From: israel@math.ubc.ca (Robert Israel)
Newsgroups: sci.math.symbolic
Subject: Re: Quick Question on Maple programming.
Date: 2 Apr 1996 20:14:39 GMT

In article <31622E05.41C6@fisher.stats.uwo.ca>, Young Ho Cheong <youngho@fisher.stats.uwo.ca> writes:

|> Let  B:=(b0,b1,b2,b3,b4.b5,.....) be a vector(or list).
|> And A[k]=sum(B[k-r]*A[r],r=0..k-1)/(2*k) where A[0]=a0.
|> Hence if I input B in procedure I want to get A:=(A[0],A[1],A[2],....)
|> which is also  vector(or list).

You can have a table T with an entry T[0], but not a vector or list.
Since you asked for a vector or list, I'll use them, but note that if 
A = [a0, a1, ...] then a0 = A[1], not A[0].  
My procedure will return either a vector or list, the same type as the 
argument B.

> myproc := proc(B:{list,vector})
    local r,k,t,n,A;
        n := linalg[vectdim](B);
        A[1] := a0;
        for k from 2 to n do
            A[k] := sum(B[k-r]*A[r+1],r = 0 .. k-2)/(2*k-2)
        od;
        t := [seq(A[k],k = 1 .. n)];
        if type(B,vector) then linalg[vector](t) else t fi
    end;
> myproc([b0, b1, b2, b3]);

                                  2
[a0, 1/2 b1 a0, 1/4 b2 a0 + 1/8 b1  a0,

                                                          2
    1/6 b3 a0 + 1/12 b2 b1 a0 + 1/6 b1 (1/4 b2 a0 + 1/8 b1  a0)]

> myproc(vector([b0,b1,b2]));
                                               2
            [ a0, 1/2 b1 a0, 1/4 b2 a0 + 1/8 b1  a0 ]

-- 
Robert Israel                            israel@math.ubc.ca
Department of Mathematics             (604) 822-3629
University of British Columbia            fax 822-6074
Vancouver, BC, Canada V6T 1Y4

==============================================================================

Date: Thu, 4 Apr 96 10:49:38 CST
From: rusin (Dave Rusin)
To: youngho@fisher.stats.uwo.ca
Subject: Re: Quick Question on Maple programming.
Newsgroups: sci.math.symbolic

In article <31622E05.41C6@fisher.stats.uwo.ca> you write:

>Now I am programming something in Maple .
>But I have in trouble in doing some kind of recurion.
>Here is what I want to get:
>
>
>Let  B:=(b0,b1,b2,b3,b4.b5,.....) be a vector(or list).
>And A[k]=sum(B[k-r]*A[r],r=0..k-1)/(2*k) where A[0]=a0.
>Hence if I input B in procedure I want to get A:=(A[0],A[1],A[2],....)
>which is also  vector(or list).

Are your ranges really right? It appears  b0  never affects your
computation.

One approach is a little more slick theory than it is computation.
Observe that if we multiply the above equation by  2*k*X^k  (where  X
is a formal variable) and sum over all  k, we get
	2*X*(sum(A[k]*k*X^(k-1), k=0..infinity) = 
	= sum(sum(B[k-r]*A[r]*X^k, k=r+1..infinity), r=0..infinity)
In the inner sum, if we let  s=k-r, we have
	A[r]*X^r*sum(B[s]*X^s, s=1..infinity)

So if we define power series	
	BigB(X) = sum(B[s]*X^s, s=1..infinity)
and
	BigA(X) = sum(A[s]*X^r, r=0..infinity)
then your defining property becomes
	2*X*diff(BigA(X),X) = BigA(X)*BigB(X)

If  BigA  and  BigB  represent _convergent_ power series, this means
BigA  is the powerseries of a solution to the equation
	f'/f = g/2x
where  g  is the function for which  BigB  is the power series. Thus
f  is a constant multiple of  exp(int(g/2x, x)) = x^(b0/2) * exp( int(h, x)),
where  h(x) = (g(x)-g(0))/(2x)  (that is, I pulled out the one term in
g/2x  which was not a positive power of  x).

Thus if the  B[s]  are known to follow some sort of pattern, one can
(sometimes) deduce a pattern for the  A[s]  by using this 
differential-equations argument.

dave