> Many thanks to all who have responded to my question. > The question I posted was actually a smaller section of what I had > originally planned on asking. I was hoping to use what I had learnt > from your answers and apply it to a larger problem but I haven't had > much success. > > What I want is to go from expression1: > -((P10 P20 P30)/Sqrt[2]) + (i P10 P20 P30)/Sqrt[2] - i P11 P20 P30 + > i P10 P21 P30 + (P11 P21 P30)/Sqrt[2] - (i P11 P21 P30)/Sqrt[2] - > P10 P20 P31 - (P11 P20 P31)/Sqrt[2] - (i P11 P20 P31)/Sqrt[2] + ( > P10 P21 P31)/Sqrt[2] + (i P10 P21 P31)/Sqrt[2] + P11 P21 P31 > > to expression2: > ((1 + Sqrt[2]) i - 1)/4*(P10 - P11) - ( > 1 + Sqrt[2] + i)/4*(P20 - P21) + ( > 1 - Sqrt[2] + i)/4*(P10 - P11)*(P30 - P31) + ( > 1 + (Sqrt[2] - 1) i)/4*(P20 - P21)*(P30 - P31) > > Given that P10 + P11=1,P20 + P21=1 and P30 + P31=1, > expression 2 becomes expression3: > ((1 + Sqrt[2]) i - 1)/4*(P10 - P11)*(P20 + P21)*(P30 + P31) - ( > 1 + Sqrt[2] + i)/4*(P10 + P11)*(P20 - P21)*(P30 + P31) + ( > 1 - Sqrt[2] + i)/4*(P10 - P11)*(P20 + P21)*(P30 - P31) + ( > 1 + (Sqrt[2] - 1) i)/4*(P10 + P11)*(P20 - P21)*(P30 - P31) > > I know that they are equal because when I use Expand[expression3], I > obtain expression1. > > I've tried forcing the simplification by introducing temporary > expressions and back substituting to go from expression1 to expression > 3 but I've realized then it doesn't work when introducing the extra > terms (P10 + P11),(P20 + P21) and (P30 + P31). Any ideas? > > Thanks, > Minh This will bring you fairly close to the form you are trying to obtain. expr = -((P10 P20 P30)/Sqrt[2]) + (i P10 P20 P30)/Sqrt[2] - i P11 P20 P30 + i P10 P21 P30 + (P11 P21 P30)/Sqrt[2] - (i P11 P21 P30)/Sqrt[2] - P10 P20 P31 - (P11 P20 P31)/Sqrt[2] - (i P11 P20 P31)/ Sqrt[2] + (P10 P21 P31)/Sqrt[2] + (i P10 P21 P31)/Sqrt[2] + P11 P21 P31; rels = {P10 + P11 - 1, P20 + P21 - 1, P30 + P31 - 1, P10 - P11 - p101, P20 - P21 - p201, P30 - P31 - p301}; vars = {P10, P11, P20, P21, P30, P31}; vars2 = {p101, p201, p301}; subs = {p101 -> P10 - P11, p201 -> P20 - P21, p301 -> P30 - P31}; gb = GroebnerBasis[rels, vars]; In[34]:= Collect[PolynomialReduce[expr, gb, vars][[2]], vars2] /. subs Out[34]= (P10 - P11) (1/4 (-1 + i + Sqrt[2] i) + 1/4 (1 - Sqrt[2] + i) (P30 - P31)) + (P20 - P21) (1/4 (-1 - Sqrt[2] - i) + 1/4 (1 - i + Sqrt[2] i) (P30 - P31)) Daniel Lichtblau Wolfram Research

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7/5/2010 10:03:26 AM

To get the 'double struck small e' type esc e e esc, or click it off the palette. In any case, it is not the Latin small case e. David Park djmp@earthlink.net http://home.earthlink.net/~djmp/ From: New Guy [mailto:NewGuy@yahoo.com] On Fri, 2 Dec 2005 11:05:41 +0000 (UTC), Jean-Marc Gulliet <jeanmarc.gulliet@gmail.com> wrote: >Now, the base of the natural logarithm is written E (capital e). Beg to differ here. The BasicInput.nb palette uses the small 'e' that is kind of double-struck. Bookreader ...

