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Hello
for linear systems
Example #1
[HP-Prime CAS] solve( 4*x+3*y=10 and 5*x-2*y=1, {x,y} ) [Enter] return { [ 1, 2 ] }
// OK, but best
[Request] tag the answers
solve( 4*x+3*y=10 and 5*x-2*y=1, {x,y} ) [Enter] return { [ x=1, y=2 ] } OK OK and if { [ x=1, and: y=2 ] } OK, OK, OK /!\ "and:" as tag (idea)
http://www.wolframalpha.com/input/?i=solve%28+4*x%2B3*y%3D10+and+5*x-2*y%3D1%2C+%7Bx%2Cy%7D%29
purpose, use the expression (output) in a program
left(x = 3) => x // to extract the variable
right(x = 3) => 3 // to extract the result
[Request] rename left to leftPart and right to rightPart
linSolve( 4*x+3*y=10 and 5*x-2*y=1, {x,y} ) [Enter] return [ 1, 2 ] OK and if [ x=1, y=2 ] OK OK, and if [ x=1, and:y=2 ] OK OK OK
Example #2
[HP-Prime CAS] solve( 4*x+6*y=10 and 2*x-3*y=6, {x,y} ) [Enter] return {}
[Request] Better a boolean output (true/false) like TI-CAS+ calculators
solve( 4*x+6*y=10 and 2*x-3*y=6, {x,y} ) [Enter] return false
purpose use the expression (output) in a program
if solve( 4*x+6*y=10 and 2*x-3*y=6, {x,y} ) == true then ... else ...
now required a "trick" use DIM( ... ) =(
if DIM( solve( 4*x+6*y=10 and 2*x-3*y=6) )== true, {x,y} ) then // ugly use DIM(
equal with linSolve
linSolve( 4*x+6*y=10 and 2*x-3*y=6, {x,y} )return [ ] ?? empty vector, ugly and nothing didactic for teaching and learning of math
best
linSolve( 4*x+6*y=10 and 2*x-3*y=6, {x,y} )return false =)
if linSolve( 4*x+6*y=10 and 2*x-3*y=6, {x,y} ) == true then ...
//
[BUG#0 solve cmds] with SIMULT
simult(
[ [ 4, 6 ] [ [10]
[ 2, 3 ]], [6]]
return [ 0 ; 1 ] ?? // Bug 4*0 + 6 * 1 = 10 => 6 = 10 => false
simult(
[ [ 4, 6 ] [ [10]
[ 2, 3 ]], [6]]
return "Not Solutions Exist" // OK // or "singular matrix" like TI-CAS+ calcs or wolframAlpha
see wolframalpha
http://www.wolframalpha.com/input/?i=solve%28+4*x%2B6*y%3D10+and+2*x%2B3*y%3D6%2C+%7Bx%2Cy%7D%29
Example #3
solve( 4*x+6*y=12 and 2*x-3*y=2, {x,y} ) [Enter] return false =)
simult(
[ [ 4, 6 ] [ [12]
[ 2, 3 ]], [6]]
return [ 1 ; 0 ] ?? // Bug
simult(
[ [ 4, 6 ] [ [12]
[ 2, 3 ]], [6]]
return "Infinite Solutions" // Super OK
Thanks
///////////////////////////////
PD: [spanish]
REQUEST Se requiere un comando para etiquetar similar a ->TAG de la HP50
ejemplo
tag( "cplx1:", 3+4*i ) [Enter] cplx1: 3+4*i
tag({ "x:", "y:", "z:" }, [ 3 4 1 ] ) [Enter] [ x:3, y:4, z:1 ]
/!\ la etiqueta no afectaría los cálculos, es muy útil por ejemplo para indicar en pantalla resultados mas bonitos sin la necesidad de convertirlos a cadenas, pues se cambiaría el tipo de dato
Edited: 6 Aug 2013, 2:57 p.m. after one or more responses were posted
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I don't understand at all why you want the result 'false' for a system of equation with no solution. For me it's a non sense. The 'system' is not false, there is no solution -> S={}. My english is limited for such things but an empty set is not 'false'
I also don't like the solution in the format (x=xx and y=xx) or (x=xx and y=yy) : it is verbious and confusing with boolean operators.
