Re: Redefining a function with a rule for coefficients
- To: mathgroup at smc.vnet.net
- Subject: [mg59386] Re: Redefining a function with a rule for coefficients
- From: "Jens-Peer Kuska" <kuska at informatik.uni-leipzig.de>
- Date: Mon, 8 Aug 2005 03:34:33 -0400 (EDT)
- Organization: Uni Leipzig
- References: <dd4fd7$htb$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Hi,
f[A_,B_][x_,y_]:= A x + B y
g[A_,B_][x_,y_]:= A x - B y
coeff = Solve[{f[A, B][1, 1] == 3, g[A, B][1, 1]
== -1}, {A, B}];
res={f[A, B], g[A, B]} /. coeff[[1]]
Plot3D[res[[1]][x, y], {x, 0, 1}, {y, 0, 1}]
Regards
Jens
"Wonseok Shin" <wssaca at gmail.com> schrieb im
Newsbeitrag news:dd4fd7$htb$1 at smc.vnet.net...
| Suppose that there are two functions f and g:
| f[x_, y_] := A x + B y;
| g[x_, y_] := A x - B y;
| where the coefficients A and B are unknown.
|
| We know some kind of boundary condition f[1, 1]
== 3 and g[1, 1] == -1.
| I want to solve for coefficients A and B using
this boundary
| condition, and to put them back into f and g.
|
| Of course, solving for A and B is very
straightforward:
| coeff = Solve[{f[1, 1] == 3, g[1, 1] == -1}, {A,
B}]
|
| This gives the solution in the form:
| {{A -> 1, B -> 2}}
| which is the rule assigned in the variable
'coeff.'
|
| The next process is, of course, to put 'coeff'
into f and g, and to
| redefine them to be x - 2y and x + 2y. What is
the most standard way
| of doing this?
|
| Here is my solution:
| f[x_, y_] = f[x, y] /. coeff[[1]];
| g[x_, y_] = g[x, y] /. coeff[[1]];
|
| Note that I used Set (=) instead of SetDelayed
(:=).
|
| It works but looks clumsy, and has a potential
error. Look at the
| following codes:
| In[1]:=
| f[x_, y_] := A x + B y;
| g[x_, y_] := A x - B y;
|
| In[3]:=
| x = 1;
| ?f
|
| Global`f
| f[x_, y_] := A x + B y
|
| (* The assignment x = 1 does not affect the
definition of f[x_, y_]. *)
|
| In[5]:=
| coeff = Solve[{f[1, 1] == 3, g[1, 1] == -1}, {A,
B}]
|
| Out[5]=
| {{A -> 1, B -> 2}}
|
| In[6]:=
| f[x_, y_] = f[x, y] /. coeff
|
| Out[6]=
| 1 + 2 y
|
| In Out[6] our expectation is x + 2 y, but since
we've assigend 1 to x,
| this specific value is used for x in the Set
procedure in In[4]. Using
| SetDelayed (:=) instead of Set (=) generates
more serious problem,
| because it causes an infinite recursion when we
evaluate f, for
| example, at (x, y) = (1, 1).
|
| Since determining coefficients of functions
using a boundary condition
| is very common situation, I believe there exists
some standard and
| elegant way to do this.
|
| Thanks,
| --
| Wonseok Shin
| wssaca at gmail.com
|