A53
===
Is there a reason for this strange behavior of TrigExpand?All
of these expressions simplify both the numerator and
denominator.(Cos[t]^2 + Sin[t]^2 is replaced by 1.)expr = (1
+ Cos[t]^2 + Sin[t]^2)/(2 + Cos[t]^2 +
Sin[t]^2);TrigExpand[expr]2/3expr = (1 + f[t] + Cos[t]^2 +
Sin[t]^2)/(2 + f[t] + Cos[t]^2 + Sin[t]^2);TrigExpand[expr](2
+ f[t])/(3 + f[t])expr = (1 + Cos[t] + Cos[t]^2 + Sin[t]^2)/(2
+ f[t] + Cos[t]^2 + Sin[t]^2);TrigExpand[expr](2 + Cos[t])/(3
+ f[t])But the following leaves the denominator
untouched.expr = (1 + Cos[t] + Cos[t]^2 + Sin[t]^2)/(2 +
Cos[t] + Cos[t]^2 + Sin[t]^2);TrigExpand[expr](2 + Cos[t])/(2
+ Cos[t] + Cos[t]^2 + Sin[t]^2)On the other hand, Simplify,
which uses the trig identities works.expr = (1 + Cos[t] +
Cos[t]^2 + Sin[t]^2)/(2 + Cos[t] + Cos[t]^2 +
Sin[t]^2);Simplify[expr](2 + Cos[t])/(3 + Cos[t])David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/
====
I'm
also using 4.2.0, and on my machine the Limit package gives 0
for both, like yours, but without the package loaded, it
leaves both forms unevaluated (unlike your experience).
Weirder and weirder.BobbyOn Sun, 26 Jan 2003 18:44:27 -0500
(EST), David W. Cantrell > I just noticed something strange,
closely related to the original > question> in this thread,
which I can't explain. Using version 4.2.0 for Windows:
Limit[(Exp[-x]-Exp[-2x])/(Exp[-x]+Exp[-2x]), x-> Infinity]
does not give an answer (although the built-in Limit function
_should_> of course be able to do so) but, merely using an
alternative notation,
Limit[(E^(-x)-E^(-2x))/(E^(-x)+E^(-2x)), x-> Infinity] yields,
incorrectly, 0 . [The correct answer is 1 .] The reason I
think of these notations as alternatives is that both> Exp[x]
and E^x have FullForm of Power[E, x]. So what's going on?
Why
does the first not give an answer, while the> second gives a
wrong answer? BTW, using the Standard Add-on Package
Calculus`Limit`, _both_ give the> incorrect answer 0 . PLEASE
do not respond with workarounds. I know several already,
the> easiest of which is to just do the problem in my head!
David Cantrell-- majort@cox-internet.comBobby R.
Treat
====
Selwyn,things aren't, what they appear to be:In[2]:=
Through[{InputForm, Floor}[#]] & /@ {3.0000000000000000,
3.00000000000000000}Out[2]= {{3., 3},
{2.999999999999999999999999999991459`17.6021, 2}}You're
right, of course, perhaps, except for
wow.--Hartmut>-----Original Message----->Sent: Saturday,
January 25, 2003 7:27 AM>To: mathgroup@smc.vnet.net>Wow. But
apparently it has nothing to do with Log.
Look:Floor[3.0000000000000000] 3Floor[3.00000000000000000]
2--->Selwyn Hollis> With Mathematica 4.1 on Windows98:>
N[Log[8]/Log[2]]> 3.> Floor[N[Log[8]/Log[2]]]> 2>
Beware!>>Reply-To: kuska@informatik.uni-leipzig.de
====
Esc pd
Esc Ctrl_ xEsc is the Escape key, Ctrl the Ctrl-Key and Ctrl_
meanthe bot keys must be pressed at the same time. Jens How do
you enter the partial derivative shown at: www.previze.com/
partialderivative.gif> Reply-To:
kuska@informatik.uni-leipzig.de
====
that is called
webMathematica -- have a look at the WRI web-site. Jens
What's the best/simplest way to encode or format a
Mathematica notebook> to make it available on a web site? (as
a downloadable source file> only, sufficiently
brief that file
compression is not needed, nothing> live or intended for
online execution, no need for Output cells or> graphics
outputs in the online file, but intended to be dowbnloaded
and> executed by users on multiple platforms, with header and
text cells kept> distinct from Input cells) --> 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)Reply-To:
kuska@informatik.uni-leipzig.de
====
Solve[Exp[-(x -
m1)/(2s1)]/Sqrt[s1] == Exp[-(x - m2)/(2s2)]/Sqrt[s2], x]??
Jens Can anyone please tell me how to find the intersection of
two gaussians?> Is there any standard method to do that?>
Vaidyanathan. --> Vaidyanathan Ramadurai> Graduate Student>
http://www4.ncsu.edu/~vramadu
====
Selwyn, Diana,You might also
(or instead of showing the MagnificationPopUp) like to set
WindowSize width -> Fit height -> FitSo that the window size
adjusts to fit the contents; otherwise if
themagnification is
made greater than 1 we need to add ResizeArea to
theWindowFrameElements so as to be able to do this
manually.Allan---------------------Allan HayesMathematica
Training and ConsultingLeicester
UKhay@haystack.demon.co.ukVoice: +44 (0)116 271 4198> Diana,
A major improvement to the palette is to set these options:
WindowElements->MagnificationPopUp>
WindowFrameElements->{CloseBox, ZoomBox} You can do this
with a text editor or with the Options Inspector. If> you do
it with a text editor, delete the cache ID line (as
instructed> in the comments at the top). To use the Options
Inspector, it's easiest if you close all notebooks> and
other
palettes first. Open the Options Inspector, and then click on>
the title bar of the palette. The title bar of the Options
Inspector> should then read Options for BasicInput.nb. Now
select Show Option> Values for notebook. Then under Notebook
Options > File Options, set Editable -> True. Next,> under
Notebook Options > Window Properties, choose the desired
values> for WindowElements and WindowFrameElements. Finally,
go back and to Notebook Options > File Options, and set>
Editable -> False. Et voila! ----> Selwyn Hollis> Folks,> I
have used Mathematica for about two months, and it appears
that the> BasicInput.nb palette had shrunk, in that the walls
of the window have> come> in. I can't, therefore, access all
of the icons that I want.> Is there a way to fix this, other
than re-installing the application?> Diana> -->
====
=================================================> God
made the integers, all else is the work of man.> L.
Kronecker, Jahresber. DMV 2, S. 19.>Reply-To:
kuska@informatik.uni-leipzig.de
====
what
manyTable[f[i,4],{i,1,4}]produce ? Jens I have a function
f[a_,b_] defined some way. i want to fix one
argument,> and
then generate an array of values where the other argument
varies. How> can I do this? For example, I want: {f[1, 4],
f[2, 4], f[3, 4], f[4, 4]} Is there an easy way to do this?>
====
>-----Original Message----->Sent: Sunday, January 26,
2003 11:23 AM>To: mathgroup@smc.vnet.net> I have a function
f[a_,b_] defined some way. i want to fix
one>argument,> and
then generate an array of values where the other >argument
varies.>How> can I do this?> For example, I want:> {f[1, 4],
f[2, 4], f[3, 4], f[4, 4]}> Is there an easy way to do this?>
Zachary,> Table[f[x,4],{x,1,4}]> {f[1, 4], f[2, 4], f[3, 4],
f[4, 4]}> Or, for irregular values or a known list of
values:> f[x,#]&/@{ 1, 2.3, 8,-3+I, Pi,a}> {f[x, 1], f[x,
2.3], f[x, 8], f[x, -3 + I], f[x, Pi],> f[x, a]}Good lord. I
didn't expect this many responses :) Anyway, >the Table
method>seems the simplest. Now I extend on this a little bit.
Is >there any way I>can apply FullSimplify to the result
before it gets inserted >into the array?>Zachary,to molest
you further...In[5]:= Thread[f[{1, 2, 3.5, 4}, 4]]Out[5]=
{f[1, 4], f[2, 4], f[3.5, 4], f[4, 4]}In[7]:= ArcTan[{1, 2,
3.5, 4}, 4]Out[7]= {ArcTan[4], ArcTan[2],
0.8519663271732721`, Pi/4}...as this didn't show up among
the
responses. The second, utmost simple,form applies to function
having the Listable attribute.As to your last question, it's
not quite clear to me, what do you meanbefore it gets
inserted into the array. If you do that, you risk to
missall simplifications with the arguments inserted.
Compare:In[17]:= FullSimplify[Thread[Log[#1]/Log[#2] &[{1, 2,
3.5, 4}, 4]]]Out[17]= {0, 1/2, 0.9036774610288021`, 1}This
simplifies the result: as a list is simple enough, all
elements aresimplified.In[16]:=
Thread[Evaluate[FullSimplify[Log[#1]/Log[#2]]] &[{1, 2, 3.5,
4},4]]Out[16]= {0, Log[2]/Log[4], 0.9036774610288021`, 1}This
simplifies the function expression before it is applied, such
not allsimplifications of interest can be done (in this case).
However it might bethat this last step ist not needed nor
desired, then this might be moreperformant, esp. if the list
is quite long.--HartmutIn[11]:=
Thread[FullSimplify[Log[#1]/Log[#2] &[{1, 2, 3.5, 4},
4]]]Out[11]= {0, 1/2, 0.9036774610288021`, 1}Reply-To:
kuska@informatik.uni-leipzig.de
====
With[{x=10}, First /@
FactorInteger[10]]may help you. Jens All, I know that I can
generate the divisors of any integer with the Divisors>
command. I would like to start with x = 10, for example,
and generate the> divisors of x, and then determine the
sum of the divisors. I would then> like to increment x, up
to 100, for example. > Diana -->
====
=================================================> God
made the integers, all else is the work of man.> L.
Kronecker, Jahresber. DMV 2, S. 19.Reply-To:
kuska@informatik.uni-leipzig.de
====
AFIK there is no
posibility to connect to a kernel thatis already listen to a
parent link. The kernel can connectto a second frontend like
program but a second frontend can notconnect to a kernel that
is already running and know nothingabout the new link. The
only way to tell the kernelthat i has to listen to a parent
link is the command linebut if the kernel is already running
ou can't send a newcommand line to it. Jens > How does one
determine whether an instance of the Mathematica Kernal is>
running via Mathlink , or even Visual Basic. I only want to
keep one> connect to this instance not start up a new
instance. All I can think of is> keep the MLINK variable as
a global and check that it does not equal zero. that will
connect to this. > --> Daniel HeneghanReply-To:
kuska@informatik.uni-leipzig.de
====
Mathematica can't do
Gouraud shading because it assign the colorsper polygon and
not per vertex. You can try to make a huge number of
PlotPoints to obtain so manypolygons that the difference is
invisible or you can use
MathGL3dhttp://phong.informatik.uni-leipzig.de/~kuska/
mathgl3dv3/id3.htm Jens i'm looking for a function to plot a
cone with a interpolated gouraud> shading in Mathematica.> I
tried with another system, but the> result is awful, because
the circle of the cone is squared. thanks.Reply-To:
kuska@informatik.uni-leipzig.de
====
without the t I want to
plot 3D Data (Curves) and export it to vrml. IÇve found
this
nice> MathGL3D Tool, but canÇt manage to change the scales
of
the axes, as one can> do with BoxRatios for ordinary
Mathematica Graphics3D. for example:
Show[Graphics3D[{Line[{{0, 0, 0}, {1, 1, 0.1}}]}, BoxRatios
-> {1, 1, 1}]] shows a scaled Plot, but
MVShow3D[Graphics3D[{Line[{{0, 0, 0}, {1, 1, 0.1}}]},
BoxRatios -> {1, 1,> 1}]] doesnÇt.> Is there an easy way
to
create scaled Plots in MathGL3D?You will not happy with it,
butMVShow3D[Graphics3D[{Line[{{0, 0, 0}, {1, 1, 0.1}}]},
BoxRatios -> {.1, .1, 1}]]should do it. How about exporting
axes and text to vrml? It is not possible with MathGL3d. >
Are there easier ways than> tool?Why not use LiveGraphics3D
http://wwwvis.informatik.uni-stuttgart.de/~kraus/
LiveGraphics3D/index.htmlby Martin Kraus orJavaView
http://www-sfb288.math.tu-berlin.de/vgp/javaview/on the
www-page. It has the advantage that the visitor don't needa
VRML browser.
====
In version 4.2.1.1 for Mac OS X both return
the same result: the original expression rewritten in two
dimensional form> I just noticed something strange, closely
related to the original > question> in this thread, which I
can't explain. Using version 4.2.0 for Windows:
Limit[(Exp[-x]-Exp[-2x])/(Exp[-x]+Exp[-2x]), x-> Infinity]
does not give an answer (although the built-in Limit function
_should_> of course be able to do so) but, merely using an
alternative notation,
Limit[(E^(-x)-E^(-2x))/(E^(-x)+E^(-2x)), x-> Infinity] yields,
incorrectly, 0 . [The correct answer is 1 .] The reason I
think of these notations as alternatives is that both> Exp[x]
and E^x have FullForm of Power[E, x]. So what's going on?
Why
does the first not give an answer, while the> second gives a
wrong answer? BTW, using the Standard Add-on Package
Calculus`Limit`, _both_ give the> incorrect answer 0 . PLEASE
do not respond with workarounds. I know several already,
the> easiest of which is to just do the problem in my head!
====
Mathematica is NOT giving wrong answers in this case.>
It is assuming (non-zero) real parameters and giving the
right> answer in that case. Whether Mathematica is or is not
giving a wrong answer in the case> result = Integrate[
Abs[Sin[k x]]^2, {x,0,1}]; N[ result /. k->I+1 ]> depends
upon whether it is or is not appropriate for Mathematica to>
make a default assumption that k is a nonzero real. IMO, that
could be> debated.It would be extremely silly to defend this
as a default assumption,since in all other cases
Mathematica goes through a lot of troubleto single out cases
like Im[k] == 0 in its answers. > But surely, as I noted
previously in this thread, Mathematica -- at> least version
4.2 for Windows -- does give a wrong answer for> result =
Integrate[ Abs[Sin[k x]]^2, {x,0,1}, Assumptions- Element[k,
Complexes]; N[ result /. k->I+1 ]. I do not see how the>
incorrectness of this can be debated (other than to say that>
Mathematica should be allowed to ignore an _explicitly
stated_> assumption!)Coming to think of it, Mathematica could
of course also ignore _any_explicitly stated fact in its input
and give the default result 42to all questions! But version
4.2 is probably still lacking one orderof magnitude in wisdom
to do this. :-)-- Jos < Jos.Bergervoet@philips .n_o_spa_m. com
>
====
I realize that we're dealing with the vagaries of
internal arithmetic, but it is highly disquieting that
3.000... (with any number of zeros) would ever be anything
but the binary ßoating-point number .11 * 2^2 !Selwyn>
Selwyn, things aren't, what they appear to be: In[2]:=
Through[{InputForm, Floor}[#]] & /@> {3.0000000000000000,
3.00000000000000000}> Out[2]= {{3., 3},>
{2.999999999999999999999999999991459`17.6021, 2}} You're
right, of course, perhaps, except for wow. --> -----Original
Message-----> Sent: Saturday, January 25, 2003 7:27 AM> To:
mathgroup@smc.vnet.net> Wow. But apparently it has nothing to
do with Log. Look:> Floor[3.0000000000000000]> 3>
Floor[3.00000000000000000]> 2> ---> Selwyn Hollis> With
Mathematica 4.1 on Windows98:> N[Log[8]/Log[2]]> 3.>
Floor[N[Log[8]/Log[2]]]> 2> Beware!>
====
Diana: Look up
DivisorSigma. Table[DivisorSigma[1,x],{x,10,100}] willgive
you what you want directly, I believe. Best, HarveyHarvey P.
DaleUniversity Professor of Philanthropy and the LawDirector,
National Center on Philanthropy and the LawNew York University
School of LawRoom 206A110 West 3rd StreetNew York, N.Y.
10012-1074-----Original Message-------God made the integers,
all else is the work of man.L. Kronecker, Jahresber. DMV 2,
S. 19.
====
Im using Mathematica 4.2 on my MacOSX notebook but
still have 4.0 on mydesktop. I have encountered a strange
thing which is causing meproblems. I have defined the
wavefunction in momentum space for a shoin 4.0 when i
evaluate for the ground state I get the answer I wouldexpect.
