Try with order [2,2], [1,2], [2,3]... and you will see what happens : in the second form, taylor expansion is continued up to terms in x^m * y^n (if the call is made with order [m,n]).
In the first, it looks like they are given for all x^i * y^j with i + j <= max(m, n) and i <= m and j <= n.
I don't know whether it's really a bug, at least the manual is incomplete. Both may be useful, but I like to simply call
taylor(f(x, y), [x, y], n) and get a taylor expansion up to order n. It's equivalent to [n,n] in Maxima.
HTH
> Message du 14/08/12 12:48
> De : "Wilhelm Haager"
> A : maxima at math.utexas.edu
> Copie ? :
> Objet : [Maxima] Multidimensional taylor series
>
> Hi,
>
> Both forms of taylor
>
> 1. taylor(expr,[x_1, x_2, . . . ], [a_1, a_2, . . . ], [n_1, n_2, . .
> . ]) and
> 2. taylor(expr,[x_1, a_1, n_1], [x_2, a_2, n_2], . . . )
>
> should yield the same result, but they appearently don't.
>
> E.g.
>
> ta1:taylor(f(x,y),[x,y],[0,0],[1,1]);
> yields
> (%o81) f(0,0)+(('at('at('diff(f(x,y),x,1),x = 0),y = 0))*x
> +('at('diff(f(0,y),y,1),y = 0))*y)
>
> tb1:taylor(f(x,y),[x,0,1],[y,0,1]);
> yields
> (%o82) f(0,0)+('at('diff(f(0,y),y,1),y = 0))*y
> +('at('at('diff(f(x,y),x,1),x = 0),y = 0)
> +('at('at('diff(f(x,y),x,1,y,1),x = 0),y = 0))*y)
> *x
>
> The difference ta1-tb1 gives Zero,
> the difference expand(ta1)-expand(tb1) does NOT give Zero.
>
> I suspect, that the 2nd form of taylor erroneously returns a mixed
> derivative.
> Is that really a bug, or have I overseen anything?
>
> Thanks in advance
> Wilhelm Haager
>
> _______________________________________________
> Maxima mailing list
> Maxima at math.utexas.edu
> http://www.math.utexas.edu/mailman/listinfo/maxima
>