A19 == Technical support has responded to my query. They say that one of the simplifications that Mathematica does when using Simplify is to Factor the expression, and factoring an expression consisting of inexact numbers leads to the type of behavior I observed. They stated that changing the behavior of Simplify to avoid the problems I noticed would make Simplify less capable, and that the majority of users would prefer to have Simplify behave the way it currently does. Of course, that doesn't answer the question about why the number of terms in the expression would cause the coefficients to go from rational to inexact to rational again. At any rate, I have another lesson learned. When using Simplify, check the result if you are mixing infinite precision and inexact numbers, as the result may not be what you wanted. In the past I always assumed that the result of using Simplify on an expression produced a result equivalent to the original expression, but now I discover that this is not true. Carl Woll Physics Dept U of Washington ----- Original Message ----- > {i, > n}]], x[1]] Now lets try CW for the first 100 values of n: > In[30]:= > Table[CW[n], {n, 100}] Out[30]= > {1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, > 0.5`12.3085, 0.5`12.0074, 0.5`11.7064, 0.5`11.4054, > 0.5`11.1043, 0.5`10.8033, 0.5`10.5023, 0.5`10.2012, > 0.5`9.9002, 0.5`9.5992, 0.5`9.2982, 0.5`8.9971, > 0.5`8.6961, 0.5`8.3951, 0.5`8.094, 0.5`7.793, 0.5`7.492, > 0.5`7.1909, 0.5`6.8899, 0.5`6.5889, 0.5`6.2879, > 0.5`5.9868, 0.5`5.6858, 0.5`5.3848, 0.5`5.0837, > 0.5`4.7827, 0.5`4.4817, 0.5`4.1806, 0.5`3.8796, > 0.5`3.5786, 0.5`3.2776, 0.5`2.9765, 0.5`2.6755, > 0.5`2.3745, 0.5`2.3745, 0.5`1.7724, 0.5`1.7724, > 0.5`1.1703, 0.5`1.1703, 0.5`0.5683, 0``0.1423, 0``0.1423, > 0``0.1423, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, > 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, > 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, > 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, > 1/2, 1/2, 1/2} In[31]:= Map[Length,Split[#]] Out[31]= > {12,43,45} This is indeed curious.The problem seems to occur for values between 13 > and 55 and then go away (for good???) At first I thought it maybe in > some fascinating way related to some properties of the integer n, but > now I am not sure. It certainly worth a careful examination. I hope > whoever discovers the cause of this will let us know. Andrzej Andrzej Kozlowski > Yokohama, Japan > http://www.mimuw.edu.pl/~akoz/ > http://platon.c.u-tokyo.ac.jp/andrzej/ To Technical Support and the Mathematica User community, I'm writing to report what I consider to be a bug. First, I want to > show a > simplified example of the problem. Consider the following expression: expr=0.22 + x[0] + (3*(-0.16+ x[1]))/4 + (9*(0.546 + x[2]))/16; When simplified I expected to get some real number plus > x[0]+3x[1]/4+9x[2]/16, but instead I get the following: Simplify[expr] > 0.407125 + x[0] + 0.75 x[1] + 0.5625 x[2] As you can see, for some reason Mathematica converted the fractions > 3/4 and > 9/16 to real machine numbers. I consider this to be a bug. Now, for an example more representative of the situation that I've been > coming across. expr12 = 1.001`17 + Sum[(x[i] - 1.001`17)/2^i, {i, 12}]; > expr13 = 1.001`17 + Sum[(x[i] - 1.001`17)/2^i, {i, 13}]; > expr55 = 1.001`17 + Sum[(x[i] - 1.001`17)/2^i, {i, 55}]; As you can see, I have replaced the real numbers by extended precision > numbers. The simplified example above demonstrates that the problem > exists > when using machine numbers. Now, we'll see what happens when we use > arbitrary precision numbers. First, let's simplify the expression with > 12 > terms. Simplify[expr12] > (1.0010000000000 + 2048 x[1] + 1024 x[2] + 512 x[3] + 256 x[4] + 128 > x[5] + 64 x[6] + 32 x[7] + 16 x[8] + 8 x[9] + 4 x[10] + 2 x[11] + x[12]) > / 4096 As you can see, a sum with 12 terms upon simplification has > coefficients > which are still integers as they should be. However, increasing the > number > of terms to 13 yields Simplify[expr13] > 0.0001221923828125 + 0.500000000000 x[1] + 0.250000000000 x[2] + 0.1250000000000 x[3] + 0.0625000000000 x[4] + 0.0312500000000 x[5] + 0.01562500000000 x[6] + 0.00781250000000 x[7] + 0.00390625000000 > x[8] + 0.001953125000000 x[9] + 0.000976562500000 x[10] + 0.000488281250000 > x[11] > + 0.000244140625000 x[12] + 0.0001220703125000 x[13] Now, all of the coefficients are converted to real numbers, > replicating the > bug from the simplified example. Finally, let's see what happens when > we > have 55 terms. Rather than putting the resulting expression here, I > will > just leave it at the end of the post. The result though is somewhat > surprising. Each of the coefficients of the x[i] are again real > numbers, but > now their precision is only 0! The proper result of course is the sum > of > some real number (with a precision close to 0 due to numerical > cancellation) > and an expression consisting of rational numbers multiplied by x[i]. > The > loss of precision of the coefficients of the x[i] is precisely what > occurred > to me. Of course, in this simple example, simply using Expand instead > of > Simplify produces the expected result, and is my workaround. I hope you > agree with me that this is a bug, and one that Wolfram needs to > correct. Carl Woll > Physics Dept > U of Washington Simplify[expr55] > -16 -1 -1 -1 > 0. 10 + 0. x[1] + 0. x[2] + 0. 10 x[3] + 0. 10 x[4] + 0. 10 > x[5] + -2 -2 -2 -3 -3 > 0. 10 x[6] + 0. 10 x[7] + 0. 10 x[8] + 0. 10 x[9] + 0. 10 > x[10] > + -3 -3 -4 -4 > -4 > 0. 10 x[11] + 0. 10 x[12] + 0. 10 x[13] + 0. 10 x[14] + 0. 10 > x[15] + -5 -5 -5 -6 > -6 > 0. 10 x[16] + 0. 10 x[17] + 0. 10 x[18] + 0. 10 x[19] + 0. 10 > x[20] + -6 -6 -7 -7 > -7 > 0. 10 x[21] + 0. 10 x[22] + 0. 10 x[23] + 0. 10 x[24] + 0. 10 > x[25] + -8 -8 -8 -9 > -9 > 0. 10 x[26] + 0. 10 x[27] + 0. 10 x[28] + 0. 10 x[29] + 0. 10 > x[30] + -9 -9 -10 -10 > 0. 10 x[31] + 0. 10 x[32] + 0. 10 x[33] + 0. 10 x[34] + -10 -11 -11 -11 > 0. 10 x[35] + 0. 10 x[36] + 0. 10 x[37] + 0. 10 x[38] + -12 -12 -12 -12 > 0. 10 x[39] + 0. 10 x[40] + 0. 10 x[41] + 0. 10 x[42] + -13 -13 -13 -14 > 0. 10 x[43] + 0. 10 x[44] + 0. 10 x[45] + 0. 10 x[46] + -14 -14 -15 -15 > 0. 10 x[47] + 0. 10 x[48] + 0. 10 x[49] + 0. 10 x[50] + -15 -15 -16 -16 > - > 16 > 0. 10 x[51] + 0. 10 x[52] + 0. 10 x[53] + 0. 10 x[54] + > 0. 10 > x[55] ==== Dear Carl You have discovered what is perhaps a bug but maybe something even mor einteresting . However, I think you stopped your investigation a little prematurely. Here is a function that just computes the coefficient of x[1] in your Simplified expression for various values of n: CW[n_] := Coefficient[Simplify[1.001`17 + Sum[(x[i] - 1.001`17)/2^i, {i, n}]], x[1]] Now lets try CW for the first 100 values of n: In[30]:= Table[CW[n], {n, 100}] Out[30]= {1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 0.5`12.3085, 0.5`12.0074, 0.5`11.7064, 0.5`11.4054, 0.5`11.1043, 0.5`10.8033, 0.5`10.5023, 0.5`10.2012, 0.5`9.9002, 0.5`9.5992, 0.5`9.2982, 0.5`8.9971, 0.5`8.6961, 0.5`8.3951, 0.5`8.094, 0.5`7.793, 0.5`7.492, 0.5`7.1909, 0.5`6.8899, 0.5`6.5889, 0.5`6.2879, 0.5`5.9868, 0.5`5.6858, 0.5`5.3848, 0.5`5.0837, 0.5`4.7827, 0.5`4.4817, 0.5`4.1806, 0.5`3.8796, 0.5`3.5786, 0.5`3.2776, 0.5`2.9765, 0.5`2.6755, 0.5`2.3745, 0.5`2.3745, 0.5`1.7724, 0.5`1.7724, 0.5`1.1703, 0.5`1.1703, 0.5`0.5683, 0``0.1423, 0``0.1423, 0``0.1423, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2} In[31]:= Map[Length,Split[#]] Out[31]= {12,43,45} This is indeed curious.The problem seems to occur for values between 13 and 55 and then go away (for good???) At first I thought it maybe in some fascinating way related to some properties of the integer n, but now I am not sure. It certainly worth a careful examination. I hope whoever discovers the cause of this will let us know. Andrzej Andrzej Kozlowski Yokohama, Japan http://www.mimuw.edu.pl/~akoz/ http://platon.c.u-tokyo.ac.jp/andrzej/ > To Technical Support and the Mathematica User community, I'm writing to report what I consider to be a bug. First, I want to > show a > simplified example of the problem. Consider the following expression: expr=0.22 + x[0] + (3*(-0.16+ x[1]))/4 + (9*(0.546 + x[2]))/16; When simplified I expected to get some real number plus > x[0]+3x[1]/4+9x[2]/16, but instead I get the following: Simplify[expr] > 0.407125 + x[0] + 0.75 x[1] + 0.5625 x[2] As you can see, for some reason Mathematica converted the fractions > 3/4 and > 9/16 to real machine numbers. I consider this to be a bug. Now, for an example more representative of the situation that I've been > coming across. expr12 = 1.001`17 + Sum[(x[i] - 1.001`17)/2^i, {i, 12}]; > expr13 = 1.001`17 + Sum[(x[i] - 1.001`17)/2^i, {i, 13}]; > expr55 = 1.001`17 + Sum[(x[i] - 1.001`17)/2^i, {i, 55}]; As you can see, I have replaced the real numbers by extended precision > numbers. The simplified example above demonstrates that the problem > exists > when using machine numbers. Now, we'll see what happens when we use > arbitrary precision numbers. First, let's simplify the expression with > 12 > terms. Simplify[expr12] > (1.0010000000000 + 2048 x[1] + 1024 x[2] + 512 x[3] + 256 x[4] + 128 > x[5] + 64 x[6] + 32 x[7] + 16 x[8] + 8 x[9] + 4 x[10] + 2 x[11] + x[12]) > / 4096 As you can see, a sum with 12 terms upon simplification has > coefficients > which are still integers as they should be. However, increasing the > number > of terms to 13 yields Simplify[expr13] > 0.0001221923828125 + 0.500000000000 x[1] + 0.250000000000 x[2] + 0.1250000000000 x[3] + 0.0625000000000 x[4] + 0.0312500000000 x[5] + 0.01562500000000 x[6] + 0.00781250000000 x[7] + 0.00390625000000 > x[8] + 0.001953125000000 x[9] + 0.000976562500000 x[10] + 0.000488281250000 > x[11] > + 0.000244140625000 x[12] + 0.0001220703125000 x[13] Now, all of the coefficients are converted to real numbers, > replicating the > bug from the simplified example. Finally, let's see what happens when > we > have 55 terms. Rather than putting the resulting expression here, I > will > just leave it at the end of the post. The result though is somewhat > surprising. Each of the coefficients of the x[i] are again real > numbers, but > now their precision is only 0! The proper result of course is the sum > of > some real number (with a precision close to 0 due to numerical > cancellation) > and an expression consisting of rational numbers multiplied by x[i]. > The > loss of precision of the coefficients of the x[i] is precisely what > occurred > to me. Of course, in this simple example, simply using Expand instead > of > Simplify produces the expected result, and is my workaround. I hope you > agree with me that this is a bug, and one that Wolfram needs to > correct. Carl Woll > Physics Dept > U of Washington Simplify[expr55] > -16 -1 -1 -1 > 0. 10 + 0. x[1] + 0. x[2] + 0. 10 x[3] + 0. 10 x[4] + 0. 10 > x[5] + -2 -2 -2 -3 -3 > 0. 10 x[6] + 0. 10 x[7] + 0. 10 x[8] + 0. 10 x[9] + 0. 10 > x[10] > + -3 -3 -4 -4 > -4 > 0. 10 x[11] + 0. 10 x[12] + 0. 10 x[13] + 0. 10 x[14] + 0. 10 > x[15] + -5 -5 -5 -6 > -6 > 0. 10 x[16] + 0. 10 x[17] + 0. 10 x[18] + 0. 10 x[19] + 0. 10 > x[20] + -6 -6 -7 -7 > -7 > 0. 10 x[21] + 0. 10 x[22] + 0. 10 x[23] + 0. 10 x[24] + 0. 10 > x[25] + -8 -8 -8 -9 > -9 > 0. 10 x[26] + 0. 10 x[27] + 0. 10 x[28] + 0. 10 x[29] + 0. 10 > x[30] + -9 -9 -10 -10 > 0. 10 x[31] + 0. 10 x[32] + 0. 10 x[33] + 0. 10 x[34] + -10 -11 -11 -11 > 0. 10 x[35] + 0. 10 x[36] + 0. 10 x[37] + 0. 10 x[38] + -12 -12 -12 -12 > 0. 10 x[39] + 0. 10 x[40] + 0. 10 x[41] + 0. 10 x[42] + -13 -13 -13 -14 > 0. 10 x[43] + 0. 10 x[44] + 0. 10 x[45] + 0. 10 x[46] + -14 -14 -15 -15 > 0. 10 x[47] + 0. 10 x[48] + 0. 10 x[49] + 0. 10 x[50] + -15 -15 -16 -16 > - > 16 > 0. 10 x[51] + 0. 10 x[52] + 0. 10 x[53] + 0. 10 x[54] + > 0. 10 > x[55] ==== I think, for input of that many numbers (and other inputs), I might use an input file that has textual names or descriptions in the first ten or twenty columns, followed by values starting at a fixed column after that. Mathematica or Java could easily read inputs from that, and a human could read it as well. If you're concerned that a human might jumble the file format -- accidentally deleting lines, etc. -- a program could key on the names or descriptions rather than trusting them to be in correct order, and point out missing values. A strategy like that allows you to start with a previous input file (not from scratch) and change only what needs to change. Also, Mathematica 4.2 adds XML support to the picture, and that might be useful. Bobby -----Original Message----- I agree with Mr. Kuska, that the system Mr Nagesh describes is not > userfriendly. But I think, the suggestions of Mr. Kuska do not make it more > userfriendly, rather the opposite is true. Mr. Nagesh asks Is any body here have expertise or information about the capability of > Mathematica as a system simulation tool? > Mr. Kuska answers: > Since the most system simulation tools are simply solving a system of > ordinary differntial equations it is simple to do this with NDSolve[]. My comment: That is: He sees the simulation system merely as a set of differential > equations. hmm, since the original poster write My refrigeration system simulation package is likely to have > approximately 60 First order Differential equations. it seems not completly wrong to assume that the system consists of > of ode's .. > Yout should not ignore the word merely. It is not enough to have a set 60 differential equations and a set of 200-250 numbers. That is not simulation system, which can be used by users with the exception, perhaps, of the programmer of the system himself. How does e.g. the user know what meaning a number in the set has, ought he count the numbers from the beginning? Your nice command shows only, if there is an input, which is not a number. But I think the user would like to know, which of the 200 elements are not numbers. The only good of your command is, that it looks nice and shows your knowledge! The question of Mr. Nagesh: My 4th Objective:- How can I program the check for correctness of the > input values supplied by the package user ? > The answer of Mr. Kuska is: > And @@ (NumericQ /@ {aListOfAllYourNumericParameters}) My comment: This is a nice command and shows the knowledge of Mr.Kuska. But does Mr. > Nagesh understand it and is it sufficient to check, if all inputs are > numerical? It seems you have a deeper knowlege about the things that Mr. Nagesh > understand. You know him personally ? It is not very polite to > make speculations *what* a other person understand. And no it is not sufficent to check that all parameters are numbers. > Typical paramters described by intervals, where the assumptions of > a model are valid. But for this checks one needs more knowlege > about the meaning of the parameters. And one needs the knowlege about > the differntial equations, to find out the eigenvalues of the jacobian > ... > Additionally I think, it is not userfriendly to see the input merely as a > set of 200-250 numbers. My suggestion is, that JLink is used, a suggestion Mr. Kusk takes into > consideration, too. That will be fast as lightning ! > But further I suggest, that classes are defined in Java, which represent the > parts of the system. That is notable nonsense! When the differntial equations should be > solved with Mathematica, the parts can't be Java classes. Mathematica's > NDSolve[] need a explicit expression to integrate the equations. That's not nonsense, the Mathematica program does not fetch the classes but the numbers in the classes (or better in the objects). in the Java classes in textform. They can then be fetched from the Mathematica program and transformed into expressions by the Command ToExpression. The aim is, to have a clear separation of the system into components, which are manageable and understandable. OK you can call a Java class member from Mathematica but this will > be incredible slow when the right hand side is evaluated 200-300 > times and every evaluation make several callbacks to the Java source. > Event handler of the Java main program (without some modification in > the event loop) while it is evaluating an other command. My idea is to fetch the values once from the Java objects at the beginning. 200-250 numbers is not so much.. > Constructors of the classes should build objects with default values. That's a great idea. If a simulation run should be documented, one > always > need the full listing of the Java source to find the actual parameter > settings. That is not my opinion. I think not every user of the simulation system should have to know Java and Mathematica. The user must look for the values in the objects. And the values are in the objects, if they come from the constructor or from the graphical user interface. I think, there should be listings of the objects including names of the variables. In the objects the values are in an meaningful environment. Graphical user interfaces > should give the opportunity to change the data fields in the objects and > check the input for correctness. *and* what has a GUI for 200-250 variables to do with Mathematica ? > BTW one has typical much less variables because many parameters > are fixed and it make no sense to change, for example, material > constants of materials that can't exchanged > The system should give the opportunity, to store the objects on harddisk > (serialization). accessed directly. > Can you be so kind, to explain *how* your posting help a person that > ask How can I build a simulation system with Mathematica ? You *can* say Don't use Mathematica, use Java! but this has nothing > to do with the question or with my reply. It is you, who proposes to solve the problem with C/C++ and not to use Mathematica (see below!) My point of view is: Use Mathematica, for what Mathemtica is good, and Java, for what Java is good. Mathematica is not so good as Java for data entry and Java is better than Mathematica to represent the simulation system (by objects). But I still would suggest to use C/C++ and a numerical > ode-solver, make a fancy GUI/Script > interface and don't use Mathematica for such a simple task. > The ode-solver is the smallest part of such a simulation system. Jens > My name is Nagesh and pursuing research studies in Refrigeration. At > present I am writing a Dynamic Refrigeration System Simulation Package. > I > am using Mathematica as a programming language for the same since last > one > year. I don't have any programming experience before this. I have > following > querries:- > 1. Is any body here have expertise or information about the capability > of > Mathematica as a system simulation tool? Since the most system simulation tools are simply solving a system of > ordinary differntial equations it is simple to do this with NDSolve[]. 2. Is is possible to program a user friendly interface for my system > simulation package with Mathematica or I have to use some other > software? Write a MathLink or J/Link frontend that launch the kernel. But you > should keep > in mind that the user interface is typical 80-90 % of your code. > If you just whant to solve some ode's it is probably easyer to > include one of the excelent ode-solvers from netlib in your C-code > than to call Mathematica to do that. As long as you dont wish to change > the ode's very often (than Mathematica is more flexible) you should > not use Mathematica. 3. My refrigeration system simulation package is likely to have > approximately 60 First order Differential equations. Is is possible to > solve these in Mathematica ? Sure. > If yes then can anybody here guide me about > this further. Write down the equations and call NDSolve[]. > I am explaining below in short about the objectives I want to fulfill > from > coding out of my main input file 1. Example from Main Input File ( this will contain about 200-250 > variables > which will be entered by the user of this package) This sounds like a *very* userfiendly interface ;-) > Below is examples of two variables entered into this file, which will be > used in other analysis files for further evaluation. 2. Example from other analysis file ( there will be about 20-25 other > such > component analysis files ) where the above mentioned variables from main > input file will be used for further evaluations:- Below is one example from this file explaining how the variables from > main > input file will be used in other files. I hope that this short information will be useful for guiding me to > solve > the following problems that I am facing. I am facing follwing problems > or > objectives:- 1. My 1st Objective:- The user of this package must be able to change > only > the value of the variable in the main input file but he must not be able > to > change the name of the variable itself. For example he must be able to > change the value of the variable but he must not be able to change > the > name of this variable itself. > Here our problem is how to achieve or program it so that our objective > will > be fullfilled. Options with defaulf values ? or something like {aParam,bParam}={ODEParameter1,ODEParameter2} /. > userRules /. > {ODEParameter1->1,ODEParameter2->2} > 2. My 2nd Objective:- How I can program the main input file so that it > will > be user friendly in terms of its visuals and satisfying the constraint > mentioned above in objective1. What is *userfiendly* in a file with 250 variables ??? > 3. My 3rd Objective:- How can I program the optional values for each > variable in the main input file ? so that there will be always a value > assigned to each variable listed in main input file whenever the user > opens > up this file. If user want to change the values of some variables then > he > can change them and run the simulation otherwise the simulation run will > be > done with optional values assigned to each variable in the input file. See above. > 4. My 4th Objective:- How can I program the check for correctness of the > input values supplied by the package user ? And @@ (NumericQ /@ {aListOfAllYourNumericParameters}) > Jens ==== hi > AFAIK, Mac OS is now BSD or something like that. That makes it almost > certain that it could support QT. As I pointed out in another post, I can > run the KDE on Windows XP. I haven't been in the trenches with the Qt > impression is, it really is 'code once, run everywhere'. i've checked the trolltech hp. qt fully supports the following OS: MS Windows 95/98/Me, NT4, 2000 and XP. Mac OS X. > This is one of the reasons I am such a Mozilla fan. Konqueror works quite > like. But Mozilla runs everywhere with more or less a uniform look and > feel. Yes, Mozilla is written with Gtk and not with Qt, but that just > shows that WRI has options. i'm not sure if it is allowed to create a closed-source product like the mathematica fronend based on a gpl'ed library like gtk. with qt you have the possibility to buy licenses for the commercial version of qt, allowing you to create closed-source apps. > I'm a KDE fan. I've used the KDE since it was in alpha 0.4. I remember > back when it would compile in a few minutes on a pentium II. Now it takes > several hours on a P4. i started with beta1 (if i remember correctly) - sure it is really big now, but i think kde is still far away from being bloated..... imagine the mathematica frontend being seamlessly integrated into the linux ui's look and feel.... > I've always hated motif. The file chooser simply stinks. And that's just a > start. the whole motif ui stinks....:-) gerald ************************************* Gerald Roth M@th Desktop Development ************************************* ==== > hi, moving the frontend over to QT would have some neat side effects: > consistent look & feel with the modern linux gui, themeability, source > antialiased > truetype fonts as QT supports Xrender and Xft (looks great - see KDE3). i > think all of those points are of value, but the most important might be > source compatibility. ONE frontend for MOST (or ALL) platforms - sounds > like a dream :-)) AFAIK, Mac OS is now BSD or something like that. That makes it almost certain that it could support QT. As I pointed out in another post, I can run the KDE on Windows XP. I haven't been in the trenches with the Qt impression is, it really is 'code once, run everywhere'. This is one of the reasons I am such a Mozilla fan. Konqueror works quite like. But Mozilla runs everywhere with more or less a uniform look and feel. Yes, Mozilla is written with Gtk and not with Qt, but that just shows that WRI has options. I'm a KDE fan. I've used the KDE since it was in alpha 0.4. I remember back when it would compile in a few minutes on a pentium II. Now it takes several hours on a P4. But if WRI wanted to go the Gtk route, they could achieve the same ends. I've always hated motif. The file chooser simply stinks. And that's just a start. gerald > STH ==== Can I solve this inequality with Mathematica? Log[x,a]+Log[a x,a]>0 I've tried all know, but I get get any good result. CeZaR ==== First thanks to all, and in particular Bobby Treat, for your help with > this question. The best solution was as follows: lst = ReadList[c:data.txt, {Number, Number}] > adjacenceMatrix[ > x:{{_, _}..}] := Module[{actors, events}, > {actors, events} = Union /@ Transpose[x]; > Array[If[MemberQ[x, {actors[[#1]], events[[#2]]}], 1, 0] & , > {Length[actors], Length[events]}]] a = adjacenceMatrix[lst]; > b = a . Transpose[a]; > c = b (1 - IdentityMatrix[Length[b]]) C is the desired symmetric matrix with off diagonal values of >=0, > indicating the number of times two actors participate in the same event. > The diagonal is set to 0. A few items in response to Bobby's message, below. While c is, in fact, > a huge matrix with lots of cells equal to zero, that is exactly how we > need it structured for our analysis and research question (not relevant > to the list, but I'd be happy to discuss off list). Processing time is > actually not too bad!! I'm running a PIII 900 with 512 SDRAM, and the > code ran a 177 x 3669 matrix in under 90 seconds. MatrixForm [c] > presented no problems in viewing in the front end, but then it's only > 177 x 177. > Tom ********************************************** > Thomas P. Moliterno > Graduate School of Management > University of California, Irvine > tmoliter@uci.edu > ********************************************** > [...] There are several ways to go about this and which is best will vary based on relative number of events vs. number of actors. Below I show three variations. The first is a minor recoding of the one above. The second iterates over all pairs of actors. The third looks at all events for common actors. I then show three examples. The first two methods have the advantage that they do not require that events be positive integers. With some extra work the third method could also get around this restriction. toAdjacency0[data:{{_, _}..}] := Module[ {actors, events, mat1, mat2}, {actors, events} = Union /@ Transpose[data]; mat1 = Array[If[MemberQ[data, {actors[[#1]], events[[#2]]}], 1, 0] & , {Length[actors], Length[events]}]; mat2 = mat1 . Transpose[mat1]; mat2 * (1-IdentityMatrix[Length[mat2]]) ] toAdjacency1[origdata_] := Module[ {data=Union[origdata], mat}, data = Map[Last, Split[data,#1[[1]]===#2[[1]]&], {2}]; mat = Table [If [j>k, Length[Intersection[data[[j]],data[[k]]]], 0], {j,Length[data]}, {k,Length[data]}]; mat+Transpose[mat] ] toAdjacency2[origdata_] := Module[ {data=Sort[Map[Reverse,Union[origdata]]], mat, len, event}, data = Map[Last, Split[data,#1[[1]]===#2[[1]]&], {2}]; dim = Length[Union[Flatten[data]]]; len = Length[data]; mat = Table[0, {dim}, {dim}]; Do [ event = data[[j]]; Do [ Do [ mat[[event[[m]],event[[k]]]] += 1; mat[[event[[k]],event[[m]]]] += 1, {m,k-1}], {k,Length[event]}], {j,len}]; mat ] data1 = Table[{Random[Integer,{1,1000}], Random[Integer,50]}, {10000}]; data2 = Table[{Random[Integer,{1,1000}], Random[Integer,100]}, {10000}]; data3 = Table[{Random[Integer,{1,1000}], Random[Integer,200]}, {10000}]; Timings are on a 1.5 GHz machine running the Mathematica 4.2 kernel In[107]:= Timing[m0 = toAdjacency0[data1];] Out[107]= {5.44 Second, Null} In[108]:= Timing[m1 = toAdjacency1[data1];] Out[108]= {10.5 Second, Null} In[109]:= Timing[m2 = toAdjacency2[data1];] Out[109]= {16.24 Second, Null} In[110]:= m0 === m1 === m2 Out[110]= True Note that for this example the result is not terrible sparse (less than 20%). In[112]:= Count[Flatten[m0], 0] Out[112]= 191374 In[115]:= Timing[m0 = toAdjacency0[data2];] Out[115]= {11.51 Second, Null} In[116]:= Timing[m1 = toAdjacency1[data2];] Out[116]= {10.92 Second, Null} In[117]:= Timing[m2 = toAdjacency2[data2];] Out[117]= {9.07 Second, Null} Curiously this was the first example I tried, and all three methods perform about the same in this case. The result, not suprisingly, is more sparse (40%) because we have the same number of actors and pairs as previously, but now with more events to spread out over the pairs. In[118]:= Count[Flatten[m0], 0] Out[118]= 403232 When we get sparser still, the third method begins to dominate and the first is relatively slower. In[119]:= Timing[m0 = toAdjacency0[data3];] Out[119]= {22.73 Second, Null} In[120]:= Timing[m1 = toAdjacency1[data3];] Out[120]= {10.88 Second, Null} In[121]:= Timing[m2 = toAdjacency2[data3];] Out[121]= {4.96 Second, Null} Now sparsity is over 60%. In[122]:= Count[Flatten[m0], 0] Out[122]= 624350 The relative speed of this last method, in this instance, is derived from the fact that individual event lists are on average half the size of the previous case. Hence the main loop is expected to improve on average by a factor of 2 (you get a factor of 4 for iterating over all pairs in each event, but lose a factor of 2 because there are twice as many event lists). My guess is that a preprocessor that assesses number of actors vs. number of events would be the best way to choose between the first and third methods (which, inexplicably, are labelled as methods 0 and 2). It is not clear to me whether the middle approach will ever dominate. I have not given much thought to concocting examples where it would because offhand I suspect they would be pathological as in dense and with large intersections. As a last remark I'll note that these might run significantly faster if coded with Compile. Whether that is viable depends on the form of the data. In the above example where everything is a machine integer that approach would certainly work. Daniel Lichtblau Wolfram Research ==== First thanks to all, and in particular Bobby Treat, for your help with this question. The best solution was as follows: lst = ReadList[c:data.txt, {Number, Number}] adjacenceMatrix[ x:{{_, _}..}] := Module[{actors, events}, {actors, events} = Union /@ Transpose[x]; Array[If[MemberQ[x, {actors[[#1]], events[[#2]]}], 1, 0] & , {Length[actors], Length[events]}]] a = adjacenceMatrix[lst]; b = a . Transpose[a]; c = b (1 - IdentityMatrix[Length[b]]) C is the desired symmetric matrix with off diagonal values of >=0, indicating the number of times two actors participate in the same event. The diagonal is set to 0. A few items in response to Bobby's message, below. While c is, in fact, a huge matrix with lots of cells equal to zero, that is exactly how we need it structured for our analysis and research question (not relevant to the list, but I'd be happy to discuss off list). Processing time is actually not too bad!! I'm running a PIII 900 with 512 SDRAM, and the code ran a 177 x 3669 matrix in under 90 seconds. MatrixForm [c] presented no problems in viewing in the front end, but then it's only 177 x 177. Tom ********************************************** Thomas P. Moliterno Graduate School of Management University of California, Irvine tmoliter@uci.edu ********************************************** -----Original Message----- columns will be numbered from 1 to the number of observed actors or events, and will correspond to actors and events in sorted order. That said, you're asking for a VERY large matrix, and most of its entries will be zero. I'll suggest another way, later. The following indicates AT MOST 13.4% of the entries could be non-zero: lst = ReadList[moliterno-test1996.txt, {Number, Number}]; {actors, events} = Union /@ Transpose[lst]; N[Length[lst]/(Length[actors]*Length[actors])] 0.13350994338800987 However, a random sample shows that less than 1% will be non-zero: Timing[ Count[(MemberQ[lst, {actors[[Random[Integer, Length[actors]]]], events[[Random[Integer, Length[events]]]]}] & ) /@ Range[10000], True]/ 10000.] {7.515999999999998*Second, 0.008} Nevertheless, the following code should build the matrices you want. I'm using a 2.2GHz P4 and 1024MB RDRAM, so if you have a slower machine, be warned. adjacenceMatrix[ x:{{_, _}..}] := Module[{actors, events}, {actors, events} = Union /@ Transpose[x]; Array[If[MemberQ[x, {actors[[#1]], events[[#2]]}], 1, 0] & , {Length[actors], Length[events]}]] Timing[a = adjacenceMatrix[lst]; ] Dimensions[a] {5.671999999999997*Second, Null} {166, 1778} Timing[b = a . Transpose[a]; ] {0.5309999999999988*Second, Null} You don't want to display a or b in MatrixForm. It will crash your anything at all from the result, and use something like b[[Range[20],Range[4]]]//MatrixForm {{47, 0, 0, 0}, {0, 3, 0, 0}, {0, 0, 7, 1}, {0, 0, 1, 59}, {0, 0, 0, 3}, {0, 0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 1}, {0, 0, 0, 0}, {0, 0, 0, 1}, {0, 0, 0, 0}, {0, 0, 0, 0}, {2, 0, 0, 1}, {0, 0, 0, 0}, {0, 1, 0, 0}, {0, 0, 0, 1}, {0, 0, 0, 2}, {0, 0, 0, 0}, {0, 0, 0, 0}} to get a glimpse at some of it. It's the 'a' matrix that's terribly sparse -- the 'b' matrix isn't unreasonable. Elements of the 'a' matrix can be quickly computed by the function aFunction = If[MemberQ[lst, {actors[[#1]], events[[#2]]}], 1, 0] &; That stores a line of code rather than all those ones and zeroes. The b matrix (call it bb this time) can be computed as: Timing[bb = (#1 . Transpose[#1] & ) [Array[aFunction, {Length[actors], Length[events]}]]; ] bb == b {6.1569999999999965*Second, Null} True Bobby Treat -----Original Message----- posting a this was the most helpful solution message to the list, but first I hoped to ask you a follow up question, if I may (and I'll capture your off-line response here in my final posting to the list). I've run the code (copied from you) below, and get the correct output for made-up data, but when I import in real data, I get an error message. Here's the input I'm running: lst = ReadList[c:test1996.txt, {Number, Number}] AdjacenceMatrix[lst : {{_, _} ..}] := Module[{actors, events, adj}, {actors, events} = Union /@ Transpose[lst]; adj = Table[0, {Length[actors]}, {Length[events]}]; Scan[(Part[adj, Sequence @@ #] = 1) &, lst /. MapIndexed[Rule[#1, First[#2]] &, events]]; adj] MatrixForm[a = AdjacenceMatrix[lst]] MatrixForm[b = a.Transpose[a]] And here's what I get for output: Set::partw : Part 300007 of <<1>> does not exist. Set::partw : Part 300007 of <<1>> does not exist. Set::partw : Part 300007 of <<1>> does not exist. General::stop : Further output of Set::partw will be suppressed during this calculation. Then I get the two matrices (a & b as per your code), but they are just filled with zeros. So it gets to about the 4th line of your code, but then doesn't fill-in from my data. Finally, I should note that 30007 is one of the actors in the data that I've read in. In case you want to run this yourself, I've attached the raw data file. There are 166 actors and 1778 events: both actors and events are coded with 6-digit numbers, actors begin with 3's, events with 2's. I'm sure this is a silly question, and that there is an easy answer ... But I sure can't find it. So I really appreciate your help and interest!!!! Tom ********************************************** Thomas P. Moliterno Graduate School of Management University of California, Irvine tmoliter@uci.edu ********************************************** -----Original Message----- If you multiply it by its transpose, you get something else that's useful: lst = {{1, A}, {1, B}, {2, B}, {3, C}, {3, D}, {1, D}, {1, C}}; AdjacenceMatrix[lst : {{_, _} ..}] := Module[{actors, events, adj}, {actors, events} = Union /@ Transpose[lst]; adj = Table[0, {Length[actors]}, {Length[events]}]; Scan[(Part[adj, Sequence @@ #] = 1) &, lst /. MapIndexed[Rule[#1, First[#2]] &, events]]; adj] MatrixForm[a = AdjacenceMatrix[lst]] MatrixForm[b = a.Transpose[a]] Matrix 'b' records how many events two actors have in common. On the diagonal, it shows the total number of events each actor is connected to. It's easy to put zeroes on the diagonal: MatrixForm[c = b (1 - IdentityMatrix[Length[b]])] To get the originally intended incidence matrix, this works: d = c /. {_?Positive -> 1} However, I think matrices 'a' and 'b' are actually more useful, and 'a' easily leads to all the others. Bobby -----Original Message----- AdjacenceMatrix[lst : {{_, _} ..}] := Module[ {actors,events adj}, {events, actors} = Union /@ Transpose[lst]; adj = Table[0, {Length[events]}, {Length[actors]}]; Scan[(Part[adj, Sequence @@ #] = 1) &, lst /. MapIndexed[Rule[#1, First[#2]] &, actors]]; adj ] you get In[]:=AdjacenceMatrix[lst] Out[]={{1, 1, 0, 1}, {0, 1, 0, 0}, {0, 0, 1, 1}} Jens I need to create an adjacency matrix from my data, which is currently in > the form of a .txt file and is basically a two column incidence list. > For example: 1 A > 1 B > 2 B > 3 C > . . > . . > . . > m n Where 1 to m represent actors and A to n represent events. My goal is to > have an (m x m) matrix where cell i,j equals 1 if two actors are > incident to the same event (in the sample above, 1 and 2 are both > incident to B) and 0 otherwise (w/ zeros on the diagonal). I'm new to Mathmatica, and so I'm on the steep part of the learning > curve ... All I've been able to figure out so far is how to get my > incidence list into the program using Import[filename.txt]. But then > what? How do I convert to the adjacency matrix? I've found the > ToAdjacencyMatrix[] command in DiscreteMath`Combinatorica`, but I can't > seem to get it to work ... > Tom ********************************************** > Thomas P. Moliterno > Graduate School of Management > University of California, Irvine > tmoliter@uci.edu > ********************************************** ==== These are working for me. If anybody wants to try this modified keymapping http://66.92.149.152/proprietary/com/wri/proprietary/com/wri/ch05.html http://public.globalsymmetry.com/proprietary/com/wri/KeyEventTranslations.tr .txt STH ==== > These are working for me. If anybody wants to try this modified > keymapping http://66.92.149.152/proprietary/com/wri/proprietary/com/wri/ch05.html > http://public.globalsymmetry.com/proprietary/com/wri/KeyEventTranslations.tr .txt STH I found a bug in my hack of the KeyEventTranslations.tr file which didn't manifest itself in 4.1, but did in 4.2. I had a trailing ',' which was causing Mathematica to hang on startup after the chched KeyEventTranslations.m was created (second launch with the buggy file.) I also discovered that the file I was trying to share had been chmod'ed to 400. I am not sure if that is a daemon running, or scp which did it. It's open now, and I'll keep an eye on it. I'm attaching it as well, just in case. I put it here on my box: ~/.Mathematica/SystemFiles/FrontEnd/TextResources STH @@resource KeyEventTranslations (* Modifiers can be Shift, Control, Command, Option For Macintosh: Command = Command Key, Option = Option Key For X11: Command = Mod1, Option = Mod2 For Windows: Command = Alt, Option = Alt NOTE: I Hacked this for my purposes. I find it more natural. There may be problems. It comes with all the warranty that GNU software does. *) EventTranslations[{ (* Evaluation *) Item[KeyEvent[Enter], EvaluateCells], Item[KeyEvent[KeypadEnter], EvaluateCells], Item[KeyEvent[Return, Modifiers -> {Shift}], EvaluateCells], Item[KeyEvent[KP_Enter], EvaluateCells], Item[KeyEvent[KeypadEnter, Modifiers -> {Shift}], EvaluateNextCell], Item[KeyEvent[KP_Enter, Modifiers -> {Shift}], EvaluateNextCell], Item[KeyEvent[Enter, Modifiers -> {Shift}], EvaluateNextCell], Item[KeyEvent[Return, Modifiers -> {Command}], Evaluate[All]], Item[KeyEvent[Return, Modifiers -> {Option}], SimilarCellBelow], Item[KeyEvent[Escape], ShortNameDelimiter], (* Cursor control *) Item[KeyEvent[Up], MovePreviousLine], Item[KeyEvent[Down], MoveNextLine], Item[KeyEvent[Left], MovePrevious], Item[KeyEvent[Right], MoveNext], Item[KeyEvent[Up, Modifiers -> {Option}], MovePreviousLine], Item[KeyEvent[Down, Modifiers -> {Option}], MoveNextLine], Item[KeyEvent[Right, Modifiers -> {Option}], MoveNextWord], Item[KeyEvent[Left, Modifiers -> {Option}], MovePreviousWord], Item[KeyEvent[Right, Modifiers->{Control}], MoveNextWord], Item[KeyEvent[Left, Modifiers->{Control}], MovePreviousWord], Item[KeyEvent[End], MoveLineEnd], Item[KeyEvent[Home], MoveLineBeginning], (* Selection *) Item[KeyEvent[Right, Modifiers -> {Shift}], SelectNext], Item[KeyEvent[Left, Modifiers -> {Shift}], SelectPrevious], Item[KeyEvent[Right, Modifiers -> {Control, Shift}], SelectNextWord], Item[KeyEvent[Left, Modifiers -> {Control, Shift}], SelectPreviousWord], Item[KeyEvent[Down, Modifiers -> {Shift}], SelectNextLine], Item[KeyEvent[Up, Modifiers -> {Shift}], SelectPreviousLine], Item[KeyEvent[Home, Modifiers -> {Shift}], SelectLineBeginning], Item[KeyEvent[End, Modifiers -> {Shift}], SelectLineEnd], Item[KeyEvent[., Modifiers -> {Control}], ExpandSelection], (* Notebook window control *) Item[KeyEvent[PageUp], ScrollPageUp], Item[KeyEvent[PageDown], ScrollPageDown], Item[KeyEvent[Prior], ScrollPageUp], Item[KeyEvent[Next], ScrollPageDown], Item[KeyEvent[Home, Modifiers -> {Control}], ScrollNotebookStart], Item[KeyEvent[End, Modifiers -> {Control}], ScrollNotebookEnd], (* Input *) Item[KeyEvent[Return], Linebreak], Item[KeyEvent[Tab], Tab], Item[KeyEvent[i, Modifiers -> {Control}], Tab], Item[KeyEvent[Backspace], DeletePrevious], Item[KeyEvent[Delete], DeleteNext], Item[KeyEvent[ForwardDelete], DeleteNext], (* Typesetting input *) Item[KeyEvent[6, Modifiers -> {Control}], Superscript], Item[KeyEvent[Keypad6, Modifiers -> {Control}], Superscript], Item[KeyEvent[^, Modifiers -> {Control}], Superscript], Item[KeyEvent[-, Modifiers -> {Control}], Subscript], Item[KeyEvent[_, Modifiers ->{Control}], Subscript], Item[KeyEvent[/, Modifiers -> {Control}], Fraction], Item[KeyEvent[KP_Divide, Modifiers -> {Control}], Fraction], Item[KeyEvent[2, Modifiers -> {Control}], Radical], Item[KeyEvent[Keypad2, Modifiers -> {Control}], Radical], Item[KeyEvent[@, Modifiers -> {Control}], Radical], Item[KeyEvent[7, Modifiers -> {Control}], Above], Item[KeyEvent[&, Modifiers -> {Control}], Above], Item[KeyEvent[Keypad7, Modifiers -> {Control}], Above], Item[KeyEvent[=, Modifiers -> {Control}], Below], Item[KeyEvent[+, Modifiers -> {Control}], Below], Item[KeyEvent[,, Modifiers -> {Control}], NewColumn], Item[KeyEvent[Return, Modifiers -> {Control}], NewRow], Item[KeyEvent[9, Modifiers -> {Control}], CreateInlineCell], Item[KeyEvent[(, Modifiers -> {Control}], CreateInlineCell], Item[KeyEvent[Keypad9, Modifiers -> {Control}], CreateInlineCell], Item[KeyEvent[), Modifiers -> {Control}], MoveNextCell], Item[KeyEvent[0, Modifiers -> {Control}], MoveNextCell], Item[KeyEvent[Keypad0, Modifiers -> {Control}], MoveNextCell], Item[KeyEvent[Left, Modifiers -> {Control}, CellClass -> BoxFormData], NudgeLeft], Item[KeyEvent[Right, Modifiers -> {Control}, CellClass -> BoxFormData], NudgeRight], Item[KeyEvent[Down, Modifiers -> {Control}, CellClass -> BoxFormData], NudgeDown], Item[KeyEvent[Up, Modifiers -> {Control}, CellClass -> BoxFormData], NudgeUp], Item[KeyEvent[8, Modifiers -> {Control}], InsertRawExpression], Item[KeyEvent[*, Modifiers -> {Control}], InsertRawExpression], Item[KeyEvent[Keypad8, Modifiers -> {Control}], InsertRawExpression], Item[KeyEvent[5, Modifiers -> {Control}, CellClass -> BoxFormData], Otherscript], Item[KeyEvent[Keypad5, Modifiers -> {Control}, CellClass -> BoxFormData], Otherscript], Item[KeyEvent[%, Modifiers -> {Control}, CellClass -> BoxFormData], Otherscript], (* Typesetting motion commands *) Item[KeyEvent[ , Modifiers -> {Control}], MoveExpressionEnd], Item[KeyEvent[Tab, Modifiers -> {Shift}, CellClass -> BoxFormData], MovePreviousPlaceHolder], Item[KeyEvent[s, Modifiers -> {Command, Control}, CellClass -> BoxFormData], MovePreviousExpression], Item[KeyEvent[S, Modifiers -> {Control, Command, Shift}, CellClass -> BoxFormData], MoveNextExpression], Item[KeyEvent[S, Modifiers -> {Control, Shift}, CellClass -> BoxFormData], DeleteNextExpression], Item[KeyEvent[s, Modifiers -> {Control}, CellClass -> BoxFormData], DeletePreviousExpression], Item[KeyEvent[k, Modifiers -> {Control}], CompleteSelection[True]], (* Miscellaneous menu commands *) Item[KeyEvent[Delete, Modifiers -> {Control}], Cut], Item[KeyEvent[Insert, Modifiers -> {Control}], Copy], Item[KeyEvent[Insert, Modifiers -> {Shift}], Paste[After]], Item[KeyEvent[z, Modifiers -> {Control}], Undo], Item[KeyEvent[c, Modifiers -> {Control}], Copy], Item[KeyEvent[x, Modifiers -> {Control}], Cut], Item[KeyEvent[v, Modifiers -> {Control}], Paste[After]], Item[KeyEvent[F1], SelectionHelpDialog] (* Unsupported features and examples *) (* Item[KeyEvent[v, Modifiers -> {Control}], SelectionSpeakSummary], *) (* Item[KeyEvent[v, Modifiers -> {Control, Shift}], SelectionSpeak] *) }] ==== >particular, the y-axis label is typically rotated by 90deg so that it >reads up the y-axis. This works fine on macs and windoze, but under >linux the label runs up the y-axis; however, the letters are rotated >so that they are the same orientation as that for the x-axis. >Printouts of the NB are fine, but it looks really stupid when using a >video projector to teach or give a talk. I have mentioned this before, >perhaps, but it is also a real problem with Mathematica. I personally >don't care why the label looks peculiar, just that it does and that >there is no easy workaround. ==== >I'm writing to report what I consider to be a bug. First, I want to >show a simplified example of the problem. Consider the following >expression: expr=0.22 + x[0] + (3*(-0.16+ x[1]))/4 + (9*(0.546 + x[2]))/16; When simplified I expected to get some real number plus >x[0]+3x[1]/4+9x[2]/16, but instead I get the following: Simplify[expr] 0.407125 + x[0] + 0.75 x[1] + 0.5625 x[2] As you can see, for some reason Mathematica converted the fractions >3/4 and 9/16 to real machine numbers. I consider this to be a bug. You really are not seeing a loss of precision here. When simplify carries out the indicated multiplication such as 9*.546/16 a machine precision number is returned because on of the arguments only has machine precision. It would be incorrect for Mathematica to return a result with greater precision than the arguements. It would also be incorrect for Mathematica to refuse to preform the required multiplications when simplifying this expression. Or said differently, if you want an exact result from Mathematica *all* of the information you supply Mathematica must also be exact. It is not sensible for Mathematica to do otherwise. ==== There is nothing imprecise about a floating point number. Mathematica's non-standard usage is to treat some numbers as intervals. This non-standard usage comes from the attitude that everything I need to know about numerical computations I learned in Freshman physics lab. Anyway, I'm just pointing this out so that you realize that you are not talking about truth and beauty, but about decisions made in the Mathematica design. For example, 0.25 in IEEE floating point format is representing the EXACT value 1/4. The rational number 1/3 does not have an exact corresponding binary floating point number. That does not prevent one from computing the closest number, which from that point on that (other) number is EXACTLY represented. An analysis of your arithmetic or simplification in which each floating point number aaaa*2^bb is changed to that exact rational number may give you more satisfactory answers that an interval-like computation in which all data is submerged in mush. RJF >>It seems to me that you are arguing that if you have an expression >>consisting of one term which is very inprecise and another term which >>is very precise or exact, then the total expression is only as precise >>as the least precise portion of the expression. >> Yes. >>This is total nonsense. >> Not exactly. >>Consider adding the following terms: >1.234567890123456`16 + 0.00000000000000001`1 >consisting of one term with precision 16 and another term with >>precision 1. By your argument, Mathematica should return an answer >>with only a single digit of precision. Of course, Mathematica does no >>such thing. >> I had not considered adding two terms with much different magnitude and much different precision. Consider a different example i.e., 1.234567890123456`16 + 0.1`1 Mathematica does not and should not return a result with 16 digits of precision ==== Bill, See my comments below. There is no logical reason to insist part of the expression to be exact when another part is inexact. You cannot gain more precision than the least precise portion of the expression. Further, there is extra processing overhead associated with maintaining exact epressions as well as additional storage requirements. For a simple expression such as your example the additional overhead is insignificant. But it increases for every exact term in the expression. It doesn't take all that many terms until the overhead associate with exact computation becomes noticeable. > It seems to me that you are arguing that if you have an expression consisting of one term which is very inprecise and another term which is very precise or exact, then the total expression is only as precise as the least precise portion of the expression. This is total nonsense. Consider adding the following terms: 1.234567890123456`16 + 0.00000000000000001`1 consisting of one term with precision 16 and another term with precision 1. By your argument, Mathematica should return an answer with only a single digit of precision. Of course, Mathematica does no such thing. >Even more troubling (to me, at least) is the >following: x[0]+3x[1]/4+9x[2]/16+.4//Simplify 0.4 + x[0] + 0.75 x[1] + 0.5625 x[2] I don't want Simplify to change my nice rational numbers to machine >number approximations. If you want exact answers you *must* have *all* terms in the expresssion exact. Simply put either an expression is exact or not. No expression can be exact unless *all* of the terms within it are exact. When did I ever say that I wanted exact answers? In the example above, I wanted Simplify to do nothing, that is, leave the expression as a sum of an inexact quantity with some exact quantities. In the work where this situation arose, the inexact quantities are typically very small and the exact quantities are large, so that the precision of the overall expression when extended precision numbers are substituted for the x[i] is typically the same as the precision of the numbers being substituted. For example, suppose x[0] and x[2] are zero, and x[1] is 1`25 10^25. Substituting these numbers into the original expression will yield a result with a precision of 25, whereas substituting these numbers into the simplified expression will only have a precision of 16. I've worked very hard to keep the precision of my numbers as high as possible, and I don't want Mathematica to arbitrarily turn those very high precision numbers into much lower precision numbers. Carl Woll Physics Dept U of Washington ==== Bill, Let's step through the expansion here. 9*(0.546+x[2]))/16 can be expanded to 9*0.546/16 + 9*x[2]/16 which becomes .307125 + 9*x[2]/16 My question was why the 9/16 gets converted to .5625, as I see no reason to do so. Even more troubling (to me, at least) is the following: x[0]+3x[1]/4+9x[2]/16+.4//Simplify 0.4 + x[0] + 0.75 x[1] + 0.5625 x[2] I don't want Simplify to change my nice rational numbers to machine number approximations. Carl Woll Physics Dept U of Washington I'm writing to report what I consider to be a bug. First, I want to >show a simplified example of the problem. Consider the following >expression: expr=0.22 + x[0] + (3*(-0.16+ x[1]))/4 + (9*(0.546 + x[2]))/16; When simplified I expected to get some real number plus >x[0]+3x[1]/4+9x[2]/16, but instead I get the following: Simplify[expr] 0.407125 + x[0] + 0.75 x[1] + 0.5625 x[2] As you can see, for some reason Mathematica converted the fractions >3/4 and 9/16 to real machine numbers. I consider this to be a bug. You really are not seeing a loss of precision