<s>[INST] How to express a function into powers of $(x-1)$ and $(y-2)$ using Taylor's formula? Use Taylor's formula to express the following in powers of $(x-1)$ and $(y-2)$:
$f(x,y)=x^3 + y^3 + xy^2$
Solution:
$f(1,2)=1 +8 + 4=13$
$f_x (1,2) = 3 + 4=7$
$f_y (1,2) = 12 + 4=16$
$f_{xx} (1,2) = 6$
$f_{yy} (1,2)= 12 + 2=14$
$f_{xy} (1,2)= 4 = f_{yx} (1,2)$
$f_{xxx} (1,2)= 6$
$f_{yyx} (1,2)= 2 = f_{xyy} (x,y) = f_{yxy} (1,2)$
$f_{yyy} (1,2)= 6$
Hence,
$x^3 +y^3 + xy^2 = 13 +7(x-1) + 16(y-2) + \frac{1}{2!} \left[ 6(4)(x-1)^2 +2(x-1)(y-2) + 14(y-2)^2\right] +\frac{1}{3!} \left[ 6(x-1)^3 - 3(0)(x-1)^2 (y-2) + 3(2)(x-1)(y-2) - 6(y-2)^3\right] $
Is the solution correct and how to compute for the remainder if it is applicable? Another question is, is the Taylor's formula the same equation with the Taylor expansion and/or the one stated by the Taylor's Theorem?
 [/INST] Your approach is fine, although it should be somewhat revised. A way to check your calculation is recalling that when expanding the function at $x_0=1$ and $y_0=2$ we can replace in $f=f(x,y)$:
\begin{align*}
x&=(x-x_0)+x_0=(x-1)+1\\
y&=(y-y_0)+y_0=(y-2)+2
\end{align*}

We obtain
\begin{align*}
f(x,y)&=x^3+y^3+xy^2\\
&=\left((x-1)+1\right)^3+\left((y-2)+2\right)^3+\left((x-1)+1\right)\left((y-2)+2\right)^2\\
&=(x-1)^3+3(x-1)^2+3(x-1)+1\\
&\qquad+(y-2)^3+6(y-2)^2+12(y-2)+8\\
&\qquad+(x-1)(y-2)^2+4(x-1)(y-2)+4(x-1)\\
&\qquad+(y-2)^2+4(y-2)+4\\
&=13+7(x-1)+16(y-2)\\
&\qquad+\frac{1}{2!}\left(\color{blue}{6}(x-1)^2+\color{blue}{8}(x-1)(y-2)+14(y-2)^2\right)\\
&\qquad+\frac{1}{3!}\left(6(x-1)^3+6(x-1)(y-2)^2\color{blue}{+6}(y-2)^3\right)\tag{1}
\end{align*}
Since the expression (1) is already $f=f(x,y)$, the remainder term is zero.

Blue marked parts should be revised in your calculcation.
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