Re: Redefining a function with a rule for coefficients
- To: mathgroup at smc.vnet.net
- Subject: [mg59391] Re: [mg59376] Redefining a function with a rule for coefficients
- From: Pratik Desai <pdesai1 at umbc.edu>
- Date: Mon, 8 Aug 2005 03:34:38 -0400 (EDT)
- References: <200508070747.DAA18011@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Wonseok Shin wrote:
>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,
>
>
As a work around, why not just introduce new function like
f[x_, y_] := A x + B y;
g[x_, y_] := A x - B y;
coeff = Solve[{f[1, 1] == 3, g[1, 1] == -1}, {A, B}] // First
f1[x_, y_] = f[x, y] /. {coeff[[1]], coeff[[2]]}
g1[x_, y_] = g[x, y] /. {coeff[[1]], coeff[[2]]}
f1[x, y]
>>x + 2 y
g1[x,y]
>>x-2y
f1[1,1]
>>3
g1[1,1]
>>-1
I hope this is what you are looking for, you may look at f and g as the
most general form of your solution and f1 and g1 as the particular
solution satisfying your BC
Best regards
Pratik
--
Pratik Desai
Graduate Student
UMBC
Department of Mechanical Engineering
Phone: 410 455 8134
- References:
- Redefining a function with a rule for coefficients
- From: Wonseok Shin <wssaca@gmail.com>
- Redefining a function with a rule for coefficients