here. When simplify carries out the indicated multiplication such as 9*.546/16 a machine precision number is returned because on of the arguments only has machine precision. It would be incorrect for Mathematica to return a result with greater precision than the arguements. It would also be incorrect for Mathematica to refuse to preform the required multiplications when simplifying this expression. Or said differently, if you want an exact result from Mathematica *all* of the information you supply Mathematica must also be exact. It is not sensible for Mathematica to do otherwise. > ==== No, Euler proved that series divergent in 1737. It's the usual theorem used to show that while the primes are sparse, they're not as sparse as the squares (as the sum of THEIR inverses converges). Bobby -----Original Message----- Sal. Oppenheim jr. & Cie. KGaA Koenigsberger Strasse 29 D-60487 Frankfurt am Main GERMANY Mobile: +49(0)172 6 74 95 77 Internet: http://www.oppenheim.de ==== Timing[Sum[1./Prime[n], {n, 1000000}]] {13.860000000000001*Second, 3.0682190480544405} Bobby -----Original Message----- Sal. Oppenheim jr. & Cie. KGaA Koenigsberger Strasse 29 D-60487 Frankfurt am Main GERMANY Mobile: +49(0)172 6 74 95 77 Internet: http://www.oppenheim.de ==== That makes about twenty posts this week detailing nothing but problems the FrontEnd). I had been really considering it. Bobby -----Original Message----- resulted in a set of animation control buttons appearing in the bottom frame of the window. I clicked on one of these buttons, but nothing happened. I looke back in the menu and saw M-y as a keyboard shorcut to run an animation. I tried that with no result. I clicked another button in graphics control set, and my X windows locked up. This included the keyboard's ability to give me another display by using Ctl+Alt+F1. I went to another system and ssh-ed in and found Mathematica had over 50% of my user resources, and was climbing. The same was true for VM. I have a gig of physical RAM. Once I killed Mathematica, my X came back to life. I've had several bad experiences with Mathematica and X. I honestly believe there help isolate and fix these. Have others had such problems? STH ==== > That makes about twenty posts this week detailing nothing but problems > the FrontEnd). I had been really considering it. Bobby > A few other points on this topic, if I may. I've found solutions to a lot of the problems I was having, and documented them. For example, I have the keyboard working the way I want it. C-x, C-v, C-c, C-z, all do what I expect them to, as does Delete. I used to have X restarts after extended periods of running Mathematica. That I haven't had recently. I may simply have thrashed X when I clicked too many buttons in sequence. I just wanted to report that it happened. The same thing might happen on XP. I've had my problems there as well. Not with Mathematica, but with other programs. I haven't even attempted Mathematica on XP. I'm sure it is a bit smoother to use. I guess I was just a bit upset at the sense that WRI are still stuck in the just completing my download of KDE 3.1 Beta 2. Things have changed. Gotta go install a few rpms and restart X. Bye, STH ==== > That makes about twenty posts this week detailing nothing but problems > the FrontEnd). I had been really considering it. Bobby Please keep in mind that this was on Mathematica 4.1, I'll have 4.2 tomorrow, and we'll see how that goes. This isn't really the result I had hoped for by I very much wish WRI would give us a dual boot license. That way, we can use Mathematica in either environment the current situation dictates as the best. with Windows. I just don't do that as a matter of course. I've been related issuse, either because they aren't sure if the problem is their that second reason one bit. Hey, if it locks up my X, I want to know if that happens to others, and I want to identify the source of the problem. What I really would like to gain from all my negative statements is solutions to what I perceive as problems. If not that, then perhaps the perspective which makes me understand that what I see as problems are really just my unfamiliarity with the product. I also want to get others platform. That's how open source works. All that being said, if you are in a situation where an X lockup would really do harm to your project, I would say that I cannot claim it won't happen to you. OTOH, I am able to break just about any system. I can't swear to you that Mathematica won't lock up your Windows system. I can say you will not have as easy a time shelling into it and trying to recover your system without rebooting. STH Reply-To: jmt@dxdydz.net ==== Sorry, I thought I had seen a demonstration, some years ago ! > This sum converges, see a math text book ! > Dear Colleagues, I calculated: Sum[1/Prime[n], {n, 15000}] // N Result: 2.74716 Now I wonder if this sum will converge or keep on growing, albeit very > slowly. > Matthias Bode > Sal. Oppenheim jr. & Cie. KGaA > Koenigsberger Strasse 29 > D-60487 Frankfurt am Main > GERMANY > Mobile: +49(0)172 6 74 95 77 > Internet: http://www.oppenheim.de > Reply-To: murray@math.umass.edu ==== It is well known that the infinite series of reciprocals of the primes DIVERGES! See, for example: http://www.utm.edu/research/primes/infinity.shtml > Dear Colleagues, I calculated: Sum[1/Prime[n], {n, 15000}] // N Result: 2.74716 Now I wonder if this sum will converge or keep on growing, albeit very > slowly. > Matthias Bode > Sal. Oppenheim jr. & Cie. KGaA > Koenigsberger Strasse 29 > D-60487 Frankfurt am Main > GERMANY > Mobile: +49(0)172 6 74 95 77 > Internet: http://www.oppenheim.de -- Murray Eisenberg murray@math.umass.edu Mathematics & Statistics Dept. Lederle Graduate Research Tower phone 413 549-1020 (H) University of Massachusetts 413 545-2859 (W) 710 North Pleasant Street Amherst, MA 01375 Reply-To: jmt@dxdydz.net ==== This sum converges, see a math text book ! > Dear Colleagues, I calculated: Sum[1/Prime[n], {n, 15000}] // N Result: 2.74716 Now I wonder if this sum will converge or keep on growing, albeit very > slowly. > Matthias Bode > Sal. Oppenheim jr. & Cie. KGaA > Koenigsberger Strasse 29 > D-60487 Frankfurt am Main > GERMANY > Mobile: +49(0)172 6 74 95 77 > Internet: http://www.oppenheim.de > ==== The fact that Sum[1/Prime[n], {n, 1,Infinity}]==Infinity is a rather famous theorem of Euler. It implies that there must be infinitely many primes (otherwise the sum would be finite), and was the beginning of a vast area of mathematics, which includes such concepts as Dirichlet series, Riemann's zeta function etc. Andrzej Kozlowski Yokohama, Japan http://www.mimuw.edu.pl/~akoz/ http://platon.c.u-tokyo.ac.jp/andrzej/ On Wednesday, October 2, 2002, at 04:31 PM, Matthias.Bode@oppenheim.de > Dear Colleagues, I calculated: Sum[1/Prime[n], {n, 15000}] // N Result: 2.74716 Now I wonder if this sum will converge or keep on growing, albeit very > slowly. > Matthias Bode > Sal. Oppenheim jr. & Cie. KGaA > Koenigsberger Strasse 29 > D-60487 Frankfurt am Main > GERMANY > Mobile: +49(0)172 6 74 95 77 > Internet: http://www.oppenheim.de ==== Here's a start. If you want the thing to be greater than zero, set it equal to z (which we'll assume is greater than zero) and solve for Log[x]: Log[x, a] + Log[a*x, a] == z /. Log[a_*b_] -> Log[a] + Log[b] {f, g} = Log[x] /. Simplify[Solve[%, Log[x]]] Log[a]/Log[x] + Log[a]/(Log[a] + Log[x]) == z {-(((-2 + z + Sqrt[4 + z^2])*Log[a])/(2*z)), ((2 - z + Sqrt[4 + z^2])* Log[a])/(2*z)} Neither solution appears to be extraneous. Now the task is to find the range of these functions over positive z. Take a look at their derivatives: D[f, z] // Simplify D[g, z] // Simplify ((-1 + 2/Sqrt[4 + z^2])* Log[a])/z^2 ((-1 - 2/Sqrt[4 + z^2])*Log[a])/z^2 A little study shows that f' and g' have their signs opposite to Log[a]. Both functions are monotone. The following limits: Outer[Limit[#1, z -> #2] &, {f, g}, {0, Infinity}] Exp@% {{-(Log[a]/2), -Log[a]}, {Infinity*Log[a], 0}} {{1/Sqrt[a], 1/a}, {Indeterminate, 1}} show that f varies from -Log[a]/2 to -Log[a] and g varies from 0 to Infinity if Log[a]>0 and 0 to -Infinity if Log[a]<0. Exponentiation gives ranges for x: -1/Sqrt[a] to 1/a for f, for instance. But what does all this mean? When a>1, either x>1 or 1/a < x < 1/Sqrt[a]. When a<1, either 0, Moderator Reply-To: ==== Try this: To the original function, add a function that's (a) zero when the imaginary part is larger than some epsilon value you choose, (b) fairly large when the imaginary part is 0 or negative, and (c) as smooth as possible. For instance, something like: f=Max[(1 - x/epsilon)^5, 0] This function has four continuous derivatives in the real components -- though not in the complex variable! If it doesn't penalize the real root enough, multiply f by a constant bigger than 1 and try again. Bobby Treat -----Original Message----- are interested, the zeros are the roots of the dispersion relation for a plasma interacting with a laser). Sometimes FindRoot picks up one of these instead of the one I want. So, I'd like to tell mathematica to look for a root only in a certain rectangular region of the complex plane. Well, if I could tell it, 'look for roots with imag. part > something', I'd be happy too. I tried specifying complex values for the start and stop points of an interval, hoping mathematica would interpret these as the corners of a rectangle. No such luck. Any help would be greatly appreciated. I'd also like to point out that this and other issues about complex roots are not clearly addressed in the built-in help files. ==== This sum converges, see a math text book ! That's nonsense http://mathworld.wolfram.com/PrimeSums.html Jens Dear Colleagues, I calculated: Sum[1/Prime[n], {n, 15000}] // N Result: 2.74716 Now I wonder if this sum will converge or keep on growing, albeit very > slowly. > Matthias Bode > Sal. Oppenheim jr. & Cie. KGaA > Koenigsberger Strasse 29 > D-60487 Frankfurt am Main > GERMANY > Mobile: +49(0)172 6 74 95 77 > Internet: http://www.oppenheim.de > ==== You need to see: http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?An um=A016088 >This sum converges, see a math text book ! > >That's nonsense http://mathworld.wolfram.com/PrimeSums.html Jens >Dear Colleagues, I calculated: Sum[1/Prime[n], {n, 15000}] // N Result: 2.74716 Now I wonder if this sum will converge or keep on growing, albeit very >slowly. >Matthias Bode >Sal. Oppenheim jr. & Cie. KGaA >Koenigsberger Strasse 29 >D-60487 Frankfurt am Main >GERMANY >Mobile: +49(0)172 6 74 95 77 >Internet: http://www.oppenheim.de > ==== I have updated my package Biokmod. It can be downloaded from: http://web.usal.es/~guillerm/biokmod.htm I developped this package thinking in biokinetic application and internal dosimetry, but it can be also applied for other kind of problems envolving System of ODE with many variables. I will appreciate any coments Guillermo Sanchez http://web.usal.es/~guillerm ==== I use BinaryImport to read a file as follows: BinaryImport[file,{Byte..}] then I use Partition to rebuild the data of dimensions 252 x 253 x 255.... This works but it takes about half an hour to read the file....it's a 16 MByte file.....any way to improve this? thanks....jerry blimbaum Reply-To: kuska@informatik.uni-leipzig.de ==== BinaryImport[] is complet unusable for this task. I tryed the same with similar data sets and run out of kernel memory (with 1.5 GByte RAM) & 3GB swap). The WRI support respondet, that BinaryImport[] is not for reading *large* binary data. You can have the beta version of MathGL3d 3.1 that has an extra C-Function for that task. Since it also include excelent volume rendering functions and a native format for reading and writing compressed volume data files you should use MathGL3d 3.1b. Contact me direct if you wish to beta test the new version. Jens I use BinaryImport to read a file as follows: BinaryImport[file,{Byte..}] then I use Partition to rebuild the data of dimensions 252 x 253 x 255.... This works but it takes about half an hour to read the file....it's a 16 > MByte file.....any way to improve this? thanks....jerry blimbaum ==== Does any one know of a way to produce Phase Plane Diagrams in mathematica? I need to draw a couple of direction field and analyse the stability of critical points for my systems of equations . Ajadi __________________________________________________ Do You Yahoo!? Everything you'll ever need on one web page http://uk.my.yahoo.com Reply-To: kuska@informatik.uni-leipzig.de ==== if you don't like PlotField[] you should try to get a copy of the book Visual DSolve http://store.wolfram.com/view/book/D0706.str and the book package. Jens > Does any one know of a way to produce Phase Plane > Diagrams in mathematica? > I need to draw a couple of direction field and > analyse the stability of critical points for my > systems of equations . > Ajadi __________________________________________________ > Do You Yahoo!? > Everything you'll ever need on one web page > http://uk.my.yahoo.com ==== Please ignore a similar question I just posted about BinaryImport...I forgot an hour to import 16 MByte file, whereas this technique took less then 5 seconds...special thanks to Mariusz... jerry blimbaum NSWC panama city, fl -----Original Message----- Mariusz Jankowski University of Southern Maine mjkcc@usm.maine.edu 207-780-5580 > The question: > What is the fastest way to read binary files in Mathematica 4.0 ? > I think the fastest is with ReadList (indeed ReadSounFile that use it, seems > to be better than BinaryImport), but when i use this command... > data = ReadList[filePath, Byte] > ....it doesn't read whole file; can somebody tell me the reason? Raf. P.S.: > I made some simple tests (@ 16 bit): << Experimental` > data = BinaryImport[fileRawPath, Table[Integer16, {4000}], > ByteOrdering -> -1]; => 10 seconds data = ReadListBinary[fileRawPath, SignedInt16, ByteOrder - LeastSignificantByteFirst]; = > 26 seconds << Miscellaneous`Audio` > data = ReadSoundfile[fileWavPath] => 1.54 seconds ==== I came across this, and thought I would share. This may be OBE if the font selection is better in 4.2. I'll that installed by this time tomorrow...I hope. http://cgm.cs.mcgill.ca/~luc/math.html I started reading through this list, and had visions of recursively downloading the entire internet looking for the font's I really need. Anybody know what fonts I should have installed on my SuSE box to satisfy Mathematica's default expectations? STH ==== In: DSolve[y*D[u[x, y],x] == x*D[u[x, y],y], u[x,y], {x, y}] Out: {{u[x, y] -> C[1][(1/2)*(x^2 + y^2)]}} Square brackets are used as grouping symbols in the result!?? :^O Somebody say it isn't so. --- Selwyn Hollis ==== It isn't so. C[1] is an arbitrary (smooth) function. After all, what you have got is a partial differential equation. For example, you can take C[1] to be Sin: In[1]:= v[x_, y_] = u[x, y] /. DSolve[y*D[u[x, y], x] == x*D[u[x, y], y], u[x, y], {x, y}] /. C[1] -> Sin Out[1]= {Sin[(1/2)*(x^2 + y^2)]} In[2]:= y*D[v[x, y], x] == x*D[v[x, y], y] Out[2]= True Andrzej Kozlowski Yokohama, Japan http://www.mimuw.edu.pl/~akoz/ http://platon.c.u-tokyo.ac.jp/andrzej/ In: DSolve[y*D[u[x, y],x] == x*D[u[x, y],y], u[x,y], {x, y}] Out: {{u[x, y] -> C[1][(1/2)*(x^2 + y^2)]}} Square brackets are used as grouping symbols in the result!?? :^O Somebody say it isn't so. --- > Selwyn Hollis Reply-To: ==== My OPINION is that extra graphics RAM is useful primarily for 3D games (which are themselves completely useless). A static 3D plot doesn't need that much graphics RAM -- it's the rapid transformation of it, to simulate live action, that may require it. On the other hand, if the manufacturer's limits are that low for graphics RAM, I would get another manufacturer. Who knows what other limitations and compromises they'll saddle you with? I would be very suspicious. Bobby -----Original Message----- Is 32 MB adequate not just now, but likely to be adequate as well for the near future (say, a 3- to 5-year equipment lifetime)? -- Murray Eisenberg murray@math.umass.edu Mathematics & Statistics Dept. Lederle Graduate Research Tower phone 413 549-1020 (H) University of Massachusetts 413 545-2859 (W) 710 North Pleasant Street Amherst, MA 01375 Reply-To: kuska@informatik.uni-leipzig.de ==== My OPINION is that extra graphics RAM is useful primarily for 3D games Or complex 3d scientific visualizations ? Textures ? Try to render the skull from a 256^3 CT scan or the output of a 3d plasma simulation and you will know *what* can be done with extra 3d RAM > (which are themselves completely useless). The major effect is, that 3d Games make 3d graphics hardware less expensive. Five years ago a SGI cost 20-30 000 $ and today you can have more 3d power for 400 $ Jens > A static 3D plot doesn't > need that much graphics RAM -- it's the rapid transformation of it, to > simulate live action, that may require it. On the other hand, if the manufacturer's limits are that low for > graphics RAM, I would get another manufacturer. Who knows what other > limitations and compromises they'll saddle you with? I would be very > suspicious. Bobby -----Original Message----- We are about to order new PCs for a university student lab in which > Mathematica will be installed. Of course they will be using 2D and 3D > graphics -- plots of surfaces, e.g. Sooner or later students will want > to rotate such plots, too. An unresolved issue is how much graphics RAM to get. On existing > machines we typically have 64MB. But for the PCs we are looking at, > manufacturer's limits on graphics RAM, rather than cost, seems to limit > us to 32 MB. Is 32 MB adequate not just now, but likely to be adequate as well for > the near future (say, a 3- to 5-year equipment lifetime)? -- > Murray Eisenberg murray@math.umass.edu > Mathematics & Statistics Dept. > Lederle Graduate Research Tower phone 413 549-1020 (H) > University of Massachusetts 413 545-2859 (W) > 710 North Pleasant Street > Amherst, MA 01375 ==== Greetings MathGroup, My name is Steve Earth, and I am a new subscriber to this list and also a new user of Mathematica; so please forgive this rather simple question... I would like to enter the quartic x^4 + x^3 + x^2 + x + 1 into Mathematica and have it be able to tell me that it factors into (x^2 + GoldenRatio x + 1) ( x^2 - 1/GoldenRatio x + 1) What instructions do I need to execute to achieve this output? -Steve Earth Harker School http://www.harker.org/ ==== Steve The notebook given after NOTEBOOK below contains functions for factoring and partial fractioning. Here is an application to your problem: the first stage avoids our needing to know anything about the answer. fc=FactorR[x^4+x^3+x^2+x+1,x] (1 - (1/2)*(-1 - Sqrt[5])*x + x^2)* (1 - (1/2)*(-1 + Sqrt[5])*x + x^2) Now we need to get rid of Sqrt[5] in terms of GoldenRatio. This is rather messy: Simplify/@(fc/. Sqrt[5][Rule]2 GoldenRatio-1) (1 + x - GoldenRatio*x + x^2)*(1 + GoldenRatio*x + x^2) Simplify/@(%/.-GoldenRatio[Rule] 1/GoldenRatio -1) (1 + x/GoldenRatio + x^2)*(1 + GoldenRatio*x + x^2) Another example PartialFractionsR[(1 + x)x/(1 - 3*x + x^2), x] 1 - (2*(-1 + 4*x))/((3 + Sqrt[5] - 2*x)*(-3 + 2*x)) + (2*(-1 + 4*x))/((-3 + 2*x)*(-3 + Sqrt[5] + 2*x)) Simplify[%] (x*(1 + x))/(1 - 3*x + x^2) NOTEBOOK: to make a notebook from the following, copy from the next line to the line preceding XXX and paste into a new Mathematica notebook. Notebook[{ Cell[CellGroupData[{ Cell[Factors and PartialFractions, Subtitle], Cell[Allan Hayes, 16 August 2001, Text], Cell[< Here are some functions for factoring and expressing in partial fractions over the reals and over the complex numbers. >, Text], Cell[BoxData[ (Quit)], Input], Cell[BoxData[{ (Off[General::spell1, General::spell]), n, ((FactorC::usage = ;)n), n, ((FactorR::usage = ;)n), n, ((PartialFractionsC::usage = ;)n), n, ((PartialFractionsR::usage = ;)), n, (On[General::spell1, General::spell])}], Input, InitializationCell->True], Cell[TextData[{ FactorC[p_, x_] := , StyleBox[(*over complex numbers*), FontFamily->Arial, FontWeight->Plain], nTimes @@ Cases[Roots[p == 0, x, n Cubics -> False], u_ == v_ -> x - v]n nFactorR[p_, x_] := , StyleBox[(*over reals, coefficients must be real*), FontFamily->Arial, FontWeight->Plain], n (Times @@ Join[Cases[#1, u_ == v_ /; Im[v] == 0 :> n x - v], Cases[#1, u_ == v_ /; Im[v] > 0 :> n x^2 - x*2*Re[v] + Abs[v]^2]] & )[n Roots[p == 0, x, Cubics -> False]] }], Input, InitializationCell->True], Cell[TextData[{ PartialFractionsC[p_, x_] := , StyleBox[(*over complex numbers*), FontFamily->Arial, FontWeight->Plain], n(#+Apart[#2/FactorC[#3,x]])&@@Flatten[{PolynomialReduce[#,#2], #2}]&[Numerator[#],Denominator[#]]&[Together[p]]n n PartialFractionsR[p_, x_] := , StyleBox[(*over reals, coefficients must be real*), FontFamily->Arial, FontWeight->Plain], n(#+Apart[#2/FactorR[#3,x]])&@@Flatten[{PolynomialReduce[#,#2], #2}]&[Numerator[#],Denominator[#]]&[Together[p]] }], Input, InitializationCell->True], Cell[CellGroupData[{ Cell[PROGRAMMING NOTES, Subsubsection], Cell[TextData[{ The option , StyleBox[Cubics->False, FontFamily->Courier], is used to keep the roots of cubics in , StyleBox[Root[....], FontFamily->Courier], form. This is better for computation.n, StyleBox[Re[v], FontFamily->Courier], and , StyleBox[Abs[v]^2, FontFamily->Courier], are used rather than , StyleBox[v+Conjugate[v] , FontFamily->Courier], and , StyleBox[v*Conjugate[v], FontFamily->Courier], to prevent , StyleBox[Apart, FontFamily->Courier], from factorising , StyleBox[x^2 - x*2*Re[v] + Abs[v]^2], FontFamily->Courier], back to complex form. }], Text] }, Closed]], Cell[CellGroupData[{ Cell[EXAMPLES, Subsubsection], Cell[pol = Expand[(x - 1)*(x + 1)^2*(x^2 + x + 1)^2*(x^2 + 4)]; , Input], Cell[CellGroupData[{ Cell[f1 = FactorC[pol, x], Input], Cell[BoxData[ ((((-1) + x)) (((-2) [ImaginaryI] + x)) ((2 [ImaginaryI] + x)) ((1 + x))^2 (((((-1)))^(1/3) + x))^2 (((-(((-1)))^(2/3)) + x))^2)], Output] }, Open ]], Cell[CellGroupData[{ Cell[f2 = FactorR[pol, x], Input], Cell[BoxData[ ((((-1) + x)) ((1 + x))^2 ((4 + x^2)) ((1 + x + x^2))^2)], Output] }, Open ]], Cell[CellGroupData[{ Cell[f3 = FactorR[x^3 + x + 1, x], Input], Cell[BoxData[ (((x - Root[1 + #1 + #1^3 &, 1])) ((x^2 - 2 x Root[(-1) + 2 #1 + 8 #1^3 &, 1] + Root[(-1) - #1^4 + #1^6 &, 2]^2)))], Output] }, Open ]], Cell[< Root objects appear because of the option Cubics->False in Roots. We can sometimes get radical forms, but notice the complication. >, Text], Cell[CellGroupData[{ Cell[ToRadicals[f3], Input], Cell[BoxData[ (((((2/(3 (((-9) + @93)))))^(1/3) - ((1/2 (((-9) + @93))))^(1/3)/3^(2/3) + x)) ((1/3 + 1/3 ((29/2 - (3 @93)/2))^(1/3) + 1/3 ((1/2 ((29 + 3 @93))))^(1/3) - 2 ((((1/2 ((9 + @93))))^(1/3)/(2 3^(2/3)) - 1/(2^(2/3) ((3 ((9 + @93))))^(1/3)))) x + x^2)))], Output] }, Open ]], Cell[Inexact forms can be found, from f3 :, Text], Cell[CellGroupData[{ Cell[N[f3], Input], Cell[BoxData[ (((((0.6823278038280193`)([InvisibleSpace])) + x)) ((((1.4655712318767682`)([InvisibleSpace])) - 0.6823278038280193` x + x^2)))], Output] }, Open ]], Cell[or directly, Text], Cell[CellGroupData[{ Cell[f3 = FactorR[x^3 + x + 1//N, x], Input], Cell[BoxData[ (((((0.6823278038280193`)([InvisibleSpace])) + x)) ((((1.4655712318767682`)([InvisibleSpace])) - 0.6823278038280193` x + x^2)))], Output] }, Open ]], Cell[Partial fractions, Text], Cell[CellGroupData[{ Cell[pf1 = PartialFractionsR[(2 + x)/pol, x], Input], Cell[BoxData[ (1/(60 (((-1) + x))) - 1/(10 ((1 + x))^2) - 39/(100 ((1 + x))) + ((-54) - 31 x)/(4225 ((4 + x^2))) + ((-1) + 3 x)/(13 ((1 + x + x^2))^2) + (44 + 193 x)/(507 ((1 + x + x^2))))], Output] }, Open ]], Cell[CellGroupData[{ Cell[pf2 = PartialFractionsR[(1 + x)x/(1 - 3*x + x^2), x], Input], Cell[< 1 - (2*(-1 + 4*x))/((3 + Sqrt[5] - 2*x)*(-3 + 2*x)) + (2*(-1 + 4*x))/((-3 + 2*x)*(-3 + Sqrt[5] + 2*x)) >, Output] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ (Simplify[%])], Input], Cell[(x*(1 + x))/(1 - 3*x + x^2), Output] }, Open ]], Cell[Partial fractions will often involve Root objects , Text], Cell[CellGroupData[{ Cell[pf3 = PartialFractionsR[(1 + x)/(x^3 - x + 1), x], Input], Cell[BoxData[ (((1 + Root[1 - #1 + #1^3 &, 1]))/((((x - Root[1 - #1 + #1^3 &, 1])) ((Root[1 - #1 + #1^3 &, 1]^2 - 2 Root[1 - #1 + #1^3 &, 1] Root[(-1) - 2 #1 + 8 #1^3 &, 1] + Root[(-1) + #1^4 + #1^6 &, 2]^2)))) + ((x + Root[1 - #1 + #1^3 &, 1] + x Root[1 - #1 + #1^3 &, 1] - 2 Root[(-1) - 2 #1 + 8 #1^3 &, 1] - Root[(-1) + #1^4 + #1^6 &, 2]^2))/(((((-x^2) + 2 x Root[(-1) - 2 #1 + 8 #1^3 &, 1] - Root[(-1) + #1^4 + #1^6 &, 2]^2)) ((Root[1 - #1 + #1^3 &, 1]^2 - 2 Root[1 - #1 + #1^3 &, 1] Root[(-1) - 2 #1 + 8 #1^3 &, 1] + Root[(-1) + #1^4 + #1^6 &, 2]^2)))))], Output] }, Open ]], Cell[This can in fact be put in radical form:, Text], Cell[CellGroupData[{ Cell[ToRadicals[pf3], Input], Cell[BoxData[ (((1 - ((2/(3 ((9 - @69)))))^(1/3) - ((1/2 ((9 - @69))))^(1/3)/3^(2/3)))/(((((-(1/3)) + 1/3 ((25/2 - (3 @69)/2))^(1/3) + 1/3 ((1/2 ((25 + 3 @69))))^(1/3) + (((-((2/(3 ((9 - @69)))))^(1/3)) - ((1/2 ((9 - @69))))^(1/3)/3^(2/3)))^2 - 2 (((-((2/(3 ((9 - @69)))))^(1/3)) - ((1/2 ((9 - @69))))^(1/3)/3^(2/3))) ((1/24 ((864 - 96 @69))^(1/3) + ((1/2 ((9 + @69))))^(1/3)/(2 3^(2/3)))))) ((((2/(3 ((9 - @69)))))^(1/3) + ((1 /2 ((9 - @69))))^(1/3)/3^(2/3) + x)))) + ((1/3 - 1/3 ((25/2 - (3 @69)/2))^(1/3) - ((2/(3 ((9 - @69)))))^(1/3) - ((1/2 ((9 - @69))))^(1/3)/3 ^(2/3) - 1/3 ((1/2 ((25 + 3 @69))))^(1/3) - 2 ((1/24 ((864 - 96 @69))^(1/3) + ((1/2 ((9 + @69))))^(1/3)/(2 3^(2/3)))) + x + (((-((2/(3 ((9 - @69)))))^(1/3)) - ((1/2 ((9 - @69))))^(1/3)/3^(2/3))) x))/(((((-(1 /3)) + 1/3 ((25/2 - (3 @69)/2))^(1/3) + 1/3 ((1/2 ((25 + 3 @69))))^(1/3) + (((-((2/(3 ((9 - @69)))))^(1/3)) - ((1/2 ((9 - @69))))^(1/3)/3^(2/3)))^2 - 2 (((-((2/(3 ((9 - @69)))))^(1/3)) - ((1/2 ((9 - @69))))^(1/3)/3^(2/3))) ((1/24 ((864 - 96 @69))^(1/3) + ((1/2 ((9 + @69))))^(1/3)/(2 3^(2/3)))))) ((1/3 - 1/3 ((25/2 - (3 @69)/2))^(1/3) - 1/3 ((1/2 ((25 + 3 @69))))^(1/3) + 2 ((1/24 ((864 - 96 @69))^(1/3) + ((1/2 ((9 + @69))))^(1/3)/(2 3^(2/3)))) x - x^2)))))], Output] }, Closed]], Cell[We could have found the inexact form directly., Text], Cell[CellGroupData[{ Cell[BoxData[ (PartialFractionsR[((1 + x))/((x^3 - x + 1)) // N, x])], Input], Cell[BoxData[ ((-(0.07614206365252976`/(((1.324717957244746`)( [InvisibleSpace])) + 1.` x))) + (((0.7982664819556426`)( [InvisibleSpace])) + 0.07614206365252976` x)/(((0.754877666246693`)([InvisibleSpace])) - 1.324717957244746` x + 1.` x^2))], Output] }, Open ]] }, Closed]] }, Open ]] }, ScreenRectangle->{{0, 1024}, {0, 709}}, AutoGeneratedPackage->None, WindowSize->{534, 628}, WindowMargins->{{199, Automatic}, {0, Automatic}}, ShowCellLabel->False, StyleDefinitions -> Default.nb ] XXX -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay@haystack.demon.co.uk Voice: +44 (0)116 271 4198 > Greetings MathGroup, My name is Steve Earth, and I am a new subscriber to this list and also a > new user of Mathematica; so please forgive this rather simple question... I would like to enter the quartic x^4 + x^3 + x^2 + x + 1 into Mathematica > and have it be able to tell me that it factors into (x^2 + GoldenRatio x + 1) ( x^2 - 1/GoldenRatio x + 1) What instructions do I need to execute to achieve this output? -Steve Earth > Harker School > http://www.harker.org/ > Reply-To: kuska@informatik.uni-leipzig.de ==== In[]:=Factor[x^4 + x^3 + x^2 + x + 1, Extension -> {GoldenRatio, 1/GoldenRatio}] Out[]=-((-3 - 2*x + Sqrt[5]*x + GoldenRatio*x - 3*x^2)* (3 + x + Sqrt[5]*x + GoldenRatio*x + 3*x^2))/9 Jens Greetings MathGroup, My name is Steve Earth, and I am a new subscriber to this list and also a > new user of Mathematica; so please forgive this rather simple question... I would like to enter the quartic x^4 + x^3 + x^2 + x + 1 into Mathematica > and have it be able to tell me that it factors into (x^2 + GoldenRatio x + 1) ( x^2 - 1/GoldenRatio x + 1) What instructions do I need to execute to achieve this output? -Steve Earth > Harker School > http://www.harker.org/ ==== > The last part of my message you are quoting was completely wrong, as > was pointed out by Allan Hayes. Mathematica does not track precision of > machine arithmetic computations. In order for Mathematica to give > reliable information about the precision of a computation you have to > explicitly set the precision of all the numerical quantities. Your own example at the bottom simply shows you have not understood the > evaluation mechanism of Mathematica. Just opposite, thanks to you and other participants, I completely > understood it. SetAccuracy just takes anything and calls it accurate. > This behavior is useless if not stupid. I am not sure I understand what you are referring to as useless if not stupid. The main purpose of SetAccuracy is to allow people who have done their own error analysis to specify the numerical error in an input or in a result. It is often possible through careful numerical analysis, for example, to come up with a better error estimate than can be given by generic rules for propogation of error. Another common use of SetAccuracy is for converting machine numbers or exact numbers into variable-precision numbers in situations when it is desired that a calculation be done using variable-precision arithmetic. Is there some aspect of this that you think is useless if not stupid, or was that remark referring to something else? Dave Withoff Wolfram Research ==== > The more I play with the example the more depressing it gets. Start > with floating point numbers but explicitely arbitrary-precision ones. In[1]:= > a=77617.00000000000000000000000000000; > b=33095.00000000000000000000000000000; In[3]:= > !(333.7500000000000000000000000000000 b^6 + a^2 ((11 a^2 > b^2 - > b^6 - 121 b^4 - 2)) + 5.500000000000000000000000000000 b^8 + > a/(2 > b)) Out[3]= > !((-4.78339168666055402578083604864320577443814`26.6715*^32)) In[4]:= > Accuracy[%] Out[4]= > -6 Due to the manual section 3.1.6: When you do calculations with arbitrary-precision numbers, as > discussed in the previous section, Mathematica always keeps track of > the precision of your results, and gives only those digits which are > known to be correct, given the precision of your input. When you do > calculations with machine-precision numbers, however, Mathematica > always gives you a machine?precision result, whether or not all the > digits in the result can, in fact, be determined to be correct on the > basis of your input. Because I started with arbitrary-precision numbers Mathematica should display > only those digits that are correct, that is none. An accuracy of -6 means that the least significant correct digit is 6 digits to the left of the decimal point. The result Out[3] in the example above has 26 significant digits to the left of that (the most sigificant digit is 26+6=32 digits to the left of the decimal point), so there are 26 correct digits to display. Was there some other result you were referring to as a result in which the number of correct digits is none? Dave Withoff Wolfram Research ==== > I came across this, and thought I would share. This may be OBE if the font > selection is better in 4.2. I'll that installed by this time tomorrow...I > hope. http://cgm.cs.mcgill.ca/~luc/math.html I started reading through this list, and had visions of recursively > downloading the entire internet looking for the font's I really need. > Anybody know what fonts I should have installed on my SuSE box to satisfy > Mathematica's default expectations? The contents of the page at this URL are interesting, but probably not relevant for your purposes. The only fonts which are absolutely needed by the front end are the fonts that supply the special glyphs for mathematical notation, grouping characters, Greek letters, etc. These fonts were developed by Wolfram Research and are installed as part of the Mathematica installation. There are two generations of these fonts. The first generation was introduced in Mathematica 3.0 in the fall of 1996 and were used through the release of Mathematica 4.1. There were five families known as Math1 - Math5. Each familiy had four variants: a proportionaly-spaced medium face, a monospaced medium, a proportional bold, and a monospaced bold. The second generation is used by Mathematica 4.2. There are seven families named Mathematica1 - Mathematica7. Aside from the math fonts, the front end should be able to function properly provided that you have a font that supports the encoding for your chosen locale. The style sheets bundled with the front end use only Times, Helvetica, and Courier. Should these fonts not be available on your system, the front end has some substitution rules. For example, the Windows front end knows to use the fonts Times New Roman, Arial, and Courier New. The X Window System ships with bitmap versions of Times, Helvetica, and Courier as well as an outline of Courier. If one of these fonts must be drawn at a size for which there are no bitmaps, outline fonts provided with Mathematica are used. Helvetica is aliased to Swiss721, and Times is aliased to Utopia through a fonts.alias file in the Mathematica fonts directory. Under MacOS and Windows, you should be albe to use whatever fonts are available on your system. Under X, things are a little more complicated. The front end can display whatever fonts are made available to your X server, but it can generate PostScript only for those fonts where an Adobe Font Metric (AFM) file is available. If you wish to display or print the PostScript, you must also make the Type 1 font file available to the rendering device. Note also that the X front end has an adjustable setting for the amount of memory to reserve for storing font data. If your system has a large number of fonts, you may need to increase this setting per this FAQ page: http://support.wolfram.com/mathematica/systems/linux/interface/fonterrors.ht ml -- User Interface Programmer paulh@wolfram.com Wolfram Research, Inc. ==== > I came across this, and thought I would share. This may be OBE if the >> font >> selection is better in 4.2. I'll that installed by this time >> tomorrow...I hope. >> http://cgm.cs.mcgill.ca/~luc/math.html >> I started reading through this list, and had visions of recursively >> downloading the entire internet looking for the font's I really need. >> Anybody know what fonts I should have installed on my SuSE box to satisfy >> Mathematica's default expectations? The contents of the page at this URL are interesting, but probably not > relevant for your purposes. That's kind'o' what I thought. [snip - history lesson - thanks] > The second generation is used by Mathematica 4.2. There are seven > families named Mathematica1 - Mathematica7. Where is the documentation for installing these? This is dated: http://support.wolfram.com/mathematica/systems/linux/interface/fonterrors.ht ml I grabbed these off the net: http://support.wolfram.com/mathematica/systems/linux/general/latestfonts.htm l http://support.wolfram.com/mathematica/systems/linux/general/MathBDF_42.tar. gz http://support.wolfram.com/mathematica/systems/linux/general/MathPCF_42.tar. gz http://support.wolfram.com/mathematica/systems/linux/general/MathT1_42.tar.g z > su - ****** # cd /usr/X11/lib/X11/fonts/ # mkdir -p local/mma # cd local/mma # tar xvfz /download/com/wri/MathBDF_42.tar.gz # tar xvfz /download/com/wri/MathPCF_42.tar.gz # tar xvfz /download/com/wri/MathT1_42.tar.gz # xemacs /etc/X11/XF86Config .... # grep FontPath /etc/X11/XF86Config FontPath /usr/X11R6/lib/X11/fonts/100dpi:unscaled FontPath /usr/X11R6/lib/X11/fonts/75dpi:unscaled FontPath /usr/X11R6/lib/X11/fonts/CID FontPath /usr/X11R6/lib/X11/fonts/Speedo FontPath /usr/X11R6/lib/X11/fonts/Type1 FontPath /usr/X11R6/lib/X11/fonts/URW FontPath /usr/X11R6/lib/X11/fonts/kwintv:unscaled FontPath /usr/X11R6/lib/X11/fonts/latin2/Type1 FontPath /usr/X11R6/lib/X11/fonts/misc:unscaled FontPath /usr/X11R6/lib/X11/fonts/misc/sgi:unscaled FontPath /usr/X11R6/lib/X11/fonts/truetype FontPath /usr/X11R6/lib/X11/fonts/uni:unscaled FontPath /usr/X11R6/lib/X11/fonts/local/mma/BDF FontPath /usr/X11R6/lib/X11/fonts/local/mma/PCF FontPath /usr/X11R6/lib/X11/fonts/local/mma/T1 # SuSEconfig .... # init 3 [assuming you are already at a TTY console] # init 5 > Aside from the math fonts, the front end should be able to function > properly provided that you have a font that supports the encoding for your > chosen locale. The style sheets bundled with the front end use only > Times, Helvetica, and Courier. Should these fonts not be available on > your system, the front end has some substitution rules. For example, the Windows front end knows to use the fonts Times New Roman, > Arial, and Courier New. The X Window System ships with bitmap versions of > Times, Helvetica, and Courier as well as an outline of Courier. If one of > these fonts must be drawn at a size for which there are no bitmaps, > outline fonts provided with Mathematica are used. Helvetica is aliased to > Swiss721, and Times is aliased to Utopia through a fonts.alias file in the > Mathematica fonts directory. This is what I find anoying. Every time Mathematica does one of these substitutions, it beeps. This is what I'm calling a font fault. It's like the boy who cried wolf. I start ignoring beeps. It also bothers me that I am not seeing what the author had intended. I *believe* it is the author of the notebook or help document who determines what fonts should be used. This is a point of confusion. When I opened a help page, and didn't like the size of the fonts, I tried to adjust them without the desired result. I now believe the proper remedy is to use magnification, not a font adjustment. I have far fewer chirps (beeps) in 4.2, but I do get them when changing the magnification. > Under MacOS and Windows, you should be albe to use whatever fonts are > available on your system. Under X, things are a little more complicated. > The front end can display whatever fonts are made available to your X > server, but it can generate PostScript only for those fonts where an Adobe > Font Metric (AFM) file is available. If you wish to display or print the > PostScript, you must also make the Type 1 font file available to the > rendering device. This is for another day. I believe I have done this in years gone by, but, for now, I just want to get the optimal behavior form the crt. [snip] Here's my question in a nutshell. When I do xlsfonts, what should be listed in order to run Mathematica 4.2 and not experiece font faults generated by content provided on the CD? Please note that I *just* finished installing the fonts off the web, so I'm not sure what, if any problems still remain. I suspect the helvetica faults will still occur. STH . ==== I'm a newbie and, of course, the first thing I want to do is apparently one of the most complicated... I have an expression that looks like this: A + B/C + D*Sqrt[E]/C = 0 A,B,C,D, & E are all polynomials in x I want it to look like this (D^2)*E = (A*C + B)^2 At that point, I'll have polynomials in x on both sides. Finally, I want the equation to be written out with terms grouped by powers of x, but I think I can do that part :) I'll be very grateful to anyone who can give me some pointers. Or, at least point me to some tutorial in the Mathematica documentation. I've been looking over the documentation and I found Appendix A.5 in The Mathematica Book, but that doesn't help me. I _need_ some examples. I did find a couple of well-written posts in this newsgroup, but not quite close enough to what I want. Troy. =-=-=-=-=-=-=-=-=-= FYI, here's the expression I'm working with. denom = Sqrt[(B^2 - r^2)^2 + 4*(r^2)*(b^2)] cnu = (2*b^2 - B^2 + r^2)/denom snu = -2*b*Sqrt[B^2 - b^2]/denom sif = 2*r*b/denom cif = (r^2 - B^2)/denom pdr = -Cos[ds]*Sin[q]*(snu*cif + cnu*sif) - Sin[ds]*(cnu*cif - snu*sif) 0 == -(B^2 - b^2)*V^2/(r^2) + (((B*V)^2)/( r^2) - 2*w*b*V*Cos[q]*Cos[ds] + (w* r)^2 - (w*r*pdr)^2)*(Cos[qr])^2 Although I said it's a polynomial in x, it's really a polynomial in b that I'm after. ==== Troy, True, interactive manipulation can be difficult. However, here is one way to do what you want. We have to do the same thing to both sides of the equation. (# - D*Sqrt[K]/C)&/@(A+B/C+D*Sqrt[K]/C[Equal]0 A + B/C == -((D*Sqrt[K])/C) Together/@% (B + A*C)/C == -((D*Sqrt[K])/C) #C&/@% B + A*C == (-D)*Sqrt[K] #^2&/@% (B + A*C)^2 == D^2*K NOTES. Here is how #C&/@ (lhs ==rhs) works: #C&/@ (lhs ==rhs) --> #C&[lhs]==#C&[rhs] --> lhs C == rhs C --> ... f/@( expr) is special for for Map[f, expr] expr& is special for Function[expr] Please look up Map and Function in the Help Browser. -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay@haystack.demon.co.uk Voice: +44 (0)116 271 4198 > I'm a newbie and, of course, the first thing I want to do is apparently > one of the most complicated... I have an expression that looks like this: A + B/C + D*Sqrt[E]/C = 0 A,B,C,D, & E are all polynomials in x > I want it to look like this (D^2)*E = (A*C + B)^2 At that point, I'll have polynomials in x on both sides. Finally, I > want the equation to be written out with terms grouped by powers of x, > but I think I can do that part :) I'll be very grateful to anyone who can give me some pointers. Or, at > least point me to some tutorial in the Mathematica documentation. I've > been looking over the documentation and I found Appendix A.5 in The > Mathematica Book, but that doesn't help me. I _need_ some examples. I > did find a couple of well-written posts in this newsgroup, but not quite > close enough to what I want. > Troy. =-=-=-=-=-=-=-=-=-= FYI, here's the expression I'm working with. > denom = Sqrt[(B^2 - r^2)^2 + 4*(r^2)*(b^2)] > cnu = (2*b^2 - B^2 + r^2)/denom > snu = -2*b*Sqrt[B^2 - b^2]/denom > sif = 2*r*b/denom > cif = (r^2 - B^2)/denom pdr = -Cos[ds]*Sin[q]*(snu*cif + > cnu*sif) - Sin[ds]*(cnu*cif - snu*sif) 0 == -(B^2 - b^2)*V^2/(r^2) + (((B*V)^2)/( > r^2) - 2*w*b*V*Cos[q]*Cos[ds] + (w* > r)^2 - (w*r*pdr)^2)*(Cos[qr])^2 Although I said it's a polynomial in x, it's really a polynomial in b > that I'm after. > ==== > Troy, > True, interactive manipulation can be difficult. > However, here is one way to do what you want. > We have to do the same thing to both sides of the equation. (# - D*Sqrt[K]/C)&/@(A+B/C+D*Sqrt[K]/C[Equal]0 A + B/C == -((D*Sqrt[K])/C) I think I have to apologize for the lack of clarity in my original post. I had tried to word it carefully, but I deceived myself. I should have said: I have an expression that can be put into this form: A + B/C + D*Sqrt[K]/C = 0 A,B,C,D, & K are all polynomials in x I need to get it into that form and, in the end, I want it to look like this (D^2)*K = (A*C + B)^2 I think I gave the impression that I have polynomials A,B,C,D, & K at my fingertips. I don't. The expression I have is given at the end of this message. I'm still trying to digest the respones I've garned so far. In the meantime, I decided to post this clarification. > I'm a newbie and, of course, the first thing I want to do is >> apparently one of the most complicated... >> I have an expression that looks like this: >> A + B/C + D*Sqrt[K]/C = 0 >> A,B,C,D, & K are all polynomials in x >> I want it to look like this >> (D^2)*K = (A*C + B)^2 >> At that point, I'll have polynomials in x on both sides. Finally, I >> want the equation to be written out with terms grouped by powers of x, >> but I think I can do that part :) >> I'll be very grateful to anyone who can give me some pointers. Or, >> at least point me to some tutorial in the Mathematica documentation. >> I've been looking over the documentation and I found Appendix A.5 in >> The Mathematica Book, but that doesn't help me. I _need_ some >> examples. I did find a couple of well-written posts in this newsgroup, >> but not quite close enough to what I want. >> Troy. >> =-=-=-=-=-=-=-=-=-= >> FYI, here's the expression I'm working with. >> denom = Sqrt[(B^2 - r^2)^2 + 4*(r^2)*(b^2)] >> cnu = (2*b^2 - B^2 + r^2)/denom >> snu = -2*b*Sqrt[B^2 - b^2]/denom >> sif = 2*r*b/denom >> cif = (r^2 - B^2)/denom >> pdr = -Cos[ds]*Sin[q]*(snu*cif + >> cnu*sif) - Sin[ds]*(cnu*cif - snu*sif) >> 0 == -(B^2 - b^2)*V^2/(r^2) + (((B*V)^2)/( >> r^2) - 2*w*b*V*Cos[q]*Cos[ds] + (w* >> r)^2 - (w*r*pdr)^2)*(Cos[qr])^2 >> Although I said it's a polynomial in x, it's really a polynomial in >> b that I'm after. >> ==== I'm only a little embarassed for not having realized what was happening. (Perhaps I should have slept on it.) Surely I'm not alone in thinking this symbolism is highly nonintuitive. But of course, for it to be otherwise would require another protected symbol... --- Selwyn >>In: DSolve[y*D[u[x, y],x] == x*D[u[x, y],y], u[x,y], {x, y}] >>Out: {{u[x, y] -> C[1][(1/2)*(x^2 + y^2)]}} >>Square brackets are used as grouping symbols in the result!?? :^O >>Somebody say it isn't so. >> > It isn't so The square bracket is not delineating a factor it is enclosing the argument > to an arbitrary function named C[1]. While the function is dependent on both > x and y the dependence only occurs in the combination (x^2+y^2). > Bob Hanlon > Reply-To: ==== It isn't so. The solution is an arbitrary function of (1/2)*(x^2 + y^2)]}}. Bobby -----Original Message----- ==== Does anyone know what happened to the < in Mathematica 4.x? In some version I know I used it, but it seems to > have gone away. It allowed for real time manipulation of 3D graphics. > Ray Gittings ==== it's still there << RealTime3D` but MathGL3d may be the better solution > http://phong.informatik.uni-leipzig.de/~kuska/mathgl3dv3/ Jens >>Does anyone know what happened to the <>in Mathematica 4.x? In some version I know I used it, but it seems to >>have gone away. It allowed for real time manipulation of 3D graphics. >>Ray Gittings ==== The more I play with the example the more > depressing it gets. Start > with floating point numbers but explicitly > arbitrary-precision ones. In[1]:= > a=77617.00000000000000000000000000000; > b=33095.00000000000000000000000000000; In[3]:= > !(333.7500000000000000000000000000000 b^6 + > a^2 ((11 a^2 > b^2 - > b^6 - 121 b^4 - 2)) + > 5.500000000000000000000000000000 b^8 + > a/(2 > b)) Out[3]= !((-4.78339168666055402578083604864320577443814`26.6715*^32)) In[4]:= > Accuracy[%] Out[4]= > -6 Due to the manual section 3.1.6: When you do calculations with arbitrary-precision > numbers, as > discussed in the previous section, Mathematica > always keeps track of > the precision of your results, and gives only > those digits which are > known to be correct, given the precision of your > input. When you do > calculations with machine-precision numbers, > however, Mathematica > always gives you a machine[CapitalEth]precision result, > whether or not all the > digits in the result can, in fact, be determined > to be correct on the > basis of your input. Because I started with arbitrary-precision numbers > Mathematica should display > only those digits that are correct, that is none. No, 26 digits are correct Here is the number: -0.8273960599468213681 Here is the same number computed by Mathematica with 26 correct digits: -4.78339168666055402578083604864320577443814[Times]10^32 It looks like I have been using some wrong definition of correct.:-) You just proved that Precision is useless as a measure how good your numerical result is. > (check Precision instead > of Accuracy to see > this). You appear to be showing output in InputForm. If you > use OutputForm or > StandardForm only 26 digits will be shown. 