Re: Going mad with pattern matching

Thanks, Martin,

As always I am helped by you!  I have two questions; maybe you can answer
them:

1)  How do I include linearity in the pattern matching rules?  At present,
the command

ztrans(2+3^n,nz)

which should return the result

2z/(z-1)+z/(z-3)

produces

(2+3^n)z/(z-1).

That is, the pattern matcher incorrectly applies the rule

zt(a | freeOf?(a,n),n,z) == a*z/(z-1)

even though 2+3^n is not free of n!

2)  How do I force answers to be returned in factored form?

Thanks,
Alasdair

On 28 May 2007 18:52:36 +0200, Martin Rubey <[email protected]>
wrote:
>
> "Alasdair McAndrew" <[email protected]> writes:
> > But this doesn't quite work.  I'd be grateful for a little help
> here!  (Then
> > I'll see if I can use the z-transform to solve some difference
> equations.)
>
> it seems that rules don't like local function definitions too much.  The
> usual
> syntax would be
>
>   (op1; op2; ...; opn)
>
> which returns the value of opn, but I had no luck with this construction.
>
> zt:=operator 'zt
>
> help(z,a) ==
>     tmp := z/(z-1)
>     for i in 1..a repeat
>         tmp:=-D(tmp,z)
>     tmp
>
> ztransrules := rule
>   zt(f+g,n,z) == ztrans(f,n,z)+ztrans(g,n,z)     -- two lines for
> linearity
>   zt((a | freeOf?(a,n)),n,z) == a*ztrans(f,n,z)
>   zt(0,n,z) == 0       -- a couple of end cases, probably not needed
>   zt(1,n,z) == z/(z-1)
>   zt(a | freeOf?(a,n),n,z) == a*z/(z-1)  -- now some standard rules
>   zt((a | freeOf?(a,n))^n,n,z) == z/(z-a)
>   zt(n,n,z) == z/(z-1)^2
>   zt(n^(a | integer?(a) and a>1),n,z) == help(z, a)
>
> ztrans(f,n,z)==ztransrules zt(f,n,z)
>
>
> Does this answer your question?  I'm a little in the dark, I must say.
>
> Martin
>
>

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