A35
===
Hi,Export[img.gif,{graph1,graph2,..},??Look in the additional
information part of the Online documentation ofExport[]
Jens
====
=Dear group, I have m balls shared in n box. What is
the probability that one box has r o more balls?If m = 4, n=3
and r = 2 the solution will be (?): In[1] :=
Needs[DiscreteMath`Combinatorica`]In[2] := list =
Compositions[4,3];In[3] :=
(Count[list,2,2]+Count[list,2,3]+Count[list,2,4])/(3
Length[list])The question is that this way works only when m
and n are small , but for values sush as m=20, n=11 and r=5
the computer is out of memory. Any
idea?GuillermoSanchez----------------------------------------
-----This message was sent using Endymion MailMan.
====
=HiHow
can I tell Mathematica to pull a particular factor out of an
expression.For example in this expression:{(a -
2*wone*Sin[a*t])/4, (a - 2*wone*Sin[a*t])/4, (a +
2*wone*Sin[a*t])/4,(a + 2*wone*Sin[a*t])/4, (-a - 2*Sqrt[a^2
+ wone^2*Sin[a*t]^2])/4, (-a - 2*Sqrt[a^2
+wone^2*Sin[a*t]^2])/4, (-a + 2*Sqrt[a^2 +
wone^2*Sin[a*t]^2])/4, (-a + 2*Sqrt[a^2
+wone^2*Sin[a*t]^2])/4}I would like to bring the a out to the
front of the expression, evenoutside the matrix.How can I ask
Mathematica to do this, and how can I stop the a justßoating
back into the main expression?YoursChris RodgersSt John's
Collegehttp://rodgers.org.uk/Reply-To:
kuska@informatik.uni-leipzig.de
====
=Hi,something
like:13];NotebookPut[nb] Jens
====
=John,There is a nice trick
for this. It works on every operating system and all versions
of Mathematica from at least 4.0 onward (and probably
earlier). It uses the undocumented function
System`Private`FindFile, but if that ever goes away we'll
have to add something to take its place because J/Link uses
this exact technique to find its application resources.Put a
line like this in your .m file: thisPackageDirectory =
DirectoryName[System`Private`FindFile[$Input]]The variable
thisPackageDirectory will get the directory name in which the
.m file resides no matter how the user reads the package, 
e.g.:
<<MyPackage.m <</full/path/to/MyPackage.m <<MyPackage`
Needs[MyPackage`] etc.This trick also works if you use the
standard Mathematica application directory structure (highly
recommended), wherein you have a directory named for your
package (say MyApp): Applications/ MyApp/ MyApp.m ... other
files and dirs ... init.m <-- this file reads 
simply:
Get[MyApp`MyApp`]This standard application layout allows
users to enter simply Needs[MyApp`]to load your MyApp.m file.
The thisPackageDirectory variable would then get Todd
GayleyWolfram Research
====
=John,Although I don't recall ever
having problems with it, I am sure it notcompletely reliable.
The usage message makes this
clear:packageSourceDirectory[package] returns the list of
directories where thesource of the first currently loaded
package containing package in its context string might be
located.If you are interested you can go
tohttp://jowett.home.cern.ch/jowett/Mathematica/and download
the notebook defining the BumpEtc utility package.Perhaps
someone can improve on this function ?John Jowett
====
=Many
thanks, this is exactly what I was hoping/looking for.
John
====
=Sure, but my question was not how *I* can find out,
but how the .m file Johngobe
====
= ...But again it is wrong! It
only is correct if a and b happento be real quantities, which
is nowhere stated!So the main question still is: Why is
Mathematica makingthese very silly errors? One could expect
it from an earlyversion of a product, but that is not what
Mathematica 4.xis!So could it be, perhaps, that Mathematica
has now evolved sofar that it has its own kind of humor, and
is making theselittle jokes to entertain us? :-))--
Jo
====
=Since I don't really expect the right answer here, let
me pose the situation as it looks from here: I have to pay
for an upgrade in order to use a patch to fix a bug. Way to
go!---Selwyn Hollis
====
=I am trying to use mathematica to
define moment generating function ofa multivariate normal
distribution with mean 0 and variance T,[N(0,T)]. I would
like to be able to find the nth moment for
thisdistribution.For simplicity I would like define m =
exp[1/2 b'T b]= f(b), where bis a q*1 vector with elements
(b1 b2 b3 b4 ... bq) (b' is 1*q), and Tis a q*q matrix with
elementst11 t12 t13 ... t1qt21 t22 t23 ... t2q . . . . . . .
. . . . .tq1 tq2 tq3 tqq I have looked through the online
manual and have only found out how todefine a matrix or vector
of specific integer size.I have found the first 
moment which
disapears at b=0 to be m*b'*T.I have also found the second
moment to be m*T + m*T*b*b'T.I am having a very hard problem
finding the next moment (let alone thenext ten). I would like
to use mathematica to get all the momnets Icare to look at. I
would be greatful for any help!!Chris
Johnsoncjque@umich.edu
====
=Perhaps if you were to give an
example of the type of problem you areinterested in solving
we could better help you.Kevin
====
=Hi Alex:The notebook below
may be of some help. Just copy and paste into a Mathematica
notebook.hw
====
====================Notebook[{Cell[
CellGroupData[{Cell[BoxData[ (TraditionalForm`Expansion in
Spherical Harmonics)], Subtitle,Cell[BoxData[
(TraditionalForm`Basics)], Subsubtitle],Cell[<Spherical
harmonics are a set of complex orthonormal functions defined
on the surface of the unit sphere. Mathematica furnishes a
built-in function for their symbolic and numerical
computation. Let [Theta] and [CurlyPhi] be the usual
spherical polar and azimuthal angles, respectively, which
sweep the entire surface for (0 < [Theta] < [Pi]) and (0 <
[CurlyPhi] < 2[Pi]). These functions may be specified in
Mathematica by Cell[BoxData[ (TraditionalForm`(Y(l_,
m_))[InvisibleApplication] (([Theta]_, [CurlyPhi]_)) :=
(Y_l%m)([Theta], [CurlyPhi]))], Input],Cell[<Here the index l
is any positive integer zero or greater while the second index
m is restricted, for a given value of l, to one or another of
the values -l <= m <= l. Spherical harmonics are
customarilydefined so that the complex congugate of Y[l,m] is
given by (-1)^m Y[l, -m], and to have the orthogonality
property that Integrate[(-1)^m Y[l , -m][[Theta],[CurlyPhi]]
Y[l', m'] Sin[[Theta]] {[Theta], 0, [Pi]}, 
{[CurlyPhi], 0,
2[Pi]}] gives zero unless both conditions l = l' and m = 
m'
hold. If these conditions hold the integral yields 1. These
are the conditions of orthonormality. After first executing
the preceding function to load the program that computes a
spherical harmonic, you might then verify the properties
described above by executing the following
cell.Text],Cell[CellGroupData[{Cell[BoxData[
(TraditionalForm`Expansion of a Function)],
Subsubtitle],Cell[<The properties given above permit a given
function of the spherical polar angles to be represented as a
(generally infinite) sum of of spherical harmonics, The {l,
m)th expansion coefficients an arbitrary function f may be
calculated byCell[BoxData[ (TraditionalForm`(coef(l_,
m_))[InvisibleApplication] ((f_)) :=
[Integral]_0%[Pi]([Integral]_0%(2 [Pi])(((-1)))^m (Y( l,
(-m)))[InvisibleApplication] (([Theta], [CurlyPhi])) f
(sin([Theta])) [DifferentialD][CurlyPhi]
[DifferentialD][Theta]))], Input],Cell[BoxData[
(TraditionalForm`2 l of set The + a coefficients for given is
l such then)], Text],Cell[BoxData[ (TraditionalForm`(coefset(
l_))[InvisibleApplication] ((f_)) := Table[(coef(l,
m))[InvisibleApplication] ((f)), {m, (-l), l}])],
Input],Cell[<For example, the expansion set for the function
Cos[[Theta]]^2 corresponding to l = 0, 1, and 2
areCell[BoxData[ (TraditionalForm`TableForm[ Table[(coefset(
l))[InvisibleApplication] (((cos^2)([Theta]))), {l, 0,
3}]])], Input],Cell[<You can check for yourself that these
coefficients for this particular function are 
sufficient to
furnish the original function exactly. This is certainly not
true in general. Try some other f[[Theta], [CurlyPhi]]. },
Open ]]}, Open ]]},]Reply-To:
kuska@informatik.uni-leipzig.de
====
=Hi,of what ? Of an
analytical known function
?Sum[a[l,m]*SphericalHarmonicY[l,m,th,phi],{l,0,Infinity},{m,-
l,l}]with a[l,m]=Integrate[
Conjugate[SphericalHarmonicY[l,m,th,phi]]f[th,phi]*Sin[th],
{th,0,Pi},{phi,0,2Pi} ] Jens
====
=How to perturb the constant f
in parametric plots? f= 1 ;' for single f 
value'xy= { 2 f
Sin[t]^2 ,Sinh[f t]+f Log[t] };ParametricPlot[ xy ,
{t,0,2}];An attemt to Map did not work.' f={ 
.5,1,1.5};''trj
= Map [ xy &, f ];'--Reply-To:
kuska@informatik.uni-leipzig.de
====
=Hi, becausePrefix[] *print*
the function f[expr] in the form f @ expr.And printing an
expression is quite the oposite of parsing/reading a
expression. Jens
====
=Hey all - I'm taking a basic calculus
course that uses Mathematica. We have been studying limits
and I have been using the Limit functionto check if my
answers are correct.We were given the following function and
asked to determine a limit:(3E^(-x) - E^(-3x)) / (E^(-3x) +
E^(-x))Usually the approach is to select the dominant terms,
factor and thendetermine the limit. My initial reason had me
select -E^(-3x) in thenumerator and E^(-3x) in the
denominator. Factoring the terms wouldand the real answer is
somewhere near 3.I then tried to use the Limit function which
is not producing ananswer (perhaps I'm not sure of the
usage).Any help is greatly appreciated, David Seruyange
====
=To
whom may concern,I have a problem with Mathematica. I have
installed Mathematica 4.2 on Windows XP. I need to run a
program in Mathematica, best regard.Hung Viet.
====
=That is
more than just a shadow. There appears to be diffraction
around theedges as well. But it is pretty cool.KevinReply-To:
kuska@informatik.uni-leipzig.de
====
=Hi,a) computing the true
shadows of the 3 light sourceswould requite the construction
of the shadow volumesand take an half hour per frameb) fake a
single shadow is easy
withNeeds[Graphics`Polyhedra`]Needs[Graphics`Shapes`]toShadow
[gray_?MatrixQ, {x1_, x2_}, {y1_, y2_}, z_] := Module[{n, m,
dx, dy, points, cgraph}, {m, n} = Dimensions[gray]; dx = (x2
- x1)/(n); dy = (y2 - y1)/(m); points = Table[{x1 + dx*i, y1
+ dy*j, z}, {j, 0, m}, {i, 0, n}]; poly = Drop [#, -1] & /@
Drop[Transpose[{points, RotateRight[points], RotateLeft /@
RotateRight[points], RotateLeft /@ points},{3, 1, 2}], -1];
{EdgeForm[], Transpose[{Map[SurfaceColor[GrayLevel[#]] &,
gray,{2}], Map[Polygon, poly, {2}]}, {3, 1, 2}]}
]makeShadow[t_, opts___?OptionQ] := Module[{sh}, sh =
DensityPlot[ 1 - Exp[-(3 + Sin[t/2])*(x^2 + y^2)], {x, -2,
2}, {y, -2, 2}, toShadow[sh[[1]], {-2, 2}, {-2, 2}, -2] ]obj
= Table[RotateShape[Icosahedron[], t, t/2, t/3], {t, 0, 8,
1/10}];MapIndexed[obj];but that is *not* the shadow that the
three colored light sources wouldproduce -- that would be
much more expensive.If some one else would like to have
(hard) shadows I can include it into Jens Ashraf,the question
is: what behaviour did you expect? An answer to that
mightsolve your problem.First, Prefix[s[a_]] is *not* the same
as f@a_ In[1]:= f@a_ := 2 aIn[2]:= ?f Global`f f[a_] := 2
aIn[3]:= f[1]Out[3]= 2 Compare this to In[4]:=
Prefix[s[a_]]Out[4]= s@a_In[5]:= % //
FullFormOut[5]//FullForm= Prefix[s[Pattern[a, Blank[]]]]So
Prefix is a wrapper for printing purposes (only). Such you
cannot makedefinitions for it (as it is protected), but also,
it is a special form notsubmitted to the standard evaluation
sequence, see:In[7]:= Prefix[s[aaa___]] ^= s[aaa]Out[7]=
s[aaa]In[8]:= s[a_] := 2 aIn[9]:= ?s Global`s
Prefix[s[aaa___]] ^= s[aaa] s[a_] := 2 aIn[10]:=
Prefix[s[2]]Out[10]= Prefix[4]So even Upvalues 
won't work. You
may however doIn[12]:= Prefix[Unevaluated[s[2]]]Out[12]= 4The
very question, however, is whether this is what you intended,
or say,what did you want to achieve at all? --Hartmut
WolfReply-To:
kuska@informatik.uni-leipzig.de
====
=Hi,number.You say it
already use Simplify[]/FullSimplify[] with a second
argument,i.e.,Unprotect[Simplify]Protect[Simplify]Convert[1
Milli Meter, Meter]Meter/1000and Notice the space between
Milli and Meter Jens
====
=First of all, thanks to David and
Bill!David Park schrieb:I agree. My yk@despammed.com is
totally functional, though. Really nice spam filtering service
there (Until they havenÇt stopped this weekend, 
IÇll check
this... no, works just fine).Again, agreed:-) I tried to do so
anyway, but got annoyed by unsimplified units in the
results.Again, thanks! I will check out your
packages...YvesReply-To:
kuska@informatik.uni-leipzig.de
====
=Hi,Show[Graphics3D[{
AbsolutePointSize[4], Point /@ Flatten[Table[a*k + b*c + v*t,
{k, 1, 2}, {b, 1, 2}, {v, 1, 2}],2]} ]]it is Faltten[_,2] that
help you. JensReply-To:
kuska@informatik.uni-leipzig.de
====
=Hi,can you tell us what a
waterfall plot is ? Jens
====
=Dear Mr. Woll,I tried to follow
your advice to make the jacobianfaster. Below you find the
optimized code. (You haveto past it into Mathematica.
