovie on Pi and the discovery of the formula.
he asked me if I would accept to talkabout my <differents>
with the Borweins. I did not wanted to go in that direction,I
should had. There was that book of Jean-Paul Delahaye (le
fascinant nombre pi)that mentioned the Plouffe algorithm and
formula because I told him part of thestory. In some way I was
afraid of revealing that enormous story.Why was I so naive ?I
had a previous collaboration with Neil Sloane and the
Encyclopedia of Integer Sequencesand the web site, this was
really a big success and Neil is the person I respect the
mostin mathematics so this is why I thought (wrongly ) that my
collaboration with theBorweins had to go well, a big
mistake.Why do I write this ?To tell the truth and also the
arrogance of those people makes me sick.Will I gain something
from this ? I don't care, I have nothing to loose.Simon
Plouffe
 [A long list of complaints.] Is this for real, or
just some troll impersonating Simon Plouffe? Thereason I am
asking is because the way the Borweins are portrayed seems
tobe all too plausible, according to my personal
experience.
=Experts,I was wondering if there is anyway in
MATLAB or using any other tool to solve TWO UNKNOWNS FROM A
SINGLE EQUATION problem by exploiting the abundance of
data...!The Eqaution is,mod (a + X/Y) = 10^b;where a and b are
real, and X and Y are complex.a & X are known, and I have 256
values of b, my problem is to 'nd Y corresponding to each
value of b.As Y is complex we have to 'nd real and imaginary
parts of it, which makes this problem as 'nding two unknowns
from asingle eqation. But the previlage in this problem is we
have abundant data (of 256 values).MY WORK:
=I have been
trying by separate X and Y into real and Imag. parts as shown
belowmod (a + (Xr + j*Xi)/(Yr + j*Yi)) = 10^b;Rationalizing
the complex partmod(a + (Xr + j*Xi)*(Yr - j*Yi)/(Yr +
j*Yi)*(Yr - j*Yi)) = 10^b;mod(a +((Xr*Yr + Xi*Yi) + j*(Xi*Yr
- Xr*Yi)/(Yr^2 + Yi^2)))= 10^b;Making a common denominator and
then taking the magnitude,Its going Crazy..into 4th
order..........!((a*(Yr^2 + Yi^2) +(Xr*Yr + Xi*Yi))/(Yr^2 +
Yi^2))^2 + ((Xi*Yr - Xr*Yi)/(Yr^2 + Yi^2))^2 = 10^b;Finally in
the above I know a,b,Xi and Xr. Got to 'ng Yi and Yr.for each
value of b.and will be greateful if anyone can suggest a
wayout,Krishna.
 Experts,> I was wondering if there is
anyway in MATLAB or using any other tool> to solve TWO UNKNOWNS
FROM A SINGLE EQUATION problem by exploiting the> abundance of
data...!> The Eqaution is,> mod (a + X/Y) = 10^b;> where
a and b are real, and X and Y are complex.> a & X are known,
and I have 256 values of b, my problem is to 'nd> Y
corresponding to each value of b.What is mod? Modulus =
absolute value?GC
>Experts,>>I was wondering if there is
anyway in MATLAB or using any other tool >to solve TWO
UNKNOWNS FROM A SINGLE EQUATION problem by exploiting the
>abundance of data...!>>The Eqaution is,>>mod (a + X/Y) =
10^b;mod == modulus (otherwise known as absolute value)?This
will not allow you to determine the arguement of Y
(otherwiseknow as the angle in the complex plane). If you
chose an arguementthen you will know Y. You will notice that
mod ( Z ) = c determinesa circle of Z values in the complex
plane so your Y will be a pointinside a circle to a point on
the circle. Perhaps you have anadditional piece of
information, like X/Y is real, that will allowyou to pick the
argueemnt of Y.>where a and b are real, and X and Y are
complex.>a & X are known, and I have 256 values of b, my
problem is to 'nd >Y corresponding to each value of b.>>As Y
is complex we have to 'nd real and imaginary parts of it,
>which makes this problem as 'nding two unknowns from 
a>single
eqation. But the previlage in this problem is we have >abundant
data (of 256 values).>>MY WORK:>
I have been trying by
separate X and Y into real and Imag. parts >as shown
below>>mod (a + (Xr + j*Xi)/(Yr + j*Yi)) =
10^b;>>Rationalizing the complex part>mod(a + (Xr +
j*Xi)*(Yr - j*Yi)/(Yr + j*Yi)*(Yr - j*Yi)) = 10^b;>>mod(a
+((Xr*Yr + Xi*Yi) + j*(Xi*Yr - Xr*Yi)/(Yr^2 + Yi^2)))=
10^b;>>Making a common denominator and then taking the
magnitude,>>Its going Crazy..into 4th
order..........!>>((a*(Yr^2 + Yi^2) +(Xr*Yr + Xi*Yi))/(Yr^2
+ Yi^2))^2 + >((Xi*Yr - Xr*Yi)/(Yr^2 + Yi^2))^2 =
10^b;>>Finally in the above I know a,b,Xi and Xr. Got to 'ng
Yi and Yr.>for each value of b.>>and will be greateful if
anyone can suggest a wayout,>Krishna.