In 4.2 the answer is the same but I get an Integer 1 appearing
whichmakes it very difficult to manipulate the result to the
form I want.I have attached the Notebook content below. This
is the 4.0 version butwith the 4.2 result copied across.Any
comments/help?DonNotebook[{Cell[BoxData[
([CurlyPhi]_[Gamma]_[ p_] :=
(([ImaginaryI]^(-[Gamma]))
(((@[Pi]2^[Gamma]
([Gamma]!))/a)))^(-(1/2))
([ExponentialE]^(((-p^2) a^2)/2))
HermiteH[[Gamma], a p])], Input],Cell[BoxData[
(In 4.0)], Input],Cell[CellGroupData[{Cell[BoxData[
([CurlyPhi]_0[p])], Input],Cell[BoxData[
(TraditionalForm`[ExponentialE]^((-(1/2))
a^2 p^2)/(@(1/a)
@[Pi]%4))], Output]}, Open ]],Cell[BoxData[
([ExponentialE]^((-(1/2)) a^2
p^2)/(@(1/a) @[Pi]%4))],
Text],Cell[BoxData[ (In 4.2)],
Input],Cell[CellGroupData[{Cell[BoxData[
(TraditionalForm`[CurlyPhi]_0[p])],
Input],Cell[BoxData[
(TraditionalForm`[ExponentialE]^((-(1/2))
a^2p^2)/@1[@[Pi]/a])], Output]},
Open ]]},ScreenRectangle->{{0, 1056}, {0,
772}},WindowSize->{520, 650},WindowMargins->{{28, Automatic},
{-30,
Automatic}},MacintoshSystemPageSetup-><00<0001804P000000
]P2:?oQon82n@960dL5:0?l0080001804P000000]P2:0010000I00000400`
<300000Gd000400@
0000000000000004P801T1T000000000000000000000000000000000000000
0000>]
====
If I enter Sum[p^i, {i, 0, Infinity}] Mathematica
says, it is 1/(1-p), but doesn't say something about the
domain for p: 1/(1-p) is only valid for-1=1 and p<=-1?PS: you can
find a
nice animation for the geometric series at
http://www.matheprisma.de/Module/Craps/summe.htm-- Frank Bu¤,
fb@frank-buss.dehttp://www.frank-buss.de,
http://www.it4-systems.de
====
hi,maybe it's a foolish
question, but i'm new to mathematica:i want to plot (using
Plot3D of course) at least 2 3D-functions into thesame
diagram. how to i have to handle this?thanx in
regard.lhuv
====
I'm relatively new using mathematica 4.2 .I
have a set of integro-differential equations to solve.Does
anyone have an idea how to proceed?
====
I'm student and using
mathematica 4.0.At the moment I have to write a report for my
university. I want to useLaTeX to write this document. Since
I do a lot of calculations withmathematica I use the
TexForm[...] (//HoldForm) command to get a latexoutput. I
just have to copy/paste the output; it works quite well.But
then I wanted to include some plots from mathematica in my
latexdocument. If I apply it to Plot[...]: e.g.:
TeXForm[Plot[Sin[x], {x, -Pi,Pi}]] I get a strange
outputs.Is it possible to get a LaTeX package, which compiles
these instructions? Oris this output just nonsense? Are there
other possibilities than using copyas bitmap to get my plot
into the latex document?Perhaps anyone had similar problems or
knows a useful link.Jochen
====
Have a look at the
Utilities`Notation` pckage which does what you want todo. You
can find it in the Help Browser under Addons -> Extras ->
Utilities.--Steve LuttrellWest Malvern, UK I would like to
define a new operator called # on mathematica with> some
properties and make theoritical calculations with it. For
example> I'd like to calculate (a # b)^n then. Is there any
means to do that ?
====
Unfortunately, # (Slot in
Mathematica, q.v.) is already taken. You mustchoose another
symbol which doesn't belong to the list of reserved
names,and
then you can do what you say. TypeInformation[#] to see the
meaning of #.Tomas GarzaMexico City----- Original Message
-----
====
*not* with # (or Slot[1])But the Notation package
may help youto define a other symbol as operator. Jens > I
would like to define a new operator called # on mathematica
with> some properties and make theoritical calculations with
it. For example> I'd like to calculate (a # b)^n then. Is
there any means to do that ?
====
Nathan,You can define your own
operators but not using #. # is a reserved symbol
inMathematica and stands for a slot in a pure
function.However, you have many other choices. For a fanciful
example...CirclePlus[a_, b_] := Plus @@ Flatten[Divisors /@
{a, b}](10[CirclePlus]12)^397336CirclePlus can also be
entered as esc c + esc.Or you can make up your own
name.zawak[a_, b_] := Plus @@ Flatten[Divisors /@ {a,
b}](10~zawak~12)^397336where ~ ~ turns zawak into an infix
operator.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/Sender:
steve@smc.vnet.netApproved: Steven M. Christensen
, Moderator
====
Friedrich Laher schrieb:>
Zachary Turner schrieb:Apparently Mathematica randomly
returns roots a root of a complex number.>Is there a way I
can write my own function that will return a set
consisting>of all n roots of a given complex number. For
example, Root[z,n] = {a set>consisting of n elements} >
LstOfRoots[z_, n_] :=> Module[{a = Abs[z]^(1/n), f}, f[i_] :=
a Root[#^n - 1, i]; Map[f, > Range[n]]]I am ashamed to admit
that the above one is only valid for real zHere is a more
general solution - assuming an natural n.The exponential form
is best for calculating roots.LstOfRoots[z_, n_] := Block[{az
= Arg[z], r = Abs[z]^(1/n), j, ang = [ImaginaryI](az +
2[Pi] j)/n}, Table[r [ExponentialE]^ang, {j, 0, n -
1}]}]I1st tried the names arg and abs ang ( angle ) and r (
radius )but that is rejected by mathematica.
====
Being at a
stage were all editing is finished, Springer's
schedule is to
publish the Programmingvolume and the Graphics volume within
the next few months andwithin three months after their
publication the Numericsvolume and the Symbolics volume.My
editor, Wayne Yuhasz might beable
to give you a more concrete publication date.Michael
Trott
====
How would you let mathematica be more verbose aka
Show work for simple series and sigma notation.Mr.
Lingwood
====
If I could have mathematica on a palmtop I would
be a truly fulfilled human being.
[setting: a cocktail party] Girl: ...so in the limit you'd
find that... Me: well let's plot that [whips out
palmtop with
mathematica] Girl: [swoons]It's clearly possible in
principle
since modern palmtops are morepowerful than desktop machines
that ran mathematica just fine severalyears ago.But has anyone
made this work in practice?Daniel-- Daniel Reeves --
http://ai.eecs.umich.edu/people/dreeves/ Sowmya: Is this guy
a mathematician?Terence: Worse, an economist. At least
mathematicians are honest about their disdain for the real
world.
====
One possibility:In[1]:=g1 = Plot[x^2, {x, 0,
5}];In[2]:=< thank you!Reply-To:
kuska@informatik.uni-leipzig.de
====
Needs[Graphics`FilledPlot
`]Block[{$DisplayFunction = Identity}, g1 = FilledPlot[x^2,
{x, 2, 4}]; g2 = Plot[x^2, {x, 0, 5}]; ]Show[g2, g1] Jens
i'm
relatively new using mathematica 4.2. the other day, i came
across what> seemed to be a simple problem, but i couldn't
figure out how to do it. say i was plotting y=x^2, and i
wanted to shade the region under the graph> only between x=2
to x=4 over a range of {x, 0, 5}, is there a way to do> this?
i tried FilledPlot, but it filled the entire area under the
graph. thank you!
====
Occasionally Mathematica 4.2 under MacOS
9.2.2 gives me the following warning:Unable to open
fileFiles:SystemFiles:FrontEnd:TextResources:Macintosh:
UnicodeLanguageFontMapping.tr[snip]However, Find shows that
the file is there, in the Mathematica 4.2 Files folder. There
are actually 5 files by the same name, all
underTextResources.The warning is harmless but annoying.
Never had it under 4.1, although exactly the same files are
there. (Kept 4.1 because the 4.2 front end is ßakier,
crashing often and leaving a running kernel)Question: how can
I turn off this warning? Is it related to the factthat both
4.1 and 4.2 are installed?
====
Todd,I have a further question
on a different type of table.I have written a program to
generate a multiplication table forZ_5[i], or the
multiplication table of the additive integer group offive
letters for complex numbers.The program is as
follows:Z5[x_,y_,z_,w_] = (x + yi)(z +
wi)Mod[Array[Z5,{5,5,5,5},{0,0,0,0}]//TableForm(The i in
the command above is actually the symbol for the
imaginarynumber i.)When I go to try to export the file to a
csv file with the
followingcommand,Export[C:Z5-i-notAField.csv,Mod[Array[
Z5,{5,5,5,5},{0,0,0,0}],I get a bunch of formating characters
in the CSV file, and can't useit.Can you tell me
how I should
change the program or restate the exportcommand?Diana> Diana,
Did you do this? << DiscreteMath`Combinatorica`> A5 =
AlternatingGroup[5]> mult[x_?PermutationQ, y_?PermutationQ]
:= Permute[x, y]> MultiplicationTable[A5, mult] // TableForm>
Export[C:tableDELETETHISLATER.csv,
MultiplicationTable[A5, mult], CSV] It works fine for me. I
hope you didn't type all that other stuff with the> quotes
in
by hand, as it appears from your posting. Do you know that
dummy = MultiplicationTable[A5, mult] // TableForm>
Export[C:tableDELETETHISLATER.csv, dummy], CSV] won't
work, but (dummy = MultiplicationTable[A5, mult] )//
TableForm> Export[C:tableDELETETHISLATER.csv, dummy],
CSV] will? Todd Rose Steve,> Well, it took me a few days to
catch on to how to extrapolate list data> from> the table
output. I was finally able to take your program, and apply it
to> the Cayley Table for A_5.> To keep the size to a minimum
for the newsgroup post, I have pasted my> notebook file to
export the Cayley Table for A_4.> The notebook command looks
like:> Export[C:cayley4table.csv, {{1, 2, 3, 4,
5, 6,> 7, 8, 9, 10, 11, 12}, {2, 3, 1,
7, 9, 8, 10,> 12, 11, 4, 5, 6}, {3,
1, 2, 10, 11, 12> , 4, 6, 5, 7, 9, 8},
{4, 6, 5, 1, 3, 2, 11,> 10> , 12, 8, 7,
9}, {5, 4, 6, 8, 7, 9, > 1, 2, 3, 11,
12, 10}, {6, 5, 4, 11, 12, 10, > 8, 9,
7, 1, 3, 2}, {7, 8, 9, 2, 1, 3, > 5,
4, 6, 12, 10, 11}, {8, 9, 7, 5, 6, 4,
12,> 11> , 10, 2, 1, 3}, {9, 7, 8, 12,
10, 11> , 2, 3, 1, 5, 6, 4}, {10, 12,
11, 3, 2, 1,> 9,> 7, 8, 6, 4, 5}, {11,
10, 12, 6, 4, 5> , 3, 1, 2, 9, 8, 7},
{12, 11, 10, 9, 8, 7, > 6, 5, 4, 3, 2,
1}}, CSV]> Diana> Here is a little example of how to
export in a format that Excel can> read:>
Export[C:table.csv, {{1, 2}, {3, 4}}, CSV]> Steve
Luttrell> Math friends,> If I program Mathematica to
calculate a Cayley Table for A_5, for> example,> and it
displays on the screen in the notebook, I have not been able
to> figure out how to paste the values into Excel without all
the extra> formatting, such as quote marks. Has someone worked
this out?> For example, the Cayley Table for A_5 copies as
follows into Excel...> !(*> TagBox[GridBox[{> {1,
2, 3, 4, 5, 6, 7, 8, 9, 10, 11,> 12,>
13,> 14, 15, 16, 17, 18, 19, 20, 21, 22,
23,> 24,> 25, 26, 27, 28, 29, 30, 31, 32,
33, 34,> 35,> 36, 37, 38, 39, 40, 41, 42,
43, 44, 45,> 46,> 47, 48, 49, 50, 51,
52, 53, 54, 55, 56,> 57,> 58, 59, 60},>
{2, 3, 1, 7, 9, 8, 10, 12, 11, 4, 5,>
6,> 14,> 15, 13, 19, 21, 20, 22, 24, 23,
16, 17,> 18,> 37, 39, 38, 40, 42, 41, 46,
47, 48, 43,> 44,> 45, 49, 51, 50, 52,
54, 53, 58, 59, 60,> 55,> 56, 57, 25,
26, 27, 28, 29, 30, 31, 32,> 33,> 34,
35, 36},> {3, 1, 2, 10, 11, 12, 4, 6,
5, 7, 9,> 8,> 15,> 13, 14, 22, 23, 24,
16, 18, 17, 19, 21,> 20,> 49, 50, 51,
52, 53, 54, 55, 56, 57, 58,> 59,> 60,
25, 27, 26, 28, 30, 29, 34, 35, 36,>
31,> 32, 33, 37, 39, 38, 40, 42, 41, 46,
47,> 48,> 43, 44, 45},> {4, 6, 5, 1, 3,
2, 11, 10, 12, 8, 7,> 9,> 25,> 27, 26,
28, 30, 29, 34, 35, 36, 31, 32,> 33,>
13, 15, 14, 16, 18, 17, 22, 23, 24, 19,>
20,> 21, 50, 49, 51, 55, 56, 57, 52, 53,
54,> 58,> 60, 59, 38, 37, 39, 43, 44,
45, 40, 41,> 42,> 46, 48, 47},> {5, 4,
6, 8, 7, 9, 1, 2, 3, 11, 12,> 10,>
26,> 25, 27, 31, 32, 33, 28, 29, 30, 34,
36,> 35,> 38, 37, 39, 43, 44, 45, 40, 41,
42, 46,> 48,> 47, 13, 14, 15, 16, 17, 18,
19, 20, 21,> 22,> 23, 24, 50, 51, 49,
55, 57, 56, 58, 60,> 59,> 52, 53, 54},>
and so on...>
====
If I enter Sum[p^i, {i, 0, Infinity}]
Mathematica says, it is 1/(1-p), but doesn't say something
about the domain for p: 1/(1-p) is only valid for-1=1 and p<=-1?PS: you can
find
a nice animation for the geometric series at
====
====> Jos R
Bergervoet result =
Integrate[ Abs[Sin[k x]]^2, {x,0,1}]> N[ result /. k->I+1 ]>
(* Analytical approach gives 0.261044 + 0.616283 I, WRONG !!!
*)> k=I+1; NIntegrate[ Abs[Sin[k x]^2], {x,0,1}]> (* Numerical
check gives 0.679391 *)> ...> What should I do to circumvent
such errors?> One thing that works in Mathematica (as well as
in the other CAS) is to> Integrate[ Abs[Sin[(a+b*I) x]]^2,
{x,0,1}].> This gives (a*Sinh[2*b] - b*Sin[2*a]) / (4*a*b),>
which agrees with your result below. But again it is wrong!
It only is correct if a and b happen> to be real quantities,
which is nowhere stated!Well, let me state it now here:I was
merely trying to provide a workaround which would give you
acorrect result in the event that your coefficient k was
complex. (I wasnot trying in any way to exonerate
Mathematica!) In that event, k may bewritten, _without loss
of generality_, as a+b*I, where a and b are bothREAL.> So the
main question still is: Why is Mathematica making> these very
silly errors? One could expect it from an early> version of a
product, but that is not what Mathematica 4.x> is!Yes, the
main question remains.David
====
Clean those wiper blades :)
For large positive x, the dominant term isE^(-x), hence 3.