32 > Out[3]= -4.7833916866605540257808360 10 InputForm is showing more because it exposes bad > digits as well as > good ones. > To relax a bit, set a new input cell to > StandardForm and type > 77617.000000000000000000000000000000000 Convert it to InputForm. You get 77616.999999999999999999999999999999999999999999952771`37.9031 Convert back to StandardForm 77616.99999999999999999999999999999999999999999976637`37.9031 Again to InputForm 77616.99999999999999999999999999999999999999999963735`37.9031 Back to StandardForm 77616.99999999999999999999999999999999999999999951376`37.9031 See what you can get if you have enough patience > or a small program. PK Agreed, it's not very pretty. I am uncertain as to > whether this > indicates a bug in StandardForm or elsewhere in the > underlying numerics > code, and will defer to our numerics experts on that > issue. My guess is > it is a bug if only because it violates the spirit > of IEEE arithmetic > wherein floats that have integer values should be > representable as such > (or something to that effect). I will point out, > however, that the two > numbers in question are equal to the specified > precision. Also it > appears to be improved in our development kernel. > Daniel Lichtblau > Wolfram Research __________________________________________________ Do you Yahoo!? http://faith.yahoo.com ==== > You are entitled to your opinion. For my applications > this behavior IS useless. > I agree that Mathematica is probably useless for you. This is however not because it is useless or useless for your application, but because to use its full power you have to study it, understand it, and in particular, for numerical work, understand the model of arithmetic it uses. Lie with mathematics they are really no shortcuts that will lead you to its full power. In addition, since it is a computer program, it has certain conventions, which may not be the same as the conventions of other programs (they all have conventions) but which you have to accept to be able to use it. Now, once you have done that, you may still not like the way Mathematica does things and there are genuine experts in numerics who indeed do not like and are quite vocal about it. But they never say it is useless, because by saying that you are either displaying your own ignorance or engaging in stupid and pointless abuse. On a more serious level, there seem to be two basic approaches to numeric computation relevant to this discussion. It seems to me (though I am no expert in this sort of thing) that there are three types of situations that one may encounter. Firstly, there is the vast majority of rather simple computations for which built-in machine floating point arithmetic , which carries no guarantee of precision at all is meant for. It clearly must be sufficient for the majority of purposes, since most general purpose and even technical software available uses not other method. The reason of course is that it is by far the fastest way to do such computations (as well as being sufficient for most situations). The second type of situation is when you actually know the precision of your input and would like the program to give you some idea about the precision of the output you might expect. This is the most likely situation in empirical science and is exactly what SetPrecision is meant for. Most reasonable people would agree that Mathematica works well in this situation. There is finally one more situation, to which the only reasonable criticism that I have read in this thread appears to be directed at. That is the situation when you actually know your input exactly, but working with exact numbers is far too slow. So what you have to do is to replace your exact numbers with inexact ones padded with 0's. In Mathematica you have to take a guess at how much padding you will need, than use SetPrecision to pad the numbers, and then check the Precision of your answer. It may turn out that you did not get as much precision as you needed, in which case you have to use more zeros. Or it may be that you used more than enough, which mans that your computation could have been done faster. I learned from Leszek Sczaniecki that there is an approach due to Oliver Aberth which lets you only specify the desired precision of your answer and the program itself will choose the correct padding for your input. It woudl ertainly be nice to have this ability, but I honestly think that it would be only of marginal advantage over making your own guess. It seems to me that the checking that the Aberth mthod must require will be time consuming and wiht a bit of practice one can probably get better results as far as speed is concerned using the Mathematica approach. But this is just pure speculation and certainly it woudl be nice if such a possibility existed. Andrzej Kozlowski Toyama International University JAPAN http://sigma.tuins.ac.jp/~andrzej/ ==== >Yes, there seems to be a lot of people who have a visceral hatred for >Microsoft and Windows. They are even willing to shed blood to avoid >Windows. But why? Windows works and you don't have to become a systems >programmer. Furthermore, I think that Steven Wolfram uses some version of Windows. >So guess which system Mathematica will be best tuned up for? If it is true Wolfram uses Mathematica on a Windows based machine my experience is it doesn't translate to Mathematica running better on Windows. I use Mathematica on WindowsNT and on a Mac (currently Mac OS X). I have found Mathematica to be more stable on a Mac than on Windows. On more than one occassion I've seen errors I made Mathematica code to crash the entire machine under WindowsNT. I've never had this happen running things on a Mac. ==== >Yes, there seems to be a lot of people who have a visceral hatred for >>Microsoft and Windows. They are even willing to shed blood to avoid >>Windows. But why? Windows works and you don't have to become a systems >>programmer. >>Furthermore, I think that Steven Wolfram uses some version of Windows. >>So guess which system Mathematica will be best tuned up for? If it is true Wolfram uses Mathematica on a Windows based machine my > experience is it doesn't translate to Mathematica running better on > Windows. I use Mathematica on WindowsNT and on a Mac (currently Mac OS X). > I have found Mathematica to be more stable on a Mac than on Windows. On > more than one occassion I've seen errors I made Mathematica code to crash > the entire machine under WindowsNT. I've never had this happen running > things on a Mac. I believe you hit the nail on the head. I suspect Dr. Wolfram is running on Mac. I have the feeling WRI is a clandestine Mac holdout. STH . ==== I ran Turbo XML http://www.tibco.com/solutions/products/extensibility/turbo_xml.jsp on ToFileName[{$TopDirectory, SystemFiles, IncludeFiles, XML}, notebookml1.dtd ] It gave me an error saying: There is more than one attribute named class. My guess is this was the intention: hattons@ljosalfr:~/.Mathematica/SystemFiles/IncludeFiles/XML/NotebookML1> diff /opt/Wolfram/Mathematica/4.2/SystemFiles/IncludeFiles/XML/NotebookML1/notebo okml.dtd /home/hattons/.Mathematica/SystemFiles/IncludeFiles/XML/NotebookML1/notebook ml.dtd 91c91 Comments? STH . ==== I've posted Mathematica notebooks and packages illustrating most of the neural networks discussed in James Anderson's book An Introduction to Neural Networks to the Brainstage Research web site (www.brainstage.com). Have fun! Don Donald Doherty, Ph.D. Brainstage Research, Inc. donald.doherty@brainstage.com ==== can I have mathematica solver things like a(over)b, (in how many ways can you pick b items from a items)? I have mathematica 4. Stefan ==== Stefan You certainly can - try Binomial[a,b]. Mark Westwood > can I have mathematica solver things like a(over)b, (in how many ways > can you pick b items from a items)? I have mathematica 4. Stefan ==== There are two basic ways, the second of which has two forms. The basic ways are: 1. Using the built in function ContourPlot, e.g.: ContourPlot[x^3y + y^3 - 9, {x, -9, 9}, {y, -27, 27}, Contours -> {0}, ContourShading -> False, Axes -> True, Frame -> False, PlotPoints -> 50, AxesOrigin -> {0, 0}] alternatively you can use a Standard package: <{0,0}] or ImplicitPlot[x^3y+y^3-==9,{x,-9,9},{y,-27,27},AxesOrigin->{0,0}] The difference between these two is that the first one gives you a smoother picture but requires the equation to be solvable (by Mathematica) for y. The second will give a picture very similar to that produced by the first method. Andrzej Kozlowski Yokohama, Japan http://www.mimuw.edu.pl/~akoz/ http://platon.c.u-tokyo.ac.jp/andrzej/ > How can I plot functions like: (x-2)^2 + 2(y-3)^2 = 6 and x^3y + y^3 = 9 using Mathematica? ==== Since you already know the answer, the simplest way is: In[51]:= Factor[x^4 + x^3 + x^2 + x + 1, Extension -> {GoldenRatio}] Out[51]= (-(-1 - x + GoldenRatio*x - x^2))*(1 + GoldenRatio*x + x^2) Andrzej Kozlowski Yokohama, Japan http://www.mimuw.edu.pl/~akoz/ http://platon.c.u-tokyo.ac.jp/andrzej/ > Greetings MathGroup, My name is Steve Earth, and I am a new subscriber to this list and > also a > new user of Mathematica; so please forgive this rather simple > question... I would like to enter the quartic x^4 + x^3 + x^2 + x + 1 into > Mathematica > and have it be able to tell me that it factors into (x^2 + GoldenRatio x + 1) ( x^2 - 1/GoldenRatio x + 1) What instructions do I need to execute to achieve this output? -Steve Earth > Harker School > http://www.harker.org/ Greetings MathGroup, My name is Steve Earth, and I am a new subscriber to this list and > also a > new user of Mathematica; so please forgive this rather simple > question... I would like to enter the quartic x^4 + x^3 + x^2 + x + 1 into > Mathematica > and have it be able to tell me that it factors into (x^2 + GoldenRatio x + 1) ( x^2 - 1/GoldenRatio x + 1) What instructions do I need to execute to achieve this output? -Steve Earth > Harker School > http://www.harker.org/ > Andrzej Kozlowski Yokohama, Japan http://www.mimuw.edu.pl/~akoz/ http://platon.c.u-tokyo.ac.jp/andrzej/ ==== I confess I am not 100% sure what you mean. Would you like to do this in steps, like you would do it by hand? In[1]:= a + b/c + d*(Sqrt[e]/c) == 0; In[2]:= Thread[(#1 - a - b/c & )[%], Equal] Out[2]= (d*Sqrt[e])/c == -a - b/c In[3]:= Thread[(#1*c & )[%], Equal] Out[3]= d*Sqrt[e] == (-a - b/c)*c In[4]:= Thread[(#1^2 & )[%], Equal] Out[4]= d^2*e == (-a - b/c)^2*c^2 In[5]:= Simplify[%] Out[5]= d^2*e == (b + a*c)^2 Of course you can combine all the steps into a single function, but I think it will be fairly complicated. My own favourite way to do this sort of thing is: In[1]:= Simplify[d^2*e == (d^2*e /. AlgebraicRules[ a + b/c + d*(Sqrt[e]/c) == 0, e])] Out[1]= d^2*e == (b + a*c)^2 However, AlgebraicRules has not been documented since version 4. It should be possible to do this using PolynomialReduce but it seems to require the sort of skill only Daniel Lichtblau possesses;) Andrzej Kozlowski Toyama International University JAPAN http://sigma.tuins.ac.jp/~andrzej/ > I'm a newbie and, of course, the first thing I want to do is apparently > one of the most complicated... I have an expression that looks like this: A + B/C + D*Sqrt[E]/C = 0 A,B,C,D, & E are all polynomials in x > I want it to look like this (D^2)*E = (A*C + B)^2 At that point, I'll have polynomials in x on both sides. Finally, I > want the equation to be written out with terms grouped by powers of x, > but I think I can do that part :) I'll be very grateful to anyone who can give me some pointers. Or, at > least point me to some tutorial in the Mathematica documentation. I've > been looking over the documentation and I found Appendix A.5 in The > Mathematica Book, but that doesn't help me. I _need_ some examples. I > did find a couple of well-written posts in this newsgroup, but not > quite > close enough to what I want. > Troy. =-=-=-=-=-=-=-=-=-= FYI, here's the expression I'm working with. > denom = Sqrt[(B^2 - r^2)^2 + 4*(r^2)*(b^2)] > cnu = (2*b^2 - B^2 + r^2)/denom > snu = -2*b*Sqrt[B^2 - b^2]/denom > sif = 2*r*b/denom > cif = (r^2 - B^2)/denom pdr = -Cos[ds]*Sin[q]*(snu*cif + > cnu*sif) - Sin[ds]*(cnu*cif - snu*sif) 0 == -(B^2 - b^2)*V^2/(r^2) + (((B*V)^2)/( > r^2) - 2*w*b*V*Cos[q]*Cos[ds] + (w* > r)^2 - (w*r*pdr)^2)*(Cos[qr])^2 Although I said it's a polynomial in x, it's really a polynomial in b > that I'm after. ==== Only on second reading I noticed the part about a,b,c,d being polynomials in x. Both methods will still work if first perform the same operation as below and finally use the replacement rule %/.{a->p[x],b->q[x],c->r[x],d->u[x],e->v[x]}, where p[x] etc are the given polynomials. Of course collecting of terms can be done with Collect[%,x]. Andrzej Kozlowski Yokohama, Japan http://www.mimuw.edu.pl/~akoz/ http://platon.c.u-tokyo.ac.jp/andrzej/ > I confess I am not 100% sure what you mean. Would you like to do this > in steps, like you would do it by hand? In[1]:= > a + b/c + d*(Sqrt[e]/c) == 0; In[2]:= > Thread[(#1 - a - b/c & )[%], Equal] Out[2]= > (d*Sqrt[e])/c == -a - b/c In[3]:= > Thread[(#1*c & )[%], Equal] Out[3]= > d*Sqrt[e] == (-a - b/c)*c In[4]:= > Thread[(#1^2 & )[%], Equal] Out[4]= > d^2*e == (-a - b/c)^2*c^2 In[5]:= > Simplify[%] Out[5]= > d^2*e == (b + a*c)^2 Of course you can combine all the steps into a single function, but I > think it will be fairly complicated. My own favourite way to do this sort of thing is: In[1]:= > Simplify[d^2*e == (d^2*e /. AlgebraicRules[ > a + b/c + d*(Sqrt[e]/c) == 0, e])] Out[1]= > d^2*e == (b + a*c)^2 However, AlgebraicRules has not been documented since version 4. It > should be possible to do this using PolynomialReduce but it seems to > require the sort of skill only Daniel Lichtblau possesses;) Andrzej Kozlowski > Toyama International University > JAPAN > http://sigma.tuins.ac.jp/~andrzej/ > I'm a newbie and, of course, the first thing I want to do is >> apparently >> one of the most complicated... >> I have an expression that looks like this: >> A + B/C + D*Sqrt[E]/C = 0 >> A,B,C,D, & E are all polynomials in x >> I want it to look like this >> (D^2)*E = (A*C + B)^2 >> At that point, I'll have polynomials in x on both sides. Finally, I >> want the equation to be written out with terms grouped by powers of x, >> but I think I can do that part :) >> I'll be very grateful to anyone who can give me some pointers. Or, at >> least point me to some tutorial in the Mathematica documentation. >> I've >> been looking over the documentation and I found Appendix A.5 in The >> Mathematica Book, but that doesn't help me. I _need_ some examples. I >> did find a couple of well-written posts in this newsgroup, but not >> quite >> close enough to what I want. >> Troy. >> =-=-=-=-=-=-=-=-=-= >> FYI, here's the expression I'm working with. >> denom = Sqrt[(B^2 - r^2)^2 + 4*(r^2)*(b^2)] >> cnu = (2*b^2 - B^2 + r^2)/denom >> snu = -2*b*Sqrt[B^2 - b^2]/denom >> sif = 2*r*b/denom >> cif = (r^2 - B^2)/denom >> pdr = -Cos[ds]*Sin[q]*(snu*cif + >> cnu*sif) - Sin[ds]*(cnu*cif - snu*sif) >> 0 == -(B^2 - b^2)*V^2/(r^2) + (((B*V)^2)/( >> r^2) - 2*w*b*V*Cos[q]*Cos[ds] + (w* >> r)^2 - (w*r*pdr)^2)*(Cos[qr])^2 >> Although I said it's a polynomial in x, it's really a polynomial in >> b >> that I'm after. > > ==== <{-1,5}] ImplicitPlot[x^3y + y^3 == 9, {x, -10, 10}] Meilleures salutations Florian Jaccard -----Message d'origine----- Envoy.8e : dim., 6. octobre 2002 11:34 è : mathgroup@smc.vnet.net Objet : Plotting ellipses and other functions How can I plot functions like: (x-2)^2 + 2(y-3)^2 = 6 and x^3y + y^3 = 9 using Mathematica? ==== >How can I plot functions like: (x-2)^2 + 2(y-3)^2 = 6 and x^3y + y^3 = 9 using Mathematica? > Needs[Graphics`ImplicitPlot`]; ImplicitPlot[(x - 2)^2 + 2(y - 3)^2 == 6, {x, -1, 5}, {y, 1, 5}]; ImplicitPlot[x^3 y + y^3 == 9, {x, -6, 6}, {y, -6, 6}]; Bob Hanlon ==== >Could somebody please inform me how to Round numbers to a >certain Accuracy using Mathematica 4.2. This is not as easy as it >sounds. >Every function that I have read Rounds the Display, and not the actual >number. myRound[x_, n_] := Round[10^n*x]/10.^n; Table[myRound[Random[], 3], {10}] {0.044, 0.019, 0.738, 0.298, 0.917, 0.171, 0.021, 0.314, 0.658, 0.153} Bob Hanlon Reply-To: tgarza01@prodigy.net.mx ==== You might use ImplicitPlot: In[1]:= << Graphics`ImplicitPlot` In[2]:= eqn1 = (x - 2)^2 + 2*(y - 3)^2 == 6; In[3]:= ImplicitPlot[eqn1, {x, -2, 6}]; In[4]:= eqn2 = x^3*y + y^3 == 9; In[5]:= ImplicitPlot[eqn2, {x, -8, 8}]; Tomas Garza Mexico City Original Message: ----------------- ==== You may use Binomial. It could also be useful to look at the AddOn package DiscreteMath`Combinatorica`, where you will find a wealth of interesting things related to that. In[1]:= Binomial[6,2] Out[1]= 15 Tomas Garza Mexico City Original Message: ----------------- ==== I feel as if I've finally had the breakthrough in intuitively understanding the Mathematica editor, or at least the basics. I want to explain to others what they really need to know about the editor to use it for basic purposes. Part of the reason I now understand the editor better is that I've since worked with LyX http://www.devel.lyx.org and XEmacs http://www.xemacs.org, and I've also become proficient with DocBook XML. When I first started using the editor, reading the Mathematica Help didn't seem to help. It seemed to tell me a whole lot more than I needed to know, and didn't tell me what I really needed to know. Now I'm in the position of being able to use it, but not being able to explain exactly what it is I've learned. Is there any documentation directed toward the beginner, which tells him or her what to do, what to expect, what quirks to be aware of, and etc.? I'm looking for something along the lines of click here to make this happen. If you see this, it may seem weird, but that's normal. Those brackedt on the left really mean... . STH . ==== >I would like to enter the quartic x^4 + x^3 + x^2 + x + 1 into Mathematica >and have it be able to tell me that it factors into (x^2 + GoldenRatio x + 1) ( x^2 - 1/GoldenRatio x + 1) What instructions do I need to execute to achieve this output? > soln = Factor[x^4 + x^3 + x^2 + x + 1, Extension -> GoldenRatio] // Simplify (x^2 - GoldenRatio*x + x + 1)*(x^2 + GoldenRatio*x + 1) soln = Simplify /@ (soln /. -GoldenRatio -> -1 - 1/GoldenRatio) (x^2 - x/GoldenRatio + 1)*(x^2 + GoldenRatio*x + 1) soln // FunctionExpand // FullSimplify x^4 + x^3 + x^2 + x + 1 Bob Hanlon ==== Stefan, I don't think I completely understand your question, but I think you are looking for the Binomial function in Mathematica. Binomial[4, 2] 6 David Park djmp@earthlink.net http://home.earthlink.net/~djmp/ Sender: steve@smc.vnet.net Approved: Steven M. Christensen , Moderator ==== Steve, You could use the Extension feature of Factor as documented in Help. expr = x^4 + x^3 + x^2 + x + 1 ans = Factor[expr, Extension -> {1/GoldenRatio}] (-(1/4))*(-2 - x + Sqrt[5]*x - 2*x^2)* (2 + x + Sqrt[5]*x + 2*x^2) You could also use... Factor[expr, Extension -> {Sqrt[5]}] It took me some effort to figure out how to manipulate the answer into your form. ans /. {x + Sqrt[5]*x -> (2*GoldenRatio)*x, -x + Sqrt[5]*x -> (2/GoldenRatio)*x} % /. (-4^(-1))*a_*b_ :> Simplify[-a/2]*Simplify[b/2] (-(1/4))*(-2 + (2*x)/GoldenRatio - 2*x^2)* (2 + 2*GoldenRatio*x + 2*x^2) (1 - x/GoldenRatio + x^2)*(1 + GoldenRatio*x + x^2) David Park djmp@earthlink.net http://home.earthlink.net/~djmp/ http://www.harker.org/ ==== David, To plot equations like that, simply use ImplicitPlot. Needs[Graphics`ImplicitPlot`] ImplicitPlot[(x - 2)^2 + 2(y - 3)^2 == 6, {x, -1, 5}, {y, 1, 5}]; ImplicitPlot[x^3*y + y^3 == 9, {x, -10, 10}, {y, -10, 10}]; You may have to fish a little to obtain the appropriate x and y ranges. Start by making them larger and then narrow down to the region that you want. I have put a new package at my web site for solving conic section problems in the plane. You can solve for complete information on any conic section and obtain a parametric representation for plotting it. The package also comes with complete Help documentation and examples. Using your first example (the second is not a conic). Needs[ConicSections`ConicSections`] eqn = (x - 2)^2 + 2(y - 3)^2 == 6; The routine ParseConic will take any quadratic equation and return the scale a, eccentricity e, a parametrization, and rotation matrix P, translation T and reflection matrix R that transforms the conic from standard position to its actual position. (In standard position the conic has its foci and verticies on the x-axis with the center at zero.) {{a, e}, curve[t_], {P, T, R}} = ParseConic[eqn] {{Sqrt[6], 1/Sqrt[2]}, {2 + Sqrt[6]*Cos[t], 3 + Sqrt[3]*Sin[t]}, {{{1, 0}, {0, 1}}, {2, 3}, {{1, 0}, {0, 1}}}} We could then plot the curve using ParametricPlot, which is more efficient and controllable. ParametricPlot[Evaluate[curve[t]], {t, -Pi, Pi}, AspectRatio -> Automatic, Frame -> True, Axes -> None, PlotLabel -> eqn]; Knowing a and e we can use the StandardConic routine to obtain all the information about the conic in standard position as a set of rules. standarddata = StandardConic[{a, e}] {conictype -> Ellipse, conicequation -> x^2/6 + y^2/3 == 1, coniccurve -> {Sqrt[6]*Cos[t], Sqrt[3]*Sin[t]}, coniccurvedomain -> {-Pi, Pi}, coniccenter -> {0, 0}, conicfocus -> {{Sqrt[3], 0}, {-Sqrt[3], 0}}, conicdirectrix -> {x == -2*Sqrt[3], x == 2*Sqrt[3]}, conicvertex -> {{Sqrt[6], 0}, {-Sqrt[6], 0}}} routine to obtain the same information for the conic in its actual position. standarddata // TransformEllipseRules[P, T, R] {conictype -> Ellipse, conicequation -> (1/6)*((-2 + x)^2 + 2*(-3 + y)^2) == 1, coniccurve -> {2 + Sqrt[6]*Cos[t], 3 + Sqrt[3]*Sin[t]}, coniccurvedomain -> {-Pi, Pi}, coniccenter -> {2, 3}, conicfocus -> {{2 + Sqrt[3], 3}, {2 - Sqrt[3], 3}}, conicdirectrix -> {2*Sqrt[3] + x == 2, x == 2*(1 + Sqrt[3])}, conicvertex -> {{2 + Sqrt[6], 3}, {2 - Sqrt[6], 3}}} David Park djmp@earthlink.net http://home.earthlink.net/~djmp/ Sender: steve@smc.vnet.net Approved: Steven M. Christensen , Moderator Reply-To: ==== Try this: A + B/C + D*(Sqrt[E]/C) == 0 (#1 - %[[1,{1, 2}]] & ) /@ % (C*#1 & ) /@ % (#1^2 & ) /@ % Simplify[%] Also, be aware that E is the natural logarithm base, reserved for that purpose. DrBob -----Original Message----- want the equation to be written out with terms grouped by powers of x, but I think I can do that part :) I'll be very grateful to anyone who can give me some pointers. Or, at least point me to some tutorial in the Mathematica documentation. I've been looking over the documentation and I found Appendix A.5 in The Mathematica Book, but that doesn't help me. I _need_ some examples. I did find a couple of well-written posts in this newsgroup, but not quite close enough to what I want. Troy. =-=-=-=-=-=-=-=-=-= FYI, here's the expression I'm working with. denom = Sqrt[(B^2 - r^2)^2 + 4*(r^2)*(b^2)] cnu = (2*b^2 - B^2 + r^2)/denom snu = -2*b*Sqrt[B^2 - b^2]/denom sif = 2*r*b/denom cif = (r^2 - B^2)/denom pdr = -Cos[ds]*Sin[q]*(snu*cif + cnu*sif) - Sin[ds]*(cnu*cif - snu*sif) 0 == -(B^2 - b^2)*V^2/(r^2) + (((B*V)^2)/( r^2) - 2*w*b*V*Cos[q]*Cos[ds] + (w* r)^2 - (w*r*pdr)^2)*(Cos[qr])^2 Although I said it's a polynomial in x, it's really a polynomial in b that I'm after. Reply-To: ==== It's a little ugly, but here's my solution: x^4 + x^3 + x^2 + x + 1 Simplify@Factor[%, Extension -> {GoldenRatio, 1/GoldenRatio}] Collect[%[[2]]/3, x]Collect[%[[3]]/3, x] % /. Sqrt[5] -> 2GoldenRatio - 1 % // Simplify Collect[#, x] & /@ % % /. {1 - GoldenRatio -> -1/GoldenRatio} FullSimplify@Expand@% (The last line is a check.) Bobby -----Original Message----- (x^2 + GoldenRatio x + 1) ( x^2 - 1/GoldenRatio x + 1) What instructions do I need to execute to achieve this output? -Steve Earth Harker School http://www.harker.org/ ==== >How can I plot functions like: (x-2)^2 + 2(y-3)^2 = 6 and x^3y + y^3 = 9 using Mathematica? Use ImplicitPlot in the package Graphics`ImplicitPlot` ==== David, <{0,0}] -----Original Message----- Sender: steve@smc.vnet.net Approved: Steven M. Christensen , Moderator Reply-To: jcd@q-e-d.org ==== I look for a way to right align entries output by MatrixForm. There is an ad hoc option, but I haven't figured out how it is supposed to work in Mathematica 4.0 (if it works at all). m={{MatrixForm[{0,0,0}],-123456789},{123456789,1}}; MatrixForm[m,TableAlignments->{Right,Top}] appears identical to MatrixForm[m] (with no SetOptions override) Strangely, Mathematica doesn't complain if you input ill options like: MatrixForm[m,TableAlignments->{WhateverIsIllegal,Right,Top,MakeSomeSpiralOfI t}] In Mathematica 2.2x, the behavior is rather strange: one option alone works, but when two are given (e.g. {Right,Top}), only the last seems to be acted upon. I didn't bother to try hard with the older version, so don't flame me if I'm wrong. At the other hand, when a TableAlignments->Right is given to TableForm, entries are right-aligned correctly. Can someone provide a way out (even a slow external module) or tell me how to use this option. I apologize for using a phony address in an attempt ==== I decided to export one of my notebooks to html/mathml http://baldur.globalsymmetry.com/proprietary/com/wri/notebooks/essential/ 66.92.149.152 baldur.globalsymmetry.com When I try to view it with Mozilla, I get a '?' in place of the imaginary number symbol. When I first load the page with Mozilla, I get an error telling me To properly display the MathML on this page you need to install the following fonts: CMSY 10, CMEX 10, Math1, Math2, Math4. For further infromation see: http://www.mozilla.org/projects/mathml/fonts I did what the instructions at the URL told me, and I still get the error. I looked through the fonts in the Mathematica font directory, and I found: Mathematica1Mono.9.bdf -wri-mathematica1mono-medium-r-normal--9-90-75-75-m-50-adobe-fontspecific Mathematica3.12.bdf -wri-mathematica3-medium-r-normal--12-120-75-75-p-70-adobe-fontspecific Mathematica3.24.bdf -wri-mathematica3-medium-r-normal--24-240-75-75-p-130-adobe-fontspecific Mathematica3.36.bdf -wri-mathematica3-medium-r-normal--36-360-75-75-p-210-adobe-fontspecific .... Mathematica7.12.bdf -wri-mathematica7-medium-r-normal--12-120-75-75-p-40-adobe-fontspecific Mathematica7.24.bdf .... -wri-mathematica6-medium-r-normal--12-120-75-75-p-30-adobe-fontspecific Mathematica6.24.bdf -wri-mathematica6-medium-r-normal--24-240-75-75-p-70-adobe-fontspecific Mathematica6.36.bdf -wri-mathematica6-medium-r-normal--15-150-75-75-p-50-adobe-fontspecific Mathematica5.12.bdf -wri-mathematica5-medium-r-normal--12-120-75-75-p-50-adobe-fontspecific Mathematica5.24.bdf -wri-mathematica5-medium-r-normal--24-240-75-75-p-110-adobe-fontspecific Mathematica5.36.bdf -wri-mathematica5-medium-r-normal--36-360-75-75-p-160-adobe-fontspecific Mathematica5.13.bdf -wri-mathematica5-medium-r-normal--13-130-75-75-p-50-adobe-fontspecific Mathematica4.18.bdf -wri-mathematica4-medium-r-normal--18-180-75-75-p-130-adobe-fontspecific Mathematica5.10.bdf .... I suspect this is what mozilla is looking for, but I don't know exactly how to tell it as much. Any ideas? STH . Reply-To: ==== I suppose my silliness is understandable, in light of all the confusion, both here and in the Browser on what significance arithmetic is, what bignums and bigfloats might be, etc. If many smart people are confused, there's a possibility --- just a possibility, mind you --- that it isn't entirely their fault. Yes? No? >>like 71 above, or -5 for Accuracy in the example that fooled me), but it's not a big deal For people who don't understand it as well as you, yes, it's a big deal. My purpose in all this is to understand the issue well enough to know how to proceed. I think that I do, now, but I doubt everybody on the list does. Daniel says this will all be clearer in the next release, and that's good! Bobby -----Original Message----- -1.180591620717411303424`71.0721*^21 > 71 71.0721 digits of precision? I don't think so!! Either I am it altogether or you are just simply beating to death the point that in case of machine arithmetic (only!) Precision and Accuracy are purely formal and essentially meaningless. One can argue whether in this case there is any point of returning any value for Precision, or Accuracy (like 71 above, or -5 for Accuracy in the example that fooled me), but it's not a big deal and it most certainly does not make SetPrecision meaningless. On the contrary, SetPrecision is very useful and in fact it is SetPrecision that can tell you that the answer above is meaningless: In[8]:= f = SetAccuracy[333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + 5.5*b^8 + a/(2*b), 50]; a=SetPrecision[77617.,$MachinePrecision]; b = SetPrecision[ 33096.,$MachinePrecision]; In[10]:= {f,Precision[f]} Out[10]= {1.19801754103509`0*^19, 0} I would say this is correct and show that SetPrecision is very useful indeed. It tells you (what of course you ought to already know in this case anyway) that machine precision will not give you a realiable answer in this case. If you know your numbers with a great deal of accuracy you can get an accurate answer: In[24]:= f = SetAccuracy[333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + 5.5*b^8 + a/(2*b), 100]; a=SetPrecision[77617.,100]; b = SetPrecision[33096.,100]; In[26]:= {f, Precision[f]} Out[26]= {-0.82739605994682136814116509547981629199903311578438481991 781484167246798617832`61.2597, 61} Again you can be pretty sure that you got an accurate answer, provided of course your original setting of precision was valid. Honestly, to say that SetPrecision and SetAccuaracy are useless is one of the silliest thing I have read on this list in years. > Andrzej Kozlowski Yokohama, Japan http://www.mimuw.edu.pl/~akoz/ http://platon.c.u-tokyo.ac.jp/andrzej/ ==== How can I plot functions like: (x-2)^2 + 2(y-3)^2 = 6 and x^3y + y^3 = 9 using Mathematica? Reply-To: kuska@informatik.uni-leipzig.de ==== look what Graphics`ImplicitPlot` does. Jens How can I plot functions like: (x-2)^2 + 2(y-3)^2 = 6 and x^3y + y^3 = 9 using Mathematica? ==== David: For a start you can try the following: Using ContourPlot ContourPlot[(x - 2)^2 + 2(y - 3)^2 - 6, {x, -1, 5}, {y, -1, 5}, Contours -> {0}, ContourShading -> False, ContourSmoothing -> 5] Or using ImplicitPlot << Graphics`ImplicitPlot` ImplicitPlot[(x - 2)^2 + 2(y - 3)^2 == 6, {x, -1, 5}] ImplicitPlot[x^3y + y^3 == 9, {x, -4, 4}] Hans > How can I plot functions like: (x-2)^2 + 2(y-3)^2 = 6 and x^3y + y^3 = 9 using Mathematica? ==== Try reading help under the keyword ImplicitPlot | How can I plot functions like: | | (x-2)^2 + 2(y-3)^2 = 6 | | and | | x^3y + y^3 = 9 | | using Mathematica? | | | | ==== David Enquire within the Help Browser for the ImplicitPlot package and all will be revealed. For example, I snipped the following lines from the documentation: << Graphics`ImplicitPlot` ImplicitPlot[x^2 + 2 y^2 == 3, {x, -2, 2}] Hope this is of sufficient help to get you started - post again if you have any further questions. Mark Westwood How can I plot functions like: (x-2)^2 + 2(y-3)^2 = 6 and x^3y + y^3 = 9 using Mathematica? ==== you can. Use: < is there any way of linking Mathematica with Excel? Lu.92s ==== I am getting the same problem (though Mathematica 4.1). Has anyone any ideas? Aron. I am facing the problem in starting the link to math kernel from within >Excel. Specifically, when I get the message link failed to open when I click >Launch button in the 'Start Mathematica Link' dialog box. I have tried using >Multilink too so that I am able to access the kernel from Mathematica and >Excel simultaneously but I still get the same message. I am currently using Mathematica 4.0 and Excel 2002 (Excel XP). The programs >Mathematica and Excel otherwise appear to work fine. I am using Windows XP >home edition on Dell 8100 laptop. I have used the Mathematica Link for MS >Excel for Excel 2000 in my installation as there were no specific files for >Excel 2002. Given that I was able to successfully add the menus within excel >I think the addin should work fine but it does not? All help would be sincerely appreciated. Sincerely, Tahir Sheikh. ==== Download the 2002 file from this page. I use it for Excel 2002 Service Pack 2. http://support.wolfram.com/applicationpacks/excel_link/excelxp.html I am getting the same problem (though Mathematica 4.1). > Has anyone any ideas? Aron. I am facing the problem in starting the link to math kernel from within >Excel. Specifically, when I get the message link failed to open when I click >Launch button in the 'Start Mathematica Link' dialog box. I have tried > using >Multilink too so that I am able to access the kernel from Mathematica > and >Excel simultaneously but I still get the same message. I am currently using Mathematica 4.0 and Excel 2002 (Excel XP). The > programs >Mathematica and Excel otherwise appear to work fine. I am using Windows > XP >home edition on Dell 8100 laptop. I have used the Mathematica Link for > MS >Excel for Excel 2000 in my installation as there were no specific files > for >Excel 2002. Given that I was able to successfully add the menus within > excel >I think the addin should work fine but it does not? All help would be sincerely appreciated. Sincerely, Tahir Sheikh. > ==== I did not request any accuracy for f. I set the accuracy of the numerical components of the expression f. You cannot request the accuracy of the result of your computation in Mathematica, you can only set the accuracy of the input and later check what accuracy of the output results form it. In my last message on this topic I tried to explain this in the plainest and simplest way I could think of. There is nothing more left for me to say. I feel like Sisyphus but unlike him I can at least give up! Andrzej Kozlowski >> [...] >> I would say this is correct and show that SetPrecision is very useful >> indeed. It tells you (what of course you ought to already know in this >> case anyway) that machine precision will not give you a realiable >> answer in this case. If you know your numbers with a great deal of >> accuracy you can get an accurate answer: >> In[24]:= >> f = SetAccuracy[333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - >> 121*b^4 - 2) + 5.5*b^8 + a/(2*b), 100]; >> a=SetPrecision[77617.,100]; b = SetPrecision[33096.,100]; >> In[26]:= >> {f, Precision[f]} >> Out[26]= >> {-0.82739605994682136814116509547981629199903311578438481991 >> 781484167246798617832`61.2597, 61} > > Congratulations! You just requested accuracy of 100 for f and got 61 ( > to convince yourself add Accuracy[f] to In[26]). If In[24] one > replaces SetAccuracy by SetPrecision the result is similar. PK > Again you can be pretty sure that you got an accurate answer, provided >> of course your original setting of precision was valid. >> Honestly, to say that SetPrecision and SetAccuaracy are useless is one >> of the silliest thing I have read on this list in years. > >> Andrzej Kozlowski >> Yokohama, Japan >> http://www.mimuw.edu.pl/~akoz/ >> http://platon.c.u-tokyo.ac.jp/andrzej/ > Andrzej Kozlowski Yokohama, Japan http://www.mimuw.edu.pl/~akoz/ http://platon.c.u-tokyo.ac.jp/andrzej/ Reply-To: ==== Here's a more intuitive method, perhaps: a + b/c + d*(Sqrt[e]/c) == 0 f = %[[1, 3]] %% /. f -> g First@Solve[%, g] f^2 == (g^2 /. %) However, it occurs to me you might want a more general method to collect a radical on one side and then square both sides. If so, here's a clumsy first attempt: expr = a + b/c + d*(Sqrt[e]/c) == 0 f = First@Cases[expr, Power[a_, Rational[b_, c_]], Infinity] power = First@Cases[f, Rational[b_, c_] -> Rational[c, b], Infinity] coefficient = First@Cases[expr, Times[a_, f] -> a, Infinity] Solve[expr /. coefficient f -> g, g][[1, 1]] g^2 == (g^2 /. %) % /. g -> coefficient f DrBob -----Original Message----- DrBob -----Original Message----- want the equation to be written out with terms grouped by powers of x, but I think I can do that part :) I'll be very grateful to anyone who can give me some pointers. Or, at least point me to some tutorial in the Mathematica documentation. I've been looking over the documentation and I found Appendix A.5 in The Mathematica Book, but that doesn't help me. I _need_ some examples. I did find a couple of well-written posts in this newsgroup, but not quite close enough to what I want. Troy. =-=-=-=-=-=-=-=-=-= FYI, here's the expression I'm working with. denom = Sqrt[(B^2 - r^2)^2 + 4*(r^2)*(b^2)] cnu = (2*b^2 - B^2 + r^2)/denom snu = -2*b*Sqrt[B^2 - b^2]/denom sif = 2*r*b/denom cif = (r^2 - B^2)/denom pdr = -Cos[ds]*Sin[q]*(snu*cif + cnu*sif) - Sin[ds]*(cnu*cif - snu*sif) 0 == -(B^2 - b^2)*V^2/(r^2) + (((B*V)^2)/( r^2) - 2*w*b*V*Cos[q]*Cos[ds] + (w* r)^2 - (w*r*pdr)^2)*(Cos[qr])^2 Although I said it's a polynomial in x, it's really a polynomial in b that I'm after. ==== On Friday, October 4, 2002, at 06:01 PM, DrBob [...] I would say this is correct and show that > SetPrecision is very useful > indeed. It tells you (what of course you ought > to already know in this > case anyway) that machine precision will not > give you a realiable > answer in this case. If you know your numbers > with a great deal of > accuracy you can get an accurate answer: In[24]:= > f = SetAccuracy[333.75*b^6 + a^2*(11*a^2*b^2 - > b^6 - > 121*b^4 - 2) + 5.5*b^8 + a/(2*b), 100]; > a=SetPrecision[77617.,100]; b = > SetPrecision[33096.,100]; > In[26]:= > {f, Precision[f]} Out[26]= {-0.82739605994682136814116509547981629199903311578438481991 > 781484167246798617832`61.2597, 61} > Congratulations! You just requested accuracy of > 100 for f and got 61 ( > to convince yourself add Accuracy[f] to In[26]). > If In[24] one > replaces SetAccuracy by SetPrecision the result is > similar. PK > [...] One has (initially) an accuracy of 100 for an > expression that contains > variables. In[25]:= Clear[a,b,f] In[26]:= f = SetAccuracy[333.75*b^6 + > a^2*(11*a^2*b^2 - b^6 - > 121*b^4 - 2) + 5.5*b^8 + a/(2*b), 100]; In[27]:= Accuracy[f] > Out[27]= 100. Now we assign values to some indeterminants in f. In[28]:= a = SetPrecision[77617.,100]; b = > SetPrecision[33096.,100]; In[29]:= {f, Precision[f], Accuracy[f]} > Out[29]= > {-0.8273960599468213681411650954798162919990331157843848199178148, 61.2599, 61.3422} The precision and accuracy has dropped. This is all > according to > standard numerical analysis regarding cancellation > error. You'll find it > in any textbook on the topic. > Assume that I want accuracy and precision of 100 for f. You advice me to make experiments to find out, what should be the initial precision and accuracy of a and b to reach the requested accuracy and precision for f. Notice, that you cannot just repeat I[26], we saw already what happens. I have to re-type I[24], I[25], I[26], I[27], I[28], and I[29] as many times as needed to get f with accuracy and precision 100. Dan, you simply advocate to do MANUAL WORK that should be done by machine. Let's suppose that in the above example I just want 60 digits not 61. Precisely, I want 60 digits and nothing or zeros afterwards. Let's see if I could use SetAccuracy. In[30]:= SetAccuracy[%, 60] Out[30]= -0.82739605994682136814116509547981629199903311578438481991781 In[31]:= % // FullForm Out[30]//FullForm= -0.827396059946821368141165095479816291999033115784384819917814841672467988` 59.9177 Oops, it did not work (as expected). Let's highlight with mouse the expression in Out[30] and copy to a new cell. Oops, we got -0.827396059946821368141165095479816291999033115784384819917814841672467988` 59.9177 again. Let's change Out[30] to a text cell and then copy. In[31]:= -0.82739605994682136814116509547981629199903311578438481991781 Out[31]= -0.82739605994682136814116509547981629199903311578438481991781 Success? Not so fast. In[32]:= % // FullForm Out[32]//FullForm= -0.8273960599468213681411650954798162919990331157843848199178099999999999986 35 08`59.2041 Dan, is there any simple way to get what I want? As I repeated already number of times, at this stage of the development of computer technology, software should do it for me (!). We both know that this is doable. Some of the textbooks that you just advised me to read describe it. As a developer of Mathematica, tell us why do you consider this to be a bad idea? Peter Kosta > As for what happens when you artificially raise > precision (or accuracy) > of machine numbers far beyond that guaranteed by > their internal > representation, that falls into to category of > garbage in, garbage out. > It is, howoever, valid to use SetPrecision to raise > precision in > (typically iterative) algorithms where significance > arithmetic might be > unduly pessimistic due to incorrect assumptions > about uncorollatedness > of numerical error. Examples of such usage have > appeared in this news > group. > Daniel Lichtblau > Wolfram Research __________________________________________________ Do you Yahoo!? http://faith.yahoo.com ==== > Are there any known issues with simpy treating the JLink.jar as a Java > extension as follows? > cp JLink.jar $JAVA_HOME/jre/lib/ext? > According to my understanding of the discussion in the Java Tutorial on > extensions, that should work: It does; starting with M4.2, J/Link 2.0 gets preinstalled and comes with a 1.4 murphee ==== Are there any known issues with simpy treating the JLink.jar as a Java extension as follows? cp JLink.jar $JAVA_HOME/jre/lib/ext? According to my understanding of the discussion in the Java Tutorial on extensions, that should work: http://java.sun.com/docs/books/tutorial/ext/basics/install.html Commants? STH . ==== >Are there any known issues with simpy treating the JLink.jar as a Java >extension as follows? >cp JLink.jar $JAVA_HOME/jre/lib/ext? According to my understanding of the discussion in the Java Tutorial on >extensions, that should work: >http://java.sun.com/docs/books/tutorial/ext/basics/install.html Commants? You should not do this. Code from the jre/lib/ext directory is trusted, so this poses a security risk from malicious applets. Leave JLink.jar where it lives in the JLink directory. If you want it to be available to all Java programs on your system, add its location to your CLASSPATH environment variable (this is not a security risk, as remote applets cannot load classes from CLASSPATH). Todd Gayley Wolfram Research ==== The key is in using the command Factor with the option Extension: In[1]:= Factor[x^4 + x^3 + x^2 + x + 1, Extension -> {GoldenRatio}] Out[1]= -((-1 - x + GoldenRatio*x - x^2)*(1 + GoldenRatio*x + x^2)) For manual verification you should keep in mind that: 1/GoldenRatio = GoldenRatio - 1 Germ.87n Buitrago ----- Original Message ----- > (x^2 + GoldenRatio x + 1) ( x^2 - 1/GoldenRatio x + 1) What instructions do I need to execute to achieve this output? -Steve Earth > Harker School > http://www.harker.org/ > ==== Actually, including 1/GoldenRatio in the extension leads to an unnecessarily complicated formula. In this case there is no real need to so, since by definition In[30]:= Unevaluated[1/GoldenRatio==GoldenRatio-1]//FullSimplify Out[30]= True If one really insists on having the answer in the form proposed in Steve's original posting one can simply do: (Collect[#1, x] & ) /@ Factor[x^4 + x^3 + x^2 + x + 1, Extension -> {GoldenRatio}] /. -1 + GoldenRatio -> 1/GoldenRatio (-(-1 + x/GoldenRatio - x^2))*(1 + GoldenRatio*x + x^2) In[]:=Factor[x^4 + x^3 + x^2 + x + 1, Extension -> {GoldenRatio, > 1/GoldenRatio}] > Out[]=-((-3 - 2*x + Sqrt[5]*x + GoldenRatio*x - 3*x^2)* > (3 + x + Sqrt[5]*x + GoldenRatio*x + 3*x^2))/9 Jens > Greetings MathGroup, >> My name is Steve Earth, and I am a new subscriber to this list and >> also a >> new user of Mathematica; so please forgive this rather simple >> question... >> I would like to enter the quartic x^4 + x^3 + x^2 + x + 1 into >> Mathematica >> and have it be able to tell me that it factors into >> (x^2 + GoldenRatio x + 1) ( x^2 - 1/GoldenRatio x + 1) >> What instructions do I need to execute to achieve this output? >> -Steve Earth >> Harker School >> http://www.harker.org/ > Andrzej Kozlowski Yokohama, Japan http://www.mimuw.edu.pl/~akoz/ http://platon.c.u-tokyo.ac.jp/andrzej/ ==== I want to apply a function to every k-th element of a long list and add the result to the k+1 element. [Actually k = 3 and I just want to multiply myList[[k]] by a constant (independent of k) and add the result to myList[[k+1]] for every value of k that's divisible by 3.] Is there a way to do this -- or in general to get at every k-th element of a list -- that's faster and more elegant than writing a brute force Do[] loop or using Mod[] operators, and that will take advantage of native List operators, but still not be too recondite? I've been thinking about multiplying a copy of myList by a mask list {0,0,1,0,0,1,..} to generate a masked copy and approaches like that. Better ways??? ==== Take[list, am, n, sa] gives elements m through n in steps of s. > I want to apply a function to every k-th element of a long list and > add the result to the k+1 element. [Actually k = 3 and I just want to multiply myList[[k]] by a > constant (independent of k) and add the result to myList[[k+1]] for > every value of k that's divisible by 3.] Is there a way to do this -- or in general to get at every k-th > element of a list -- that's faster and more elegant than writing a brute > force Do[] loop or using Mod[] operators, and that will take > advantage of native List operators, but still not be too recondite? I've been thinking about multiplying a copy of myList by a mask list > {0,0,1,0,0,1,..} to generate a masked copy and approaches like that. > Better ways??? ==== lst= Range[14] {1,2,3,4,5,6,7,8,9,10,11,12,13,14} A list of positions in lst ( for your purpose Range[1, Length[lst], 3] will do) ps= {3,5,10}; The following applies h to each ps element in lst and adds the result to the following element (lst[[ps+1]]=h/@lst[[ps]]+lst[[ps+1]];lst) {1,2,3,4+h[3],5,6+h[5],7,8,9,10,11+h[10],12,13,14} -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay@haystack.demon.co.uk Voice: +44 (0)116 271 4198 > I want to apply a function to every k-th element of a long list and > add the result to the k+1 element. [Actually k = 3 and I just want to multiply myList[[k]] by a > constant (independent of k) and add the result to myList[[k+1]] for > every value of k that's divisible by 3.] Is there a way to do this -- or in general to get at every k-th > element of a list -- that's faster and more elegant than writing a brute > force Do[] loop or using Mod[] operators, and that will take > advantage of native List operators, but still not be too recondite? I've been thinking about multiplying a copy of myList by a mask list > {0,0,1,0,0,1,..} to generate a masked copy and approaches like that. > Better ways??? > ==== Consider the following approach, whish uses the MapAt command, that is Map with 'mapping-position' control. dummyFun={#,trueFun@#}& list={a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p} spec=Partition[Range[3,Length[list],3],1] MapAt[dummyFun,list,spec] %//Flatten Hope that is what you want, Borut | I want to apply a function to every k-th element of a long list and | add the result to the k+1 element. | | [Actually k = 3 and I just want to multiply myList[[k]] by a | constant (independent of k) and add the result to myList[[k+1]] for | every value of k that's divisible by 3.] | | Is there a way to do this -- or in general to get at every k-th | element of a list -- that's faster and more elegant than writing a brute | force Do[] loop or using Mod[] operators, and that will take | advantage of native List operators, but still not be too recondite? | | I've been thinking about multiplying a copy of myList by a mask list | {0,0,1,0,0,1,..} to generate a masked copy and approaches like that. | Better ways??? | Reply-To: kuska@informatik.uni-leipzig.de ==== something like: With[{k=3}, Flatten[ {#[[2]] + c*#[[1]], #[[3]]} & /@ Partition[lst, k, k, {1, 1}], 1] ] ?? Jens I want to apply a function to every k-th element of a long list and > add the result to the k+1 element. [Actually k = 3 and I just want to multiply myList[[k]] by a > constant (independent of k) and add the result to myList[[k+1]] for > every value of k that's divisible by 3.] Is there a way to do this -- or in general to get at every k-th > element of a list -- that's faster and more elegant than writing a brute > force Do[] loop or using Mod[] operators, and that will take > advantage of native List operators, but still not be too recondite? I've been thinking about multiplying a copy of myList by a mask list > {0,0,1,0,0,1,..} to generate a masked copy and approaches like that. > Better ways??? Reply-To: ==== Daniel, >>The precision/accuracy tracking mechanism will generally let you know, in some fashion, that you have no trustworthy digits. But it is up to the user to check that sort of thing. In this case Mathematica did NOT let us know, in any fashion, that we had no trustworthy digits. Precision and Accuracy outputs were completely misleading. (16 and -5 respectively.) Even Andrzej Kozlowski, who's adept in Mathematica, thought that would be meaningful, and never came up with a better way to check (other than using infinite precision for numbers that probably aren't known that exactly). Peter Kosta demonstrated that he could get a completely erroneous answer with Infinite precision. I blame the problem primarily, and I don't think there's any way to make the answer meaningful. That's not Mathematica's fault at all, and users need to be aware of that old maxim: garbage in, garbage out. comes up with a 22-digit result, it doesn't take much sophistication to realize the answer can't have 16-digit precision. Here's an even more extreme result: f = SetAccuracy[333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + 5.5*b^8 + a/(2*b), 50]; a = 77617.; b = 33096.; f Precision[f] -1.180591620717411303424`71.0721*^21 71 71.0721 digits of precision? I don't think so!! We can do the following instead: x = Interval[333.75]; y = Interval[5.5]; a = Interval[77617.]; b = Interval[33096.]; x*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + y*b^8 + a/(2*b) Interval[{-4.486248158726164*^22, 4.2501298345826815*^22}] and that looks like the right answer, finally!! I like that method. However, that doesn't change the fact that Accuracy, Precision, and SetAccuracy appear to be completely useless. I haven't seen an example in which they did what anyone (but you) thought they should do. Bobby -----Original Message----- > you're not aware there's a problem, it lets you go on your merry way, > working with noise. Bobby Mathematica is not a mind reader. But the evaluation sequence, while complicated, is reasonably well documented. If you perform machine arithmetic, or for that matter significance arithmetic, and there is massive cancellation error, no use of SetAccuracy after the fact will fix it. The precision/accuracy tracking mechanism will generally let you know, in some fashion, that you have no trustworthy digits. But it is up to the user to check that sort of thing. It is not obvious to me what sort of error the software might notice to report. If you have a concise example of input, and expected output, I can look further. I've not seen anything in this thread that struck me as a failure of the software to warn the user, but maybe I missed something. Daniel ==== GentleBeings I have a straightforward implementation of successive approximations but I cannot seem to froce the code to find the correct solution when I have trig or exponentials involved. Can the assembled wisdom point to straghtforward fixes I know FindRoot works the object is to teach programming and successive approx, tho. kenf Below is the code Clear[f, g, gi, lim, r, rr, fr, gir, a, b, c, d, conv]; Plot[{x * ((x + 3)), 10*Sin[x]}, {x, 0.01, 2.4}, PlotStyle -> {{RGBColor[1, 0, 0], Thickness[ .006]}, {RGBColor[0, 0, 1], Thickness[ .006]}} ]; rr = FindRoot[x * ((x + 3)) == 10*Sin[x], {x, 2, 0.01, 2.4}]; f[a_] := a * ((a + 3)) /; a > 0; g[b_] := 10. * Sin[b] /; b > 0; gi[c_] := ArcSin[0.1*c] /; c > 0; Print[Actual root is , rr]; lim = 10; r = 2.0; conv = 10^-4; For[i = 1, i < lim, i++, { fr = f[r]; gir = gi[fr]; d = Abs[N[gir] - r]; i If[d < conv, Break[]]; r = gir; Print[The value of x = , r, found after , i, iterations,, with a tolerence , d, n] } ] Print[The value of x = , r, found after , i, iterations,, with a tolerence , d, n] Every man, woman and responsible child has an unalienable individual, civil, Constitutional and human right to obtain, own, and carry, openly or concealed, any weapon -- rifle, shotgun, handgun, machine gun, anything -- any time, any place, without asking anyone's permission. L. Neil Smith ==== > I want to apply a function to every k-th element of a long list and > add the result to the k+1 element. [Actually k = 3 and I just want to multiply myList[[k]] by a > constant (independent of k) and add the result to myList[[k+1]] for > every value of k that's divisible by 3.] Is there a way to do this -- or in general to get at every k-th > element of a list -- that's faster and more elegant than writing a > brute > force Do[] loop or using Mod[] operators, and that will take > advantage of native List operators, but still not be too recondite? I've been thinking about multiplying a copy of myList by a mask > list > {0,0,1,0,0,1,..} to generate a masked copy and approaches like that. > Better ways??? Here is a generalization of what you've asked: f[l_List,c_,k_Integer,p_Integer]:=Flatten[Block[{r=#[[k]] c},Join[Take[#,k-1],{r,#[[k+1]]+ r},Drop[#,k+1]]]&/@Partition[l,p]]/;(Mod[Length[l],p]==0&&k (Expand[#1] == Expand[#2] & )] {(1 + (1/2)*(1 - Sqrt[5])*x + x^2)* (1 + (1/2)*(1 + Sqrt[5])*x + x^2)} Andrzej Kozlowski Toyama International University JAPAN http://sigma.tuins.ac.jp/~andrzej/ > Steve > The notebook given after NOTEBOOK below contains functions for > factoring and > partial fractioning. > Here is an application to your problem: the first stage avoids our > needing > to know anything about the answer. fc=FactorR[x^4+x^3+x^2+x+1,x] (1 - (1/2)*(-1 - Sqrt[5])*x + x^2)* > (1 - (1/2)*(-1 + Sqrt[5])*x + x^2) Now we need to get rid of Sqrt[5] in terms of GoldenRatio. > This is rather messy: Simplify/@(fc/. Sqrt[5][Rule]2 GoldenRatio-1) (1 + x - GoldenRatio*x + x^2)*(1 + GoldenRatio*x + x^2) Simplify/@(%/.-GoldenRatio[Rule] 1/GoldenRatio -1) (1 + x/GoldenRatio + x^2)*(1 + GoldenRatio*x + x^2) > Another example PartialFractionsR[(1 + x)x/(1 - 3*x + x^2), x] 1 - (2*(-1 + 4*x))/((3 + Sqrt[5] - 2*x)*(-3 + 2*x)) + > (2*(-1 + 4*x))/((-3 + 2*x)*(-3 + Sqrt[5] + 2*x)) Simplify[%] (x*(1 + x))/(1 - 3*x + x^2) NOTEBOOK: to make a notebook from the following, copy from the next > line to > the line preceding XXX and paste into a new Mathematica notebook. Notebook[{ Cell[CellGroupData[{ > Cell[Factors and PartialFractions, Subtitle], Cell[Allan Hayes, 16 August 2001, Text], Cell[< > Here are some functions for factoring and expressing in partial > fractions over the reals and over the complex numbers. >, Text], Cell[BoxData[ > (Quit)], Input], Cell[BoxData[{ > (Off[General::spell1, General::spell]), n, > ((FactorC::usage = polynomial in x with complex coefficients, gives its factorization > over the complex numbers.n > The output may include Root objects which may be evaluated with > ToRadicals or N.>;)n), n, > ((FactorR::usage = polynomial in x with real coefficients, gives its factorization over > the reals.n > The output may include Root objects which may be evaluated with > ToRadicals or N.>;)n), n, > ((PartialFractionsC::usage = where ratl is a rational in x with complex coefficients, gives its > factorization over the complex numbers.n > The output may include Root objects which may be evaluated with > ToRadicals or N.>;)n), n, > ((PartialFractionsR::usage = where ratl is a rational in x with real coefficients, gives its > factorization over the real numbers.n > The output may include Root objects which may be evaluated with > ToRadicals or N.