Robertas Vilkas

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12 years, 157 days

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These are answers submitted by Robertas Vilkas

Thank you, unapply indeed works here (even with x[k]).
(somehow I misused it before)

> f:=unapply(cat(x,1),[seq(cat(x,k),k=1..3)]);
> f(x3,x2,x1);
                       f := (x1, x2, x3) -> x1
                                  x3
> g:=unapply(x[1],[seq(x[k],k=1..3)]);
> g(x[3],x[2],x[1]);
                     g := (x_1, x_2, x_3) -> x_1
                                 x[3]
> f(x3,x2,x1)-g(x3,x2,x1);
                                  0

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