Otherwise it's unreadable).Now I would be really surprised 
if
it could be improvedfurther, since the code is really short
and makes extensiveuse of the listable attribute and matrix
algebra operators.Nevertheless, evaluation time has increased
only by a smallamount.BUT MY ORIGINAL QUESTION IS OPEN:
MATHEMATICA STILLREMAINS to be slower by a factor of 12 than
its competitors. Johannes!((pdf = Function[{x},
Exp[(-(x^2/2))]/@(2 [Pi])];)[IndentingNewLine] (cdf =
Function[{x}, 0.5 ((1. +
Erf[x/Sqrt[2.0]]))];)[IndentingNewLine] SetAttributes[pdf,
Listable]; SetAttributes[cdf, Listable];)!(LogLikJacC =
Compile[{{y, _Integer, 1}, {x, _Real, 2}, {b, _Real, 1}},
Module[{p, c, z = Map[If[# [Equal] 1, 1, (-1)] &, y] x . b},
[IndentingNewLine]p = pdf[z]; c = cdf[z]; (-Transpose[ x] .
((x (( p c z)(+)(p^2)( ))/c^2)))]])Johannes LudsteckEconomics
DepartmentUniversity of RegensburgUniversitaetsstrasse 3193053
RegensburgHi all,I have 2 simple questions about animated gif
picture file generation.1.) I would like to export the
graphics which I generated in Mathematica 4.1, in animated
gif format and paste it in PowerPoint 2000. I tried copy as
but there is no gif format to be selected. Only wmf, bitmap
and wave are showed.2.) Is there any way to control the
pictures ßicking speed ?Does anyone have any ideas?Shz
Shz
====
=======================================================

====
=====================of the individual or entity to which
they are addressed. Any disclosure, copying,distribution and
diversion contrary to the applicable export control laws
andregulations including US Export Administration Regulations
is strictly prohibited.and do not disclose it to others.
Please notify the postmaster@hitachi.com.myof the delivery
error by replying to this message and then delete it from
your
====
======================================================

====
==========================The basic problem is the
function above is constant for negative values except at x =
-2. So, when Mathematica samples the function, there is
nothing to force the adaptive sampling algorithm to subdivide
unless the original sample just happens to select x = -2 as a
sample point.A good discussion of why Mathematica plot
routines give misleading results can be found in the book The
Mathematica Graphics Guidebook by Cameron Smith and Nancy
Blachman.By experimenting with the number of PlotPoints and
the PlotRange, you might find a combination that would cause
Mathematica to sample the function at -2 which would show the
problem.
====
=not sure if anyone is still reading this but i
found: For z = StartRow(w) To EndRow(w) x = Sheet1.Cells(z,
xCol).Value Ôeach xi y = Sheet1.Cells(z, yCol).Value 
Ôeach yi
Sxxx = Sxxx + x ^ 3 Syyy = Syyy + y ^ 3 Sxxy = Sxxy + (x ^ 2 *
y) Sxyy = Sxyy + (x * y ^ 2) Sxx = Sxx + x ^ 2 Syy = Syy + y ^
2 Sxy = Sxy + (x * y) Sx = Sx + x Sy = Sy + y Next z A = N *
Sxx - Sx ^ 2 B = N * Sxy - Sx * Sy C = N * Syy - Sy ^ 2 D =
0.5 * (N * Sxyy - Sx * Syy + N * Sxxx - Sx * Sxx) E = 0.5 *
(N * Sxxy - Sy * Sxx + N * Syyy - Sy * Syy) Xc = (D * C - B *
E) / (A * C - B ^ 2) Yc = (A * E - B * D) / (A * C - B ^ 2) R
= Sqr((Sxx - 2 * Xc * Sx + N * Xc * Xc + Syy - 2 * Yc * Sy +N
* Yc * Yc) / N)to be really useful given all xi, yi on a
circle it spits back the radius (R) and center(Xc, Yc)more
info
athttp://groups.google.com/groups?hl=en&lr=&ie=UTF-8&threadm=
35f59745.37468454%40news.wwnet.net&rnum=1&prev=/groups%3Fhl%
3Den%26lr%3D%26ie%3DISO-8859-1%26q%3Dbest%2Bfit%2Bcircle%
26meta%3Dfrom a IEEE PAMI paper by Thomas and
Chanhttp://www.cs.bsu.edu/homepages/math/people/regfac/kjones
/circles.pdf'A Few Methods for Fitting Circles to 
Data' by
Dale Umbach and KerryN. Jones from Ball State
universitymelanie
====
=I am using Mathematica 4.2 on Mac OS X
(Ver 10.2.3). I have tried the userFile =
Experimental`FileBrowse[] as P. Hinton recommends, and I get
a file save dialog; all the filenames are greyed 
out, and the
dialog is asking me for a file name tosave. I 
can't figure out
how to get it to return the file name.
====
=Dear All,Probably a
stupid question, but:Can I set the size of the font used in
text cells at 13 ?The options on the menu go: ...9, 10, 12,
14, 16, ...But 12 seems too small to read easily, and 14
seems too big.I want my external examiner to have as
comfortable a time reading my thesisas possible !Reply-To:
kuska@informatik.uni-leipzig.de
====
=Hi,Export[] the *list* of
pictures into a GIF, something
likeExport[myanaimation.gif,{graphics1,graphics2,graphics3,..
.},GIF]should do that. Jens
====
=Strange. That doesn't work for
me (using version 4.2 for Windows).I did exactly what you said
(except with Assumptions instead ofAssumption) yet got the
complex result reported by JRB.DavidReply-To:
kuska@informatik.uni-leipzig.de
====
=Hi,I have never had
problems with Windows 200 (SP3) and mathematica4.2 and every
thing work as it should. Jens
====
=I am surprised too, because
I have been using Mathematica 4.0.x,Mathematica 4.1.2, and
Mathematica 4.2.x on my Windows 2000 sp3computer with no
problems.Has anyone else here experienced this kind of
problem?--Bhuvanesh,Wolfram Research.
====
=Gareth Russell had
an expression such as
Gamma[a]*Gamma[b]/Beta[a,b]*Gamma[a+b]and wanted to make a
new expression where for every subexpression with headGamma,
if the arguments include b, each b is replaced by b+1. He
wanted a functionthat would make this change. My solution is
below.-----------------------In[1]:=SubPosnQ[list1_,list2_]:=
(Length[list2]<Length[list1])&&(Take[list1,Length[list2]]
====

list2)In[2]:=Modify[expr_]:= With[{ posn = Intersection[
Position[expr, b], Position[expr, _Gamma], ReplacePart[ expr,
b+1, posn ] ]-----------------------Example:In[3]:=
ex=(20*Gamma[a]^2*Gamma[1+a]^2*Gamma[3+b]^3*Gamma[4+b])/
(Beta[a,b]^3*Gamma[3+a+b]*Gamma[5+a+b]);In[4]:=
Modify[ex]Out[4]=
(20*Gamma[a]^2*Gamma[1+a]^2*Gamma[4+b]^3*Gamma[5+b])
(Beta[a,b]^3*Gamma[4+a+b]*Gamma[6+a+b])----------------------
--In the above solution I tried to make it concise and easy
to read. If speed is more important you could make SubPosnQ a
compiled function, or use a pure function instead of SubPosnQ.
You could also avoid using Intesection to make (posn) since
it's rather slow when a non-default SameTest is
used.------------------------ Ted ErsekDownload my collection
of Mathematica tricks from:
http://www.verbeia.com/mathematica/tips/Tricks.htmland
http://www.verbeia.com/mathematica/tips/GraphicsTricks.html==
== Ealier I sent a solution to Gareth Russell's problem. He
had an expression such as
Gamma[a]*Gamma[b]/Beta[a,b]*Gamma[a+b] and wanted to make a
new expression where for every subexpressionwith head Gamma,
if the arguments include b, each b is replaced by b+1. I
provided afunction that would make this change. I have a much
improved solution below. Withmy new version you can specify
the pattern and replacement rule to use! Here it is:
--------------------------------In[1]:=
PatternReplaceAll::usage= PatternReplaceAll[expr, pattern,
rule] usesReplaceAll to make the changes specified by rule,
but the replacements are only madeto subexpressions of expr
that match pattern. PatternReplaceAll[expr,pattern, {rules}]
uses one or more rules.; PatternReplaceAll[expr_, pattn_,
{rules:(_Rule|_RuleDelayed)..} ]:= With[
{posn=Reverse@Sort@Position[Hold@@{expr}, pattn]},
ReleaseHold[Fold[MapAt[ Function[a, a/.{rules}], #1,
#2]&,Hold@@{expr}, posn ]] ] PatternReplaceAll[ expr_,
pattn_, SingleRule:(_Rule| _RuleDelayed) ] :=
PatternReplaceAll[ expr, pattn, {SingleRule}
](*----------------------- (1) (Reverse@Sort...) ensures the
later positions in (posn) still applyeven after folding
several times. (2) We can't let the replacements take effect
until after Fold is donebecause the position of things may
change. To solve that I use Hold@@{expr}.False. Instead I use
rules:(_Rule|_RuleDelayed).----------------------------*)
Examples:In[4]:=ex=b*Gamma[1+a]^2*Gamma[1+b]/Gamma[2,
4+2b]*PolyGamma[2*b]*(b+2)^2;In[5]:=Out[5]=(b*(2 +
b)^2*Gamma[1 + a]^2*Gamma[1 + bb]*PolyGamma[0, 2*b])/Gamma[2,
4 +2*bb]------------------Next I use a named pattern in the
replacement rule!In[6]:=Out[6]:=(b*(2 + b)^2*Gamma[3 +
a]^2*Gamma[3 + b]*PolyGamma[0, 2*b])/Gamma[4, 6
+4*b]------------------Next I use a list of replacement
rules.In[7]:=Out[7]=(b*(2 + b)^2*Gamma[3 + a]^2*Gamma[3 +
bb]*PolyGamma[0, 2*b])/Gamma[4, 6
+4*bb]------------------------ Ted Ersek Download my
collection of Mathematica tricks from:
http://www.verbeia.com/mathematica/tips/Tricks.html and
http://www.verbeia.com/mathematica/tips/GraphicsTricks.html

====
=How to get multiple plots on same x-y graph for given x
domain forstraight line y=m*x+C ? matrix (m,c) is given, each
with a uniformincrement.Please indicate code lines, a
cluttered output notwithstanding.
====
=You're making the common
mistake of thinking that Mathematica is a magic black box that
can solve anything. I'm not sure what Jens-P.K. was 
attempting
to say about Mathematica 4.2 or solving the equation by hand,
so let me try to say something here that humans can
understand.The appropriate technique for equations such as
the one you have (having no explicit t-dependence) is to
reduce the order by substitutingr' = v and 
r'' = v dv/dr.That
gives you a separable equation for v in terms of y:v dv ==
-1/r drIntegration of that, and use of the initial conditions
gives you v^2 = geometric argument) you have the first-order
equationr' == -Sqrt[ -2 Log[r] ]This is not very pleasant,
but it is separable. Separating and integrating results in
this implicit solution:-Integrate[ 1/Sqrt[ -2 Log[y] ],
{y,1,r}] == tMathematica will express the integral here in
terms of the special function Erfi, but that's 
little more
than notation. After some massaging, you getSqrt[Pi/2]
Erfi[Sqrt[Log[1/r]]] == tPlot the left side over {r,0,1} and
you'll see the graph of the solution with the axes
reversed.Hope this helps.----Selwyn
Hollishttp://www.math.armstrong.edu/faculty/hollis/
mmadeReply-To: murray@math.umass.edu
====
=A notebook has the
input-output cell pair:1I do Save As Special XML
(XHTML+MathML) to file limit.xml. When viewed in a browser
supporting native MathML (e.g., Mozilla 1.1), the input cell
includes a spurious [Rule] while otherwise displaying
everything appropriately sized brackets:Specifically, the
MathML source of the erroneoulsy generated input cell is:--
Murray Eisenberg murray@math.umass.eduMathematics &
Statistics Dept.Lederle Graduate Research Tower phone 413
549-1020 (H)University of Massachusetts 413 545-2859 (W)710
North Pleasant StreetAmherst, MA 01375
====
=Adalbert,I hope
that the following may be of help - I am using the default
stylesheet.Define the multiplier to convert 
printers' points
to mm. mm = 72/25.4;Set, for image size, isx = 100*mm; isy =
150*mm;Consider,10}},The frame, with aspect ratio 1/2, is
made as big as possible inside, andcentered in, the bounding
box (shown by clicking the picture).1) For the frame to fill
the bounding box, and so have the same dimensions,we must
give it the same aspect ratio as the bounding box
(alternatively2) Unfortunately, the printed form will not
have the dimensions {isx, isy},it will be 0.8 of this. This
is because in the style sheet Styles forwith this style sheet
print with the correct dimensions (for safety importa private
style sheet).expressions of the graphics cells that we want
to print to correctdimensions.3) There remains one more
problem. You will see below that although theframe's aspect
ratio stays the same (1/2}, it has now to make way for
thelabels. This could require more
experimenting.False},should have a width of xi
inches.want:
====
=Further to my earlier posting in this thread.
