Le forum de XCAS

Adding new rules to the Xcas kernel should be an easy job in words but in practice it requires many hours of work and movement of neurons that very few have. There are very few CAS programmers in the world and my brain is not at that level. I would like to have the power to access the source code an...

Having a smaller number of cases for further processing is good from a computational point of view, but not from the point of view of teaching and learning mathematics, because textual mathematics follows writing rules. In addition to spending more time, with current binary processors, that time to ...

Hello

solve((5<(-2*x+7))AND ((-2*x+7)≤11),x,'=' ); [Enter] returns (x≥-2) and (x<1)
but -2≤x<1 [Enter] returns (x>=-2) and (1>x)


is it possible to improve the results of mathematical expressions as they appear in textbooks?
(x>=-2) and (1>x) => (x>=-2) and (x<1)

Hello

a possible problem is when an expression is applied in an inequality.

autosimplify(2);

(((x-2)/(x+2))<2)*(x+2) [Enter] ([(6+x)/(2+x),(12+2*x)/(2+x)])>0

is expected => (x-2) < (2*(x+2))

I have noticed that depending on the simplification flag, an assignment is performed.
You can improve the xcas interpreter by adding a flag to the system that allows you to decide whether you want the equal sign to always act as an assignment or not,

Another possible problem is when an expression is applied in an inequality.

(((x-2)/(x+2))<2)*(x+2) [Enter] ([(6+x)/(2+x),(12+2*x)/(2+x)])>0

=> (x-2) < (2*(x+2))

Tested on version 1.99.0 - 990 and it works excellently

ifactors(18,'*') [Enter] Returns 2*3*3

Hello with autosimplify flag: none ((a=b)*-1 [enter] returns -a=(-b) ((a>b)*-1 [enter] returns (-b)>(-a) ((a=b)*-1)|([a = 4,b = 3]) [enter] returns (-4=-3) ((a>b)*-1)|([a = 4,b = 3]) [enter] returns (true)*-1 => -1 ? // Shouldn't the left side be resolved first? ((a>b)*-1)|([a = 4,b = 3]) => (-b)>(-...

Hello I believe that a 'CAS' to be used in learning and teaching mathematics should have a form in which the user tells the machine what to do, so that the machine does not return automatic processes that are meaningless to the learner. This implies that the input is not interpreted except for trivi...

Hello
It is important to show the complete factorization of a number, so is there a possibility of obtaining the product of factors without powers?

ifactor(18) [enter] 2^1*3^2
ifactor(18, '*') [enter] '2*3*3'

Greetings Mr. Parisse, it is good that the GUI can be contracted with third parties, that is, seek sponsorship to be able to pay for these services. The Xcas online user interface works well, but in underdeveloped countries like Colombia we do not have guaranteed internet service during a class sess...

Hello Based on order powexpand(a^(m-n)) returns a^m/a^n In step-by-step algebraic manipulation I believe that a command is needed to expand an exponent, in a base that can be a product, division or expression. I don't know if baseexpand is a good name 0: powexpand(a^(m-n)) returns a^m/a^n 1: baseexp...

very good morning Within algebraic manipulation it is important to identify the terms or addends of an algebraic expression, there is already a command to separate the factors split(), I think that Xcas should provide us with a new command terms() so that students identify the terms after a manual p...

Is it possible to add in the [M] eq menu an option to convert two-dimensional notation to one-dimensional and live in the outputs of each line?

(1D)
a^(b^c)

(2D)
__ c
_ b
a

Hello I am a professor of algorithmic logic and programming in first university level courses, in these last two years I have been dictating the class sessions with c#, but I would like the algorithmic thinking to go more of the numerical, this is symbolic numerical. I am testing the xcasjs interfac...