I'm not sure why Mathematica cannot find the
limit directlyof
your expression, but if you Simplify first, you do indeed get
3.Tom Burton > Hey all - I'm taking a basic calculus course
that uses Mathematica.> We have been studying limits and I
have been using the Limit function> to 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 then> determine the limit. My initial reason had me
select -E^(-3x) in the> numerator and E^(-3x) in the
denominator. Factoring the terms would> yield -1, thus the
limit for x->infinity. But I plotted the function> and the
real answer is somewhere near 3.> I then tried to use the
Limit function which is not producing an> answer (perhaps
I'm
not sure of the usage). Any help is greatly appreciated, David
Seruyange
====
If you multiplicate your function by 1 in an
intelligent way, hereE^(3x)/E^(3x), you will more easily be
able to study your limits by hand.Mathematica also
appreciates the simplification before you ask the limit !Here
with Mathematica :In[7]:=f[x_] = Simplify[(3/E^x -
E^(-3*x))/(E^(-3*x) + E^(-x))]Out[7]=(-1 + 3*E^(2*x))/(1 +
E^(2*x))In[8]:=Limit[f[x], x ->
Infinity]Out[8]=3In[9]:=Limit[f[x], x ->
-Infinity]Out[9]=-1The
Plot confirms this facts !In[5]:=Plot[(3/E^x -
E^(-3*x))/(E^(-3*x) + E^(-x)), {x, -100, 100}]Meilleures
salutationsFlorian Jaccardprofesseur de
Math.8ematiquesEICN-HES-----Message d'origine-----Envoy.8e
: mar., 21. janvier 2003 13:40è : mathgroup@smc.vnet.netObjet
: Problem with Limits; basic calculusHey 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 wouldyield
-1, thus the limit for x->infinity. But I plotted the
functionand 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
====
I strongly suggest that you
buy the following book:Mathematical Statistics with
Mathematica Authors Colin Rose Murray D. Smith there you'll
find the answer you looking for and
more.cheersyannis-----Mensaje original-----Enviado el: martes
21 de enero de 2003 13:40Para: mathgroup@smc.vnet.netAsunto:
Moments of the multivariate normal distributionI 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.eduReply-To: Mark Coleman
====
I'm working one some
webMathematica applications for finance and
finance that I
will soon be putting online. In the course of designing my
web site, I came to realize that I need an interesting
mathematically-oriented logo. Can anyone point me to an
interesting/cool Mathematica graphic (that is not
copyrighted), or better yet, the code for such a graphic?
Ideally 3-D...Many thanks!Mark
====
>-----Original
Message----->Sent: Tuesday, January 21, 2003 1:40 PM>To:
mathgroup@smc.vnet.net>Hey all - I'm taking a basic calculus
course that uses Mathematica. >We have been studying limits
and I have been using the Limit function>to 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 then>determine the limit. My initial reason had me
select -E^(-3x) in the>numerator and E^(-3x) in the
denominator. Factoring the terms would>yield -1, thus the
limit for x->infinity. But I plotted the function>and the real
answer is somewhere near 3.>I then tried to use the Limit
function which is not producing an>answer (perhaps I'm not
sure of the usage).Any help is greatly appreciated, David
Seruyange>David,as you told, factor and determine the
limit:In[1]:= expr = (3E^(-x) - E^(-3x))/(E^(-3x) +
E^(-x));In[9]:= Limit[expr, x -> -Infinity]Out[9]= -1In[10]:=
Limit[Factor[expr], x -> Infinity]Out[10]= 3You might be
interested in the identityIn[11]:= Factor[expr] ===
Together[TrigToExp[2Tanh[x] + 1]]Out[11]= TrueWhat you have
to look for is the appropriate form to
simplify.--HartmutReply-To:
kuska@informatik.uni-leipzig.de
====
since xy is a varaible and
not a function and you can onlymap functions xy[f_, t_] := {2
f Sin[t]^2, Sinh[f t] + f
Log[t]};ParametricPlot[Evaluate[xy[1, t]], {t, 0, 2}];andf =
{.5, 1, 1.5};trj = Map[xy[#, t] &, f]will work. 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.> Ô f={ .5,1,1.5};'> Ôtrj =
Map [ xy &, f ];'>
--Reply-To: kuska@informatik.uni-leipzig.de
====
a) as long as
Times[] has the Listable attribute it will not work to get a
expression like a*{x,y,z} because Times will scatter a across
the vector elementsb) what is withm = {(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};a*Hold[Evaluate[Apart[#/a] & /@ m]] /.
a_*Hold[b_] :> Hold[a*b] Jens > How 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, even> outside the matrix. How
can I ask Mathematica to do this, and how can I stop the a
just> ßoating back into the main expression? Yours Chris
Rodgers> St John's College> http://rodgers.org.uk/
====
One way
to factor a from expr isexpr2 = a
HoldForm@@{Simplify[expr/a]}The HoldForm is hidden in
OutputForm, StandardForm, and TraditionalForm. Adrawback is
that the HoldForm may complicate manipulation of expr2.Tom
Burton
====
>-----Original Message----->Sent: Tuesday, January
21, 2003 1:38 PM>To: mathgroup@smc.vnet.net>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
];'>--Preferably, I don't press everthing into
one
line:In[1]:= f = 1;Now, as f has a certain value, we must
block it off, when defining xy:In[2]:= Block[{f}, xy = {2 f
Sin[t]^2, Sinh[f t] + f Log[t]}]Out[2]= {2 Sin[t]^2, Log[t] +
Sinh[t]}f seemingly disappeared, butIn[3]:= ?xy Global`xy xy =
{2 f Sin[t]^2, f Log[t] + Sinh[f t]}Rest is
easy:In[4]:=ParametricPlot[ Evaluate[Block[{f = #}, xy] & /@
{.5, 1, 1.5}], {t, 0, 3}, PlotStyle -> {Hue[0], Hue[.3],
Hue[.6]}]--Hartmut
====
MatchQ[x^1, x^_Integer] I got False
result. I think I should get True for this. Could anyone
explain this ?
====
There seems to be some confusion on the
terms undefined andindeterminate. First of all,
karthik's
statement that 0/0 cannot beindeterminate because it has to
take the form (0x1)/0 is in error. That's just another name
for 0/0. We could easily rewrite 0/0 in manydifferent
forms.Should we say that it is all numbers? Let's consider a
differentexample:What is 6/2?Well, by reducing, we get 6/2=3.
How do we know that is true? We canmultiply 3 by 2 to get 6:
3*2=6. Now, let's apply that back to theoriginal question.If
we look at 0/0, what times zero gives us zero?0*0=0,
therefore 0/0=00*1=0, therefore 0/0=10*2=0, therefore 0/0=2
... and on and on to the point any number can satisfy the
question:What is 0/0?Now, is it undefined? Let's
take a look
at that.6/0=undefined. Why? The answer is in the question we
answeredbefore. What multiplied by zero gives 6? The answer
is, we don'tknow. There is noting defined to
give us the
number zero.0/0 cannot be undefined because there are plenty
of definitions forwhat it gives. It must therefore be
indeterminate.
====
It's actually pretty easy to modify the
menu to include whatever sizes you want. (Although
technically it might violate the license agreement... ??)
Just find the file called MenuSetup.tr. On my Mac
it's in
SystemFiles : FrontEnd : TextResources : Macintosh.Once you
open it and scroll down a bit, it will be obvious what to do.
(But, just in case, you should make a back-up
first.)-----Selwyn
Hollishttp://www.math.armstrong.edu/faculty/hollis> 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 > thesis> as possible
!
====
Danke schoen Jens! That's just what I needed. After some
refinement, here's the result as a QuickTime
movie (140k):
http://www.appliedsymbols.com/mma/icosaspin.mov(If WRI had
the same mediocre standards as their competition, they'd
feature this on their website.. :-)----Selwyn Hollis a)
computing the true shadows of the 3 light sources> would
requite the construction of the shadow volumes> and take an
half hour per frame b) fake a single shadow is easy with
Needs[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},>
DisplayFunction -> Identity, opts];> toShadow[sh[[1]], {-2,
2}, {-2, 2}, -2]> ]> obj = Table[RotateShape[Icosahedron[],
t, t/2, t/3], {t, 0, 8, 1/10}];>> MapIndexed[>
Show[Graphics3D[{makeShadow[First[#2], PlotPoints -> 30],
#1},> Boxed -> False, PlotRange -> {{-2, 2}, {-2, 2}, {-2,
2}}]] &,> obj]; but that is *not* the shadow that the three
colored light sources would> produce -- that would be much
more expensive. If some one else would like to have (hard)
shadows I can include it > into Jens> I'm interestied in
using Mathematica to do something similar to this:>
http://www.mapleapps.com/powertools/logos/appess3.gif> Does
anyone have a guess at what kind of rotation is being used?>
And how one might get a similar shadow effect? (Without
resorting to a> different rendering engine.)> Selwyn To:
mathgroup@smc.vnet.net> -----Original Message-----> To:
mathgroup@smc.vnet.net> Sent: Saturday, January 18, 2003 6:39
AM> To: mathgroup@smc.vnet.net> Does anyone why f@a_=2 a gives
a proper answer for f[1]> and not in the case of
Prefix[s[a_]]=
2 a eventhough that> Prefix[s[a_]]=> f@a_???> In[1]:=> f@a_=2
a> f[1]> Out[1]=> 2 a> Out[2]=> 2> In[3]:=> Prefix[s[a_]]>
Prefix[s[a_]]=2 a> s[1]> Out[3]=> s@a_> Set::write: Tag
Prefix
in s@a_ is Protected.> Out[4]=> 2 a> Out[5]=> s[1]>> Ashraf,
the question is: what behaviour did you expect? An answer to
that might> solve your problem. First, Prefix[s[a_]] is *not*
the same as f@a_ In[1]:= f@a_ := 2 a> In[2]:= ?f> Global`f>
f[a_] := 2 a In[3]:= f[1]> Out[3]= 2 Compare this to In[4]:=
Prefix[s[a_]]> Out[4]= s@a_ In[5]:= % // FullForm>
Out[5]//FullForm=> Prefix[s[Pattern[a, Blank[]]]] So
Prefix is
a wrapper for printing purposes (only). Such you cannot >
make> definitions for it (as it is protected), but also, it is
a special > form not> submitted to the standard evaluation
sequence, see: In[7]:= Prefix[s[aaa___]] ^= s[aaa]> Out[7]=
s[aaa] In[8]:= s[a_] := 2 a> In[9]:= ?s> Global`s>
Prefix[s[aaa___]] ^= s[aaa]> s[a_] := 2 a> In[10]:=
Prefix[s[2]]> Out[10]= Prefix[4] So even Upvalues
won't work.
You may however do In[12]:= Prefix[Unevaluated[s[2]]]>
Out[12]= 4 The very question, however, is whether this is
what you intended, or > say,> what did you want to achieve at
all? --
====
Since Exp[-x] goes to zero as x goes to +infinity,
the dominant termis not Exp[-3x]==Exp[-x]^3 but Exp[-x], so 3
is the real value of thelimit.Using Simplify on the original
expression yields:-1 + 3 Exp[2x]---------------1+ 2 Exp[2x]It
is clear (use L'Hospital rule for formal proof) that the
limit
isindeed 3.Finally, by applying FullSimplify we get 1 +
2Tanh[x] and sinceLimit[Tanh[x],x->inf]==1 we also get
3.Orestis> Hey all - I'm taking a basic calculus course that
uses Mathematica. > We have been studying limits and I have
been using the Limit function> to 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 then> determine the limit. My initial reason had me
select -E^(-3x) in the> numerator and E^(-3x) in the
denominator. Factoring the terms would> yield -1, thus the
limit for x->infinity. But I plotted the function> and the
real answer is somewhere near 3.> I then tried to use the
Limit function which is not producing an> answer (perhaps
I'm
not sure of the usage). Any help is greatly appreciated, David
SeruyangeReply-To: kuska@informatik.uni-leipzig.de
====
if you
useexpr = (3E^(-x) - E^(-3x))/(E^(-3x) + E^(-x));Cancel[expr
/. {Exp[-x] -> z, Exp[-3x] -> z^3}]z->0 when x->Infinity and
from the limit of(3 - z^2)/(1 + z^2)is obvious 3 for z->0
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 function> to 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 then> determine the limit. My initial reason had me
select -E^(-3x) in the> numerator and E^(-3x) in the
denominator. Factoring the terms would> yield -1, thus the
limit for x->infinity. But I plotted the function> and the
real answer is somewhere near 3.> I then tried to use the
Limit function which is not producing an> answer (perhaps
I'm
not sure of the usage). Any help is greatly appreciated, David
Seruyange
====
> Hey all - I'm taking a basic calculus course
that uses Mathematica.> We have been studying limits and I
have been using the Limit function> to 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 then> determine the limit. My initial reason had me
select -E^(-3x) in the> numerator and E^(-3x) in the
denominator. Factoring the terms would> yield -1, thus the
limit for x->infinity. But I plotted the function> and the
real answer is somewhere near 3.> I then tried to use the
Limit function which is not producing an> answer (perhaps
I'm
not sure of the usage).Your idea of considering the dominant
terms is good, but you thought thatthe wrong ones were
dominant. Since you are working with x -> Infinity,the
dominant terms in numerator and denominator are 3E^(-x) and
E^(-x),resp. Thus, their ratio approaches 3, just as your
graph indicated.I hate to report, however, that your usage of
Mathematica was correct.That its Limit function could not give
an answer for such a simpleproblem is ludicrous. I had thought
that perhaps you just needed to usethe Standard Add-on Package
Calculus`Limit`. But I hate to report thatusing it is even
worse! Rather than merely giving no answer, it then givesthe
incorrect answer 0. Even more ludicrous.There is a simple
workaround, however. Just Simplify your functionfirst, before
asking for its limit. Mathematica
gives(-1+3E^(2x))/(1+E^(2x)) as the simplification. Then,
using either thebuilt-in Limit function or the one in the
add-on package, the correctanswer, 3, is obtained.David
Cantrell
====
On 1/21/03 at 7:38 AM,
google.news.invalid@web2news.net (Narasimham>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];' Try trj = {
2 # Sin[t]^2 ,Sinh[f t]+# Log[t] }&/@{ .5,1,1.5};
====
If an
object is made in ParametricPlot3D, is there a way of
includingcode lines to click/drag the mouse so as to rotate
it ( say about avertcal axis) and to be able to see it from
the other side ? --
====
we are searching all solutions where
the function f results null. f[x_] := -7500 * Ceiling[(0.5 *
x) / 880] + (5 * x)Solve[{f[x] == 0}, x]As f contains jump
discontinuities, we recieved the following
error:InverseFunction::ifun: Inverse functions are being
used. Values may be lost for multivalued
inverses.Solve::tdep: The equations appear to involve the
variables to be solved for in an essentially non-algebraic
way.We would be pleased if anybody could help us.Philipp
BurkertCarsten Siegmund
====
All:I'm striving to stay away from
the mouse, using keyboard shortcutsinstead, so that I can move
more quickly.Here's one that's irritating me
(on windows):You
are in the midst of editing in a cell. You decide you want
tochange the style of the cell (e.g. from text to small
text). There arethe Alt- shortcuts, but those only seem to
work when the cell itself isselected. And so, the question:
What keystrokes will select a cellitself, from someplace
within the cell?
====
>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 then>determine the limit. My initial reason had me
select -E^(-3x) in the>numerator and E^(-3x) in the
denominator.E^(-x) > E^(-3x) for x > 0. So, try multiplying
the numerator and denominator by E^x
====
I was wondering if
someone here would be nice enough to convert this C
toMathematica code. A module would be best I suppose, so
users can just callit just like Random[].I'd like to
experiment with this RNG in Mathematica. In fact, IIRC
Mathematica uses some variant of SWC also.By the way, the
following words are from another thread in another
NGregarding a RNG.*** Words below from Dr. Marsaglia ***Here
is a MWC example with period 3056868392^33216-1, a
mere10^4005 times as long as that of the Mersenne twister,
yet faster,far simpler, and seemingly at least as
well-performing in testsof randomness:static unsigned long
Q[1038],c=362436;unsigned long MWC1038(void){static unsigned
long i=1037;unsigned long long t, a=611373678LL;t=a*Q[i]+c;
c=(t>32);if(--i) return(Q[i]=t);i=1037; return(Q[0]=t); }You
need to assign random 32-bits seeds to the static array
Q[1038]and to the initial carry c, with 0<=c<61137367.George
Marsaglia
====
Turtle-Interpretation - has been used for
describing structures in 3D spaceand you probably need to get
the .m file from the original location whereyou acquired the
notebook using turtleinterpretation.When procuring notebooks
from external sources you might want to check for*.m files as
well, as some notebooks will create it if it is missing
andsome will not.Steven shippee@jcs.mil> To whom may concern,
I have a problem with Mathematica. I have installed
Mathematica 4.2 onWindows XP. I need to run a program in
Mathematica, Get[turtleinterpretation.m], this instruction
could not be compiledfind this package in Mathematica and also
in the Internet. Could you help me?> best regard. Hung
Viet.Reply-To: kuska@informatik.uni-leipzig.de
====
no, because
MathSourcehttp://library.wolfram.com/databasedon't know it
and
every user of mathematica can choose anyname for packages we
can not know *who* has written the package. Jens 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.
