Taylor series of $f(x,y) = x \cdot e^{\cos{xy}-1}$ around the point $(1,0)$ , my attempt so far:
Obviously just by checking first and second partial derivatives it's only getting more complicated. I'm not sure how it is possible to determine the general form of $D^k f(1,0)((x-1,y),(x-1,y)...(x-1,y))$ (where there are k pairs of $(x-1,y)$). And without it I'm not sure how to determine the infinite series needed in this problem.
I'm not sure if I'm looking for the wrong thing or something.
So far, all of the problems dealing with multivariable functions I've had were about finding the second or third order of the Taylor expansion, but never the entire thing.
Any hints would be appreciated!
(Sorry: The following is valid for the point $(0,0)$ instead of $(1,0)$.)
Consider the auxiliary function $$g(z):=e^{\cos z-1}$$ of one variable $z$. This function satisfies the IVP $$g'(z)=-\sin z\> g(z),\quad g(0)=1\ .\tag{1}$$ The known series for $\sin$ allows to obtain a finitary recursive formula for the coefficients $a_n$ of the Taylor series of $g$, so that you in principle have $$g(z)=\sum_{n=1}^\infty a_n z^n$$with known rational coefficients $a_n$. Mathematica says that $$g(z)=1-{z^2\over2}+{z^4\over6}-{31z^6\over720}+{379 z^8\over 40\,320}-{1639z^{10}\over 907\,200}+ \ ?z^{12}\ .$$ Given this you can write $$f(x,y)= x\sum_{n=0}^\infty a_n (x\, y)^n\ ,$$ which is the "entire thing" you wanted.