>;)), n, > (On[General::spell1, General::spell])}], Input, > InitializationCell->True], Cell[TextData[{ > FactorC[p_, x_] := , > StyleBox[(*over complex numbers*), > FontFamily->Arial, > FontWeight->Plain], > nTimes @@ Cases[Roots[p == 0, x, n Cubics -> False], u_ == > v_ -> x - v]n nFactorR[p_, x_] := , > StyleBox[(*over reals, coefficients must be real*), > FontFamily->Arial, > FontWeight->Plain], > n (Times @@ Join[Cases[#1, u_ == v_ /; Im[v] == 0 :> n x > - v], Cases[#1, u_ == v_ /; Im[v] > 0 :> n x^2 - x*2*Re[v] + > Abs[v]^2]] & )[n Roots[p == 0, x, Cubics -> False]] > }], Input, > InitializationCell->True], Cell[TextData[{ > PartialFractionsC[p_, x_] := , > StyleBox[(*over complex numbers*), > FontFamily->Arial, > FontWeight->Plain], > n(#+Apart[#2/FactorC[#3,x]])&@@Flatten[{PolynomialReduce[#,#2], > #2}]&[Numerator[#],Denominator[#]]&[Together[p]]n n > PartialFractionsR[p_, x_] := , > StyleBox[(*over reals, coefficients must be real*), > FontFamily->Arial, > FontWeight->Plain], > n(#+Apart[#2/FactorR[#3,x]])&@@Flatten[{PolynomialReduce[#,#2], > #2}]&[Numerator[#],Denominator[#]]&[Together[p]] > }], Input, > InitializationCell->True], Cell[CellGroupData[{ Cell[PROGRAMMING NOTES, Subsubsection], Cell[TextData[{ > The option , > StyleBox[Cubics->False, > FontFamily->Courier], > is used to keep the roots of cubics in , > StyleBox[Root[....], > FontFamily->Courier], > form. This is better for computation.n, > StyleBox[Re[v], > FontFamily->Courier], > and , > StyleBox[Abs[v]^2, > FontFamily->Courier], > are used rather than , > StyleBox[v+Conjugate[v] , > FontFamily->Courier], > and , > StyleBox[v*Conjugate[v], > FontFamily->Courier], > to prevent , > StyleBox[Apart, > FontFamily->Courier], > from factorising , > StyleBox[x^2 - x*2*Re[v] + Abs[v]^2], > FontFamily->Courier], > back to complex form. > }], Text] > }, Closed]], Cell[CellGroupData[{ Cell[EXAMPLES, Subsubsection], Cell[pol = Expand[(x - 1)*(x + 1)^2*(x^2 + x + 1)^2*(x^2 + 4)]; , > Input], Cell[CellGroupData[{ Cell[f1 = FactorC[pol, x], Input], Cell[BoxData[ > ((((-1) + x)) (((-2) [ImaginaryI] + > x)) ((2 [ImaginaryI] + > x)) ((1 + x))^2 (((((-1)))^(1/3) + x))^2 > (((-(((-1)))^(2/3)) + x))^2)], Output] > }, Open ]], Cell[CellGroupData[{ Cell[f2 = FactorR[pol, x], Input], Cell[BoxData[ > ((((-1) + x)) ((1 + x))^2 ((4 + > x^2)) ((1 + x + x^2))^2)], Output] > }, Open ]], Cell[CellGroupData[{ Cell[f3 = FactorR[x^3 + x + 1, x], Input], Cell[BoxData[ > (((x - Root[1 + #1 + #1^3 &, 1])) ((x^2 - > 2 x Root[(-1) + 2 #1 + 8 #1^3 &, 1] + > Root[(-1) - #1^4 + #1^6 &, 2]^2)))], Output] > }, Open ]], Cell[< > Root objects appear because of the option Cubics->False in Roots. > We can sometimes get radical forms, but notice the complication. >, Text], Cell[CellGroupData[{ Cell[ToRadicals[f3], Input], Cell[BoxData[ > (((((2/(3 (((-9) + @93)))))^(1/3) - ((1/2 > (((-9) + @93))))^(1/3)/3^(2/3) + x)) ((1/3 + > 1/3 ((29/2 - (3 @93)/2))^(1/3) + > 1/3 ((1/2 ((29 + 3 @93))))^(1/3) - > 2 ((((1/2 ((9 + @93))))^(1/3)/(2 > 3^(2/3)) - > 1/(2^(2/3) ((3 ((9 + > @93))))^(1/3)))) x + x^2)))], Output] > }, Open ]], Cell[Inexact forms can be found, from f3 :, Text], Cell[CellGroupData[{ Cell[N[f3], Input], Cell[BoxData[ > (((((0.6823278038280193`)([InvisibleSpace])) + > x)) ((((1.4655712318767682`)([InvisibleSpace])) - > 0.6823278038280193` x + x^2)))], Output] > }, Open ]], Cell[or directly, Text], Cell[CellGroupData[{ Cell[f3 = FactorR[x^3 + x + 1//N, x], Input], Cell[BoxData[ > (((((0.6823278038280193`)([InvisibleSpace])) + > x)) ((((1.4655712318767682`)([InvisibleSpace])) - > 0.6823278038280193` x + x^2)))], Output] > }, Open ]], Cell[Partial fractions, Text], Cell[CellGroupData[{ Cell[pf1 = PartialFractionsR[(2 + x)/pol, x], Input], Cell[BoxData[ > (1/(60 (((-1) + x))) - 1/(10 ((1 + x))^2) - > 39/(100 ((1 + x))) + ((-54) - 31 x)/(4225 ((4 + > x^2))) + ((-1) + 3 x)/(13 ((1 + x + x^2))^2) + (44 + > 193 x)/(507 ((1 + x + x^2))))], Output] > }, Open ]], Cell[CellGroupData[{ Cell[pf2 = PartialFractionsR[(1 + x)x/(1 - 3*x + x^2), x], > Input], Cell[< > 1 - (2*(-1 + 4*x))/((3 + Sqrt[5] - 2*x)*(-3 + 2*x)) + > (2*(-1 + 4*x))/((-3 + 2*x)*(-3 + Sqrt[5] + 2*x)) >, Output] > }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ > (Simplify[%])], Input], Cell[(x*(1 + x))/(1 - 3*x + x^2), Output] > }, Open ]], Cell[Partial fractions will often involve Root objects , Text], Cell[CellGroupData[{ Cell[pf3 = PartialFractionsR[(1 + x)/(x^3 - x + 1), x], Input], Cell[BoxData[ > (((1 + > Root[1 - #1 + #1^3 &, > 1]))/((((x - > Root[1 - #1 + #1^3 &, > 1])) ((Root[1 - #1 + #1^3 &, 1]^2 - > 2 Root[1 - #1 + #1^3 &, > 1] Root[(-1) - 2 #1 + 8 #1^3 &, 1] + > Root[(-1) + #1^4 + #1^6 &, 2]^2)))) + ((x + > Root[1 - #1 + #1^3 &, 1] + > x Root[1 - #1 + #1^3 &, 1] - > 2 Root[(-1) - 2 #1 + 8 #1^3 &, 1] - > Root[(-1) + #1^4 + #1^6 &, 2]^2))/(((((-x^2) + > 2 x Root[(-1) - 2 #1 + 8 #1^3 &, 1] - > Root[(-1) + #1^4 + #1^6 &, 2]^2)) ((Root[1 - > #1 + #1^3 &, 1]^2 - > 2 Root[1 - #1 + #1^3 &, > 1] Root[(-1) - 2 #1 + 8 #1^3 &, 1] + > Root[(-1) + #1^4 + #1^6 &, 2]^2)))))], > Output] > }, Open ]], Cell[This can in fact be put in radical form:, Text], Cell[CellGroupData[{ Cell[ToRadicals[pf3], Input], Cell[BoxData[ > (((1 - ((2/(3 ((9 - @69)))))^(1/3) - ((1/2 ((9 > - @69))))^(1/3)/3^(2/3)))/(((((-(1/3)) + > 1/3 ((25/2 - (3 @69)/2))^(1/3) + > 1/3 ((1/2 ((25 + 3 @69))))^(1/3) + > (((-((2/(3 ((9 - @69)))))^(1/3)) - ((1/2 ((9 - > @69))))^(1/3)/3^(2/3)))^2 - > 2 (((-((2/(3 ((9 - @69)))))^(1/3)) - > ((1/2 ((9 - @69))))^(1/3)/3^(2/3))) ((1/24 ((864 > - 96 @69))^(1/3) + ((1/2 ((9 + @69))))^(1/3)/(2 > 3^(2/3)))))) ((((2/(3 ((9 - @69)))))^(1/3) + ((1 > /2 ((9 - @69))))^(1/3)/3^(2/3) + x)))) + ((1/3 - > 1/3 ((25/2 - (3 @69)/2))^(1/3) - ((2/(3 > ((9 - @69)))))^(1/3) - ((1/2 ((9 - @69))))^(1/3)/3 > ^(2/3) - 1/3 ((1/2 ((25 + 3 @69))))^(1/3) - > 2 ((1/24 ((864 - 96 @69))^(1/3) + ((1/2 > ((9 + @69))))^(1/3)/(2 3^(2/3)))) + > x + (((-((2/(3 ((9 - @69)))))^(1/3)) - > ((1/2 ((9 - @69))))^(1/3)/3^(2/3))) x))/(((((-(1 > /3)) + 1/3 ((25/2 - (3 @69)/2))^(1/3) + > 1/3 ((1/2 ((25 + 3 @69))))^(1/3) + > (((-((2/(3 ((9 - @69)))))^(1/3)) - ((1/2 ((9 - > @69))))^(1/3)/3^(2/3)))^2 - > 2 (((-((2/(3 ((9 - @69)))))^(1/3)) - > ((1/2 ((9 - @69))))^(1/3)/3^(2/3))) ((1/24 ((864 > - 96 @69))^(1/3) + ((1/2 ((9 + @69))))^(1/3)/(2 > 3^(2/3)))))) ((1/3 - > 1/3 ((25/2 - (3 @69)/2))^(1/3) - > 1/3 ((1/2 ((25 + 3 @69))))^(1/3) + > 2 ((1/24 ((864 - 96 @69))^(1/3) + ((1/2 > ((9 + @69))))^(1/3)/(2 3^(2/3)))) x - > x^2)))))], Output] > }, Closed]], Cell[We could have found the inexact form directly., Text], Cell[CellGroupData[{ Cell[BoxData[ > (PartialFractionsR[((1 + x))/((x^3 - x + 1)) // N, > x])], Input], Cell[BoxData[ > ((-(0.07614206365252976`/(((1.324717957244746`)( > [InvisibleSpace])) + > 1.` x))) + (((0.7982664819556426`)( > [InvisibleSpace])) + 0.07614206365252976` > x)/(((0.754877666246693`)([InvisibleSpace])) - > 1.324717957244746` x + 1.` x^2))], Output] > }, Open ]] > }, Closed]] > }, Open ]] > }, > ScreenRectangle->{{0, 1024}, {0, 709}}, > AutoGeneratedPackage->None, > WindowSize->{534, 628}, > WindowMargins->{{199, Automatic}, {0, Automatic}}, > ShowCellLabel->False, > StyleDefinitions -> Default.nb > ] XXX > -- > Allan --------------------- > Allan Hayes > Mathematica Training and Consulting > Leicester UK > www.haystack.demon.co.uk > hay@haystack.demon.co.uk > Voice: +44 (0)116 271 4198 >> Greetings MathGroup, >> My name is Steve Earth, and I am a new subscriber to this list and >> also a >> new user of Mathematica; so please forgive this rather simple >> question... >> I would like to enter the quartic x^4 + x^3 + x^2 + x + 1 into >> Mathematica > and have it be able to tell me that it factors into >> (x^2 + GoldenRatio x + 1) ( x^2 - 1/GoldenRatio x + 1) >> What instructions do I need to execute to achieve this output? >> -Steve Earth >> Harker School >> http://www.harker.org/ > > ==== In my earlier posting I used Union and SameTest to replace two equivalent answers (arising form the symmetry of the equation) by a single one. However, the way I did, while givign the right answer, it makes little logical sense since in general applying Expand would make any factorizations the same leaving us always with just a single one. WIthout using Union at all we get: (a + b*x + x^2)*(c + d*x + x^2) /. Select[SolveAlways[x^4 + x^3 + x^2 + x + 1 == (a + b*x + x^2)*(c + d*x + x^2), x], FreeQ[#1, I] & ] {(1 + (1/2 - Sqrt[5]/2)*x + x^2)* (1 + (1/2)*(1 + Sqrt[5])*x + x^2), (1 + (1/2)*(1 - Sqrt[5])*x + x^2)* (1 + (1/2)*(1 + Sqrt[5])*x + x^2)} Since having two identical answers differing only in the way they are written out is a bit of a nuisance, a way to get rid of one of them which does not suffer from the illogicality of my first approach is: Union[(a + b*x + x^2)*(c + d*x + x^2) /. Select[SolveAlways[x^4 + x^3 + x^2 + x + 1 == (a + b*x + x^2)*(c + d*x + x^2), x], FreeQ[#1, I] & ], SameTest -> (Simplify[First[#1] == First[#2]] & )] {(1 + (1/2)*(1 - Sqrt[5])*x + x^2)* (1 + (1/2)*(1 + Sqrt[5])*x + x^2)} > There is an equivalent approach that will give the answer without > knowing it in advance. All we need to know is that any quartic can be > factored over the reals as a product of two quadratics, so: > Union[(a + b*x + x^2)*(c + d*x + x^2) /. > Select[SolveAlways[x^4 + x^3 + x^2 + x + 1 == > (a + b*x + x^2)*(c + d*x + x^2), x], FreeQ[#1, I] & ], > SameTest -> (Expand[#1] == Expand[#2] & )] > {(1 + (1/2)*(1 - Sqrt[5])*x + x^2)* > (1 + (1/2)*(1 + Sqrt[5])*x + x^2)} > Andrzej Kozlowski > Toyama International University > JAPAN > http://sigma.tuins.ac.jp/~andrzej/ > Steve >> The notebook given after NOTEBOOK below contains functions for >> factoring and >> partial fractioning. >> Here is an application to your problem: the first stage avoids our >> needing >> to know anything about the answer. >> fc=FactorR[x^4+x^3+x^2+x+1,x] >> (1 - (1/2)*(-1 - Sqrt[5])*x + x^2)* >> (1 - (1/2)*(-1 + Sqrt[5])*x + x^2) >> Now we need to get rid of Sqrt[5] in terms of GoldenRatio. >> This is rather messy: >> Simplify/@(fc/. Sqrt[5][Rule]2 GoldenRatio-1) >> (1 + x - GoldenRatio*x + x^2)*(1 + GoldenRatio*x + x^2) >> Simplify/@(%/.-GoldenRatio[Rule] 1/GoldenRatio -1) >> (1 + x/GoldenRatio + x^2)*(1 + GoldenRatio*x + x^2) >> Another example >> PartialFractionsR[(1 + x)x/(1 - 3*x + x^2), x] >> 1 - (2*(-1 + 4*x))/((3 + Sqrt[5] - 2*x)*(-3 + 2*x)) + >> (2*(-1 + 4*x))/((-3 + 2*x)*(-3 + Sqrt[5] + 2*x)) >> Simplify[%] >> (x*(1 + x))/(1 - 3*x + x^2) >> NOTEBOOK: to make a notebook from the following, copy from the next >> line to >> the line preceding XXX and paste into a new Mathematica notebook. >> Notebook[{ >> Cell[CellGroupData[{ >> Cell[Factors and PartialFractions, Subtitle], >> Cell[Allan Hayes, 16 August 2001, Text], >> Cell[< >> Here are some functions for factoring and expressing in partial >> fractions over the reals and over the complex numbers. >>, Text], >> Cell[BoxData[ >> (Quit)], Input], >> Cell[BoxData[{ >> (Off[General::spell1, General::spell]), n, >> ((FactorC::usage = > polynomial in x with complex coefficients, gives its factorization >> over the complex numbers.n >> The output may include Root objects which may be evaluated with >> ToRadicals or N.>;)n), n, >> ((FactorR::usage = > polynomial in x with real coefficients, gives its factorization over >> the reals.n >> The output may include Root objects which may be evaluated with >> ToRadicals or N.>;)n), n, >> ((PartialFractionsC::usage = > where ratl is a rational in x with complex coefficients, gives its >> factorization over the complex numbers.n >> The output may include Root objects which may be evaluated with >> ToRadicals or N.>;)n), n, >> ((PartialFractionsR::usage = > where ratl is a rational in x with real coefficients, gives its >> factorization over the real numbers.n >> The output may include Root objects which may be evaluated with >> ToRadicals or N.>;)), n, >> (On[General::spell1, General::spell])}], Input, >> InitializationCell->True], >> Cell[TextData[{ >> FactorC[p_, x_] := , >> StyleBox[(*over complex numbers*), >> FontFamily->Arial, >> FontWeight->Plain], >> nTimes @@ Cases[Roots[p == 0, x, n Cubics -> False], u_ == >> v_ -> x - v]n nFactorR[p_, x_] := , >> StyleBox[(*over reals, coefficients must be real*), >> FontFamily->Arial, >> FontWeight->Plain], >> n (Times @@ Join[Cases[#1, u_ == v_ /; Im[v] == 0 :> n x >> - v], Cases[#1, u_ == v_ /; Im[v] > 0 :> n x^2 - x*2*Re[v] + >> Abs[v]^2]] & )[n Roots[p == 0, x, Cubics -> False]] >> }], Input, >> InitializationCell->True], >> Cell[TextData[{ >> PartialFractionsC[p_, x_] := , >> StyleBox[(*over complex numbers*), >> FontFamily->Arial, >> FontWeight->Plain], >> n(#+Apart[#2/FactorC[#3,x]])&@@Flatten[{PolynomialReduce[#,#2], >> #2}]&[Numerator[#],Denominator[#]]&[Together[p]]n n >> PartialFractionsR[p_, x_] := , >> StyleBox[(*over reals, coefficients must be real*), >> FontFamily->Arial, >> FontWeight->Plain], >> n(#+Apart[#2/FactorR[#3,x]])&@@Flatten[{PolynomialReduce[#,#2], >> #2}]&[Numerator[#],Denominator[#]]&[Together[p]] >> }], Input, >> InitializationCell->True], >> Cell[CellGroupData[{ >> Cell[PROGRAMMING NOTES, Subsubsection], >> Cell[TextData[{ >> The option , >> StyleBox[Cubics->False, >> FontFamily->Courier], >> is used to keep the roots of cubics in , >> StyleBox[Root[....], >> FontFamily->Courier], >> form. This is better for computation.n, >> StyleBox[Re[v], >> FontFamily->Courier], >> and , >> StyleBox[Abs[v]^2, >> FontFamily->Courier], >> are used rather than , >> StyleBox[v+Conjugate[v] , >> FontFamily->Courier], >> and , >> StyleBox[v*Conjugate[v], >> FontFamily->Courier], >> to prevent , >> StyleBox[Apart, >> FontFamily->Courier], >> from factorising , >> StyleBox[x^2 - x*2*Re[v] + Abs[v]^2], >> FontFamily->Courier], >> back to complex form. >> }], Text] >> }, Closed]], >> Cell[CellGroupData[{ >> Cell[EXAMPLES, Subsubsection], >> Cell[pol = Expand[(x - 1)*(x + 1)^2*(x^2 + x + 1)^2*(x^2 + 4)]; , >> Input], >> Cell[CellGroupData[{ >> Cell[f1 = FactorC[pol, x], Input], >> Cell[BoxData[ >> ((((-1) + x)) (((-2) [ImaginaryI] + >> x)) ((2 [ImaginaryI] + >> x)) ((1 + x))^2 (((((-1)))^(1/3) + x))^2 >> (((-(((-1)))^(2/3)) + x))^2)], Output] >> }, Open ]], >> Cell[CellGroupData[{ >> Cell[f2 = FactorR[pol, x], Input], >> Cell[BoxData[ >> ((((-1) + x)) ((1 + x))^2 ((4 + >> x^2)) ((1 + x + x^2))^2)], Output] >> }, Open ]], >> Cell[CellGroupData[{ >> Cell[f3 = FactorR[x^3 + x + 1, x], Input], >> Cell[BoxData[ >> (((x - Root[1 + #1 + #1^3 &, 1])) ((x^2 - >> 2 x Root[(-1) + 2 #1 + 8 #1^3 &, 1] + >> Root[(-1) - #1^4 + #1^6 &, 2]^2)))], Output] >> }, Open ]], >> Cell[< >> Root objects appear because of the option Cubics->False in Roots. >> We can sometimes get radical forms, but notice the complication. >>, Text], >> Cell[CellGroupData[{ >> Cell[ToRadicals[f3], Input], >> Cell[BoxData[ >> (((((2/(3 (((-9) + @93)))))^(1/3) - ((1/2 >> (((-9) + @93))))^(1/3)/3^(2/3) + x)) ((1/3 + >> 1/3 ((29/2 - (3 @93)/2))^(1/3) + >> 1/3 ((1/2 ((29 + 3 @93))))^(1/3) - >> 2 ((((1/2 ((9 + @93))))^(1/3)/(2 >> 3^(2/3)) - >> 1/(2^(2/3) ((3 ((9 + >> @93))))^(1/3)))) x + x^2)))], Output] >> }, Open ]], >> Cell[Inexact forms can be found, from f3 :, Text], >> Cell[CellGroupData[{ >> Cell[N[f3], Input], >> Cell[BoxData[ >> (((((0.6823278038280193`)([InvisibleSpace])) + >> x)) ((((1.4655712318767682`)([InvisibleSpace])) - >> 0.6823278038280193` x + x^2)))], Output] >> }, Open ]], >> Cell[or directly, Text], >> Cell[CellGroupData[{ >> Cell[f3 = FactorR[x^3 + x + 1//N, x], Input], >> Cell[BoxData[ >> (((((0.6823278038280193`)([InvisibleSpace])) + >> x)) ((((1.4655712318767682`)([InvisibleSpace])) - >> 0.6823278038280193` x + x^2)))], Output] >> }, Open ]], >> Cell[Partial fractions, Text], >> Cell[CellGroupData[{ >> Cell[pf1 = PartialFractionsR[(2 + x)/pol, x], Input], >> Cell[BoxData[ >> (1/(60 (((-1) + x))) - 1/(10 ((1 + x))^2) - >> 39/(100 ((1 + x))) + ((-54) - 31 x)/(4225 ((4 + >> x^2))) + ((-1) + 3 x)/(13 ((1 + x + x^2))^2) + (44 + >> 193 x)/(507 ((1 + x + x^2))))], Output] >> }, Open ]], >> Cell[CellGroupData[{ >> Cell[pf2 = PartialFractionsR[(1 + x)x/(1 - 3*x + x^2), x], >> Input], >> Cell[< >> 1 - (2*(-1 + 4*x))/((3 + Sqrt[5] - 2*x)*(-3 + 2*x)) + >> (2*(-1 + 4*x))/((-3 + 2*x)*(-3 + Sqrt[5] + 2*x)) >>, Output] >> }, Open ]], >> Cell[CellGroupData[{ >> Cell[BoxData[ >> (Simplify[%])], Input], >> Cell[(x*(1 + x))/(1 - 3*x + x^2), Output] >> }, Open ]], >> Cell[Partial fractions will often involve Root objects , Text], >> Cell[CellGroupData[{ >> Cell[pf3 = PartialFractionsR[(1 + x)/(x^3 - x + 1), x], Input], >> Cell[BoxData[ >> (((1 + >> Root[1 - #1 + #1^3 &, >> 1]))/((((x - >> Root[1 - #1 + #1^3 &, >> 1])) ((Root[1 - #1 + #1^3 &, 1]^2 - >> 2 Root[1 - #1 + #1^3 &, >> 1] Root[(-1) - 2 #1 + 8 #1^3 &, 1] + >> Root[(-1) + #1^4 + #1^6 &, 2]^2)))) + ((x + >> Root[1 - #1 + #1^3 &, 1] + >> x Root[1 - #1 + #1^3 &, 1] - >> 2 Root[(-1) - 2 #1 + 8 #1^3 &, 1] - >> Root[(-1) + #1^4 + #1^6 &, 2]^2))/(((((-x^2) + >> 2 x Root[(-1) - 2 #1 + 8 #1^3 &, 1] - >> Root[(-1) + #1^4 + #1^6 &, 2]^2)) ((Root[1 - >> #1 + #1^3 &, 1]^2 - >> 2 Root[1 - #1 + #1^3 &, >> 1] Root[(-1) - 2 #1 + 8 #1^3 &, 1] + >> Root[(-1) + #1^4 + #1^6 &, 2]^2)))))], >> Output] >> }, Open ]], >> Cell[This can in fact be put in radical form:, Text], >> Cell[CellGroupData[{ >> Cell[ToRadicals[pf3], Input], >> Cell[BoxData[ >> (((1 - ((2/(3 ((9 - @69)))))^(1/3) - ((1/2 ((9 >> - @69))))^(1/3)/3^(2/3)))/(((((-(1/3)) + >> 1/3 ((25/2 - (3 @69)/2))^(1/3) + >> 1/3 ((1/2 ((25 + 3 @69))))^(1/3) + >> (((-((2/(3 ((9 - @69)))))^(1/3)) - ((1/2 ((9 - >> @69))))^(1/3)/3^(2/3)))^2 - >> 2 (((-((2/(3 ((9 - @69)))))^(1/3)) - >> ((1/2 ((9 - @69))))^(1/3)/3^(2/3))) ((1/24 ((864 >> - 96 @69))^(1/3) + ((1/2 ((9 + @69))))^(1/3)/(2 >> 3^(2/3)))))) ((((2/(3 ((9 - @69)))))^(1/3) + ((1 >> /2 ((9 - @69))))^(1/3)/3^(2/3) + x)))) + ((1/3 - >> 1/3 ((25/2 - (3 @69)/2))^(1/3) - ((2/(3 >> ((9 - @69)))))^(1/3) - ((1/2 ((9 - @69))))^(1/3)/3 >> ^(2/3) - 1/3 ((1/2 ((25 + 3 @69))))^(1/3) - >> 2 ((1/24 ((864 - 96 @69))^(1/3) + ((1/2 >> ((9 + @69))))^(1/3)/(2 3^(2/3)))) + >> x + (((-((2/(3 ((9 - @69)))))^(1/3)) - >> ((1/2 ((9 - @69))))^(1/3)/3^(2/3))) x))/(((((-(1 >> /3)) + 1/3 ((25/2 - (3 @69)/2))^(1/3) + >> 1/3 ((1/2 ((25 + 3 @69))))^(1/3) + >> (((-((2/(3 ((9 - @69)))))^(1/3)) - ((1/2 ((9 - >> @69))))^(1/3)/3^(2/3)))^2 - >> 2 (((-((2/(3 ((9 - @69)))))^(1/3)) - >> ((1/2 ((9 - @69))))^(1/3)/3^(2/3))) ((1/24 ((864 >> - 96 @69))^(1/3) + ((1/2 ((9 + @69))))^(1/3)/(2 >> 3^(2/3)))))) ((1/3 - >> 1/3 ((25/2 - (3 @69)/2))^(1/3) - >> 1/3 ((1/2 ((25 + 3 @69))))^(1/3) + >> 2 ((1/24 ((864 - 96 @69))^(1/3) + ((1/2 >> ((9 + @69))))^(1/3)/(2 3^(2/3)))) x - >> x^2)))))], Output] >> }, Closed]], >> Cell[We could have found the inexact form directly., Text], >> Cell[CellGroupData[{ >> Cell[BoxData[ >> (PartialFractionsR[((1 + x))/((x^3 - x + 1)) // N, >> x])], Input], >> Cell[BoxData[ >> ((-(0.07614206365252976`/(((1.324717957244746`)( >> [InvisibleSpace])) + >> 1.` x))) + (((0.7982664819556426`)( >> [InvisibleSpace])) + 0.07614206365252976` >> x)/(((0.754877666246693`)([InvisibleSpace])) - >> 1.324717957244746` x + 1.` x^2))], Output] >> }, Open ]] >> }, Closed]] >> }, Open ]] >> }, >> ScreenRectangle->{{0, 1024}, {0, 709}}, >> AutoGeneratedPackage->None, >> WindowSize->{534, 628}, >> WindowMargins->{{199, Automatic}, {0, Automatic}}, >> ShowCellLabel->False, >> StyleDefinitions -> Default.nb >> ] >> XXX >> -- >> Allan >> --------------------- >> Allan Hayes >> Mathematica Training and Consulting >> Leicester UK >> www.haystack.demon.co.uk >> hay@haystack.demon.co.uk >> Voice: +44 (0)116 271 4198 > Greetings MathGroup, My name is Steve Earth, and I am a new subscriber to this list and > also a > new user of Mathematica; so please forgive this rather simple > question... I would like to enter the quartic x^4 + x^3 + x^2 + x + 1 into > Mathematica > and have it be able to tell me that it factors into (x^2 + GoldenRatio x + 1) ( x^2 - 1/GoldenRatio x + 1) What instructions do I need to execute to achieve this output? -Steve Earth > Harker School > http://www.harker.org/ > Andrzej Kozlowski Yokohama, Japan http://www.mimuw.edu.pl/~akoz/ http://platon.c.u-tokyo.ac.jp/andrzej/ Reply-To: ==== Let's be realistic. If you want 60 digits of precision, too bad! -- in the real world. There's nothing we can measure that closely. Drug concentrations in clinical trials are generally measured within 15%, for instance. Even machine precision is more than can be realistically expected in any application I can think of. Even getting a satellite to Jupiter probably involves more error in the final result than machine precision. (If not, it's because we rely on ongoing corrections and natural factors that put the satellite where it should be, such as gravity drawing it toward each rendezvous -- not on that kind of precision in propulsion or guidance.) So... unless all numerics in a problem have a theoretical origin, and could be represented in Mathematica as Infinite precision expressions... all this talk of higher-precision computation seems futile. The realistic question is this: given that I have confidence, say, in 6 digits of precision for the inputs of an expression, how many digits can I trust in the end result? Giving inputs MORE precision than they deserve isn't the best way to answer that question. Here are two methods of answering it in Mathematica. One uses bignums and the other uses Intervals. Repetitious trial and error are NOT required either way. BIGNUMS: ClearAll[a, b, f, x, y] f = x*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + y*b^8 + a/(2*b); a = SetPrecision[77617, 6]; b = SetPrecision[33096, 6]; x = SetPrecision[33375/100, 6]; y = SetPrecision[55/10, 6]; InputForm[f] -4.1396`-12.5121*^19 Several previous solutions have set the precision or accuracy of f before giving a and b (and possibly x and y) values. That results in making the exponents imprecise along with all coefficients (not just x and y), which may or may not be what we want. INTERVALS: I'll first digress to figure out what Interval is equivalent to 6-digit precision. You might not actually do this if you like the Interval method, but you have to decide SOMEHOW what Interval width to use. nums = {77617, 33096, 33375/100, 55/10}; (Interval[SetPrecision[#1, 6]] & ) /@ nums /. Interval[{a_, b_}] :> 2*((b - a)/(b + a)); InputForm[%] {3.2209438653903`0.207*^-6, 3.7768914672467`0.2761*^-6, 2.9260299625467`0.1652*^-6, 2.7743252840909`0.1421*^-6} For these numbers, # + Interval[{-1,1}]*#/630000& gets us very close, so I'll use that. The second method therefore is: g = #1 + Interval[{-1, 1}]*(#1/630000) & ; a = g[77617]; b = g[33096]; x = g[333.75]; y = g[5.5]; f = x*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + y*b^8 + a/(2*b) (Min[f] + Max[f])/2 Interval[{-2.136361928005054*^32, 2.1363651195928296*^32}] 1.5957938878075177*^26 Either method shows the answer has no trustworthy digits, but I think the second result is far easier to interpret. Here's another example, using the Sin function, whose derivative is Cos, whose magnitude is bounded by one. The precision of the Sin of a number should be GREATER than the precision of the number itself, especially when Cos is small. a = SetPrecision[Pi/2, 6]; InputForm[Sin[a]] 0.9999999999990905052982256654`11.6078 a = g[N[Pi/2]]; Sin[a] - 1 Interval[{-3.1084024243455137*^-12, 0}] I hope this was worthwhile to someone. DrBob -----Original Message----- >[...] I would say this is correct and show that > SetPrecision is very useful > indeed. It tells you (what of course you ought > to already know in this > case anyway) that machine precision will not > give you a realiable > answer in this case. If you know your numbers > with a great deal of > accuracy you can get an accurate answer: In[24]:= > f = SetAccuracy[333.75*b^6 + a^2*(11*a^2*b^2 - > b^6 - > 121*b^4 - 2) + 5.5*b^8 + a/(2*b), 100]; > a=SetPrecision[77617.,100]; b = > SetPrecision[33096.,100]; > In[26]:= > {f, Precision[f]} Out[26]= {-0.82739605994682136814116509547981629199903311578438481991 > 781484167246798617832`61.2597, 61} > Congratulations! You just requested accuracy of > 100 for f and got 61 ( > to convince yourself add Accuracy[f] to In[26]). > If In[24] one > replaces SetAccuracy by SetPrecision the result is > similar. PK > [...] One has (initially) an accuracy of 100 for an > expression that contains > variables. In[25]:= Clear[a,b,f] In[26]:= f = SetAccuracy[333.75*b^6 + > a^2*(11*a^2*b^2 - b^6 - > 121*b^4 - 2) + 5.5*b^8 + a/(2*b), 100]; In[27]:= Accuracy[f] > Out[27]= 100. Now we assign values to some indeterminants in f. In[28]:= a = SetPrecision[77617.,100]; b = > SetPrecision[33096.,100]; In[29]:= {f, Precision[f], Accuracy[f]} > Out[29]= > {-0.8273960599468213681411650954798162919990331157843848199178148, 61.2599, 61.3422} The precision and accuracy has dropped. This is all > according to > standard numerical analysis regarding cancellation > error. You'll find it > in any textbook on the topic. > Assume that I want accuracy and precision of 100 for f. You advice me to make experiments to find out, what should be the initial precision and accuracy of a and b to reach the requested accuracy and precision for f. Notice, that you cannot just repeat I[26], we saw already what happens. I have to re-type I[24], I[25], I[26], I[27], I[28], and I[29] as many times as needed to get f with accuracy and precision 100. Dan, you simply advocate to do MANUAL WORK that should be done by machine. Let's suppose that in the above example I just want 60 digits not 61. Precisely, I want 60 digits and nothing or zeros afterwards. Let's see if I could use SetAccuracy. In[30]:= SetAccuracy[%, 60] Out[30]= -0.82739605994682136814116509547981629199903311578438481991781 In[31]:= % // FullForm Out[30]//FullForm= -0.827396059946821368141165095479816291999033115784384819917814841672467 988` 59.9177 Oops, it did not work (as expected). Let's highlight with mouse the expression in Out[30] and copy to a new cell. Oops, we got -0.827396059946821368141165095479816291999033115784384819917814841672467 988` 59.9177 again. Let's change Out[30] to a text cell and then copy. In[31]:= -0.82739605994682136814116509547981629199903311578438481991781 Out[31]= -0.82739605994682136814116509547981629199903311578438481991781 Success? Not so fast. In[32]:= % // FullForm Out[32]//FullForm= -0.827396059946821368141165095479816291999033115784384819917809999999999 998635 08`59.2041 Dan, is there any simple way to get what I want? As I repeated already number of times, at this stage of the development of computer technology, software should do it for me (!). We both know that this is doable. Some of the textbooks that you just advised me to read describe it. As a developer of Mathematica, tell us why do you consider this to be a bad idea? Peter Kosta > As for what happens when you artificially raise > precision (or accuracy) > of machine numbers far beyond that guaranteed by > their internal > representation, that falls into to category of > garbage in, garbage out. > It is, howoever, valid to use SetPrecision to raise > precision in > (typically iterative) algorithms where significance > arithmetic might be > unduly pessimistic due to incorrect assumptions > about uncorollatedness > of numerical error. Examples of such usage have > appeared in this news > group. > Daniel Lichtblau > Wolfram Research __________________________________________________ Do you Yahoo!? http://faith.yahoo.com ==== Bobby, The example that I gave in my previous posting does not make my point, or it makes it in rather a hidden way, but it does show something interesting about computation with bigfloats. I'll explain what I mean by this and then give an example that does make the point directly. First, the previous example: Sin[#1]*10^#1*Log[1+10^(-#1)]&[15.9] -0.336629 Sin[#1]*10^#1*Log[1+10^(-#1)]&[SetPrecision[15.9,20]] Precision[%] -0.190858581374189370 17.7558 Sin[#1]*10^#1*Log[1+10^(-#1)]&[SetPrecision[15.9,7]] Precision[%] -0.19086 4.70309 (**) It looks as if the internal computations must be to a higher precision than 4 and that they start at SetPrecision[15.9,7]//FullForm 15.9000000000000003553`7 With MaxError = 10^-Accuracy[sp]//FullForm 1.590000000000001`*^-6 Roughly speaking, not more than the first seven digits are asserted to be correct. Now, the new example (taken from Stan Wagon, Programming Tips, Mathematica in Education and Research Volume 7, Number 2, 1988 p50) Clear[x] ser= Normal[Series[Cos[x],{x,0,200}]]; x= 75.0; ser//FullForm -2.7019882604300525`*^15 Probably not reliable. Set precision to 20: x= SetPrecision[75.0, 20]; (a=ser)//FullForm -16928.799183047`1.4688 MaxError= 10^-Accuracy[a] 575.263 Not good enough. Raise the precision to x= SetAccuracy[75.0,40]; (a=ser)//FullForm 1.0807905977573169155`7.7627 MaxError = 10^-Accuracy[a] !(1.3999657487996298`*^-6) That is 1.3999657487996298 10^-6 Good enough Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay@haystack.demon.co.uk Voice: +44 (0)116 271 4198 ==== Bobby, You note some important points but there are situations in which raising accuracy is a practical necessity. 1) It may be that a calculation (perhaps describing a real world phenonenon) necessitates our raising the precision of inuts so as to work to high precision internally even when the output is not very sensitive to changes in input. Compare the following graphs pts = Table[({#1, Sin[#1]*10^#1*Log[1 + 10^(-#1)]} & )[x], {x, 15., 20., 0.1}]; ListPlot[pts, PlotJoined -> True] pts = Table[({#1, Sin[#1]*10^#1*Log[1 + 10^(-#1)]} & )[ SetAccuracy[x, 20]], {x, 15., 20., 0.1}]; ListPlot[pts, PlotJoined -> True] 2) Another situation in which high precision numbers are needed is when a computation with exact numbers would be slow or might use up all the memory available. We then need to replace the exact numbers with high precision bigfloats which causes the number of digits used internally to be restricted and the maximum error in the output to be reported. The N function will raise the precision of the exact numbers to try and reach the requested precision. There are also concerns with, for example, how the replacing bigfloats are related to the original numbers. How important this depends on the use to which the calculation is being put and how sensitive the output is to changes in inputs - in the plotting above it is not important. -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay@haystack.demon.co.uk Voice: +44 (0)116 271 4198 > Let's be realistic. If you want 60 digits of precision, too bad! -- in > the real world. There's nothing we can measure that closely. Drug > concentrations in clinical trials are generally measured within 15%, for > instance. Even machine precision is more than can be realistically > expected in any application I can think of. Even getting a satellite to > Jupiter probably involves more error in the final result than machine > precision. (If not, it's because we rely on ongoing corrections and > natural factors that put the satellite where it should be, such as > gravity drawing it toward each rendezvous -- not on that kind of > precision in propulsion or guidance.) So... unless all numerics in a problem have a theoretical origin, and > could be represented in Mathematica as Infinite precision expressions... > all this talk of higher-precision computation seems futile. The realistic question is this: given that I have confidence, say, in 6 > digits of precision for the inputs of an expression, how many digits can > I trust in the end result? Giving inputs MORE precision than they > deserve isn't the best way to answer that question. Here are two > methods of answering it in Mathematica. One uses bignums and the > other uses Intervals. Repetitious trial and error are NOT required either way. BIGNUMS: ClearAll[a, b, f, x, y] > f = x*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + y*b^8 + a/(2*b); > a = SetPrecision[77617, 6]; > b = SetPrecision[33096, 6]; > x = SetPrecision[33375/100, 6]; > y = SetPrecision[55/10, 6]; > InputForm[f] -4.1396`-12.5121*^19 Several previous solutions have set the precision or accuracy of f > before giving a and b (and possibly x and y) values. That results in > making the exponents imprecise along with all coefficients (not just x > and y), which may or may not be what we want. INTERVALS: I'll first digress to figure out what Interval is equivalent to 6-digit > precision. You might not actually do this if you like the Interval > method, but you have to decide SOMEHOW what Interval width to use. nums = {77617, 33096, 33375/100, 55/10}; > (Interval[SetPrecision[#1, 6]] & ) /@ nums /. Interval[{a_, b_}] : 2*((b - a)/(b + a)); > InputForm[%] {3.2209438653903`0.207*^-6, > 3.7768914672467`0.2761*^-6, > 2.9260299625467`0.1652*^-6, > 2.7743252840909`0.1421*^-6} For these numbers, # + Interval[{-1,1}]*#/630000& gets us very close, so > I'll use that. The second method therefore is: g = #1 + Interval[{-1, 1}]*(#1/630000) & ; > a = g[77617]; > b = g[33096]; > x = g[333.75]; > y = g[5.5]; > f = x*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + y*b^8 + a/(2*b) > (Min[f] + Max[f])/2 Interval[{-2.136361928005054*^32, 2.1363651195928296*^32}] 1.5957938878075177*^26 Either method shows the answer has no trustworthy digits, but I think > the second result is far easier to interpret. Here's another example, using the Sin function, whose derivative is Cos, > whose magnitude is bounded by one. The precision of the Sin of a number > should be GREATER than the precision of the number itself, especially > when Cos is small. a = SetPrecision[Pi/2, 6]; > InputForm[Sin[a]] 0.9999999999990905052982256654`11.6078 a = g[N[Pi/2]]; > Sin[a] - 1 Interval[{-3.1084024243455137*^-12, 0}] I hope this was worthwhile to someone. DrBob -----Original Message----- > On Friday, October 4, 2002, at 06:01 PM, DrBob [...] I would say this is correct and show that > SetPrecision is very useful > indeed. It tells you (what of course you ought > to already know in this > case anyway) that machine precision will not > give you a realiable > answer in this case. If you know your numbers > with a great deal of > accuracy you can get an accurate answer: In[24]:= > f = SetAccuracy[333.75*b^6 + a^2*(11*a^2*b^2 - > b^6 - > 121*b^4 - 2) + 5.5*b^8 + a/(2*b), 100]; > a=SetPrecision[77617.,100]; b = > SetPrecision[33096.,100]; > In[26]:= > {f, Precision[f]} Out[26]= > {-0.82739605994682136814116509547981629199903311578438481991 > 781484167246798617832`61.2597, 61} > Congratulations! You just requested accuracy of > 100 for f and got 61 ( > to convince yourself add Accuracy[f] to In[26]). > If In[24] one > replaces SetAccuracy by SetPrecision the result is > similar. PK > [...] One has (initially) an accuracy of 100 for an > expression that contains > variables. In[25]:= Clear[a,b,f] In[26]:= f = SetAccuracy[333.75*b^6 + > a^2*(11*a^2*b^2 - b^6 - > 121*b^4 - 2) + 5.5*b^8 + a/(2*b), 100]; In[27]:= Accuracy[f] > Out[27]= 100. Now we assign values to some indeterminants in f. In[28]:= a = SetPrecision[77617.,100]; b = > SetPrecision[33096.,100]; In[29]:= {f, Precision[f], Accuracy[f]} > Out[29]= {-0.8273960599468213681411650954798162919990331157843848199178148, 61.2599, 61.3422} The precision and accuracy has dropped. This is all > according to > standard numerical analysis regarding cancellation > error. You'll find it > in any textbook on the topic. > Assume that I want accuracy and precision of 100 for > f. You advice me to make experiments to find out, what > should be the initial precision and accuracy of a and > b to reach the requested accuracy and precision for f. > Notice, that you cannot just repeat I[26], we saw > already what happens. I have to re-type I[24], I[25], > I[26], I[27], I[28], and I[29] as many times as needed > to get f with accuracy and precision 100. Dan, you simply advocate to do MANUAL WORK that should > be done by machine. Let's suppose that in the above example I just want 60 > digits not 61. Precisely, I want 60 digits and nothing > or zeros afterwards. Let's see if I could use > SetAccuracy. In[30]:= > SetAccuracy[%, 60] Out[30]= > -0.82739605994682136814116509547981629199903311578438481991781 In[31]:= > % // FullForm Out[30]//FullForm= > -0.827396059946821368141165095479816291999033115784384819917814841672467 > 988` > 59.9177 Oops, it did not work (as expected). Let's highlight > with mouse the expression in Out[30] and copy to a new > cell. Oops, we got > -0.827396059946821368141165095479816291999033115784384819917814841672467 > 988` > 59.9177 > again. Let's change Out[30] to a text cell and then > copy. In[31]:= > -0.82739605994682136814116509547981629199903311578438481991781 Out[31]= > -0.82739605994682136814116509547981629199903311578438481991781 Success? Not so fast. In[32]:= > % // FullForm Out[32]//FullForm= > -0.827396059946821368141165095479816291999033115784384819917809999999999 > 998635 > 08`59.2041 Dan, is there any simple way to get what I want? As I repeated already number of times, at this stage > of the development of computer technology, software > should do it for me (!). We both know that this is > doable. Some of the textbooks that you just advised me > to read describe it. As a developer of Mathematica, > tell us why do you consider this to be a bad idea? Peter Kosta As for what happens when you artificially raise > precision (or accuracy) > of machine numbers far beyond that guaranteed by > their internal > representation, that falls into to category of > garbage in, garbage out. > It is, howoever, valid to use SetPrecision to raise > precision in > (typically iterative) algorithms where significance > arithmetic might be > unduly pessimistic due to incorrect assumptions > about uncorollatedness > of numerical error. Examples of such usage have > appeared in this news > group. > Daniel Lichtblau > Wolfram Research > __________________________________________________ > Do you Yahoo!? > http://faith.yahoo.com Reply-To: ==== Here's a start, with a more complicated function: lst = Range[50]; f = #1^2 & ; k = 7; r = Range[k + 1, Length[lst], k]; lst[[r]] = lst[[r]] + f /@ lst[[r - 1]]; lst {1, 2, 3, 4, 5, 6, 7, 57, 9, 10, 11, 12, 13, 14, 211, 16, 17, 18, 19, 20, 21, 463, 23, 24, 25, 26, 27, 28, 813, 30, 31, 32, 33, 34, 35, 1261, 37, 38, 39, 40, 41, 42, 1807, 44, 45, 46, 47, 48, 49, 2451} or: g[lst_List, k_Integer?Positive, f_Function] := Module[ {result = lst, r = Range[k + 1, Length[lst], k]}, result[[r]] = result[[r]] + f /@ lst[[r - 1]]; result ] lst = Range[50]; lst = g[lst, 7, #1^2 & ] or: lst = Range[50]; lst + Drop[Prepend[MapIndexed[If[Mod[First@#2, k] == 0, f@#1, 0] &, lst], 0], -1] DrBob -----Original Message----- advantage of native List operators, but still not be too recondite? I've been thinking about multiplying a copy of myList by a mask list {0,0,1,0,0,1,..} to generate a masked copy and approaches like that. Better ways??? ==== I'm playing around with Mathematica just trying to see what happens if... One thing I came up with was lst1={a,b,c}; lst2={d,e,f}; Map[lst1,lst2} which resulted in the following rather unusual looking expression(?): {{a, b, c}[d], {a, b, c}[e], {a, b, c}[f]} I'm wondering if such a List represents something 'meaningful'. Any opinions? STH . ==== > I'm playing around with Mathematica just trying to see what happens if... One thing > I came up with was lst1={a,b,c}; lst2={d,e,f}; Map[lst1,lst2} which > resulted in the following rather unusual looking expression(?): {{a, b, c}[d], {a, b, c}[e], {a, b, c}[f]} I'm wondering if such a List represents something 'meaningful'. Any > opinions? Through /@ {{a, b, c}[d], {a, b, c}[e], {a, b, c}[f]} {{a[d], b[d], c[d]}, {a[e], b[e], c[e]}, {a[f], b[f], c[f]}} which might be useful. -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay@haystack.demon.co.uk Voice: +44 (0)116 271 4198 ==== >I want to apply a function to every k-th element of a long list and >add the result to the k+1 element. [Actually k = 3 and I just want to multiply myList[[k]] by a >constant (independent of k) and add the result to myList[[k+1]] for >every value of k that's divisible by 3.] lst = Table[Random[], {20}] fact = 2 Here's a matrix method. f represents an indexed element of a matrix. f[x_,x_] := If[Mod[x,3]==0, fact, 1] f[x_, y_] := If[Mod[x,3]==0&&y[Equal]x+1, 2, 0] Create a matrix from the elements. arr = Array[f,{Length[lst], Length[lst]}]; Then matrix multiply. newlst1 = lst.arr Here's another way with the highly underrated MapIndexed. Create pairs of the nth and n-1th values. pairs = Transpose[{lst, Prepend[Drop[lst,-1],0]}] Create a function that takes the nth value (val), the n-1th value (prevval), and the index (num). multlst[{val_, prevval_}, {num_}] := Switch[Mod[num, 3], 0, fact*val, 1,val+ fact*prevval, 2, val ] Then, MapIndexed across the pairs. newlst2=MapIndexed[multlst, pairs] -------------------------------------------------------------- Omega Consulting The final answer to your Mathematica needs Spend less time searching and more time finding. http://www.wz.com/internet/Mathematica.html ==== I have two equations that I have solved for: x[n_] := 2331 + 8 n y[n_] := -3108 - 11n I want to include only solutions which are non-negative, that is x >= 0 and y >= 0. In this example we can do 2331 + 8n > = 0 and solve for n, n >= -291.375 and -3108 - 11 n >= 0 and solve for n, n <= -282.545 So we have -291.375 <= n <= -282.545. The integer solution set here is for n = {-290, -289, -288, -287, -286, -285, -284, -283}. So in this case we have 8 non-negative solutions. Given that I can supply x[n] and y[n], how do I go about finding the set n? ==== Here is a way of doing what you want (output cells are indented): <=0,y[n]>=0},n] -(2331/8) <= n <= -(3108/11) FullForm[soln] Inequality[Rational[-2331,8],LessEqual,n,LessEqual,Rational[-3108,11]] Range[Apply[Sequence,{Ceiling[soln[[1]]],Floor[soln[[-1]]]}]] {-291, -290, -289, -288, -287, -286, -285, -284, -283} -- Steve Luttrell West Malvern, UK I have two equations that I have solved for: x[n_] := 2331 + 8 n > y[n_] := -3108 - 11n I want to include only solutions which are non-negative, that is x >= 0 and > y >= 0. In this example we can do 2331 + 8n > = 0 and solve for n, n >= -291.375 > and -3108 - 11 n >= 0 and solve for n, n <= -282.545 So we have -291.375 <= n <= -282.545. The integer solution set here is for n = > {-290, -289, -288, -287, -286, -285, -284, -283}. So in this case we have 8 non-negative solutions. Given that I can supply x[n] and y[n], how do I go about finding the set n? ==== Let's first consider your original problem and take a small list as an example: mylist = {a, b, c, d, e, f, g}; As you pointed out there is a rather obvious and natural way to do it using the Do loop. In[5]:= Do[mylist[[3*i + 1]] = k*mylist[[3*i]] + mylist[[3*i + 1]], {i, 1, Length[mylist]/3}]; mylist Out[5]= {a, b, c, d + c*k, e, f, g + f*k} One can also do it using something like your second approach. Notice the need to re-set mylist which got changed by the Do loop: In[6]:= mylist={a,b,c,d,e,f,g}; In[7]:= RotateRight[Table[If[Mod[i,3]==0,k, 0],{i,1,Length[mylist]}]*mylist]+mylist Out[7]= {a,b,c,d+c k,e,f,g+f k} There is an ambiguity that appears if the length of the list is exactly divisible by 3. In that case k times the last element should be added to the next element. In this case the first approach (using Do) will produce an error message while the second will interpret the next element to mean the first element of the list. One can fix that but I shan't bother to do so and assume that the length of the list is not divisible by 3. Now let's take a large list and compare the performance of the two approaches: In[8]:= k = 2; In[9]:= mylist = Table[Random[Integer, {1, 9}], {10000}]; In[11]:= mylist1 = mylist; In[12]:= Timing[list1 = (Do[mylist[[3*i + 1]] = k*mylist[[3*i]] + mylist[[3*i + 1]], {i, 1, Length[mylist]/3}]; mylist); ] Out[12]= {0.21999999999999997*Second, Null} In[13]:= mylist = mylist1; In[14]:= Timing[list2 = RotateRight[Table[If[Mod[i, 3] == 0, k, 0], {i, 1, Length[mylist]}]*mylist] + mylist; ] Out[14]= {0.020000000000000018*Second, Null} In[15]:= list1 == list2 Out[15]= True So you were right, at least in this implementation the second approach turns out to be much faster. Note also however, that when you change the problem to the one that you originally stated the two approaches will no longer give the same answer. The reason is that your statement is ambiguous: apply a function to every k-th element of a long list and add the result to the k+1 element can either mean that you want to take the original k-th element of the original list, multiply by a constant and add to the next one, or do the same with the already altered k-th element (by the previous step of the procedure). The Do loop approach will do the latter and the other the former. Andrzej Kozlowski Yokohama, Japan http://www.mimuw.edu.pl/~akoz/ http://platon.c.u-tokyo.ac.jp/andrzej/ > I want to apply a function to every k-th element of a long list and > add the result to the k+1 element. [Actually k = 3 and I just want to multiply myList[[k]] by a > constant (independent of k) and add the result to myList[[k+1]] for > every value of k that's divisible by 3.] Is there a way to do this -- or in general to get at every k-th > element of a list -- that's faster and more elegant than writing a > brute > force Do[] loop or using Mod[] operators, and that will take > advantage of native List operators, but still not be too recondite? I've been thinking about multiplying a copy of myList by a mask > list > {0,0,1,0,0,1,..} to generate a masked copy and approaches like that. > Better ways??? > ==== Can Mathematica factor the polynomial p1=x^6+9/14*x^5+9/28*x^4+3/35*x^3+9/700*x^2+9/8750*x+3/87500; without a priori knowledge of the Extension field? ==== Carlos, Futher to my previous posting (which gave the code for the function FactorR used below), here is a complete factorisation by radicals. I also test that the product of the factors gives the original polynomial. We want to factor the polynomial p1 = x^6 + (9/14)*x^5 + (9/28)*x^4 + (3/35)*x^3 + (9/700)*x^2 + (9/8750)*x + 3/87500; in radicals. We can't expect this to be easy or even possible in terms of radicals (the general quintic is not solvable interms of radicals). But, using the function FactorR given in my posting, Re:factoring quartic over radicals, sent a few days ago (08/012/02) , we get p2 = FactorR[p1, x] (x^2 - 2*x*Root[3 + 45*#1 + 225*#1^2 + 700*#1^3 & , 1] + Root[-3 + 225*#1^2 - 5625*#1^4 + 87500*#1^6 & , 2]^2)* (x^2 - 2*x*Root[1827 + 65340*#1 + 974700*#1^2 + 7824000*#1^3 + 36360000*#1^4 + 100800000*#1^5 + 156800000*#1^6 & , 2] + Root[9 - 1350*#1^2 + 84375*#1^4 - 3056250*#1^6 - 11250000*#1^8 - 984375000*#1^10 + 7656250000*#1^12 & , 3]^2)* (x^2 - 2*x*Root[1827 + 65340*#1 + 974700*#1^2 + 7824000*#1^3 + 36360000*#1^4 + 100800000*#1^5 + 156800000*#1^6 & , 1] + Root[9 - 1350*#1^2 + 84375*#1^4 - 3056250*#1^6 - 11250000*#1^8 - 984375000*#1^10 + 7656250000*#1^12 & , 4]^2) Try to change the root objects to radical form: p3 = p2 /. r_Root :> ToRadicals[r] (3/140 + (1/140)*(13/5)^(2/3)*3^(1/3) - (1/140)*(13/5)^(1/3)*3^(2/3) - 2*(-(3/28) - (1/28)*(13/5)^(2/3)*3^(1/3) + (1/28)*(13/5)^(1/3)*3^(2/3))*x + x^2)*(x^2 - 2*x*Root[1827 + 65340*#1 + 974700*#1^2 + 7824000*#1^3 + 36360000*#1^4 + 100800000*#1^5 + 156800000*#1^6 & , 2] + Root[9 - 1350*#1 + 84375*#1^2 - 3056250*#1^3 - 11250000*#1^4 - 984375000*#1^5 + 7656250000*#1^6 & , 1])* (x^2 - 2*x*Root[1827 + 65340*#1 + 974700*#1^2 + 7824000*#1^3 + 36360000*#1^4 + 100800000*#1^5 + 156800000*#1^6 & , 1] + Root[9 - 1350*#1 + 84375*#1^2 - 3056250*#1^3 - 11250000*#1^4 - 984375000*#1^5 + 7656250000*#1^6 & , 2]) We succeeded with the first factor: f1 = p3[[1]] 3/140 + (1/140)*(13/5)^(2/3)*3^(1/3) - (1/140)*(13/5)^(1/3)*3^(2/3) - 2*(-(3/28) - (1/28)*(13/5)^(2/3)*3^(1/3) + (1/28)*(13/5)^(1/3)*3^(2/3))*x + x^2 The product of the other two factors, in a form avoiding root objects, is easily found by division: q = PolynomialQuotient[p1, f1, x] 3/3500 + ((13/5)^(1/3)*3^(2/3))/3500 + (3/175 + (1/175)*(13/5)^(1/3)*3^(2/3))*x + (9/70 - (1/140)*(13/5)^(2/3)*3^(1/3) + (1/28)*(13/5)^(1/3)*3^(2/3))*x^2 + (3/7 - (1/14)*(13/5)^(2/3)*3^(1/3) + (1/14)*(13/5)^(1/3)*3^(2/3))*x^3 + x^4 Try FactorR on this f23 = FactorR[q, x] (x^2 + ((-(1/2))*Sqrt[117/1960 + (9/784)*(13/5)^(2/3)*3^(1/3) + (39*(13/5)^(1/3)*3^(2/3))/3920] + Im[(1/2)*Sqrt[117/1960 + (9/784)*(13/5)^(2/3)*3^(1/3) + (39*(13/5)^(1/3)*3^(2/3))/3920 + (-2250 + 25*13^(2/3)*15^(1/3) - 125*13^(1/3)*15^(2/3))/8750 + (3*(7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)*15^(2/3))^2)/ 1225000000 - (-((2*(300 + 20*13^(1/3)*15^(2/3)))/4375) + (1/17500)*((7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)*15^(2/3))* ((2250 - 25*13^(2/3)*15^(1/3) + 125*13^(1/3)*15^(2/3))/4375 - (7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)*15^(2/3))^2/ 306250000)))/(4*Sqrt[-(117/1960) - (9/784)*(13/5)^(2/3)*3^(1/3) - (39*(13/5)^(1/3)*3^(2/3))/3920])]])^2 - 2*x*((-7500 + 250*13^(2/3)*15^(1/3) - 250*13^(1/3)*15^(2/3))/70000 + Re[(1/2)*Sqrt[117/1960 + (9/784)*(13/5)^(2/3)*3^(1/3) + (39*(13/5)^(1/3)*3^(2/3))/3920 + (-2250 + 25*13^(2/3)*15^(1/3) - 125*13^(1/3)*15^(2/3))/8750 + (3*(7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)*15^(2/3))^2)/ 1225000000 - (-((2*(300 + 20*13^(1/3)*15^(2/3)))/4375) + (1/17500)*((7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)*15^(2/3))* ((2250 - 25*13^(2/3)*15^(1/3) + 125*13^(1/3)*15^(2/3))/4375 - (7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)*15^(2/3))^2/ 306250000)))/(4*Sqrt[-(117/1960) - (9/784)*(13/5)^(2/3)*3^(1/3) - (39*(13/5)^(1/3)*3^(2/3))/3920])]]) + ((-7500 + 250*13^(2/3)*15^(1/3) - 250*13^(1/3)*15^(2/3))/70000 + Re[(1/2)*Sqrt[117/1960 + (9/784)*(13/5)^(2/3)*3^(1/3) + (39*(13/5)^(1/3)*3^(2/3))/3920 + (-2250 + 25*13^(2/3)*15^(1/3) - 125*13^(1/3)*15^(2/3))/8750 + (3*(7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)*15^(2/3))^2)/ 1225000000 - (-((2*(300 + 20*13^(1/3)*15^(2/3)))/4375) + (1/17500)*((7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)*15^(2/3))* ((2250 - 25*13^(2/3)*15^(1/3) + 125*13^(1/3)*15^(2/3))/4375 - (7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)*15^(2/3))^2/ 306250000)))/(4*Sqrt[-(117/1960) - (9/784)*(13/5)^(2/3)*3^(1/3) - (39*(13/5)^(1/3)*3^(2/3))/3920])]])^2)* (x^2 + ((1/2)*Sqrt[117/1960 + (9/784)*(13/5)^(2/3)*3^(1/3) + (39*(13/5)^(1/3)*3^(2/3))/3920] + Im[(-(1/2))*Sqrt[117/1960 + (9/784)*(13/5)^(2/3)*3^(1/3) + (39*(13/5)^(1/3)*3^(2/3))/3920 + (-2250 + 25*13^(2/3)*15^(1/3) - 125*13^(1/3)*15^(2/3))/8750 + (3*(7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)*15^(2/3))^2)/ 1225000000 + (-((2*(300 + 20*13^(1/3)*15^(2/3)))/4375) + (1/17500)*((7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)*15^(2/3))* ((2250 - 25*13^(2/3)*15^(1/3) + 125*13^(1/3)*15^(2/3))/4375 - (7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)*15^(2/3))^2/ 306250000)))/(4*Sqrt[-(117/1960) - (9/784)*(13/5)^(2/3)*3^(1/3) - (39*(13/5)^(1/3)*3^(2/3))/3920])]])^2 - 2*x*((-7500 + 250*13^(2/3)*15^(1/3) - 250*13^(1/3)*15^(2/3))/70000 + Re[(-(1/2))*Sqrt[117/1960 + (9/784)*(13/5)^(2/3)*3^(1/3) + (39*(13/5)^(1/3)*3^(2/3))/3920 + (-2250 + 25*13^(2/3)*15^(1/3) - 125*13^(1/3)*15^(2/3))/8750 + (3*(7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)*15^(2/3))^2)/ 1225000000 + (-((2*(300 + 20*13^(1/3)*15^(2/3)))/4375) + (1/17500)*((7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)*15^(2/3))* ((2250 - 25*13^(2/3)*15^(1/3) + 125*13^(1/3)*15^(2/3))/4375 - (7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)*15^(2/3))^2/ 306250000)))/(4*Sqrt[-(117/1960) - (9/784)*(13/5)^(2/3)*3^(1/3) - (39*(13/5)^(1/3)*3^(2/3))/3920])]]) + ((-7500 + 250*13^(2/3)*15^(1/3) - 250*13^(1/3)*15^(2/3))/70000 + Re[(-(1/2))*Sqrt[117/1960 + (9/784)*(13/5)^(2/3)*3^(1/3) + (39*(13/5)^(1/3)*3^(2/3))/3920 + (-2250 + 25*13^(2/3)*15^(1/3) - 125*13^(1/3)*15^(2/3))/8750 + (3*(7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)*15^(2/3))^2)/ 1225000000 + (-((2*(300 + 20*13^(1/3)*15^(2/3)))/4375) + (1/17500)*((7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)*15^(2/3))* ((2250 - 25*13^(2/3)*15^(1/3) + 125*13^(1/3)*15^(2/3))/4375 - (7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)*15^(2/3))^2/ 306250000)))/(4*Sqrt[-(117/1960) - (9/784)*(13/5)^(2/3)*3^(1/3) - (39*(13/5)^(1/3)*3^(2/3))/3920])]])^2) We try to get rid of the parts Re[.] and Im[.]:, f231 = f23 /. z:(_Re | _Im) :> ToRadicals[FullSimplify[z]] (((-7500 + 250*13^(2/3)*15^(1/3) - 250*13^(1/3)*15^(2/3))/70000 - (1/280)*Sqrt[3*(-390 + 13*13^(2/3)*15^(1/3) + 15*13^(1/3)*15^(2/3))])^2 + ((1/2)*Sqrt[117/1960 + (9/784)*(13/5)^(2/3)*3^(1/3) + (39*(13/5)^(1/3)*3^(2/3))/3920] + (1/280)*Sqrt[1170 + 73*13^(2/3)*15^(1/3) + 67*13^(1/3)*15^(2/3)])^2 - 2*((-7500 + 250*13^(2/3)*15^(1/3) - 250*13^(1/3)*15^(2/3))/70000 - (1/280)*Sqrt[3*(-390 + 13*13^(2/3)*15^(1/3) + 15*13^(1/3)*15^(2/3))])*x + x^2)*(((-7500 + 250*13^(2/3)*15^(1/3) - 250*13^(1/3)*15^(2/3))/70000 + (1/280)*Sqrt[3*(-390 + 13*13^(2/3)*15^(1/3) + 15*13^(1/3)*15^(2/3))])^2 + ((-(1/2))*Sqrt[117/1960 + (9/784)*(13/5)^(2/3)*3^(1/3) + (39*(13/5)^(1/3)*3^(2/3))/3920] + (1/280)*Sqrt[1170 + 73*13^(2/3)*15^(1/3) + 67*13^(1/3)*15^(2/3)])^2 - 2*((-7500 + 250*13^(2/3)*15^(1/3) - 250*13^(1/3)*15^(2/3))/70000 + (1/280)*Sqrt[3*(-390 + 13*13^(2/3)*15^(1/3) + 15*13^(1/3)*15^(2/3))])*x + x^2) We now have the ramaining two factors in radical form, but a little simplification helps: f232 = f231 /. (n_)?NumericQ :> Simplify[n] ((1/78400)*(30 - 13^(2/3)*15^(1/3) + 13^(1/3)*15^(2/3) + Sqrt[-1170 + 39*13^(2/3)*15^(1/3) + 45*13^(1/3)*15^(2/3)])^2 + (1/78400)*(Sqrt[1170 + 45*13^(2/3)*15^(1/3) + 39*13^(1/3)*15^(2/3)] + Sqrt[1170 + 73*13^(2/3)*15^(1/3) + 67*13^(1/3)*15^(2/3)])^2 - (1/140)*(-30 + 13^(2/3)*15^(1/3) - 13^(1/3)*15^(2/3) - Sqrt[-1170 + 39*13^(2/3)*15^(1/3) + 45*13^(1/3)*15^(2/3)])*x + x^2)* ((1/78400)*(-30 + 13^(2/3)*15^(1/3) - 13^(1/3)*15^(2/3) + Sqrt[-1170 + 39*13^(2/3)*15^(1/3) + 45*13^(1/3)*15^(2/3)])^2 + (1/78400)*(Sqrt[1170 + 45*13^(2/3)*15^(1/3) + 39*13^(1/3)*15^(2/3)] - Sqrt[1170 + 73*13^(2/3)*15^(1/3) + 67*13^(1/3)*15^(2/3)])^2 - (1/140)*(-30 + 13^(2/3)*15^(1/3) - 13^(1/3)*15^(2/3) + Sqrt[-1170 + 39*13^(2/3)*15^(1/3) + 45*13^(1/3)*15^(2/3)])*x + x^2) TEST Test if the product of the factors is equal to p1: prd1 = Collect[Expand[f232*f1], x] 172077/3841600000 - (4959*(13/5)^(2/3)*3^(1/3))/768320000 + (117*(13/5)^(1/3)*3^(2/3))/27440000 - (1/1920800000)* (3*Sqrt[1170 + 45*13^(2/3)*15^(1/3) + 39*13^(1/3)*15^(2/3)]* Sqrt[-1170 + 39*13^(2/3)*15^(1/3) + 45*13^(1/3)*15^(2/3)]* Sqrt[1170 + 73*13^(2/3)*15^(1/3) + 67*13^(1/3)*15^(2/3)]) + (1/3841600000)*((13/5)^(1/3)*3^(2/3)*Sqrt[1170 + 45*13^(2/3)*15^(1/3) + 39*13^(1/3)*15^(2/3)]*Sqrt[-1170 + 39*13^(2/3)*15^(1/3) + 45*13^(1/3)*15^(2/3)]*Sqrt[1170 + 73*13^(2/3)*15^(1/3) + 67*13^(1/3)*15^(2/3)]) + (491193/384160000 - (7731*(13/5)^(2/3)*3^(1/3))/76832000 + (117*(13/5)^(1/3)*3^(2/3))/2744000 - (1/384160000)* (9*Sqrt[1170 + 45*13^(2/3)*15^(1/3) + 39*13^(1/3)*15^(2/3)]* Sqrt[-1170 + 39*13^(2/3)*15^(1/3) + 45*13^(1/3)*15^(2/3)]* Sqrt[1170 + 73*13^(2/3)*15^(1/3) + 67*13^(1/3)*15^(2/3)]) - (1/384160000)*((13/5)^(2/3)*3^(1/3)*Sqrt[1170 + 45*13^(2/3)*15^(1/3) + 39*13^(1/3)*15^(2/3)]*Sqrt[-1170 + 39*13^(2/3)*15^(1/3) + 45*13^(1/3)*15^(2/3)]*Sqrt[1170 + 73*13^(2/3)*15^(1/3) + 67*13^(1/3)*15^(2/3)]) + (1/192080000)*((13/5)^(1/3)*3^(2/3)* Sqrt[1170 + 45*13^(2/3)*15^(1/3) + 39*13^(1/3)*15^(2/3)]* Sqrt[-1170 + 39*13^(2/3)*15^(1/3) + 45*13^(1/3)*15^(2/3)]* Sqrt[1170 + 73*13^(2/3)*15^(1/3) + 67*13^(1/3)*15^(2/3)]))*x + (1087173/76832000 - (12771*(13/5)^(2/3)*3^(1/3))/30732800 + (5967*(13/5)^(1/3)*3^(2/3))/30732800 - (1/76832000)* (9*Sqrt[1170 + 45*13^(2/3)*15^(1/3) + 39*13^(1/3)*15^(2/3)]* Sqrt[-1170 + 39*13^(2/3)*15^(1/3) + 45*13^(1/3)*15^(2/3)]* Sqrt[1170 + 73*13^(2/3)*15^(1/3) + 67*13^(1/3)*15^(2/3)]) - (1/153664000)*(3*(13/5)^(2/3)*3^(1/3)*Sqrt[1170 + 45*13^(2/3)*15^(1/3) + 39*13^(1/3)*15^(2/3)]*Sqrt[-1170 + 39*13^(2/3)*15^(1/3) + 45*13^(1/3)*15^(2/3)]*Sqrt[1170 + 73*13^(2/3)*15^(1/3) + 67*13^(1/3)*15^(2/3)]) + (1/153664000)*(3*(13/5)^(1/3)*3^(2/3)* Sqrt[1170 + 45*13^(2/3)*15^(1/3) + 39*13^(1/3)*15^(2/3)]* Sqrt[-1170 + 39*13^(2/3)*15^(1/3) + 45*13^(1/3)*15^(2/3)]* Sqrt[1170 + 73*13^(2/3)*15^(1/3) + 67*13^(1/3)*15^(2/3)]))*x^2 + (23871/274400 - (99*(13/5)^(2/3)*3^(1/3))/109760 + (663*(13/5)^(1/3)*3^(2/3))/ 548800 - (1/2744000)*(Sqrt[1170 + 45*13^(2/3)*15^(1/3) + 39*13^(1/3)*15^(2/3)]*Sqrt[-1170 + 39*13^(2/3)*15^(1/3) + 45*13^(1/3)*15^(2/3)]*Sqrt[1170 + 73*13^(2/3)*15^(1/3) + 67*13^(1/3)*15^(2/3)]))*x^3 + (9*x^4)/28 + (9*x^5)/14 + x^6 prd1 /. (n_)?NumericQ :> ToRadicals[FullSimplify[n]] 172077/3841600000 - (4959*(13/5)^(2/3)*3^(1/3))/768320000 + (117*(13/5)^(1/3)*3^(2/3))/27440000 - (9*(234 + 221*(13/5)^(1/3)*3^(2/3) - 33*13^(2/3)*15^(1/3)))/384160000 + (117*(-165 + 17*13^(2/3)*15^(1/3) + 6*13^(1/3)*15^(2/3)))/3841600000 + (9*x)/8750 + (9*x^2)/700 + (3*x^3)/35 + (9*x^4)/28 + (9*x^5)/14 + x^6 Together[%] (3 + 90*x + 1125*x^2 + 7500*x^3 + 28125*x^4 + 56250*x^5 + 87500*x^6)/87500 Apart[%] 3/87500 + (9*x)/8750 + (9*x^2)/700 + (3*x^3)/35 + (9*x^4)/28 + (9*x^5)/14 + x^6 This is p1: p1 3/87500 + (9*x)/8750 + (9*x^2)/700 + (3*x^3)/35 + (9*x^4)/28 + (9*x^5)/14 + x^6 ------------------ It is ususlly better to try to reduce a difference to zero than to reduce one form to another tst1 = Collect[Expand[f232*f1 - p1], x] tst2 = tst1 /. (n_)?NumericQ :> ToRadicals[FullSimplify[n]] 8073/768320000 - (4959*(13/5)^(2/3)*3^(1/3))/768320000 + (117*(13/5)^(1/3)*3^(2/3))/27440000 - (9*(234 + 221*(13/5)^(1/3)*3^(2/3) - 33*13^(2/3)*15^(1/3)))/384160000 + (117*(-165 + 17*13^(2/3)*15^(1/3) + 6*13^(1/3)*15^(2/3)))/3841600000 Together[%] 0 -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay@haystack.demon.co.uk Voice: +44 (0)116 271 4198 > Can Mathematica factor the polynomial p1=x^6+9/14*x^5+9/28*x^4+3/35*x^3+9/700*x^2+9/8750*x+3/87500; without a priori knowledge of the Extension field? > ==== Carlos, p1=x^6+9/14*x^5+9/28*x^4+3/35*x^3+9/700*x^2+9/8750*x+3/87500; We can't expect this to be easy or even possible in terms of radicals (the general quintic is not solvable interms of radicals). But, using the function FactorR given in my posting, Re:factoring quartic over radicals, sent a few days (08/012/02) ago, we get p2=FactorR[p1,x] (x^2 - 2*x*Root[3 + 45*#1 + 225*#1^2 + 700*#1^3 & , 1] + Root[-3 + 225*#1^2 - 5625*#1^4 + 87500*#1^6 & , 2]^2)* (x^2 - 2*x*Root[1827 + 65340*#1 + 974700*#1^2 + 7824000*#1^3 + 36360000*#1^4 + 100800000*#1^5 + 156800000*#1^6 & , 2] + Root[9 - 1350*#1^2 + 84375*#1^4 - 3056250*#1^6 - 11250000*#1^8 - 984375000*#1^10 + 7656250000*#1^12 & , 3]^2)* (x^2 - 2*x*Root[1827 + 65340*#1 + 974700*#1^2 + 7824000*#1^3 + 36360000*#1^4 + 100800000*#1^5 + 156800000*#1^6 & , 1] + Root[9 - 1350*#1^2 + 84375*#1^4 - 3056250*#1^6 - 11250000*#1^8 - 984375000*#1^10 + 7656250000*#1^12 & , 4]^2) Try to express the factors in terms of radicals p3=ToRadicals/@p2 (3/140 + (1/140)*(13/5)^(2/3)*3^(1/3) - (1/140)*(13/5)^(1/3)*3^(2/3) - 2*(-(3/28) - (1/28)*(13/5)^(2/3)*3^(1/3) + (1/28)*(13/5)^(1/3)*3^(2/3))*x + x^2)* (x^2 - 2*x*Root[1827 + 65340*#1 + 974700*#1^2 + 7824000*#1^3 + 36360000*#1^4 + 100800000*#1^5 + 156800000*#1^6 & , 2] + Root[9 - 1350*#1 + 84375*#1^2 - 3056250*#1^3 - 11250000*#1^4 - 984375000*#1^5 + 7656250000*#1^6 & , 1])* (x^2 - 2*x*Root[1827 + 65340*#1 + 974700*#1^2 + 7824000*#1^3 + 36360000*#1^4 + 100800000*#1^5 + 156800000*#1^6 & , 1] + Root[9 - 1350*#1 + 84375*#1^2 - 3056250*#1^3 - 11250000*#1^4 - 984375000*#1^5 + 7656250000*#1^6 & , 2]) We succeeded with the first factor f1=p3[[1]] 3/140 + (1/140)*(13/5)^(2/3)*3^(1/3) - (1/140)*(13/5)^(1/3)*3^(2/3) - 2*(-(3/28) - (1/28)*(13/5)^(2/3)*3^(1/3) + (1/28)*(13/5)^(1/3)*3^(2/3))*x + x^2 The product of the other two factors, in a form avoiding root objects, is easily found: q=PolynomialQuotient[p1,f1,x] 3/3500 + ((13/5)^(1/3)*3^(2/3))/3500 + (3/175 + (1/175)*(13/5)^(1/3)*3^(2/3))*x + (9/70 - (1/140)*(13/5)^(2/3)*3^(1/3) + (1/28)*(13/5)^(1/3)*3^(2/3))*x^2 + (3/7 - (1/14)*(13/5)^(2/3)*3^(1/3) + (1/14)*(13/5)^(1/3)*3^(2/3))*x^3 + x^4 5*(3 + 10*x*(6 + 5*x*(9 + 10*x*(3 + 7*x))))) Try FactorR on this f23=FactorR[q,x] (x^2 + ((-(1/2))*Sqrt[117/1960 + (9/784)*(13/5)^(2/3)* 3^(1/3) + (39*(13/5)^(1/3)*3^(2/3))/3920] + Im[(1/2)*Sqrt[117/1960 + (9/784)*(13/5)^(2/3)* 3^(1/3) + (39*(13/5)^(1/3)*3^(2/3))/3920 + (-2250 + 25*13^(2/3)*15^(1/3) - 125*13^(1/3)* 15^(2/3))/8750 + (3*(7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)* 15^(2/3))^2)/1225000000 - (-((2*(300 + 20*13^(1/3)*15^(2/3)))/4375) + (1/17500)*((7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)*15^(2/3))*((2250 - 25*13^(2/3)* 15^(1/3) + 125*13^(1/3)*15^(2/3))/4375 - (7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)*15^(2/3))^2/306250000)))/ (4*Sqrt[-(117/1960) - (9/784)*(13/5)^(2/3)* 3^(1/3) - (39*(13/5)^(1/3)*3^(2/3))/ 3920])]])^2 - 2*x*((-7500 + 250*13^(2/3)*15^(1/3) - 250*13^(1/3)*15^(2/3))/70000 + Re[(1/2)*Sqrt[117/1960 + (9/784)*(13/5)^(2/3)* 3^(1/3) + (39*(13/5)^(1/3)*3^(2/3))/3920 + (-2250 + 25*13^(2/3)*15^(1/3) - 125*13^(1/3)* 15^(2/3))/8750 + (3*(7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)* 15^(2/3))^2)/1225000000 - (-((2*(300 + 20*13^(1/3)*15^(2/3)))/4375) + (1/17500)*((7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)*15^(2/3))*((2250 - 25*13^(2/3)* 15^(1/3) + 125*13^(1/3)*15^(2/3))/4375 - (7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)*15^(2/3))^2/306250000)))/ (4*Sqrt[-(117/1960) - (9/784)*(13/5)^(2/3)* 3^(1/3) - (39*(13/5)^(1/3)*3^(2/3))/ 3920])]]) + ((-7500 + 250*13^(2/3)*15^(1/3) - 250*13^(1/3)* 15^(2/3))/70000 + Re[(1/2)*Sqrt[117/1960 + (9/784)*(13/5)^(2/3)* 3^(1/3) + (39*(13/5)^(1/3)*3^(2/3))/3920 + (-2250 + 25*13^(2/3)*15^(1/3) - 125*13^(1/3)* 15^(2/3))/8750 + (3*(7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)* 15^(2/3))^2)/1225000000 - (-((2*(300 + 20*13^(1/3)*15^(2/3)))/4375) + (1/17500)*((7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)*15^(2/3))*((2250 - 25*13^(2/3)* 15^(1/3) + 125*13^(1/3)*15^(2/3))/4375 - (7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)*15^(2/3))^2/306250000)))/ (4*Sqrt[-(117/1960) - (9/784)*(13/5)^(2/3)* 3^(1/3) - (39*(13/5)^(1/3)*3^(2/3))/ 3920])]])^2)* (x^2 + ((1/2)*Sqrt[117/1960 + (9/784)*(13/5)^(2/3)* 3^(1/3) + (39*(13/5)^(1/3)*3^(2/3))/3920] + Im[(-(1/2))*Sqrt[117/1960 + (9/784)*(13/5)^(2/3)* 3^(1/3) + (39*(13/5)^(1/3)*3^(2/3))/3920 + (-2250 + 25*13^(2/3)*15^(1/3) - 125*13^(1/3)* 15^(2/3))/8750 + (3*(7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)* 15^(2/3))^2)/1225000000 + (-((2*(300 + 20*13^(1/3)*15^(2/3)))/4375) + (1/17500)*((7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)*15^(2/3))*((2250 - 25*13^(2/3)* 15^(1/3) + 125*13^(1/3)*15^(2/3))/4375 - (7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)*15^(2/3))^2/306250000)))/ (4*Sqrt[-(117/1960) - (9/784)*(13/5)^(2/3)* 3^(1/3) - (39*(13/5)^(1/3)*3^(2/3))/ 3920])]])^2 - 2*x*((-7500 + 250*13^(2/3)*15^(1/3) - 250*13^(1/3)*15^(2/3))/70000 + Re[(-(1/2))*Sqrt[117/1960 + (9/784)*(13/5)^(2/3)* 3^(1/3) + (39*(13/5)^(1/3)*3^(2/3))/3920 + (-2250 + 25*13^(2/3)*15^(1/3) - 125*13^(1/3)* 15^(2/3))/8750 + (3*(7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)* 15^(2/3))^2)/1225000000 + (-((2*(300 + 20*13^(1/3)*15^(2/3)))/4375) + (1/17500)*((7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)*15^(2/3))*((2250 - 25*13^(2/3)* 15^(1/3) + 125*13^(1/3)*15^(2/3))/4375 - (7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)*15^(2/3))^2/306250000)))/ (4*Sqrt[-(117/1960) - (9/784)*(13/5)^(2/3)* 3^(1/3) - (39*(13/5)^(1/3)*3^(2/3))/ 3920])]]) + ((-7500 + 250*13^(2/3)*15^(1/3) - 250*13^(1/3)* 15^(2/3))/70000 + Re[(-(1/2))*Sqrt[117/1960 + (9/784)*(13/5)^(2/3)* 3^(1/3) + (39*(13/5)^(1/3)*3^(2/3))/3920 + (-2250 + 25*13^(2/3)*15^(1/3) - 125*13^(1/3)* 15^(2/3))/8750 + (3*(7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)* 15^(2/3))^2)/1225000000 + (-((2*(300 + 20*13^(1/3)*15^(2/3)))/4375) + (1/17500)*((7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)*15^(2/3))*((2250 - 25*13^(2/3)* 15^(1/3) + 125*13^(1/3)*15^(2/3))/4375 - (7500 - 250*13^(2/3)*15^(1/3) + 250*13^(1/3)*15^(2/3))^2/306250000)))/ (4*Sqrt[-(117/1960) - (9/784)*(13/5)^(2/3)* 3^(1/3) - (39*(13/5)^(1/3)*3^(2/3))/ 3920])]])^2) We still use Re and Im -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay@haystack.demon.co.uk Voice: +44 (0)116 271 4198 > Can Mathematica factor the polynomial p1=x^6+9/14*x^5+9/28*x^4+3/35*x^3+9/700*x^2+9/8750*x+3/87500; without a priori knowledge of the Extension field? > ==== Dear MathGroup Experts, Is there a slick way to use Mathematica to plot the phase portrait of a pair of (straightforward) differential equations? I'd also like to see a plot of the associated vector/direction field. I found a good package called DynPac, but it's very cumbersome to use, and I was hoping there's another tool out there that my students would feel more comfortable using. Jason -- Jason Miller, Ph.D. Division of Mathematics and Computer Science Truman State University 100 East Normal St. Kirksville, MO 63501 http://vh216801.truman.edu 660.785.7430 ==== >I have two equations that I have solved for: x[n_] := 2331 + 8 n >y[n_] := -3108 - 11n I want to include only solutions which are non-negative, that is x >= 0 >and >y >= 0. In this example we can do 2331 + 8n > = 0 and solve for n, n >= -291.375 >and -3108 - 11 n >= 0 and solve for n, n <= -282.545 So we have -291.375 <= n <= -282.545. The integer solution set here is for n = >{-290, -289, -288, -287, -286, -285, -284, -283}. So in this case we have 8 non-negative solutions. Given that I can supply x[n] and y[n], how do I go about finding the set >n? > Needs[Algebra`InequalitySolve`]; x[n_] := 2331 + 8n; y[n_] := -3108 - 11n; rng = InequalitySolve[{x[n] >= 0, y[n] >= 0}, n]; soln = Range[Ceiling[rng[[1]]], rng[[-1]]] {-291, -290, -289, -288, -287, -286, -285, -284, -283} Length[soln] 9 Bob Hanlon ==== Factoring without specifying the extension does not really make sense. Of course Mathematica can easily factor yur polynomial into linear factors over the complex numbers (with the help of Solve), but I suspect you are really asking for is factoring over the reals. This is harder and needs more human input. But anyway, Mathematica can do this, or at least I have done it using Mathematica. In fact if you are satisfied with a numerical answer Mathematica can do alone and in seconds: In[1]:= Simplify[N[x^6 + (9/14)*x^5 + (9/28)*x^4 + (3/35)*x^3 + (9/700)*x^2 + (9/8750)*x + 3/87500]] Out[1]= 1.*(0.010974992601737198 + 0.20255610310498295*x + x^2)*(0.020476912388332692 + 0.2047691238833268*x + x^2)*(0.15256133957420948 + 0.23553191586883315*x + x^2) But I have in fact been foolish enough to compute the exact answer too. I do not propose to post it here for it's absolutely horrible (expressed in terms of Root objects) and quite useless. However if you really want to see it I can send it to you privately. Andrzej Kozlowski Yokohama, Japan http://www.mimuw.edu.pl/~akoz/ http://platon.c.u-tokyo.ac.jp/andrzej/ > To: mathgroup@smc.vnet.net Can Mathematica factor the polynomial p1=x^6+9/14*x^5+9/28*x^4+3/35*x^3+9/700*x^2+9/8750*x+3/87500; without a priori knowledge of the Extension field? > ==== Can Mathematica factor the polynomial p1=x^6+9/14*x^5+9/28*x^4+3/35*x^3+9/700*x^2+9/8750*x+3/87500; without a priori knowledge of the Extension field? The endless sci.math.symbolic thread strikes MathGroup! Not exactly possible with no prior knowledge. Factor must work with a given field, and the default is the rationals. You might direct it, say by using the discriminant of the polynomial (as pointed out by Peter Montgomery and Stephen Forrest on sci.math.symbolic. One may do this in Mathematica as: p1 = x^6+9/14*x^5+9/28*x^4+3/35*x^3+9/700*x^2+9/8750*x+3/87500; I cribbed code for Discriminant right from www.mathworld.com: Discriminant[p_?PolynomialQ,x_] := With[{n = Exponent[p,x]}, Cancel[((-1)^(n(n-1)/2)Resultant[p,D[p,x],x])/Coefficient[p,x,n]^(2n-1)]] In[3]:= InputForm[Factor[p1, Extension->Sqrt[Discriminant[p1,x]]]] Out[3]//InputForm= ((-15*I + Sqrt[195] - (225*I)*x + 15*Sqrt[195]*x - (1125*I)*x^2 + 75*Sqrt[195]*x^2 - (3500*I)*x^3)*(15*I + Sqrt[195] + (225*I)*x + 15*Sqrt[195]*x + (1125*I)*x^2 + 75*Sqrt[195]*x^2 + (3500*I)*x^3))/12250000 Daniel Lichtblau Wolfram Research ==== On second thoughts: any one who really wants to see what the exact answer is, can evaluate the following: sols = Select[{a, b} /. SolveAlways[ x^6 + (9/14)*x^5 + (9/28)*x^4 + (3/35)*x^3 + (9/700)*x^2 + (9/8750)*x + 3/87500 == (x^2 + a*x + b)*(x^4 + c*x^3 + d*x^2 + e*x + f), x], FreeQ[N[#1], _Complex] & ] Times @@ (Map[x^2 + {x, 1}.# &, sols]) N[%] (0.010974992601737203 + 0.20255610310498245*x + x^2)* (0.020476912388332696 + 0.20476912388332683*x + x^2)* (0.15256133957420942 + 0.23553164887424147*x + x^2) Andrzej Kozlowski Yokohama, Japan http://www.mimuw.edu.pl/~akoz/ http://platon.c.u-tokyo.ac.jp/andrzej/ > Factoring without specifying the extension does not really make sense. > Of course Mathematica can easily factor yur polynomial into linear > factors over the complex numbers (with the help of Solve), but I > suspect you are really asking for is factoring over the reals. This is > harder and needs more human input. But anyway, Mathematica can do > this, or at least I have done it using Mathematica. In fact if you are > satisfied with a numerical answer Mathematica can do alone and in > seconds: In[1]:= > Simplify[N[x^6 + (9/14)*x^5 + (9/28)*x^4 + (3/35)*x^3 + (9/700)*x^2 + > (9/8750)*x + 3/87500]] Out[1]= > 1.*(0.010974992601737198 + 0.20255610310498295*x + > x^2)*(0.020476912388332692 + > 0.2047691238833268*x + x^2)*(0.15256133957420948 + > 0.23553191586883315*x + x^2) But I have in fact been foolish enough to compute the exact answer > too. I do not propose to post it here for it's absolutely horrible > (expressed in terms of Root objects) and quite useless. However if you > really want to see it I can send it to you privately. Andrzej Kozlowski > Yokohama, Japan > http://www.mimuw.edu.pl/~akoz/ > http://platon.c.u-tokyo.ac.jp/andrzej/ >> To: mathgroup@smc.vnet.net >> Can Mathematica factor the polynomial >> p1=x^6+9/14*x^5+9/28*x^4+3/35*x^3+9/700*x^2+9/8750*x+3/87500; >> without a priori knowledge of the Extension field? > > Reply-To: ==== Those aren't equations; they're functions, and we don't solve functions, we solve equations. What equations do you want to solve? DrBob -----Original Message----- and -3108 - 11 n >= 0 and solve for n, n <= -282.545 So we have -291.375 <= n <= -282.545. The integer solution set here is for n = {-290, -289, -288, -287, -286, -285, -284, -283}. So in this case we have 8 non-negative solutions. Given that I can supply x[n] and y[n], how do I go about finding the set n? ==== Is there a logically fundamental difference between functional and procedural programming? What I mean to ask is, can we do exactly the same thing with purely functional approaches as we can with purely procedural approaches? Is this basically the recursive verses iterative distinction? Why would one chose one approach over the other? STH . ==== > Let's be realistic. If you want 60 digits of precision, too bad! -- in > the real world. There's nothing we can measure that closely. Drug > concentrations in clinical trials are generally measured within 15%, for > instance. Even machine precision is more than can be realistically > expected in any application I can think of. Even getting a satellite to > Jupiter probably involves more error in the final result than machine > precision. (If not, it's because we rely on ongoing corrections and > natural factors that put the satellite where it should be, such as > gravity drawing it toward each rendezvous -- not on that kind of > precision in propulsion or guidance.) So... unless all numerics in a problem have a theoretical origin, and > could be represented in Mathematica as Infinite precision expressions... > all this talk of higher-precision computation seems futile. ...Except in the very rare instance when one needs to do intermediate calculations with, e.g., 60 digits of precision in order to get only a few correct digits of the final answer. The length of this thread is surely proof of the need for a definitive reference on the topic. Has there been a Mathematica-centered numerical analysis book published since Skeel & Keiper? --- Selwyn Hollis ==== > In[24]:= > f = SetAccuracy[333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - > 121*b^4 - 2) + 5.5*b^8 + a/(2*b), 100]; > a=SetPrecision[77617.,100]; b = SetPrecision[33096.,100]; In[26]:= > {f, Precision[f]} Out[26]= > {-0.82739605994682136814116509547981629199903311578438481991 > 781484167246798617832`61.2597, 61} Congratulations! You just requested accuracy of > 100 for f and got 61 ( > to convince yourself add Accuracy[f] to In[26]). There is no request in that example for an accuracy of 100 in the result. The only request is for an accuracy of 100 in the input. > In[26]:= f = SetAccuracy[333.75*b^6 + > a^2*(11*a^2*b^2 - b^6 - > 121*b^4 - 2) + 5.5*b^8 + a/(2*b), 100]; In[27]:= Accuracy[f] > Out[27]= 100. Now we assign values to some indeterminants in f. In[28]:= a = SetPrecision[77617.,100]; b = SetPrecision[33096.,100]; In[29]:= {f, Precision[f], Accuracy[f]} > Out[29]= > {-0.8273960599468213681411650954798162919990331157843848199178148, > 61.2599, 61.3422} The precision and accuracy has dropped. This is all > according to > standard numerical analysis regarding cancellation > error. You'll find it in any textbook on the topic. > Assume that I want accuracy and precision of 100 for > f. You advice me to make experiments to find out, what > should be the initial precision and accuracy of a and > b to reach the requested accuracy and precision for f. > Notice, that you cannot just repeat I[26], we saw > already what happens. I have to re-type I[24], I[25], > I[26], I[27], I[28], and I[29] as many times as needed > to get f with accuracy and precision 100. Dan, you simply advocate to do MANUAL WORK that should > be done by machine. You do not have to do any of this manually. The machine (Mathematica) will do all of this, usually using built-in functions. The N function, for example, will automatically adjust the working precision to give you a precision that you request, provided that doing so doesn't involve making up arbitrary digits. The example above starts out with machine numbers (333.75, 5.5, etc.), uses SetPrecision and SetAccuracy to make up arbitrary digits to pad those numbers out to some specified number of digits, and then does some simple arithmetic. If the goal is to get some specified number of digits in the result, and it is ok to make up arbitrary digits like this to achieve that goal, then the only manual work required to achive that goal is to apply SetAccuracy or SetPrecision to the result, to tell the computer that that is what you want. > Let's suppose that in the above example I just want 60 > digits not 61. Precisely, I want 60 digits and nothing > or zeros afterwards. Let's see if I could use > SetAccuracy. In[30]:= SetAccuracy[%, 60] Out[30]= -0.82739605994682136814116509547981629199903311578438481991781 In[31]:= % // FullForm Out[30]//FullForm= > -0.827396059946821368141165095479816291999033115784384819917814841672467988` > 59.9177 Oops, it did not work (as expected). If you could explain what you were expecting I am sure there are many contributors to this group who could explain to you why it did not do that. > Let's highlight > with mouse the expression in Out[30] and copy to a new > cell. Oops, we got > -0.827396059946821368141165095479816291999033115784384819917814841672467988` > 59.9177 again. Let's change Out[30] to a text cell and then copy. In[31]:= -0.82739605994682136814116509547981629199903311578438481991781 Out[31]= -0.82739605994682136814116509547981629199903311578438481991781 Success? Not so fast. If you could describe what you were trying to achieve with all of that copying and pasting and such I am again sure that there are many contributors to this group who could describe how to do it. It is very unlikely that the process will involve any copying and pasting or detours through text cells. > In[32]:= > % // FullForm Out[32]//FullForm= > -0.8273960599468213681411650954798162919990331157843848199178099999999999986 3 5 > 08`59.2041 Dan, is there any simple way to get what I want? Probably the answer is yes, but you will have to describe more clearly what you want. > As I repeated already number of times, at this stage > of the development of computer technology, software > should do it for me (!). If what you want to do is get a certain number of digits in the result, and it is ok to make up arbitrary digits as in the examples above, then you can do that by simply applying SetPrecision or SetAccuracy to the result. If you want the computer to automatically adjust the working precision to give a certain precision in the result, you can do that using N. If you want something else, and you can describe what that is, then probably someone can describe how to get Mathematica to do that for you. Dave Withoff Wolfram Research ==== Putting in Print[Some header text, TableForm[ Table[----------]]] puts the header text and the Table in the same cell, but with a largish gap (3 lines?) between the text and the Table's header lines. Hardly a serious problem, pretty trivial in fact, but a bit ugly and seems as if it really shouldn't work this way. Any way to get rid of this? siegman@stanford.edu Reply-To: ==== The first problem is that ArcSin[x/10] isn't a left-inverse of 10 Sin[x] in the region of the root. That's why your code converges on another root. A better inverse for your purpose is the gi below: f = #(# + 3) &; g = 10Sin@# &; gi = Pi - ArcSin[#/10] &; Using that inverse, there's still a problem, though. Your approximation of the root with r = gi@f@r fails because the derivative of gi@f at the root is gi'[f@rr]f'[rr] -2.02342 That's greater than one in magnitude, so distance from the root is magnified rather than diminished. Instead, try r = fi@g@r Where fi[y_] := Evaluate[x /. Last@Solve[f@x == y, x]] This should work, since fi'[g@rr]g'[rr] -0.49421355442685166 Sure enough, it does work: Values = NestList[fi[g[#1]] & , 2., 12] {2., 1.8679332339369226, 1.9368280058437026, 1.9040490698559083, 1.9205018812387413, 1.9124384783620436, 1.916439313493727, 1.9144659935649395, 1.9154421880638148, 1.914959973607326, 1.915198347545482, 1.9150805538598723, 1.9151387724997768} (f[#1] - g[#1] & ) /@ values {0.9070257317431825, -0.4688124734947827, 0.2242366717647286, -0.11228304957089463, 0.05509675095190758, -0.02732121417963107, 0.0134795615740817, -0.006667318794729482, 0.003293718665990042, -0.0016281317372435211, 0.0008045637255396088, -0.00039764607942949226, 0.0001965172491082967} Absolute error is cut in half at each iteration, as expected with a derivative for fi@g near -1/2. The negative sign causes the error to alternate in sign. We can also look at the log-absolute value of the error as follows: (Log[Abs[f[#1] - g[#1]]] & ) /@ values {-0.09758445910053692, -0.7575524337892089, -1.4950532145264737, -2.18673236744676, -2.8986645309519163, -3.6000918023816326, -4.306580698204814, -5.010537479670821, -5.715738058891756, -6.420322094992857, -7.125210383308283, -7.829948195959684, -8.534760348785936} Rest[%] - Drop[%, -1] {-0.659967974688672, -0.7375007807372648, -0.6916791529202861, -0.7119321635051565, -0.7014272714297163, -0.7064888958231816, -0.7039567814660064, -0.7052005792209357, -0.7045840361011004, -0.7048882883154262, -0.704737812651401, -0.7048121528262516} The difference in logarithm of the absolute error approaches Log@Abs[fi'[g@rr]g'[rr]] -0.7047875587967565 If you want to use this for teaching, try to use as little code as possible -- as I have above -- and always try to avoid Do loops in Mathematica on principle. DrBob -----Original Message----- kenf Below is the code Clear[f, g, gi, lim, r, rr, fr, gir, a, b, c, d, conv]; Plot[{x * ((x + 3)), 10*Sin[x]}, {x, 0.01, 2.4}, PlotStyle -> {{RGBColor[1, 0, 0], Thickness[ .006]}, {RGBColor[0, 0, 1], Thickness[ .006]}} ]; rr = FindRoot[x * ((x + 3)) == 10*Sin[x], {x, 2, 0.01, 2.4}]; f[a_] := a * ((a + 3)) /; a > 0; g[b_] := 10. * Sin[b] /; b > 0; gi[c_] := ArcSin[0.1*c] /; c > 0; Print[Actual root is , rr]; lim = 10; r = 2.0; conv = 10^-4; For[i = 1, i < lim, i++, { fr = f[r]; gir = gi[fr]; d = Abs[N[gir] - r]; i If[d < conv, Break[]]; r = gir; Print[The value of x = , r, found after , i, iterations,, with a tolerence , d, n] } ] Print[The value of x = , r, found after , i, iterations,, with a tolerence , d, n] Every man, woman and responsible child has an unalienable individual, civil, Constitutional and human right to obtain, own, and carry, openly or concealed, any weapon -- rifle, shotgun, handgun, machine gun, anything -- any time, any place, without asking anyone's permission. L. Neil Smith ==== f[t_] = {t + Sqrt[3] Sin[t], 2 Cos[t], t Sqrt[3] - Sin[t]} So It's basically a vector whose coordinates are determined based on the values you pass in. Then I took the derivative by just typing f', which outputs {1 + Sqrt[3] Cos[#1], -2 Sin[#1], Sqrt[3] - Cos[#1]}& What I'd like to do is have Mathematica calculate the norm of this as it would any vector, so that I can play with the norm function. As it turns out, the norm in this case is identical to Sqrt[8], so it would be nice if Mathematica could figure that out. Is it possible to do this? ==== [My previous reply seems to have gone astray - at least it has not come back to me. Here is a slightly edited repeat] You have computed f[t_] = {t + Sqrt[3] Sin[t], 2 Cos[t], t Sqrt[3] - Sin[t]}; and then found f' {1 + Sqrt[3]*Cos[#1], -2*Sin[#1], Sqrt[3] - Cos[#1]} & To get the function for the norm of the derivative we can use norm = Evaluate/@(Simplify/@(Sqrt[#.#]&/@(f'))) 2*Sqrt[2] & We map the usual functions for calclulating and simplifying the norm inside Function[.] (which is the full form of (.)& and then map the function Evaluation to make the result evaluate -- this is needed since Function has the attribute HoldAll. Please note that the parentheses are essential. -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay@haystack.demon.co.uk Voice: +44 (0)116 271 4198 f[t_] = {t + Sqrt[3] Sin[t], 2 Cos[t], t Sqrt[3] - Sin[t]} So It's basically a vector whose coordinates are determined based on the > values you pass in. Then I took the derivative by just typing f', which outputs {1 + Sqrt[3] Cos[#1], -2 Sin[#1], Sqrt[3] - Cos[#1]}& > What I'd like to do is have Mathematica calculate the norm of this as it > would any vector, so that I can play with the norm function. As it turns > out, the norm in this case is identical to Sqrt[8], so it would be nice if > Mathematica could figure that out. Is it possible to do this? > ==== > f[t_] = {t + Sqrt[3] Sin[t], 2 Cos[t], t Sqrt[3] - Sin[t]} So It's basically a vector whose coordinates are determined based on the > values you pass in. Then I took the derivative by just typing f', which outputs {1 + Sqrt[3] Cos[#1], -2 Sin[#1], Sqrt[3] - Cos[#1]}& > What I'd like to do is have Mathematica calculate the norm of this as it > would any vector, so that I can play with the norm function. I don't think whether the definition is prompt (=) or delayed (:=) is as important as supplied the argument [t] in In := Simplify[Sqrt[f'[t] . f'[t]]] Out = 2 Sqrt[2] Or you can construct a new pure function with In := Evaluate[Simplify[Sqrt[f'[#] . f'[#]]]]& Out = 2 Sqrt[2] & Of course, there is not much left to play with :) Tom Burton ==== >I want to apply a function to every k-th element of a long list and >add the result to the k+1 element. [Actually k = 3 and I just want to multiply myList[[k]] by a >constant (independent of k) and add the result to myList[[k+1]] for >every value of k that's divisible by 3.] If I understand what you are trying to do correctly, the following should work (f[#[[1]]+#[[2]])&/@Partition[Drop[list,k-1],2,k] Here f is the function you want applied to the k-th element ==== MathGroup/Mathematica Newsgroup users - in the past 24 hours or so and have not seen your post, please resend/repost it. Disk drive problems caused some of the posts in the queue to be lost. This has been fixed. Sorry for the delay. Steve Christensen Moderator ==== >Please allow me to summarize what I've learned in the recent discussion, and >retract my claim that Accuracy, Precision, and SetAccuracy are useless. >Numbers come in three varieties The technical term is flavors. >- machine precision, Infinite precision, >and bignum or bigfloat. Bignums and bigfloats (synonymous?) Actually bignums can also refer to integers too large to represent as machine integers. But I tend to use bignum when I really mean bigfloat, and I suspect this sloppy practice may be common. >aren't called that in the Help Browser, but they're the result of using >N[expr,k] or >SetAccuracy[expr,k] where k is bigger than machine precision. >If k <= machine precision, the result is a machine precision number, even >if you know the expression isn't that precise. >If, when you use N or SetAccuracy as described above, the expression >contains undefined symbols, you get an expression with all its numerics >replaced by bignums of the indicated precision. When the symbols are >defined later, if ANY of them are machine precision, the expression is >computed with machine arithmetic - with the side-effect that coefficients >that originally were Infinite-precision are now only machine precision. >That is, x^2 might have become x^2.0000000000000000000000000000000000 >but later became x^2., for instance. I think this is correct in cases where all symbolic stuff gets replaced by numeric values. In general there is a sort of coercion to lowest precision, with the caveat that machine floats pollute everything. >If all the symbols have been set to bignum or Infinite precision values, >the computation will be done taking precision into account, and the result >has a Precision or Accuracy that makes sense. In all other cases, >Precision returns Infinity for entirely Infinite-precision expressions >and 16 for everything else. I'm not sure I understand this last sentence. My interpretation: Computations that are exact will have infinite precision. Computations in machine arithmetic will claim a precision of 16. If that is what you are claiming, then yes, that's what Mathematica is doing (but see my last remarks). >When one of the experts says significance arithmetic that's what they >mean - using SetAccuracy or N to give things more than 16 digits, leaving >no machine precision numbers anywhere in the expression, and using Accuracy >or Precision, which ARE meaningful in that case, to judge the result. >(It's meaningful if all your inputs really have more than 16 digits of >precision, that is.) I'm as guilty as anyone else in this thread, perhaps more so, of being too loose with the technical jargon. Also I am not certain what version 4 makes of SetAccuracy/SetPrecision in terms of significance arithmetic. In the development kernel they will force everything in sight to have the indicated precision, whether justified or not. This may well introduce error even with exact input, e.g. in cases where intermediate computations would require higher precision in order to get an end result with the requested precision or accuracy. N[], on the other hand, will handle that and, except in pathological circumstances, will give a result with the correct precision. As another minor point, arbitrary precision numbers are simply (tautologically?) numbers that may have arbitrarily large precision (subject to software limitations). Significance arithmetic refers to a particular model of manipulating such numbers with a mechanism for tracking precision. There are other models, in particular fixed precision arithmetic; that we use the former, by default, is an occasional source of sturm und drang in this news group. I'm sure the distinction between arbitrary precision numbers and significance arithmetic has at least minor relevance to this thread, and I imagine I've helped to confuse the issue in some places by using the terms almost interchangeably. >You can't use significance arithmetic to determine how much precision a >result has if your inputs have 16 or 15 or 2 digits of precision. One can, if the numbers are really bignums (of low precision, naturally). What one cannot do at present is create such low precision numbers via N[]. >In the example we've been looking at, you can give the inputs MORE accuracy >than you really believe they have, and still get back 0 digits from >Precision at the end, so there are clearly no trustworthy digits when >you use the original inputs either. If an expression is on the razor's >edge, and has lost only a few digits of precision, that wouldn't work >so well. >Oddly enough, significance arithmetic in the Browser doesn't take you >to any of that. Instead, it takes you to Interval arithmetic, a more >sophisticated method, which may give a more accurate gauge of how much >precision you really have, and WILL deal with machine precision numbers >and numbers with even less precision. It does a very good job on the >example. However, it isn't very suitable for Complex numbers, matrices, >etc. NSolve and NIntegrate probably can't handle it, either. I have filed a suggestion in-house that the documentation on significance arithmetic take one to the section on arbitrary precision numbers (3.1.5), as that would be more appropriate. Note that that section, while primarily concerned with the significance arithmetic model, also briefly mentions fixed precision bignum arithmetic. >Daniel Lichtblau promises that all this will be clearer in the next release. I'm not sure I'd go that far. What I will claim is that the distinction between machine numbers and bignums will be more transparent to users. At present if one does, say, N[number,precision] then one will get a machine number if prec<=$MachinePrecision. We have made a change so that this will no longer be the case. I am not prepared to go into details at this time (sorry). Perhaps more important for everyday use, and certainly more pertinent for this thread, Precision[] will distinguish between bignums of 16 digits precision and machine numbers. Again, I have to defer on details. At the very least I think the pitfall of believing a claim of 16 digits precision for machine numbers will be removed. Daniel Lichtblau Wolfram Research ==== Bobby, One point: >.... bigfloats ... [are] the result of using N[expr,k] or SetAccuracy[expr,k] where k is bigger than machine precision. If k <= > machine > precision, the result is a machine precision number. We get bigfloats with k<=machine precision with SetAccuracy and SetPrecision but not with N: Example a=SetPrecision[2.3,5] 2.3000 Precision[a] 5. Precision[a^2000] 1.69897 Also, of course, when more than machine precision significant digits are given a=1.01234567891234500; Precision[a] 17.301 Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay@haystack.demon.co.uk Voice: +44 (0)116 271 4198 Reply-To: ==== Please allow me to summarize what I've learned in the recent discussion, and retract my claim that Accuracy, Precision, and SetAccuracy are useless. Numbers come in three varieties - machine precision, Infinite precision, and bignum or bigfloat. Bignums and bigfloats (synonymous?) aren't called that in the Help Browser, but they're the result of using N[expr,k] or SetAccuracy[expr,k] where k is bigger than machine precision. If k <= machine precision, the result is a machine precision number, even if you know the expression isn't that precise. If, when you use N or SetAccuracy as described above, the expression contains undefined symbols, you get an expression with all its numerics replaced by bignums of the indicated precision. When the symbols are defined later, if ANY of them are machine precision, the expression is computed with machine arithmetic - with the side-effect that coefficients that originally were Infinite-precision are now only machine precision. That is, x^2 might have become x^2.0000000000000000000000000000000000 but later became x^2., for instance. If all the symbols have been set to bignum or Infinite precision values, the computation will be done taking precision into account, and the result has a Precision or Accuracy that makes sense. In all other cases, Precision returns Infinity for entirely Infinite-precision expressions and 16 for everything else. When one of the experts says significance arithmetic that's what they mean - using SetAccuracy or N to give things more than 16 digits, leaving no machine precision numbers anywhere in the expression, and using Accuracy or Precision, which ARE meaningful in that case, to judge the result. (It's meaningful if all your inputs really have more than 16 digits of precision, that is.) You can't use significance arithmetic to determine how much precision a result has if your inputs have 16 or 15 or 2 digits of precision. In the example we've been looking at, you can give the inputs MORE accuracy than you really believe they have, and still get back 0 digits from Precision at the end, so there are clearly no trustworthy digits when you use the original inputs either. If an expression is on the razor's edge, and has lost only a few digits of precision, that wouldn't work so well. Oddly enough, significance arithmetic in the Browser doesn't take you to any of that. Instead, it takes you to Interval arithmetic, a more sophisticated method, which may give a more accurate gauge of how much precision you really have, and WILL deal with machine precision numbers and numbers with even less precision. It does a very good job on the example. However, it isn't very suitable for Complex numbers, matrices, etc. NSolve and NIntegrate probably can't handle it, either. Daniel Lichtblau promises that all this will be clearer in the next release. DrBob ==== Deleting the browserindex.nb does not correct the online help e.g the online mathematica book is not fully accessable etc.Why are some of the *.nb files corrupted in the first place?It seems to occur only on win2k. Try deleting ... >Mathematica4.2DocumentationEnglishMainBookBrowserIndex.nb ==== > Is there a logically fundamental difference between functional and > procedural programming? What I mean to ask is, can we do exactly the > same > thing with purely functional approaches as we can with purely > procedural > approaches? > There is a fundamental difference, a strictly functional language does statement alone then it is obvious that there are some things that can be done in a procedural (nowadays people prefer to call them imperative) language that cannot be done in a functional language. On the other hand in imperative languages without pointers to procedures on cannot define a function which takes another function as an argument which functional languages allow. What you are probably really interested in is whether there exist a class of programs in the sense of a transformation on some input set into an output set which can be expressed in only one paradigm. Because nearly all major programming languages (SQL being the glaring exception) can implement a Turing Machine any algorithm which can be performed by a Turing Machine can be performed by any programming language, imperative or functional. So in principle functional and imperative languages are equivalently powerful although in practice it is easier to express some concepts in one paradigm or another. > Is this basically the recursive verses iterative distinction? > No. > Why would one chose one approach over the other? > In software engineering circles there is some evidence that the functional programming paradigm allows for efficient implementation of large programming projects, see http://www.math.chalmers.se/~rjmh/Papers/whyfp.html. Also it is possible to prove a strictly functional program correct because it cannot have side effects (no destructive updates of global states) which caused some excitement in academia but has not been a big selling point in industry. That said, by now you must realize that Mathematica is not a strictly functional language although you can use it as such. Actually Mathematica has a lot in common with LISP which is sometimes lumped together with the functional languages since it admits that programming style but both Mathematica and LISP are not pure (the desirability of purity being left to personal preference). One last advantage of functional programs though (which is related to the fact that they admit correctness proofs) is that it is relatively easy for a compiler/interpreter to optimize a functional program. This is why implicit iteration (Map in Mathematica) tends to outperform imperative style loops (Do, For, While etc.), which have to do a destructive update of the loop counter variable and therefore cannot reorganize a loop since statements in the loop may also destructively update the loop counter. C et al. get around this problem by being closely matched to the hardware. Ssezi ==== Hey folks, I have been working on a problem that seems to not lend itself to a solution. The following Mathematica code begins with the expression that I am trying to solve. For the curious, it's a degree 2 zonal and sectoral harmonics problem where I am trying to calculate and plot the geoid of earth as compared to an ellipse to see how well the geoid is approximated as an ellipse. In any case, we have the following relation ship, U =GM/r( 1 - (ae/r)^2 ( J2 (3/2 Sin[t]^2 - 1/2) - 3 Cos[t]^2 (C22 Cos[2 x] + S22 Sin[2 x])); Ur =1/2 we^2 (r Cos[t])^2; W[x_] =U + Ur; In trying to reorder W to become a function r wrt t, that is r[t_], I tried, among others, Solve[W[t], r] which returned ({{r -> Root[ae^2 GM J2 + 6 ae^2 C22 GM Cos[ [t]] ^2 - 3 ae^2 GM J2 Sin[ [t]] ^2 + 2 GM #1 ^2 - 2 W0 #1 ^3 + we^2 Cos[ [t]] ^2 #1 ^5 &, 1]}, {r -> Root[ae^2 GM J2 + 6 ae^2 C22 GM Cos[ [t]] ^2 - 3 ae^2 GM J2 Sin[ [t]] ^2 + 2 GM #1 ^2 - 2 W0 #1 ^3 + we^2 Cos[ [t]] ^2 #1 ^5 &, 2]}, {r -> Root[ae^2 GM J2 + 6 ae^2 C22 GM Cos[ [t]] ^2 - 3 ae^2 GM J2 Sin[ [t]] ^2 + 2 GM #1 ^2 - 2 W0 #1 ^3 + we^2 Cos[ [t]] ^2 #1 ^5 &, 3]}, {r -> Root[ae^2 GM J2 + 6 ae^2 C22 GM Cos[ [t]] ^2 - 3 ae^2 GM J2 Sin[ [t]] ^2 + 2 GM #1 ^2 - 2 W0 #1 ^3 + we^2 Cos[ [t]] ^2 #1 ^5 &, 4]}, {r -> Root[ae^2 GM J2 + 6 ae^2 C22 GM Cos[ [t]] ^2 - 3 ae^2 GM J2 Sin[ [t]] ^2 + 2 GM #1 ^2 - 2 W0 #1 ^3 + we^2 Cos[ [t]] ^2 #1 ^5 &, 5]}} ) which wasn't too much help, though it is a list of 5 Root functions. But in order to plot, I need a function r(t) so I can plot r wrt t...right? ParametricPlot[r[t], {t, 0, Pi}] So, I guess my questions are as follows: 1. How do I get Solve[ ] to output numbers, as //N and NSolve did nothing to Solve[r[t], ...] to get any numbers instead of just r -> Root[...]? 2. Is there a way to use ParametricPlot[ W[t], {t, 0.0, Pi}] instead of using r[t] and negating the whole issue of solving W[t] for r[t]? I have read that Solve only works for up to 4th order polynomials. I have been unable to find anything that works on my problem, having tried SolveAlways[ ] and other, and combination of others. Any help is welcome. I'll be glad to forward my Notebook if someone jdhouse4@mac.com Ph.D. Graduate Student Aerospace Engineering University of Texas at Austin 512-784-3205 ==== Try can use ImplicitPlot from the Graphics package. Janusz. > Hey folks, I have been working on a problem that seems to not lend itself to a > solution. The following Mathematica code begins with the expression > that I am trying to solve. For the curious, it's a degree 2 zonal and > sectoral harmonics problem where I am trying to calculate and plot the > geoid of earth as compared to an ellipse to see how well the geoid is > approximated as an ellipse. In any case, we have the following relation > ship, U =GM/r( 1 - (ae/r)^2 ( J2 (3/2 Sin[t]^2 - 1/2) - 3 Cos[t]^2 (C22 > Cos[2 x] + S22 Sin[2 x])); > Ur =1/2 we^2 (r Cos[t])^2; > W[x_] =U + Ur; In trying to reorder W to become a function r wrt t, that is r[t_], I > tried, among others, Solve[W[t], r] which returned ({{r - Root[ae^2 GM J2 + 6 ae^2 C22 GM Cos[ [t]] ^2 - > 3 ae^2 GM J2 Sin[ [t]] ^2 + 2 GM #1 ^2 - > 2 W0 #1 ^3 + we^2 Cos[ [t]] ^2 #1 ^5 &, > 1]}, {r - Root[ae^2 GM J2 + 6 ae^2 C22 GM Cos[ [t]] ^2 - > 3 ae^2 GM J2 Sin[ [t]] ^2 + 2 GM #1 ^2 - > 2 W0 #1 ^3 + we^2 Cos[ [t]] ^2 #1 ^5 &, > 2]}, {r - Root[ae^2 GM J2 + 6 ae^2 C22 GM Cos[ [t]] ^2 - > 3 ae^2 GM J2 Sin[ [t]] ^2 + 2 GM #1 ^2 - > 2 W0 #1 ^3 + we^2 Cos[ [t]] ^2 #1 ^5 &, > 3]}, {r - Root[ae^2 GM J2 + 6 ae^2 C22 GM Cos[ [t]] ^2 - > 3 ae^2 GM J2 Sin[ [t]] ^2 + 2 GM #1 ^2 - > 2 W0 #1 ^3 + we^2 Cos[ [t]] ^2 #1 ^5 &, > 4]}, {r - Root[ae^2 GM J2 + 6 ae^2 C22 GM Cos[ [t]] ^2 - > 3 ae^2 GM J2 Sin[ [t]] ^2 + 2 GM #1 ^2 - > 2 W0 #1 ^3 + we^2 Cos[ [t]] ^2 #1 ^5 &, 5]}} ) which wasn't too much help, though it is a list of 5 Root functions. > But in order to plot, I need a function r(t) so I can plot r wrt > t...right? ParametricPlot[r[t], {t, 0, Pi}] So, I guess my questions are as follows: > 1. How do I get Solve[ ] to output numbers, as //N and NSolve did > nothing to Solve[r[t], ...] to get any numbers instead of just r - Root[...]? 