From symbolic to proper differentiation in Maxima
I'm struggling to find a way to switch from a symbolic declaration of the differential operator to its implementation
I give you an example.
F: (10-'diff(x(t),t)^2 -2*x(t)*'diff(x(t),t) -5*x(t)^2)*%e^(-t);
E: ratsimp(diff(F, x(t)) - diff(diff(F, 'diff(x(t),t)), t));
sol: ode2(E, x(t), t);
sol: ev(sol, [%k1 = C1, %k2=C2]);
trans_cond: diff(F, 'diff(x(t), t));
trans_cond: ev(trans_cond, sol);
trans_cond: at(trans_cond, [t=1]);
The corresponding output maintains the symbolic notation whereas I would like to evaluate the diff() obtained after the last substitution.
Giving the result:
% 4*C1-C2^(-2)
maxima differentiation
asked Nov 25 '18 at 16:35
Marco RepettoMarco Repetto
467
add a comment |
I'm struggling to find a way to switch from a symbolic declaration of the differential operator to its implementation
I give you an example.
F: (10-'diff(x(t),t)^2 -2*x(t)*'diff(x(t),t) -5*x(t)^2)*%e^(-t);
E: ratsimp(diff(F, x(t)) - diff(diff(F, 'diff(x(t),t)), t));
sol: ode2(E, x(t), t);
sol: ev(sol, [%k1 = C1, %k2=C2]);
trans_cond: diff(F, 'diff(x(t), t));
trans_cond: ev(trans_cond, sol);
trans_cond: at(trans_cond, [t=1]);
The corresponding output maintains the symbolic notation whereas I would like to evaluate the diff() obtained after the last substitution.
Giving the result:
% 4*C1-C2^(-2)
maxima differentiation
asked Nov 25 '18 at 16:35
Marco RepettoMarco Repetto
467
add a comment |
I'm struggling to find a way to switch from a symbolic declaration of the differential operator to its implementation
I give you an example.
F: (10-'diff(x(t),t)^2 -2*x(t)*'diff(x(t),t) -5*x(t)^2)*%e^(-t);
E: ratsimp(diff(F, x(t)) - diff(diff(F, 'diff(x(t),t)), t));
sol: ode2(E, x(t), t);
sol: ev(sol, [%k1 = C1, %k2=C2]);
trans_cond: diff(F, 'diff(x(t), t));
trans_cond: ev(trans_cond, sol);
trans_cond: at(trans_cond, [t=1]);
The corresponding output maintains the symbolic notation whereas I would like to evaluate the diff() obtained after the last substitution.
Giving the result:
% 4*C1-C2^(-2)
maxima differentiation
asked Nov 25 '18 at 16:35
Marco RepettoMarco Repetto
467
I'm struggling to find a way to switch from a symbolic declaration of the differential operator to its implementation
I give you an example.
F: (10-'diff(x(t),t)^2 -2*x(t)*'diff(x(t),t) -5*x(t)^2)*%e^(-t);
E: ratsimp(diff(F, x(t)) - diff(diff(F, 'diff(x(t),t)), t));
sol: ode2(E, x(t), t);
sol: ev(sol, [%k1 = C1, %k2=C2]);
trans_cond: diff(F, 'diff(x(t), t));
trans_cond: ev(trans_cond, sol);
trans_cond: at(trans_cond, [t=1]);
The corresponding output maintains the symbolic notation whereas I would like to evaluate the diff() obtained after the last substitution.
Giving the result:
% 4*C1-C2^(-2)
maxima differentiation
maxima differentiation
asked Nov 25 '18 at 16:35
Marco RepettoMarco Repetto
467
asked Nov 25 '18 at 16:35
Marco RepettoMarco Repetto
467
asked Nov 25 '18 at 16:35
Marco RepettoMarco Repetto
467
asked Nov 25 '18 at 16:35
Marco RepettoMarco Repetto
467
asked Nov 25 '18 at 16:35
Marco RepettoMarco Repetto
467
467
add a comment |
add a comment |
2 Answers
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Another solution. The nouns option for ev causes the evaluation of symbolic derivatives, and also any other noun expressions such as symbolic integrals, symbolic summations, etc.
(%i2) 'diff(4*x^2, x);
d 2
(%o2) -- (4 x )
dx
(%i3) ev (%o2, nouns);
(%o3) 8 x
A shorter form of ev(..., nouns) is recognized by the interactive console. You can input ..., nouns instead.
(%i5) %o2, nouns;
(%o5) 8 x
Here is ev(..., nouns) applied to symbolic integral:
(%i6) 'integrate (x^2, x);
/
[ 2
(%o6) I x dx
]
/
(%i7) %, nouns;
3
x
(%o7) --
3
and here, to a symbolic summation:
(%i8) 'sum (f(k), k, 1, 3);
3
====
(%o8) > f(k)
/
====
k = 1
(%i9) %, nouns;
(%o9) f(3) + f(2) + f(1)
answered Nov 26 '18 at 3:02
Robert DodierRobert Dodier
11.3k11834
add a comment |
Found the answer, ev() carries the option diff which solves all the symbolic differentiation in the expression.
answered Nov 25 '18 at 16:45
Marco RepettoMarco Repetto
467
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Another solution. The nouns option for ev causes the evaluation of symbolic derivatives, and also any other noun expressions such as symbolic integrals, symbolic summations, etc.
