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Exploiting the results of an optimization

The code bellow work perfectly but if i get ride of show(), it doesn't display in LaTeX secondly I would like to display not Simply show(x1) but "x=show(x1)". How can I do that ?

(The question continue after the code)

%display latex var('A, x, y, l, alpha, beta, R, p_x, p_y'); U= Ax^(alpha)y^(beta); show(U) D = p_xx + p_yy; show(D) show(U) solve(D==R, y) L = U-l(D-R) show(L) L_x= L.diff(x) show(L_x) L_y= L.diff(y) show(L_y) L_l= L.diff(l) %display latex var('A, x, y, l, alpha, beta, R, p_x, p_y'); U= Ax^(alpha)y^(beta); show(U) D = p_xx + p_yy; show(D) show(U) solve(D==R, y) L = U-l(D-R) show(L) L_x= L.diff(x) show(L_x) L_y= L.diff(y) show(L_y) L_l= L.diff(l) show(L_l) z=solve([L_x==0, L_y==0, L_l==0,], x, y, l) show(z[0]) x1=z[0][0].right() show(x1) y1=z[0][1].right() show(y1) l1=z[0][2].right().canonicalize_radical() show(l1) U1=U.subs(x=x1,y=y1).canonicalize_radical() show(U1) latex(U1)

Now yesterday some one shows me how to do the same thing with a dictionary

z=solve([L_x==0, L_y==0, L_l==0,], x, y, l, solution_dict=True) show(z)

how can I extact $x= solution$ and so on.

Exploiting the results of an optimization

The code bellow work perfectly but if i get ride of show(), it doesn't display in LaTeX secondly I would like to display not Simply show(x1) but "x=show(x1)". How can I do that ?

(The question continue after the code)

%display latex
var('A, x, y, l, alpha, beta, R, p_x, p_y');
U= Ax^(alpha)y^(beta);
A*x^(alpha)*y^(beta);
show(U)
D = p_xx p_x*x + p_yy;
p_y*y;
show(D)
show(U)
solve(D==R, y)
L = U-l(D-R)
U-l*(D-R)
show(L)
L_x= L.diff(x)
show(L_x)
L_y= L.diff(y)
show(L_y)
L_l= L.diff(l)
%display latex
var('A, x, y, l, alpha, beta, R, p_x, p_y');
U= Ax^(alpha)y^(beta);
A*x^(alpha)*y^(beta);
show(U)
D = p_xx p_x*x + p_yy;
p_y*y;
show(D)
show(U)
solve(D==R, y)
L = U-l(D-R)
U-l*(D-R)
show(L)
L_x= L.diff(x)
show(L_x)
L_y= L.diff(y)
show(L_y)
L_l= L.diff(l)
show(L_l)
z=solve([L_x==0, L_y==0, L_l==0,], x, y, l)
show(z[0])
x1=z[0][0].right()
show(x1)
y1=z[0][1].right()
show(y1)
l1=z[0][2].right().canonicalize_radical()
show(l1)
U1=U.subs(x=x1,y=y1).canonicalize_radical()
show(U1)
latex(U1)

latex(U1)

Now yesterday some one shows me how to do the same thing with a dictionary

z=solve([L_x==0, L_y==0, L_l==0,], x, y, l, solution_dict=True)
show(z)

show(z)

how can I extact $x= solution$ and so on.