What do you mean by "automatic"? Using 'simiplfy' I get
...
eq1:=subst(%,z = 1/(2*n));
Type: Equation(Expression(Integer))
s1:=solve(%,f1);
Type: List(Equation(Expression(Integer)))
simplify eval(f1,s1)
e1
(13) ------------------
2 2 2 2
(n + m + l )%pi
Type: Expression(Integer)
(14) ->
Is this what you expect?
Regards,
Bill Page.
On Fri, Aug 14, 2009 at 11:42 AM, Dan Hatton<[email protected]> wrote:
>
> Dear All,
>
> Appended below is an Axiom input script whose last two lines (the
> subst() and solve()) produce output involving cos(%pi/2), sin(%pi/2),
> and sin(%pi), which I'd naively expect to be automatically simplfied
> to 0, 1, and 0 respectively. How can I get this automatic
> simplification to happen, please?
>
> Thanks,
>
> Dan
>
> l:Union(Variable l,Integer)
> m:Union(Variable m,Integer)
> n:Union(Variable n,Integer)
>
> A := a*sin(l*%pi*x)*cos(m*%pi*y)*cos(n*%pi*z)
> B := b*cos(l*%pi*x)*sin(m*%pi*y)*cos(n*%pi*z)
> C := c*cos(l*%pi*x)*cos(m*%pi*y)*sin(n*%pi*z)
> D := d*cos(l*%pi*x)*cos(m*%pi*y)*sin(n*%pi*z)
>
> E :=
> e1*cos(l*%pi*x)*cos(m*%pi*y)*sin(n*%pi*z)+(e2+e3*cos(2*l*%pi*x)+e4*cos(2*m*%pi*y))*sin(2*n*%pi*z)
> F :=
> f1*cos(l*%pi*x)*cos(m*%pi*y)*sin(n*%pi*z)+(f2+f3*cos(2*l*%pi*x)+f4*cos(2*m*%pi*y))*sin(2*n*%pi*z)
>
> A*differentiate(D,x)+B*differentiate(D,y)+C*differentiate(D,z)-E-differentiate(F,x,2)-differentiate(F,y,2)-differentiate(F,z,2)
> = 0
>
> subst(%,z = 1/(2*n))
>
> solve(%,f1)
>
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