Finding second-order taylor's series

Question

Find second-order Taylor’s series approximation of the function $$F(x,y)=e^y(x-1)^2 + x$$ at the points (a) $(0,0)$ and (b) $(1,1)$.

I appreciate for any help.

I'm not confidence with my solve so I want some others solves.

Answer 1

$$F(x,y)=e^y(x-1)^2 + x$$ Now $F(0,0)=1\qquad \text{and}\qquad F(1,1)=1$

The formula for 2nd order Taylor approximation can be simplified to

$$f(a+h_1,b+h_2)\approx f(a,b) + f_x(a,b)h_1 + f_y(a,b)h_2 + \frac{1}{2}(h_1^2f_{xx}+h_1h_2f_{xy}+h_1h_2f_{yx}+h_2^2f_{yy})$$

where $h_1 = x-a$ and $h_2= y-b$

Here $$F_x=1+2(x-1)e^y\quad, \quad F_y=e^y(x-1)^2\quad, \quad F_{xx}=2e^y\quad, \quad F_{xy}=F_{yx}=2(x-1)e^y\quad, \quad F_{yy}=e^y(x-1)^2$$

Now $$F(x,y)\approx F(0,0)+F_x(0,0)(x-0)+F_y(0,0)(y-0)+\frac{1}{2}\{(x-0)^2F_{xx}(0,0)+(x-0)(y-0)F_{xy}(0,0)+(x-0)(y-0)F_{yx}(0,0)+(y-0)^2F_{yy}(0,0)\}$$

Hence $$F_x(0,0)=-1\quad, \quad F_y(0,0)=1\quad, \quad F_{xx}(0,0)=2\quad, \quad F_{xy}(0,0)=F_{yx}(0,0)=-2\quad, \quad F_{yy}(0,0)=1$$

So

$$F(x,y)\approx 1 -x+y+\frac{1}{2}\{2x^2-4xy+y^2\}$$

${}$

Again $$F(x,y)\approx F(1,1)+F_x(1,1)(x-1)+F_y(1,1)(y-1)+\frac{1}{2}\{(x-1)^2F_{xx}(1,1)+(x-1)(y-1)F_{xy}(1,1)+(x-1)(y-1)F_{yx}(1,1)+(y-1)^2F_{yy}(1,1)\}$$

Hence $$F_x(1,1)=1\quad, \quad F_y(1,1)=0\quad, \quad F_{xx}(1,1)=2e\quad, \quad F_{xy}(1,1)=F_{yx}(1,1)=0\quad, \quad F_{yy}(1,1)=0$$

So

$$F(x,y)\approx x+e(x-1)^2$$

Answer 2

Hint: Look up multivariable Taylor series. You will need the matrices of first and second partial derivatives (aka the Jacobian and the Hessian).

Let me help. $F_x(x,y)=2e^y(x-1)+1, F_y(x,y)=e^y(x-1)^2, F_{xx}(x,y)=2e^y, F_{yy}(x,y)=e^y(x-1)^2$ and $F_{xy}(x,y)=2xe^y-2e^y$.

Now plug in the two points: $F_x(0,0)=-1, F_y(0,0)=1, F_{xx}(0,0)=2, F_{yy}(0,0)=1$ and $F_{xy}(0,0)=-2$.

Now let's finish $a$: $p_2(x,y)=1+\begin {pmatrix}-1&1\end{pmatrix}\begin {pmatrix}x\\y\end{pmatrix}+\dfrac12\begin {pmatrix}x&y\end{pmatrix}\begin{pmatrix}2&-2\\-2&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=1+y-x+\dfrac12\begin {pmatrix}x&y\end{pmatrix}\begin{pmatrix}2x-2y\\y-2x\end{pmatrix}=1+y-x+x^2-2xy+\frac{y^2}2 $.

I leave the second order Taylor polynomial at $(1,1)$ to you.