Ok, Final tally is completed. I get 2559 functions in Mathematica 7 (Kernel and AddOns) 1920 functions in the kernel, and the rest in the AddOns packages. A table showing each context and which functions in it is now on my page below http://12000.org/my_notes/compare_mathematica/index.htm in the version 7.0 row, look for the column before last. Determining which are the functions in the packages was bit more tricky since the InformationSytanx[] can't be used there. So had to do it the old way, i.e. get the usage information for each symbol and parse it out lookin...

expr = Expand[((1 + Sqrt[2]) i - 1)/4*(P10 - P11) - (1 + Sqrt[2] + i)/4*(P20 - P21)]; Your initial form is not what Mathematica considers the simplest form expr // Simplify (1/4)*((Sqrt[2]*i + i - 1)*P10 - (Sqrt[2]*i + i - 1)*P11 - (i + Sqrt[2] + 1)*(P20 - P21)) You need to force it (expr /. {P10 -> x1 + P11, P20 -> x2 + P21} // Simplify) /. {x1 -> (P10 - P11), x2 -> (P20 - P21)} (1/4)*((Sqrt[2]*i + i - 1)*(P10 - P11) - (i + Sqrt[2] + 1)* (P20 - P21)) This is almost your initial form Bob Hanlon ---- Minh <dminhle@gmail.c...

You missed the last closing parenthesis in your expression. The command should end with two parentheses before the comma: ... + (Z*U)^2)),Z] ...

expr1 = -(P10 P20 P30)/Sqrt[2] + (i P10 P20 P30)/Sqrt[2] - i P11 P20 P30 + i P10 P21 P30 + (P11 P21 P30)/Sqrt[2] - (i P11 P21 P30)/Sqrt[2] - P10 P20 P31 - (P11 P20 P31)/Sqrt[2] - (i P11 P20 P31)/Sqrt[2] + (P10 P21 P31)/Sqrt[2] + (i P10 P21 P31)/Sqrt[2] + P11 P21 P31; expr2 = ((1 + Sqrt[2]) i - 1)/4* (P10 - P11) - (1 + Sqrt[2] + i)/4* (P20 - P21) + (1 - Sqrt[2] + i)/4* (P10 - P11)*(P30 - P31) + (1 + (Sqrt[2] - 1) i)/4* (P20 - P21)*(P30 - P31); To get the form of expr2 from expr1 expr22 = Simplify[expr1, { ...

Sir, thanks for the tip; wasn't aware of NLimit. It speeds things up. My original goal was to get this limit to d -> Infinity but 1000 was about as high as I could go. Now I got it up to 10^6 in steps and the result doesn't appear to be converging at all. At 10^7 I get errors. I had to use the Scale->1000 option to get above d-> 1000. Perhaps my NLimit attempt is futile; thanks for your help. Rob On Dec 27, 1:03 pm, "w...@wavebounce.com" <w...@wavebounce.com> wrote: > Sir, thanks for the tip; wasn't aware of NLimit. > > It speeds things u...

> -----Original Message----- > From: Steven T. Hatton [mailto:hattons@globalsymmetry.com] > Sent: 15 December 2005 10:30 > Subject: Re: Mathematica Programmer vs. Programming > in Mathematica > ....... > > I wonder what value there woudl be in trying to explain what makes > > Mathematica "functions" different from functions in > languages such as > > C in a book addressed to readers most of whom have no > knowledge of C > > and are not particualry interested in getting it? > > I suspect you will not find very many people who have never > programmed in Java, C#, or C++ and are likely to use > Mathematica extensively. I'm not really saying the > introduction should use these languages extensively as a > means of contrast. I'm saying that Mathematica should be > introduced very differently from the way these general > purpose languages are introduced. Hang on for a minute - our entire corridor is full of scientists who use Mathematica every day. And only a couple have used even C. Most scientists have grown up with Fortran, right? (at least judging from the way they use Mathematica! ;) So your statement that most Mathematica people start from Java or C is incorrect. > "Everything is an expression" should be explained in terms of > recursive data structures, which they are. Obviously, such a > presentation should i...

I'd like to add the ComplexAnalysis package at my web site below: This package contains complex analysis routines and complex graphics routines. There are routines that convert the regular 2D Graphics into equivalent complex forms. For example ComplexLine[{z1,z2,z3...}] takes complex numbers for the point coordinates. There are routines for producing one or two panel plots or animations of complex functions. Each panel may be one of the following plot types. 1) Cartesian/PolarSurface - Plots the surface s[f[z]] where f is a complex function and s is a real function. 2) Cartesian...

> In doing a program in Mathematica that receives an > expression called f and converts it to a function. > However the command Function[x,f] doesn't work, what > do I do? > In[1]:= f = 3 x; In[2]:= g = Function[x, f // Evaluate] Out[2]= Function[x, 3 x] ...

