Find the form of the quadratic approximation of a function $f$ at a certain point $a$ and find its euclidean normal form.
$f(x,y)=\cos(x)\cos(y)$ and $a=\left(-\frac{\pi}{2}\,\,\,-\frac{\pi}{2}\right)^T$
I know that the quadratic approximation is:
$$z=T_2(f,x,a)$$
So I calculated $T_2(f,x,a)=xy+\frac{\pi }{2}x+\frac{\pi }{2}y+\frac{\pi^2 }{4}$
Which gives the approximation of$(x:=x_1,y:=x_2,z:=x_3)$:
$$\mathcal{Q}:=\lbrace (x_1,x_2,x_3):x_1x_2+\frac{\pi }{2}x_1+\frac{\pi }{2}x_2+\frac{\pi^2 }{4}-x_3=0\rbrace$$
We can write this as:
$$\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}^T\begin{pmatrix} 0 & \frac{1}{2}&0\\ \frac{1}{2} & 0&0\\ 0 &0&0 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}+2\begin{pmatrix} \frac{\pi}{4} \\ \frac{\pi}{4} \\ -\frac{1}{2} \end{pmatrix}^T\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}+\frac{\pi^2 }{4}=0$$
Now we rotate our quadratic approximation so that we get only pure quadratic components:
$$A=\begin{pmatrix} 0 & \frac{1}{2}&0\\ \frac{1}{2} & 0&0\\ 0 &0&0 \end{pmatrix}\longrightarrow A-\lambda=\begin{pmatrix} -\lambda & \frac{1}{2}&0\\ \frac{1}{2} & -\lambda&0\\ 0 &0&-\lambda \end{pmatrix}$$
$\det(A-\lambda)=-\lambda^3+\lambda\frac{1}{4}\Longrightarrow \lambda_{1,2,3}=0;-\frac{1}{2};\frac{1}{2}$
and our Transformationmatrix $T$ build by the normed eigenvectors is:
$$T=\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}&0\\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}&0\\ 0 &0&1 \end{pmatrix}=T^{-1}$$
After proper rotation we get:
$$\begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix}^T\begin{pmatrix} \frac{1}{2} & 0&0\\ 0 & -\frac{1}{2}&0\\ 0 &0&0 \end{pmatrix}\begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix}+2\left(\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}&0\\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}&0\\ 0 &0&1 \end{pmatrix}\begin{pmatrix} \frac{\pi}{4} \\ \frac{\pi}{4} \\ -\frac{1}{2} \end{pmatrix}\right)^T\begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix}+\frac{\pi^2 }{4}=0$$
$$\Longleftrightarrow \frac{1}{2}y_1^2-\frac{1}{2}y_2^2+\frac{\pi}{\sqrt{2}}y_1-y_3+\frac{\pi^2 }{4}=0$$
Now we can move against the linear part by first rewriting:
$$\left(\frac{1}{\sqrt{2}}y_1+\frac{\pi}{2}\right)^2-\frac{1}{2}y_2^2-y_3=0\Longleftrightarrow\frac{1}{2}\left(y_1+\frac{\sqrt{2}\pi}{2}\right)^2-\frac{1}{2}y_2^2-y_3=0$$
And now defining:$\,\,\,\,z_1:=y_1+\frac{\sqrt{2}\pi}{2},\,\,\,z_2:=y_2,\,\,\,z_3:=y_3$
$\longrightarrow$
$$\frac{1}{2}z_1^2-\frac{1}{2}z_2^2-z_3=0$$
$$\Longleftrightarrow -z_1^2+z_2^2+2z_3=0$$
which is a hyperbolic paraboloid
Could someone check for me if every step was correct, or if somewhere I went wrong? That would be great :)