 The Eqaution is,>
mod (a + X/Y) = 10^b;>> where a and b are real, and X and Y
are complex.> a & X are known, and I have 256 values of b, my
problem is to 'nd> Y corresponding to each value of
b.>Presuming mod x means abs value x, notation |x|10^b = |(ay
+ x)/y| = |ay + x|/|y||y| 10^b = |ay + x|. Taking squares of
both sides and expanding100^b ((y_r)^2 + (y_i)^2) = (ay_r +
x_r)^2 + (ay_i + x_i)^2This results in a relation p(y_r, y_i)
= 0 given a,b and x are knowwhere p is a quadratic polynomial
in two variables and real coef'cients.(y_r)^2 (100^b - a^2) +
(y_i)^2 (100^b - a^2) + 2*100^b y_r y_i - 2a x_r y_r - 2a x_i
y_i - (x_r)^2 - (x_i)^2 = 0> I have been trying by separate X
and Y into real and Imag. parts> mod (a + (Xr + j*Xi)/(Yr +
j*Yi)) = 10^b;<snip>> Making a common denominator and then
taking the magnitude,> Its going Crazy..into 4th
order..........!Use my second order results.> Finally in the
above I know a,b,Xi and Xr. Got to 'ng Yi and Yr.> for each
value of b.>You can't as there's not enuf 
information in the
orginal equationexcept to limit y_r and y_i to a quadratic
relation or conic section.Only under exceptional circumstances
would the quadratic relationproduce a unique y_r, y_i.
=Yeah,
Mod in my equation is COMPLEX MODULUSi.e. Absolute value of the
complex number inside it.I am sorry, I didn't make it 
clear.I
too had to agree with William after alot of exertion.to see If
I can 'nd any other relation between the parameters, So that
the problem will turn out to be a Two Eqs and Two Unknowns.
Krishna> The Eqaution is,> mod (a + X/Y) = 10^b;>>
where a and b are real, and X and Y are complex.> a & X are
known, and I have 256 values of b, my problem is to 'nd> Y
corresponding to each value of b.>> Presuming mod x means
abs value x, notation |x|> 10^b = |(ay + x)/y| = |ay + x|/|y| |y| 10^b = |ay + x|. Taking squares of both sides and
expanding> 100^b ((y_r)^2 + (y_i)^2) = (ay_r + x_r)^2 + (ay_i
+ x_i)^2> This results in a relation p(y_r, y_i) = 0 given
a,b and x are know> where p is a quadratic polynomial in two
variables and real coef'cients.> (y_r)^2 (100^b - a^2) +
(y_i)^2 (100^b - a^2) + 2*100^b y_r y_i> - 2a x_r y_r - 2a x_i
y_i - (x_r)^2 - (x_i)^2 = 0> I have been trying by separate
X and Y into real and Imag. parts> mod (a + (Xr + j*Xi)/(Yr +
j*Yi)) = 10^b;> <snip>> Making a common denominator and then
taking the magnitude,> Its going Crazy..into 4th
order..........!> Use my second order results.> Finally
in the above I know a,b,Xi and Xr. Got to 'ng Yi and Yr.>
for each value of b.>> You can't as there's 
not enuf
information in the orginal equation> except to limit y_r and
y_i to a quadratic relation or conic section.> Only under
exceptional circumstances would the quadratic relation>
produce a unique y_r, y_i.
