Given was the $a=(0\,\,\, \pi)^T$, a function $f:\mathbb{R}^2\longrightarrow\mathbb{R}:(x,y)^T\mapsto \sin(x)\sin(y)\,\,\,$and $v=(\frac{\pi}{2}\,\,\,-\frac{\pi}{2})^T$
I had to compare the error estimation for
$$|f(a+v)-T_1(f,a+v,a)|$$
to the actual error.
First of all I computed:
$T_1(f,(x,y)^T,a)=f((0\,\,\, \pi))+(x\,\,\,y-\pi)^T\bullet\triangledown f((0\,\,\, \pi))=0+(x\,\,\,y-\pi)^T\begin{pmatrix} \cos(0)\sin(\pi) \\ \ \cos(\pi)\sin(0) \end{pmatrix}=0$
so the actual error is: $|f((\frac{\pi}{2}\,\,\,\frac{\pi}{2}))|=1$
Now to the error estimation:
First of all I computed:
$$\frac{1}{2}\partial_{(\frac{\pi}{2}\,\,\,-\frac{\pi}{2})}^2f(a+\vartheta v), \,\,\,\vartheta\in[0,1]$$
$$\frac{1}{2}\left(\frac{\pi}{2}\,\,\,-\frac{\pi}{2}\right)\begin{pmatrix} \cos(\vartheta\frac{\pi}{2})\cos(\pi-\vartheta\frac{\pi}{2}) & -\sin(\vartheta\frac{\pi}{2})\sin(\pi-\vartheta\frac{\pi}{2})\\ -\sin(\vartheta\frac{\pi}{2})\sin(\pi-\vartheta\frac{\pi}{2}) & \cos(\vartheta\frac{\pi}{2})\cos(\pi-\vartheta\frac{\pi}{2}) \end{pmatrix}\left(\frac{\pi}{2}\,\,\,-\frac{\pi}{2}\right)^T$$
$$=\frac{1}{2}\left(\frac{\pi}{2}\,\,\,-\frac{\pi}{2}\right)\begin{pmatrix} \frac{\pi}{2}\cos(\vartheta\frac{\pi}{2})\cos(\pi-\vartheta\frac{\pi}{2})+\frac{\pi}{2}\sin(\vartheta\frac{\pi}{2})\sin(\pi-\vartheta\frac{\pi}{2}) \\ -(\frac{\pi}{2}\cos(\vartheta\frac{\pi}{2})\cos(\pi-\vartheta\frac{\pi}{2})+\frac{\pi}{2}\sin(\vartheta\frac{\pi}{2})\sin(\pi-\vartheta\frac{\pi}{2})) \end{pmatrix}$$
$$=\frac{1}{2}\left(\frac{\pi}{2}\,\,\,-\frac{\pi}{2}\right)\begin{pmatrix} \frac{\pi}{2}\cos(\vartheta\frac{\pi}{2}-\pi+\vartheta\frac{\pi}{2}) \\ -\frac{\pi}{2}\cos(\vartheta\frac{\pi}{2}-\pi+\vartheta\frac{\pi}{2}) \end{pmatrix}=\frac{1}{2}\left(\frac{\pi}{2}\,\,\,-\frac{\pi}{2}\right)\begin{pmatrix} \frac{\pi}{2}\cos(\vartheta\pi-\pi) \\ -\frac{\pi}{2}\cos(\vartheta\pi-\pi) \end{pmatrix}$$
$$=\frac{1}{2}\left(\frac{\pi^2}{4}\cos(\vartheta\pi-\pi)+\frac{\pi^2}{4}\cos(\vartheta\pi-\pi)\right)=\frac{\pi^2}{4}\cos(\vartheta\pi-\pi)\le\frac{\pi^2}{4}$$
So the maximum error is $\frac{\pi^2}{4}$
Is this correct?