To futher complicate thestory: the cell option ImageRegion
defines rectangle relative to thebounding box and the graphics
option PlotRegion defines a rectanglerelative to this. The
latter rectangle (default the image box) is actuallythat in
which the plot range ticks and labels are located as
describedpreviously for the image
box.Allan---------------------Allan HayesMathematica Training
and ConsultingLeicester
UKwww.haystack.demon.co.ukhay@haystack.demon.co.ukVoice: +44
(0)116 271 4198
====
=You can use the OPLinearAlgebra
package(http://library.wolfram.com/database/MathSource/4271/)
to solve this type ofnoncommutative algebra problem.--Steve
LuttrellWest Malvern, UK{B11,
====
=Well when solving any
problem you make use of what you have
available.Mike
====
=Actually I'm surprised that you would need
an SOR for Black-Scholes. It isanalogous to
diffusion-reaction equation in electrochemistry and that can
besolved using a tridiagonal solver.Mike
====
=I am trying to
solve system of 3 coupled PDE's but getting
errorNDSolve::ndnum: Encountered non-numerical value for a
derivative at t==0apparentely the problem is in inital
conditions (derivate of one functionthat go into the
equationfor another is not specified explicitely in the
initial moment of time)any que how to go around this would be
greatly appreciatedalexanderfollowing system of 3 PDEeq11 =
D[a0[t,x], {t, 2}] - D[a0[t, x], {x, 2}] ==
I*e*(Conjugate[f[t,x]]*D[f[t, x], {t, 1}] - f[t,
x]*Conjugate[D[f[t, x], {t, 1}]]) + e^2*(a0[t,x] +
el*x)*(Abs[f[t, x]])^2;eq12 = D[ax[t, x], {t, 2}] - D[ax[t,
x], {x, 2}] ==I*e*(Conjugate[f[t,x]]*D[f[t, x], {x, 1}] -
f[t, x]*Conjugate[D[f[t, x], {x, 1}]]) +e^2*(ax[t,x] -
el*t)*(Abs[f[t, x]])^2;eq13 = D[f[t, x], {t, 2}] - D[f[t, x],
{x, 2}] + (m^2 - e^2*((a0[t, x] +el*x)^2))*f[t, x]
-2*I*e*((a0[t, x])*D[f[t, x], {t, 1}]) == 0;with the initial
and boundary conditionseq21 = a0[0, x] == 0; eq22 =
Derivative[1, 0][a0][0, x] == 0; eq23 =a0[t, -x0] == 0; eq24
= a0[t, x0] == 0;eq31 = ax[0, x] == 0; eq32 = Derivative[1,
0][ax][0, x] == 0; eq33 =ax[t, -x0] == 0; eq34 = ax[t, x0] ==
0;eq41 = f[0, x] == p; eq42 = Derivative[1, 0][f][0, x] == p1;
eq43 =f[t, -x0] == 0; eq44 = f[t, x0] == 0;wherep =
NIntegrate[Exp[I*k*x - k^2/2*a^2], {k, 0, Infinity}];p1 =
NIntegrate[-I*Sqrt[m^2 + k^2]*Exp[I*k*x - k^2/2*a^2], {k,
0,Infinity}];when I try to solve it numerically withres =
NDSolve[{eq11, eq12, eq13, eq21, eq22, eq23, eq24, eq31,
eq32, eq33,eq34, eq41, eq42, eq43, eq44},{a0[t, x], ax[t, x],
f[t, x]}, {t, 0, 10},it gives meNDSolve::ndnum: Encountered
non-numerical value for a derivative at t==0and quitsi nailed
it down to presence of D[f[t, x], {x, 1}] derivative in
eq12equation which was not specified at t==0 in the initial
conditionsexplicitlybut should be known because f[0,x] is
specified.
====
=Hallo,I would like to use the units of
Miscellaneous Units (Mathematica 4.2 Windows).1) Why does
Simplify[Sqrt[x*Meter^2]] giveSqrt[Meter^2*x] ? The result of
the simplification is o.k. when I assume not seem to be the
right way.Am I missing something? Is there a way to get the
units properly simplified?2) There is Meter and Centimeter
predefined. Where have the Millimeters gone? I did not succeed
in prepending any of the SI unit prefixes. Could I simply use
MilliMeter? Did not really work for me...Yves
====
=Dear
Mathematica user,I have a very simple problem to solve, but
no elegant solution for it.I deal with NxN matrices of
integral entries where rows and columns are allproportional,
e.g. for 3x3: ( 4 8 12 ) ( )M = ( 5 10 15 ) ( ) ( 6 12 18 )I
need to decompose these matrices as a (Times) product: M =
{{1,2,3},{1,2,3},{1,2,3}} * {{4,4,4},{5,5,5},{6,6,6}}there is
no error: * is Times, not Dot.Finding the absolute values of
components of the product M is easy, usingGcd on rows of M
then of Transpose[M].I MUST be blind, but what I don't see 
is
how to restore correct signsefficiently without resorting to
ugly code. For instance: ( -4 -8 12 ) ( )M = ( 5 10 -15 ) ( )
( 6 12 -18 ) M = {{1,2,-3},{1,2,-3},{1,2,-3}} *
{{-4,-4,-4},{5,5,5},{6,6,6}}There always are two solutions
with global sign change, since ifShame on me for asking such
an elementary question!
====
=Dear Selwyn,unfortunately you are
not right here. I didn'tsince it is more complicated (and
almost unreadable in ASCII form, but nice if you paste it
into Mathematica).THE SPEED GAINS ASSOCIATED WITH COMPILATION
ARE NEGLECTIBLE HERE!Just compare the evaluation timings (I
append the compiled version below).The LogLik and LogLikGrad
functions are not the problem:Computation times do not differ
much from Gauss and Stata.It is plausible that compilation
increases speed here only marginally, since the uncompiled
function makes extensive use of the Listable Attribute (the
arguments of PDF and CDF are long Lists!) and the package
functions areoptimized for list arguments.Try out the
compiled version of LogLikGradC below and you will see. Thus
the original question remains unanswered. By the way: The
answer is really important for me, since a negative answer
means that I simply cannot use Mathematica forreal world
statistical problems (e.g. problems requiringmaximum
likelihood or any other minimum estimators) on largedatasets
(with 5000 observations and more). I would be surprised if it
were possible to increase computation time of the following
function.!((LogLikGradC = Compile[{{y, _Integer, 1}, {x,
_Real, 2}, {b, _Real, 1}}, [IndentingNewLine]With[{s = (If[#
[Equal] 1, 1, (-1)] &) /@ y}, [IndentingNewLine]Tr /@
Transpose[(((s Exp[(-(#^2/2))])/((@(( )((([Pi])( ))/2))) ((1.
+ Erf[#/@2])))) x &)[s x . b]]]];))If you try FindMinimum
using LogLikC and LogLikGradC (as Gradient function) you will
see that providing a symbolic gradient function again does not
speed up things.By the way: I would be very happy if a Wolfram
Research guy would respond, since Wolfram obtained large
amounts of money for the updates to the fast numerics version
and thefast statistics packages, but Mathematica apparently is
notreally usable for computationally demanding
applications.Most examples in the statistics packages are
toy-examples with50-200 data points.My most important problem
is that the documentation on algorithms of functions like
FindMinimum and FindRoot is very thin. THE EVALUATION TIMES
OF LOGLIKGRAD DIFFER ONLY BY A SMALL AMOUNT IN MATHEMATICA
AND GAUSS OR STATA, BUT NOBODY CAN TELL ME WHY FINDMINIMUM IS
SO MUCH SLOWER THAN THE GAUSS AND STATA MINIMIZATION
PROCEDURES. Johannes LudsteckJohannes LudsteckInstitut fuer
VolkswirtschaftslehreLehrstuhl Prof. Dr. MoellerUniversitaet
RegensburgUniversitaetsstrasse 3193053 Regensburg
====
=The last
time I needed to work with an animated gif, I used a
utilitycalled GifBuilder (developed by Yves Piguet) with good
results. This wasfor a Mac, pre-OS X.Ken LevasseurMath.
Sci.UMass Lowell
====
=I'd like, in a .m file for a package, to
be able to know what directorythe .m file is being loaded
from, so I can search in the same place(or subdirectories)
for related files. I've looked through 
everythingto do with
directories I can find in The Book, but I can't 
find
anythingthat looks as though it would help. John
====
=John,Why
is this patch specifically for Mathematica 4.2? Or does it
patch 4.1 as well?---Selwyn Hollis
====
=Studentversion 4.2,KDEI
generated a Plot,right-clicked into it,and tryed some of the
conversionswhen suddenly the only reaction of my systemwas to
mouse-MOVEMENTS .Not even Cntrl-Alt-Del got me a rebootof my
machine,I had to use the resetbutton.That's not very
nice.
====
=Alexander,In your definition of p and p1, there are
parameters a and m, which you haven't specified. 
What should
those be?---Selwyn Hollis
====
=a={2,3,5}c={3,5,3}t={-1,3,5}I am
using Table[a*k+b*c+v*t,{k,1,2},{b,1,2},{v,1,2}]to generate
some points. After this I need to plot the points by using
3Dcommands like Scatter3DPlot.How can I get the points from
Table in the form of a list so that I can
plotthem?.
====
=Searches of past discussion yielded a couple of
threads on the topic ofwaterfall plots a couple of years ago,
but nothing too helpful.I'm just looking for hints as to
which fundamental plot command to startfrom.
====
=Ashraf,I
think it is because Prefix is a formatting function that
affects how theexpression is displayed. But the prefix header
is still there.Prefix[s[a_]]% //
FullForms@a_Prefix[s[Pattern[a, Blank[]]]]So when you try to
make an assignment with it, Mathematica objects becausePrefix
is protected.On the other hand...s@a_% //
FullForms[a_]s[Pattern[a, Blank[]]]and you can make an
assignment because s is not protected.I have to admit that
this is a little tricky and confused me at first.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/2
aOut[2]=2In[3]:=Prefix[s[a_]]Prefix[s[a_]]=2
as[1]Out[3]=s@a_Set::write: Tag Prefix in s@a_ is
Protected.
====
=====Sounds like something my students would
write. Anyway, maybe it's something like this:Given vectors 
y
and z, find the vector x (in some subspace?) such that
??---Selwyn HollisReply-To:
kuska@informatik.uni-leipzig.de
====
=Hi,why not sayQuit[]in the
last input cell of your notebook ? Jens
====
=I think the real
solution is to put a statement in the very first cell (andmake
that cell an initialization cell) to clear all variables. The
neteffect is the same, namely the next time you use that
notebook, allvariables are cleared. Or have I missed some
subtlety?- Eric.--

====
==========================================================

====
=====There is never a single right solution. There are
always multiplewrong ones,
though.
====
===================================================

====
============ ====As Jens-Peer Kuska and Andrzej Kozlowski
both hypothesized, unpacking isthe issue. Specifically, this
is done by Inner. One can improve speedfor this sort of
computation by using operations that do not need
tounpack.In[3]:= x = Table[2.,{i,100000}];In[4]:= y =
Table[If[i<1000000,2.,i],{i,100000}];In[5]:=
Timing[Inner[Divide,x,x]]Out[5]= {0.27 Second,
100000.}In[6]:= Timing[Inner[Divide,y,y]]Out[6]= {0.16
Second, 100000.}In[7]:= Timing[Apply[Plus,x/x]]Out[7]= {0.06
Second, 100000.}In[8]:= Timing[Apply[Plus,y/y]]Out[8]= {0.12
Second, 100000.}Daniel LichtblauWolfram ResearchReply-To:
kuska@informatik.uni-leipzig.de
====
=Hi,if InputForm[] doe not
help you, you canUnprotect[Times]Format[Times[a_, b_],
StandardForm] := SequenceForm[a, *, b]Protect[Times]to get
the * symbol back. Jens
====
=Jonathan Mann wanted to have a
muliplication sign used when the result ofthe following is
displayed.In[1]:= 40 (8)^tOut[1] 5 8^(1+t)--------------You
can evaluate the following to make things work the way you
want.In[2]:= MakeBoxes[n_Integer*m_Integer^pow_,form_]:=
RowBox[{ToString[n],*,SuperscriptBox[ToString[m],MakeBoxes[
pow,form]]}]-------------If you put this line of code in your
(init.m) file it will evaluate everytime Mathematica is
launched. Ted ErsekDownload my collection of Mathematica
tricks from:
http://www.verbeia.com/mathematica/tips/Tricks.htmland
http://www.verbeia.com/mathematica/tips/GraphicsTricks.html==
==In particular, Ted Ersek said something that interested me.
However, I tried to put it in my init.m file and it 
didn't
work. Could someone please elaborate on the location of this
file and suggestions on the best line of code to do the
job.Jonathan Mannmtheory@msn.com
====
=Dear extrabyteI have read
your post to MathGroup and, to give a glance to the matter, I
would like to receive the notebook your web page was
generated from. Please, can you send it to me as an attach ?
