Use Taylor's theorem to compute a quadratic approximation of $f : \mathbb{R}^2 \rightarrow \mathbb{R}$ I was given this question.
"Use Taylor's theorem to compute a quadratic approximation of $f : \mathbb{R}^2 \rightarrow \mathbb{R}$ defined as $$f(x) = x_1^4 - 2x_1^3x_2 - 4x_1^2x_2^2 +5x_1x_2^3 +2x_2^4$$
about the point $x_0 = (1, −1)^T.$
Evaluate the series for $p = (0.1, 0.01)^T$ and compare the value with the value of $f(x_0 + p).$"
I'm a little unfamiliar with how to compute a Taylor approximation of $f : \mathbb{R}^2 \rightarrow \mathbb{R}$. 
Looking over my notes I found an equation I thought was supposed to apply to this question 
$$f(y)=f(x) + \nabla f(x)^T(y-x) +\frac12 (y-x)^T \nabla^2f(x)(y-x) + o\Arrowvert y-x \Arrowvert^2$$
I've found both $\nabla f(x)$ and $\nabla ^2 f(x)$, but I'm unsure of how exactly to apply this equation to the problem. For example I don't know what in the equation corresponds to what in the problem. Any help would be appreciated.
 A: Since you found both $\nabla f$ and $\nabla^2 f$, you should now calculate these two matrices at $x_0=(x_1,x_2)=(1,-1)$; they are the first and second derivative of your scalar field, and $(1,-1)$ is the point about which you're writing your Taylor expansion of $f$. 
The next steps would be:
(1) to calculate the approximate value of $f$ you find by the Taylor series using $y=(1,-1)+(0.1,0.01)=(1.1,-0.99)$, so $y-x=p$; 
(2) to find $f(1.1,-0.99)$ by using the full expression of $f$; and finally 
(3) to compare these two values in order to see how good an approximation you have. 
Overall, the formula you have from your notes is correct; it leads to a second-order Taylor approximation of $f$ about $x_0$. In the formula, $x$ is the point about which you're making the approximation ($x_0$ in your case), and $y$ is the point you're approximating your function at (for you, $x_0+p$); in this way, $y-x$ quantifies essentially "how far" from the approximation point you are. The further you go, the worse your approximation becomes (quadratically) as it can be seen from the error term. 
