Re: Puzzled over (un)changing argument symbols in functions
- To: mathgroup at smc.vnet.net
- Subject: [mg27977] Re: Puzzled over (un)changing argument symbols in functions
- From: "Allan Hayes" <hay at haystack.demon.co.uk>
- Date: Tue, 27 Mar 2001 01:26:13 -0500 (EST)
- References: <99n5t5$ihf@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
fSoln = RSolve[{a[n] == (2 + b)*a[n - 1] - a[n - 2],
a[3] == 1, a[-3] == 1}, a[n], n] /.
n - 3 -> True;
With
f2[n_,b_]:=(a[n]/.fSoln[[1]]);
We see that we are storing the rule for
?f2
Global`f2
f2[n_, b_] := a[n] /. fSoln[[1]]
Note that there is no b on the right, so the b_ on the left has nowhere to
send to.
But if we use
f2[n_,b_]=(a[n]/.fSoln[[1]]);
Then
?f2
Global`f2
f2[n_, b_] = If[n >= -3, ((2 + b - Sqrt[b]*Sqrt[4 + b])^
n + (2 + b + Sqrt[b]*Sqrt[4 + b])^n)/2^n, 0]/
((2 + b)*(1 + 4*b + b^2))
and f2 behaves as follows
f2[n,b]
If[n >= -3, ((2 + b - Sqrt[b]*Sqrt[4 + b])^n +
(2 + b + Sqrt[b]*Sqrt[4 + b])^n)/2^n, 0]/
((2 + b)*(1 + 4*b + b^2))
f2[n,b1]
If[n >= -3, ((2 + b1 - Sqrt[b1]*Sqrt[4 + b1])^n +
(2 + b1 + Sqrt[b1]*Sqrt[4 + b1])^n)/2^n, 0]/
((2 + b1)*(1 + 4*b1 + b1^2))
--
Allan
---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565
"A. E. Siegman" <siegman at stanford.edu> wrote in message
news:99n5t5$ihf at smc.vnet.net...
> Here are three cells just to confirm that if I define a trivial function
> f1[y,z], then substitute y1 and z1 for y and z, the result is what you'd
> think it would be
>
> In[1] := f1[y_, z_] := y - z;
>
> In[2] := f1[y, z]
>
> Out[2] = y - z
>
> In[3] := f1[y1, z1]
>
> Out[3] = y1 - z1
>
> Now I define a slightly more complex but still algebraic function
> f2[n,b] using RSolve. For simplicity I haven't printed the Outputs
> below, but the essential result is that b is *not* replaced by b1 in
> Output[] -- Outputs [7] and [8] are identical:
>
> In[4] := << DiscreteMath`RSolve`;
>
> In[5] := fSoln = RSolve[ { a[n] == (2 + b)a[n - 1] - a[n - 2],
> a[3] == 1, a[-3] == 1}, a[n], n ] /. (n ? -3) ->True;
>
> In[6] := f2[n_, b_] := (a[n] /. fSoln[[1]]);
>
> In[7]:= f2[n, b]
>
> In[8]:= f2[n, b1]
>
> Why doesn't f2[n,b] behave like f1[y,x] did?
>