That's for avoid retyping it. Which Mathemathica version 
have
you got the trouble from ?Scusa, ho il sospetto che tu sia
italiano. Mi sbaglio ?Nicola
Avaglianonicolaus_ava@yahoo.itReply-To:
Anthony.C.Iannetti@grc.nasa.gov
====
=Dear Mathematica users, I
do I find the jacobian resulting from the Newton method for
solvingsimultaneous nonlinear equations? I realize the
FindRoot[] can solvethe system for me, but I would like to
study the Jacobian resulting fromthe linearization. If
somebody has an example, I would appreciate it.Tony
====
=I fear
that this is a FAQ but after a couple of hours
unsuccessfullysearching, I dare to post the question in the
forum (sorry when imissed an obvious FAQ entry).When I type
some charachters into the Mathematica FrontEnd I only get
emptyboxes instead of the typed characters. This effect came
up when II already followed some bugfixes. I corrected the
MaxForXListFontsline in the file Specific.tr. The 
wri fonts can
be seen withxlsfonts furthermore the fontpath countains links
to the Type1 andX directories. However, the problem still
persists.Is there any instruction in the web which describes
a solution for theproblem?Norbert
====
=hido you have the NUM
LOCK key activated? turn it off, otherwise the wonderful and
soo stable :-)))) mathematica frontend for linux will not
work properly...you can easily verify if it is this problem
that bugs you if the ENTER key of the numpad does not work
properly.gerald-- *************************************Gerald
RothM@th Desktop
Development*************************************
====
=I am not
sure what is your problem (which characters do you type?),
butcheck if the NumLock key is ON. It should be OFF when you
type in theFrontEnd, and upgrading your system has maybe
changed the deafult settingsfor the NumLock key upon starting
X Window System. Marko----- Original Message -----
====
=Dear
Andrzej KozlowskiAt first, I am so grateful to your reply
which helped me with some stuffs.Anyway, tracing the inputs,
I found something strange to me.I can't understand 
it.Jungkyu
ChoiSeoul, South-Korea.----- Original Message ----- 
====
=I am
not sure xactly what you are asking for. However, it is easy
to replicate with Mathematica your computation of the number
of irreducible polynomials of the form:x^50 +x^m + x^n + x^k+
1In[1]:=testQ[{m_, n_, k_}] := PlusIn[2]:=t =
Flatten[Table[{m, n, k}, {m, 49}, {n, m - 1}, {k, n - 1}],
2];In[3]:=Timing[ans = Select[t, testQ];
]Out[3]={75.39*Second,
Null}In[4]:=Length[ans]Out[4]=1421Andrzej KozlowskiYokohama,
Japanhttp://www.mimuw.edu.pl/~akoz/http://
platon.c.u-tokyo.ac.jp/andrzej/
====
=Is this what you
mean:<<Algebra`FiniteFields`FieldIrreducible[GF[2^50],x]1 +
x^46 + x^47 + x^48 + x^50?Andrzej KozlowskiYokohama,
Japanhttp://www.mimuw.edu.pl/~akoz/http://
platon.c.u-tokyo.ac.jp/andrzej/Reply-To:
kuska@informatik.uni-leipzig.de
====
=Hi,SetAttributes[BackSet,
HoldAll]$backsetList = {}BackSet[a_, b_] := (a = b;
AppendTo[$backsetList, {Hold[a],
Hold[b]}];)Unprotect[Set]Set[a_, b_] /; Module[{pos, what},
pos = Position[$backsetList, a]; Print[Part[$backsetList,
Sequence @@ pos[[1]]]]; what = Part[$backsetList, Sequence @@
pos[[1]]]; Block[{$backsetList = {}}, Evaluate[what] = b; a =
b] ]andBackSet[status, {len1, len2}];status = {3, 2}will
work. Jens
====
=I have a matrix A which is 600x600 and I want
to projectit into semantic space using matrix B (which is
529x400)is it possible to do this with out trimming A to a
600x400matrix?thanks in advance,David The hunger for
knowledge only grows as one knows more.....David R. Hardoon
David.R.Hardoon@cs.rhul.ac.ukDept of Computer Science, Royal
Holloway, University of LondonMobile: +44 7967 634954 www:
http://g4.cs.rhul.ac.uk
====
=I am having trouble figuring out
how to change the color and thickness ofErrorBars. Can anyone
help me with this problem? ALso, how do you changethe font and
font size of an AxesLabel?Kenf
====
=One potential advantage to
using List as your head is thatThat's useful advice - though
advice I can't follow here.I don't use List as 
the head for a
fundamental reason:I'm relying on the head to indicate what
type of item I have.Some such indication is fundamental to my
quest to make programs that areeasy to UNDERSTAND and easy to
CHANGE.In this specific example, note that the functions have
argument patternsthat say they only act on STACKs.This is
essential. Without it - no polymorphism.It is powerful and
elegant to represent all kinds of data in lists - butthose
lists should be tagged in some way.IMO, it is important that
Mathematica not discourage the use of customheads, by
performance issues. Fortunately, once I had a defined symbol
as ahead, performance was indistinguishable from a List head,
for the operationsthat I need.-- Steve S.
====
=Found out what I
was doing wrong.I defined node AFTER the function
definitions.As a result, the function patterns 
didn't match
-the definitions were to any undefined symbol,but 
my data was
using a string node as the head.Mathematica was just building
a big structure of node and add chunks,without actually
calling any functions. Changed the code by moving the
linenode = nodeAHEAD of the add[...] definitions.Now I get
time values around O^1.24, for 20,000 iterations.This looks
consistent with an O(n*Log[n]) algorithm.Unfortunately, now
that it is really doing the work, the performance is
notgreat.gives practical performance in Mathematica.-- Steve
S.
====
=Reinhard,I presume that you are _not_ asking whether
you can pass a CompiledFunction to Java and then execute it
entirely within the Java runtime. That is not possible--to
call the CompiledFunction with some arguments you must call
back to Mathematica.To pass a CompiledFunction to Java, use
J/Link's Expr class, which can store any Mathematica
expression. If you have a method that takes an Expr argument,
then you can pass anything for that argument slot from
Mathematica. Say you have a Java method like this: private
Expr storedExpr; public void storeThisExpr(Expr e) {
storedExpr = e; }You can call this from Mathematica like
this:In[100]:= cf = Compile[....];In[101]:=
obj@storeThisExpr[cf]If you are manually reading the
CompiledFunction from a link, use the getExpr() method: Expr
e = ml.getExpr();No matter which way you get the
CompiledFunction into Java as an Expr, to call back to
Mathematica to execute it you have to use a relatively
obscure sequence of MathLink calls. You cannot use
putFunction() because the head is not a symbol--it's the
expression CompiledFunction[...]. You must manually build the
expression from a head and arguments. Here is the code: //
Assume e is an Expr containing a CompiledFunction that // you
want to call with the argument 42.
ml.putNext(MathLink.MLTKFUNC); ml.putArgCount(1); // Next we
put the head ml.put(e); // Then the arguments ml.put(42);
ml.waitForAnswer(); // Now read the answer ...Having said all
this, do you really need to pass the CompiledFunction itself
to Java? Why not just send its name to Java (i.e., assign the
function to a symbol and then pass the symbol name to Java)?
Let the CompiledFunction live only in Mathematica and just
refer to it by name from your Java code.Todd GayleyWolfram
ResearchReply-To: kuska@informatik.uni-leipzig.de
====
=Hi,no
because Mathematica store a rectangular grid of
functionvalues in SurfaceGraphics[] objects. But how
isOff[Throw::argt]Off[Plot3D::plnc]Off[Plot3D::gval]Plot3D[
Catch[If[y < 1 - x^2, x + y, Throw[]]], {y, 0, 1}, {x, -1,
1}, Jens
====
=With help from David Park I can state an answer
to my own question now:We want to plot the part of the
surface z=f(x,y) that satisfiesa<x<b and g(x)<y<h(x).Let
y(x,w)=(1-w)g(x)+w.h(x) and p(x,w)=f(x,y(w,x)).Now
plotParametericPlot3D[{x,(1-w)g(x)+w.h(x),p(x,w)},{x,a,b},{w,
0,1}]Simple comme dire bonjour!Frans de Boer
====
=Mihajlo
Vanevic was kind enough to point out where my system was
kaput.He pointed to Cell/DefaultOutputFormatType which was
set on to InputForm.I changed it to StandardForm and now my
system is (temporarily at least)at peace. No telling how many
years ago I screwed that up and only nowI noticed it.Many
thanks to you all for helping.Rob
====
=Ray and others: thanks
for the suggestions but these commands are just repeatedin
the output as if my version 4.0 doesn't understand
them.InputForm[result] gives50/5263InputForm[N[result]]
gives0.009500285008550257(no quotes)andNumberForm[N[result],
4] givesNumberForm[0.009500285008550257, 4]Someday I'll
update to ver. 4.2 and see if this starts working.Rob
====
=Hi
All.Does anyone know if the following sum has a solution
(hint of how to solve it is welcome too:)) ):T[n_, p_] :=
Sum[Binomial[2 j, j+1] p^j, {j, 1, n}]Constantine
Elster.Constantine ElsterComputer Science Dept.Technion
I.I.T.Office: Taub 411Reply-To:
kuska@informatik.uni-leipzig.de
====
=Hi,Mathematica 4.2
sayIn[]:=Sum[Binomial[2 j, j + 1] p^j, {j, 1, n}] //
FullSimplify //InputFormOut[]= -(-1 + Sqrt[1 - 4*p] +
2*p)/(2*Sqrt[1 - 4*p]*p) - (p^(1 + n)*Gamma[3 + 2*n]* ((1 +
n)*(3 + n)*Hypergeometric2F1[1, 3/2 + n, 3 + n, 4*p] + 2*(3 +
2*n)*p*Hypergeometric2F1[2, 5/2 + n, 4 + n, 4*p]))/(Gamma[2 +
n]*Gamma[4 + n]) Jens
====
=Ashraf,You should read up on Solve,
here is what you want to doSolve[Y==C+M+v*Y,Y]the Mathematica
response isA couple of things to note:1) the equals in the
equation is a double = sign. could also put a space between
the v and Y instead: v Y3) the answer is a Rule. You will
need to read up on those as well.Here is one way to use the
result:soln=Solve[Y==C+M+v*Y,Y]y=Y/.soln[[1]]you will now
have y = (C+M)/(1-v)Good luck,Kevin
====
=I assume that by your
use of the term dynamic models you mean that this isa matrix
equation? If so, you're pretty much out of luck. This is, to
meat least, probably the biggest on-going disappointment in
Mathematica. Manyphysics and engineering problems, and
especially in the teaching of physicsand engineeering,
involve the general solution of matrix equations. I'veyet to
find a package that allows for matrix equation solutions as
we've allseen done on the blackboard in college. (Perhaps
I've not looked in all theright places or I misunderstand
some package that's out there. If so, I'dbe 
overjoyed to be
proven wrong.)- Eric.--

====
==========================================================

====
=====What if the ÔHokey-Pokey' really IS what 
it's all
about??