S={s1; s2; ...} is far better (a set of solution)
or
S={} if there is no solution (an empty set)
You also not use the correct syntax for linsolve ( [] and not {} ), and solve ( idem) . The AND syntax to solve a system of equation is curious for me and not documented in on line help of the Prime
Edited: 5 Aug 2013, 2:47 a.m.
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>> I don't understand at all why you want the result 'false' for a system of equation with no solution
can also be infinite solutions, in a linear system
a student think when see an empty list?
while you would think if you see "false" or "true" or when return the same input?
all these details are think by TI-CAS developers
>> I also don't like the solution in the format (x=... and y=...) or (x=... and y=...) : it is verbious and confusing with boolean operators.
[Castillian]
todo lo contrario, es mucho mas "inteligente" esta respuesta, me dice que soluciones son condicionales, decir, si las dos soluciones satisfacen la ecuación o si alguna de ellas satisface la ecuación
[Eng] sorry, Google translator
on the contrary, is much "smarter" this response ((x=... and y=...) or (x=... and y=...)), I said that solutions are conditional, ie if the two or more solutions satisfy the equation or if any of them satisfies the equation
Please watch the following equation with 4 solutions
[HP-Prime CAS] solve ( x^2-10*x+y^2=-8 and x*y^2+x+10*y=-8, {x,y} ) [Enter] return
{
[1, 1],
[2.19 , 3.02 ]
}
[HP-Prime CAS++] solve ( x^2-10*x+y^2=-8 and x*y^2+x+10*y=-8, {x,y} ) [Enter] return
{
[x:1, and y:1],
[x: 2.19, and y:3.02 ]
}
(idea or request) "x:", "y:" & "and y:" as TAG (->TAG CMD HP50G), then to operate the label disappears
>> S={} if there is no solution (an empty set)
When the HP-Prime can not calculate, returns {} [BUG], should return the same expression otherwise you would think that there are no solutions or infinite solutions, etc ...
>> The AND syntax to solve a system of equation is curious for me and not documented in on line help of the Prime
[HP-Prime CAS] solve ( x^2-10*x+y^2=-8 AND x*y^2+x+10*y=-8, {x,y} )
{ x^2-10*x+y^2=-8 , x*y^2+x+10*y=-8 } == x^2-10*x+y^2=-8 AND x*y^2+x+10*y=-8 // =) TI-CAS and others CAS computer sofware style =]
and if apply this input form, why not in the output?
Edited: 5 Aug 2013, 9:51 a.m.
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I don't agree with you here
In the same way, when the TI returns 'false' with solve or linsolve command it is an error: There are no such things as a 'false' equation or 'false' system of equations
5=5 is true, but it's not an equation, it's an equality
6=5 is false
but the system of equation :
/
| x-y=0
| x-y-4=0
\
is not 'false'. S={}
I you search the intersection of 2 straight lines wich are // , the system is not 'false'. There is no solution and that's a different thing
The same with 2 circles you can get : {} or {[a,b]} or {[a,b],[c,d]} or the equation of circle if they are the same. But no 'false' results nor false system
EDIT :
"a student think when see an empty list?"
An emply list is a 'void set' (ensemble vide) = no solution
An infinity of solution can be :
y=f(x) ex : {y=2.x}
or All (x,y) € |R² in very special cases
Edit :in this case,Prime returns {[x,y]}
If an infinity of solution results as {}, there is a bug
The only bug I see here is (I cant test):
solve(sqrt(x^x))=x^sqrt(x))
EDIT : I get "unable to isolate x"...for this one
You can find the obvious solutions {1.,4.} with the numeric solver
And always the same thing : the Prime dont verify matrix dimensions with your simult with bad parameters
EDIT 3
Quote:
if solve( 4*x+6*y=10 and 2*x-3*y=6, {x,y} ) == true then ... else ...
now required a "trick" use DIM( ... ) =(
if DIM( solve( 4*x+6*y=10 and 2*x-3*y=6) )== true, {x,y} ) then // ugly use DIM(
you can use if solve([eq1,eq2],[x,y])<>{}
Edited: 5 Aug 2013, 2:00 p.m.