====
>-----Original
Message----->Sent: Tuesday, January 21, 2003 1:40 PM>To:
mathgroup@smc.vnet.netHow 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, even>outside the matrix.How can
I ask Mathematica to do this, and how can I stop the a
just>ßoating back into the main expression?YoursChris
Rodgers>St John's College>http://rodgers.org.uk/>Chris,The
problem comes from the Listable attribute of Plus. If you
want to seeyour expression (which I call alist for a while)
as you want, you can clearthat Attribute for a while:
ClearAttributes[Plus, Listable] Thread[Apart[alist],
Plus]Now, I think, you see what you want. The procedure is
dangerous however,since the attributes of Plus should be
restored, before you make any otherevaluation.
SetAttributes[Plus, Listable]Assuming you need that form only
for printing purposes, it might be moreconvenient just to wrap
it into HoldForm: Block[{Plus}, HoldForm[#] &[Thread[Sort /@
Apart[alist], Plus]]]The Block temporarily inactivates Plus
(and its attributes), HoldFormprevents threading back to the
original form, after the expression havingleft
Block.--Hartmut
====
> Has anybody encountered problems with
Windows 2000 Service Pack 3 ? It> appears to change the way
Mathlink works in Mathematica 4.0.> <...snipped... >
Additionally, Mathematica 4.2 installation will not complete
(after> asking whether or not I want to enter password,
Mathematica fails to> launch). After starting Mathematica
manually, and having entered pwd> information, JLink fails
Needs[JLink`] InstallJava[] causes a file browser to pop-up
and asks to select some non otherwise> specified
file. I have
reported the problems toram support which seemed surprised>
by this ... Fulvio 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 sp3> computer with no problems. Has
anyone else here experienced this kind of
problem?Bhuvanesh,Did you take the steps of installing
Mathematica 4.2 on a machinewhich had Whidnows 2000 + SP3 ? I
would like to reiterate that theprevious problems was observed
on three completely different machines,all running Microsoft
Windows 2000, so I am hard pressed to believethat this is a
ßuke. Two of the machines, the ones that had all ofthe latest
Windows Update, failed installation (it never completed)while
the third, which had not been updated after SP2, worked fine.
Itmay very well be that if you update windows to SP3 AFTER
Mathematica4.2 is installed you may not have (or notice) a
problem {a problem, itseems worth repeating, which is
noticeable ONLY with Mathlink andJLink, everything else seems
OK ). You should try installingMathematica 4.2 on a computer
running Windows 2000 with all of the SPand updates, and then
you may give us an update/comment on whether ornot you have a
problem. As it stands, I must conclude that thesoftware does
not run under Windows 2000.Fulvio
====
something
likehttp://phong.informatik.uni-leipzig.de/~kuska/mathgl3dv3/
id15.htmhttp://phong.informatik.uni-leipzig.de/~kuska/
mathgl3dv3/id16.htmBut I have hundreds more of it ... Jens >
I'm working one some webMathematica applications for
finance
and finance that I> will soon be putting online. In the
course of designing my web site, I> came to realize that I
need an interesting mathematically-oriented logo.> Can anyone
point me to an interesting/cool Mathematica graphic (that is
not> copyrighted), or better yet, the code for such a
graphic? Ideally> 3-D... Many thanks! > Mark
====
x^1 evaluates
to just x which is no longer of the form (=FullForm)
Power[x,1]. Since the matching is syntactic and not semantic
you do not get a match. However you do get one if instead you
evaluate:In[1]:=MatchQ[Unevaluated[x^1],x^_Integer]Out[1]=
Trueor:In[2]:=MatchQ[x^1,x^_Integer:1]Out[2]=TrueAndrzej
Kozlowski> MatchQ[x^1, x^_Integer]> I got False result. I
think I should get True for this.> Could anyone explain
this ?Andrzej KozlowskiYokohama,
Japanhttp://www.mimuw.edu.pl/~akoz/http://
platon.c.u-tokyo.ac.jp/andrzej/Reply-To:
kuska@informatik.uni-leipzig.de
====
Aber bitte -- ist mir
immer eine Freude.The true/ not faked animation with all
shadows can be found
athttp://phong.informatik.uni-leipzig.de/~kuska/icosshad.mov
Jens Danke schoen Jens! That's just what I needed. After
some
refinement,> here's the result as a QuickTime
movie (140k):
http://www.appliedsymbols.com/mma/icosaspin.mov (If WRI had
the same mediocre standards as their competition, they'd>
feature this on their website.. :-) ----> Selwyn Hollis > a)
computing the true shadows of the 3 light sources> would
requite the construction of the shadow volumes> and take an
half hour per frame> b) fake a single shadow is easy with>
Needs[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},>
DisplayFunction -> Identity, opts];> toShadow[sh[[1]], {-2,
2}, {-2, 2}, -2]> ]> obj = Table[RotateShape[Icosahedron[],
t, t/2, t/3], {t, 0, 8, 1/10}];> MapIndexed[>
Show[Graphics3D[{makeShadow[First[#2], PlotPoints -> 30],
#1},> Boxed -> False, PlotRange -> {{-2, 2}, {-2, 2}, {-2,
2}}]] &,> obj];> but that is *not* the shadow that the three
colored light sources would> produce -- that would be much
more expensive.> If some one else would like to have (hard)
shadows I can include it> into> Jens> I'm interestied in
using Mathematica to do something similar to this:>
http://www.mapleapps.com/powertools/logos/appess3.gif> Does
anyone have a guess at what kind of rotation is being used?>
And how one might get a similar shadow effect? (Without
resorting to a> different rendering engine.)> Selwyn> To:
mathgroup@smc.vnet.net> -----Original Message-----> To:
mathgroup@smc.vnet.net> Sent: Saturday, January 18, 2003 6:39
AM> To: mathgroup@smc.vnet.net> Does anyone why f@a_=2 a gives
a proper answer for f[1]> and not in the case of
Prefix[s[a_]]=
2 a eventhough that> Prefix[s[a_]]=> f@a_???> In[1]:=> f@a_=2
a> f[1]> Out[1]=> 2 a> Out[2]=> 2> In[3]:=> Prefix[s[a_]]>
Prefix[s[a_]]=2 a> s[1]> Out[3]=> s@a_> Set::write: Tag
Prefix
in s@a_ is Protected.> Out[4]=> 2 a> Out[5]=> s[1]> Ashraf,>
the question is: what behaviour did you expect? An answer to
that might> solve your problem.> First, Prefix[s[a_]] is *not*
the same as f@a_> In[1]:= f@a_ := 2 a> In[2]:= ?f> Global`f>
f[a_] := 2 a> In[3]:= f[1]> Out[3]= 2> Compare this to>
In[4]:= Prefix[s[a_]]> Out[4]= s@a_> In[5]:= % // FullForm>
Out[5]//FullForm=> Prefix[s[Pattern[a, Blank[]]]]> So
Prefix is
a wrapper for printing purposes (only). Such you cannot> make>
definitions for it (as it is protected), but also, it is a
special> form not> submitted to the standard evaluation
sequence, see:> In[7]:= Prefix[s[aaa___]] ^= s[aaa]> Out[7]=
s[aaa]> In[8]:= s[a_] := 2 a> In[9]:= ?s> Global`s>
Prefix[s[aaa___]] ^= s[aaa]> s[a_] := 2 a> In[10]:=
Prefix[s[2]]> Out[10]= Prefix[4]> So even Upvalues
won't work.
You may however do> In[12]:= Prefix[Unevaluated[s[2]]]>
Out[12]= 4> The very question, however, is whether this is
what you intended, or> say,> what did you want to achieve at
all?> -->
====
>How to perturb the constant f in parametric
plots?> Preferably, I don't press everthing into one line:>
f
= 1;> Now, as f has a certain value, we must block it off,
when defining xy: Block[{f}, xy = {2 f Sin[t]^2, Sinh[f t] + f
Log[t]}] ParametricPlot[ Evaluate[Block[{f = #}, xy] & /@ {.5,
1, 1.5}], {t, 0, 3}]Actually got earlier some plot by
ParamaeterPlot3D where [ x,y,z=0] =F(f,t) where f,t, are
parameters and chose a vertically up View point,but this one
gives a better ßat plot by perturbation rather than
takingspecial 3D case that I really wanted and thank you for
this. --
====
G.L.,There are a number of ways. Your could
install Jens-Peer Kuska's MathGLpackage available from his
web site.http://phong.informatik.uni-leipzig.de/~kuska/You
could use the RealTime3D` package that comes with
Mathematica. (It workson Windows, I'm not certain about
other platforms.)plot1 =
ParametricPlot3D[......]< 0.1, AnimationDirection ->
Forward]}]You only need the SpinShow statement. The other
statements select the cells,close them and start the
animation. The above will run the animation for 200seconds,
with each frame being displayed for 0.1 second.
TheAnimationDirection can also be set to Backward or
ForwardBackward.In running animations like SpinShow, you can
use the cursor keys (up ordown) to stop the animation and
advance one frame at a time. This is often avery convenient
method for viewing animations. You can use the numeric keysto
control the speed.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/Sender:
steve@smc.vnet.netApproved: Steven M. Christensen
, Moderator
====
>-----Original
Message----->Sent: Wednesday, January 22, 2003 12:14 PM>To:
mathgroup@smc.vnet.net>discontinuitieswe are searching all
solutions where the function f results null. f[x_] := -7500 *
Ceiling[(0.5 * x) / 880] + (5 * x)>Solve[{f[x] == 0}, x]As f
contains jump discontinuities, we recieved the following
error:InverseFunction::ifun: Inverse functions are being
used. >Values may be >lost for multivalued
inverses.Solve::tdep: The equations appear to involve the
variables >to be solved >for in an essentially non-algebraic
way.We would be pleased if anybody could help us.Philipp
Burkert>Carsten Siegmund>Philipp, Carsten,there seems to be
no method to solve problems of this kind in general. But
Ican tackle your case: Instead of looking at the zeros for
your risingsawtooth, we solve for a corresponding family of
functions, and the sort outthose solutions which are not on
any tooth.We replace the Ceiling part of the definition by
constants:As Ceiling[(0.5*x)/880] are Integers we define xx =
Range[-10, 20] * 880 / 0.5for a certain range and make f
depended on a second parameter: f[x_, xx_] :=
-7500*(0.5*xx)/880 + (5*x)We now solve f[x, xx] for every xx,
but keep only those solutions whichqualify. res = With[{x0 =
First[x /. Solve[f[x, #] == 0, x]]}, If[f[x0] == 0, {x ->
x0}, Unevaluated[Sequence[]]]] & /@ xxIn[33]:= xxres = x /.
resOut[33]= {0, 1500., 3000., 4500., 6000., 7500., 9000.}We
check: Plot[f[x], {x, -2000, 12000}, PlotPoints -> 500,
Epilog -> {PointSize[0.01], Hue[0], Point[{#, f[#]}] & /@
xxres}]Perhaps you can adapt the method to your real
Problem.--Hartmut
====
> I was wondering if someone here would
be nice enough to convert this C to> Mathematica code. A
module would be best I suppose, so users can just call> it
just like Random[]. I'd like to experiment with this RNG in
Mathematica. In fact, IIRC Mathematica uses some variant of
SWC also. By the way, the following words are from another
thread in another NG> regarding a RNG. *** Words below from
Dr. Marsaglia *** Here is a MWC example with period
3056868392^33216-1, a mere> 10^4005 times as long as that of
the Mersenne twister, yet faster,> far simpler, and seemingly
at least as well-performing in tests> of randomness: static
unsigned long Q[1038],c=362436; unsigned long MWC1038(void){>
static unsigned long i=1037;> unsigned long long t,
a=611373678LL;> t=a*Q[i]+c; c=(t>32);> if(--i)
return(Q[i]=t);> i=1037; return(Q[0]=t);> } You need to
assign random 32-bits seeds to the static array Q[1038]> and
to the initial carry c, with 0<=c<61137367. George MarsagliaI
think the code below does what is intended. This assumes a
long longis 64 bits and a long is 32 bits. These are both
quite platformdependent for C; maybe Marsaglia specified a
particular platform? (a URLto his note would be helpful). For
seeding the q and c variables I usebuilt-in Random. I also
changed the variable names to avoid singlecharacters and
capitalized names.SeedRandom[1111];qarray =
Table[Random[Integer,{0,2^32-1}], {1038}];cc =
Random[Integer, {0,61137367}];index = 1037;aconst =
611373678;mwc1038[] := Module[ {}, tt =
aconst*qarray[[index]] + cc; cc =
Developer`BitShiftRight[tt,32]; (*Could instead use Quotient
*) tt = Mod[tt,2^32]; index--; If [index!=0, qarray[[index]]
= tt; Return[tt]]; index = 1037; qarray[[0]] = tt
]Example:Table[mwc1038[], {10}]Out[48]= {1392539486,
2580953375, 977837705, 1393978643, 2243451427, 374027900,
2689334407, 2953422566, 41924440, 2262165557}Daniel
LichtblauWolfram Research
====
Dear newsgroupIt is possible to
calculate an Laplace-Operator with Mathematica???For each
coordinate systems in each dimensions??If yes, how?AxelPs.: I
mean as Laplace-Operator Nabla^2 not
theLaplace-Beltrami-operator (it is a extention of the
Laplace-Operator)
====
Hendrik van Hees (
http://theory.gsi.de/~vanhees/ ) gave me the solution
forthis problem:For example: 2Dim - Polar
coordinates------------x = r Sin[a]y = r Cos[b]xvec = {x,y}q
= {r,a}jacobian =
Table[D[xvec[[mu]],q[[nu]]],{mu,1,2},{nu,1,2}]gcov =
FullSimplify[Transpose[jacobian].jacobian]gcontra =
Inverse[gcov]g = Det[gcov]grad =
Table[D[phi[q[[1]],q[[2]]],q[[k]]],{k,1,2}];Lapl =
FullSimplify[1/Sqrt[g] Sum[D
[Sqrt[g](gcontra.grad)[[j]],q[[j]]],{j,1,2}]]-----------It is
going on for each coordinate systems in each dimensions.many
thanks to HendrikAxelAxel Ligon schrieb im
Newsbeitrag> Dear newsgroup It is possible to calculate an
Laplace-Operator with Mathematica???> For each coordinate
systems in each dimensions??> If yes, how? Axel Ps.: I mean
as Laplace-Operator Nabla^2 not the>
Laplace-Beltrami-operator (it is a extention of the
Laplace-Operator)>
====
UËytkownik Axel Ligon
napisañ w wiadomo.a6ci> Dear newsgroup It
is possible to calculate an Laplace-Operator with
Mathematica???> For each coordinate systems in each
dimensions??> If yes, how? Axel Ps.: I mean as
Laplace-Operator Nabla^2 not the> Laplace-Beltrami-operator
(it is a extention of the Laplace-Operator)>
Calculus`VectorAnalysis`GartekReply-To:
kuska@informatik.uni-leipzig.de
====
what maythe Lapacian[]
function from< For each coordinate systems in each
dimensions??> If yes, how? Axel Ps.: I mean as
Laplace-Operator Nabla^2 not the> Laplace-Beltrami-operator
(it is a extention of the Laplace-Operator)
====
Dear
MathGroup,Steven Shippee asked me about methods of solving
the following problem.Find the numbers n such that n^2 is a
five digit number with 0 in the seconddigit and 1 in the last
digit. (i.e., x0xx1 where x is a digit 0..9).We know how to
solve the problem by testing a list of all possible
candidatenumbers. But is there a method that uses Solve and
modular arithimetic orsome other clever method? If one looks
at the answers one would think theremust be.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/
====
>
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?Two brute
force-type methods that use functions in combinatoria:1)
Given m and n, initialize a Composition c={m,0,...,0} so it
is aset of Length[c]=n. Now repeatedly call
c=NextComposition[c], testingeach iteration to see if it
meets your criteria. Stop when you getc={0,...,0,m}.2)
Repeatedly call RandomComposition[m,n] and test, using a
samplesize such that you have confidence that the probability
calculated viaMonte Carlo is close enough.