====
===================================================

====
============ ====Ashraf,Solve[Y == C + M + v Y, Y]You must
use == and not = in the equation, and you must leave a
spacebetween v and Y, as is all clearly explained in The
Mathematica Book. Youmust also be a little wary of using
capital letter symbols as some of themare reserved by
Mathematica.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/
====
=You
can use the OPLinearAlgebra
package(http://library.wolfram.com/database/MathSource/4271/)
to solve this type ofnoncommutative algebra problem. Here is
what the solution looks like (The[Diamond] is a
non-commutative multiply.):<< OPLinearAlgebra`SetOperators[Y,
C, M, v]{C, M, v, Y}OPSolve[{Y == C + M + v [Diamond] Y},
{Y}]Calling OPLinearEqnToMatrices ... CollectPower is
On.Calling OPLinearSolve to give the result ... --Steve
LuttrellWest Malvern, UKReply-To:
kuska@informatik.uni-leipzig.de
====
=Hi,Table[ Block[{t}, y[t]
/. DSolve[{y'[t] == -y[t], y[0] == 1}, y[t], t] ], {t, 0,
10}] JensReply-To: kuska@informatik.uni-leipzig.de
====
=Hi,,
ist used for Times[] and not for spaces
andUnprotect[Times]Format[Times[a_, b_], TeXForm] :=
SequenceForm[a, , b]Protect[Times]does it. Jens
====
=You may
(or may not) be aware that for a little over a year we've
been writing a Mathematica ezine for WZ.com.With WZ.com no
longer active, we've decided to continue the ezine on our
own. It will remain a free ezine, but in a slightly different
format. We hope to keep the quick-and-easy feel of the past,
combined with some in-depth material. And you'll no longer
see any ads for any of the items WZ.com was selling. Plus, it
is an opportunity to take a more active role. You can find
previous issues at
http://omegaconsultinggroup.com/Services/wz.html. To
subscribe to the line Subscribe me to the ezine.Concerned
about your privacy? Aren't we all! Take a look at our 
privacy
policy at
http://omegaconsultinggroup.com/Services/privacy.html. In a
address for anything other than sending you the
ezine.-------------------------------------------------------
-------
====
=de==== the SystemFiles/Include directory include
already the map fiels for pdßatex and dvips.The Math1 <->
Mathematica1 ... Math4 <-> Mathematica4 aremostly unchanged
but the Math5 font has the gothic, the double struckand the
caligraphic alphabet inside. Now Mathematica5 hold
thecaligraphiccharacters, Mathematica6 the ghotic characters
and Mathematica7 thedouble struck characters. And so your
trick work for greek lettersbut not for ghotic, double struck
and some of the caligraphicletters. Jens> Dear MathGroupers,>
Mathematica 4.2 comes with seven font families, compared with
five in the> previous version. I am wondering if anyone knows
how compatible these fonts> are.> The reason I ask is that I
have a large number of graphics exported from> Mathematica
4.1 under Mac OS 9, which now need to be incorporated into a>
(pdf)LaTeX document under Mac OS X. TeXShop/teTeX under Mac OS
X chokes> badly on the OS9 versions of the old font sets, but
if I edit my LaTeX .map> file to tell pdfLaTeX to embed
Mathematica1.pfa instead of Math1(.pfa),> everything seems to
work. Ditto for Mathematica2/Math2. So I'm getting rhos> and
thetas where I should be getting rhos and thetas. This is a
relief, but> I'm wondering whether this trick will work in
general for Mathematica> graphics generated under from
previous versions. What's in the new fonts> 
that's not in the
old fonts? The documentation just says New characters for>
double struck, script, and Gothic numbers, the planets, and
musical notes.> Does this mean that Math[Bold,Mono][1-5] =
Mathematica[Bold,Mono][1-5]?> Luci EllisReply-To:
kuska@informatik.uni-leipzig.de
====
=look at the online-help
for Graphics[] ! What is listed first ??The following graphics
primitives can be used:....andThe following graphics
directives can be used:What else can be done ? Should the
text color red ?and ßashing ? Jens> So today I'm doing 
some
graphics. Oh: there are _options_ and _directives_.> So I try
to find out what are all the graphics primatives?> When you
look in the index of the 1500 page book, or type Graphics
Primatives> in the on-line Help, you don't find 
it. You have
to look under Graphics or> Graphics3D. In the book, it's in
the appendix.> And the entry in the appendix is not quite
complete:> It says Graphics[primatives, options] represents a
two-dimensional graphical> image.> It should say returns a
graphis object The reason is that the appendix is> the place
for the complete and formal description of the functions and
all> their arguments and returns. No longer the place for
chatty-teachy.> (represents is chatty-teachy)> To understand
the system, I need to be able to trace the sequence>
Line[...] is a graphics _primative_> Graphics[ ... graphics
primatives ....] is a graphics _object_> Show [... graphics
object...] puts a picture on the screen.> I don't know if
this sounds dumb on my part, but I do know that I have just>
wasted a bunch of time on this, and that having to go back
and forth just to> find out the type of the return of a
function (to use old fashioned> nomenclature) .... which
people knew thirty years ago was part of documenting a>
system .... seems pretty dumb on the part of Mathematica.>
Dave GolberReply-To: kuska@informatik.uni-leipzig.de
====
=>
What this example shows is that the description>
Graphics[primatives,options] represents ...> is inconsistent.
The first written argument primatives must be an actual> 
single
argument. If you want to submit several graphics primatives,
you must> combine them into a List.If there are multiple
elemts allowed the documentation would
writeGraphics[primitive1,primitive2,...,options]as to see in
the documetation for Union[]> On the other hand, the second
written argument may be multiple actual> arguments, or may be
a List.> The documentation is inconsistent.> I see no
explanation of this any part of the Appendix. And it's
particularly> disappointing because systems were able to
document these sorts of things> decades ago.> Sorting this
kind of thing out takes hours. Every time I do something
slightly> new with Mathematica, I find I spend a lot of time
because of sloppy> documentation.Hmm, sloppy documentation ?
realy ? My Windows XP came with a 12 pagesbooklet*that* is a
sloppy documentation ! The book should be splitted into3 or 4
parts and it is to heavy to read it in a bathtub but it isnot
sloppy.> I'd really like to hear Mathematica personnel 
report
on (1) Do they agree there> is an error here, and (2) When do
they expect to have corrections available.> And by the way:
another thing that needs to be corrected in the
documentation> of Graphics is how graphics directives get
used . The header line is:> Graphics[primatives,options]>
Unfortunately, a graphics directive is neither a primative or
an option. And> sure enough, when we read a little farther on,
we see The following graphics> primatives can be used This is
vague. Used how? Remember that the appendix> the Appendix.)>
(The fact that the body can say nothing better than the vague
statement that> directives can be used is of course a result
of the fact that the form> Graphics[primatives,options] does
not allow directives.)There is a much harder inconsistence in
Mathematicas Graphic[], thegraphics directives should be
options to the primitives and notstandalonefunctions,
i.e.,Line[{{0,0},{1,1}}, Color->RGBColor[1,0,0]]instead
of{RGBColor[1,0,0],Line[{{0,0},{1,1}}]}May be that you save a
lot of time when you
readhttp://www.amazon.com/exec/obidos/ASIN/0387940472/qid=
1038213121/sr=2-1/ref=sr_2_1/002-6128995-6365638andhttp://
www.math.ubc.ca/people/faculty/cass/graphics/text/www/
index.html Jens
====
=This will take you pretty
close:FullSimplify[ComplexExpand[ d*Pi + I*d*(Log[M] -
2*Log[(-I)*Sqrt[(M*r)/d] + Sqrt[M - (M*r)/d]]),
TargetFunctions -> {Re, Im}], M > 0 && 0 < r < d]d*(Pi +
2*ArcTan[Sqrt[M - (M*r)/d], -Sqrt[(M*r)/d]])Still, there
appears to be something missing among the rules FullSimplify
is applying.It seems related to the following behaviour, that
certianly suggests a bug:FullSimplify[ArcTan[x, y], {x > 0, y
< 0}]ArcTan[x, y]FullSimplify[ArcTan[2*x, 2*y], {x > 0, y <
0}]ArcTan[y/x]>> I'm trying to simplify (M, r, d are 
positive
reals with r<d)> d*Pi + I*d*(Log[M] - 2*Log[(-I)*Sqrt[(M*r)/d]
+ Sqrt[M - (M*r)/d]])>> to>> 2*d*ArcCos[Sqrt[r/d]]>> I've
tried applying>> FullSimplify[#,M>0 && 0<r<d]&>> and>>
ExpToTrig>> but Mathematica won't get rid of the Log and the
I and give me the > ArcCos.>> Does anyone know which
Mathematica functions to apply to my expression > to> give me
the simplification I want?>>>>Andrzej KozlowskiYokohama,
Japanhttp://www.mimuw.edu.pl/~akoz/http://
platon.c.u-tokyo.ac.jp/andrzej/
====
=Dear Borut,The problem is
that Mathematica does not export proprietary fonts. The
solution depends on which Mathematica version you use.If you
use Mathematica 4.0 or 4.1, you can download paul hinton's
emmathfnt tool which manages inclusion of the fonts or you
can configure ghostscript so that it puts Mathematica fonts
into the graphics files.If you use 4.2.0, give it up, since
font export isn't possible any more for the unexperienced
user. This is because the emmathfnt tool does not work with
the newfont naming scheme. And neither the emmathfnt tool nor
hinton who explained me the issues involved in a friendly and
clear reply and referred me to the wolfram support team. Now
I am waiting about one week for an answer from
support@wolfram.com.If you are an experienced user (with a c
compiler available), try to adapt the source code of
emmathfnt and recompile it. I tried to do this but I failed,
since I don't understand the font naming scheme and other
details.If you use version 4.2.1, you can configure font
export in the preferences dialog (or give it as an
option).[Wolfram research seems to have realized *finally*
that eps font export is an important feature demanded by many
users. It is surprising that this process lasted some years,
since EPS font export questions were a FAQ in MathGroup.] If
you have premier service, simply obtain the updatedversion.
Johannes Ludsteck> As I am generating Mathematica images for
inserting them into TeX, problems> occur, like> f(t) seen as
fHtL on EPS image in GSView, FontFamily->Trebuchet MS (or>
Helvetica) not taken into account, ...> Can you direct me to
any relevant material on the internet, book about ways> to
tackle this problems?> Borut Levart>
<><><><><><><><><><><><><><><><><><>Johannes LudsteckInstitut
fuer VolkswirtschaftslehreLehrstuhl Prof. Dr.
MoellerUniversitaet RegensburgUniversitaetsstrasse 3193053
Regensburg
====
=>> 1. If I want to print out the elements of S
which have order 15, do I > use a> nested If or Do loop?I
think here by print out you mean just make a printable list
of the elements? In that case you do not need any loops or
nested Ifs. Mathematica's programming language is quite
different form most others. It's more like Lisp and Prolog
than any more usual ones (you can also program in that style,
called imperative, but doing so in Mathematica is 
inefficient).
So here is how to make a list of elements which have order
15.S = Flatten[Outer[List, Range[0, 89], Range[0, 35]],
1];mult[{a_, b_}, {c_, d_}] := {Mod[a + c, 90], Mod[b + d,
36]}ord[p_] := Length[NestWhileList[mult[p, #1] & , p, #1 !=
{0, 0} & ]]orders = Map[ord,S];So now S is the list of
elements and orders a list of orders. You can get all
elements of order 15
with:Extract[S,Position[orders,15]]{{6,0},{6,12},{6,24},{12,0
},{12,12},{12,24},{18,12},{18,
24},{24,0},{24,12},{24,24},{36,12},{36,
24},{42,0},{42,12},{42,24},{48,0},{48,12},{48,24},{54,12},{54
,
24},{66,0},{66,12},{66,24},{72,12},{72,24},{78,0},{78,12},{78
, 24},{84,0},{84,12},{84,24}}>> 2. I have just ordered the
Mathematica Link to EXCEL package. I will > use it> able to>
find a way to get all of the output of the table to print on >
Mathematica. I> have tried page wrap of various sorts. Is
there a way?I think here by print out you mean print out on a
printer? This sounds like a problem with your particular
operating system which I most likely cannot solve. I use
Mathematica on Mac OS X and a PostScript printer any have
never had any problems with printing. If you really have a
printing problem you should describe precisely the hardware
and the operating system you are using and someone will
probably be able to help you.If on the other hand you are
referring to printing out a table on the screen: just use the
command TableForm to format your output as a table. It takes
various options that can make it look nicer. But I think you
were referring to real printing.Andrzej> Andrzej,> Two
questions:>> 1. If I want to print out the elements of S
which have order 15, do I > use a> nested If or Do loop?>> 2.
I have just ordered the Mathematica Link to EXCEL package. I
will > use it> able to> find a way to get all of the output of
the table to print on > Mathematica. I> have tried page wrap
of various sorts. Is there a way?>> much> more fun.>> Diana>
Mathematica groupies,>> I just got my copy of Mathematica
tonight. I understand permutations >> of> S_5>> and A_5,
etc., but I am having a little difficulty figuring 
out how to>>
calculate the order of elements of Z_90 and Z_36, and the
order of> elements>> of the external direct product of Z_90
(+) Z_36.>> I am wanting to calculate the number of cyclic
subgroups of order 15 >> in> Z_90>> (+) Z_36 with
Mathematica.>> Z_90 is the additive group {0, 1, 2, 3, ...,
89}. Z_36 is the additive> group>> {0, 1, 2, 3, ..., 36}.>>
So, elements of Z_90 (+) Z_36 would be 2-tuples of the form:
(a, b), >> where> a>> is an element of Z_90, and b is an
element of Z_36. If you add (a, b) >> to>> itself 15 times,
you will get {0, 0}. Note that the operation of >> adding>
the>> first component of the 2-tuple is modulo 90, and the
operation of >> adding> the>> second component of the 2-tuple
is modulo 36.>> Can someone help?>> Diana M.>>>>Andrzej
KozlowskiYokohama,
Japanhttp://www.mimuw.edu.pl/~akoz/http://
platon.c.u-tokyo.ac.jp/andrzej/
====
=Not a function call but a
pattern with Head tree. Actually pretty basic Mathematica
programming. Look up the Mathematica book under Head etc. Oh,
by the way, one more thing. Knowing too much about other
programming languages is actually a bit of a handicap as far
as Mathematica is concerned, unless they are Lisp and Prolog
(and even then you have to be pretty careful for things tend
to be not quite what they seem...)Andrzej KozlowskiYokohama,
Japanhttp://www.mimuw.edu.pl/~akoz/http://
platon.c.u-tokyo.ac.jp/andrzej/> I found this:>
rightTree[tree[_, _, right_]] := right>> in Dr.
Mç.94ç[Times]der's Computer Science with 
Mathematica. I
understand the > spirit of> the statement, but it looks
rather strange to me. Is there a way of> describing, in a
language such as English, exactly what this definition> means?
The part that I find confusing is the inclusion of what looks 
>
like a> function call in the parameter list. This simply 
defies
my sense of > how> programming languages work. That doesn't
mean it's a bad thing. It > just> means that I 
don't feel
comfortable with the syntax.>> Is there somewhere in the
Mathematica Book , or Help which discusses > such> arcana?>>
-- > STH> Hatton's Law:> There is only One inviolable
Law.>>
====
=With Mathematica 4.1 (and previous versions, I'm
pretty sure), I would oftendisplay a graphics image in a
notebook, grab it with the mouse and enlargeit to a size
often larger than the notebook window or even my screen.
Itwas possible to scroll the window to the right to see parts
of the graphicat high magnification.This is no longer possible
with 4.2. Parts of the image that fall outsidethe width of
the notebook window are not displayed and you cannot
scrollright to see them. Thus there is a limit when the
notebook window ismaximised to take the whole screen width.I
suspected that the ImageRegion option had something to do
with this andtried changing it in the Option Inspector but to
no avail.Does anyone know how to get around this annoying
problem?John Jowett
====
=> I found this:> rightTree[tree[_, _,
right_]] := right> in Dr. 
