expand $f(x,y)= x^y$ in $(x-1)$ and $(y-1)$

Question

Useing Taylor expansion and partial derivatives I got to this point $$\begin{align}f(x,y) &= x^y\\ &= f(1,1)\\ &=f(1+(x-1),1+(y-1))\end{align}$$ how do I proceed further.

Answer 1

You can also solve this without Taylor.

Notice that, setting $u:=x-1,v:=y-1$,

$$x^y=x(1+u)^{v}$$

and using the Generalized Binomial formula,

$$x^y= \\(1+u)\left(1+vu+\frac{v(v-1)}{2!}u^2+\frac{v(v-1)(v-2)}{3!}u^3+\cdots\right).$$

Then you have to expand the falling factorials

$$v(v-1)(v-2)=v^3-3v^2+2v$$ and the like.

The coefficients of these polynomials are the Stirling numbers of the first kind, with alternating signs. To deal with the factor $(1+u)$, you need to add the coefficients in successive pairs.