2. Is there a way to use ParametricPlot[ W[t], {t, 0.0, Pi}] instead of > using r[t] and negating the whole issue of solving W[t] for r[t]? I have read that Solve only works for up to 4th order polynomials. I > have been unable to find anything that works on my problem, having > tried SolveAlways[ ] and other, and combination of others. Any help is welcome. I'll be glad to forward my Notebook if someone jdhouse4@mac.com Ph.D. Graduate Student > Aerospace Engineering > University of Texas at Austin 512-784-3205 ==== Dear friends, I have build a table with this pattern: Flatten[{{{d, X, Y, Z}}, Table[{t, x[t], y[t], z[t]}, {t, 1, 10}], Table[{t, x[t], y[t], z[t]}, {t, 20, 100, 10}], Table[{t, x[t], y[t], z[t]}, {t, 200, 1000, 100}]}, 1] // TableForm I would like obtain the same Output in a more elegant way. In other word, how Can I avoid write Table[{t, x[t], y[t], z[t]} a few times. Thans Guillermo Sanchez --------------------------------------------- This message was sent using Endymion MailMan. Reply-To: kuska@informatik.uni-leipzig.de ==== Flatten[{{{d, X, Y, Z}}, Flatten[Table[{t, x[t], y[t], z[t]}, Evaluate[{t, Sequence @@ #}]] & /@ {{1, 10}, {20, 100, 10}, {200, 1000, 100}}, 1]}, 1] // TableForm ??? Jens Dear friends, > I have build a table with this pattern: Flatten[{{{d, X, Y, Z}}, > Table[{t, x[t], y[t], z[t]}, {t, 1, 10}], > Table[{t, x[t], y[t], z[t]}, {t, 20, 100, 10}], > Table[{t, x[t], y[t], z[t]}, {t, 200, 1000, 100}]}, 1] // TableForm I would like obtain the same Output in a more elegant way. In other word, how > Can I avoid write Table[{t, x[t], y[t], z[t]} a few times. Thans Guillermo > Sanchez --------------------------------------------- > This message was sent using Endymion MailMan. ==== > I would like obtain the same Output in a more elegant way. In other word, how > Can I avoid write Table[{t, x[t], y[t], z[t]} a few times. Table[{t=10^k; t, x[t], y[t], z[t]}, {k, 0, 3, 0.2}] or something similar. ==== Guillermo, Instead of Flatten[{{{d,X,Y,Z}},Table[{t,x[t],y[t],z[t]},{t,1,10}], Table[{t,x[t],y[t],z[t]},{t,20,100,10}], Table[{t,x[t],y[t],z[t]},{t,200,1000,100}]},1]; we can use Flatten[{{{{d,X,Y,Z}}}, Table[{t,x[t],y[t],z[t]},#]&/@{{t,1,10},{t,20,100,10},{t,200,1000, 100}}},2]; Test %===%% True -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay@haystack.demon.co.uk Voice: +44 (0)116 271 4198 > Dear friends, > I have build a table with this pattern: Flatten[{{{d, X, Y, Z}}, > Table[{t, x[t], y[t], z[t]}, {t, 1, 10}], > Table[{t, x[t], y[t], z[t]}, {t, 20, 100, 10}], > Table[{t, x[t], y[t], z[t]}, {t, 200, 1000, 100}]}, 1] // TableForm I would like obtain the same Output in a more elegant way. In other word, how > Can I avoid write Table[{t, x[t], y[t], z[t]} a few times. Thans Guillermo > Sanchez --------------------------------------------- > This message was sent using Endymion MailMan. > ==== I have two seperate list questions that I was hoping to get help with. Question 1. I have a variable length list similar to that generated by FactorInteger, that is {number, exponent} pairs. An example follows. lista = {{2,3},{3,1},{5,1}} ... this is the number 2^3 * 3^1 * 5^1 I want to generate a list of all the products of numbers from this list. I can tell that I get a total (3+1)*(1+1)*(1+1) = 4*2*2 = 16, products and I want a list showing all of those. These would be: 2^3 can generate {2^0, 2^1, 2^2, 2^3} = {1, 2, 4 ,8} 3^1 can generate {3^0, 3^1} = {3} ... we dont care about the duplicate 1 5^1 can generate {5^0, 5^1} = {5} ... we dont care about the duplicate 1 Hence the 4*2*2 = 16 (the product of one more of the exponents) above. Next we should get 16 products (from these lists), namely (I left them as products below to show what I am after): {1, 2, 4, 8, 1*3, 2*3, 4*3, 8*3, 1*5, 2 * 5, 4* 5, 8* 5, 1*3*5, 2*3*5, 4*3*5, 8*3*5} If the list were lista = {{2,4}, {3,2}, {5, 3},{7^5}}, we would have (4+1)(2+1)(3+1)(5+1) = 360 products, for example and the return values should be a single list showing all of those. Question 2. I have two lists and want to generate two new lists from them. These two lists are {number, exponent} pairs. In the first list, I want the minimum intersection of {number, exponent} pairs. In the second list, I want the maximum union of {number, exponent} pairs. Let me show an example: Input: list1 = {{2, 3}, {3, 4}, {5, 6}, {7, 2}, {17, 5}} list2 = {{2, 5}, {3, 2}, {5, 1}, {7, 3}} Output: minint = {{2, 3}, {3, 2}, {5, 1}, {7, 2}} Note: In this example we only kept those pairs where the intersection of the number exists and also keep the min power of those. maxint = {{2, 5}, {3, 4}, {5, 6}, {7, 3}, {17, 5}} Note: In this example we kept the union of lists and also keep the max power of each. Flip ==== When using Parametric Plot to make closed Lissajous curves, it's best not to take the parameter too far. for instance the code ParametricPlot[{Sin[t], Sin[2t]}, {t, 0, 10Pi}] looks much the same if you go to 20Pi or 30 Pi. But if you go too far, say, 1000Pi, the curve will stray so much that it can appear, deceptively, to be an open Lissajous curve, filling the rectangle. I don't understand how it's doing this. I assume the reason has to do with machine precision, but can anyone tell me in a little detail what's happening here? thanks, -- _______________ Steve Story Polymer Research Group 411B Cox North Carolina State University 1-919-515-8147 _______________ ==== I'm attempting to identify the essential aspects of Mathematica. I believe the place to start is with the 'functional operations'. I'm seeking the 'basis' of Mathematica. Kind of the orthonormal subset of functionality which can be used to derive all the other. I'm also trying to be pragmatic. I'm not trying to reinvent Mathematica, I'm just trying to understand the invention that already exists. If anyone is interested in seeing what I've gathered so far, I have a notebook in both HTML, and Mathematica (4.2, if that matters) notebook format available here: http://public.globalsymmetry.com/proprietary/com/wri/notebooks/with-gif/esse ntial/essential.nb http://public.globalsymmetry.com/proprietary/com/wri/notebooks/with-gif/esse ntial/index.html Please let me know if you have any problems accessing these. The sysadmin is kind of new at running a DNS server. The server may go down for a while at Ideally that should take about an hour... and then there's reality... What I'm hoping for is some constructive feedback regarding my selection of functions. I am aware that I've neglected the more advanced use of these functions such as level specification. I'm trying to keep things as simple as possible. I'm not looking for the obfuscated Mathematica challenge. Not just yet. I'm seeking the examples which, if correctly understood, will make other Mathematica functionality fall into place. I believe there is also a set of 'procedural operators' which can properly be treated separately. I hope to get to them soon. I know I've received some very good feedback on other questions I've asked on the news group. I owe people responses to their thoughtful input. I hope to address these soon. I am very grateful to all who have responded to my questions. STH . ==== I would like to mark representative samples of t on a parametric plot, where t is the third parameter. For example, how could I mark the 8 values t=0, t=Pi/4, ..., t=7Pi/4 on the plot generated by: ParametricPlot[{Sin[t], Cos[t]}, {t, 0, 2Pi}] Reply-To: kuska@informatik.uni-leipzig.de ==== ff[t_] := {Sin[t], Cos[t]} ParametricPlot[Evaluate[ff[t]], {t, 0, 2Pi}, Epilog -> {PointSize[0.025], (Point[ff[#]] & /@ Table[phi, {phi, 0, 2Pi, Pi/4}])}] Jens I would like to mark representative samples of t on a parametric plot, > where t is the third parameter. For example, how could I mark the 8 > values t=0, t=Pi/4, ..., t=7Pi/4 on the plot generated by: > ParametricPlot[{Sin[t], Cos[t]}, {t, 0, 2Pi}] > ==== How do I evaluate this product in mathematica: f[s_]=Product[a+b*Exp[s*x[i]],{i,1,n}] ? D[f[s],s] does not evaluate the terms. ==== I want to perform this calculation: In[1]:=z1 = a1 + b1 I Out[1]=a1 + [ImaginaryI] b1 In[3]:=z2 = a2 + b2 I Out[3]=a2 + [ImaginaryI] b2 In[19]:=Abs[(z1 - z2)/(1 - z1 Conjugate[z2])] This should output 1! But it doesn't work... Also, Abs[a1+b1 I] doesn't get the right result. Any ideeas? CeZaR ==== Of course one can use standard programming techniques to answer this and it will in fact be the most efficient method. But as you will probably get lots of answers of this kind, I will do it in another way: by exploiting a few standard built-in number theoretic functions which are very closely connected with your problems. Question 1: In[1]:= funct1[l_List]:=Outer[Times,Sequence@@(Divisors/@Power@@@l)]//Flatten In[2]:= funct1[{{2,3},{3,1},{5,1}}] Out[2]= {1,5,3,15,2,10,6,30,4,20,12,60,8,40,24,120} Note what we did. We first converted your pairs {a,b} back into powers a^b then found all the divisors using the built in Divisors function, then found all the products using Outer. Question 2. In[3]:= minint[list1_,list2_]:=GCD[ Times@@Power@@@list1,Times@@Power@@@list2]//FactorInteger In[4]:= maxint[list1list1_,list2_]:=LCM[Times@@Power@@@list1,Times@@Power@@@list 2] //FactorInteger e.g. In[5]:= list1 = {{2, 3}, {3, 4}, {5, 6}, {7, 2}, {17, 5}}; In[6]:= list2 = {{2, 5}, {3, 2}, {5, 1}, {7, 3}}; In[7]:= minint[list1,list2] Out[7]= {{2,3},{3,2},{5,1},{7,2}} In[8]:= maxint[list1,list2] Out[8]= {{2,5},{3,4},{5,6},{7,3},{17,5}} Basically all we did was to use the built in functions GCD and LCM after converting your lists of powers to numbers. Then we factored them again. In this case to finally factor an integer, which guarantees the programs to be inefficient for large numbers. However if your original list of pairs were indeed the result of using FactorInteger, then you should of course use versions of the above programs that can be applied to the original un-factored integers. Indeed, in that case this is the only efficient way to proceed. Andrzej Kozlowski Yokohama, Japan http://www.mimuw.edu.pl/~akoz/ http://platon.c.u-tokyo.ac.jp/andrzej/ I have two seperate list questions that I was hoping to get help with. Question 1. I have a variable length list similar to that generated by > FactorInteger, > that is {number, exponent} pairs. An example follows. lista = {{2,3},{3,1},{5,1}} ... this is the number 2^3 * 3^1 * 5^1 I want to generate a list of all the products of numbers from this > list. I can tell that I get a total (3+1)*(1+1)*(1+1) = 4*2*2 = 16, products > and I > want a list showing all of those. These would be: 2^3 can generate {2^0, 2^1, 2^2, 2^3} = {1, 2, 4 ,8} > 3^1 can generate {3^0, 3^1} = {3} ... we dont care about the > duplicate 1 > 5^1 can generate {5^0, 5^1} = {5} ... we dont care about the duplicate > 1 Hence the 4*2*2 = 16 (the product of one more of the exponents) above. Next we should get 16 products (from these lists), namely (I left them > as > products below to show what I am after): {1, 2, 4, 8, 1*3, 2*3, 4*3, 8*3, 1*5, 2 * 5, 4* 5, 8* 5, 1*3*5, > 2*3*5, 4*3*5, 8*3*5} If the list were lista = {{2,4}, {3,2}, {5, 3},{7^5}}, we would have > (4+1)(2+1)(3+1)(5+1) = 360 products, for example and the return values > should be a single list showing all of those. Question 2. I have two lists and want to generate two new lists from them. These > two > lists are {number, exponent} pairs. In the first list, I want the minimum intersection of {number, > exponent} > pairs. In the second list, I want the maximum union of {number, exponent} > pairs. Let me show an example: Input: list1 = {{2, 3}, {3, 4}, {5, 6}, {7, 2}, {17, 5}} list2 = {{2, 5}, {3, 2}, {5, 1}, {7, 3}} Output: minint = {{2, 3}, {3, 2}, {5, 1}, {7, 2}} Note: In this example we only kept those pairs where the intersection > of the > number exists and also keep the min power of those. maxint = {{2, 5}, {3, 4}, {5, 6}, {7, 3}, {17, 5}} Note: In this example we kept the union of lists and also keep the max > power > of each. > Flip > ==== To start with, what you are saying is simply not true. A simple example: In[1]:= Abs[(z1 - z2)/(1 - z1*Conjugate[z2])] /. {z1 -> 1 + I, z2 -> 1 - I} Out[1]= 2/Sqrt[5] Presumably you meant Abs[(z1 - z2)/(z1 - Conjugate[z2]) in which case: In[1]:= ComplexExpand[Abs[(z1-z2)/(z1- Conjugate[z2])],TargetFunctions->{Im,Re}] Out[1]= 1 Andrzej Kozlowski Yokohama, Japan http://www.mimuw.edu.pl/~akoz/ http://platon.c.u-tokyo.ac.jp/andrzej/ I want to perform this calculation: In[1]:=z1 = a1 + b1 I > Out[1]=a1 + [ImaginaryI] b1 > In[3]:=z2 = a2 + b2 I > Out[3]=a2 + [ImaginaryI] b2 > In[19]:=Abs[(z1 - z2)/(1 - z1 Conjugate[z2])] This should output 1! But it doesn't work... Also, Abs[a1+b1 I] doesn't get the right result. > Any ideeas? CeZaR > ==== Charles, One method is to use Epilog as follows. For simplicity I define the parametrization of the curve, the list of t values and the list of associated points. It is also nice to add some color to the plot. Needs[Graphics`Colors`] curve[t_] := {Sin[t], Cos[t]} tvals = Pi/4Range[0, 7]; pts = curve /@ tvals; ParametricPlot[Evaluate[curve[t]], {t, 0, 2Pi}, PlotStyle -> Blue, Epilog -> {Black, AbsolutePointSize[5], Point /@ pts, MapThread[Text[#1, 1.1 #2] &, {tvals, pts}]}, Axes -> None, AspectRatio -> Automatic, PlotRange -> All, Background -> Linen, ImageSize -> 430]; If you use the DrawGraphics package at my web site, this can be done slightly easier, without using PlotStyle or Epilog. ParametricDraw just extracts the graphics primitives, actually the Line, from ParametricPlot without a side plot. Then we can just combine it with the Point and Text primitives. Draw2D is a shortcut for Show[Graphics[...],opts]; Needs[DrawGraphics`DrawingMaster`] Draw2D[ {Blue, ParametricDraw[Evaluate[curve[t]], {t, 0, 2Pi}], Black, AbsolutePointSize[5], Point /@ pts, MapThread[Text[#1, 1.1 #2] &, {tvals, pts}]}, AspectRatio -> Automatic, Background -> Linen, ImageSize -> 430]; David Park djmp@earthlink.net http://home.earthlink.net/~djmp/ Sender: steve@smc.vnet.net Approved: Steven M. Christensen , Moderator ==== I think a better place to look for the essential aspects of Mathematica are in rule application, pattern matching, and mathematical knowledge. I will be interested in hearing what other people have to say. Richard Palmer > I'm attempting to identify the essential aspects of Mathematica. I believe > the > place to start is with the 'functional operations'. I'm seeking the > 'basis' of Mathematica. Kind of the orthonormal subset of functionality > which can be used to derive all the other. I'm also trying to be > pragmatic. I'm not trying to reinvent Mathematica, I'm just trying to > understand > the invention that already exists. If anyone is interested in seeing what I've gathered so far, I have a > notebook in both HTML, and Mathematica (4.2, if that matters) notebook format > available here: http://public.globalsymmetry.com/proprietary/com/wri/notebooks/with-gif/esse n t > ial/essential.nb > http://public.globalsymmetry.com/proprietary/com/wri/notebooks/with-gif/esse n t > ial/index.html Please let me know if you have any problems accessing these. The sysadmin is > kind of new at running a DNS server. The server may go down for a while at > Ideally that should take about an hour... and then there's reality... What I'm hoping for is some constructive feedback regarding my selection of > functions. I am aware that I've neglected the more advanced use of these > functions such as level specification. I'm trying to keep things as simple > as possible. I'm not looking for the obfuscated Mathematica challenge. Not > just > yet. I'm seeking the examples which, if correctly understood, will make > other Mathematica functionality fall into place. I believe there is also a > set of > 'procedural operators' which can properly be treated separately. I hope to > get to them soon. I know I've received some very good feedback on other questions I've asked > on the news group. I owe people responses to their thoughtful input. I > hope to address these soon. I am very grateful to all who have responded > to my questions. STH ==== >I have build a table with this pattern: Flatten[{{{d, X, Y, Z}}, > Table[{t, x[t], y[t], z[t]}, {t, 1, 10}], > Table[{t, x[t], y[t], z[t]}, {t, 20, 100, 10}], > Table[{t, x[t], y[t], z[t]}, {t, 200, 1000, 100}]}, 1] // TableForm I would like obtain the same Output in a more elegant way. In other word, >how >Can I avoid write Table[{t, x[t], y[t], z[t]} a few times. > Fit[{1, 20, 200}, {1, n, n^2}, n] Flatten[ Prepend[ Table[{t, x[t], y[t], z[t]}, {n, 3}, {t, (161*n^2)/2 - (445*n)/2 + 143, 10^n, 10^(n - 1)}], {{d, X, Y, Z}}], 1] Bob Hanlon ==== >I would like to mark representative samples of t on a parametric plot, >where t is the third parameter. For example, how could I mark the 8 >values t=0, t=Pi/4, ..., t=7Pi/4 on the plot generated by: >ParametricPlot[{Sin[t], Cos[t]}, {t, 0, 2Pi}] > ParametricPlot[{Sin[t], Cos[t]}, {t, 0, 2Pi}, AspectRatio -> Automatic, Epilog -> {AbsolutePointSize[5], RGBColor[1, 0, 0], Table[Point[{Sin[t], Cos[t]}], {t, 0, 7Pi/4, Pi/4}]}]; Bob Hanlon ==== > I would like to mark representative samples of t on a parametric plot, > where t is the third parameter. For example, how could I mark the 8 > values t=0, t=Pi/4, ..., t=7Pi/4 on the plot generated by: > ParametricPlot[{Sin[t], Cos[t]}, {t, 0, 2Pi}] One way is to use Epilog: ParametricPlot[{Sin[t], Cos[t]}, {t, 0, 2*Pi}, Epilog -> {PointSize[0.02], Table[With[{p = {Sin[t], Cos[t]}}, {Point[p],Text[t, p, -1.5*p]}], {t, Pi/4, 7*(Pi/4), Pi/4}]}, PlotRange -> All] ==== >I would like to mark representative samples of t on a parametric plot, >where t is the third parameter. For example, how could I mark the 8 >values t=0, t=Pi/4, ..., t=7Pi/4 on the plot generated by: >ParametricPlot[{Sin[t], Cos[t]}, {t, 0, 2Pi}] There are a number of possibilities depending on exactly how you want things to appear You could use Ticks->{Cos[# Pi/4]&/@Range[0,7],Sin[# Pi/4]&/@Range[0,7]} combined with Axes->True and seting AxesOrigin to an appropriate location Another option would be using Epilog->MapThread[Point[{#1,#2}]&,Cos[# Pi/4]&/@Range[0,7],Sin[# Pi/4]&/@Range[0,7]}] to plot points at the key locations. In order to get a satisfactory display you may also need to adjust either PointSize or PlotStyle. Other options include making plots of the marks and the parametric plot as separate graphics then combining them with the Show command ==== >I have two seperate list questions that I was hoping to get help with. Question 1. I have a variable length list similar to that generated by FactorInteger, >that is {number, exponent} pairs. An example follows. lista = {{2,3},{3,1},{5,1}} ... this is the number 2^3 * 3^1 * 5^1 I want to generate a list of all the products of numbers from this list. I can tell that I get a total (3+1)*(1+1)*(1+1) = 4*2*2 = 16, products >and I >want a list showing all of those. These would be: 2^3 can generate {2^0, 2^1, 2^2, 2^3} = {1, 2, 4 ,8} >3^1 can generate {3^0, 3^1} = {3} ... we dont care about the duplicate >1 >5^1 can generate {5^0, 5^1} = {5} ... we dont care about the duplicate >1 Hence the 4*2*2 = 16 (the product of one more of the exponents) above. Next we should get 16 products (from these lists), namely (I left them >as >products below to show what I am after): {1, 2, 4, 8, 1*3, 2*3, 4*3, 8*3, 1*5, 2 * 5, 4* 5, 8* 5, 1*3*5, >2*3*5, 4*3*5, 8*3*5} If the list were lista = {{2,4}, {3,2}, {5, 3},{7^5}}, we would have >(4+1)(2+1)(3+1)(5+1) = 360 products, for example and the return values >should be a single list showing all of those. Question 2. I have two lists and want to generate two new lists from them. These two >lists are {number, exponent} pairs. In the first list, I want the minimum intersection of {number, exponent} >pairs. In the second list, I want the maximum union of {number, exponent} pairs. Let me show an example: Input: list1 = {{2, 3}, {3, 4}, {5, 6}, {7, 2}, {17, 5}} list2 = {{2, 5}, {3, 2}, {5, 1}, {7, 3}} Output: minint = {{2, 3}, {3, 2}, {5, 1}, {7, 2}} Note: In this example we only kept those pairs where the intersection of >the >number exists and also keep the min power of those. maxint = {{2, 5}, {3, 4}, {5, 6}, {7, 3}, {17, 5}} Note: In this example we kept the union of lists and also keep the max >power >of each. > allProducts[x_] := Module[{sx = Sort[x, #2[[2]] < #1[[2]] &], f}, Union[Flatten[ Outer[f, Sequence @@ (PadRight[#, sx[[1, -1]] + 1, 1] & /@ (#[[1]]^ Range[0, #[[2]]] & /@ sx))]] /. f -> Times]]; lista = {{2, 3}, {3, 1}, {5, 1}}; allProducts[lista] {1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120} minInt[x_, y_] := FactorInteger[GCD[ Times @@ (#[[1]]^#[[2]] & /@ x), Times @@ (#[[1]]^#[[2]] & /@ y)]]; maxInt[x_, y_] := Union[x, y] //. {s___, {b_, e1_}, {b_, e2_}, r___} :> {s, {b, Max[e1, e2]}, r} list1 = {{2, 3}, {3, 4}, {5, 6}, {7, 2}, {17, 5}}; list2 = {{2, 5}, {3, 2}, {5, 1}, {7, 3}}; minInt[list1, list2] {{2, 3}, {3, 2}, {5, 1}, {7, 2}} maxInt[list1, list2] {{2, 5}, {3, 4}, {5, 6}, {7, 3}, {17, 5}} Bob Hanlon ==== >I want to perform this calculation: In[1]:=z1 = a1 + b1 I Out[1]=a1 + [ImaginaryI] b1 In[3]:=z2 = a2 + b2 >I Out[3]=a2 + [ImaginaryI] b2 In[19]:=Abs[(z1 - z2)/(1 - z1 >Conjugate[z2])] This should output 1! But it doesn't work... Also, Abs[a1+b1 I] doesn't get the right result. Any ideeas? What do you mean by doesn't work and doesn't get the right result? Do you mean Mathematica returns an unevaluated expression? If so, have you assigned values to the symbols a1, b1, a2 and b2? If you haven't assigned values, how is Mathematica to know these symbols do not take on complex values? ==== I want to perform this calculation: In[1]:=z1 = a1 + b1 I Out[1]=a1 + [ImaginaryI] b1 In[3]:=z2 = a2 + b2 >I Out[3]=a2 + [ImaginaryI] b2 In[19]:=Abs[(z1 - z2)/(1 - z1 >Conjugate[z2])] This should output 1! But it doesn't work... Also, Abs[a1+b1 I] doesn't get the right result. Any ideeas? What do you mean by doesn't work and doesn't get the right result? Do you mean Mathematica returns an unevaluated expression? If so, have you assigned values to the symbols a1, b1, a2 and b2? If you haven't assigned values, how is Mathematica to know these symbols do not take on complex values? Yes, it returns an unevaluated expression. CeZaR ==== Steve, Increase the PlotPoints. ParametricPlot[{Sin[t], Sin[2t]}, {t, 0, 1000Pi}, PlotPoints -> 2000]; But, generally one wouldn't want to make the t domain greater than required to plot the complete figure, in this case 0 to 2 Pi. David Park djmp@earthlink.net http://home.earthlink.net/~djmp/ -- _______________ Steve Story Polymer Research Group 411B Cox North Carolina State University 1-919-515-8147 _______________ ==== Guillermo, This might be considered slightly better.. TableForm[{#, x[#], y[#], z[#]} & /@ Join[Range[10], Range[20, 100, 10], Range[200, 1000, 100]], TableHeadings -> {None, {d, X, Y, Z}}] David Park djmp@earthlink.net http://home.earthlink.net/~djmp/ Thans Guillermo Sanchez --------------------------------------------- This message was sent using Endymion MailMan. ==== Luca, It looks like Mathematica doesn't implement the Leibniz rule on indefinite products. Here is my attempt to implement it. I hope I haven't slipped up. You may get some other better answers. I am not going to paste in all the output because you can duplicate it in your own notebook. First I define your f, separating the parameters from the variable s. f[a_, b_, n_][s_] := Product[a + b*Exp[s*x[i]], {i, 1, n}] You could differentiate with respect to s by f[a,b,n]'[s] but Mathematica doesn't evaluate. However, if you specify an integer for n, Mathematica will evaluate. f[a,b,3]'[s] LeibnizD::usage = LeibnizD[product, x] will differentiate an indefinite product expression with respect to x using the Leibniz rule. The product must be of the form Product[factor, {iter, min, max}].; LeibnizD[p_Product, x_] := Module[{factor, term, piter, pmin, pmax}, {piter, pmin, pmax} = Rest[p]; factor = First[p]; term = (D[factor, x])/factor ; p Sum[term // Evaluate, {piter, pmin, pmax} // Evaluate]] Then the following gives the differentiation in terms of the original product times a sum. LeibnizD[f[a, b, n][s], s] Checking one case.. f[a, b, 3]'[s] == (LeibnizD[f[a, b, n][s], s] /. n -> 3) // Simplify True And if it correct for one case it must be correct for all. Right? David Park djmp@earthlink.net http://home.earthlink.net/~djmp/ Sender: steve@smc.vnet.net Approved: Steven M. Christensen , Moderator ==== I noticed a rather curious trick that can be used to avoid having to use FactorInteger in the code I sent earlier for minint and maxint. Here is the new version: In[2]:= list1 = {{2, 3}, {3, 4}, {5, 6}, {7, 2}, {17, 5}}; In[3]:= list2 = {{2, 5}, {3, 2}, {5, 1}, {7, 3}}; In[4]:= minint[list1_List, list2_List] := ToExpression[PolynomialGCD[Times @@ Apply[ToString[#1]^#2 & , list1, {1}], Times @@ Apply[ToString[#1]^#2 & , list2, {1}]] /. {Times -> List, Power -> List}] In[5]:= maxint[list1_List, list2_List] := ToExpression[PolynomialLCM[Times @@ Apply[ToString[#1]^#2 & , list1, {1}], Times @@ Apply[ToString[#1]^#2 & , list2, {1}]] /. {Times -> List, Power -> List}] In[6]:= minint[list1, list2] Out[6]= {{2, 3}, {3, 2}, 5, {7, 2}} In[7]:= maxint[list1, list2] Out[7]= {{17, 5}, {2, 5}, {3, 4}, {5, 6}, {7, 3}} The idea is to do exactly the same thing as before but now we find the GCD and LCM of expressions like 2^3 *5^4 and 2^4*5^3 etc. Note that the base in a^b is a string and not a number so we need to use PolynomialGCD and PoynomialLCM instead of GCD and LCM. Using such algebraic functions will usually make the program slower, but on the other hand we need not use FactorInteger so when the numbers one gets are large enough the present version should be faster. In any case it seems a curious idea so I thought it worth posting (even though it is easy to write conventional programs to do the same thing that ought to be much more efficient). > Of course one can use standard programming techniques to answer this > and it will in fact be the most efficient method. But as you will > probably get lots of answers of this kind, I will do it in another > way: by exploiting a few standard built-in number theoretic functions > which are very closely connected with your problems. Question 1: In[1]:= > funct1[l_List]:=Outer[Times,Sequence@@(Divisors/@Power@@@l)]//Flatten In[2]:= > funct1[{{2,3},{3,1},{5,1}}] Out[2]= > {1,5,3,15,2,10,6,30,4,20,12,60,8,40,24,120} Note what we did. We first converted your pairs {a,b} back into > powers a^b then found all the divisors using the built in Divisors > function, then found all the products using Outer. Question 2. In[3]:= > minint[list1_,list2_]:=GCD[ > Times@@Power@@@list1,Times@@Power@@@list2]//FactorInteger In[4]:= > maxint[list1list1_,list2_]:=LCM[Times@@Power@@@list1,Times@@Power@@@lis > t2] > //FactorInteger e.g. In[5]:= > list1 = {{2, 3}, {3, 4}, {5, 6}, {7, 2}, {17, 5}}; > In[6]:= > list2 = {{2, 5}, {3, 2}, {5, 1}, {7, 3}}; > In[7]:= > minint[list1,list2] Out[7]= > {{2,3},{3,2},{5,1},{7,2}} In[8]:= > maxint[list1,list2] Out[8]= > {{2,5},{3,4},{5,6},{7,3},{17,5}} Basically all we did was to use the built in functions GCD and LCM > after converting your lists of powers to numbers. Then we factored > them again. In this case to finally factor an integer, which guarantees the > programs to be inefficient for large numbers. However if your original > list of pairs were indeed the result of using FactorInteger, then you > should of course use versions of the above programs that can be > applied to the original un-factored integers. Indeed, in that case > this is the only efficient way to proceed. Andrzej Kozlowski > Yokohama, Japan > http://www.mimuw.edu.pl/~akoz/ > http://platon.c.u-tokyo.ac.jp/andrzej/ > I have two seperate list questions that I was hoping to get help with. >> Question 1. >> I have a variable length list similar to that generated by >> FactorInteger, >> that is {number, exponent} pairs. An example follows. >> lista = {{2,3},{3,1},{5,1}} ... this is the number 2^3 * 3^1 * 5^1 >> I want to generate a list of all the products of numbers from this >> list. >> I can tell that I get a total (3+1)*(1+1)*(1+1) = 4*2*2 = 16, >> products and I >> want a list showing all of those. >> These would be: >> 2^3 can generate {2^0, 2^1, 2^2, 2^3} = {1, 2, 4 ,8} >> 3^1 can generate {3^0, 3^1} = {3} ... we dont care about the >> duplicate 1 >> 5^1 can generate {5^0, 5^1} = {5} ... we dont care about the >> duplicate 1 >> Hence the 4*2*2 = 16 (the product of one more of the exponents) above. >> Next we should get 16 products (from these lists), namely (I left >> them as >> products below to show what I am after): >> {1, 2, 4, 8, 1*3, 2*3, 4*3, 8*3, 1*5, 2 * 5, 4* 5, 8* 5, 1*3*5, >> 2*3*5, 4*3*5, 8*3*5} >> If the list were lista = {{2,4}, {3,2}, {5, 3},{7^5}}, we would have >> (4+1)(2+1)(3+1)(5+1) = 360 products, for example and the return values >> should be a single list showing all of those. >> Question 2. >> I have two lists and want to generate two new lists from them. These >> two >> lists are {number, exponent} pairs. >> In the first list, I want the minimum intersection of {number, >> exponent} >> pairs. >> In the second list, I want the maximum union of {number, exponent} >> pairs. >> Let me show an example: >> Input: >> list1 = {{2, 3}, {3, 4}, {5, 6}, {7, 2}, {17, 5}} >> list2 = {{2, 5}, {3, 2}, {5, 1}, {7, 3}} >> Output: >> minint = {{2, 3}, {3, 2}, {5, 1}, {7, 2}} >> Note: In this example we only kept those pairs where the intersection >> of the >> number exists and also keep the min power of those. >> maxint = {{2, 5}, {3, 4}, {5, 6}, {7, 3}, {17, 5}} >> Note: In this example we kept the union of lists and also keep the >> max power >> of each. >> Flip > Andrzej Kozlowski > Yokohama, Japan > http://www.mimuw.edu.pl/~akoz/ > http://platon.c.u-tokyo.ac.jp/andrzej/ Andrzej Kozlowski Yokohama, Japan http://www.mimuw.edu.pl/~akoz/ http://platon.c.u-tokyo.ac.jp/andrzej/ ==== If you use SetDelayed rather than Set, and then Simplify, Mathematica returns the answer you expect: f[t_] := {t + Sqrt[3] Sin[t], 2 Cos[t], t Sqrt[3] - Sin[t]} ; Simplify[Sqrt[f'[t] .f'[t]]] Out[2]= 2 Sqrt[2] >Sqrt[3] Sin[t], 2 Cos[t], t Sqrt[3] - Sin[t]} So It's basically a vector >whose coordinates are determined based on the values you pass in. Then I >took the derivative by just typing f', which outputs {1 + Sqrt[3] Cos[#1], >-2 Sin[#1], Sqrt[3] - Cos[#1]}& What I'd like to do is have Mathematica >calculate the norm of this as it would any vector, so that I can play with >the norm function. As it turns out, the norm in this case is identical to >Sqrt[8], so it would be nice if Mathematica could figure that out. Is it Christopher J. Purcell Defence R&D Canada .9a Atlantic 9 Grove St., PO Box 1012 Dartmouth NS Canada B2Y 3Z7 ==== >I wanted to do list manipulation methods explicitly on the lists without >using GCD and FactoInteger specifically. I am doing things with numbers >that I know the factorization for, but the methods in Mathematica (or anything >else for that matter) don't suffice. Is there an easy way to get rid of the GCD and FactorInteger in the second >method. > minInt[x_, y_] := Select[Union[x, y], MemberQ[First /@ Intersection[x, y, SameTest -> (#1[[1]] == #2[[1]] &)], #[[1]]] &] //. {s___, {b_, e1_}, {b_, e2_}, r___} :> {s, {b, Min[e1, e2]}, r}; list1 = {{2, 3}, {3, 4}, {5, 6}, {7, 2}, {17, 5}}; list2 = {{2, 5}, {3, 2}, {5, 1}, {7, 3}}; minInt[list1, list2] {{2, 3}, {3, 2}, {5, 1}, {7, 2}} Bob Hanlon ==== I'm using mozilla I built a few hours ago. The classical mechanics site look very nice. I did, however receive the errors described here: http://baldur.globalsymmetry.com/proprietary/com/wri/ch08.html To properly display the MathML on this page you need the following fonts: CMSY10, CMEX10, Math1, Math2, Math4. For further infromation see: http://www.mozilla.org/projects/mathml/fonts Among the fonts which I seem to be missin is the one used to display the imaginary coeficient in the MathML discussed below. You will find more discussion of my experiences with XML and Mathematica on the page I linked to above. I believe there is something which still needs to be done with my fonts, but I haven't had time to research it. If anyone knows the solution, I would really appreciate knowing. I can think of installed. The display is a bit better on the XP system, but for the most part, the results are similar. STH > Steven > I have done a great deal with MathML and esp with Mathematica 4.2 > a) you need a modern browser NS 7.0 or Mozilla 1 or Amaya > b) IE needs a plugin www.dessci.com has mathplayer but there are other > issues > c) The rendering is done by the browser as is the XML parsing > d) an on line example is at http://core.ecu.edu/phys/flurchickk/Classes/CM4226/classicalMechanics2-1.xm >l e) you can test the browser with the w3c test site > http://www.w3.org/Math/testsuite/ > -----Original Message----- > Sent: Saturday, October 12, 2002 5:05 AM > To: mathgroup@smc.vnet.net > I'm trying to get a handle on Mathematica's XML capabilities. I'm finding > a few > things to be a bit confusing. One of these is where the > http://www.wolfram.com/XML/DTD/2001/NBMLwMathML.dtd really is. Another > point of confusion is how exactly the rendering in the browser is expected > to take place. If needs be, I can spin up my own CSS. Does Mathematica > provide > CSS for MathML or NBML? I've character mentioned entity references before, > but I still haven't found an answer. What I've found here seems > inconsistent with what Mathematica chose for the character entity reference > for an > imaginary number: http://www.bitjungle.com/~isoent/ This is what Mathematica produced for a complex number:

cpx = 1066 + 42⁢ⅈ ;

 If I understand the http://www.bitjungle.com/~isoent/ent.xml, the imaginary > number symbol should be ࠿ which is a black letter capital 'I'. That > doesn't seem correct to me. I'm currently stumbling around in here looking > for a possible clue as to what I should expect: > http://www.physiome.org.nz/Docs/web-tech/specs/mathML20/chapter3.html Has anybody worked with this? STH ==== I'm trying to get a handle on Mathematica's XML capabilities. I'm finding a few things to be a bit confusing. One of these is where the http://www.wolfram.com/XML/DTD/2001/NBMLwMathML.dtd really is. Another point of confusion is how exactly the rendering in the browser is expected to take place. If needs be, I can spin up my own CSS. Does Mathematica provide CSS for MathML or NBML? I've character mentioned entity references before, but I still haven't found an answer. What I've found here seems inconsistent with what Mathematica chose for the character entity reference for an imaginary number: http://www.bitjungle.com/~isoent/ This is what Mathematica produced for a complex number:

cpx = 1066 + 42 ⁢ ⅈ ;

If I understand the http://www.bitjungle.com/~isoent/ent.xml, the imaginary number symbol should be ࠿ which is a black letter capital 'I'. That doesn't seem correct to me. I'm currently stumbling around in here looking for a possible clue as to what I should expect: http://www.physiome.org.nz/Docs/web-tech/specs/mathML20/chapter3.html Has anybody worked with this? STH . Reply-To: Lester Ingber ==== If you have very strong credentials for the position described below, Lester Ingber Director R&D DUNN Capital Management Stuart FL Some recent press on DUNN can be seen on http://www.businessweek.com/magazine/content/02_39/b3801113.htm http://www.businessweek.com/magazine/content/02_39/b3801114.htm Financial Engineer A disciplined, quantitative, analytic individual proficient in prototyping and coding (such as C/C++, Maple/Mathematica, or Visual Basic, etc.) is sought for financial engineering/risk:reward optimization research position with established Florida hedge fund (over two decades in the business and $1 billion in assets under management). A PhD in a mathematical science, such as physics, statistics, math, or computer-science, is preferred. Hands-on experience in the financial industry is required. Emphasis is on applying state-of-the-art methods to financial time-series of various frequencies. Ability to work with a team to transform ideas/models into robust, intelligible code is key. Salary: commensurate with experience, with bonuses tied to the individual's and the firm's performance. Status of Selection Process All applicants will be reviewed, and a long list will be generated for phone interviews. Other applicants will not be contacted further. Information on the status of this process will be available in http://www.ingber.com/open_positions.html face-to-face interviews. During the visit for the physical interview a small coding exam will be given. Start date for this position may range anywhere from immediately to six months thereafter, depending on both the candidate's and the firm's needs. -- Prof. Lester Ingber ingber@ingber.com ingber@alumni.caltech.edu www.ingber.com www.alumni.caltech.edu/~ingber ==== Sorry, my message was nonsense. The reason why it appeared to give the answer 1 is that I forgot to evaluate z1 an z2: In[7]:= z1 = a1 + b1*I; In[8]:= z2 = a2 + b2*I; In[9]:= ComplexExpand[Abs[(z1 - z2)/(z1 - Conjugate[z2])], TargetFunctions -> {Im, Re}] Out[9]= Sqrt[(a1 - a2)^2 + (b1 - b2)^2]/ Sqrt[(a1 - a2)^2 + (b1 + b2)^2] The answer is clearly not unless b2==0. What exactly did you have in mind? > To start with, what you are saying is simply not true. A simple > example: In[1]:= > Abs[(z1 - z2)/(1 - z1*Conjugate[z2])] /. > {z1 -> 1 + I, z2 -> 1 - I} Out[1]= > 2/Sqrt[5] Presumably you meant Abs[(z1 - z2)/(z1 - Conjugate[z2]) in which case: In[1]:= > ComplexExpand[Abs[(z1-z2)/(z1- > Conjugate[z2])],TargetFunctions->{Im,Re}] Out[1]= > 1 Andrzej Kozlowski > Yokohama, Japan > http://www.mimuw.edu.pl/~akoz/ > http://platon.c.u-tokyo.ac.jp/andrzej/ > I want to perform this calculation: >> In[1]:=z1 = a1 + b1 I >> Out[1]=a1 + [ImaginaryI] b1 >> In[3]:=z2 = a2 + b2 I >> Out[3]=a2 + [ImaginaryI] b2 >> In[19]:=Abs[(z1 - z2)/(1 - z1 Conjugate[z2])] >> This should output 1! But it doesn't work... >> Also, Abs[a1+b1 I] doesn't get the right result. >> Any ideeas? >> CeZaR > > Andrzej Kozlowski Yokohama, Japan http://www.mimuw.edu.pl/~akoz/ http://platon.c.u-tokyo.ac.jp/andrzej/ ==== Dear Listers, I find myself defining functions in terms of differentiation. For example, f[x_,t_]:=Sin[x*t] dfx[x_,t]:=D[Sin[y,t],y]/.y->x This works well, but it seems to me that there should be a better way to do this. That is, there should be a better way to define a 'derivative' of a previous function without going through the replacement contortions. I can't find the answer in the archive. Can someone tell me the most straightforward way to do this? Will it work to define a gradient vector or Jacobian matrix? A Hessian matrix? -- Jason Miller, Ph.D. Division of Mathematics and Computer Science Truman State University 100 East Normal St. Kirksville, MO 63501 http://vh216801.truman.edu 660.785.7430 ==== Jason, We have dfx[x_,t_]= D[f[x,t],x] t Cos[t x] One advantage of using = rather than := is that it differentiates once, when the definition is stored, Definition[dfx] t Cos[t x] With := we get Clear[dfx] dfx[x_,t_]:= D[f[x,t],x] Definition[dfx] dfx[x_, t_] := D[f[x, t], x] So the differentiation is done each time that the function dfx is evaluated. -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay@haystack.demon.co.uk Voice: +44 (0)116 271 4198 > Dear Listers, I find myself defining functions in terms of differentiation. For example, f[x_,t_]:=Sin[x*t] > dfx[x_,t]:=D[Sin[y,t],y]/.y->x This works well, but it seems to me that there should be a better way > to do this. That is, there should be a better way to define a > 'derivative' of a previous function without going through the > replacement contortions. I can't find the answer in the archive. > Can someone tell me the most straightforward way to do this? Will it > work to define a gradient vector or Jacobian matrix? A Hessian > matrix? -- > Jason Miller, Ph.D. > Division of Mathematics and Computer Science > Truman State University > 100 East Normal St. > Kirksville, MO 63501 > http://vh216801.truman.edu > 660.785.7430 > Reply-To: kuska@informatik.uni-leipzig.de ==== f[x_, t_] := Sin[x*t] dfx[x_, t_] := Module[{y, df}, df = D[f[y, t], y]; Block[{y = x}, df ] ] Jens Dear Listers, I find myself defining functions in terms of differentiation. For example, f[x_,t_]:=Sin[x*t] > dfx[x_,t]:=D[Sin[y,t],y]/.y->x This works well, but it seems to me that there should be a better way > to do this. That is, there should be a better way to define a > 'derivative' of a previous function without going through the > replacement contortions. I can't find the answer in the archive. > Can someone tell me the most straightforward way to do this? Will it > work to define a gradient vector or Jacobian matrix? A Hessian > matrix? -- > Jason Miller, Ph.D. > Division of Mathematics and Computer Science > Truman State University > 100 East Normal St. > Kirksville, MO 63501 > http://vh216801.truman.edu > 660.785.7430 ==== I was trying to write a function, when given a list, say, {a,b,c,d}, the output is a op b op c op d, where op is (in LaTeX) bigotimes, or esc c * esc ([CircleTimes]) in Mathematica. f[{x_}]:=x; f[{x_, y_, z___}]:=f[Join[{x[CircleTimes]y},{z}]]; However, the output was not exactly what I expected, it looked like: ((a op b) op c) op d It seems when doing the Join operation, a pair of parenthesis was added. Can someone let me know how I can get rid of these parenthesis? JT _________________________________________________________________ Chat with friends online, try MSN Messenger: http://messenger.msn.com ==== > I was trying to write a function, when given a list, say, {a,b,c,d}, the > output is a op b op c op d, where op is (in LaTeX) bigotimes, or esc c * > esc ([CircleTimes]) in Mathematica. f[{x_}]:=x; > f[{x_, y_, z___}]:=f[Join[{x[CircleTimes]y},{z}]]; However, the output was not exactly what I expected, it looked like: ((a op b) op c) op d It seems when doing the Join operation, a pair of parenthesis was added. Can someone let me know how I can get rid of these parenthesis? > JT > _________________________________________________________________ > Chat with friends online, try MSN Messenger: http://messenger.msn.com You can be very effective with a simple trick f[MyList_]:=Drop[Flatten[Transpose[{MyList,Table[op,Length[MyList]]}]],-1 ] then you can apply on it a conversion to string then save it to a file to be processed by TeX/LateX ==== JJJ, Try: f[{x_}]:=x; f[{x_,y__}]:=x[CircleTimes]y Test f[{a,b,c,d}] a[CircleTimes]b[CircleTimes]c[CircleTimes]d -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay@haystack.demon.co.uk Voice: +44 (0)116 271 4198 > I was trying to write a function, when given a list, say, {a,b,c,d}, the > output is a op b op c op d, where op is (in LaTeX) bigotimes, or esc c * > esc ([CircleTimes]) in Mathematica. f[{x_}]:=x; > f[{x_, y_, z___}]:=f[Join[{x[CircleTimes]y},{z}]]; However, the output was not exactly what I expected, it looked like: ((a op b) op c) op d It seems when doing the Join operation, a pair of parenthesis was added. Can someone let me know how I can get rid of these parenthesis? > JT > _________________________________________________________________ > Chat with friends online, try MSN Messenger: http://messenger.msn.com Reply-To: kuska@informatik.uni-leipzig.de ==== SetAttributes[CircleTimes, Flat] f[{x_}] := x; f[{x_, y_, z___}] := f[Join[{x[CircleTimes]y}, {z}]]; Jens I was trying to write a function, when given a list, say, {a,b,c,d}, the > output is a op b op c op d, where op is (in LaTeX) bigotimes, or esc c * > esc ([CircleTimes]) in Mathematica. f[{x_}]:=x; > f[{x_, y_, z___}]:=f[Join[{x[CircleTimes]y},{z}]]; However, the output was not exactly what I expected, it looked like: ((a op b) op c) op d It seems when doing the Join operation, a pair of parenthesis was added. Can someone let me know how I can get rid of these parenthesis? > JT _________________________________________________________________ > Chat with friends online, try MSN Messenger: http://messenger.msn.com ==== the least of which was a power failure. For some strange reason Mathematica stopped producing the pretty html+MathML it had been. I started trouble shooting, and blew away what I had pointed to. I violated Tom Jackson's (IBM & UMUC) first rule of holes: when you're in one, stop digging. Here's the parent directory. http://public.globalsymmetry.com/proprietary/com/wri/notebooks/ As you can see (If I haven't fixed it by the time you look again,) Mathematica is outputting the conversion command, rather than the actual MathML I really don't know what happened. I removed the init.m and deleted the cache, and it still refuses to work correctly. It's kind of frustrating. There are a lot of powerful XML features in Mathematica, but they are not easy to use. STH > cannot access... > thanks, > Steven Taracevicz > PO Box 1752 > Santa Monica, CA 90406-1752 310.396.4001 > 310.388.3265 fax ==== > I'm attempting to identify the essential aspects of Mathematica. I believe > the > place to start is with the 'functional operations'. I'm seeking the > 'basis' of Mathematica. Kind of the orthonormal subset of functionality > which can be used to derive all the other. I'm also trying to be > pragmatic. I'm not trying to reinvent Mathematica, I'm just trying to > understand > the invention that already exists... I apologize in advance for this pessimistic response, but here goes: I have attempted something similar on two occasions, but with very specific audiences in mind. Even so, my efforts were little better than useless. (I would be willing to send you my most recent attempt, but as I said, it didn't work well.) I doubt that you can find agreement among the sea of users of Mathematica as to what is essential or basic or an orthonormal subset. The kernel is well described in the Mathematica book (for those of us who are careful readers or are refreshing memories) and in other books (for the rest of us). Would it not be a better use of your time to refer to selected passages in existing books? Documentation of the front end is uneven, with vast uncharted areas (I'm not even sure when it's round or flat!). Original writing about the front end might be more fruitful. Tom Burton ==== The way I learned XML, XML is for content, and CSS is for display. It seems that MathML violates that separation between style and content. There is little chance that I'm the first one to mention this. Is there a history of discussion on this topic somewhere? I don't want to get too far into MathML on this list, but, since I'm skinning my knuckles trying to learn to work with MathML and Mathematica, I figure a bit of discussion on this topic in this context is in order. STH ==== How can I plot functions like: (x-2)^2 + 2(y-3)^2 = 6 and x^3y + y^3 = 9 using Mathematica? David, x /. Solve[x^3 y + y^3 == 9, x]; g=Plot[Sign[(9 - y^3)/y]* Abs[(9 - y^3)/y]^(1/3) , {y, -7, 7} , AspectRatio -> Automatic]; Show[g /. {x_?NumberQ, y_?NumberQ} -> {y, x}, Epilog -> {RGBColor[1,0,0],Line[{{-(3^(7/9)/2^(2/9)), -5}, {-(3^(7/9)/2^(2/9)), 5}}]}]; ==== Is there an SGML catalog for the WRI DTDs on the CD? I'm not sure if I should expect psgml with xemacs to handle namespaces correctly. Currently, I'm having problems using the xml generated by Mathematica with psgml. I'm not sure exactly how to set up an sgml catalog to support these. I really don't want to try to reinvent the wheel, if there is already one available. I'm thouroughly preplexed by the overall behavior of the XML support. Things which were working have now stopped working, or have significantly changed their behavior. This is a very exciting area, but it seems very difficult to get started. I believe the dtd to include in a catalog is this: /opt/Wolfram/Mathematica/4.2/SystemFiles/IncludeFiles/XML/xhtml-math11-f.dtd I'm just not sure what the public identifier should be. STH -- Hatton's Law: There is only One inviolable Law. ==== Yes, there seems to be a lot of people who have a visceral hatred for Microsoft and Windows. They are even willing to shed blood to avoid Windows. But why? Windows works and you don't have to become a systems programmer. Furthermore, I think that Steven Wolfram uses some version of Windows. So guess which system Mathematica will be best tuned up for? I have no problems with Mathematica and Windows on my single computer. There may be reasons for using a non-Microsoft operating system. But if you are going to do it, make certain that they are good reasons. David Park djmp@earthlink.net http://home.earthlink.net/~djmp/ ==== > Yes, there seems to be a lot of people who have a visceral hatred for > Microsoft and Windows. They are even willing to shed blood to avoid > Windows. But why? Windows works and you don't have to become a systems > programmer. Furthermore, I think that Steven Wolfram uses some version of Windows. So > guess which system Mathematica will be best tuned up for? I have no problems with Mathematica and Windows on my single computer. > There may be reasons for using a non-Microsoft operating system. But if > you are going to do it, make certain that they are good reasons. that of Windows XP by orders of magnitude. I recall when I first started been using Windows NT since October of 1992. (Yes, I know it hadn't been paper on the architecture of NT. In 1997 I was well on my way to being an MCSE. no stinkin' GUI' to 'have a look at the KDE project'. I took the latter route. The KDE has gone from a simple graphical desktop with a few more features than the CDE, (and a lot more glitches) to being the best desktop available. It's growth seems to be exponential. Windows seems, at best, to be linear. All of these are usability issues. There is another reason I don't like using Microsoft products. I've also been using Mozilla since 1995. (Yes, it has always been called Mozilla.) I was one of the original beta testers for the Netscape line of internet servers. When I saw what Netscape Communications were aiming for, Windows quickly lost its luster. Netscape products were designed from the ground up with portability in mind. They were striving for uniform functionality across all platforms. I also saw what Microsoft did to undermine Netscape's R & D resources. Microsoft would condescend to having not competition in their market. Where I come from, people don't put up with that. Where do I come from? I was born in Illinois. I'm obviously not of the opinion that closed source is unacceptable. I wouldn't be using Mathematica if I were. I suspect one day Mathematica will face a real open source challeng. Her name is Charolette. She is the mother of Mozilla. That will probably be years from now. WRI need to be prepared to adjust to that eventuality when it comes. > David Park > djmp@earthlink.net > http://home.earthlink.net/~djmp/ STH . ==== >In: DSolve[y*D[u[x, y],x] == x*D[u[x, y],y], u[x,y], {x, y}] Out: {{u[x, y] -> C[1][(1/2)*(x^2 + y^2)]}} Square brackets are used as grouping symbols in the result!?? :^O Somebody say it isn't so. > It isn't so The square bracket is not delineating a factor it is enclosing the argument to an arbitrary function named C[1]. While the function is dependent on both x and y the dependence only occurs in the combination (x^2+y^2). Bob Hanlon Reply-To: Mark Coleman ==== Greetings, I hope this inquiry is not off-topic for the list. I was wondering if anyone knew of Mathematica-based education resources (either for students or teachers) that would be suitable for teaching *elementary school* students (kindergartden - 5th grade) some basic fundamentals of mathematics, e.g., simple algebra, geometry, numbers,etc. I know there is a solid body of work directed towards high school and college students, but I am not sure of resources for younger students. Mark ==== > f[{x_}]:=x; > f[{x_, y_, z___}]:=f[Join[{x[CircleTimes]y},{z}]]; However, the output was not exactly what I expected, it looked like: ((a op b) op c) op d It seems when doing the Join operation, a pair of parenthesis was added. Can someone let me know how I can get rid of these parenthesis? SetAttributes[CircleTimes,Flat] f[{a,b,c,d}] a[CircleTimes]b[CircleTimes]c[CircleTimes]d Tom Burton ==== The Illinois Institute of Technology's Stuart Graduate School of Business in collaboration with Wolfram Research, Inc. is offering a seminar series on applications of the Mathematica software system in mathematical and computational finance. The weekly seminars will begin Friday, October 25, 2002, and will be held at the Stuart Graduate School of Business in Chicago. The speakers will illustrate the built-in functionality of Mathematica as well as the extensive Mathematica applications available from both Wolfram Research and independent developers. They will also give detailed synopses of applications that solve a wide range of financial problems. Advance registration is required. You can register by writing to seminar@wolfram.com. For more information, visit the seminar website at: http://www.wolfram.com/services/seminars/chicago2002/ ==== Allan, I would like to add something to this , something which duzznt deal with the question directly but with the answer...since I've seen it many times on the mathgroup forum, I feel i'd like to make a philosophical comment here...when I look at the reply and the solution , which might be perfectly good, but it makes me wonder...of what value is it to a Newbie????......quite often I seen questions posted Who can do this or that the fastest?....How about who can come up with a solution that is the easiest to comprehend?....Allan, I realize that you are an advanced user, etc..and that you think along these lines...but for me, and perhaps many others, I cant think along these lines even though I have read mathgroup for many years now...with that said, naturally i'm quite thankful to the 'gurus' who provide answers.... jerry blimbaum NSWC panama city, fl -----Original Message----- To get the function for the norm of the derivative we can use norm = Evaluate/@(Simplify/@(Sqrt[#.#]&/@(f'))) 2*Sqrt[2] & We map the usual functions for calclulating and simplifying the norm inside Function[.] (which is the full form of (.)