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Example #4
solve( x1-3*x2+x3=1 and 2*x1-x2-2*x3=-1 and x1+2*x2-3*x3=-1, {x1,x2, x3})
x1-(7/5)*x3 = 0 => x1 =7/5*x3
x2 -(4/5)x3 = 0 => x2 = (4/5)x3
0*x3 = 1 // => 0 = 1 inconsistent system (FALSE equation)
source
http://www.cengage.com/cgi-wadsworth/course_products_wp.pl?fid=M20bI&flag=instructor&product_isbn_issn=9780618783762&disciplinenumber=1
[request]
factor( [ t; 4*t; (1/2)*t ] ) => t * [ 1; 4 ; 1/2 ]
transpose([[ 1 2 ] [ a, b ]]) => [[ 1 2 ] [ a, b ]]T
concat( M1, M2) => M1|M2
Edited: 6 Aug 2013, 3:10 p.m.
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"0*x3 = 1 // => 0 = 1 inconsistent system (FALSE equation) "
This '0*x3=1 => 0=1' hurts my eyes ;)
0*x^3=1 is an equation without solution (altough in some case x=oo )
0=1 is a false equality, a mathematical non sense
0*x3=1 => S={} as there is no x€|R that verify this.
Is 'a*x=1' a sometimes true / sometimes false equation ?
Another example :
x/x=1 is not a 'True' equation
S=|R* or if you prefer
S=|R - {0}
That means that x/x=1 is not the same than x=x wich gives S=|R
Edited: 7 Aug 2013, 6:27 a.m.
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[HP-prime CAS 0.8.3] solve (sqrt(x^x))=x^sqrt(x), x) [Enter] {} // BUG
[HP-prime CAS 1.1.0] solve (sqrt(x^x))=x^sqrt(x), x) [Enter]
["Unable to isolate x in exp(1/2*x*ln(x))-exp(exp(1/2*ln(x))*ln(x))"] // OK // I hope in the coming months, that CAS can solve this =)
... [Spn] =(
translate.google.en
Se ha corregido MAL un BUG que reporte recientemente, el cual es que si un sistema lineal no tiene solución, o infinitas soluciones SIMULT, retornaba como salida el vector columna final de la operación RREF, este resultado o vector no satisface las ecuaciones, por lo tanto como lo hacen otros CAS, debe arrojar un mensaje de error, ahora la HP-Prime CAS, retorna un BAD mensaje de error
M1: = [[1,-3,1],[2,-1,-2],[1,2,-3]]
V1:= [[1],[2],[-1]] => // [Request] Me gustaría que el CAS aceptara V1:= [ 1; 2; -1 ]
[HP-prime CAS 0.8.3] simult( M1, V1 ) => [ ] // BUG
[HP-prime CAS 1.1.0] simult( M1, V1 ) =>
Error: Bad Argument Value" <= ???
simult( M1, V1 ) =>
[ solution ], "Infinitas soluciones" or "no hay solución" dependiendo de las ecuaciones finales// OK
"Infinitas soluciones" o "no hay solución" pero no "Bad Argument Value" como que "Mal Valor de Argumento"??, cual el primero o el segundo argumento es un mal valor?, Acá no hay error en los argumentos, por lo tanto lo considero nuevamente como un BUG
Es muy fácil detectar si hay "Infinitas soluciones" o "no hay solución" se realiza RREF( CONCAT( M1,V1 )) y se analiza las ecuaciones de salida, para determinar si hay "solución única", "Infinitas soluciones" or "no hay solución", yo programe esto hace mucho tiempo en la HP48GX
[Request] permitir el operador | como concatenación de matrices/vectores
CONCAT( M1,V1 ) == M1|V1, otros CAS usan barra vertical punteada
RREF( CONCAT( M1,V1 )) => RREF( M1 | V1 )
Edited: 7 Aug 2013, 6:45 p.m.
