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MathGroup Archive 1999

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Re: Simple recursive assignment

  • To: mathgroup at smc.vnet.net
  • Subject: [mg16028] Re: [mg15996] Simple recursive assignment
  • From: "Wolf, Hartmut" <hwolf at debis.com>
  • Date: Sat, 20 Feb 1999 02:52:12 -0500
  • Organization: debis Systemhaus
  • References: <199902190827.DAA18900@smc.vnet.net.>
  • Sender: owner-wri-mathgroup at wolfram.com

Ciao Peltio,

Peltio schrieb:
> 
> Can someone spare a little time for a trivial question?
> 
> The assignement
>     a[0]=1.;
>     a[1]=2.;
>     a[n_]:=a[n]=a[n-1]-a[n-2]/n
> works fine, all right...
>     a[5]
>         0.291667
> 

equally fine will work that


 In[4]:=b[0]=1.;
        b[1]=2.;

 In[5]:=expr[n_]:= b[n-1]-b[n-2]/n

 In[6]:=b[n_]:=b[n]=expr[n]

 In[7]:=b[5]
 Out[7]=0.291667

but not

>     expr=b[n-1]-b[n-2]/n;
> 
>     b[0]=1;
>     b[1]=2;
>     b[n_]:=b[n]=expr
> 
>     b[3]
>         $RecursionLimit::reclim: etc.
> 

Why? From where shall expr get it's value for n? Case 1: n is just a
Symbol with no value. Then if you call b[3] then you call expr, and that
calls b[n-1] and b[n-2] and n hasn't got a value! which means calling
expr, which means calling b[n-1] and b[n-2] with the *same* n! (if n had
a value before, say 5, then expr is defined as b[4]-b[3]/5 and b[4] is
the same as expr as is also b[3]) ad infinitum!

But if you do the following, you will succeed:

 In[9]:= Remove[b, expr, n]
 In[10]:= b[0]=1.;
          b[1]=2.;
 In[11]:= expr= b[n-1]-b[n-2]/n;

 In[12]:= b[nn_]:= b[nn]=Block[{n=nn}, expr]

 In[13]:= b[5]
 Out[13]= 0.291667

(Just a remark: if you use (n=nn; expr) instead of Block[...] the you
will also terminate, but get a wrong result. Can you see why?)

ciao, Hartmut

H. Wolf, debis Systemhaus, Darmstadt, Germania
=======[mailto:hwolf at debis.com]=======



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