On 12/4/05 at 5:57 AM, NewGuy@yahoo.com (New Guy) wrote: >On Fri, 2 Dec 2005 11:05:41 +0000 (UTC), Jean-Marc Gulliet ><jeanmarc.gulliet@gmail.com> wrote: >>Now, the base of the natural logarithm is written E (capital e). >Beg to differ here. The BasicInput.nb palette uses the small 'e' >that is kind of double-struck. The BasicInput palette displays the constants Pi and E in TraditionalForm. If you type E//TraditionalForm into an empty cell, the corresponding output cell will be the same character you get after clicking the entry button on the BasicInput pallete. The point is, to enter the constant E from the keyboard, you enter it as a capital E. -- To reply via email subtract one hundred and four ...

Thank you Bob for the excellent suggestions. I've never used Grid before, so I got side-tracked on learning about this function. I know there are many variations one can do along this theme. So, here's a variation that incorporated some of your ideas. I added a column that attempts to find a simpler version if it finds one. What is considered simpler is of course subjective, so this is only an attempt. :>) // ===== Main Function ===== TruthTable[equ_,steps_List:{}]:=Module[{v,data,min,hdg}, v = BooleanVariables[equ]; AppendTo[v,{steps,equ}]; (* If shorter versi...

On 5/7/09 at 6:33 AM, siegman@stanford.edu (AES) wrote: >Bob F (author of the original post appended below) and I are >evidently very much in agreement about the value of PDF >documentation. Let me just add one note. >Split windows are certainly good, particularly if you have either >large (or multiple) monitors and/or excellent eyesight. >An equally good, maybe even better, alternative is to be able to >jump back and forth (preferably with a single keystroke or >mouseclick) between two different full-screen "environments" or >"views&qu...

On 2/4/04, Andrzej Kozlowski wrote: >Actually one can use Paul's argument to prove >the following stronger statement: > >Sum[(Subscript[x,i]/(1 + Sum[Subscript[x,j]^2, {j, i}]))^2, {i, n}] < 1 > >for every positive integer n. > >It is easy to see that this implies the >inequality in the original problem (use >Schwarz's inequality). MathWorld only gives the integral form at http://mathworld.wolfram.com/SchwarzsInequality.html The required form of the inequality is at http://mathworld.wolfram.com/CauchysInequality.html ...

hi, What I want to do is to : Plot Reflectivity: Refelectivity is an expression ( or a function ) depending on nfilm which in turn depends on epsmodel( which in itself is a function of 3 parameters and a variable ) therefore : epsmodel = f(a,b,c,d,x) (* result is a comlpex number *) nfilm = sqrt[epsmodel] reflectivity = ((nfilm - 1) / (nfilm +1))^2 I just want to plot reflectivity and see the variation in it when (a,b,c,x) varies ( using the Manipulate[] tool ) Please let me know how to do this in term of these functions ? the exact formalism for which I got the error : ...

Log[1/E^.5] -0.5 Log[1/E^-.5] � .5 True Refine[1/E^x, 0<x<1] E^-x Or if you need to, Refine[Log[1/E^x], 0<x<1] -x Kris -------------- Original message -------------- > On Sun, 27 Nov 2005 08:24:53 +0000 (UTC), Bob Hanlon > wrote: > > >The output that you want is "unstable", i.e., Mathematica automatically > >converts it. > > > > Thanks. > > I have another question if you don't mind. > > I'm trying to just check and see if I got a whole bunch of > even-numbered study exercises r...

Hi Mukhtar <snippage> > Kernel window bears no function (except may be to force shut > down evaluation when mathematica stops responding to ALT+.) and I do > not see why it would be kept on desktop. May be it is beautiful. The kernel is quite useful in and of itself. For one, I can evaluate everything straight away -- I don't have to start the front end, hit "Evaluate", then wait for the kernel to load and my results to appear. In addition it is quite useful, provided your path is set accordingly, as a calculator in a console window. Rega...

On Fri, 22 May 2009 01:50:51 -0400 (EDT), skkaul wrote: > On May 10, 5:13 am, John Fultz <jfu...@wolfram.com> wrote: > >> <install= >> directory>/SystemFiles/FrontEnd/TextResources/UnicodeFontMapping.tr >> > Besides the comments, is there any documentation on this file? In > particular, what are type V and H entries, and what font is referenced > by -2? > > Thanks, > Shiva It's not documented because it's not intended for user consumption, although being able to edit it very rarely allows working around certain issues...

It would seem that by the time Exp[x^2]*Erfc[x] runs into exponent trouble, that the approximation 1/(x Sqrt[Pi]) would be sufficiently accurate for most numeric calculations. Thus, you might form a function like scaledErfc[ x_ ] := If[ x < 25000, Exp[x^2]*Erfc[x] , 1/(x Sqrt[Pi]) ] -----Original Message----- From: Peter Pein [mailto:petsie@dordos.net] Subject: Re: scaled complementary error function in Mathematica? Tanim Islam schrieb: > Hi: > > I was wondering if anyone had a package or notebook file that computes > the scaled complementary error functi...

