Solve for diff(e2, x, 1) after x and d have been reassigned new values after e4 at e5.

Burton writes.

>(%o5)?[y?=?0]

 

>Now?e4?has?been?reassigned?new
values?at?e5,?x=50?and?d=50.?If?I?write?e6:?solve(e5,x);?then?I?get?[].

 

I do not understand what "new values at e5" means. Currently e5 is [y=0]
(see output %o5). That makes input (%6) equivalent to solve([y=0], x).
Because there are no values for x that satisfy the equation y=0, Maxima
gives the empty list (not an error) as output in (%o6). 

 

 >(%i6)?e6:?solve(e5,x);

 >(%o6)???????????????????????????????

 >[]

 

I don't see a problem. What am I missing?

--Barton

 

Maxima is new to me so that is part of my problem.

To start over again the issue is to Solve for diff(e2,x,1) after x and d
have been reassigned new values after e4 at e5.

 

(%i1) e1: x^2+d*y*4+d^2*(-4);

                                            2      2

(%o1)                          4 d y + x  - 4 d

(%i2) e2: solve(e1, y);

                                                2      2

                                              x  - 4 d

(%o2)                          [y = - ---------]

                                                 4 d

(%i3) e3: subst([y=0, d=5], e1);

                                          2

(%o3)                              x  - 100

(%i4) e4: solve(e3, x);

(%o4)                         [x = - 10, x = 10]

(%i5) e5: subst([x=50, d=25], e2);

(%o5)                               [y = 0]

 

So at this point x=-10 and 10 when y=0.

Now e4 has been reassigned new values at e5, x=50 and d=50. If I write e6:
solve(e5,x); which gives me []. Now when I continue to solve and assign as
follows.

 

(%i6) e6: solve(e5,x);

(%o6)                                []

(%i7) e7: diff(e2,x,1);

                                                 x

(%o7)                            [0 = - ---]

                                               2 d

(%i8) %phi[1] : e7;

                                                 x

(%o8)                            [0 = - ---]

                                               2 d

(%i9) %phi[2] : e7;

                                                 x

(%o9)                            [0 = - ---]

                                               2 d

Now when I execute.

(%i10) subst([e6[1], d=25],%phi[1]); 

(%i11) subst([e6[2], d=25],%phi[2]);

I get "Improper index to list or matrix - an error.

 

The values I should have for (%i10) and (%i11) should be 1.0 and -1.0. The
purpose of this set of expressions is to be able to reassign values to x and
d and wind up with the derivative values of (%i2).

 

William Porter

wporter at omegapar.com