 Experts,> I was wondering if
there is anyway in MATLAB or using any other tool > to solve
TWO UNKNOWNS FROM A SINGLE EQUATION problem by exploiting the
> abundance of data...!> <SNIP>> As Y is complex we have to
'nd real and imaginary parts of it, > which makes this 
problem
as 'nding two unknowns from a> single eqation. But the
previlage in this problem is we have > abundant data (of 256
values).> I have just published a page http://wehner.org/euler
involves a NEGATION FUNCTION -1 to the power of X.-1 to the
power of X resolves into Cos(PiX) + i Sin(PiX)but it also
resolves into Cos(PiX) - i Sin(PiX)There is therefore
ambiguity.Nature has shown from the reciprocal of the Gamma
function the REALpart of these two equations.Nature REFUSES TO
GET INVOLVED in deciding between the two.The reason is that the
starting data (argument) is one-dimensional (onanother post I
called it MONAL) whilst the 'nishing data (result)
isTWO-DIMENSIONAL, which I called BINAL.The problem is
therefore insoluble, as a dimension is missing.Exactly the
same thing happens with Euler's Exp(iX) = Cos(X) + 
iSin(X).The
INVERSE of this is LOGe(Cos(X) + i Sin(X)) = iX which is
WRONG.The correct inverse is LOGe(Cos(X) + i Sin(X)) = (iX
MODULO 2Pi)I suggested on another newsgroup that one can
consider the number-lineiX:
__________________________________broken up by Euler's
equation into lengths of 2PI: ______ ______ ______ ______
______Nature now has TWO freedoms of movement - the JUMP from
one cycle toanother and the GLIDE within a module.Nature
considers all modules to be identical. This makes the
JUMPpointless.Considering all modules identical, nature turns
a typical one into acircle in the iY plane.The iY complex
plane requires two axes of freedom, but these areavailable.So
when you have ONE axis of freedom and want to extract TWO
pieces ofaxis-data from it, you will 'nd nature pushing you
into the world ofrepetitive MODULAR arithmetic.Charles Douglas
Wehner
 calculator, math handbook and computer algebra
system.>> www.mathHandbook.com>
=Please forgive my
limited knowledge in mathematics and hopefully you would
understand my question.Suppose f(x) is a function of degree n,
what is the expression (representation) of the nth derivative
of f(x)?How about the nth derivative of f(g(x))?Tom
Please
forgive my limited knowledge in mathematics and hopefully you
>would understand my question.>>Suppose f(x) is a function of
degree n, what is the expression >(representation) of the nth
derivative of f(x)?It looks like you copied the homework
problem wrong. At this levelfunctions do not have degrees. My
guess is that the real problemis about a polynomial of degree
n.I would suggest you look at f'(x) when f(x) = 
ax+bf''(x)
when f(x) = ax^2+bx+cf'''(x) 
when f(x) = ax^3+bx^2+cx+dand see
if can spot a trend.> How about the nth derivative of
f(g(x))?Again, I think you left out some information given to
you in the problem.--
http://www.math.fsu.edu/~bellenotbellenot <At/> math.fsu.edu
i;@/WO(?;[KC9sW;wG/4@H[_VFFH4?QHJ#O(?m}7fQMrJ,]0THA'|e-EPG_>
56Mi}_RRhBS'a2}u_7jm)0_+'=$V#E2r4#IQE/d)yMv3_4
@hl<)mA&*tDN/
=In sci.math.symbolic,
Tom<unknown@unknown.com><4rMLa.8100$gu6.4940@fe03.atl2.
webusenet.com>:> Please forgive my limited knowledge in
mathematics and hopefully you > would understand my question. Suppose f(x) is a function of degree n, what is the
expression > (representation) of the nth derivative of f(x)? How about the nth derivative of f(g(x))?> Tom> One
notation I've seen is D_n(f(x)), where 
ï_' indicates a
subscript.Another is the Liebnitz Notation: d^{n}f/dx^n.
Someone elsealready mentioned the prime notation; one
extension of that isf^{(n)}(x) -- which should not be confused
with f^{n}(x), whichis simply equal to f(x)^n.(The {} are
necessary for LaTeX in math mode.)-- #191,
ewill3@earthlink.netIt's still legal to go
.sigless.
=X-No-archive: yesI am doing research that
requires me to reason about operations on randomvariables, for
example, products/quotients, max/min,...etc. Assumeall my
r.v.'s are normally distributed. I realize there are not
closedform expressions for the results that I need, but I'm
wondering if anyof Mathematica, Maple, Matlab or anything else
out there can even plot say f=x*y or f=min(x,y) where both x,y
are normally distributed r.v's (with a given mean and 
variance
of course).I am hoping some tool out there or toolbox can help
me get to grips withthese operations. I have a lot of research
material on the topic and I'll settle for any numerical
implementation to just get a feel for what I'm looking at.
I've come across statool but that is rather primitive andI
need to a programming environment around it.Much
requires me to reason about operations on random>variables,
for example, products/quotients, max/min,...etc. Assume>all my
r.v.'s are normally distributed. I realize there are not
closed>form expressions for the results that I need, but I'm
wondering if any>of Mathematica, Maple, Matlab or anything
else out there can even plot >say f=x*y or f=min(x,y) where
both x,y are normally distributed r.v's >(with a given mean
and variance of course).>>I am hoping some tool out there or
toolbox can help me get to grips with>these operations. I have
a lot of research material on the topic and I'll >settle for
any numerical implementation to just get a feel for what I'm
>looking at. I've come across statool but that is rather
primitive and>I need to a programming environment around it.>I
am not sure what you are looking for, but maybe this will
work:http://www.r-project.org/A.L.
 I am doing research
that requires me to reason about operations on random>
variables, for example, products/quotients, max/min,...etc.
Assume> all my r.v.'s are normally distributed. I realize
there are not closed> form expressions for the results that I
need, but I'm wondering if any> of Mathematica, Maple, 
Matlab
or anything else out there can even plot > say f=x*y or
f=min(x,y) where both x,y are normally distributed r.v's >
(with a given mean and variance of course).The mathStatica
software suite http://www.mathStatica.com(which builds on top
of the symbolic engine of Mathematica) was designed to handle
manipulations of rv's, including automatingTransformations 
of
random variables, such as products, ratios etc. In fact, for
examples of the type you list, closed-form symbolicsolutions
often exist and can usually be derived in a couple of
lineswith mathStatica.Colin