On 8/1/10 at 4:58 AM, camille.segarra@gmail.com (Camille) wrote: >I am fairly new to mathematica. I am stuck with a problem I cannot >solve. I would like to call a previous equation into the function >Function. If a type directly the expression or copy paste it, it >works. However, when I call the expression by its name it does not. >Here is the code for a more precise explanation: > >In: ll Out: d + a x + h x^2 + b y + e x y + c z + j z^2 >In: Function[##, a x + b y + c z + d + e x y + h x^2 + j z^2] & @@ >{Listp} Out: Function[{a, b, c, d, e, h, ...

On Tue, 5 Apr 2005, dh wrote: > Hi APC, > I tried to simplify the problem a bit. There is definitly a bug that > Wolfram should take notice. It would be nice if WRI could give an answer . > > The folllowing is obviously correct: > Sum[Sin[k]*Cos[k + 1], {k, 1, 1}] > Out: Cos[2] Sin[1] > > When we do the same with indefinite summation: > Sum[Sin[k]*Cos[k + 1], {k, 1, n}] /. n -> 1 // Simplify // Expand > Out: Sin[1]/2 + Cos[2] Sin[1] > > we get an additional term: Sin[1]/2 !!! > > Sincerely, Daniel > > APC wrote: &g...

On 3/23/10 at 4:23 AM, hemmecke@gmail.com (hemmecke) wrote: >Does somebody know why I get different behaviour for the following >General::dupsym: The symbol Array with context A` already exists. >message? Why does that message appear at all? If A`Array exists, >then Mathematica should just use it, shouldn't it? >According to >http://reference.wolfram.com/mathematica/tutorial/Contexts.html we >have: >`name a symbol in the current context >So why is Mathematica complaining? The message is not telling you the symbol exists in the current cont...

Given that: Expand[((1 + Sqrt[2]) i - 1)/4*(P10 - P11) - (1 + Sqrt[2] + i)/ 4*(P20 - P21)] will output -(P10/4) + (i P10)/4 + (i P10)/(2 Sqrt[2]) + P11/4 - (i P11)/4 - ( i P11)/(2 Sqrt[2]) - P20/4 - P20/(2 Sqrt[2]) - ( i P20)/4 + P21/4 + P21/(2 Sqrt[2]) + (i P21)/4 How do I get from: -(P10/4) + (i P10)/4 + (i P10)/(2 Sqrt[2]) + P11/4 - (i P11)/4 - ( i P11)/(2 Sqrt[2]) - P20/4 - P20/(2 Sqrt[2]) - ( i P20)/4 + P21/4 + P21/(2 Sqrt[2]) + (i P21)/4 back to ((1 + Sqrt[2]) i - 1)/4*(P10 - P11) - (1 + Sqrt[2] + i)/ 4*(P20 - P21) I've tried using the Collect function...

Hi all, I would like to use the collect function for this: (1 + 2 Z + Z Z.U)^2 + (1 + 2 Z + Z Z.U) (Z (1 + Z) + Z.U (Z + Z Z.U)) + (1 + Z + Z.U + (Z.U)^2) (Z (1 + Z) + Z.U (Z + Z Z.U)) + (Z (1 + Z) + Z.U (Z + Z Z.U)) (Z (1 + Z.U) + Z.U (Z + (Z.U)^2)) This is what I tried in Mathematica: Collect[(1 + 2*Z + Z* Z*U)^2 + (1 + 2*Z + Z* Z*U)* (Z* (1 + Z) + Z*U* (Z + Z* Z*U)) + (1 + Z + Z*U + (Z*U)^2)* (Z *(1 + Z) + Z*U (Z + Z *Z*U)) + (Z* (1 + Z) + Z*U* (Z + Z* Z*U)) *(Z* (1 + Z*U) + Z*U *(Z + (Z*U)^2), Z] But it does not work However, it works when I tried this one: Collect[...

You have to use the correct Mathematica syntax. E and Log instead of e and LN. You don't even have to use Simplif in this case. Log[1/E^2] -2 David Park djmp@earthlink.net http://home.earthlink.net/~djmp/ From: New Guy [mailto:NewGuy@yahoo.com] On Sun, 27 Nov 2005 08:24:53 +0000 (UTC), Bob Hanlon <hanlonr@cox.net> wrote: >The output that you want is "unstable", i.e., Mathematica automatically >converts it. > Thanks. I have another question if you don't mind. I'm trying to just check and see if I got a whole bunch of even-nu...

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