Mí¡.89[Hyphen]åder's Computer
Science with Mathematica. I understand the spirit of> the
statement, but it looks rather strange to me. Is there a way
of> describing, in a language such as English, exactly what
this definition> means? The part that I find 
confusing is the
inclusion of what looks like a> function call in the
parameter list. This simply defies my sense of how>
programming languages work. That doesn't mean 
it's a bad
thing. It just> means that I don't feel comfortable with the
syntax.> Is there somewhere in the Mathematica Book , or Help
which discusses such> arcana?How about rightTree returns the
third element of a three-element expressionheaded by
Ôtree'?Think Lisp (I date myself :-). Lower your 
mental wall
between functions anddata. Everything is an expression [The
Mathematica Book section 2.1.1]. Inthe example above,
rightTree resembles a function, and tree resembles
astructure, but these apparent roles could (in principle)
reverse elsewhere.If tree also evaluates as a function, then
the expression above would stillwork if rightTree has the
HoldAll or HoldFirst attribute.Tom Burton
====
=Steve,The header
tree is not a function call but a wrapper for three objects.
Isthe following code as much of a
mystery?pick3rdOnListOf3[List[_, _, third_]]:= thirdDavid
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/means
that I don't feel comfortable with the syntax.Is there
somewhere in the Mathematica Book , or Help which discusses
sucharcana?--STHHatton's Law:There is only One inviolable
Law.
====
=1.
ConsiderIn[1]:=Function[y,Module[{x=1},y+x]][x]Module[{x=1},#
+x]&[x]Out[1]=1+xOut[2]=2The first example always works as a
function returning its argument plusone (unless we pass it
Ôunallowed' argument x$); the second 
exampledoesn't.I think
the unnamed Slot parameters should internally be treated
thisway:In[3]:=Function[slot1,Module[{x=1},slot1+x]][x]Out[3]
=1+xThis works consistently for any arguments passed to the
function,doesn't interfere with the Module local variables
and gives the sameresults as thedefinition using named
parameters.2.
ConsiderIn[4]:=x=1;Function[x,Evaluate@x]Module[{x=2},
Function[x,Evaluate@x]]x=.Out[5]=Function[x,1]Out[6]=Function
[x,1]The reason why the second example works this way is that
Function makesthe global x variable local, not the Module's
x$n.This means we can't define functions that 
depend on
Module's localvariables unless we use some special
tricks.Wouldn't it be better if Function made 
Module's
variables local insteadso thatModule[{x},
Function[x,fx]]becameModule[{x}, Function[x$n,fx]]?This
behaviour would be useful in conjunction with the
modificationdiscussed below. The same goes for Rule and Set
inside Module.3.
ConsiderIn[8]:=y=x;Function[x,Evaluate[2x+y]]y=.Module[{y=x},
Function[x,Evaluate[2x+y]]]Out[9]=Function[x,3
x]Out[11]=Function[x$,x+2 x$]problem is that it makes the
task of creating functions inside Modulesmuch more
complicated;if we want to do something like Unapply[fx, x],
there is a simplesolution Evaluate[fx/.x->#]& but it doesn't
work for fx=#+x&.Is it possible to change the evaluation
rules so that inFunction[x,Evaluate[fx]] fx is evaluated
outside the Function scope?Then the result is passed to the
Function; in the second example aboveit would work as
follows:Evaluate[2x+y]->3x->Function[x,3x],or, in the case of
x being local to the
Module,Evaluate[2x$n+y]->3x$n->Function[x$n,3x$n].The same
goes for the right-hand side of Rule and Set.Maxim Rytin
====
=>
I found this:> rightTree[tree[_, _, right_]] := right>> in Dr.
Mç.94ç[Times]der's Computer Science with 
Mathematica. I
understand the > spirit of> the statement, but it looks
rather strange to me. Is there a way of> describing, in a
language such as English, exactly what this definition> means?
The part that I find confusing is the inclusion of what looks 
>
like a> function call in the parameter list. This simply 
defies
my sense of > how> programming languages work. That doesn't
mean it's a bad thing. It > just> means that I 
don't feel
comfortable with the syntax.>It's best to forget about
function calls, think about expression evaluation. What I
mean is Mathematica operates using a read-eval-print loop
similarly to LISP (again, it really helps to have some
exposure to LISP and Prolog when dealing with programming
issues in Mathematica). So you enter an expression,
Mathematica reads it, then passes it off to be evaluated then
finally prints the resulting expression. Function call implies
all sorts of arcana about putting references or copies of
arguments on a call stack, jumping to a subroutine, restoring
the call stack etc. that really isn't a factor in 
Mathematica.
What does this have to do with the above expression?First
let's get rid of the infix notation
rightTree[tree[_,_,right_]]:=right is equivalent to
SetDelayed[rightTree[tree[_,_,right_]],right] when the
evaluator sees the SetDelayed it means means create a rule
associating the DownValue of the first argument (held) with
the second (also held). Since all the arguments are held
there is no more evaluation.That's pretty simple, the real
magic happens when you evaluate the expression
rightTree[(some expression with head tree and three
arguments, let's call it t for now)]. The evaluator searches
the rule list of rightTree for the most specific match, since
t has head tree (and assuming no more specific rule has been
written) that's the rule that transforms the expression. 
That
rule says that the third argument of the expression
tree[_,_,_] is to be called right and that is the value of
the expression rightTree[tree[_,_,right_]]. So the expression
represented by the pattern right_ is passed to the evaluator.
If that expression has no transformation rules associated
with it then evaluation stops and that value is returned.This
isn't really a question of syntax (since the LISP syntax 
looks
something like (defun rightTree (tree) (car (nthcdr 2 tree))),
I've omitted the type checking) it's a 
question of semantics.
Incidentally I believe the Prolog syntax actually looks a lot
more like the Mathematica, it's been a while since I dealt
with Prolog but I think rightTree isisaTree(Node) :-
atom(Node)isaTree([Left,Node,Right]) :-
isatree(Left),atom(Node),isatree(Right).rightTree(tree) :-
isaTree(tree),Right.This version has the type checking but a
real Prolog programmer would probably use guards.> Is there
somewhere in the Mathematica Book , or Help which discusses >
such> arcana?> This is all explained in section 2.5.4 of the
Mathematica book. The key is to understand that at the most
basic level all Mathematica does is take expressions, match
them to rules (as expressed by patterns), apply the
transformation and repeat the above process until no more
transformations can be done.Ssezi
====
=Steven,Mathematica may
look like it uses functions but it is actually to do withthe
action of replacement rules on expressions.Thus,
rightTree[tree[_, _, right_]] := right( or for that matter
rightTree[tree[_, _, r_]] := r)means: Store a rule so that
when during the evaluation process an expression of the form
rightTree[tree[_, _, x_]] is found, it is replaced by xThe
rule is stored in DownValues[rightTree] DownValues[rightTree]
{HoldPattern[rightTree[tree[_,_,right_]]]:>right}The
evaluation process is described in Appendix 4 (A.4) of the
Book.--Allan---------------------Allan HayesMathematica
Training and ConsultingLeicester
UKwww.haystack.demon.co.ukhay@haystack.demon.co.ukVoice: +44
(0)116 271 4198> I found this:> rightTree[tree[_, _, right_]]
:= right>> in Dr. Míóder's Computer Science 
with Mathematica.
I understand thespirit of> the statement, but it looks rather
strange to me. Is there a way of> describing, in a language
such as English, exactly what this definition> means? The part
that I find confusing is the inclusion of what looks likea>
function call in the parameter list. This simply defies my
sense of how> programming languages work. That doesn't mean
it's a bad thing. It just> means that I don't 
feel
comfortable with the syntax.>> Is there somewhere in the
Mathematica Book , or Help which discusses such> arcana?>>
--> STH> Hatton's Law:> There is only One inviolable
Law.>>Reply-To: kuska@informatik.uni-leipzig.de
====
=this mean:
if someone wish the right branch of a tree[]it does not matter
what is in the left branch of thetree or in the tree's node.
Jens> I found this:> rightTree[tree[_, _, right_]] := right>
in Dr. Míóder's Computer Science with 
Mathematica. I
understand the spirit of> the statement, but it looks rather
strange to me. Is there a way of> describing, in a language
such as English, exactly what this definition> means? The part
that I find confusing is the inclusion of what looks like a>
function call in the parameter list. This simply defies my
sense of how> programming languages work. That doesn't mean
it's a bad thing. It just> means that I don't 
feel
comfortable with the syntax.> Is there somewhere in the
Mathematica Book , or Help which discusses such> arcana?> -->
STH> Hatton's Law:> There is only One inviolable Law.
====
=> I
think, when Mathematica was created, OO was not yet existent
in> programming languages.You could easily learn about the
history of OO. Simula (1967), Smalltalk,the variety of very
sophisticated lisp OO systems: ßavors, LOOPS, CLOS,all
predate Mathematica.Wolfram claims Mathematica unifies various
programming paradigms,and if you don't look too closely, it
does have many resemblances.However, much of it is really
implemented on top of pattern matchingand a kind of
fixed-point evaluation. Thus details like scope, vital
infunctional programming, were wrong initially (Block), and
repaired so they arebetter (Module) but still not quite
right.You also misunderstand the nature of language
implementation anddesign. Mathematica or Java or Lisp.. are
designed for various reasons.Any of them can be implemented
by an interpreter or via compilation.Or both. If you mean
Mathematica is interactive, sure. But thatdoes not mean it
must be interpreted. In fact there is a compiler usedby
numerical evaluation in plotting and a few other spots.I
think that it would be useful for you to learn about
programming languagesinstead of believing that all you need
to know about programming languagesis that subset that
appears in Mathematica or Java. There are many many
textbookson the subject of programming languages. I suggest
you find The Art of theMeta Object Protocol by Bobrow and
Kiczales, if you wish to understand thevarieties of OO
possible.RJF I am astonished about the OO-like features,
which are> built into Mathematica despite of this. Into older
languages OO must be> built after they were created, and this
is also done.> In my opinion it is not the main criterion,
that a program runs fast in that> programming language. If
this was the case you should program in assembler> language.
The fact, that Mathematica is an interpreted language shows,
that> other considerations were prominent, when the language
was designed.> Compiled languages are faster.> The main
criterion is in my opion is, that the program can be created
easily> and yet more important, that it can be easily
understood and modified> afterwards.> I think, Mathematica is
well suited, to integrate an OO-System.> Hermann Schmitt.>
----- Original Message ----->>Well, it simply means that I do
not agree with some of the things you>>seem to think, that's
all. Basically it amounts to the fact that>>Mathematica was
not meant to be used as an OO-langauge by Wolfram and>>was
deliberately not constructed in this way, unlike those
languages>>that are designed from ground up to be OO. So when
you write in>>OO-style you are simply dressing-up Mathematica
language in a foreign>>paradigm and usually end up with less
efficient code. Unlike what you>>seem to think, I think there
is a big difference between built-in>>abilities and those
added by users, and it shows itself in performance.>>In fact
being aware of this difference is the most important factor
in>>efficient Mathematica programming. This distinguishes
Mathematica from>>most other languages.>>I don't really want
to enter into any arguments which after all are>>only a
matter of opinion, but I would propose the
following>>objective test of good OO package in Mathematica.
The test is that>>a proficient user cannot write easily a more
efficient and equivalent>>code without the package. So far I
have not seen any OOP in>>Mathematica that does anything in a
way that I cannot do equally or>>(usually) more efficiently
myself without it. Unless an OOP package can>>satisfy this
requirement it will be used only by users who learned>>their
programming on other languages and have not adapted their
style>>to Mathematica.>>By the way, I have a good reason to
believe that Wolfram will introduce>>some OOP into
Mathematica, that will actually enhance rather than>>reduce
performance. I will then think of it as real OOP rather
than>>pseudo-OOP.>>Andrzej Kozlowski>>>>>what do you mean
with pseudo-OOP?>Hermann Schmitt>----- Original Message
----->To: <mathgroup@smc.vnet.net>>Sent: Tuesday, November
05, 2002 11:00 AM>>>>>Various have people have sent
interesting examples of OO programming>>in>>Mathematica, as a
proof that it can be done. However, let me clarify>>one
possible confusion: I for one have never doubted that it can
be>>done. (In fact I have occasionally tried it myself). The
more>>doubtful>>issue is is it worth doing?, or is it really
the best way to>>program>>in Mathematica?. I can imagine that
OOP may be useful for dealing>>with large structures and there
may be (as I have been told) some>>performance advantages in
introducing some OOP into Mathemaitca. But>>none the examples
I have seen here seems to me to gain much from>>pseudo-OOP,
which is in fact also the trouble with Maeder's package.