(%i2) 'diff(4*x^2, x);
d 2
(%o2) -- (4 x )
dx
(%i3) ev (%o2, nouns);
(%o3) 8 x
A shorter form of ev(..., nouns) is recognized by the interactive console. You can input ..., nouns instead.
(%i5) %o2, nouns;
(%o5) 8 x
Here is ev(..., nouns) applied to symbolic integral:
(%i6) 'integrate (x^2, x);
/
[ 2
(%o6) I x dx
]
/
(%i7) %, nouns;
3
x
(%o7) --
3
and here, to a symbolic summation:
(%i8) 'sum (f(k), k, 1, 3);
3
====
(%o8) > f(k)
/
====
k = 1
(%i9) %, nouns;
(%o9) f(3) + f(2) + f(1)
answered Nov 26 '18 at 3:02
Robert DodierRobert Dodier
11.3k11834
add a comment |
Another solution. The nouns option for ev causes the evaluation of symbolic derivatives, and also any other noun expressions such as symbolic integrals, symbolic summations, etc.
(%i2) 'diff(4*x^2, x);
d 2
(%o2) -- (4 x )
dx
(%i3) ev (%o2, nouns);
(%o3) 8 x
A shorter form of ev(..., nouns) is recognized by the interactive console. You can input ..., nouns instead.
(%i5) %o2, nouns;
(%o5) 8 x
Here is ev(..., nouns) applied to symbolic integral:
(%i6) 'integrate (x^2, x);
/
[ 2
(%o6) I x dx
]
/
(%i7) %, nouns;
3
x
(%o7) --
3
and here, to a symbolic summation:
(%i8) 'sum (f(k), k, 1, 3);
3
====
(%o8) > f(k)
/
====
k = 1
(%i9) %, nouns;
(%o9) f(3) + f(2) + f(1)
answered Nov 26 '18 at 3:02
Robert DodierRobert Dodier
11.3k11834
add a comment |
Another solution. The nouns option for ev causes the evaluation of symbolic derivatives, and also any other noun expressions such as symbolic integrals, symbolic summations, etc.
(%i2) 'diff(4*x^2, x);
d 2
(%o2) -- (4 x )
dx
(%i3) ev (%o2, nouns);
(%o3) 8 x
A shorter form of ev(..., nouns) is recognized by the interactive console. You can input ..., nouns instead.
(%i5) %o2, nouns;
(%o5) 8 x
Here is ev(..., nouns) applied to symbolic integral:
(%i6) 'integrate (x^2, x);
/
[ 2
(%o6) I x dx
]
/
(%i7) %, nouns;
3
x
(%o7) --
3
and here, to a symbolic summation:
(%i8) 'sum (f(k), k, 1, 3);
3
====
(%o8) > f(k)
/
====
k = 1
(%i9) %, nouns;
(%o9) f(3) + f(2) + f(1)
answered Nov 26 '18 at 3:02
Robert DodierRobert Dodier
11.3k11834
Another solution. The nouns option for ev causes the evaluation of symbolic derivatives, and also any other noun expressions such as symbolic integrals, symbolic summations, etc.
(%i2) 'diff(4*x^2, x);
d 2
(%o2) -- (4 x )
dx
(%i3) ev (%o2, nouns);
(%o3) 8 x
A shorter form of ev(..., nouns) is recognized by the interactive console. You can input ..., nouns instead.
(%i5) %o2, nouns;
(%o5) 8 x
Here is ev(..., nouns) applied to symbolic integral:
(%i6) 'integrate (x^2, x);
/
[ 2
(%o6) I x dx
]
/
(%i7) %, nouns;
3
x
(%o7) --
3
and here, to a symbolic summation:
(%i8) 'sum (f(k), k, 1, 3);
3
====
(%o8) > f(k)
/
====
k = 1
(%i9) %, nouns;
(%o9) f(3) + f(2) + f(1)
answered Nov 26 '18 at 3:02
Robert DodierRobert Dodier
11.3k11834
answered Nov 26 '18 at 3:02
Robert DodierRobert Dodier
11.3k11834
answered Nov 26 '18 at 3:02
Robert DodierRobert Dodier
11.3k11834
answered Nov 26 '18 at 3:02
Robert DodierRobert Dodier
11.3k11834
11.3k11834
add a comment |
add a comment |
Found the answer, ev() carries the option diff which solves all the symbolic differentiation in the expression.
answered Nov 25 '18 at 16:45
Marco RepettoMarco Repetto
467
add a comment |
Found the answer, ev() carries the option diff which solves all the symbolic differentiation in the expression.
answered Nov 25 '18 at 16:45
Marco RepettoMarco Repetto
467
add a comment |
Found the answer, ev() carries the option diff which solves all the symbolic differentiation in the expression.
answered Nov 25 '18 at 16:45
Marco RepettoMarco Repetto
467
Found the answer, ev() carries the option diff which solves all the symbolic differentiation in the expression.
answered Nov 25 '18 at 16:45
Marco RepettoMarco Repetto
467
answered Nov 25 '18 at 16:45
Marco RepettoMarco Repetto
467
answered Nov 25 '18 at 16:45
Marco RepettoMarco Repetto
467
answered Nov 25 '18 at 16:45
Marco RepettoMarco Repetto
467
467
add a comment |
add a comment |
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