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El TI-CAS es un CAS mas "inteligente" que el HP-CAS, por que se aproxima mas a como un ser humano lo haría, mientras que el HP-CAS sus respuestas no son tan intuitivas y como es para un mercado estudiantil pre-universitario, dudo que tenga buena acogida, casi siempre en la HP-Prime hay que estar re-escribiendo las expresiones para obtener algo mas facil de manipular
Ejemplo: Sistemas de ecuaciones lineales con "Soluciones infinitas"
x1-3*x2+x3 = 1
2*x1-x2-2*x3 = -1
2*x1-6*x2+2*x3 = 2
en forma A*x=b
[[1,-3,1,1], [ x1; x2; x3 ] = [ 1; -1; 2 ]
[2,-1,-2,-1], *
[2,-6,2,2]]
M1:=
[[1,-3,1],
[2,-1,-2],
[2,-6,2]]
x:= [[ x1],[x2][x3]]
b: = [[1], [-1],[2]]
sol:={}
Aplicando RREF (reduced row echelon form)
rref( concat( M1, b ) ) [Enter] =>
[[1,0,-7/5,-4/5],
[0,1,-4/5,-3/5],
[0,0,0,0]]
x1=(7/5)*x3-4/5
x2=(4/5)*x3-3/5
x3=x3
[TI-CAS] solve( x1 -3*x2 +x3 = 1 and 2*x1 -x2 -2*x3 = -1 and 2*x1 -6*x2 +2*x3 = 3), { x1, x2, x3 } )
:)
x1=(7* c1 -4)/5 and
x2=(4*c1-3 /5) and
x3=c1
sol:= Ans
sol:= x1=(7* c1 -4)/5 and x2=(4*c1-3 /5) and x3=c1
reemplaza x3 = c1, en la parte derecha de las expresiones, salida en función de un parámetro, el cual tiene un propósito evaluar las expresiones sin afectar la variable parametro
x1=(7* c1 -4)/5 and x2=(4*c1-3 /5) and x3=c1 | (c1 =2) =>
x1=1 and x2 = 1 and x3=2
pueden observar con mas detalle la solución en:
http://www.math.odu.edu/~bogacki/cgi-bin/lat.cgi?cmd=sys;m1_1=1;m1_2=-3;m1_3=1;m1_4=1;m2_1=2;m2_2=-1;m2_3=-2;m2_4=-1;m3_1=2;m3_2=-6;m3_3=2;m3_4=2;curn=3;curn=4;vdat=3;vdat=R3;whid=V;whid=1;submit21=Submit;hideops=1
[HP-CAS] solve( x1 -3*x2 +x3 = 1 and 2*x1 -x2 -2*x3 = -1 and 2*x1 -6*x2 +2*x3 = 3), { x1, x2, x3 } )
:(
{ [ x1, -1/35*(-20*x1+5), -1/4*(5/35*(-20*x1+5)-3) ] }
Comparen las dos respuestas anteriores, por cual calculadora se inclinarían?