I>>have seen a couple fairly large project written using it
(one is the>>Object-Oriented Graph Theory in Chapter III of
Grey's Mastering>>Mathematica), but all of them can be
programmed with no more effort>>using the natural Mathematica
functional-pattern based style.>>>>Andrzej
Kozlowski>>>>>>>Andrzej Kozlowski>>Yokohama,
Japan>>http://www.mimuw.edu.pl/~akoz/>>http://
platon.c.u-tokyo.ac.jp/andrzej/> In-reply-To:
<200211250656.BAA14527@smc.vnet.net>
====
=D> What does it mean
when I submit a differential equation, and the sameD>
equation appears as the output?Mathematica cannot find the
answer.D> I have tried this equation:D> DSolve[y'[x] == 
(-x^2
+ 2 y[x]^(-3))/(2 x y[x] - 3 x^2), y[x], x]D> Is it too
complicated for the program?In the version you tried,
yes.Please note that y'[x] = dy/dx = 1/(dx/dy)= 
1/x'[y](This
trick only works with the first derivative, 
don't jump to a
conclusionthat y''[x] = 
1/x''[y], that's totally wrong - 
by
the way, why? ;-)(* DSolve[y'[x] == (-x^2 + 2 y[x]^(-3))/(2 
x
y[x] - 3 x^2),y[x], x] *) In[1] := DSolve[x'[y] == (-x[y]^2 
+
2 y^(-3))/(2 x[y] y - 3 x[y]^2), x[y], y] Out[1] = {{x[y] ->
y/3 - (2^(1/3)*y^4)/(3*(-27*y^4 - 2*y^9 + 27*y^6*C[1] +
Sqrt[-4*y^18 + (-27*y^4 - 2*y^9 + 27*y^6*C[1])^2])^(1/3)) -
(-27*y^4 - 2*y^9 + 27*y^6*C[1] + Sqrt[-4*y^18 + (-27*y^4 -
2*y^9 + 27*y^6*C[1])^2])^(1/3)/(3*2^(1/3)*y^2)}, {x[y] -> y/3
+ ((1 + I*Sqrt[3])*y^4)/(3*2^(2/3)*(-27*y^4 - 2*y^9 +
27*y^6*C[1] + Sqrt[-4*y^18 + (-27*y^4 - 2*y^9 +
27*y^6*C[1])^2])^(1/3)) + ((1 - I*Sqrt[3])*(-27*y^4 - 2*y^9 +
27*y^6*C[1] + Sqrt[-4*y^18 + (-27*y^4 - 2*y^9 +
27*y^6*C[1])^2])^(1/3))/(6*2^(1/3)*y^2)}, {x[y] -> y/3 + ((1
- I*Sqrt[3])*y^4)/(3*2^(2/3)*(-27*y^4 - 2*y^9 + 27*y^6*C[1] +
Sqrt[-4*y^18 + (-27*y^4 - 2*y^9 + 27*y^6*C[1])^2])^(1/3)) +
((1 + I*Sqrt[3])*(-27*y^4 - 2*y^9 + 27*y^6*C[1] +
Sqrt[-4*y^18 + (-27*y^4 - 2*y^9 +
27*y^6*C[1])^2])^(1/3))/(6*2^(1/3)*y^2)}}That is you have 3
implicit equations connecting x and y. Your ODE is
solved.Best wishes,Vladimir BondarenkoMathematical and
Production DirectorSymbolic Testing GroupMail : 76 Zalesskaya
Str, Simferopol, Crimea, Ukraine
====
=> What does it mean when I
submit a differential equation, and the same> equation appears
as the output?> I have tried this equation:> DSolve[y'[x] ==
(-x^2 + 2 y[x]^(-3))/(2 x y[x] - 3 x^2), y[x], x]> Is it too
complicated for the program?Yes, in general, if Mathematica
returns input unprocessed without comment,you may conclude
that Mathematica cannot process it, for whatever
reason.(Sometimes Mathematica is not sure why, so better to
say nothing. In thiscase, perhaps a hint would be nice.) This
differential equation, though notcomplicated, is strongly
nonlinear. Few such DE's admit analyticalsolutions. Try
NDSolve.Tom Burton
====
=> As I am generating Mathematica images
for inserting them into TeX, problems> occur, like> f(t) seen
as fHtL on EPS image in GSView, FontFamily->Trebuchet MS (or>
Helvetica) not taken into account, ...> Can you direct me to
any relevant material on the internet, book about ways> to
tackle this problems?Note to Wolfram:As messages about this
particular gotcha keep appearing and reappearing time after
time after time (from me among others), perhaps it's time to
implement a more long-term solution to this whole Operator
Substitution hassle than just repeatedly replying to these
newsgroup messages.It may be too late to change the default
setting on the Operator Substitution option -- although my
view would be that if a user explicitly specifies
FontFamily->xxx, the user has stated what she wants, and
Operator Substitution should definitely be automatically
turned off. In addition:1) Add at least a brief notation on
this issue in the Help messages for TextStyle and other
places, so useres will at least learn of the Operator
Substitution problem and have a pointer to look for more
detailed info.2) *Much* upgraded documentation within the
Help system of the (presently very mysterious) PrivateFont
and OperatorSubstitution optionsPower tends to corrupt.
Absolute power corrupts absolutely. Lord Acton
(1834-1902)Dependence on advertising tends to corrupt. Total
dependence on advertising corrupts totally. (today's
equivalent) Reply-To: kuska@informatik.uni-leipzig.de
====
=look
into
the$TopDirectory/SystemFiles/IncludeFiles/TeX/texmf/doc/
wolframdirectory and read the msymdoc.pdf it include the
instructions tomake the Mathematica fonts work with
TeX/Ghostscript .. Jens> As I am generating Mathematica
images for inserting them into TeX, problems> occur, like>
f(t) seen as fHtL on EPS image in GSView,
FontFamily->Trebuchet MS (or> Helvetica) not taken into
account, ...> Can you direct me to any relevant material on
the internet, book about ways> to tackle this problems?>
Borut Levart
====
=In response to David Park
<djmp@earthlink.net>:> 1) [How to] Save and restore the
default options.> saveListPlotOptions = Options[ListPlot];>
SetOptions[ListPlot, Sequence @@ saveListPlotOptions]; 
That's
useful -- and it makes the point that in a situation like this
you have to use the Sequence command in front of the list of
options. But, I'm still surprised there's no 
way in
Mathematic to, in essence, reset the kernel to its initial
default startup state -- to, in essence, reload the kernel
from disk or, in essence, *really* Remove everything that's
happened since the most recent startup. There used to be the
CleanSlate package to do something akin to this, but it
wasn't actually all that clean, and I don't 
know if it's
still around.> 2) Within Mathematica and a single plot you
can use the TextStyle option.> TextStyle -> {FontFamily ->
Helvetica, FontSize -> 12,> FontWeight -> Bold}];> or for a
whole notebook> $TextStyle -> {FontFamily -> Helvetica,
FontSize -> 12, FontWeight ->> Bold} But neither of these
solves the turn off Operator Substitution problem, do they?
-- and that's the aspect that I'm after.> 
3)Use the AxesFront
option from the FilledPlot package.> I have always maintained
that AxesFront should be a general Graphics option> and not
buried in the FilledPlot package. I'd agree with that!Power
tends to corrupt. Absolute power corrupts absolutely. Lord
Acton (1834-1902)Dependence on advertising tends to corrupt.
Total dependence on advertising corrupts totally. (today's
equivalent) 
====
=I have a Mathematica program that uses
DatabaseAccess package toretrieve large amount of data from
Sybase database. DatabaseAccess isavailable for Windows only
since it uses odbc. Now, I wanted to runaccess. At the end of
this message I'll insert the code that
emulatesDataSourceEvaluate and OpenDataSource from
DatabaseAccess but usesjdbc. The implementation is simple and
straightforward. The problem isthe code works MUCH slower than
the equivalent odbc-based one on NT. Iretrieve A LOT of data
so this is really annoying. So I have twoquestions:1. Is it
one of the Mathematica functions that make it slow
(theconversion of the result of the sql query into
Mathematica table) oris it some fundamental java-related
problem (JLink, jdbc etc)?2. How can I make it run
faster?AlexanderInstallJava[];AddToClassPath[ jconn2.jar,
jdbc2_0-stdext.jar];LoadJavaClass[java.sql.Types];
LoadJavaClass[java.util.Calendar];Clear[GetColumn];GetColumn[
i_, type_, rs_] := Switch[type, Types`CHAR,rs@getString[i],
Types`DOUBLE,rs@getDouble[i], Types`INTEGER,rs@getInt[i],
Types`VARCHAR,rs@getString[i], Types`TIMESTAMP,JavaBlock[
ToString[cal@get[Calendar`MONTH] + 1]<>/<>
ToString[cal@get[Calendar`DATE]]<>/<>
ToString[cal@get[Calendar`YEAR]]] ],
Types`NUMERIC,rs@getInt[i]]ClearAll[DataSourceEvaluate];
DataSourceEvaluate[conn_,sql_String] := JavaBlock[ Module[{st
= conn@createStatement[],rs,md, types,res}, res = If
[st@execute[sql], rs = st@getResultSet[]; md =
rs@getMetaData[]; types =
Array[md@getColumnType[#]&,{md@getColumnCount[]}]; Rest[
NestWhileList[ Array[GetColumn[#,types[[#]], rs]
&,{Length@types}] &, 1, rs@next[] &]], {}]; st@close[];
res]]Clear[OpenDataSource];OpenDataSource[database_String,
user_String,password_String] := Module[{driver =
JavaNew[com.sybase.jdbc2.jdbc.SybDriver], dbUrl =
jdbc:sybase:Tds:<>$DBServer<>:<>$DBPort<>/<>database, prop =
JavaNew[java.util.Properties]}, prop@setProperty[user,user];
prop@setProperty[password,password];
driver@connect[dbUrl,prop]];
====
=I argued over the more
appropriate definition of (or algorithm for) theinverse
cotangent function with someone, based on what math we have
learned.(A) arccot x = Pi/2 - arctan x(B) arccot x = arctan
1/xMy calculus book used the definition A when it talked about
the derivativesof inverse trig functions. I think it said that
definition B is also used insome situations, but I 
don't have
the book now.It's obvious why my calculus book and most
calculus study guides found onthe Internet use the definition
A--arccot would be continuous on its domain,and the
complementary angle relationship is preserved. But
Mathematica usesdefinition B, or a definition 
closer to B.
Seehttp://mathworld.wolfram.com/InverseCotangent.htmlSo, is
there any merit of definition B, besides convenience? Should
wesacrifice the continuity of inverse cotangent just to
satisfy the simplex-reciprocal relationship?And this
pagehttp://mathworld.wolfram.com/
InverseTrigonometricFunctions.htmlsays the domain (my math
teacher would say range) of ArcCot is (0, Pi/2)or (-Pi,
-Pi/2). Where does that come from? It's not even consistent
withthe graph on Wolfram's own inverse cotangent page! The
range for ArcCsclisted there is strange too. Also, Wolfram
doesn't say that arccot is notdifferentiable because of
discontinuity at x=0 when it lists the firstderivative of
ArcCot[z], while definition B would require that. Is
itsomething in more advanced mathematics that I don't know
yet?When I argued over the two definitions few weeks ago, my
opinion was thatdefinition B was acceptable, or even should be
authoritative, becauseWolfram and possibly other systems use
it. But after my opponent pointed out thatinconsistency, I
didn't know what to say. Still, I believe there is a
reasonwhy definition B not only survived but also is used in
these authoritativeYaohan Chen
====
= > As I am generating
Mathematica images for inserting them into > TeX, problems>
occur, like> f(t) seen as fHtL on EPS image in GSView, >
FontFamily->Trebuchet MS (or> Helvetica) not taken into
account, ...> Can you direct me to any relevant material on
the internet, > book about ways> to tackle this problems?This
happens because the application you're inserting the graphic
in dosen'tunderstand where to find the 
particular font that
Mathematica uses to renderthe characters ( & ) (& more
besides ...).emmathfnt for this purpose.It's explained in
reasonable detail
athttp://support.wolfram.com/mathematica/graphics/export/
includefonts.htmlwhere you can find a link to the above
program. Dave.
====
======================================= Dr.
David Annetts EM Modelling Analyst Australia
David.Annetts@csiro.au
====
====================================

====
=====> I have the following data of the following form.>
{123-456-7899, John Smith,2223334444}> How can I convert this
to > {123,456,7899, John,Smith,222,333,4444}> and then to>
{1234567899johnsmith00000)> 1. The - are removed> 2. John
Smith is concerted to lower case> 3. The space between John
and Smith is removed (ie john and smith is > concatenated)>
4. 123454567899johnsmith is padded right so the the total
number of > characters is 24.> John> Did you have a look at
operations on strings in the book (chapter 2.7.2 in version
4)?StringJoin, StringTake, StringReplace should get you
going.eg (should do most of the
work):StringReplace[abcdabcdaabcabcd, {abc -> Y, d ->
XXX}]YXXXYXXXaYYXXXHope that helps.Marc-- Marc Heusserremove
the obvious CHEERS and MERCIAL... from the reply
addressReply-To: selwynh@earthlink.net
====
=John,Here's one
approach. I'm assuming that you're starting 
out with
strings.data1 = {123-456-7899, John Smith, 2223334444};{l1,
l2, l3} = Characters /@ data1;data2 = Flatten[{
StringJoin[DeleteCases[#1,-]]&/@ Partition[l1, 4],
StringJoin[#1]&@ l2[[Range@@ Prepend[Flatten[Position[l2,
]]-1, 1]]], StringJoin[#1]&@ l2[[Range@@
Append[Flatten[Position[l2, ]]+1, Length[l2]]]],
StringJoin[#1]&/@ {l3[[{1,2,3}]], l3[[{4,5,6}]],
l3[[{7,8,9,10}]]}}] {123, 456, 7899, John, Smith, 222, 333,
4444}data3 = StringJoin[data3 =
StringJoin[data2[[Range[1,5]]]], StringJoin[Table[0, {24 -
Length[Characters[data3]]}]]]
1234567899JohnSmith00000---Selwyn Hollis>I have the following
data of the following form.>>{123-456-7899, John
Smith,2223334444}>>How can I convert this to >{123,456,7899,
John,Smith,222,333,4444}>and then
to>>{1234567899johnsmith00000)>>1. The - are removed>2. John
Smith is concerted to lower case>3. The space between John
and Smith is removed (ie john and smith is >concatenated)>4.