====
> 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?Two ideas, both
of which are brute force, and use some similar functions...1)
Use the NextComposition[ ] function, starting with list
{m,0,0,...,0}] anditerating through to {0,...,0,m}, counting
the ones that meet your constraint.Divide this by
NumberOfComposition[m,n]. This gives exact answer.2) Run
RandomComposition[m,n] a fixed number of times, counting the
ones thatmeet your constraint, divided by number of times
run. This gives approximateanswer. You could monitor the
run, and stop on some convergence property.
====
Why doesn't
this multiplication equal zero ?In[1]:= DiracDelta[2 + t]
DiracDelta[t]Out[1]= DiracDelta[t] DiracDelta[2 + t]A delta
multiplied by a delayed delta = 0 ... right ?Also like this
:In[2]:= DiracDelta[2 + t] DiracDelta[t] //
SimplifyDiracDeltaOut[2]= SimplifyDiracDelta[DiracDelta[t]
DiracDelta[2 + t]]thanksMattReply-To:
kuska@informatik.uni-leipzig.de
====
because Mathematica has no
build-in rule to handleproduct of delta-Functionsbut now it
hasUnprotect[DiracDelta]DiracDelta /:
DiracDelta[a_]^n_Integer := DiracDelta[a]DiracDelta /:
DiracDelta[a_]* DiracDelta[b_] /; a =!= b :=
0Protect[DiracDelta]But you should keep in mind that only
expressionIntegrate[a_*DiracDelta[t],{t,-Infinity,In[CapitalTh
orn]nity}]/;
FreeQ[a,_DiracDelta]are covered by the theory of
distributions. Jens Why doesn't this multiplication equal
zero ? In[1]:= DiracDelta[2 + t] DiracDelta[t]> Out[1]=
DiracDelta[t] DiracDelta[2 + t] A delta multiplied by a
delayed delta = 0 ... right ? Also like this : In[2]:=
DiracDelta[2 + t] DiracDelta[t] // SimplifyDiracDelta>
Out[2]= SimplifyDiracDelta[DiracDelta[t] DiracDelta[2 + t]]
thanks> Matt
====
I don't really see the point in all this, but
here is my opinion:0/0 is indeterminate and not undefined
because it can NOT be defined!It is in a sense
Ôundefinable',
hence indeterminate.If you don't know anything about complex
numbers, it appears to youthat Sqrt[-1] has no meaning, just
as 0/0. It is not the same though:Sqrt[-1] can be defined to
have a single (non-real) value I, so thatthe whole construct
that results has an internal consistency. Since itcan
actually be defined, it is undefined(if you stick
to real
numbers)and not indeterminate.0/0 is indeterminate for
exactly the opposite reason: if someone triesto give it a
reasonable value, it can be proven that ANY number wouldbe
equal to it and the resulting construct would collapse
immediately.Therefore 0/0 is indeterminate, eg.
undefinable.Orestis Vantzos> There seems to be some confusion
on the terms undefined and> indeterminate. First of all,
karthik's statement that 0/0 cannot be> indeterminate
because
it has to take the form (0x1)/0 is in error. > That's just
another name for 0/0. We could easily rewrite 0/0 in many>
different forms. Should we say that it is all numbers? Let's
consider a different> example: What is 6/2? Well, by
reducing, we get 6/2=3. How do we know that is true? We can>
multiply 3 by 2 to get 6: 3*2=6. Now, let's apply that back
to the> original question. If we look at 0/0, what times zero
gives us zero? 0*0=0, therefore 0/0=0> 0*1=0, therefore 0/0=1>
0*2=0, therefore 0/0=2 ... and on and on to the point any
number can satisfy the question:> What is 0/0? Now, is it
undefined? Let's take a look at that.
6/0=undefined. Why? The
answer is in the question we answered> before. What
multiplied by zero gives 6? The answer is, we don't> know.
There is noting defined to give us the number zero. 0/0 cannot
be undefined because there are plenty of
definitions for> what
it gives. It must therefore be indeterminate.
====
How do I
wrap and call a C subroutine that should return nothing back
toMathematica. I want to somehow specify a :ReturnType: of
None or Void,but mprep doesn't recognize those tokens.
If I specify Manual then myMathematica session just hangs
waiting for an answer from my MathLinkprogram. Does MathLink
have a function like MLPutEmptyPacket() that justtells
Mathematica Expect nothing from this MathLink function. It's
OK tostop waiting and return to your command loop.This is
(part of) the template file I am using::Begin::Function:
SetValue:Pattern: SetValue[ value_?NumericQ ]:Arguments:
{value}:ArgumentTypes: {Real}:ReturnType: Manual:End:void
SetValue ( double value ){ /* * Do something useful with
value. * It doesn't make sense to return anything. */ /*
What
goes here to keep Mathematica from hanging? */}How do I
specify a true MathLink subroutine that doesn't return a
value toReply-To: kuska@informatik.uni-leipzig.de
====
end you
function withMLPutSymbol(stdlink,Null);and not with
nothing. Mathematica expect alwaysa result form a function
call, because it isa functional language. Jens How do I wrap
and call a C subroutine that should return nothing back to>
Mathematica. I want to somehow specify a :ReturnType: of
None or Void,> but mprep doesn't recognize those tokens.
If I specify Manual then my> Mathematica session just hangs
waiting for an answer from my MathLink> program. Does
MathLink have a function like MLPutEmptyPacket() that just>
tells Mathematica Expect nothing from this MathLink
function. It's OK to> stop waiting and return to your
command loop. This is (part of) the template file I am using:
:Begin:> :Function: SetValue> :Pattern: SetValue[
value_?NumericQ ]> :Arguments: {value}> :ArgumentTypes:
{Real}> :ReturnType: Manual> :End: void SetValue ( double
value )> {> /*> * Do something useful with value.> * It
doesn't make sense to return anything.> */ /* What goes here
to keep Mathematica from hanging? */> } How do I specify a
true MathLink subroutine that doesn't return a value to>
====
While playing variations on Hartmut's elegant solution I
came across thefollowing illustration of the perils of
blocking basic operations like Plusexpr = {(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}Block[{Plus},HoldForm[#]
&[FullSimplify[expr]]]//InputForm HoldForm[{(4 + a)/4, (4 +
a)/4, (4 + a)/4, (4 + a)/4, (-a - 2*Sqrt[1 +a^2])/4, (-a -
2*Sqrt[1 + a^2])/4, (-a + 2*Sqrt[1 + a^2])/4, (-a + 2*Sqrt[1+
a^2])/4}]Allan---------------------Allan HayesMathematica
Training and ConsultingLeicester
UKwww.haystack.demon.co.ukhay@haystack.demon.co.ukVoice: +44
(0)116 271 4198 >-----Original Message----->Sent: Tuesday,
January 21, 2003 1:40 PM>To: mathgroup@smc.vnet.net>How 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, even>outside the matrix.>How can
I ask Mathematica to do this, and how can I stop the a
just>ßoating back into the main expression? The problem comes
from the Listable attribute of Plus. If you want to see> your
expression (which I call alist for a while) as you want, you
canclear> that Attribute for a while: ClearAttributes[Plus,
Listable] Thread[Apart[alist], Plus]> Now, I think, you see
what you want. The procedure is dangerous however,> since the
attributes of Plus should be restored, before you make any
other> evaluation. SetAttributes[Plus, Listable]> Assuming
you need that form only for printing purposes, it might be
more> convenient just to wrap it into HoldForm:
Block[{Plus},> HoldForm[#] &[Thread[Sort /@ Apart[alist],
Plus]]] The Block temporarily inactivates Plus (and its
attributes), HoldForm> prevents threading back to the
original form, after the expression having> left Block.
--
====
> MatchQ[x^1, x^_Integer]> I got False result. I
think I should get True for this.> Could anyone explain
this ?Kimberly,x^1 evaluates to x before the match is
tested.But we have: MatchQ[Unevaluated[x^1], x^_Integer]
True--Allan---------------------Allan HayesMathematica
Training and ConsultingLeicester
UKwww.haystack.demon.co.ukhay@haystack.demon.co.ukVoice: +44
(0)116 271 4198
====
MatchQ[x^1, x^_Integer] gives
FalsebutMatchQ[Unevaluated[x^1], x^_Integer] gives TrueThe
first result is because x^1 evaluates to x which
doesn't match
thepattern x^_Integer. The second result is because
Unevaluated prevents thisevaluation from occurring before the
pattern matching.--Steve LuttrellWest Malvern, UK> MatchQ[x^1,
x^_Integer]> I got False result. I think I should get True
for this.> Could anyone explain this ?
====
MatchQ evaluates x^1
to get x ... then x is compared with the patternand does not
match, since syntactically it is not of the formx^_Integer
(it doesn't even have a Power head.)In[2]:= 1
_IntegerTrace[MatchQ[x , x ]]Out[2]= 1xx _IntegerMatchQ[x, x
]FalseOn the other hand MatchQ[x, x^_.] will match... see
Default in theHelp Browser.Orestis Vantzos> MatchQ[x^1,
x^_Integer] > I got False result. I think I should get
True for this. > Could anyone explain this ?> Reply-To:
kuska@informatik.uni-leipzig.de
====
tryMatchQ[x^1,
x^_.]orMatchQ[x^1, x^n_./; IntegerQ[n]] Jens MatchQ[x^1,
x^_Integer]> I got False result. I think I should get
True for this.> Could anyone explain this ?> Reply-To:
k.chourdakis@qmul.ac.uk
====
Dear all,I am trying to simulate a
multivariate t-distribution,and encounter the following
problem:(* First load the packages
*)<we are
searching all solutions where the function f results
null.>f[x_] := -7500 * Ceiling[(0.5 * x) / 880] + (5 * x) By
inspection f[0]=0Eliminating Ceiling to get f=-7500 * (0.5 *
x) / 880 + (5 * x) gives a result that is linear in x. Since
it is linear in x, there can only be one root. So, clearly
the only root to the original equation is x = 0.
====
hiI want
to solove u_{t}=u_{x} * u ^ 2 + u_{xx} * (1-u) and u_{t}
=u_{xx} + exp (-u )+ exp (-2u) by using Method Of Lineplease
help me.Note: Please give your explain by using computer code
or computer algorithm(numerical)Reply-To:
kuska@informatik.uni-leipzig.de
====
it looks pretty much like
a student home work.Ask your instructor/tutor. Jens hi I want
to solove u_{t}=u_{x} * u ^ 2 + u_{xx} * (1-u) and u_{t}
=u_{xx} + exp (-u )+ exp (-2u) by using Method Of Line
please help me. Note: Please give your explain by using
computer code or computer algorithm(numerical)Reply-To:
kuska@informatik.uni-leipzig.de
====
a) you can use MathGL3d
form
http://phong.informatik.uni-leipzig.de/~kuska/mathgl3dv3/id3.
htmb) you can load < code lines to
click/drag the mouse so as to rotate it ( say about a>
vertcal axis) and to be able to see it from the other side ?>
--
====
>If an object is made in ParametricPlot3D, is there a
way of including>code lines to click/drag the mouse so as to
rotate it ( say about a>vertcal axis) and to be able to see
it from the other side ?>--This can be done using JLink and
the LiveGraphics3D applet. For an example,
seehttp://omegaconsultinggroup.com/Services/ezine19.nborhttp
://omegaconsultinggroup.com/Services/ezine19.htmlunder the
section
QuickTip.--------------------------------------------------
------------Omega ConsultingThe final answer to your
Mathematica
needshttp://omegaconsultinggroup.com
====
Kirk,This question
has come up before, and you could search the archives
atgoogle orram to find an answer.At any rate, the simplest way
to select a cell while the cursor is in thecell is to
repeatedly hit Ctrl-. until the vertical bar is highlighted.
Themain problem with this approach is that if you hit Ctrl-.
one too manytimes, you will select a whole bunch of cells. A
smaller problem is that ifyou are deeply buried within a
cell, it may take many Ctrl-.s to select thecell.A more
complicated method to do what you want which avoids the
aboveproblems is to create a new hot key by editing the
KeyEventTranslations.trsystem file. If you are interested in
doing this, you can search thearchives for discussions about
this file, or you can fiddle with the
file(after you save a copy
of it). One thing to watch out for is to avoid usinghot keys
that have already been defined for the menu system. The menu
hotkeys are defined in the file MenuSetup.tr.If
you want to
pursue the second option above, feel free to ask for
moreadvice.Carl WollPhysics DeptU of Washington> All: I'm
striving to stay away from the mouse, using keyboard
shortcuts> instead, so that I can move more quickly. Here's
one that's irritating me (on windows): You are in the midst
of editing in a cell. You decide you want to> change the
style of the cell (e.g. from text to small text). There are>
the Alt- shortcuts, but those only seem to work when the cell
itself is> selected. And so, the question: What keystrokes
will select a cell> itself, from someplace within the
cell?
====
> All: I'm striving to stay away from the mouse,
using keyboard shortcuts> instead, so that I can move more
quickly. Here's one that's irritating me (on
windows): You
are in the midst of editing in a cell. You decide you want
to> change the style of the cell (e.g. from text to small
text). There are> the Alt- shortcuts, but those only seem to
work when the cell itself is> selected. And so, the question:
What keystrokes will select a cell> itself, from someplace
within the cell?> Control-period expands the current
selection. If you have the entirecontents of a cell selected,
the selection is expanded to be the cellbracket. So,
typically, you would have to press Control-period twoto four
times (but if your selection is in a deeply nested
expression,perhaps more) to select the cell bracket.A related
mouse trick (I know you didn't ask, but since
I'm on
thesoapbox, I thought I'd mention it) can duplicate this
keystroke withthe mouse. If you double-click on some text,
that's the same ashaving the selection there and pressing
Control-period once. Triple-clicking is like pressing
control-period twice. And so on (except thatyou cannot select
the cell bracket using the mouse via this means).This is
particularly useful in input/output cells where the
selectioncan be expanded by knowledge of the expression
structure. This meansthat you can, for example, easily select
a whole string, or the entirecontents of a set of square
brackets by either triple-clicking insidethe string/square
brackets or by using Control-period twice when insidethe
string/square brackets.Sincerely,John
Fultzjfultz@wolfram.comUser Interface GroupWolfram Research,
Inc.
====
Suppose you have an initial value (value of assets
Vo) and you know (like in a binomial option pricing model)
that this asset value can either go up (Vo u) or down (Vo d),
where u is 1.1 (let's assume this) and d==1/u. This process
continues defining a binomial tree for the evolution of this
asset value. Now suppose that, in step 3 (M1), you have to
subtract P1 from the asset value at that moment. You have to
subtract this amount P1 to each node (in step 3 you will have
4 nodes). Then continue with the binomial tree using u and d
as defined above. I really have a problem here. My
implementation is an algorith that produces a n-step binomial
tree starting with value Vo and with given u & d parameters.