sol:= Ans
sol:= { [ x1, -1/35*(-20*x1+5), -1/4*(5/35*(-20*x1+5)-3) ] }
re-escribiendo las expresiones anterior
sol := simplify( Sol )
sol := simplify( { [ x1, -1/35*(-20*x1+5), -1/4*(5/35*(-20*x1+5)-3)] } =>
sol := [ x1, (4*x1-1)/7, (5*x1+4)/7]
reescribiendo para que sea útil dentro de un programa
x := sol // fail
transpose(x) := sol // fail
vx : = [ x1, x2, x3 ] // se requiere una nueva definición del vector x como vector fila
sol: = (vx = sol) =>
sol:= ([ x1, x2, x3 ] = [ x1, (4*x1-1)/7, (5*x1+4)/7 ])
sol:=([ x1 = x1, x2 = (4*x1-1)/7, x3=(5*x1+4)/7 ] )
todos los pasos anteriores para poder tener una respuesta mejor :(
en resulmen en la HP-Prime
[ x1, x2, x3 ] = simplify ( solve ( x1 -3*x2 +x3 = 1 and 2*x1 -x2 -2*x3 = -1 and 2*x1 -6*x2 +2*x3 = 3), { x1, x2, x3 } ) )
[ x1 = x1, x2 = (4*x1-1)/7, x3=(5*x1+4)/7 ]
mientras que en la TI-CAS
solve ( x1 -3*x2 +x3 = 1 and 2*x1 -x2 -2*x3 = -1 and 2*x1 -6*x2 +2*x3 = 3), { x1, x2, x3 } )
x1=(7* c1 -4)/5 and x2=(4*c1-3 /5) and x3=c1
con el comando LInSolve las cosas son mejores
[ x1, x2, x3 ] = solve ( x1 -3*x2 +x3 = 1 and 2*x1 -x2 -2*x3 = -1 and 2*x1 -6*x2 +2*x3 = 3), { x1, x2, x3 } )
[ x1=(7/5* x3 -4), x2=(4/5*x3-3 ), x3=x3 ]
pero hay un problema [ x1=(7/5* x3 -4), x2=(4/5*x3-3 ), x3=x3 ] | x=3 =>
[ x1 = 2, x2=1, 2=2 ] ver tercer elemento ( 2 =2 )
Si linsolve arrojara en función de un parámetro seria fabuloso y si incluyera las variables, se igualria al CAS de las TI-calculators
linSolve ( x1 -3*x2 +x3 = 1 and 2*x1 -x2 -2*x3 = -1 and 2*x1 -6*x2 +2*x3 = 3), { x1, x2, x3 } )
[ x1=(7/5* t -4), x2=(4/5*t-3 ), x3=t ] // Super CAS
[ x1=(7/5* t -4), x2=(4/5*t-3 ), x3=t ] | t = 2
[ x1 = 2, x2=1, x3=2 ]
Edited: 8 Aug 2013, 10:11 a.m.
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Quote:
[TI-CAS] solve( x1 -3*x2 +x3 = 1 and 2*x1 -x2 -2*x3 = -1 and 2*x1 -6*x2 +2*x3 = 3), { x1, x2, x3 } )
:)
x1=(7* c1 -4)/5 and
x2=(4*c1-3 /5) and
x3=c1
sol:= Ans
sol:= x1=(7* c1 -4)/5 and x2=(4*c1-3 /5) and x3=c1
reemplaza x3 = c1, en la parte derecha de las expresiones, salida en función de un parámetro, el cual tiene un propósito evaluar las expresiones sin afectar la variable parametro
x1=(7* c1 -4)/5 and x2=(4*c1-3 /5) and x3=c1 | (c1 =2) =>
x1=1 and x2 = 1 and x3=2
What a complex way to do !
Why not :
s:=linsolve([x1-3*x2+x3=1,2*x1-x2-2*x3=-1,2*x1-6*x2+2*x3=2],[x1,x2,x3])
s|x3=2
Edit : Your system of equation is not the same in the beginning and the end of your post. I take the first one.
To HP -> The spaces between equations is too small in this release
Edited: 8 Aug 2013, 12:32 p.m. after one or more responses were posted
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I think as you would do in a program, not in the entry line
Please can create a program that made the steps above, detecting the parameter
Edited: 8 Aug 2013, 12:33 p.m.
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Sorry but I don't understand spanish, I dont'see see any program
Cheers,
EDIT :
For what I understand of your TI example, to do the same on the HP
linsolve([x1-3*x2+x3=1,2*x1-x2-2*x3=-1,2*x1-6*x2+2*x3=2],[x1,x2,x3])
STO> s
s|x3=c1
STO> s
s|c1=2
But I don't see any interest to replace x3 with c1
The last rev has auto-simplification options in CAS setup
Edited: 8 Aug 2013, 5:11 p.m.
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