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