However my problem is that this algorithm subtracts P1 for
all the values of the tree after step 3. What the algorithm
its expected to do is to subtract P1 only for the nodes of
step 3 and then go on with the calculation. Ignacio
====
the
package< hi I want to solove u_{t}=u_{x} * u ^ 2 +
u_{xx} * (1-u) and u_{t} =u_{xx} + exp (-u )+ exp (-2u) by
using Method Of Line please help me. Note: Please give your
explain by using computer code or computer
algorithm(numerical)
====
> Mathematica is NOT giving wrong
answers in this case. It is assuming (non-zero) real
parameters and giving the right> answer in that case.Whether
Mathematica is or is not giving a wrong answer in the
caseresult = Integrate[ Abs[Sin[k x]]^2, {x,0,1}]; N[ result
/. k->I+1 ]depends upon whether it is or is not appropriate
for Mathematica tomake a default assumption that k is a
nonzero real. IMO, that could bedebated.But surely, as I
noted previously in this thread, Mathematica -- atleast
version 4.2 for Windows -- does give a wrong answer forresult
= Integrate[ Abs[Sin[k x]]^2, {x,0,1}, Assumptions->Element[k,
Complexes]; N[ result /. k->I+1 ]. I do not see how
theincorrectness of this can be debated (other than to say
thatMathematica should be allowed to ignore an _explicitly
stated_assumption!)David> On Wed, 22 Jan 2003 06:09:24 -0500
(EST), David W. Cantrell > Jos R Bergervoet
result = Integrate[
Abs[Sin[k x]]^2, {x,0,1}]> N[ result /. k->I+1 ]> (*
Analytical approach gives 0.261044 + 0.616283 I, WRONG!!! *)>
k=I+1; NIntegrate[ Abs[Sin[k x]^2], {x,0,1}]> (* Numerical
check gives 0.679391 *)> ...> What should I do to circumvent
such errors?> One thing that works in Mathematica (as well as
in the other> CAS) is to> Integrate[ Abs[Sin[(a+b*I) x]]^2,
{x,0,1}].> This gives (a*Sinh[2*b] - b*Sin[2*a]) / (4*a*b),>
which agrees with your result below.> But again it is wrong!
It only is correct if a and b happen> to be real quantities,
which is nowhere stated!> Well, let me state it now here:> I
was merely trying to provide a workaround which would give>
you a correct result in the event that your coefficient k
wascomplex.> (I was not trying in any way to exonerate
Mathematica!) In that> event, k may be written, _without loss
of generality_, as a+b*I,> where a and b are both REAL.> So
the main question still is: Why is Mathematica making> these
very silly errors? One could expect it from an early> version
of a product, but that is not what Mathematica 4.x> is!> Yes,
the main question remains.
====
Mathematica is NOT giving wrong
answers in this case.It is assuming (non-zero) real parameters
and giving the right answer in that case.BobbyOn Wed, 22 Jan
2003 06:09:24 -0500 (EST), David W. Cantrell > Jos R
Bergervoet result =
Integrate[ Abs[Sin[k x]]^2, {x,0,1}]> N[ result /. k->I+1 ]>
(* Analytical approach gives 0.261044 + 0.616283 I, WRONG !!!
*)> k=I+1; NIntegrate[ Abs[Sin[k x]^2], {x,0,1}]> (* Numerical
check gives 0.679391 *)> ...> What should I do to circumvent
such errors?> One thing that works in Mathematica (as well as
in the other CAS) is > to> Integrate[ Abs[Sin[(a+b*I) x]]^2,
{x,0,1}].> This gives (a*Sinh[2*b] - b*Sin[2*a]) / (4*a*b),>
which agrees with your result below.> But again it is wrong!
It only is correct if a and b happen> to be real quantities,
which is nowhere stated! Well, let me state it now here: I
was merely trying to provide a workaround which would give
you a> correct result in the event that your coefficient k was
complex. (I was> not trying in any way to exonerate
Mathematica!) In that event, k may be> written, _without loss
of generality_, as a+b*I, where a and b are both> REAL.> So
the main question still is: Why is Mathematica making> these
very silly errors? One could expect it from an early> version
of a product, but that is not what Mathematica 4.x> is! Yes,
the main question remains. David--
majort@cox-internet.comBobby R. Treat
====
Kimberly,Because
Mathematica evaluates and simplifies x^1 to x, which does not
matchyour form.x^1 // FullFormxBut this
works...MatchQ[Unevaluated[x^1], x^_Integer]TrueUnderstanding
and controlling order of evaluation in Mathematica is
probablyone of its toughest aspects. Fortunately, most of
the time things work theway you want.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/Sender:
steve@smc.vnet.netApproved: Steven M. Christensen
, Moderator
====
Does anyone have
experience of using Mathematica or any other math application
to teach a blind student?I am interested of any knowlegde no
matter level, ie grade 7 to university. My student will next
semester be in grade 7.SincerelyTruls Cronberg
====
Is this
what you did (essentially)? Flatten[ Outer[List, Range[9],
{0}, Range[0, 9], Range[0, 9], {1}], 4], _Integer]%*%{101,
201, 301}{10201, 40401, 90601}I haven't thought of a much
smarter method, as yet, but here's a far less exhaustive
search:nSquare /. List /@ Thread[c -> {1, 9}]Flatten[% /.
List /@ Thread[a -> {1, 2, 3}]]At this point, it's clear
that
b=0, c=1, a=1, 2, or 3 gives three valid answers and that b
can't be anything but 0 when a = 3, which narrows the
field
considerably.nSquare /. List /@ Thread[c -> {1, 9}]Flatten[%
/. List /@ Thread[a -> {1, 2}]]Flatten[% /. List /@ Thread[b
-> Range[9]]]Visual inspection shows no solutions among
these, which can be verified as follows.Cases[%, q_ /; 10^4 <
q < 10^5]Cases[%, q_ /; Mod[Floor[q/10^3], 10] ==
0]Alternatively, the following isn't too exhaustive,
either:nSquare /. List /@ Thread[c -> {1, 9}]Flatten[% /.
List /@ Thread[a -> {1, 2, 3}]]Flatten[% /. List /@ Thread[b
-> Range[0, 9]]]Cases[%, q_ /; 10^4 < q < 10^5]Cases[%, q_ /;
Mod[Floor[q/10^3], 10] == 0]BobbyOn Thu, 23 Jan 2003 08:04:38
-0500 (EST), David Park second> digit and 1 in the last
digit. (i.e., x0xx1 where x is a digit 0..9). We know how to
solve the problem by testing a list of all possible >
candidate> numbers. But is there a method that uses Solve and
modular arithimetic or> some other clever method? If one looks
at the answers one would think > there> must be. David Park>
djmp@earthlink.net> http://home.earthlink.net/~djmp/ --
majort@cox-internet.comBobby R. Treat
====
David: If n squared
yields a five-digit answer, n must be 100 orgreater but not
more than 316. Only n's ending in 1 or 9 will produce
asquare
ending in 1. Join[Range[101,316,10],
Range[109,316,10]]generates all (43) such n's. It is simple
to select from those the few(3) that yield 0 as the second
digit in their squares:
Select[,IntegerDigits[#^2][[2]]== 0 &] They are: 101,
201, and 301. Givenhowfew candidates there are to test (43),
I wonder why this approach, eventhough it involves testing a
list of all possible candidate numbers,isn't the fastest and
easiest route. Best, Harvey-----Original Message-----orsome
other clever method? If one looks at the answers one would
thinktheremust be.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/
Reply-To: no spam please
====
i'm looking
for a function to plot a cone with a interpolated
gouraudshading in Mathematica.I tried with another system,
but theresult is awful, because the circle of the cone is
squared.thanks.
====
With Mathematica 4.1 on
Windows98:N[Log[8]/Log[2]]3.Floor[N[Log[8]/Log[2]]]2Beware!==
==How does one determine whether an instance of the
Mathematica Kernal isrunning via Mathlink , or even Visual
Basic. I only want to keep oneconnect to this instance not
start up a new instance. All I can think of iskeep the MLINK
variable as a global and check that it does not equal
zero.that will connect to this.--Daniel Heneghan
====
Here's a
useful plot:f[x_] := 5(x - 1500 Ceiling[x/1760])plot1 =
Block[{$DisplayFunction = Identity}, Plot[f@x, {x, -2000,
15000}]];Show[plot1, Graphics@{ AbsolutePointSize@5,
Point[{#, f@#}] & /@ Range[0, 9000, 1500]} ];Bobby> You
missed two roots (0 and 1500). Here's the case Mod[x, 1760]
>
0: f[x_] := 5(x - 1500 Ceiling[x/1760])> xRule = x -> 1760 k +
y;> Simplify[f@x /. xRule, {0 < y < 1760, k
.89ÌÌ
Integers}]> % /. Ceiling[a_] -> 1> yRule = First@Solve[% ==
0, y]> Flatten[k /. Solve[y == # /. yRule, k] & /@ {1,
1759}]> kValues = Range[Ceiling@Min@%, Floor@Max@%]> yValues
= y /. (yRule /. List /@ Thread[k -> kValues])> 1760kValues +
yValues> f /@ % and here's the case Mod[x, 1760] == 0:>>
xRule
= x -> 1760 k + y;> Simplify[5(x - 1500 Ceiling[x/1760]) //.
{xRule, y -> 0}, {k .89ÌÌ Integers}]> yRule =
First@Solve[%
== 0, k]> 1760k /. %> f@% Bobby On Thu, 23 Jan 2003 08:05:15
-0500 (EST), Orestis Vantzos > Your function simplifies to:>
f[x_]:=5(x - 1500 Ceiling[x/1760])> Now assume that [First
Case] x==1760 k + y , 0 Then
Ceiling[x/1760]== Ceiling[k + y/1760]== k+1> so that f[x]==5
(-1500 + 260 k + y)> If f[x]==0 then 5 (-1500 + 260 k + y)==0
and we solve for y:> y== 1500-260 k> 0 0< 1500-260k
<=1760> -1500<-260 k <= 260> 5.77 > k >= 1> So k ranges from 1
to 5 and since x==1500(k+1) the roots are:> Table[1500(k +
1),{k,1,5}]> {3000, 4500, 6000, 7500, 9000}> Orestis Vantzos>
we are searching all solutions where the function f results
null.> f[x_] := -7500 * Ceiling[(0.5 * x) / 880] + (5 * x)>
Solve[{f[x] == 0}, x]> As f contains jump discontinuities, we
recieved the following error:> InverseFunction::ifun:
Inverse functions are being used. Values may > be > lost
for multivalued inverses.> Solve::tdep: The equations
appear to involve the variables to be > solved > for in an
essentially non-algebraic way.> We would be pleased if
anybody could help us. -- majort@cox-internet.comBobby R.
Treat
====
You missed two roots (0 and 1500). Here's the case
Mod[x, 1760] > 0:f[x_] := 5(x - 1500 Ceiling[x/1760])xRule =
x -> 1760 k + y;Simplify[f@x /. xRule, {0 < y < 1760, k
.89ÌÌ Integers}]% /. Ceiling[a_] -> 1yRule =
First@Solve[%
== 0, y]Flatten[k /. Solve[y == # /. yRule, k] & /@ {1,
1759}]kValues = Range[Ceiling@Min@%, Floor@Max@%]yValues = y
/. (yRule /. List /@ Thread[k -> kValues])1760kValues +
yValuesf /@ %and here's the case Mod[x, 1760] == 0:xRule = x
-> 1760 k + y;Simplify[5(x - 1500 Ceiling[x/1760]) //.
{xRule, y -> 0}, {k .89ÌÌ Integers}]yRule =
First@Solve[%
== 0, k]1760k /. %f@%BobbyOn Thu, 23 Jan 2003 08:05:15 -0500
(EST), Orestis Vantzos > Your function simplifies to:>
f[x_]:=5(x - 1500 Ceiling[x/1760]) Now assume that [First
Case] x==1760 k + y , 0 Then
Ceiling[x/1760]== Ceiling[k + y/1760]== k+1> so that f[x]==5
(-1500 + 260 k + y) If f[x]==0 then 5 (-1500 + 260 k + y)==0
and we solve for y:> y== 1500-260 k 0 -1500<-260 k <= 260> 5.77 > k >= 1 So k ranges from 1
to 5 and since x==1500(k+1) the roots are:> Table[1500(k +
1),{k,1,5}]> {3000, 4500, 6000, 7500, 9000} Orestis Vantzos>
we are searching all solutions where the function f results
null.> f[x_] := -7500 * Ceiling[(0.5 * x) / 880] + (5 * x)>
Solve[{f[x] == 0}, x]> As f contains jump discontinuities, we
recieved the following error:> InverseFunction::ifun:
Inverse functions are being used. Values may > be > lost
for multivalued inverses.> Solve::tdep: The equations
appear to involve the variables to be > solved > for in an
essentially non-algebraic way.> We would be pleased if
anybody could help us.> Philipp Burkert> Carsten Siegmund--
majort@cox-internet.comBobby R. Treat
====
we are searching
all solutions where the function f results null. f[x_] :=
-7500 * Ceiling[(0.5 * x) / 880] + (5 * x)> Solve[{f[x] ==
0}, x]Sometimes, it is helpful to try plotting the function
on a macro scale andthen to find regions of interest.Once
regions of interest are found, one can then use other methods
to locatethe roots.In your problem, do:Plot[f[x], {x, -4000,
15000}]and see where the area of interest is.Can you now
define the region?What other commands are available to you
when you can define regions?Note: I haven't
looked at other
root finding methods to know if this can besolved, but there
probably is something in Mathematica, just some
recommendationswhen all seems lost to canned methods.HTH,
Flip
====
Your function simplifies to:f[x_]:=5(x - 1500
Ceiling[x/1760])Now assume that [First Case] x==1760 k + y ,
00< 1500-260k <=1760-1500<-260 k <= 2605.77 > k >=
1So k ranges from 1 to 5 and since x==1500(k+1) the roots
are:Table[1500(k + 1),{k,1,5}]{3000, 4500, 6000, 7500,
9000}Orestis Vantzos we are searching all solutions where the
function f results null. f[x_] := -7500 * Ceiling[(0.5 * x) /
880] + (5 * x)> Solve[{f[x] == 0}, x] As f contains jump
discontinuities, we recieved the following error:
InverseFunction::ifun: Inverse functions are being used.
Values may be > lost for multivalued inverses.
Solve::tdep: The equations appear to involve the variables
to be solved > for in an essentially non-algebraic way. We
would be pleased if anybody could help us. Philipp Burkert>
Carsten Siegmund
====
> we are searching all solutions where
the function f results null. f[x_] := -7500 * Ceiling[(0.5 *
x) / 880] + (5 * x)> Solve[{f[x] == 0}, x] As f contains jump
discontinuities, we recieved the following error:
InverseFunction::ifun: Inverse functions are being used.
Values may be> lost for multivalued inverses.
Solve::tdep: The equations appear to involve the variables
to be> solved for in an essentially non-algebraic way. We
would be pleased if anybody could help us.Your equation has 7
roots. To see them, let me suggest that you firstPlot[f[x],
{-2000, 15000}]. (BTW, please ignore the spurious
verticalsegments!) As you can then guess from the graph, the
solutions are 0,1500, 3000, 4500, 6000, 7500, and 9000.You
discovered that Solve will not help you. However, FindRoot
can be used.For example, try FindRoot[f[x]==0, {x, 7000,
8000}]. It does find a root,but not the one expected! (It
gives the root 9000, rather than 7500.) Soeven with FindRoot,
you must be savvy, not depending on Mathematica
tooheavily.David
====
It's a SAMPLE variance, so almost
anything can happen ONCE.I'm not getting sample variances
anything like that, though, so there could be a bug in your
version of Mathematica. I'm using version 4.2 with
WinXP.Also, you may as well look at the simulated covariance
matrix, CovarianceMatrix@Transpose@ET, rather than just the
variance.BobbyOn Thu, 23 Jan 2003 08:05:54 -0500 (EST),
KyriakosChourdakis > Dear all, I am trying to simulate a
multivariate t-distribution,> and encounter the following
problem: (* First load the packages *)>
<
< df = 20; (* the degrees of freedom - Since the
number> is large, it should make the t similar to a normal *)
(* Create a random variance-covariance matrix *)> rS0 =
RandomArray[UniformDistribution[-0.30,> 0.80],{NS, NS}];> rS0
= rS0.Transpose[rS0];> rS0 = (rS0/Max[rS0]) (1 -
IdentityMatrix[NS]) + IdentityMatrix[NS]; (* the var-covar
matrix is *)> rS0> {{1, -0.290755},> {-0.290755, 1 }} (*
create 10000 draws of 2 dimensional vectors *)> ET =>
Transpose[RandomArray[MultivariateTDistribution[rS0,> df],
10000]]; (* the theoretical cov matrix is ok, with>
var=df/[df-2]... *)> ST =
N[CovarianceMatrix[MultivariateTDistribution[rS0,> df]]]>
{{1.11111, -0.323061}, {-0.323061, 1.11111}} (* ... But the
simulated variance is not*)> Mean /@ ET> Variance /@ ET>
{-0.00426186, 0.00824351}> {0.067534, 0.0670659} (* On the
other hand, if I draw from the corresponding> multinormal *)>
EN = Transpose[RandomArray[MultinormalDistribution[{0,> 0},
ST], 10000]]; (* Everything is OK *)> Mean /@ EN> Variance /@
EN> {0.0196956, -0.00675908} (* Should be 0 *)> {1.1362,
1.1099} (* Should be 1.111 *) I cannot understand why the
simulated variance is> 0.067 instead of 1.111/[20-2].