& and then map the function Evaluation to make the result evaluate -- this is needed since Function has the attribute HoldAll. Please note that the parentheses are essential. -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay@haystack.demon.co.uk Voice: +44 (0)116 271 4198 f[t_] = {t + Sqrt[3] Sin[t], 2 Cos[t], t Sqrt[3] - Sin[t]} So It's basically a vector whose coordinates are determined based on the > values you pass in. Then I took the derivative by just typing f', which outputs {1 + Sqrt[3] Cos[#1], -2 Sin[#1], Sqrt[3] - Cos[#1]}& > What I'd like to do is have Mathematica calculate the norm of this as it > would any vector, so that I can play with the norm function. As it turns > out, the norm in this case is identical to Sqrt[8], so it would be nice if > Mathematica could figure that out. Is it possible to do this? > ==== Dear Netters, I am also looking for Automated formal vector analysis. I am currently doing it with Mathematica and its package VectorAnalysis but it does not allow DIRECT algebra on tensors and vectors (or else I have not found how to do it) Can anyone else help ? Nicolas Fressengeas -- ________________________________________________________ Dr. Nicolas Fressengeas - - - http://www.ese-metz.fr/~fresseng Sup.8elec / Laboratoire Mat.8eriaux Optiques, Photonique et Syst.8fmes 2 rue E.Belin, 57070 METZ Cedex Plan d'acc.8fs: http://www.iti.fr/PlanPerso/23704/1 When everything else fails, read the instructions... Alexey Skoblikov a .8ecrit dans le message > Is there any tool for dealing with DIRECT tensor algebra, i.e. when tensor > is not a matrix of components, but considered to be the invariant object? ==== funcList = {Exp[x], Sin[y], z^3}; varList = {x, y, z}; MapThread[Plot[#1, {#2, -5, 5}] &, {funcList, varList}]; Inner[Plot[#1, {#2, -5, 5}] &, funcList, varList]; Bob Hanlon >how can I solve the following problem: The task is to successively plot the functions given in FuncList with the accociated variable >FuncList = {Exp[x], Sin[y], z^3}; VarList = {x, y, z}; Do[Plot[FuncList[LeftDoubleBracket] > i[RightDoubleBracket], {VarList[LeftDoubleBracket] > i[RightDoubleBracket], -5, 5}], {i, 1, Length[VarList]}]] ==== All you need is Evaluate in your code: Do[Plot[Evaluate[FuncList[[i]], {VarList[[i]], -5, 5}]], {i, 1, Length[VarList]}] or you might prefer: Plot[#1, {#2, -5, 5}] & @@@ Transpose[{FuncList, VarList}] Andrzej Kozlowski Yokohama, Japan http://www.mimuw.edu.pl/~akoz/ http://platon.c.u-tokyo.ac.jp/andrzej/ > Dear colleagues, how can I solve the following problem: The task is to successively plot > the functions given in FuncList with the accociated variable > FuncList = {Exp[x], Sin[y], z^3}; VarList = {x, y, z}; Do[Plot[FuncList[LeftDoubleBracket] > i[RightDoubleBracket], {VarList[LeftDoubleBracket] > i[RightDoubleBracket], -5, 5}], {i, 1, Length[VarList]}]] Frank Brand > ==== (Somehow the original posting never reached me). But what's wrong with dfx[x_, t_] := Derivative[1, 0][f][x, t] ? Andrzej Yokohama, Japan http://www.mimuw.edu.pl/~akoz/ http://platon.c.u-tokyo.ac.jp/andrzej/ f[x_, t_] := Sin[x*t] > dfx[x_, t_] := Module[{y, df}, > df = D[f[y, t], y]; > Block[{y = x}, > df > ] > ] Jens > Dear Listers, >> I find myself defining functions in terms of differentiation. For >> example, >> f[x_,t_]:=Sin[x*t] >> dfx[x_,t]:=D[Sin[y,t],y]/.y->x >> This works well, but it seems to me that there should be a better way >> to do this. That is, there should be a better way to define a >> 'derivative' of a previous function without going through the >> replacement contortions. I can't find the answer in the archive. >> Can someone tell me the most straightforward way to do this? Will it >> work to define a gradient vector or Jacobian matrix? A Hessian >> matrix? >> -- >> Jason Miller, Ph.D. >> Division of Mathematics and Computer Science >> Truman State University >> 100 East Normal St. >> Kirksville, MO 63501 >> http://vh216801.truman.edu >> 660.785.7430 > ==== It seems that I am having problems in embedding the eps fonts in Mathematica generated files using the program emmathfnt from Mathsource. Currently, I am using version 4.2 and with the previous version 4.1, I did not have any problems with emmathfnt. This is what I am doing: In[90]:= gr=Plot[Sin[x], {x, 0, Pi}, FrameLabel[Rule] { [Alpha] , [Beta]}]; scratchFile=Export[Close[OpenTemporary[]],gr,EPS]; Run[C:WINNTemmathfnt,-o,temp,-d, C:Program FilesWolfram ResearchMathematica4.2SystemFilesFontsType1,scratc hFile] Out[92]= 0 The file is generated but the fonts are not included. Can some one tell me what I am doing wrong? Note the executable file emmathfnt is the same found on Mathsource and the location of the file is in C:WINNT. Also since I still have version 4.0 fonts, I tried the above Run command without the -d flag and I got the same result. Finally, one can not help but wonder, why there is no option in export that allow the fonts to be included in Mathematica generated figures for better portability. Wissam AlSaidi ==== Frank, The first method is just to Evaluate the iterator. FuncList = {Exp[x], Sin[y], z^3}; VarList = {x, y, z}; Do[Plot[FuncList[[i]], Evaluate[{VarList[[i]], -5, 5}]], {i, 1, Length[VarList]}] But a simpler method is to use functional programming... MapThread[Plot[#1, {#2, -5, 5}] &, {FuncList, VarList}]; David Park djmp@earthlink.net http://home.earthlink.net/~djmp/ Frank Brand ==== IÇm new, so IÇm sorry if the question is so easy, I really think that is easy, but I donÇt know how to do it. I have this: b=n-m a=x-b Y=3a + 4a^2 and the program show me this: 3(x-n+m) + 4(x-n+m)^2 or, something like that, the problem is that I want the program show me Y in function of b, or sometimes in function of a, something like this: Y=3(x-b) + 4(x-b)^2 or Y=3a + 4a^2 ==== Forget about procedural programming, i.e., no more Do's ever again. Use MapThread: In[1]:= FuncList={Exp[x],Sin[y],z^3}; VarList={x,y,z}; In[3]:= f[a_,b_]:=Plot[a,{b,-5,5}] In[4]:= MapThread[f,{FuncList,VarList}]; Oh, and try not to use uppercase letters in the first position of your functions and variable names. These are reserved for Mathematica built-in functions. Tomas Garza Mexico City ----- Original Message ----- > VarList = {x, y, z}; Do[Plot[FuncList[LeftDoubleBracket] > i[RightDoubleBracket], {VarList[LeftDoubleBracket] > i[RightDoubleBracket], -5, 5}], {i, 1, Length[VarList]}]] Frank Brand > ==== The solution is using Evaluate. In[1]:= FuncList = {Exp[x], Sin[y], z^3}; In[2]:= VarList = {x, y, z}; In[3]:= Do[Plot[Evaluate[FuncList[[i]], {VarList[[i]], -5, 5}]], {i, 1, Length[VarList]}] Greetings, Germ.87n Buitrago A. ----- Original Message ----- > VarList = {x, y, z}; Do[Plot[FuncList[LeftDoubleBracket] > i[RightDoubleBracket], {VarList[LeftDoubleBracket] > i[RightDoubleBracket], -5, 5}], {i, 1, Length[VarList]}]] Frank Brand > ==== Dear colleagues, how can I solve the following problem: The task is to successively plot the functions given in FuncList with the accociated variable FuncList = {Exp[x], Sin[y], z^3}; VarList = {x, y, z}; Do[Plot[FuncList[LeftDoubleBracket] i[RightDoubleBracket], {VarList[LeftDoubleBracket] i[RightDoubleBracket], -5, 5}], {i, 1, Length[VarList]}]] Frank Brand ==== Try something less clumsy, eg : Plot[#1,{#2,-5,5}]&@@@Transpose[{FuncList,VarList}] or why not have a particular symbol for the independent variable - then just map the plot command over the FuncList : Plot[#,{t,-5,5}]&/@{Exp[t],Sin[t],t^3} bye, Borut | Dear colleagues, | | how can I solve the following problem: The task is to successively plot | the functions given in FuncList with the accociated variable | | | FuncList = {Exp[x], Sin[y], z^3}; | | VarList = {x, y, z}; | | Do[Plot[FuncList[LeftDoubleBracket] | i[RightDoubleBracket], {VarList[LeftDoubleBracket] | i[RightDoubleBracket], -5, 5}], {i, 1, Length[VarList]}]] | | Frank Brand | | | ==== Frank, FuncList = {Exp[x], Sin[y], z^3}; VarList = {x, y, z}; MapThread[Plot[#1, {#2, -5, 5}] &, {FuncList, VarList}] -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay@haystack.demon.co.uk Voice: +44 (0)116 271 4198 > Dear colleagues, how can I solve the following problem: The task is to successively plot > the functions given in FuncList with the accociated variable > FuncList = {Exp[x], Sin[y], z^3}; VarList = {x, y, z}; Do[Plot[FuncList[LeftDoubleBracket] > i[RightDoubleBracket], {VarList[LeftDoubleBracket] > i[RightDoubleBracket], -5, 5}], {i, 1, Length[VarList]}]] Frank Brand > Reply-To: kuska@informatik.uni-leipzig.de ==== FuncList = {Exp[x], Sin[y], z^3}; VarList = {x, y, z}; Plot[Evaluate[#[[1]], {#[[2]], -5, 5}]] & /@ Transpose[{FuncList, VarList}] Jens Dear colleagues, how can I solve the following problem: The task is to successively plot > the functions given in FuncList with the accociated variable FuncList = {Exp[x], Sin[y], z^3}; VarList = {x, y, z}; Do[Plot[FuncList[LeftDoubleBracket] > i[RightDoubleBracket], {VarList[LeftDoubleBracket] > i[RightDoubleBracket], -5, 5}], {i, 1, Length[VarList]}]] Frank Brand ==== Greetings This problem can be solved by conventional programming, but I wonder if there is an elegant Mathematica solution ? A list contains pairs of values, with each pair representing the lower and upper edge of a sub-range. Some of the sub-ranges partially overlap, some fully overlap, others don't overlap at all. The problem is to produce a second list that contains the overall upper and lower edges of the overlapping sub-ranges. A simple example : {{100,200},{150,250},{120,270},{300,400}} would result in {{100,270},{300,400}}. In the real case, the input list has several hundred elements and the output list typically has five elements. I have a working solution based on loops, but there must be a more elegant one. I would be very grateful for any suggestions. John Leary ==== John The simplest solution to your trial problem that I could come up with was the following: Apply[List, Apply[Interval, {{100, 200}, {150, 250}, {120, 270}, {300, 400}}]] I hope you can figure it out. I didn't test it on a larger input dataset - my usual experience is that built-in functions are faster for operations on lists than fiddling around with loops. Mark Westwood Greetings This problem can be solved by conventional programming, but I wonder if > there is an elegant Mathematica solution ? A list contains pairs of values, with each pair representing the lower and > upper edge of a sub-range. Some of the sub-ranges partially overlap, some > fully overlap, others don't overlap at all. The problem is to produce a > second list that contains the overall upper and lower edges of the > overlapping sub-ranges. A simple example : {{100,200},{150,250},{120,270},{300,400}} would result > in {{100,270},{300,400}}. In the real case, the input list has several hundred elements and the > output list typically has five elements. I have a working solution based on loops, but there must be a more elegant > one. I would be very grateful for any suggestions. > John Leary ==== I remember there being a similar thread on this subject a while ago, but I am too lazy to check it out. The simplest solution seems to be using Interval, although it may not be very fast. For example In[3]:= Interval[{100,200},{150,250},{120,270},{300,400}] Out[3]= Interval[{100, 270}, {300, 400}] If the above is not sufficiently fast for you, then you may want to search the archives for the thread I mentioned above. Carl Woll Physics Dept U of Washington > Greetings This problem can be solved by conventional programming, but I wonder if > there is an elegant Mathematica solution ? A list contains pairs of values, with each pair representing the lower and > upper edge of a sub-range. Some of the sub-ranges partially overlap, some > fully overlap, others don't overlap at all. The problem is to produce a > second list that contains the overall upper and lower edges of the > overlapping sub-ranges. A simple example : {{100,200},{150,250},{120,270},{300,400}} would result > in {{100,270},{300,400}}. In the real case, the input list has several hundred elements and the > output list typically has five elements. I have a working solution based on loops, but there must be a more elegant > one. I would be very grateful for any suggestions. John Leary ==== >Greetings This problem can be solved by conventional programming, but I wonder if >there is an elegant Mathematica solution ? A list contains pairs of values, with each pair representing the lower and >upper edge of a sub-range. Some of the sub-ranges partially overlap, some >fully overlap, others don't overlap at all. The problem is to produce a >second list that contains the overall upper and lower edges of the >overlapping sub-ranges. A simple example : {{100,200},{150,250},{120,270},{300,400}} would result >in {{100,270},{300,400}}. In the real case, the input list has several hundred elements and the >output list typically has five elements. I have a working solution based on loops, but there must be a more elegant >one. I would be very grateful for any suggestions. Block[{data = {{100,200},{150,250},{120,270}, {300,400}}}, data = Sort[data,#[[1]]<#2[[1]]&]; {{data[[1,1]], Fold[If[#<#2[[1]],#,Max[#,#2[[2]]]]&, data[[1,2]],Rest[data]]}, {Fold[If[#>#2[[2]],#,Min[#,#2[[1]]]]&, (data=Reverse@data)[[1,1]],Rest[data]], data[[1,2]]}}] --> {{100, 270}, {300, 400}} I haven't tested too extensively. The so-called Mathematica way is illustrated by the use of Fold function to process data, without which you must resort to conventional looping. DH ==== John, The solution using Split that I previously supplied may unnecessarily sort more than once -- here is a correction -- it should be slightly quicker.. lst = Table[{#, # + Random[Integer, {0, 9}]} &[ Random[Integer, {0, 1000}]], {1000}]; FixedPoint[{Min[#],Max[#]}&/@ Split[Sort[#], #1[[2]][GreaterEqual]#2[[1]]&]&,lst]//Timing {0.22 Second,{{1,64},{66,66},{67,175},{177,363},{365,548},{551,853},{857, 857},{858,938},{940,1003}}} FixedPoint[{Min[#],Max[#]}&/@Split[#, #1[[2]][GreaterEqual]#2[[1]]&]&, Sort[lst]]//Timing {0.16 Second,{{1,64},{66,66},{67,175},{177,363},{365,548},{551,853},{857, 857},{858,938},{940,1003}}} Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay@haystack.demon.co.uk Voice: +44 (0)116 271 4198 > Greetings This problem can be solved by conventional programming, but I wonder if > there is an elegant Mathematica solution ? A list contains pairs of values, with each pair representing the lower and > upper edge of a sub-range. Some of the sub-ranges partially overlap, some > fully overlap, others don't overlap at all. The problem is to produce a > second list that contains the overall upper and lower edges of the > overlapping sub-ranges. A simple example : {{100,200},{150,250},{120,270},{300,400}} would result > in {{100,270},{300,400}}. In the real case, the input list has several hundred elements and the > output list typically has five elements. I have a working solution based on loops, but there must be a more elegant > one. I would be very grateful for any suggestions. John Leary ==== ............. > A list contains pairs of values, with each pair representing the lower and > upper edge of a sub-range. Some of the sub-ranges partially overlap, some > fully overlap, others don't overlap at all. The problem is to produce a > second list that contains the overall upper and lower edges of the > overlapping sub-ranges. A simple example : {{100,200},{150,250},{120,270},{300,400}} would result > in {{100,270},{300,400}}. John, Generate a list of pairs: lst=Table[{#,#+Random[Integer,{0,9}]}&[Random[Integer,{0,1000}]],{1000}]; A slow solution Sort[lst]//. {x___,{a_,b_},{c_,d_},y___}/;c<=b:>{x,{a,Max[b,d]},y}//Timing {5. Second,{{0,219},{221,431},{432,568},{569,599},{600,697},{699,1005}}} A faster one FixedPoint[{Min[#],Max[#]}&/@Split[Sort[#], #1[[2]]>=#2[[1]]&]&, lst]//Timing {0.22 Second,{{0,219},{221,431},{432,568},{569,599},{600,697},{699,1005}}} -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay@haystack.demon.co.uk Voice: +44 (0)116 271 4198 ==== >This problem can be solved by conventional programming, but I wonder if >there is an elegant Mathematica solution ? >A list contains pairs of values, with each pair representing the lower and >upper edge of a sub-range. Some of the sub-ranges partially overlap, some >fully overlap, others don't overlap at all. The problem is to produce a >second list that contains the overall upper and lower edges of the >overlapping sub-ranges. >A simple example : {{100,200},{150,250},{120,270},{300,400}} would result >in {{100,270},{300,400}}. >In the real case, the input list has several hundred elements and the >output list typically has five elements. If the numbers are all integers less than 401, as in your example, then you could start with the list {1,2,3,...400} and compare each number to every pair in your list-of-pairs. If there exists no pair that the given number falls between (inclusive), replace the number with zero. Converting the resulting list to the output you want is an exercise for the reader (clever replacement rules will do it easily). -- Tim Dellinger www.ews.uiuc.edu/~tdelling tdelling@uiuc.edu ==== Interval[{100, 200}, {150, 250}, {120, 270}, {300, 400}] does what you want. -- Steve Luttrell West Malvern, UK > Greetings This problem can be solved by conventional programming, but I wonder if > there is an elegant Mathematica solution ? A list contains pairs of values, with each pair representing the lower and > upper edge of a sub-range. Some of the sub-ranges partially overlap, some > fully overlap, others don't overlap at all. The problem is to produce a > second list that contains the overall upper and lower edges of the > overlapping sub-ranges. A simple example : {{100,200},{150,250},{120,270},{300,400}} would result > in {{100,270},{300,400}}. In the real case, the input list has several hundred elements and the > output list typically has five elements. I have a working solution based on loops, but there must be a more elegant > one. I would be very grateful for any suggestions. John Leary ==== Mathematica Training Course Whether you.89re a beginner or seasoned professional, our training services can help you improve your Mathematica skills. We offer public and private training. Mathematica Intermediate & Programming Course (2 days) ----------------- Amsterdam, December 19-20 Mathematica is an exhaustive, powerful, and user-friendly software package. It is easy to perform basic calculations right way, but when you really want to explore and use the real power of Mathematica an investment is necessary. 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As I work in the fields of finance and economics, where we feel > ourselves blessed if we get three digits of accuracy, I'm curious as to what > scientific endeavors require 50+ digits of precision? As I recall there are > some areas, such as high energy physics and some elements of astronomy, that > might require so many digits in some circumstances. Are there others? > -Mark I'm not sure what scientific endeavors might directly benefit. I can say that alot of computational endeavors need high precision, and of course some of these are used behind the scenes in scientific computation. Below I list a few. (i) Exact computation that uses approximation behind the scenes. Even someting so mundane as integer division can fall into this category. (ii) Finding relations among real or integer values can be done with high precision arithmetic. For example, one can implement LLL lattice reduction via approximate arithmetic. it's cousin, PSLQ, is entirely an approximate arithmetic procedure and at times it requires high precision. Applications of these would include cryptography and the like, hence this lies at least partly in the realm of applied math. (iii) Some algorithms may have low precision input but require higher precision at intermediate steps. An example is the method we use for solving systems of multivariate polynomial equations. You can regard the input as representing a family of problems (parametrized by the range of fuzz implied by the low precision input). Clearly the raising of precision in such circumstances is in some sense artificial, insofar as what we obtain is a solution to a particular member of the family (actually to a narrow subfamily). However we also make some attempt to detect ill conditioning; if the problem is well conditioned then solutions to all members of the family will be reasonably near to the one we obtain. (iv) Some statistical functions may require fairly high precision behind the scenes in order to obtain reasonable results for inputs that are not outrageous. This can often be mollified by changing the algorithm used but sometimes high precision is the simplest way to proceed. (v) Computational geometry problems frequently become nongeneric arithmetic can help to handle cases wherein nongenericity makes the problem pathological (often perturbation or similar devices are also needed). integration to counter various ills such as cancellation error. I'm sure there are oodles of other computational examples wherein high precision saves the day. The moral is that, while scientific examples rarely provide high precison input, methods of computation required by scientists may well still require high precision arithmetic. Also note that while financial forecasting may be blessed to get three digits, other aspects of the financial world require much more. Around 10 years ago a bank investigated purchasing Mathematica. Apparently they wanted to be certain they had amounts figured to better than the nearest penny (or so I heard). When working with exchange rates I suppose this could be important; crude rounding might allow for those weird secrets work. Wanna buy a Euro from me? Daniel Lichtblau Wolfram Research ==== > The more I play with the example the more > depressing it gets. Start > with floating point numbers but explicitly > arbitrary-precision ones. In[1]:= > a=77617.00000000000000000000000000000; > b=33095.00000000000000000000000000000; In[3]:= > !(333.7500000000000000000000000000000 b^6 + > a^2 ((11 a^2 > b^2 - > b^6 - 121 b^4 - 2)) + > 5.500000000000000000000000000000 b^8 + > a/(2 > b)) Out[3]= > !((-4.78339168666055402578083604864320577443814`26.6715*^32)) In[4]:= > Accuracy[%] Out[4]= > -6 Due to the manual section 3.1.6: When you do calculations with arbitrary-precision > numbers, as > discussed in the previous section, Mathematica > always keeps track of > the precision of your results, and gives only > those digits which are > known to be correct, given the precision of your > input. When you do > calculations with machine-precision numbers, > however, Mathematica > always gives you a machine[CapitalEth]precision result, > whether or not all the > digits in the result can, in fact, be determined > to be correct on the > basis of your input. Because I started with arbitrary-precision numbers > Mathematica should display > only those digits that are correct, that is none. No, 26 digits are correct Here is the number: > -0.8273960599468213681 Here is the same number computed by Mathematica with 26 > correct digits: > -4.78339168666055402578083604864320577443814[Times]10^32 It looks like I have been using some wrong definition > of correct.:-) You just proved that Precision is useless as a measure > how good your numerical result is. [...] I rather hope I proved nothing of the sort. Also I'm afraid Mathematica kept better track of the numbers than you did. As for definitions of correct, rather than remark on yours I'll just expose what I meant with specific numerical examples below. You did not actually say why you thought -0.8273960599468213681 would be the appropriate result. So I'll go through the computation in exact arithmetic (I have to admit I am puzzled as to why you did not do this). First I'll rewrite your expression using more variables. a = 77617.00000000000000000000000000000; b = 33095.00000000000000000000000000000; c = 333.7500000000000000000000000000000; d = 5.500000000000000000000000000000; In[6]:= InputForm[val = c*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + d*b^8 + a/(2*b)] Out[6]//InputForm= -4.78339168666055402578083604864320577443814`26.6715*^32 This is what I got. Now we'll redo in exact arithmetic. rata = Rationalize[a]; ratb = Rationalize[b]; ratc = Rationalize[c]; ratd = Rationalize[d]; In[12]:= InputForm[exactval = ratc*ratb^6 + rata^2* (11*rata^2*ratb^2 - ratb^6 -121*ratb^4 - 2) + ratd*ratb^8 + rata/(2*ratb)] Out[12]//InputForm= -63322539148012414193286707611938758031/132380 In[13]:= InputForm[N[exactval]] Out[13]//InputForm= -4.783391686660554*^32 I think it is reasonable to claim that these agree. I dug through some early posts on the topic and came across a value for b at one time that was one larger (this supports my long-held suspicion that numbers slowly decay in discrete increments...). In[14]:= b2 = 33096.00000000000000000000000000000; In[15]:= InputForm[val2 = c*b2^6 + a^2*(11*a^2*b2^2 - b2^6 - 121*b2^4 - 2) + d*b2^8 + a/(2*b2)] Out[15]//InputForm= -0.827469148`-0.3833 What this reveals is a number that Mathematica claims has NO trustworty digits. The InputForm also reveals the untrustworthy digits, and, sure enough, it comes close to what you have called the correct value, and specifically they agree to four places. My opinion is that the Mathematica significance arithmetic is doing, shall I say, precisely as it ought. Daniel Lichtblau Wolfram Research Reply-To: Mark Coleman ==== Greetings, I have read with great interest this lively debate on numerical prcesion and accuracy. As I work in the fields of finance and economics, where we feel ourselves blessed if we get three digits of accuracy, I'm curious as to what scientific endeavors require 50+ digits of precision? As I recall there are some areas, such as high energy physics and some elements of astronomy, that might require so many digits in some circumstances. Are there others? -Mark ==== > Greetings, I have read with great interest this lively debate on numerical prcesion and > accuracy. As I work in the fields of finance and economics, where we feel > ourselves blessed if we get three digits of accuracy, I'm curious as to what > scientific endeavors require 50+ digits of precision? As I recall there are > some areas, such as high energy physics and some elements of astronomy, that > might require so many digits in some circumstances. Are there others? > -Mark Mark, There may be occasions when the outcome of a real process is so sensitive to changes in input that unless we know very precisely what the input is then we can know very little about the outcome - chaotic processes are of this kind. The difficulty is real and no amount of computer power or clever progamming will do much about it. Another situation is when the the process is not so sensitive but calculating with our formula or programme introduces accumulates significant errors. Here is a very artificial example of the latter (I time the computation and find the MaximumMemory used in the session as we go through the example): ser=Normal[Series[Cos[#],{#,0,200}]]; MaxMemoryUsed[] 1714248 Calculating with machine number does not show much of a pattern ( I have deleted the graphics - please evaluate the code), pts= With[{ss=ser},Table[ {#,ss}&[x], {x,50.,70., .1}]];//Timing ListPlot[pts, PlotJoined->True]; MaxMemoryUsed[] {5.11 Second,Null} 1723840 Using bigfloat inputs with precision 20 shows some pattern: pts= With[{ss=ser},Table[ {#,ss}&[SetPrecision[x,20]], {x,50.,70., .1}]];//Timing ListPlot[pts, PlotJoined->True,PlotRange[Rule]All]; MaxMemoryUsed[] {17.52 Second,Null} 1759664 Precision 40 does very well: pts= With[{ss=ser},Table[ {#,ss}&[SetPrecision[x,40]], {x,50.,70., .1}]];//Timing ListPlot[pts, PlotJoined->True,PlotRange[Rule]All]; MaxMemoryUsed[] {19.38 Second,Null} 1797072 Now we might think the correct outcomes are showing up, and use an interpolating function for further , and faster, calculation. f=Interpolation[pts] InterpolatingFunction[{{50.000000,70.00000}},<>] pts= Table[ f[x],{x,50, 70, .1}];//Timing ListPlot[pts, PlotJoined->True,PlotRange[Rule]All]; MaxMemoryUsed[] {0.33 Second,Null} As a matter of interest, this is what happens if we substitute exact numbers (rationals and integers) for reals-- the computation takes an excessively long time and quite a bit more memory. pts= With[{ss=ser},Table[ {#,ss}&[SetPrecision[x,Infinity]], {x,50.,70., .1}]];//Timing ListPlot[pts, PlotJoined->True,PlotRange[Rule]All]; MaxMemoryUsed[] {992.28 Second,Null} 2413808 This also shows that we may in fact want to replace exact inputs with bigfloats. I should be interested to hear of other example, really real one in particular. I imagine that there are many situations where trends and shapes are more important than specific values. -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay@haystack.demon.co.uk Voice: +44 (0)116 271 4198 > Greetings, I have read with great interest this lively debate on numerical prcesion > and > accuracy. As I work in the fields of finance and economics, where we feel > ourselves blessed if we get three digits of accuracy, I'm curious as to > what > scientific endeavors require 50+ digits of precision? As I recall there > are > some areas, such as high energy physics and some elements of astronomy, > that > might require so many digits in some circumstances. Are there others? > -Mark . ==== > [...] This subthread gets difficult to follow with all the indenting and the like so I will edit a bit for clarity. Here is some input. f = SetPrecision[333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + 5.5*b^8 + a/(2*b), 100]; One question that arose is, as best I can phrase it, How might I set values for a and b and retain 100 digits of precision (or perhaps accuracy)? There are a few answers and which you like will depend on what you really want to do. (i) You can use a = SetPrecision[77617.,100]; b = SetPrecision[33096.,100]; As has been pointed out a few times, this will give a number with 100 digits of precision. But that number will not resemble: Clear[a,b] a = SetPrecision[77617,100]; b = SetPrecision[33096,100]; 33375/100*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + 11/2*b^8 + a/(2*b) or any of the equivalent variations that have been presented in this thread. (ii) You can make input exact before doing any approximate arithmetic wherein canellation error might arise. Using Rationalize, I show this below. Clear[a,b] f = 333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + 5.5*b^8 + a/(2*b) f = Rationalize[f] In[39]:= Precision[N[f /. {a->77617,b->33096}, 100]] Out[39]= 100. (iii) You cannot do what you want because you work with low precision input and want a high precision result. > Assume that I want accuracy and precision of 100 for > f. You advice me to make experiments to find out, what > should be the initial precision and accuracy of a and > b to reach the requested accuracy and precision for f. > Notice, that you cannot just repeat I[26], we saw > already what happens. I have to re-type I[24], I[25], > I[26], I[27], I[28], and I[29] as many times as needed > to get f with accuracy and precision 100. I am curious to know how exactly you defined I[...]. It involves a protected symbol. In[54]:= I[a_] := a^2 SetDelayed::write: Tag Complex in I[a_] is Protected. Out[54]= $Failed As for obtaining requested accuracy or precision, I gave no such advice. If you begin with exact input this can be done via N as per response (ii) above. I did not do any experimenting but simply directed N[] to find the result to 100 digits precision. > Dan, you simply advocate to do MANUAL WORK that should > be done by machine. See my above remark. > Let's suppose that in the above example I just want 60 > digits not 61. Precisely, I want 60 digits and nothing > or zeros afterwards. Let's see if I could use > SetAccuracy. In[30]:= > SetAccuracy[%, 60] Out[30]= > -0.82739605994682136814116509547981629199903311578438481991781 In[31]:= > % // FullForm Out[30]//FullForm= > -0.827396059946821368141165095479816291999033115784384819917814841672467988` > 59.9177 Oops, it did not work (as expected). Actually it did. You have 59.9177 digits of precision. If you check you will find that you have 60. digits of accuracy as you had requested. > [...] > Dan, is there any simple way to get what I want? If what you wanted was a number with 60 digits accuracy (which certainly was what you requested), then indeed you got it. > As I repeated already number of times, at this stage > of the development of computer technology, software > should do it for me (!). We both know that this is > doable. Some of the textbooks that you just advised me > to read describe it. As a developer of Mathematica, > tell us why do you consider this to be a bad idea? Peter Kosta First I must request references since I am not certain what exactly you have in mind that the software will do. Then I'll comment on what, as best I can assess, you seem to want the software to do. Based on prior notes from you in this thread, it appears that you want it to treat a number such as 1.2 as an exact entity 5/6. This can be done by exact methods, e.g. preprocessing so that all numbers get rationalized. It cannot be done by our numerical engine (or any other), as that will not rationalize for you. It can also be done by you working with exact input. This would be a simple expedient, but quite effective. I would consider it to be a terrible idea to automatically rationalize every approximate number input and then to work with exact arithmetic. While I am not certain this is what you advocate, it is the only interpretation I can find with would allow for the sort of result you seem to expect. It is a terrible idea because it bypasses solid numerical methods that have been developed over several decades for handling computations accurately and reliably (subject to appropriate input!), and in reasonable time. It would entirely disable any functionality for which no exact methods are known, e.g. solving many ODEs, optimization, and the like. I could go on for a while but I my four typing fingers will get sore. I'll just finish by noting that it is a terrible idea because it would punish all users of numeric computation in Mathematica. I fail to see any useful purpose in that. Mind you, while I regard it as a terrible idea, I am still not certain that this was in fact what you propose. Again, it would be helpful to software) so I can see what it is you really want. Daniel Lichtblau Wolfram Research ==== > Here's an even more extreme result: f = SetAccuracy[333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - > 121*b^4 - 2) + 5.5*b^8 + a/(2*b), 50]; > a = 77617.; b = 33096.; > f > Precision[f] -1.180591620717411303424`71.0721*^21 > 71 71.0721 digits of precision? I don't think so!! Either I am it altogether or you are just simply beating to death the point that in case of machine arithmetic (only!) Precision and Accuracy are purely formal and essentially meaningless. One can argue whether in this case there is any point of returning any value for Precision, or Accuracy (like 71 above, or -5 for Accuracy in the example that fooled me), but it's not a big deal and it most certainly does not make SetPrecision meaningless. On the contrary, SetPrecision is very useful and in fact it is SetPrecision that can tell you that the answer above is meaningless: In[8]:= f = SetAccuracy[333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + 5.5*b^8 + a/(2*b), 50]; a=SetPrecision[77617.,$MachinePrecision]; b = SetPrecision[ 33096.,$MachinePrecision]; In[10]:= {f,Precision[f]} Out[10]= {1.19801754103509`0*^19, 0} I would say this is correct and show that SetPrecision is very useful indeed. It tells you (what of course you ought to already know in this case anyway) that machine precision will not give you a realiable answer in this case. If you know your numbers with a great deal of accuracy you can get an accurate answer: In[24]:= f = SetAccuracy[333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + 5.5*b^8 + a/(2*b), 100]; a=SetPrecision[77617.,100]; b = SetPrecision[33096.,100]; In[26]:= {f, Precision[f]} Out[26]= {-0.82739605994682136814116509547981629199903311578438481991 781484167246798617832`61.2597, 61} Again you can be pretty sure that you got an accurate answer, provided of course your original setting of precision was valid. Honestly, to say that SetPrecision and SetAccuaracy are useless is one of the silliest thing I have read on this list in years. > Andrzej Kozlowski Yokohama, Japan http://www.mimuw.edu.pl/~akoz/ http://platon.c.u-tokyo.ac.jp/andrzej/ ==== >[...] I would say this is correct and show that SetPrecision is very useful > indeed. It tells you (what of course you ought to already know in this > case anyway) that machine precision will not give you a realiable > answer in this case. If you know your numbers with a great deal of > accuracy you can get an accurate answer: In[24]:= > f = SetAccuracy[333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - > 121*b^4 - 2) + 5.5*b^8 + a/(2*b), 100]; > a=SetPrecision[77617.,100]; b = SetPrecision[33096.,100]; > In[26]:= > {f, Precision[f]} Out[26]= > {-0.82739605994682136814116509547981629199903311578438481991 > 781484167246798617832`61.2597, 61} > Congratulations! You just requested accuracy of 100 for f and got 61 ( > to convince yourself add Accuracy[f] to In[26]). If In[24] one > replaces SetAccuracy by SetPrecision the result is similar. PK > [...] One has (initially) an accuracy of 100 for an expression that contains variables. In[25]:= Clear[a,b,f] In[26]:= f = SetAccuracy[333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + 5.5*b^8 + a/(2*b), 100]; In[27]:= Accuracy[f] Out[27]= 100. Now we assign values to some indeterminants in f. In[28]:= a = SetPrecision[77617.,100]; b = SetPrecision[33096.,100]; In[29]:= {f, Precision[f], Accuracy[f]} Out[29]= {-0.8273960599468213681411650954798162919990331157843848199178148, 61.2599, 61.3422} The precision and accuracy has dropped. This is all according to standard numerical analysis regarding cancellation error. You'll find it in any textbook on the topic. As for what happens when you artificially raise precision (or accuracy) of machine numbers far beyond that guaranteed by their internal representation, that falls into to category of garbage in, garbage out. It is, howoever, valid to use SetPrecision to raise precision in (typically iterative) algorithms where significance arithmetic might be unduly pessimistic due to incorrect assumptions about uncorollatedness of numerical error. Examples of such usage have appeared in this news group. Daniel Lichtblau Wolfram Research ==== [...] I would say this is correct and show that SetPrecision is very useful > indeed. It tells you (what of course you ought to already know in this > case anyway) that machine precision will not give you a realiable > answer in this case. If you know your numbers with a great deal of > accuracy you can get an accurate answer: In[24]:= > f = SetAccuracy[333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - > 121*b^4 - 2) + 5.5*b^8 + a/(2*b), 100]; > a=SetPrecision[77617.,100]; b = SetPrecision[33096.,100]; > In[26]:= > {f, Precision[f]} Out[26]= > {-0.82739605994682136814116509547981629199903311578438481991 > 781484167246798617832`61.2597, 61} > Congratulations! You just requested accuracy of 100 for f and got 61 ( to convince yourself add Accuracy[f] to In[26]). If In[24] one replaces SetAccuracy by SetPrecision the result is similar. PK > Again you can be pretty sure that you got an accurate answer, provided > of course your original setting of precision was valid. Honestly, to say that SetPrecision and SetAccuaracy are useless is one > of the silliest thing I have read on this list in years. Andrzej Kozlowski > Yokohama, Japan > http://www.mimuw.edu.pl/~akoz/ > http://platon.c.u-tokyo.ac.jp/andrzej/ ==== I've been trying to use PlotVectorField for the following differential equation: dy/dt = 0.08*y*(1-y/1000) but I haven't been successful yet. I tried to do the following: f[t_, y_] := {1, 0.08*y*(1 - y/1000)} < 1]; -- Steve Luttrell West Malvern, UK > I've been trying to use PlotVectorField for the > following differential equation: > dy/dt = 0.08*y*(1-y/1000) > but I haven't been successful yet. > I tried to do the following: > f[t_, y_] := {1, 0.08*y*(1 - y/1000)} > < PlotVectorField[f[t, y], {t, 0, 80}, {y, 0, 1400}]; > but I'm getting a meaningless plot so I'd appreciate > if someone could tell me what is what I'm doing wrong. > Ruben __________________________________________________ > Do you Yahoo!? > http://faith.yahoo.com > ==== Dear group, I have the following question regarding a lengthy calculation using Mathematica: For given w points in x direction and h points in y direction, I can construct all the points using h=10; w=8; points=Flatten[Transpose[Outer[List,Range[w],Range[h]]],1] Next, I need to find all the possible pairs of point including points themselves, i.e., pair AA. I can use pairs=Outer[List,points,points,1] Then, I have to clear those pairs that repeat themselves, i.e., pair AB and pair BA. Also, when w and h are of the order of 1000s, the computation takes a very long time. Is there a better way of doing the second part of Sincerely Cheng ==== ================================================ Cheng Liu, Ph.D. MST-8, Structure/Property Relations Materials Science and Technology Division Los Alamos National Laboratory Los Alamos, New Mexico 87545 ==== ================================================ ==== Cheng, If you have h and w on the order of 1000, then your points list will have 10^6 points in it. If you then want to have a list of every possible pair of points, that list will consist of 10^12 pairs. Each pair of points consists of 4 integers, so that means your pairs list will have 4 10^12 integers in it. Even if Mathematica could store each integer using 4 bytes, that would require over 10^13 bytes, or 10000 gigabytes of storage. In order for Mathematica to function efficiently, the above storage must be in memory and not in virtual memory on the hard drive. In other words, creating such a pairs list is impossible at the present time, and is probably impossible for the forseeable future. If you truly need such a pairs list, then you may be able to work with h and w on the order of 100 if you have a large amount of memory. What in the world are you trying to do? I doubt that creating such a pairs list is necessary for you to accomplish whatever it is you are trying to do. Carl Woll Physics Dept U of Washington > Dear group, I have the following question regarding a lengthy calculation > using Mathematica: For given w points in x direction and h points in y direction, I can > construct all the points using h=10; w=8; > points=Flatten[Transpose[Outer[List,Range[w],Range[h]]],1] Next, I need to find all the possible pairs of point including points > themselves, i.e., pair AA. I can use pairs=Outer[List,points,points,1] Then, I have to clear those pairs that repeat themselves, i.e., pair AB and > pair BA. Also, when w and h are of the order of 1000s, the computation > takes a very long time. Is there a better way of doing the second part of Sincerely Cheng > ==== ================================================ > Cheng Liu, Ph.D. > MST-8, Structure/Property Relations > Materials Science and Technology Division > Los Alamos National Laboratory > Los Alamos, New Mexico 87545 ==== ================================================ ==== Here is a way with symbolic lists. (* Here are some sample lists*) lst1={A,B}; lst2={a,b,c}; (*Join into one list*) list1=Join[lst1,lst2]; (*Do the outer product to get all possible ordered pairs*) list2=Partition[Flatten[Outer[List,list1,list1]],2] {{A, A}, {A, B}, {A, a}, {A, b}, {A, c}, {B, A}, {B, B}, {B, a}, {B, b}, {B, c}, {a, A}, {a, B}, {a, a}, {a, b}, {a, c}, {b, A}, {b, B}, {b, a}, {b, b}, {b, c}, {c, A}, {c, B}, {c, a}, {c, b}, {c, c}} (*Turn the pairs into products*) list3=list2/.{x_,y_}¬{x y} {{A^2}, {A*B}, {a*A}, {A*b}, {A*c}, {A*B}, {B^2}, {a*B}, {b*B}, {B*c}, {a*A}, {a*B}, {a^2}, {a*b}, {a*c}, {A*b}, {b*B}, {a*b}, {b^2}, {b*c}, {A*c}, {B*c}, {a*c}, {b*c}, {c^2}} (*Union weeds out repeats*) list4=Union[list3,list3] {{a^2}, {a*A}, {A^2}, {a*b}, {A*b}, {b^2}, {a*B}, {A*B}, {b*B}, {B^2}, {a*c}, {A*c}, {b*c}, {B*c}, {c^2}} (*Now turn the products back into pairs*) (*This is the step the requires symbols*) list5=list4/.{x_^2}¬{x,x}/.{x_*y_}¬{x,y} {{a, a}, {a, A}, {A, A}, {a, b}, {A, b}, {b, b}, {a, B}, {A, B}, {b, B}, {B, B}, {a, c}, {A, c}, {b, c}, {B, c}, {c, c}} (*List it out*) (*If you have numbers, you can now use a Rule to replace*) Sort[lst]//TableForm -- Kevin J. McCann Joint Center for Earth Systems Technology (JCET) Department of Physics UMBC Baltimore MD 21250 > Dear group, I have the following question regarding a lengthy calculation > using Mathematica: For given w points in x direction and h points in y direction, I can > construct all the points using h=10; w=8; > points=Flatten[Transpose[Outer[List,Range[w],Range[h]]],1] Next, I need to find all the possible pairs of point including points > themselves, i.e., pair AA. I can use pairs=Outer[List,points,points,1] Then, I have to clear those pairs that repeat themselves, i.e., pair AB and > pair BA. Also, when w and h are of the order of 1000s, the computation > takes a very long time. Is there a better way of doing the second part of Sincerely Cheng > ==== ================================================ > Cheng Liu, Ph.D. > MST-8, Structure/Property Relations > Materials Science and Technology Division > Los Alamos National Laboratory > Los Alamos, New Mexico 87545 ==== ================================================ ==== Oops! I didn't read the question properly. I hope I have got it right this time. Union[Map[Sort, pairs]] does what you want. -- Steve Luttrell West Malvern, UK > Dear group, I have the following question regarding a lengthy calculation > using Mathematica: For given w points in x direction and h points in y direction, I can > construct all the points using h=10; w=8; > points=Flatten[Transpose[Outer[List,Range[w],Range[h]]],1] Next, I need to find all the possible pairs of point including points > themselves, i.e., pair AA. I can use pairs=Outer[List,points,points,1] Then, I have to clear those pairs that repeat themselves, i.e., pair AB and > pair BA. Also, when w and h are of the order of 1000s, the computation > takes a very long time. Is there a better way of doing the second part of Sincerely Cheng > ==== ================================================ > Cheng Liu, Ph.D. > MST-8, Structure/Property Relations > Materials Science and Technology Division > Los Alamos National Laboratory > Los Alamos, New Mexico 87545 ==== ================================================ ==== Cases[pairs, _?(#[[1]] != #[[2]] &)] does what you want. -- Steve Luttrell West Malvern, UK > Dear group, I have the following question regarding a lengthy calculation > using Mathematica: For given w points in x direction and h points in y direction, I can > construct all the points using h=10; w=8; > points=Flatten[Transpose[Outer[List,Range[w],Range[h]]],1] Next, I need to find all the possible pairs of point including points > themselves, i.e., pair AA. I can use pairs=Outer[List,points,points,1] Then, I have to clear those pairs that repeat themselves, i.e., pair AB and > pair BA. Also, when w and h are of the order of 1000s, the computation > takes a very long time. Is there a better way of doing the second part of Sincerely Cheng > ==== ================================================ > Cheng Liu, Ph.D. > MST-8, Structure/Property Relations > Materials Science and Technology Division > Los Alamos National Laboratory > Los Alamos, New Mexico 87545 ==== ================================================ Reply-To: ==== If we have inaccurately known parameters, I think Interval arithmetic does a far better job of assessing the situation. As for impossible demands on memory and time, the computation that took 992.3 seconds for you took 32.8 seconds for me. Anyway, it can be done faster AND more accurately without bignums: Timing[pts = With[{ss = ser}, Table[({#1, ss} & )[x], {x, 50, 70, 1/10}]]; ] ListPlot[pts, PlotJoined -> True, PlotRange -> All]; MaxMemoryUsed[] {10.640999999999998*Second, Null} In any case, we spent far more time writing code and evaluating results than waiting on execution. If anything, your examples suggest only that machine precision AND bignum computations are suspect. The results may or may not be worth the pixels they take up on my screen, and unless I compute in some alternative way instead -- or use progressively more digits in bignums until things settle down -- I can only guess at their reliability. For an application such as your example, I think the best solution is to use infinite precision for a limited number of points, and then Interpolation. It's safer than using SetPrecision because it doesn't involve guessing how many digits of precision to use, and it's far faster because it doesn't involve testing higher and higher levels of precision. The choice of points for exact computation may be tricky, but there are adaptive algorithms for that. Here's an interesting way to proceed, for instance: ser = Normal[Series[Cos[x], {x, 0, 200}]]; Timing[pts = Table[{x, ser}, {x, 50, 70, 1/2}];] f = Interpolation[pts]; Timing[plot1 = Plot[f[x], {x, 50, 70}, PlotPoints -> 30, PlotDivision -> 3];] Cases[plot1, Line[a__] -> a, Infinity][[1, All, 1]]; Timing[newPts = Union[pts, ({x, ser} /. x -> #) & /@ (Rationalize[#, 1/100] & /@ %)];] g = Interpolation[newPts, InterpolationOrder -> 5]; plot1 = Plot[Cos[x] - g[x], {x, 50, 70}, PlotRange -> All]; {1.703000000000003*Second, Null} {0.1560000000000059*Second, Null} {4.968999999999994*Second, Null} {0.546999999999997*Second, Null} Length[pts] Length[newPts] 41 124 I used only a few points for the first plot and it already looked good. Just to be sure, I used Plot to select more points, and used infinite precision computation again for those points. The final Plot shows error limited to about 10^-6. Increasing InterpolationOrder decreases errors significantly, too, at fairly small cost. Bobby -----Original Message----- >However, if the coefficients and powers of your example series were not > perfectly known, what then? We need to distinguish between having an exact value which is imperfectly known and not having an exact value - having a range of values. If I may speculate a little more on real uses: - If we have inaccurately known parameters that do have definite values we may still want to calculae accurately over possible ranges of the parameter; and if the definite values give distinctive outcomes then testing with high accuracy inputs is a way of getting a more accurate determination of the real value - rather like using an inverse function. - if parameters do not have a definite value then we are into statistics, however we might still need to know the outcomes of inputing accurate values to get an idea of the behaviour of the process. Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay@haystack.demon.co.uk Voice: +44 (0)116 271 4198 ----- Original Message ----- > adding arbitrary digits serves no purpose. Yes, plots may get smoother > as more digits are added, but they would not converge to a correct > result -- merely to a precise one. (In the chemistry industry where my wife works, the difference between > accuracy and precision is well known. Precision means getting the same > answer over and over --- whether it's right or not. Accuracy means > getting the right answer --- whether it's precise or not. It's low > variance versus small bias.) Modify your example like this: ser = N@Normal[Series[Cos[#], {#, 0, 200}]]; > Timing[pts = With[{ss = > ser}, Table[SetPrecision[{#, ss}, 80] &@x, {x, 50., 70., .1}]];] > ListPlot[pts, PlotJoined -> True, PlotRange -> All]; > MaxMemoryUsed[] Once the series coefficients have lost precision, you can't get it back > again. Furthermore, in using SetPrecision, there's a danger that one > could THINK he has regained it. Bobby -----Original Message----- > Sent: Tuesday, October 15, 2002 3:18 AM > To: mathgroup@smc.vnet.net > Greetings, I have read with great interest this lively debate on numerical > prcesion > and > accuracy. As I work in the fields of finance and economics, where we > feel > ourselves blessed if we get three digits of accuracy, I'm curious as > to > what > scientific endeavors require 50+ digits of precision? As I recall > there > are > some areas, such as high energy physics and some elements of > astronomy, > that > might require so many digits in some circumstances. Are there others? > -Mark > Mark, There may be occasions when the outcome of a real process is so > sensitive > to changes in input that unless we know very precisely what the input is > then we can know very little about the outcome - chaotic processes are > of > this kind. The difficulty is real and no amount of computer power or > clever > progamming will do much about it. Another situation is when the the process is not so sensitive but > calculating with our formula or programme introduces accumulates > significant > errors. Here is a very artificial example of the latter (I time the computation > and > find the MaximumMemory used in the session as we go through the > example): ser=Normal[Series[Cos[#],{#,0,200}]]; MaxMemoryUsed[] 1714248 Calculating with machine number does not show much of a pattern ( I > have > deleted the graphics - please evaluate the code), > pts= With[{ss=ser},Table[ {#,ss}&[x], > {x,50.,70., .1}]];//Timing > ListPlot[pts, PlotJoined->True]; > MaxMemoryUsed[] {5.11 Second,Null} 1723840 Using bigfloat inputs with precision 20 shows some pattern: pts= With[{ss=ser},Table[ {#,ss}&[SetPrecision[x,20]], > {x,50.,70., .1}]];//Timing > ListPlot[pts, PlotJoined->True,PlotRange[Rule]All]; > MaxMemoryUsed[] {17.52 Second,Null} 1759664 > Precision 40 does very well: pts= With[{ss=ser},Table[ {#,ss}&[SetPrecision[x,40]], > {x,50.,70., .1}]];//Timing > ListPlot[pts, PlotJoined->True,PlotRange[Rule]All]; > MaxMemoryUsed[] {19.38 Second,Null} 1797072 Now we might think the correct outcomes are showing up, and use an > interpolating function for further , and faster, calculation. f=Interpolation[pts] InterpolatingFunction[{{50.000000,70.00000}},<>] pts= Table[ f[x],{x,50, 70, .1}];//Timing > ListPlot[pts, PlotJoined->True,PlotRange[Rule]All]; > MaxMemoryUsed[] {0.33 Second,Null} > As a matter of interest, this is what happens if we substitute exact > numbers > (rationals and integers) for reals-- > the computation takes an excessively long time and quite a bit more > memory. pts= With[{ss=ser},Table[ {#,ss}&[SetPrecision[x,Infinity]], > {x,50.,70., .1}]];//Timing > ListPlot[pts, PlotJoined->True,PlotRange[Rule]All]; > MaxMemoryUsed[] {992.28 Second,Null} 2413808 This also shows that we may in fact want to replace exact inputs with > bigfloats. > I should be interested to hear of other example, really real one in > particular. I imagine that there are many situations where trends and > shapes > are more important than specific values. -- > Allan --------------------- > Allan Hayes > Mathematica Training and Consulting > Leicester UK > www.haystack.demon.co.uk > hay@haystack.demon.co.uk > Voice: +44 (0)116 271 4198 > ==== Greetings, Is anyone aware of Mathematica code implementing statistical cluster analys= is, e.g., k-means method, etc.? -Mark ==== Bobby, You rightly point out that care should be exercised when using (high precision) bigfloats, but this should not obscure the proper use of them. I have suggested some uses that are valid subject to circumstances (raising precision) or essential (converting exact numbers to bigfloats to avoid impossible demands on memory and time) - Daniel Lichtblau gave others. >However, if the coefficients and powers of your example series were not > perfectly known, what then? We need to distinguish between having an exact value which is imperfectly known and not having an exact value - having a range of values. If I may speculate a little more on real uses: - If we have inaccurately known parameters that do have definite values we may still want to calculae accurately over possible ranges of the parameter; and if the definite values give distinctive outcomes then testing with high accuracy inputs is a way of getting a more accurate determination of the real value - rather like using an inverse function. - if parameters do not have a definite value then we are into statistics, however we might still need to know the outcomes of inputing accurate values to get an idea of the behaviour of the process. Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay@haystack.demon.co.uk Voice: +44 (0)116 271 4198 ----- Original Message ----- > result -- merely to a precise one. (In the chemistry industry where my wife works, the difference between > accuracy and precision is well known. Precision means getting the same > answer over and over --- whether it's right or not. Accuracy means > getting the right answer --- whether it's precise or not. It's low > variance versus small bias.) Modify your example like this: ser = N@Normal[Series[Cos[#], {#, 0, 200}]]; > Timing[pts = With[{ss = > ser}, Table[SetPrecision[{#, ss}, 80] &@x, {x, 50., 70., .1}]];] > ListPlot[pts, PlotJoined -> True, PlotRange -> All]; > MaxMemoryUsed[] Once the series coefficients have lost precision, you can't get it back > again. Furthermore, in using SetPrecision, there's a danger that one > could THINK he has regained it. Bobby -----Original Message----- > Sent: Tuesday, October 15, 2002 3:18 AM > To: mathgroup@smc.vnet.net > Greetings, I have read with great interest this lively debate on numerical > prcesion > and > accuracy. As I work in the fields of finance and economics, where we > feel > ourselves blessed if we get three digits of accuracy, I'm curious as > to > what > scientific endeavors require 50+ digits of precision? As I recall > there > are > some areas, such as high energy physics and some elements of > astronomy, > that > might require so many digits in some circumstances. Are there others? > -Mark > Mark, There may be occasions when the outcome of a real process is so > sensitive > to changes in input that unless we know very precisely what the input is > then we can know very little about the outcome - chaotic processes are > of > this kind. The difficulty is real and no amount of computer power or > clever > progamming will do much about it. Another situation is when the the process is not so sensitive but > calculating with our formula or programme introduces accumulates > significant > errors. Here is a very artificial example of the latter (I time the computation > and > find the MaximumMemory used in the session as we go through the > example): ser=Normal[Series[Cos[#],{#,0,200}]]; MaxMemoryUsed[] 1714248 Calculating with machine number does not show much of a pattern ( I > have > deleted the graphics - please evaluate the code), > pts= With[{ss=ser},Table[ {#,ss}&[x], > {x,50.,70., .1}]];//Timing > ListPlot[pts, PlotJoined->True]; > MaxMemoryUsed[] {5.11 Second,Null} 1723840 Using bigfloat inputs with precision 20 shows some pattern: pts= With[{ss=ser},Table[ {#,ss}&[SetPrecision[x,20]], > {x,50.,70., .1}]];//Timing > ListPlot[pts, PlotJoined->True,PlotRange[Rule]All]; > MaxMemoryUsed[] {17.52 Second,Null} 1759664 > Precision 40 does very well: pts= With[{ss=ser},Table[ {#,ss}&[SetPrecision[x,40]], > {x,50.,70., .1}]];//Timing > ListPlot[pts, PlotJoined->True,PlotRange[Rule]All]; > MaxMemoryUsed[] {19.38 Second,Null} 1797072 Now we might think the correct outcomes are showing up, and use an > interpolating function for further , and faster, calculation. f=Interpolation[pts] InterpolatingFunction[{{50.000000,70.00000}},<>] pts= Table[ f[x],{x,50, 70, .1}];//Timing > ListPlot[pts, PlotJoined->True,PlotRange[Rule]All]; > MaxMemoryUsed[] {0.33 Second,Null} > As a matter of interest, this is what happens if we substitute exact > numbers > (rationals and integers) for reals-- > the computation takes an excessively long time and quite a bit more > memory. pts= With[{ss=ser},Table[ {#,ss}&[SetPrecision[x,Infinity]], > {x,50.,70., .1}]];//Timing > ListPlot[pts, PlotJoined->True,PlotRange[Rule]All]; > MaxMemoryUsed[] {992.28 Second,Null} 2413808 This also shows that we may in fact want to replace exact inputs with > bigfloats. > I should be interested to hear of other example, really real one in > particular. I imagine that there are many situations where trends and > shapes > are more important than specific values. -- > Allan --------------------- > Allan Hayes > Mathematica Training and Consulting > Leicester UK > www.haystack.demon.co.uk > hay@haystack.demon.co.uk > Voice: +44 (0)116 271 4198 > Reply-To: ==== You're using SetPrecision when infinite precision is a meaningful option -- when there's no doubt about the coefficients and powers in the series. Bignums clearly make the computation faster in that case. However, if the coefficients and powers of your example series were not perfectly known, what then? If they begin life as machine numbers, adding arbitrary digits serves no purpose. Yes, plots may get smoother as more digits are added, but they would not converge to a correct result -- merely to a precise one. (In the chemistry industry where my wife works, the difference between accuracy and precision is well known. Precision means getting the same answer over and over --- whether it's right or not. Accuracy means getting the right answer --- whether it's precise or not. It's low variance versus small bias.) Modify your example like this: ser = N@Normal[Series[Cos[#], {#, 0, 200}]]; Timing[pts = With[{ss = ser}, Table[SetPrecision[{#, ss}, 80] &@x, {x, 50., 70., .1}]];] ListPlot[pts, PlotJoined -> True, PlotRange -> All]; MaxMemoryUsed[] Once the series coefficients have lost precision, you can't get it back again. Furthermore, in using SetPrecision, there's a danger that one could THINK he has regained it. Bobby -----Original Message----- feel > ourselves blessed if we get three digits of accuracy, I'm curious as to what > scientific endeavors require 50+ digits of precision? As I recall there are > some areas, such as high energy physics and some elements of astronomy, that > might require so many digits in some circumstances. Are there others? > -Mark Mark, There may be occasions when the outcome of a real process is so sensitive to changes in input that unless we know very precisely what the input is then we can know very little about the outcome - chaotic processes are of this kind. The difficulty is real and no amount of computer power or clever progamming will do much about it. Another situation is when the the process is not so sensitive but calculating with our formula or programme introduces accumulates significant errors. Here is a very artificial example of the latter (I time the computation and find the MaximumMemory used in the session as we go through the example): ser=Normal[Series[Cos[#],{#,0,200}]]; MaxMemoryUsed[] 1714248 Calculating with machine number does not show much of a pattern ( I have deleted the graphics - please evaluate the code), pts= With[{ss=ser},Table[ {#,ss}&[x], {x,50.,70., .1}]];//Timing ListPlot[pts, PlotJoined->True]; MaxMemoryUsed[] {5.11 Second,Null} 1723840 Using bigfloat inputs with precision 20 shows some pattern: pts= With[{ss=ser},Table[ {#,ss}&[SetPrecision[x,20]], {x,50.,70., .1}]];//Timing ListPlot[pts, PlotJoined->True,PlotRange[Rule]All]; MaxMemoryUsed[] {17.52 Second,Null} 1759664 Precision 40 does very well: pts= With[{ss=ser},Table[ {#,ss}&[SetPrecision[x,40]], {x,50.,70., .1}]];//Timing ListPlot[pts, PlotJoined->True,PlotRange[Rule]All]; MaxMemoryUsed[] {19.38 Second,Null} 1797072 Now we might think the correct outcomes are showing up, and use an interpolating function for further , and faster, calculation. f=Interpolation[pts] InterpolatingFunction[{{50.000000,70.00000}},<>] pts= Table[ f[x],{x,50, 70, .1}];//Timing ListPlot[pts, PlotJoined->True,PlotRange[Rule]All]; MaxMemoryUsed[] {0.33 Second,Null} As a matter of interest, this is what happens if we substitute exact numbers (rationals and integers) for reals-- the computation takes an excessively long time and quite a bit more memory. pts= With[{ss=ser},Table[ {#,ss}&[SetPrecision[x,Infinity]], {x,50.,70., .1}]];//Timing ListPlot[pts, PlotJoined->True,PlotRange[Rule]All]; MaxMemoryUsed[] {992.28 Second,Null} 2413808 This also shows that we may in fact want to replace exact inputs with bigfloats. I should be interested to hear of other example, really real one in particular. I imagine that there are many situations where trends and shapes are more important than specific values. -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay@haystack.demon.co.uk Voice: +44 (0)116 271 4198 > ==== Here's a very simple solution: v = {{100, 200}, {150, 250}, {120, 270}, {300, 400}}; Interval @@ v List @@ % Interval[{100, 270}, {300, 400}] {{100, 270}, {300, 400}} DrBob -----Original Message----- second list that contains the overall upper and lower edges of the overlapping sub-ranges. A simple example : {{100,200},{150,250},{120,270},{300,400}} would result in {{100,270},{300,400}}. In the real case, the input list has several hundred elements and the output list typically has five elements. I have a working solution based on loops, but there must be a more elegant one. I would be very grateful for any suggestions. John Leary ==== Dear Fellows in MathGroup, I have a list of 17,000+ {x,y} pairs of data each x value is a positive integer from 1 to 100+ each y value is a positive real number As a *short* example, let's consider: data = {{3,1},{4,3},{3,2},{1,10},{4,2},{1,6},{5,2},{2,5},{7,1}} I want to group the data by the x value and report the arithmetic average > of the y values in each group. For the example, i want to report: output = {{1,8},{2,5},{3,1.5},{4,2.5},{5,2},{6,0},{7,1}} In this example, x=6 does not occur so i report the average y[6] = 0. Can anyone suggest a way to do this efficiently?/ many thanks > dave +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ > David E. Burmaster, Ph.D. > Alceon Corporation > POBox 382069 (new Box number effective 1 Sep 2001) > Harvard Square Station > Cambridge, MA 02238-2069 (new ZIP code effective 1 Sep 2001) Voice 617-864-4300 Web http://www.Alceon.com > +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ Probably most efficient would be to iterate over the list, bin, and then average the bins. averageByBin[data:{{_,_}..}] := Module[ {len, binsizes, averages}, len = Max[Map[First,data]]; binsizes = Table[0,{len}]; averages = Table[0,{len}]; Map [ (binsizes[[#[[1]]]]++; averages[[#[[1]]]] += #[[2]]) &, data]; Do [If[binsizes[[j]]==0, binsizes[[j]]++], {j,len}]; Transpose[{Range[len],N[averages]/binsizes}] ] In[30]:= averageByBin[data] Out[30]= {{1, 8.}, {2, 5.}, {3, 1.5}, {4, 2.5}, {5, 2.}, {6, 0.}, {7, 1.}} If you separate out integer first values from real second values in the pairs, you can enter two separate lists and take advantage of Compile to make it faster still. Daniel Lichtblau Wolfram Research ==== Dear Fellows in MathGroup, I have a list of 17,000+ {x,y} pairs of data each x value is a positive integer from 1 to 100+ each y value is a positive real number As a *short* example, let's consider: data = {{3,1},{4,3},{3,2},{1,10},{4,2},{1,6},{5,2},{2,5},{7,1}} I want to group the data by the x value and report the arithmetic average of the y values in each group. For the example, i want to report: output = {{1,8},{2,5},{3,1.5},{4,2.5},{5,2},{6,0},{7,1}} In this example, x=6 does not occur so i report the average y[6] = 0. Can anyone suggest a way to do this efficiently?/ many thanks dave +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ David E. Burmaster, Ph.D. Alceon Corporation POBox 382069 (new Box number effective 1 Sep 2001) Harvard Square Station Cambridge, MA 02238-2069 (new ZIP code effective 1 Sep 2001) Voice 617-864-4300 Web http://www.Alceon.com +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ ==== One typical idea for your application is to sort the data and then use Split. For example, ({#1[[1,1]], Plus @@ #1[[All,2]]/Length[#1]} & ) /@ Split[Sort[data], #1[[1]] == #2[[1]] & ]; The only difference between the above function and your desired result is that when there is no data for a particular integer, the average for that integer does not appear in the answer. Carl Woll Physics Dept U of Washington > Dear Fellows in MathGroup, I have a list of 17,000+ {x,y} pairs of data each x value is a positive integer from 1 to 100+ each y value is a positive real number As a *short* example, let's consider: data = {{3,1},{4,3},{3,2},{1,10},{4,2},{1,6},{5,2},{2,5},{7,1}} I want to group the data by the x value and report the arithmetic average > of the y values in each group. For the example, i want to report: output = {{1,8},{2,5},{3,1.5},{4,2.5},{5,2},{6,0},{7,1}} In this example, x=6 does not occur so i report the average y[6] = 0. Can anyone suggest a way to do this efficiently?/ many thanks > dave +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ > David E. Burmaster, Ph.D. > Alceon Corporation > POBox 382069 (new Box number effective 1 Sep 2001) > Harvard Square Station > Cambridge, MA 02238-2069 (new ZIP code effective 1 Sep 2001) Voice 617-864-4300 Web http://www.Alceon.com > +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ ==== >Dear Fellows in MathGroup, I have a list of 17,000+ {x,y} pairs of data each x value is a positive integer from 1 to 100+ each y value is a positive real number As a *short* example, let's consider: data = {{3,1},{4,3},{3,2},{1,10},{4,2},{1,6},{5,2},{2,5},{7,1}} I want to group the data by the x value and report the arithmetic average >of the y values in each group. For the example, i want to report: output = {{1,8},{2,5},{3,1.5},{4,2.5},{5,2},{6,0},{7,1}} In this example, x=6 does not occur so i report the average y[6] = 0. Can anyone suggest a way to do this efficiently?/ Block[{data = {{3,1},{4,3},{3,2},{1,10},{4,2},{1,6}, {5,2},{2,5},{7,1}}, sd = Table[{0,0.},{10}]}, Off[Infinity::indet]; Off[General::dbyz]; (sd[[#[[1]],2]]++;sd[[#[[1]],1]]+=#[[2]])&/@data; sd=MapIndexed[{#2[[1]],Divide@@#}&,sd]; On[Infinity::indet]; On[General::dbyz]; sd/.Indeterminate->0.] --> {{1, 8.}, {2, 5.}, {3, 1.5}, {4, 2.5}, {5, 2.}, {6, 0.}, {7, 1.}, {8, 0.}, {9, 0.}, {10, 0.}} This makes 3 linear passes overall on the data. For large data sets, that may be a problem. DH ==== > Dear Fellows in MathGroup, I have a list of 17,000+ {x,y} pairs of data each x value is a positive integer from 1 to 100+ each y value is a positive real number As a *short* example, let's consider: data = {{3,1},{4,3},{3,2},{1,10},{4,2},{1,6},{5,2},{2,5},{7,1}} I want to group the data by the x value and report the arithmetic average > of the y values in each group. For the example, i want to report: output = {{1,8},{2,5},{3,1.5},{4,2.5},{5,2},{6,0},{7,1}} In this example, x=6 does not occur so i report the average y[6] = 0. Can anyone suggest a way to do this efficiently?/ many thanks > dave Dave, One way: data={{3,1},{4,3},{3,2},{1,10},{4,2},{1,6},{5,2},{2,5},{7,1}}; f[x_] = 0; ((f[#1[[1,1]]] = Plus @@ #1[[All,2]]/Length[#1]) & ) /@ Split[Sort[data], #1[[1]] == #2[[1]] & ] {8, 5, 3/2, 5/2, 2, 1} Table[f[i], {i, 1, 8}] {8, 5, 3/2, 5/2, 2, 0, 1, 0} -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay@haystack.demon.co.uk Voice: +44 (0)116 271 4198 +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ > David E. Burmaster, Ph.D. > Alceon Corporation > POBox 382069 (new Box number effective 1 Sep 2001) > Harvard Square Station > Cambridge, MA 02238-2069 (new ZIP code effective 1 Sep 2001) Voice 617-864-4300 Web http://www.Alceon.com > +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ ==== Perhaps Numerical Recipes in C++ ----- Original Message ----- constants > and addition for example. I currently do this all with Mathematica but for > large symbolic calculations involving many operations Mathematica becomes quite > slow. (at least with my coding ability ;) ) Does anyone have any > recommendations for a c++ book along these lines? Simply put, I basically > want to write c++ code that functions similarly to mathematica but that > is specialized to particlar types of algebraic problems and therefore > runs faster. Any suggestions? -chris ==== I'm looking for a book recommendation. I've been using mathematica and I love it. I'd like to speed up some of my Mathematica codes though. What I really want to do is code some basic algrebraic steps in a lower level language. I need to somehow set up c/c++ code for operators, how they multiply, and also deal with multiplicative constants and addition for example. I currently do this all with Mathematica but for large symbolic calculations involving many operations Mathematica becomes quite slow. (at least with my coding ability ;) ) Does anyone have any recommendations for a c++ book along these lines? Simply put, I basically want to write c++ code that functions similarly to mathematica but that is specialized to particlar types of algebraic problems and therefore runs faster. Any suggestions? -chris ==== > I've been trying to use PlotVectorField for the > following differential equation: > dy/dt = 0.08*y*(1-y/1000) > but I haven't been successful yet. > I tried to do the following: > f[t_, y_] := {1, 0.08*y*(1 - y/1000)} > < PlotVectorField[f[t, y], {t, 0, 80}, {y, 0, 1400}]; > but I'm getting a meaningless plot so I'd appreciate > if someone could tell me what is what I'm doing wrong. > Ruben Just a thought, but have you checked your function for typos? As you've written it, f is only a function of y .... Dave. ==== > Greetings This problem can be solved by conventional programming, but I wonder if > there is an elegant Mathematica solution ? A list contains pairs of values, with each pair representing the lower > and > upper edge of a sub-range. Some of the sub-ranges partially overlap, > some > fully overlap, others don't overlap at all. The problem is to produce > a > second list that contains the overall upper and lower edges of the > overlapping sub-ranges. A simple example : {{100,200},{150,250},{120,270},{300,400}} would > result > in {{100,270},{300,400}}. In the real case, the input list has several hundred elements and the > output list typically has five elements. I have a working solution based on loops, but there must be a more > elegant > one. I would be very grateful for any suggestions. I'm not sure about elegance, but I will tell you my approach to the problem. The general algorithm would seem to be: sort the ranges such that the first range has a left edge smaller than all other ranges. If two ranges have matching left edges, sort them according to their right edge. While there are two or more ranges in the list, operate on the rest of the list, compare the first range to the result of operating on the rest of the list. If the right edge of the first edge is larger than the left edge of the first element of the result return a list with the first element being a range with the left edge of the first range and the right being the larger of the right edge of the first element or the right edge of the first range in the result of operating on the list. That statement probably isn't very clear, it's a good thing I'm not employed as a teacher. Here is code which is probably easier to follow: (*Handle some degenerate cases here *) compress[{}] := {}; compress[lst : {{_?NumericQ, _?NumericQ}}] := lst (*This pattern is the stopping point of the recursion*) compress[rng : {_?NumericQ, _?NumericQ}, {}] := {rng} (*This function operates on the rest of the list then creates the first element appropriately*) compress[rng : {_? NumericQ, _?NumericQ}, lst : {{_?NumericQ, _?NumericQ} ..}] := With[{tl = compress[First[lst], Rest[lst]]}, If[rng[[2]] > Last[tl][[2]], {rng}, If[rng[[2]] > tl[[1, 1]], {{rng[[1]], If[tl[[1, 2]] > rng[[2]], tl[[1, 2]], rng[[2]]]}, Sequence @@ Rest[tl]}, {rng, Sequence @@ tl}]]] (*This function sorts the list properly then starts the recursion*) compress[lst : {{_?NumericQ, _?NumericQ} ..}] := With[{s = Sort[lst]}, compress[First[s], Rest[s]]] You will probably have to increase $RecursionLimit. This algorithm ran in 2 seconds on a list of 1000 elements on a 1GHz G4 PowerMac. There are probably optimizations that can be made though. Ssezi ==== Ruben, Your plot looks bad because PlotVectorField uses AspectRatio->Automatic, which means that the t and y axes are at the same scale. This makes the plot very high and narrow. Try the following... PlotVectorField[f[t, y], {t, 0, 80}, {y, 0, 1400}, AspectRatio -> 1, Frame -> True, ImageSize -> 450]; David Park djmp@earthlink.net http://home.earthlink.net/~djmp/ but I'm getting a meaningless plot so I'd appreciate if someone could tell me what is what I'm doing wrong. Ruben __________________________________________________ Do you Yahoo!? http://faith.yahoo.com Reply-To: ==== You'll get more efficient methods from others, but I think the following is instructive: data = {{3, 1}, {4, 3}, {3, 2}, {1, 10}, {4, 2}, {1, 6}, {5, 2}, {2, 5}, {7, 1}}; ClearAll[total, count] total[x_] := 0 count[x_] := 0 {total[#[[1]]] += #[[2]], count[#[[1]]]++} & /@ data; ?total ?count {#, total[#]/count[#]} & /@ Union[data[[All, 1]]] DrBob -----Original Message----- of the y values in each group. For the example, i want to report: output = {{1,8},{2,5},{3,1.5},{4,2.5},{5,2},{6,0},{7,1}} In this example, x=6 does not occur so i report the average y[6] = 0. Can anyone suggest a way to do this efficiently?/ many thanks dave +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ David E. Burmaster, Ph.D. Alceon Corporation POBox 382069 (new Box number effective 1 Sep 2001) Harvard Square Station Cambridge, MA 02238-2069 (new ZIP code effective 1 Sep 2001) Voice 617-864-4300 Web http://www.Alceon.com +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ Reply-To: ==== It's not clear what you mean by pairs that repeat themselves, but does this do what you want? pairs = Flatten[Outer[List, points, points, 1], 1]; Map[Union, Map[Sort, pairs, 2], 1] DrBob -----Original Message----- themselves, i.e., pair AA. I can use pairs=Outer[List,points,points,1] Then, I have to clear those pairs that repeat themselves, i.e., pair AB and pair BA. Also, when w and h are of the order of 1000s, the computation takes a very long time. Is there a better way of doing the second part of Sincerely Cheng ==== ================================================ Cheng Liu, Ph.D. MST-8, Structure/Property Relations Materials Science and Technology Division Los Alamos National Laboratory Los Alamos, New Mexico 87545 ==== ================================================ ==== David, You will probably get a lot of answers for this. Here is my entry. data = {{3, 1}, {4, 3}, {3, 2}, {1, 10}, {4, 2}, {1, 6}, {5, 2}, {2, 5}, {7, 1}}; First I will show it step-by-step. nmax = 10; Union[Join[data, Table[{i, 0}, {i, 1, nmax}]]] Split[%, #1[[1]] == #2[[1]] & ] Map[Last, %, {2}] (Plus @@ #1/Length[#1] & ) /@ % Transpose[{Range[nmax], %}] giving {{1, 0}, {1, 6}, {1, 10}, {2, 0}, {2, 5}, {3, 0}, {3, 1}, {3, 2}, {4, 0}, {4, 2}, {4, 3}, {5, 0}, {5, 2}, {6, 0}, {7, 0}, {7, 1}, {8, 0}, {9, 0}, {10, 0}} {{{1, 0}, {1, 6}, {1, 10}}, {{2, 0}, {2, 5}}, {{3, 0}, {3, 1}, {3, 2}}, {{4, 0}, {4, 2}, {4, 3}}, {{5, 0}, {5, 2}}, {{6, 0}}, {{7, 0}, {7, 1}}, {{8, 0}}, {{9, 0}}, {{10, 0}}} {{0, 6, 10}, {0, 5}, {0, 1, 2}, {0, 2, 3}, {0, 2}, {0}, {0, 1}, {0}, {0}, {0}} {16/3, 5/2, 1, 5/3, 1, 0, 1/2, 0, 0, 0} {{1, 16/3}, {2, 5/2}, {3, 1}, {4, 5/3}, {5, 1}, {6, 0}, {7, 1/2}, {8, 0}, {9, 0}, {10, 0}} This wraps it into one statement. nmax = 10; Transpose[{Range[nmax], (Plus @@ #1/Length[#1] & ) /@ Map[Last, Split[Union[Join[data, Table[{i, 0}, {i, 1, nmax}]]], #1[[1]] == #2[[1]] & ], {2}]}] {{1, 16/3}, {2, 5/2}, {3, 1}, {4, 5/3}, {5, 1}, {6, 0}, {7, 1/2}, {8, 0}, {9, 0}, {10, 0}} This times a case of 20000 pairs on an 800MHz machine. data2 = Table[{Random[Integer, {1, 100}], Random[Real, {0, 5}]}, {20000}]; nmax = 100; data = data2; Timing[Transpose[{Range[nmax], (Plus @@ #1/Length[#1] & ) /@ Map[Last, Split[Union[Join[data, Table[{i, 0}, {i, 1, nmax}]]], #1[[1]] == #2[[1]] & ], {2}]}]; ] {0.55 Second, Null} David Park djmp@earthlink.net http://home.earthlink.net/~djmp/ output = {{1,8},{2,5},{3,1.5},{4,2.5},{5,2},{6,0},{7,1}} In this example, x=6 does not occur so i report the average y[6] = 0. Can anyone suggest a way to do this efficiently?/ many thanks dave +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ David E. Burmaster, Ph.D. Alceon Corporation POBox 382069 (new Box number effective 1 Sep 2001) Harvard Square Station Cambridge, MA 02238-2069 (new ZIP code effective 1 Sep 2001) Voice 617-864-4300 Web http://www.Alceon.com +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ ==== I'm not 100% sure of this, but how about sortout[intervals_] := ({First[#1], Last[#1]} & ) /@ Sort /@ Flatten /@ Split[Sort[intervals], #1[[2]] >= #2[[1]] & ] ? Example: intrvls = Table[x = Random[Integer, {0, 30}]; {x, x + 2}, {10}] {{18, 20}, {2, 4}, {14, 16}, {1, 3}, {0, 2}, {16, 18}, {7, 9}, {19, 21}, {6, 8}, {6, 8}} sortout[intrvls] {{0, 4}, {6, 9}, {14, 21}} --- Selwyn Hollis > Greetings This problem can be solved by conventional programming, but I wonder if > there is an elegant Mathematica solution ? A list contains pairs of values, with each pair representing the lower and > upper edge of a sub-range. Some of the sub-ranges partially overlap, some > fully overlap, others don't overlap at all. The problem is to produce a > second list that contains the overall upper and lower edges of the > overlapping sub-ranges. A simple example : {{100,200},{150,250},{120,270},{300,400}} would result > in {{100,270},{300,400}}. In the real case, the input list has several hundred elements and the > output list typically has five elements. I have a working solution based on loops, but there must be a more elegant > one. I would be very grateful for any suggestions. John Leary > ==== John, Use the Interval routine. Interval @@ {{100, 200}, {150, 250}, {120, 270}, {300, 400}} List @@ % giving... Interval[{100, 270}, {300, 400}] {{100, 270}, {300, 400}} David Park djmp@earthlink.net http://home.earthlink.net/~djmp/ in {{100,270},{300,400}}. In the real case, the input list has several hundred elements and the output list typically has five elements. I have a working solution based on loops, but there must be a more elegant one. I would be very grateful for any suggestions. John Leary ==== >I have a list of 17,000+ {x,y} pairs of data each x value is a positive integer from 1 to 100+ each y value is a positive real number As a *short* example, let's consider: data = {{3,1},{4,3},{3,2},{1,10},{4,2},{1,6},{5,2},{2,5},{7,1}} I want to group the data by the x value and report the arithmetic average >of the y values in each group. For the example, i want to report: output = {{1,8},{2,5},{3,1.5},{4,2.5},{5,2},{6,0},{7,1}} In this example, x=6 does not occur so i report the average y[6] = 0. Can anyone suggest a way to do this efficiently?/ > The basic approach is (Plus @@ #)/Length[#] & /@ Split[Sort[data], #1[[1]] == #2[[1]] &] {{1, 8}, {2, 5}, {3, 3/2}, {4, 5/2}, {5, 2}, {7, 1}} To fill in the gaps: dataAvg[data_] := Module[{val = First /@ data, xData}, xData = Join[data, {#, 0} & /@ Complement[Range[Max[val]], val]]; (Plus @@ #)/Length[#] & /@ Split[Sort[xData], #1[[1]] == #2[[1]] &]]; dataAvg[data] {{1, 8}, {2, 5}, {3, 3/2}, {4, 5/2}, {5, 2}, {6, 0}, {7, 1}} Bob Hanlon ==== >IÇm new, so IÇm sorry if the question is so easy, I really think that >is easy, but I donÇt know how to do it. I have this: b=n-m a=x-b Y=3a + 4a^2 and the program show me this: >3(x-n+m) + 4(x-n+m)^2 or, something like that, the problem is that I want the program show >me Y in function of b, or sometimes in function of a, something like >this: Y=3(x-b) + 4(x-b)^2 or Y=3a + 4a^2 > Solve[{b == n - m, a == x - b, Y == 3a + 4a^2}, Y] {{Y -> a*(4*a + 3)}} Solve[{b == n - m, a == x - b, Y == 3a + 4a^2}, Y, a] // FullSimplify {{Y -> (4*b - 4*x - 3)*(b - x)}} Bob Hanlon ==== Boole (defined in the AddOn package Calculus`Integration`) is not written to deal with symbolic parameters. So you may do better to use better to use the built-in UnitStep function for which Mathematica knows more rules. In fact, if you do not mind getting an error message you can get something like your answer by mixing UnitStep and Boole: << Calculus`Integration` In[2]:= FullSimplify[Integrate[Boole[0 < x < y < 1]*UnitStep[z - y], {z, 0, 1}],y>0] Integrate::region: The region defined by 0ç.8czç.8c1&&00] Out[3]= -(-1+y) UnitStep[x,1-y,-x+y] Note that this expresses exactly the same condition as (1-y)Boole[0 Suppose f(x,y,z)=Boole[0 over, say, z, i.e. Integrate[f[x,y,z],{z,0,1}]. One would expect to see > the > output like this (1-y)Boole[0 as an argument in the Boole function). Instead, an error appears > (warning) that the integration cannot be performed. How to resolve this issue so it produces a desired answer? Janusz. > ==== >This problem can be solved by conventional programming, but I wonder if there is an elegant Mathematica solution ? A list contains pairs of values, with each pair representing the lower >and >upper edge of a sub-range. Some of the sub-ranges partially overlap, some fully overlap, others don't overlap at all. The problem is to produce >a >second list that contains the overall upper and lower edges of the >overlapping sub-ranges. A simple example : {{100,200},{150,250},{120,270},{300,400}} would result in {{100,270},{300,400}}. In the real case, the input list has several hundred elements and the >output list typically has five elements. I have a working solution based on loops, but there must be a more elegant one. I would be very grateful for any suggestions. > lst = {{100, 200}, {150, 250}, {120, 270}, {300, 400}}; List @@ Union @@ Interval /@ lst {{100, 270}, {300, 400}} List @@ Interval[Sequence @@ lst] {{100, 270}, {300, 400}} Bob Hanlon ==== In my opinion, the best way to learn C++ and code algorithms is to simply start using it. There is oodles of source on the web and at www.sf.net to help you along too. Matt -- http://mffm.darktech.org WSOLA TimeScale Audio Mod : http://mffmtimescale.sourceforge.net/ FFTw C++ : http://mffmfftwrapper.sourceforge.net/ Vector Bass : http://mffmvectorbass.sourceforge.net/ Multimedia Time Code : http://mffmtimecode.sourceforge.net/ ==== do any of you have a working qsum author indentification workbook you are willing to share? thanks ==== >-----Original Message----- >Sent: Wednesday, October 16, 2002 8:26 PM >To: mathgroup@smc.vnet.net >Dear group, I have the following question regarding a lengthy calculation >using Mathematica: For given w points in x direction and h points in y direction, I can >construct all the points using h=10; w=8; > points=Flatten[Transpose[Outer[List,Range[w],Range[h]]],1] Next, I need to find all the possible pairs of point including points >themselves, i.e., pair AA. I can use pairs=Outer[List,points,points,1] Then, I have to clear those pairs that repeat themselves, >i.e., pair AB and >pair BA. Also, when w and h are of the order of 1000s, the >computation >takes a very long time. Is there a better way of doing the >second part of Sincerely Cheng > ==== ================================================ >Cheng Liu, Ph.D. >MST-8, Structure/Property Relations >Materials Science and Technology Division >Los Alamos National Laboratory >Los Alamos, New Mexico 87545 ==== ================================================ Cheng, you didn't tell about a specific order of your resulting pairs, so you not need Transpose in your first line: points = Flatten[Outer[List, Range[w], Range[h]], 1] Now build the pairs (for sake of clarity , let me call the points points = {p1, p2, p3, p4, p5}, of course you don't do that): Flatten[ With[{l = Length[points]}, Array[If[#1>#2, Unevaluated[Sequence[]], points[[{#1, #2}]]]&, {l, l}]], 1] {{p1, p1}, {p1, p2}, {p1, p3}, {p1, p4}, {p1, p5}, {p2, p2}, {p2, p3}, {p2, p4}, {p2, p5}, {p3, p3}, {p3, p4}, {p3, p5}, {p4, p4}, {p4, p5}, {p5, p5}} Alternatively you might do pairs = Outer[List, points, points, 1] Flatten[MapIndexed[Drop[#1, First[#2] - 1] &, pairs], 1] {{p1, p1}, {p1, p2}, {p1, p3}, {p1, p4}, {p1, p5}, {p2, p2}, {p2, p3}, {p2, p4}, {p2, p5}, {p3, p3}, {p3, p4}, {p3, p5}, {p4, p4}, {p4, p5}, {p5, p5}} ...try out which is faster. -- Hartmut Wolf ==== When I first sent my answer I thought there was no interaction between Boole and UnitStep at all,and that one could safely use UnitStep after loading the Calculus`Integration` package, but there seems to be more to it than I had assumed and it is not necessarily for the best. Consider first the following: In[1]:= FullSimplify[Integrate[UnitStep[x,y-x,z-y,1-z],{z,0,1}],y>0] Out[1]= -(-1+y) UnitStep[x,1-y,-x+y] No surprises here. Now let's load the package: In[2]:= <0] Integrate::region: The region defined by 0.89.81óz.89.81ó1&&x.89 .81«0&&-x+y.89.81«0&&1-z.89.8126 40&&-y+z.89.81«0 could not be broken down into cylinders. Integrate::region: The region defined by y.89.81óz.89.81ó1&&x.89 .81«0&&-x+y.89.81«0&&1-z.89.8126 40 could not be broken down into cylinders. Integrate::region: The region defined by -1.89.81óz.89.81ó-y&&x21 1.81«0&&-x+y.89.81«0 could not be broken down into cylinders. General::stop: Further output of Integrate::region will be suppressed during this calculation. Out[3]= -(-1+y) UnitStep[x,1-y,-x+y] Clearly an attempt was made to decompose this into cylinders with respect to z (using CAD) which of course failed. Fortunately we still get the right answer. Secondly, the package actually contains some interesting functions which have been commented out and were apparently intended for future development. On of them is: removeUnitStep[expr_] := ReplaceRepeated[expr, UnitStep[e__] :> Boole[Apply[And, Map[(# .89.81« 0) &, {e}]]]]; Using it we get: removeUnitStep[FullSimplify[Integrate[UnitStep[x,y-x,z-y,1- z],{z,0,1}],y>0]] (error messages removed) -(-1+y) Boole[x.89.81«0&&x.89.81óy&&y.89[ CapitalARing]ó1] Mathematica can't convert this into (1-y)Boole[0<=x<=y<=1] but it can do the converse: Map[LogicalExpand,(1-y)Boole[0.89.81óx.89[Capita lARing]óy.89.81ó1],{2}] (1-y) Boole[0.89.81óx&&x.89.81ó y&&y.89.81ó1] Andrzej Kozlowski Yokohama, Japan http://www.mimuw.edu.pl/~akoz/ http://platon.c.u-tokyo.ac.jp/andrzej/ > Boole (defined in the AddOn package Calculus`Integration`) is not > written to deal with symbolic parameters. So you may do better to use > better to use the built-in UnitStep function for which Mathematica > knows more rules. In fact, if you do not mind getting an error message > you can get something like your answer by mixing UnitStep and Boole: << Calculus`Integration` In[2]:= > FullSimplify[Integrate[Boole[0 < x < y < 1]*UnitStep[z - y], > {z, 0, 1}],y>0] Integrate::region: The region defined by 0.8cÁ.8cåz.8cÁ .8cå1&&0 could > not be > broken down into cylinders. Out[2]= > -(-1+y) Boole[0 already included in Boole, but the Integration package does not have > rules for combining Bool and UnitStep (don't forget that Boole is not a > built in function!). If you dispense with Bool altogether you get: In[3]:= > FullSimplify[Integrate[UnitStep[x,y-x,z-y,1-z],{z,0,1}],y>0] Out[3]= > -(-1+y) UnitStep[x,1-y,-x+y] Note that this expresses exactly the same condition as > (1-y)Boole[0 Yokohama, Japan > http://www.mimuw.edu.pl/~akoz/ > http://platon.c.u-tokyo.ac.jp/andrzej/ > and also try to use proper Mathematica >> Suppose f(x,y,z)=Boole[0> over, say, z, i.e. Integrate[f[x,y,z],{z,0,1}]. One would expect to >> see >> the >> output like this (1-y)Boole[0> as an argument in the Boole function). Instead, an error appears >> (warning) that the integration cannot be performed. >> How to resolve this issue so it produces a desired answer? >> Janusz. > > Reply-To: ==== --- not as a poor a showing for my simple-minded method as I feared. data2 = Table[{Random[Integer, {1, 100}], Random[Real, {0, 5}]}, {20000}]; nmax = 100; data = data2; Timing[dave = Transpose[{Range[nmax], (Plus @@ #1/Length[#1] &) /@ Map[Last, Split[Union[Join[data, Table[{i, 0}, {i, 1, nmax}]]], #1[[1]] == #2[[1]] &], {2}]}];] ClearAll[total, count] total[x_] := 0 count[x_] := 0 Timing[{total@#[[1]] += #[[2]], count[#[[1]]]++} & /@ data; brt = {#, total[#]/count[#]} & /@ Union[data[[All, 1]]]; ] {0.20299999999999985*Second, Null} {0.5619999999999998*Second, Null} DrBob -----Original Message----- Split[%, #1[[1]] == #2[[1]] & ] Map[Last, %, {2}] (Plus @@ #1/Length[#1] & ) /@ % Transpose[{Range[nmax], %}] giving {{1, 0}, {1, 6}, {1, 10}, {2, 0}, {2, 5}, {3, 0}, {3, 1}, {3, 2}, {4, 0}, {4, 2}, {4, 3}, {5, 0}, {5, 2}, {6, 0}, {7, 0}, {7, 1}, {8, 0}, {9, 0}, {10, 0}} {{{1, 0}, {1, 6}, {1, 10}}, {{2, 0}, {2, 5}}, {{3, 0}, {3, 1}, {3, 2}}, {{4, 0}, {4, 2}, {4, 3}}, {{5, 0}, {5, 2}}, {{6, 0}}, {{7, 0}, {7, 1}}, {{8, 0}}, {{9, 0}}, {{10, 0}}} {{0, 6, 10}, {0, 5}, {0, 1, 2}, {0, 2, 3}, {0, 2}, {0}, {0, 1}, {0}, {0}, {0}} {16/3, 5/2, 1, 5/3, 1, 0, 1/2, 0, 0, 0} {{1, 16/3}, {2, 5/2}, {3, 1}, {4, 5/3}, {5, 1}, {6, 0}, {7, 1/2}, {8, 0}, {9, 0}, {10, 0}} This wraps it into one statement. nmax = 10; Transpose[{Range[nmax], (Plus @@ #1/Length[#1] & ) /@ Map[Last, Split[Union[Join[data, Table[{i, 0}, {i, 1, nmax}]]], #1[[1]] == #2[[1]] & ], {2}]}] {{1, 16/3}, {2, 5/2}, {3, 1}, {4, 5/3}, {5, 1}, {6, 0}, {7, 1/2}, {8, 0}, {9, 0}, {10, 0}} This times a case of 20000 pairs on an 800MHz machine. data2 = Table[{Random[Integer, {1, 100}], Random[Real, {0, 5}]}, {20000}]; nmax = 100; data = data2; Timing[Transpose[{Range[nmax], (Plus @@ #1/Length[#1] & ) /@ Map[Last, Split[Union[Join[data, Table[{i, 0}, {i, 1, nmax}]]], #1[[1]] == #2[[1]] & ], {2}]}]; ] {0.55 Second, Null} David Park djmp@earthlink.net http://home.earthlink.net/~djmp/ For the example, i want to report: output = {{1,8},{2,5},{3,1.5},{4,2.5},{5,2},{6,0},{7,1}} In this example, x=6 does not occur so i report the average y[6] = 0. Can anyone suggest a way to do this efficiently?/ many thanks dave +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ David E. Burmaster, Ph.D. Alceon Corporation POBox 382069 (new Box number effective 1 Sep 2001) Harvard Square Station Cambridge, MA 02238-2069 (new ZIP code effective 1 Sep 2001) Voice 617-864-4300 Web http://www.Alceon.com +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ ==== >-----Original Message----- >Sent: Wednesday, October 16, 2002 8:26 PM >To: mathgroup@smc.vnet.net >Greetings This problem can be solved by conventional programming, but I >wonder if >there is an elegant Mathematica solution ? A list contains pairs of values, with each pair representing >the lower and >upper edge of a sub-range. Some of the sub-ranges partially >overlap, some >fully overlap, others don't overlap at all. The problem is to >produce a >second list that contains the overall upper and lower edges of the >overlapping sub-ranges. A simple example : {{100,200},{150,250},{120,270},{300,400}} >would result >in {{100,270},{300,400}}. In the real case, the input list has several hundred elements and the >output list typically has five elements. I have a working solution based on loops, but there must be a >more elegant >one. I would be very grateful for any suggestions. John Leary John, several proposals (without any attempt to moduralize): (1) use IntervalUnion: List @@ IntervalUnion @@ Interval /@ list (2) use Split (it's a little bit tricky to be correct): high = Sequence[]; {#[[1, 1]], Max[#[[All, -1]]]} & /@ Split[Sort[list], (high = Max[high,Last[#1]]) >= First[#2] || (high = Last[#1])&] (3) procedural programming: maxExtends[list_] := (sl = Sort[list]; length = Length[sl]; r = collect[]; i = 1; While[i <= length, {low, high} = sl[[i]]; If[++i <= length, {curlow, curhigh} = sl[[i]]; While[high >= curlow && (high = Max[high, curhigh]; ++i <= length), {curlow, curhigh} = sl[[i]] ]]; r = collect[r, {low, high}] ]; List @@ Flatten[r]) Let's do some benchmarks: 10,000 Intervals: list = {# - Random[], # + Random[]} & /@ NestList[# + Random[] &, 0, 10000]; List @@ IntervalUnion @@ Interval /@ list // Length // Timing {2.503 Second, 1181} high = Sequence[]; {#[[1, 1]], Max[#[[All, -1]]]} & /@ Split[Sort[ list], (high = Max[high, Last[#1]]) >= First[#2] || (high = Last[#1]) &] // Length // Timing {2.934 Second, 1181} maxExtends[list] // Length // Timing {3.926 Second, 1181} The corresponding results for 100,000 Intervals: {27.329 Second, 11266} {30.234 Second, 11266} {35.791 Second, 11266} and for 500,000 Intervals {144.058 Second, 56728} {154.782 Second, 56728} {181.111 Second, 56728} To look at scaling behaviour I just collected the prior results IntervalUnion: {%355, %345 , %350}[[All, 1, 1]] {2.503, 27.329, 144.058} % // {#[[2]]/(10*#[[1]]), #[[3]]/(5*#[[2]])} & {1.09185, 1.05425} Split: {%357, %347, %352}[[All, 1, 1]] {2.934, 30.234, 154.782} % // {#[[2]]/(10*#[[1]]), #[[3]]/(5*#[[2]])} & {1.03047, 1.02389} Procedural: {%358, %348, %353}[[All, 1, 1]] {3.926, 35.791, 181.111} % // {#[[2]]/(10*#[[1]]), #[[3]]/(5*#[[2]])} & {0.91164, 1.01205} Due to Sort, the Split and the Procedural versions should behave as O[n log n], I'm not shure whether IntervalUnion does (seems to be a little bit more progressive at costs). -- Hartmut Wolf ==== I use NDSolve to approximate a 6-dimensional and highly non-linear dynamic system numerically. QUESTION: What are usual (and simple) techniques to check the reliability of the resulting numerical solution? So far I tried to use another solution algorithm by switching the Option Method from Method->Automatic to Method->RungeKutta. Apart from numerical differences, the solution qualitatively remained unaffected (I am mainly interested in the qualitative characteristics of the solution). I would be grateful for any hint! ==== Here is a version which compresses the list going forward, it does not seem to be significantly faster than the other version but the memory use should be slightly smaller: (*Handle some degenerate cases here *) fcompress[{}]:={}; (*This pattern is the stopping point of the recursion*) fcompress[lst:{{_?NumericQ,_?NumericQ}}]:=lst (*This function needs to be called with a sorted list*) fcompress[lst:{{_?NumericQ,_?NumericQ}..}]:=With[{rng1=lst[[1]],rng2=lst [[2]]},If[rng1[[2]] lst = {{100, 200}, {150, 250}, {120, 270}, {300, 400}}; List @@ Union @@ Interval /@ lst {{100, 270}, {300, 400}} List @@ Interval[Sequence @@ lst] {{100, 270}, {300, 400}} > Forgot about the Interval function, but to continue the pattern List@@Interval@@lst also works. I personally prefer not to use Sequence objects if I can avoid them. Sseziwa PS - It is much faster than my previous posts. ==== Dear group, very clear. Suppose that I have a list of points {p1,p2, ..., pn}, I try to find all possible pairs of them. The pairs may include {pi,pi}, but {pi,pj} and {pj,pi} are considered the same and only one is counted. That said. After some try and error, I came to the following way: h=4;w=5; points=Flatten[Outer[List,Range[w],Range[h]],1]; pairs=Flatten[Map[Outer[List,{#},Drop[points,Position[points,#][[1,1]]-1],1] [ [1]]&,points],1]; The speed of the above calculation is reasonably fast. But I run into the memory problem. For example, for h=64 and w=64, the length of the list pairs will be w*h (w*h+1)/2 = 8,390,656. In my case, The numbers for h and w will be a lot larger than 64. How can I get around this memory problem or that's the dead end for my calculation (I do have 1 GB physical mem in Cheng >Dear group, I have the following question regarding a lengthy calculation >using Mathematica: For given w points in x direction and h points in y direction, I can >construct all the points using h=10; w=8; > points=Flatten[Transpose[Outer[List,Range[w],Range[h]]],1] Next, I need to find all the possible pairs of point including points >themselves, i.e., pair AA. I can use pairs=Outer[List,points,points,1] Then, I have to clear those pairs that repeat themselves, i.e., pair AB and >pair BA. Also, when w and h are of the order of 1000s, the computation >takes a very long time. Is there a better way of doing the second part of Sincerely Cheng > ==== ================================================ >Cheng Liu, Ph.D. >MST-8, Structure/Property Relations >Materials Science and Technology Division >Los Alamos National Laboratory >Los Alamos, New Mexico 87545 ==== ================================================ ==== ================================================ Cheng Liu, Ph.D. MST-8, Structure/Property Relations Materials Science and Technology Division Los Alamos National Laboratory Los Alamos, New Mexico 87545 ==== ================================================ ==== Just two small (maybe relevant) notes: ----- Original Message ----- even some more than just today's machine precision. > Sorry, but that's not as profound as it sounds. The speed of light is > indeed a very specific number, but that doesn't mean we can measure it > precisely. Instead, like E or Pi or 2 or Sqrt[7], it's a defined > constant and -- unlike E or Pi or Sqrt[7] -- the definition doesn't > allow us to compute it with arbitrary precision. Yes, it's defined now > so that it can pretend to unlimited precision -- but that only means > meters (or seconds, take your pick) aren't defined precisely. The speed of light is *postulated* (rather then defined) constant, and (since 1983.) in the international metric (aka SI) system of units one meter is *defined* to be the distance traveled by light during (exactely) 1 / 299792458 seconds (second is defined some other way). So it follows that the (exact) value for the speed of light is 299792458 m / s. However, the problem is not in the correctness of this particular value (it is actually just a convinient convention), but in the postulate that this value is the same for all observers. To be specific, if there are two observers moving relative to eachother and measuring the speed of light using two (in principle) _identical_ experimental devices, the question is whether they get the same result, ie. c' = c. Experimentally speaking, one wihses to know how much (if at all) the quantity (c'-c)/c differs from zero. If someone measures a nonzero value, that _would_ actually be for a Nobel... Marko > -----Original Message----- > To: mathgroup@smc.vnet.net > In the real world of physics there are several subatomic level > processes > which can only be distinguished by small changes in the n-th decimal > place. > But there is one example which is fairly easy to comprehend, and that is > the > constancy of the speed of light in a vacuum regardless of reference > frame, > as proposed in Einstein's special theory of relativity. If this were > true > only to the 9th or 10th decimal place, or, for that matter, to the > 50th > place, then whoever managed to show that it was not really a constant > would > certainly be in Nobel Prize territory, and much of modern physics would > need > a rewrite. Kevin Greetings, I have read with great interest this lively debate on numerical > prcesion > and > accuracy. As I work in the fields of finance and economics, where we > feel > ourselves blessed if we get three digits of accuracy, I'm curious as > to > what > scientific endeavors require 50+ digits of precision? As I recall > there > are > some areas, such as high energy physics and some elements of > astronomy, > that > might require so many digits in some circumstances. Are there > others? > -Mark > Reply-To: ==== Sorry, but that's not as profound as it sounds. The speed of light is indeed a very specific number, but that doesn't mean we can measure it precisely. Instead, like E or Pi or 2 or Sqrt[7], it's a defined constant and -- unlike E or Pi or Sqrt[7] -- the definition doesn't allow us to compute it with arbitrary precision. Yes, it's defined now so that it can pretend to unlimited precision -- but that only means meters (or seconds, take your pick) aren't defined precisely. For anything we can measure (or even COUNT, in the real world), I suspect 16-digit machine precision is more than enough. Bobby -----Original Message----- as proposed in Einstein's special theory of relativity. If this were true only to the 9th or 10th decimal place, or, for that matter, to the 50th place, then whoever managed to show that it was not really a constant would certainly be in Nobel Prize territory, and much of modern physics would need a rewrite. Kevin > Greetings, I have read with great interest this lively debate on numerical prcesion > and > accuracy. As I work in the fields of finance and economics, where we feel > ourselves blessed if we get three digits of accuracy, I'm curious as to > what > scientific endeavors require 50+ digits of precision? As I recall there > are > some areas, such as high energy physics and some elements of astronomy, > that > might require so many digits in some circumstances. Are there others? > -Mark ==== I have 2 or more separate Plots which have different y but the same x axes. Like: Plot[Sin[x],{x,0,10}]; Plot[1000Sin[x],{x,0,10}]; At the display and printout, the y axes are not aligned. Even using the PlotRegion and ImageSize Options doesn't help. The only way I found was to align it manually with the mouse. Is there a package that solves the problem? The problem is, I have many, many plots to align... I use Mathematica 4.0.2.0 on a Windows 2000 PC. Who can help? Max ==== I have 2 or more separate Plots which have different y but the same x axes. Like: Plot[Sin[x],{x,0,10}]; Plot[1000Sin[x],{x,0,10}]; At the display and printout, the y axes are not aligned. Even using the PlotRegion and ImageSize Options doesn't help. The only way I found was to align it manually with the mouse. Is there a package that solves the problem? The problem is, I have many, many plots to align... I use Mathematica 4.0.2.0 on a Windows 2000 PC. Who can help? Max Answers please to: ulbrich@biochem.mpg.de ==== > A simple example : {{100,200},{150,250},{120,270},{300,400}} would result > in {{100,270},{300,400}}. In the real case, the input list has several hundred elements and the > output list typically has five elements. I have a working solution based on loops, but there must be a more elegant > one. I would be very grateful for any suggestions. Until recently, this could have been tedious, but now, tada!: In[31] := data = {{100, 200}, {150, 250}, {120, 270}, {300, 400}}; List @@ IntervalUnion @@ Interval /@ data Out[31] = {{100, 270}, {300, 400}} Tom Burton ==== >-----Original Message----- >Sent: Wednesday, October 16, 2002 8:26 PM >To: mathgroup@smc.vnet.net >Dear Fellows in MathGroup, I have a list of 17,000+ {x,y} pairs of data each x value is a positive integer from 1 to 100+ each y value is a positive real number As a *short* example, let's consider: data = {{3,1},{4,3},{3,2},{1,10},{4,2},{1,6},{5,2},{2,5},{7,1}} I want to group the data by the x value and report the >arithmetic average >of the y values in each group. For the example, i want to report: output = {{1,8},{2,5},{3,1.5},{4,2.5},{5,2},{6,0},{7,1}} In this example, x=6 does not occur so i report the average y[6] = 0. Can anyone suggest a way to do this efficiently?/ many thanks >dave +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ >David E. Burmaster, Ph.D. >Alceon Corporation >POBox 382069 (new Box number effective 1 Sep 2001) >Harvard Square Station >Cambridge, MA 02238-2069 (new ZIP code effective 1 Sep 2001) Voice 617-864-4300 Web http://www.Alceon.com >+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ Dave, my first attempt used the same reasoning as Daniel Lichtblau proposed (and came out only slightly faster than his). However the Sort/Split idea as brought forward by Bob Hanlon and Allan Hayes is much faster; Bobby Treat's version turns out to be slower (than Daniel's and mine). To reach comparable results, I slightly modified Allan's solution (which was fastest): data = Table[{Random[Integer, {1, 98}], Random[]}, {20000}]; (f[x_] = 0; ((f[#1[[1, 1]]] = Plus @@ #1[[All, 2]]/Length[#1]) &) /@ Split[Sort[data], #1[[1]] == #2[[1]] &]; r4 = {#, f[#]} & /@ Range[98];) // Timing {3.045 Second, Null} So I reconsidered that idea and found a solution which is nearly twice as fast: binnedAverage2[data_, max_] := Module[{v, i, ix, ixx, ixxx}, {i, v} = With[{rr = Range[max]}, Transpose[Sort[Join[data, Transpose[{rr, rr - rr}]]]]]; ix = Split[i]; ixx = FoldList[Plus[#1, Length[#2]] &, 0, ix]; ixxx = Transpose[Transpose[Partition[ixx, 2, 1]] + {1, 0}]; Transpose[{First /@ ix, Plus @@ #/Max[Length[#] - 1, 1] &[Take[v, #]] & /@ ixxx}]] (r7 = binnedAverage2[data, 98]); // Timing {1.612 Second, Null} r7 == r4 True -- Hartmut Wolf ==== greetings: can someone suggest a method for capturing what is printed via ?Global`* to a list? also i would like ?@ in a list. michael ==== I need to color the surface of a regular polyhedron (an icosahedron, specifically) according to a relatively simple function of the spatial coordinates of the surface. I easily got a nice icosahedron of the appropriate size, but thus far I have been unable to resolve the coloring issue. Any advice is appreciated. Francis ==== gg={{ob1},{ob2},{ob3}}; However, the output of f having a form like: f[gg] gives {{{ob11}, {ob12}, {ob13}},{{ob21}, {ob22}, {ob23}, {ob24}}, ... ,{{obn1}, {obn2}, {obn3}}} where the number of objects in each layer can vary. The additional intermediate layer in the output of f prevents the feedback to f, when use function like nest. I looked up and tried several method, and it seems to be easy to get rid of all inner layers, or the innerest layer, or the outmost layer (use Sequence). Is there a way to get rid of the middle layer as describe above? Sincerely, JT _________________________________________________________________ Surf the Web without missing calls!æGet MSN Broadband. http://resourcecenter.msn.com/access/plans/freeactivation.asp