Le forum de XCAS

Question

The acronym Xcas is the union of X (Multi multiplatform OS) with Computer Algebra System?

1: ∂(v(t),t) [enter] returns diff(v(t),t,1)
2: ∫(entry(-1),x) [enter] Xcas is closed

now
1: ∂(v(t),t) [enter] returns diff(v(t),t,1)
2: ∫(ans(-1),t) [enter] returns 0, The expected result is v(t)

Hello, to make the script of steps an easier to understand I propose that the ans command can accept an optional second argument, extract the x element from the previous list ans(row pos)(element) -> ans(row pos, list element) current: with ans(arg1) x^2 = 2*x + 3 returns x^2=(2*x+3) ans(-1) - (2*x ...

But on line#4 left(ans(-1)) [enter] [enter] should call to the line previous line#3 (x-3)*(x+1)=0 and not on your own answer, because the input must be recalculated when modified, for example the expression is badly written or you want to change it

ok ^ The next 4 steps work perfectly ans(-1) = ans(previous) entry#1: x^2 = 2*x + 3 [enter] answer#1 x^2=(2*x+3) entry#2: ans(-1) - (2*x + 3) [enter] answer#2 x^2-2*x-3=0 entry#3: factor(ans(-1)) [enter] answer#3 (x-3)*(x+1)=0 entry#4: left(ans(-1)) [enter] answer#4 (x+1)*(x-3) entry#5: But if I giv...

Hi, if I have the following entry

#1: a
a

#2:b
b

#3:c
c

#4:d
d

#5:

Please someone please explain me how to call for example the entry line# 2 and output line# 3 from line #5 with ANS() y ENTRY() CMDS

thanks

Hello, Mr. Bernard, The following sequence sentences in the hp-prime ENTRY (2*x^2-3*x+4) > (3*x^2-2*x-2); Ans + -2*x^2; simplify(Ans); Ans + 3*x; simplify(Ans); Ans + -4; simplify(Ans); returns (2*x^2-3*x+4) > (3*x^2-2*x-2) (2*x^2-3*x+4-2*x^2) > (3*x^2-2*x-2-2*x^2) (-3*x+4) > (x^2-2*x-2) (-3*x+4+3*x...