Alternatively, why> should I multiply all values with
sqrt(1.111/0.067)=4. Best, Kyriakos>
__________________________________________________> Do You
Yahoo!?> Everything you'll ever need on one web page>
http://uk.my.yahoo.com-- majort@cox-internet.comBobby R.
Treat
====
Even when the ReturnType is Manual, your C function
needs to send something. It can really be anything, as long
as it's a complete expression. For example, you could do/*
send Print[Done] to kernel */MLPutFunction(stdlink,
Print, 1);MLPutString(stdlink, Done);Now SetValue will
print Done. But since you probably don't want anything
from SetValue, you should send the symbol
Null.MLPutSymbol(stdlink, Null);Null is an output that's
automatically suppressed, so you won't see an output
cell.>How do I wrap and call a C subroutine that should
return nothing back to>Mathematica. I want to somehow specify
a :ReturnType: of None or Void,>but mprep doesn't
recognize those tokens. If I specify Manual then
my>Mathematica session just hangs waiting for an answer from
my MathLink>program. Does MathLink have a function like
MLPutEmptyPacket() that just>tells Mathematica Expect
nothing from this MathLink function. It's OK to>stop waiting
and return to your command loop.This is (part of) the
template file I am using:>:Begin:>:Function:
SetValue>:Pattern: SetValue[ value_?NumericQ ]>:Arguments:
{value}>:ArgumentTypes: {Real}>:ReturnType: Manual>:End:void
SetValue ( double value )>{> /*> * Do something useful with
value.> * It doesn't make sense to return anything.> */ /*
What goes here to keep Mathematica from hanging? */>}>How do
I specify a true MathLink subroutine that doesn't return a
value
to>----------------------------------------------------------
----Omega ConsultingThe final answer to your Mathematica
needshttp://omegaconsultinggroup.com
====
What's the
best/simplest way to encode or format a Mathematica notebook
to make it available on a web site? (as a downloadable
source file only, sufficiently brief that
file compression is
not needed, nothing live or intended for online execution,
no need for Output cells or graphics outputs in the online
file, but intended to be dowbnloaded and executed by users on
multiple platforms, with header and text cells kept distinct
from Input cells)-- 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)
====
How do you enter
the partial derivative shown at:www.previze.com/
partialderivative.gif
====
You might want to use
VarianceTest to give a measure of how far off the sample
variance is. Without something like this, the sample
variance is wrong doesn't mean anything.NS = 2;df = 20; rS0
= RandomArray[UniformDistribution[-0.30, 0.80], {NS, NS}];rS0
= rS0.Transpose[rS0];rS0 = (rS0/Max[rS0]) (1 -
IdentityMatrix[NS]) + IdentityMatrix[NS];rS0ET =
Transpose[RandomArray[MultivariateTDistribution[rS0, df],
10000]];ST =
N[CovarianceMatrix[MultivariateTDistribution[rS0,
df]]]VarianceTest[Transpose@ET, 20/(20 - 2), TwoSided ->
True, FullReport -> True]{{1, 0.0799201}, {0.0799201,
1}}{{1.11111, 0.0888001}, {0.0888001, 1.11111}}{FullReport ->
TableForm[ {{{1.0781668250596432, 1.083761196394562},
{9702.531075394236, 9752.875382474302},
ChiSquareDistribution[ 9999]}}, TableHeadings -> {None,
{Variance, TestStat, Distribution}}], TwoSidedPValue ->
{0.0346164820576073098234084515`10.8823,
0.0800810808224253545167676788`10.8566}}Bobby> Dr. Bob, It
is true that anything can happen once, but this is> not the
case since it is happening consistently. In any way, when a
sample of 10000 is drawn and the> degrees of freedom is so
high [making all first> moments to exist and implying near
normality], one> should not be so unlucky and get both
variances wrong. I am starting to think about the bug
scenario...> I am using version 4.0 on WinME. If anyone with>
similar specs could try it I would be grateful. To find a path
around the problem, I could try to use> univariate t-random
numbers [which work fine] to> create the multivariate series,
but I am not sure how.> Is the standard X = rho Y +
sqrt(1-rho^2) W trick working when Y and W are
[standardized]> t-distributed with df degrees of freedom ? My
guess is> not since t=sqrt(Z1^2+...+Zdf^2). What would be the>
correct way? Best, Kyriakos It's a SAMPLE variance, so
almost
anything can happen> ONCE. I'm not getting sample variances
anything like that,> though, so there could> be a bug in your
version of Mathematica. I'm using> version 4.2 with WinXP.
Also, you may as well look at the simulated covariance>
matrix,> CovarianceMatrix@Transpose@ET, rather than just the>
variance. Bobby On Thu, 23 Jan 2003 08:05:54 -0500 (EST),>
KyriakosChourdakis> Dear all,> I am trying to simulate a
multivariate> t-distribution,> and encounter the following
problem:> (* First load the packages *)>
<
< NS = 2; (* the number of
series *)> df = 20; (* the degrees of freedom - Since the>
number> is large, it should make the t similar to a normal>
*)> (* Create a random variance-covariance matrix *)> rS0 =
RandomArray[UniformDistribution[-0.30,> 0.80],{NS, NS}];> rS0
= rS0.Transpose[rS0];> rS0 = (rS0/Max[rS0]) (1 -
IdentityMatrix[NS]) +> IdentityMatrix[NS];> (* the var-covar
matrix is *)> rS0> {{1, -0.290755},> {-0.290755, 1 }}> (*
create 10000 draws of 2 dimensional vectors *)> ET =>
Transpose[RandomArray[MultivariateTDistribution[rS0,> df],
10000]];> (* the theoretical cov matrix is ok, with>
var=df/[df-2]... *)> ST =>
N[CovarianceMatrix[MultivariateTDistribution[rS0,> df]]]>
{{1.11111, -0.323061}, {-0.323061, 1.11111}}> (* ... But the
simulated variance is not*)> Mean /@ ET> Variance /@ ET>
{-0.00426186, 0.00824351}> {0.067534, 0.0670659}> (* On the
other hand, if I draw from the> corresponding> multinormal
*)> EN => Transpose[RandomArray[MultinormalDistribution[{0,>
0}, ST], 10000]];> (* Everything is OK *)> Mean /@ EN>
Variance /@ EN> {0.0196956, -0.00675908} (* Should be 0 *)>
{1.1362, 1.1099} (* Should be 1.111 *)> I cannot understand
why the simulated variance is> 0.067 instead of 1.111/[20-2].
Alternatively, why> should I multiply all values with>
sqrt(1.111/0.067)=4.> Best,> Kyriakos>
__________________________________________________> Do You
Yahoo!?> Everything you'll ever need on one web page>
http://uk.my.yahoo.com>> --> majort@cox-internet.com> Bobby
R. Treat __________________________________________________>
Do You Yahoo!?> Everything you'll ever need on one web page>
http://uk.my.yahoo.com>-- majort@cox-internet.comBobby R.
TreatReply-To: k.chourdakis@qmul.ac.uk
====
Dr. Bob,It is true
that anything can happen once, but this isnot the case since
it is happening consistently.In any way, when a sample of
10000 is drawn and thedegrees of freedom is so high [making
all firstmoments to exist and implying near normality],
oneshould not be so unlucky and get both variances wrong.I am
starting to think about the bug scenario...I am using version
4.0 on WinME. If anyone withsimilar specs could try it I
would be grateful.To find a path around the problem, I could
try to useunivariate t-random numbers [which work fine]
tocreate the multivariate series, but I am not sure how.Is
the standardX = rho Y + sqrt(1-rho^2) Wtrick working when Y
and W are [standardized]t-distributed with df degrees of
freedom ? My guess isnot since t=sqrt(Z1^2+...+Zdf^2). What
would be thecorrect way?Best,KyriakosIt's a SAMPLE variance,
so almost anything can happenONCE.I'm not getting sample
variances anything like that,though, so there couldbe a bug
in your version of Mathematica. I'm usingversion 4.2 with
WinXP.Also, you may as well look at the simulated
covariancematrix,CovarianceMatrix@Transpose@ET, rather than
just thevariance.BobbyOn Thu, 23 Jan 2003 08:05:54 -0500
(EST),KyriakosChourdakis> Dear all, I am trying to simulate a
multivariatet-distribution,> and encounter the following
problem: (* First load the packages *)>
<
< df = 20; (* the degrees of freedom - Since
thenumber> is large, it should make the t similar to a
normal*) (* Create a random variance-covariance matrix *)>
rS0 = RandomArray[UniformDistribution[-0.30,> 0.80],{NS,
NS}];> rS0 = rS0.Transpose[rS0];> rS0 = (rS0/Max[rS0]) (1 -
IdentityMatrix[NS]) +IdentityMatrix[NS]; (* the var-covar
matrix is *)> rS0> {{1, -0.290755},> {-0.290755, 1 }} (*
create 10000 draws of 2 dimensional vectors *)> ET =>
Transpose[RandomArray[MultivariateTDistribution[rS0,> df],
10000]]; (* the theoretical cov matrix is ok, with>
var=df/[df-2]... *)> ST
=N[CovarianceMatrix[MultivariateTDistribution[rS0,> df]]]>
{{1.11111, -0.323061}, {-0.323061, 1.11111}} (* ... But the
simulated variance is not*)> Mean /@ ET> Variance /@ ET>
{-0.00426186, 0.00824351}> {0.067534, 0.0670659} (* On the
other hand, if I draw from thecorresponding> multinormal *)>
EN =Transpose[RandomArray[MultinormalDistribution[{0,> 0},
ST], 10000]]; (* Everything is OK *)> Mean /@ EN> Variance /@
EN> {0.0196956, -0.00675908} (* Should be 0 *)> {1.1362,
1.1099} (* Should be 1.111 *) I cannot understand why the
simulated variance is> 0.067 instead of 1.111/[20-2].
Alternatively, why> should I multiply all values
withsqrt(1.111/0.067)=4. Best, Kyriakos>
__________________________________________________> Do You
Yahoo!?> Everything you'll ever need on one web page>
http://uk.my.yahoo.com--majort@cox-internet.comBobby R.
Treat__________________________________________________Do You
Yahoo!?Everything you'll ever need on one web
pagehttp://uk.my.yahoo.com
====
I am trying to minimize a
nonlinear function of 15 variables using FindMinimum. The
function is basically a CHIsquare regression for a problem I
am studying that looks something like
this:CHIsquare=Abs[Sum(Sum(A_m * x_m * E^(-i * B_n *
y_m)))]^2 m nwhere: Sums go over all m's (# of independent
variables,15 of them) and n's (# of my data points)
respectively x's and y's are the variables
w.r.t which I want
to minimize the function <-- I must assign them COMPLEX
starting values A's and B's are COMPLEX
valued
constantsBecasue of Abs and ^2, my CHIsquare will always be
REAL but when I try using all this, it does not work - I get
a message: Starting value is not a real numberfor the first
variable to which I gave complex starting value.Is there some
way to minimize my function that allows for complex values of
variables?VesnaReply-To: k.chourdakis@qmul.ac.uk
====
A
variance test gives p-values of order 10^(-4000)which
apparently reject the null at all confidencelevels. I am
pretty certain it is bug too.K.<<
====
========>You might want
to use VarianceTest to give a measureof how far off the
sample variance is. Withoutsomething like this, the sample
variance is wrongdoesn't mean
anything.__________________________________________________Do
You Yahoo!?Everything you'll ever need on one web
pagehttp://uk.my.yahoo.com
====
I have the following problem in
a do loop. I want to repeat a procedure that involves many
functions many times, but I need to store the values of the
functions in a matrix after each iteration.Dimitirs
Psychoyios
====
I have a function f[a_,b_] defined some way. i
want to fix one argument,and then generate an array of values
where the other argument varies. Howcan I do this?For
example, I want:{f[1, 4], f[2, 4], f[3, 4], f[4, 4]}Is there
an easy way to do this?Cancel-Lock:
sha1:clUpO7TgOxEfmU//aL/sajnEP6s
====
=> I have a function
f[a_,b_] defined some way. i want to fix one
argument,> and
then generate an array of values where the other argument
varies. > For example, I want: {f[1, 4], f[2, 4], f[3, 4],
f[4, 4]}Table[ f[i,4], {i,4} ]-Nevin
====
may be there is a
simpler waythe following workesFor[(i = 0; l = {}), i < 5,
(i++; l = {Flatten[l], f[i, 4]})]; l = Flatten[l]Zachary
Turner schrieb:> I have a function f[a_,b_] defined some way.
i want to fix one argument,> and then generate an array of
values where the other argument varies. How> can I do this?
For example, I want: {f[1, 4], f[2, 4], f[3, 4], f[4, 4]} Is
there an easy way to do this? >
====
The following will do
it:f[#, b]&/@a,where a is an array of values and b is a fixed
value.Kezhao> I have a function f[a_,b_] defined some way. i
want to fix one argument,> and then generate an array of
values where the other argument varies. How> can I do this?
For example, I want: {f[1, 4], f[2, 4], f[3, 4], f[4, 4]}>
====
> I have a function f[a_,b_] defined some way. i want to
fix one argument,> and then generate an array of values where
the other argument varies. How> can I do this? For example,
I want: {f[1, 4], f[2, 4], f[3, 4], f[4, 4]} Is there an easy
way to do this?>Zachary, Table[f[x,4],{x,1,4}] {f[1, 4], f[2,
4], f[3, 4], f[4, 4]}Or, for irregular values or a known list
of values: f[x,#]&/@{ 1, 2.3, 8,-3+I, Pi,a} {f[x, 1], f[x,
2.3], f[x, 8], f[x, -3 + I], f[x, Pi], f[x,
a]}--Allan---------------------Allan HayesMathematica
Training and ConsultingLeicester
UKwww.haystack.demon.co.ukhay@haystack.demon.co.ukVoice: +44
(0)116 271 4198
====
> I have a function f[a_,b_] defined some
way. i want to fix oneargument,> and then generate an array of
values where the other argument varies.How> can I do this?>
For example, I want:> {f[1, 4], f[2, 4], f[3, 4], f[4, 4]}>
Is there an easy way to do this?> Zachary,
Table[f[x,4],{x,1,4}] {f[1, 4], f[2, 4], f[3, 4], f[4, 4]}
Or, for irregular values or a known list of values:
f[x,#]&/@{ 1, 2.3, 8,-3+I, Pi,a} {f[x, 1], f[x, 2.3], f[x,
8], f[x, -3 + I], f[x, Pi],> f[x, a]}Good lord. I didn't
expect this many responses :) Anyway, the Table methodseems
the simplest. Now I extend on this a little bit. Is there any
way Ican apply FullSimplify to the result before it gets
inserted into the array?
====
Perhaps Fold, FoldList, Nest,
NestList and NestWhileList may help.Yas I have the following
problem in a do loop. I want to repeat a procedure> that
involves many functions many times, but I need to store the
values> of the functions in a matrix after each iteration.>
Dimitirs Psychoyios
====
Zachary: Table[f[i,4],{i,4}] Best,
HarveyHarvey P. DaleUniversity Professor of Philanthropy and
the LawDirector, National Center on Philanthropy and the
LawNew York University School of LawRoom 206A110 West 3rd
StreetNew York, N.Y. 10012-1074-----Original
Message-----
====
Zachary,First, don't overlook the humble
Table statement.Table[f[x, 4], {x, 1, 4}]{f[1, 4], f[2, 4],
f[3, 4], f[4, 4]}Other methods to generate evenly spaced
arguments...Array[f[#, 4] &, 4]{f[1, 4], f[2, 4], f[3, 4],
f[4, 4]}f[#, 4] & /@ Range[4]{f[1, 4], f[2, 4], f[3, 4], f[4,
4]}For an oddball list of first arguments.f[#, 4] & /@ {1, 3,
4, 7, x}{f[1, 4], f[3, 4], f[4, 4], f[7, 4], f[x,
4]}MapThread[f[#, 4] &, {{1, 3, 4, 7, x}}]{f[1, 4], f[3, 4],
f[4, 4], f[7, 4], f[x, 4]}Or to generate an oddball values
for both arguments.MapThread[f[#1, #2] &, {{1, 3, 4, 7, x},
{2, 3, 4, 9, x}}]{f[1, 2], f[3, 3], f[4, 4], f[7, 9], f[x,
x]}David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/
====
It's not rejected at ALL confidence levels ---
just the
ones with alpha > 10^(-4000).Nevertheless, that's good
evidence of a bug.BobbyOn Fri, 24 Jan 2003 05:08:06 -0500
(EST), Kyriakos Chourdakis > A variance test gives p-values
of order 10^(-4000)> which apparently reject the null at all
confidence> levels. I am pretty certain it is bug too. K.
