Hello,
I don't understand the result (or maybe only the notation) of
multivariate taylor expansion of a formal function. I would expect the
taylor expansion of f(x,y) to order 1 to be
f(0,0) + x (df(x,y)/dx)(0,0) + y (df(x,y)/dy)(0,0)
but what I understand of maxima's answer is
f(0,0) + x (df(x,y)/dx)(0,y) + y (df(0,y)/dy)(0)
which is not the same in general (due to terms of higher order).
However, the taylor expansion of a "concrete" function seems ok.
I didn't succeed to plug a concrete function in a formal taylor expansion.
Could somebody explain what happens below ?
How to handle ?%at(...) ?
(%i1) build_info () ; display2d:false$
Maxima version: 5.14.0
Maxima build date: 17:10 2/5/2008
host type: i686-pc-linux-gnu
lisp-implementation-type: CLISP
lisp-implementation-version: 2.41 (2006-10-13) (built 3374673657)
(memory 3411216640)
(%o1)
(%i3) taylor(f(x,y),[x,y],[0,0],[1,1])$ define(g(x,y),%);
(%o4) g(x,y):=f(0,0)
+((?%at('diff(f(x,y),x,1),x = 0))*x+(?%at('diff(f(0,y),y,1),y =
0))*y)
(%i5) f(x,y):=cos(x+y)$ g(x,y);
(%o6) (?%at('diff(cos(y+x),x,1),x = 0))*x+(?%at('diff(cos(y),y,1),y =
0))*y+1
(%i7) diff(cos(x+y),x,1)$ a:at(%,x=0)$ diff(cos(y),y,1)$
b:at(%,y=0)$ a*x+b*y+1;
(%o11) 1-x*sin(y)
(%i12) taylor(cos(x+y),[x,y],[0,0],[1,1]);
(%o12) +1
I expected (%o11) to be a taylor expansion, i.e. a polynomial in x and
y, precisely the same as (%o12).
By the way, when display2d is false then
(%i2) taylor(-x,[x,y],[0,0],[1,1]);
(%o2) +(-x)
the output is not simplified. Should it be so ?
Eric Reyssat