123454567899johnsmith is padded right so the the total number
of >characters is 24.>John>>
====
=John,Here is a long journey to
your forms - I am assuming that we start with
{123-456-7899,John Smith,2223334444}and I show input forms to
distinguish strings. (fm1=MapAt[ StringReplace[ StringTake[
{2,-2}], ->]&, StringReplace[{123-456-7899,John
Smith,2223334444},{ -->,, ->,}],
{-1}])//InputForm{123,456,7899, John,Smith,
222,333,4444}(fm2=StringJoin@@ PadRight[
Characters[StringJoin@@StringReplace[ToLowerCase/@Drop[fm1,-1
],,->]],24,
0])//InputForm1234567899johnsmith00000--Allan----------------
-----Allan HayesMathematica Training and ConsultingLeicester
UKwww.haystack.demon.co.ukhay@haystack.demon.co.ukVoice: +44
(0)116 271 4198>> I have the following data of the following
form.>> {123-456-7899, John Smith,2223334444}>> How can I
convert this to> {123,456,7899, John,Smith,222,333,4444}> and
then to>> {1234567899johnsmith00000)>> 1. The - are removed>
2. John Smith is concerted to lower case> 3. The space
between John and Smith is removed (ie john and smith is>
concatenated)> 4. 123454567899johnsmith is padded right so
the the total number of> characters is 24.> John>
====
=>> I
have the following data of the following form.>>
{123-456-7899, John Smith,2223334444}>> How can I convert
this to> {123,456,7899, John,Smith,222,333,4444}> and then
to>> {1234567899johnsmith00000)>> 1. The - are removed> 2.
John Smith is concerted to lower case> 3. The space between
John and Smith is removed (ie john and smith is>
concatenated)> 4. 123454567899johnsmith is padded right so
the the total number of> characters is 24.> John>>G'day
John,robust.Some notes and function foobar[file_String]
follows;1. Import the file as Word(s), not String2. Replace
the hyphen first to avoid confusion with minus3. WRT l3 in
foobar, first step takes 222333444 and inserts a comma
andspace in the 4th and 7th positions. Second step takes the
remainingelements and makes a list. Then convert to string
and then to expression.The space after the comma just got
inserted as one character I think andyou can't do much with
that.4. WRT to l4, get rid of the last three entries in the
expression and addthe zeroes, convert to string and replace
the unwanted characters withwanted
characters.In[1]:=foobar[file_String] := Module[{l1, l2, l3,
l4}, l1 = ReadList[file, Word, RecordSeparators -> {,}]; l2 =
StringReplace[l1, {- -> ,}]; l3 = {Drop[l2, -1],
StringInsert[Last[l2], , , {4, 7}]} // ToString//ToExpression
// Flatten; l4 = StringReplace[{Drop[l3, -3], {0, 0, 0, 0, 0}}
// Flatten // ToString, {, -> , -> , J -> j, S -> s}]
]In[2]:=foobar[~/Desktop/data.txt]Out[2]={
1234567899johnsmith00000}YasReply-To:
kuska@informatik.uni-leipzig.de
====
=tryIn[]:=expr = d*Pi +
I*d*(Log[M] - 2*Log[(-I)*Sqrt[(M*r)/d] + Sqrt[M
-(M*r)/d]]);In[]:=expr1=expr /. a_.*M + b_*M :> M*(a + b) /.
Sqrt[M*a_] :> Sqrt[M]*Sqrt[a] /. a_.Log[b_] + c_.*Log[d_] :>
Log[b^a*d^c] /. Sqrt[M]*a_ + Sqrt[M]*b_ :> Sqrt[M]*(a + b) /.
Log[a_^b_] :> b*Log[a] /. Log[-I*a_ + b_] :> Log[-I*(a + I*b)]
/. Log[a_*b_Plus] :> Log[a] +
Log[b]//ExpandOut[]=(-2*I)*d*Log[Sqrt[r/d] + I*Sqrt[1 -
r/d]]In[]:=FullSimplify[expr1, Element[{r, d}, Reals] && r <
d] Out[]=2*d*ArcSin[Sqrt[1 - r/d]]ups... Jens> I'm trying to
simplify (M, r, d are positive reals with r<d)> d*Pi +
I*d*(Log[M] - 2*Log[(-I)*Sqrt[(M*r)/d] + Sqrt[M - (M*r)/d]])>
to> 2*d*ArcCos[Sqrt[r/d]]> I've tried applying>
FullSimplify[#,M>0 && 0<r<d]&> and> ExpToTrig> but
Mathematica won't get rid of the Log and the I and give me
the ArcCos.> Does anyone know which Mathematica functions to
apply to my expression to> give me the simplification I want?>

====
=According to the default complexity measure ArcTan[y/x]is
not simpler than ArcTan[x, y].In[1]:= LeafCount /@ {ArcTan[x,
y], ArcTan[y/x]}Out[1]= {3, 6}One can use ComplexityFunction
to specify that two-argumentArcTan should be replaced
whenever possible.In[2]:= f=1000 Count[#, ArcTan[_, _], {0,
Infinity}]+LeafCount[#]&;In[3]:= FullSimplify[ArcTan[x, y], {x
> 0, y < 0}, ComplexityFunction->f] yOut[3]= ArcTan[-] xWith
this ComplexityFunction we get a somewhat better resultfor
the original problem, though still not the desired
form.In[4]:= FullSimplify[ComplexExpand[ d*Pi + I*d*(Log[M] -
2*Log[(-I)*Sqrt[(M*r)/d] + Sqrt[M - (M*r)/d]]),
TargetFunctions -> {Re, Im}], M > 0 && 0 < r < d,
ComplexityFunction->f] 1Out[4]= d (Pi - 2
ArcTan[------------]) d Sqrt[-1 + -] rAdam StrzebonskiWolfram
Research> This will take you pretty close:>
FullSimplify[ComplexExpand[> d*Pi + I*d*(Log[M] -
2*Log[(-I)*Sqrt[(M*r)/d] +> Sqrt[M - (M*r)/d]]),
TargetFunctions -> {Re, Im}],> M > 0 && 0 < r < d]> d*(Pi +
2*ArcTan[Sqrt[M - (M*r)/d], -Sqrt[(M*r)/d]])> Still, there
appears to be something missing among the rules >
FullSimplify is applying.> It seems related to the following
behaviour, that certianly suggests a bug:>
FullSimplify[ArcTan[x, y], {x > 0, y < 0}]> ArcTan[x, y]>
FullSimplify[ArcTan[2*x, 2*y], {x > 0, y < 0}]> ArcTan[y/x]>>
I'm trying to simplify (M, r, d are positive reals with 
r<d)>>
d*Pi + I*d*(Log[M] - 2*Log[(-I)*Sqrt[(M*r)/d] + Sqrt[M -
(M*r)/d]])>> to>> 2*d*ArcCos[Sqrt[r/d]]>> I've tried
applying>> FullSimplify[#,M>0 && 0<r<d]&>> and>> ExpToTrig>>
but Mathematica won't get rid of the Log and the I and give
me the >> ArcCos.>> Does anyone know which Mathematica
functions to apply to my expression to>> give me the
simplification I want?>>>>>>> Andrzej Kozlowski> Yokohama,
Japan> http://www.mimuw.edu.pl/~akoz/>
http://platon.c.u-tokyo.ac.jp/andrzej/> 
====
=Dave we have, e =
d*Pi + I*d*(Log[M] - 2*Log[(-I)*Sqrt[(M*r)/d] + Sqrt[M -
(M*r)/d]]); e2 = ComplexExpand[e, TargetFunctions -> {Re,
Im}]; e3 = Simplify[e2, M > 0 && 0 < r < d] d*(Pi +
2*ArcTan[Sqrt[M - (M*r)/d], -Sqrt[(M*r)/d]]) e4 =
Simplify[e3, M > 0 && 0 < r < d, ComplexityFunction ->
(Count[#1, M, Infinity] & )] d*(Pi - 2*ArcTan[Sqrt[r]/Sqrt[d -
r]])Here I have to be interactive:a = ArcTan[Sqrt[r]/Sqrt[d -
r]] is in (0, Pi/2),.So b = Pi - 2a is in (0,Pi).Consequently
b/2 is ArcCos[Cos[b/2]].That is FullSimplify[Cos[(Pi -
2*ArcTan[Sqrt[r]/Sqrt[d - r]])/2], 0 < r < d]
1/Sqrt[d/r]--Allan---------------------Allan HayesMathematica
Training and ConsultingLeicester
UKwww.haystack.demon.co.ukhay@haystack.demon.co.ukVoice: +44
(0)116 271 4198>> I'm trying to simplify (M, r, d are
positive reals with r<d)> d*Pi + I*d*(Log[M] -
2*Log[(-I)*Sqrt[(M*r)/d] + Sqrt[M - (M*r)/d]])>> to>>
2*d*ArcCos[Sqrt[r/d]]>> I've tried applying>>
FullSimplify[#,M>0 && 0<r<d]&>> and>> ExpToTrig>> but
Mathematica won't get rid of the Log and the I and give me
the ArcCos.>> Does anyone know which Mathematica functions to
apply to my expression to> give me the simplification I
want?>>
====
=>> I'm trying to simplify (M, r, d are positive
reals with r<d)> d*Pi + I*d*(Log[M] -
2*Log[(-I)*Sqrt[(M*r)/d] + Sqrt[M - (M*r)/d]])>> to>>
2*d*ArcCos[Sqrt[r/d]]>> I've tried applying>>
FullSimplify[#,M>0 && 0<r<d]&>> and>> ExpToTrig>> but
Mathematica won't get rid of the Log and the I and give me
the > ArcCos.>> Does anyone know which Mathematica functions
to apply to my expression > to> give me the simplification I
want?>>I can't get the form you want
butFullSimplify[ComplexExpand[d*Pi + I*d*(Log[M] -
2*Log[(-I)*Sqrt[(M*r)/d] + Sqrt[M -
(M*r)/d]])],{M>0&&r>0&&d>0&&r<d}]will get you tod*(Pi+2
Arg[-I*Sqrt[M*r/d]+Sqrt[M-M*r/d]])SseziReply-To:
rolf@mertig.com
====
=one possibility is:CloseAniSequenceTest :=
(SelectionMove[EvaluationNotebook[], After, EvaluationCell];
SelectionMove[EvaluationNotebook[], Next, Cell];
SelectionMove[ EvaluationNotebook[], Next, GeneratedCell];
FrontEndExecute[ {FrontEndToken[EvaluationNotebook[],
OpenCloseGroup]}])Needs[Graphics`Animation`];
Graphics`Animation`Animate[Plot[Sin[x*t], {x, -Pi, Pi}], {t,
1, 3, 1}]; CloseAniSequenceTestRolf Mertig--Mertig
Consulting, Berlin, GermanySpecialized Software &
Traininghttp://www.mertig.com
====
=> I want to evaluate the
integral:>> f[u_] = Integrate[E^x/Cos[x], {x, 0, u},>
Assumptions -> {x [Element] Reals, u < Pi/2}]> (1 - I)*E^((1
+ I)*u)*Hypergeometric2F1[> 1/2 - I/2, 1, 3/2 - I/2,
-E^(2*I*u)] +> (1/2)*((-I)*PolyGamma[0, 1/4 - I/4] +>
I*PolyGamma[0, 3/4 - I/4])>> so far so good. This integral
should be real-valued, at least in the> Range 0<=u<Pi/2.Yes,
and Mathematica's result is real-valued.> And that is my
problem. How can I get a real result> with no imaginary I's.
Neither FunctionExpand or FullSimplify did it.If you mean
that you want the result expressed without the imaginary
Iappearing anywhere in it, please realize that such might be
impossible.Whether it's impossible in this particular case 
or
not, I don't know.Think, for example, of expressing the real
roots of a cubic polynomialin terms of nothing more
sophisticated than radicals. In the so-called_casus
irreducibilis_, it is well known that, even though the roots
arereal, they must, in general, be expressed using the
imaginary I. Yourintegral result might be similar, requiring
complex numbers for itsexpression in Mathematica, despite the
fact that it is real.David-- --------------------
http://NewsReader.Com/ --------------------Usenet Newsgroup
Service New Rate! $9.95/Month 50GB
====
=> I want to evaluate
the integral:> f[u_] = Integrate[E^x/Cos[x], {x, 0, u},>
Assumptions -> {x [Element] Reals, u < Pi/2}]> (1 - I)*E^((1
+ I)*u)*Hypergeometric2F1[> 1/2 - I/2, 1, 3/2 - I/2,
-E^(2*I*u)] +> (1/2)*((-I)*PolyGamma[0, 1/4 - I/4] +>
I*PolyGamma[0, 3/4 - I/4])> so far so good. This integral
should be real-valued, at least in the> Range 0<=u<Pi/2. And
that is my problem. How can I get a real result> with no
imaginary I's. Neither FunctionExpand or FullSimplify did
it.As far as I can see the imaginary parts of the numerical
evaluation of the above expression are numerical artifacts.
They are around <10^(-15). So you should use Chop[f[u]] to
suppress these numerical uncertainties.-- Hendrik van Hees
Fakult.8at f.9fr Physik http://theory.gsi.de/~vanhees/
D-33615 Bielefeld