<<
====
======== You might want to use VarianceTest to give a
measure> of how far off the sample variance is. Without>
something like this, the sample variance is wrong> doesn't
mean anything.
__________________________________________________> Do You
Yahoo!?> Everything you'll ever need on one web page>
http://uk.my.yahoo.com-- majort@cox-internet.comBobby R.
Treat
====
I have substituted r for rho. The greek letter rho
may be entered as esc-r-esc or [rho] .The expression below
will compute (giving zero unless you define z in terms of r
elsewhere).Abs[D[z,{r,2}]/(1+D[z,r]^2)^(3/2)]If it is a
question of typesetting the formula, you can enter the above
expression in a text cell, select the cell bracket and follow
the menu path Cell > Convert To > TraditionalForm.The result
below is the Expression version.!(*FormBox[
RowBox[{[LeftBracketingBar], FractionBox[
FractionBox[(=A1=D3^2 z), (=A1=D3r^2),
MultilineFunction->None], SuperscriptBox[ RowBox[{(,
RowBox[{ SuperscriptBox[ RowBox[{(,
FractionBox[(=A1=D3z), (=A1=D3r),
MultilineFunction->None], )}], 2], +, 1}], )}],
(3/2)]], [RightBracketingBar] }],
TraditionalForm])> How do you enter the partial derivative
shown at:
HREF=www.previze.com/partialdervative.gif>www.previze.com/>
partialderivative.gifGarry HelzerDepartment of
MathematicsUniversity of Maryland1303 Math BldgCollege Park,
MD 20742-4015
====
Andr.8e:Realize
thatIn[1]:=FullForm[N[Log[8]/Log[2]]]Out[1]//FullForm=
2.9999999999999996`Under these conditions, and keeping in
mind thatIn[2]:=?FloorIn[2]:=Floor[x] gives the greatest
integer less than or equal to x.It is obvious
thatIn[3]:=Floor[N[Log[8]/Log[2]]]Out[3]=2In[4]:=Ceiling[N[
Log[8]/Log[2]]]Out[4]=3Greetings,Germ.87n Buitrago-----
Original Message ----->
====
> With Mathematica 4.1 on
Windows98:> N[Log[8]/Log[2]]> 3.> Floor[N[Log[8]/Log[2]]]> 2>
Beware!I don't find that surprising. After all,
you asked for a
numericalapproximation first!Several comments (based on using
version 4.2 for Windows):(1) N[Log[8]/Log[2]] has FullForm
2.9999999999999996`, which, being lessthan 3, must have its
Floor equal to 2 of course.(2) Floor[Log[8]/Log[2]] returns
unevaluated, accompanied by a complaintabout uncertainty,
while Floor[Log[2, 8]] yields 3, just as desire.(3)
N[Floor[Log[8]/Log[2]]] evaluates to 2., but again
accompanied by acomplaint about uncertainty. (Such complaints
are valuable.)(4) FullSimplify[Floor[Log[8]/Log[2]]] evaluates
to 3, just as we desire.(5) It's simply a bad idea to ask
for
numerical approximations when youare dealing with something
in the vicinity of a jump discontinuity.(Indeed, I think it
could be argued that things like Floor[3.] shouldactually be
Indeterminate in Mathematica.)David
Cantrell
====
>-----Original Message----->Sent: Friday, January
24, 2003 11:06 AM>To: mathgroup@smc.vnet.net>With Mathematica
4.1 on
Windows98:>N[Log[8]/Log[2]]>3.>Floor[N[Log[8]/Log[2]]]>2>
Beware! ,that's not a problem of Mathematica, but we have to
pose the right question:In[75]:=
FullSimplify[Log[8]/Log[2]]Out[75]= 3The exact answer!You
might be interested to observe (e.g. set n = 1000):Length /@
Split@Table[Floor@(Divide @@ Log@SetPrecision[{8, 2}, p]),
{p, n}]Or else, look at:In[92]:= Divide @@
Log@SetPrecision[{8, 2}, 60] // InputFormOut[92]//InputForm=
2.99999999999999999999999999999999999999999999999999999
9999999999999740815`59.7159In[93]:= Divide @@
Log@SetPrecision[{8, 2}, 61] // InputFormOut[93]//InputForm=
3.00000000000000000000000000000000000000000000000000000
000000000000000000609051`60.7159We shouldn't expect
something
much different from this.--Hartmut
====
There are a lot of ways
!Here one way for your example :In[18]:=MapThread[f,
Transpose[({#1, 4} & )/@Range[4]]]Out[18]={f[1, 4], f[2, 4],
f[3, 4], f[4, 4]}Or, more generally :In[21]:=MapThread[f,
Transpose[({#1, fix} & ) /@ {a, b, c, d, e}]]Out[21]={f[a,
fix], f[b, fix], f[c, fix], f[d,
fix], f[e, fix]}Meilleures
salutationsFlorian Jaccard-----Message
d'origine-----Envoy.8e : ven., 24. janvier 2003
11:07è :
mathgroup@smc.vnet.netObjet : Functional programmingI have a
function f[a_,b_] defined some way. i want to fix
one
argument,and then generate an array of values where the other
argument varies. Howcan I do this?For example, I want:{f[1,
4], f[2, 4], f[3, 4], f[4, 4]}Is there an easy way to do
this?Reply-To: Diana
====
All,I know that I can
generate the divisors of any integer with the
Divisorscommand. I would like to start with x = 10, for
example, and generate thedivisors of x, and then determine
the sum of the divisors. I would thenlike to increment x, up
to 100, for
example.Diana--
====
==========================================
====
===God made the integers, all else is the work of
man.L. Kronecker, Jahresber. DMV 2, S. 19.
====
> All, I know
that I can generate the divisors of any integer with the
Divisors> command. I would like to start with x = 10,
for example, and generatethe> divisors of x, and then
determine the sum of the divisors. I would then> like to
increment x, up to 100, for example.> Diana -->
====
=================================================> God
made the integers, all else is the work of man.> L.
Kronecker, Jahresber. DMV 2, S. 19.Some tools that may be
useful:Diana,Plus@@Divisors[#]&/@Range[10,100]{
18,12,28,14,24,24,31,18,39,20,42,32,36,24,60,31,42,40,56,30,72
,32,63,48,54,
48,91,38,60,56,90,42,96,44,84,78,72,48,124,57,93,72,98,54,120,
72,120,80,90,60,
168,62,96,104,127,84,144,68,126,96,144,72,195,74,114,124,140,9
6,168,80,186,
121,126,84,224,108,132,120,180,90,234,112,168,128,144,120,252,
98,171,156,217}Rest[NestWhileList[Plus@@Divisors[1+#]&,9,#<=
100&,1,
Infinity,-1]]{18,20,32,48,57,90}Allan---------------------
Allan HayesMathematica Training and ConsultingLeicester
UKwww.haystack.demon.co.ukhay@haystack.demon.co.ukVoice: +44
(0)116 271 4198
====
{#1, #1^2, #1^3} &@Range[1,manytimes]
might be an example of how toapproach this. Or NestList often
is useful for this sort of operation. I have the following
problem in a do loop. I want to repeat a procedure> that
involves many functions many times, but I need to store the
values> of the functions in a matrix after each iteration. >
Dimitirs Psychoyios
====
I have the following problem in a do
loop. I want to repeat a procedure> that involves many
functions many times, but I need to store the values> of the
functions in a matrix after each iteration.> Dimitirs
PsychoyiosDimitris,Someway you need to include the
instruction to modify the matrix and assignit to a a symbol-
here is a simple illustration: lst={}; Do[lst=
Append[lst,n^2],{n,5}];lst
{1,4,9,16,25}--Allan---------------------Allan
HayesMathematica Training and ConsultingLeicester
UKwww.haystack.demon.co.ukhay@haystack.demon.co.ukVoice: +44
(0)116 271 4198
====
I do not understand your problem fully,
but I think you may use theiteration construct For instead of
Do. For[iter=start,.......and use a matrix e.g. with name:
mtr[iter], where iter is the value, whichchanges in the
loop.Hermann Schmitt----- Original Message ----->
====
Wow. But
apparently it has nothing to do with Log.
Look:Floor[3.0000000000000000] 3Floor[3.00000000000000000]
2---Selwyn Hollis> With Mathematica 4.1 on Windows98:>
N[Log[8]/Log[2]]> 3.> Floor[N[Log[8]/Log[2]]]> 2>
Beware!
====
>I have a function f[a_,b_] defined some way. i
want to fix one>argument, and then generate an array of values
where the other>argument varies. How can I do this?>For
example, I want:>{f[1, 4], f[2, 4], f[3, 4], f[4, 4]}>Is
there an easy way to do this?Try f[#,4]&/@Range[1,4]
====
>With
Mathematica 4.1 on Windows98:>N[Log[8]/Log[2]]
>3.>Floor[N[Log[8]/Log[2]]] >2This occurs since the
N[Log[8]/Log[2]] gives you a machine precision number that
is slightly less than 3With Mathematica 4.2 on a TiBook
running MacOS 10.2.3Sign[N[Log[8]/Log[2]] - 3]-1A better way
to express Log[8]/Log[2] would be Log[2,8] which gives an
exact result of 3
====
If I want to create a matrix A mod 2,
which is a matrix of 0's and 1's, howdo I do
that?So, if I
want all 2x2 matrices, A mod 2, there would be 2^4 = 16
suchmatrices consisting only of 0's and
1's.These will
be:{{{0, 0}, {0, 0}}, {{1, 0}, {0, 0}}, ..., {{1, 1}, {1,
1}}}I looked at permutations, discrete and combinatoria, but
didn't see anythingobvoius (but I just missed it
probably).Laslty, I'd like to be to do any size matrix (that
is, 2x2, 3x3, where it isuser selectable) and any mod
n.Flip
====
If I want to create a matrix A mod 2, which is a
matrix of 0's and 1's, how> do I do that?
Flip> For a matrix
of order n first create a list of 0s and 1s with
IntegerDigits[k,2,n^2].This operation represent the integer k
as an n^2 bit number and then separates the bits into a length
n^2 list.Next divide the list into n sublists, voila, an nxn
matrix. By letting krange from 0 to 2^(n^2) - 1 you can
generate all binary matrices ofthat order.WeshReply-To:
Matthew Senn
====
i'm relatively new
using mathematica 4.2. the other day, i came across
whatseemed to be a simple problem, but i couldn't
figure out
how to do it.say i was plotting y=x^2, and i wanted to shade
the region under the graphonly between x=2 to x=4 over a
range of {x, 0, 5}, is there a way to dothis? i tried
FilledPlot, but it filled the entire area under the
graph.thank you!
====
like
this:FilledRange[fun_,{x_,x1_,x2_},{f1_,f2_}]:=
Module[{p1,p2}, Block[{$DisplayFunction=Identity},
p1=Plot[fun,{x,x1,x2}]; p2=FilledPlot[fun,{x,f1,f2}]; ];
Show[p1,p2] ] Your example isFilledRange[x^2,{x,0,5},{2,4}]
And my example isFilledRange[Cos[x] Exp[-.1
x],{x,0,20},{4,14}] That helps? Borut> i'm relatively new
using mathematica 4.2. the other day, i came acrosswhat>
seemed to be a simple problem, but i couldn't
figure out how
to do it. say i was plotting y=x^2, and i wanted to shade the
region under the graph> only between x=2 to x=4 over a range
of {x, 0, 5}, is there a way to do> this? i tried FilledPlot,
but it filled the entire area under the graph. thank
you!>
====
Sorry for the earlier post, something went wrong in
the cut and pastedepartment.For Matthew Senn's
filled plot,
here is an example one might cut and pastedirectly into
mathematica as a starting
point?< i'm
relatively new using
mathematica 4.2. the other day, i came acrosswhat> seemed to
be a simple problem, but i couldn't figure out
how to do it.
say i was plotting y=x^2, and i wanted to shade the region
under the graph> only between x=2 to x=4 over a range of {x,
0, 5}, is there a way to do> this? i tried FilledPlot, but it
filled the entire area under the graph.Steven ShippeeHome:
(360) 493-8353Work: (360) 902-5817Reply-To: Matthew Senn
====
solution:DisplayTogether[
FilledPlot[eq1[x], {x, 2, 4}, Fills -> GrayLevel[.9]],
Plot[eq1[x], {x, 0, 4.5}]]and by the way, i'm very impressed
with the speed at which i received theresponses. you are all
very helpful, and thank you again!-- matt> i'm relatively
new
using mathematica 4.2. the other day, i came acrosswhat>
seemed to be a simple problem, but i couldn't
figure out how
to do it. say i was plotting y=x^2, and i wanted to shade the
region under the graph> only between x=2 to x=4 over a range
of {x, 0, 5}, is there a way to do> this? i tried FilledPlot,
but it filled the entire area under the graph. thank you!>
====
>
i'm relatively new using mathematica 4.2. the other day, i
came acrosswhat> seemed to be a simple problem, but i
couldn't figure out how to do it. say i was
plotting y=x^2,
and i wanted to shade the region under the graph> only
between x=2 to x=4 over a range of {x, 0, 5}, is there a way
to do> this? i tried FilledPlot, but it filled the entire area
under the graph. thank you!>Matthew,A third way: <<
Graphics`Graphics` DisplayTogether[ FilledPlot[{0, x^2}, {x,
2, 4}], Plot[x^2, {x, 0, 5}]]If we use x^2 instead of {0,x^2}
the fill is gray.--Allan---------------------Allan
HayesMathematica Training and ConsultingLeicester
UKhay@haystack.demon.co.ukVoice: +44 (0)116 271 4198
====
> i'm
relatively new using mathematica 4.2. the other day, i came
acrosswhat> seemed to be a simple problem, but i couldn't
figure out how to do it. say i was plotting y=x^2, and i
wanted to shade the region under the graph> only between x=2
to x=4 over a range of {x, 0, 5}, is there a way to do> this?
i tried FilledPlot, but it filled the entire area under the
graph. thank you!>Matthew,Two suggestions: <<
Graphics`FilledPlot` FilledPlot[{x^2, If[2 <= x <= 4, 0,
x^2]}, {x, 0, 5}];Below, I use Off[Plot::plnr] to suppress
messages complaining ofnon-real values. Off[Plot::plnr]
FilledPlot[{x^2, If[2 <= x <= 4, 0]}, {x, 0, 5}]
On[Plot::plnr]Allan---------------------Allan
HayesMathematica Training and ConsultingLeicester
UKhay@haystack.demon.co.ukVoice: +44 (0)116 271 4198
====
f[a_,
b_] := a + bWith[{b = 4}, HoldForm[f[#, b]] & /@ {1, 2, 3,
4}]{f[1, 4], f[2, 4], f[3, 4], f[4, 4]}Yas> I have a function
f[a_,b_] defined some way. i want to fix one
argument,> and then
generate an array of values where the other argument varies.
How> can I do this? For example, I want: {f[1, 4], f[2, 4],
f[3, 4], f[4, 4]} Is there an easy way to do
this?
====
MapThread[f,{{1,2,3,4},{4,4,4,4}}]Will
produce{f[1,4],f[2,4],f[3,4],f[4,4]}You can check the
link:http://mathematica.co.kr/mathcomm/20_definitionviz/
index.htmlVery good tutorial on functional
programming.Ignacio> I have a function f