How do I write the Taylor expansion of this function $z = f(x,y)$

Question

I have an assignment where I need to write the Taylor expansion for the function $$ z = f(x,y)$$

which is given by the following formula:

$$ z^3 − 2xz + y = 0$$

for the point $A(1,1)$.

I know how to perform the procedure if I was given something like this:

$$f(x,y) = x^3 + 2xy + y $$

But the formula for the function I have makes me really confused! I don't know how to proceed. The solution is

$$T_2 (x,y) = 1 + 2(x − 1) − (y − 1) − 8(x − 1)^2 + 10(x − 1)(y − 1) − 3(y − 1)^ 2$$ and as you see there are no $z$ in the solution!

Can anyone help?

Answer 1

This requires the implicit function theorem. It basically states that the given equation defines $z$ as a function of $x, y$ around $A=(1, 1)$, if the function $F(x, y, z)=z^3-2xz+y$ has a non-zero partial derivative with respect to $z$ at $A$. Indeed, $F_z(x, y, z)=3z^2-2x$ so $F_z(1, 1, 1)=1\ne 0$. Therefore, around $A=(1, 1)$, we can write $z=f(x, y)$ for some $f$, and $f$ has continuous partial derivatives around $(1, 1)$. We now have the following equation: $$ f(x, y)^3-2xf(x, y)+y=0 $$ Let's differentiate both sides with respect to $x$. We get: $$ 3f(x, y)^2f_x(x, y)-2f(x, y)-2xf_x(x, y)=0 $$ At $(1, 1)$, we get: $$ 3f(1, 1)^2f_x(1, 1)-2f(1, 1)-2f_x(1, 1)=0 $$ But we can easily see $f(1, 1)=1$, and then this yields $f_x(1, 1)=2$. Similarly, we can differentiate with respect to $y$ and get $f_y(1, 1)=-1$. Then, we can differentiate the formula again, with respect to an appropriate variable each time, to get some system relating $f_{xx}, f_{xy}, f_{yy}$. After we have those (you'll get, at $(1, 1)$, $f_x=2, f_y=-1, f_{xx}=-8, f_{xy}=10, f_{yy}=-3$) we have the following equation for a general Taylor polynomial: $$ P_n(x, y)=\sum_{i+j\le n}\frac{f_{x^iy^j}(x_0, y_0)}{i!j!}(x-x_0)^i(y-y_0)^j $$ Where $f_{x^iy^j}$ denotes differentiating $f$, $i$ times with respect to $x$ and $j$ times with respect to $y$ (the order doesn't matter, under certain conditions!). For $n=2, x_0=y_0=1$, we then get precisely what you put in your question.

Answer 2

When $x=y=1$ we obtain the equation $z^3-2z+1=0$ for $z$, having the solutions $1$ and ${1\over2}\bigl(-1\pm\sqrt{5}\bigr)$. Your source chose the solution $z=1$. We therefore introduce new variables $u$, $v$, $w$ via $$x=1+u,\quad y=1+v,\quad z=1+ w\ .$$ The given equation $z^3-2xz+y=0$ then appears as $$w+3w^2+w^3-2u -2uw+v=0\ .\tag{1}$$ We have to find $w$ as a function of $u$ and $v$ in the form $$w= a_1 u+a_2 v+ b_1u^2 +b_2 uv+ b_3 v^2+ c_1u^3+c_2 u^2 v+c_3 uv^2+c_4 v^3+\ldots\quad.$$ From $(1)$ we immediately obtain $w=2u-v\>+$ higher terms, so that we obtain $a_1=2$, $a_2=-1$. We now compute $(1)$ inclusive all quadratic terms in $u$ and $v$. Comparing the coefficients at $u^2$, $uv$, and $v^2$ then will lead to three equations for the $b_i$. Etcetera.