1
$\begingroup$

I have two generic functions :

ucdu[x : {__}] := Times @@ (x)

and 

cb[x : {__}, p : {__}] := Plus @@ (x p)

For instance, they can be applied to vectors in $\mathbb{R}^3$

nv = 3;
xx = Table[Subscript[x, i], {i, 1, nv}](*{Subscript[x, 1], ...., Subscript[x, nv]}*);
pp = Table[Subscript[p, i], {i, 1, nv}](*{Subscript[p, 1], ...., Subscript[p, nv]}*);
ucdu[xx]
cb[xx, pp]

which give :

$x_1 x_2 x_3$ and $x_1 p_1+ x_2 p_2 + x_3 p_3$

Now, I want to write a new function, say a Lagrangian, with one more variable say $\mathcal{L}[xx, pp, \lambda] = ucdu[xx]- \lambda (cb[xx, pp]- R)$. $\mathcal{L}$ must be generic enough to be called on other functions --- for instance, $ucdg[xx,\alpha\alpha]$...

I have tried 

ll[u_Symbol, cb_Symbol, λ_, xx_, pp_] := u - λ (cb - R)

but ll[ucdu[xx], cb[xx, pp], λ] returns $ll[x_1 x_2 x_3, x_1 p_1 + x_2 p_2+ x_3 p_3, \lambda,\{x_1, x_2, x_3\},\{p_1, p_2, p_3\}]$ not $x_1 x_2 x_3 - \lambda(-R + x_1 p_1 + x_2 p_2+ x_3 p_3)$

What am I doing wrong? Later I want to write the Gradient with respect of Join[xx,{\lambda}]

Improve this question
$\endgroup$
2
  • $\begingroup$ Well, for one you are not calling your function ll with enough arguments. If you define it to take five arguments, of course it will return unevaluated if you call it with three. Also, notice that the return value of ucdu[xx] has head Times, so it won't match the _Symbol pattern restriction you put in the definition. If you redefine ll without the pattern restrictions, and include any expression for xx_ and pp_ (their value doesn't seem to matter, since you don't use them in the definition of ll), then it will evaluate. Whether the return value is correct though, I am not sure. $\endgroup$
    – MarcoB
    Commented Feb 6, 2017 at 16:55
  • $\begingroup$ You might also be interested in How can I implement the method of Lagrange multipliers?. $\endgroup$
    – MarcoB
    Commented Feb 6, 2017 at 17:16

2 Answers 2

Reset to default
2
$\begingroup$

With your function definitions in place, and using indexed variables rather than subscripts:

nv = 3;
xx = Table[x[i], {i, 1, nv}];
pp = Table[p[i], {i, 1, nv}];

Clear[ll]
ll[f_, g_, xvar_, pvar_, lambda_] := f[xvar] - lambda (g[xvar, pvar] - r)

ll[ucdu, cb, xx, pp, lambda]

(* Out: x[1] x[2] x[3] - lambda (-r + p[1] x[1] + p[2] x[2] + p[3] x[3]) *)

That seems to be your intended result.

Improve this answer
$\endgroup$
1
  • $\begingroup$ MarcoB Thanks. It works also with indexes. $\endgroup$
    – cyrille.piatecki
    Commented Feb 6, 2017 at 18:21
1
$\begingroup$ 
ClearAll[ll2]
ll2[u_, cb_,  λ_] := u - λ(cb - r)

ll2[ucdu[xx], cb[xx, pp], λ]

Mathematica graphics

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.