Taylorpolynom/Zweite Ordnung/e^(sin x-cos y)/0, pi halbe/Aufgabe/Lösung

Die relevanten Ableitungen sind

${}{\frac {\partial f}{\partial x}}=\cos xe^{\sin x-\cos y}\,,$

${}{\frac {\partial f}{\partial y}}=\sin ye^{\sin x-\cos y}\,,$

${}{\frac {\partial ^{2}f}{\partial ^{2}x}}={\left(\cos ^{2}x-\sin x\right)}e^{\sin x-\cos y}\,,$

${}{\frac {\partial ^{2}f}{\partial ^{2}y}}={\left(\sin ^{2}y+\cos y\right)}e^{\sin x-\cos y}\,,$

${}{\frac {\partial }{\partial y}}{\frac {\partial f}{\partial x}}=\cos x\sin ye^{\sin x-\cos y}\,.$

Somit sind die Werte der relevanten Ableitungen im Punkt ${}\left(0,\,{\frac {\pi }{2}}\right)$ gleich

${}f\left(0,\,{\frac {\pi }{2}}\right)=e^{0}=1\,,$

${}{\frac {\partial f}{\partial x}}\left(0,\,{\frac {\pi }{2}}\right)=1\,,$

${}{\frac {\partial f}{\partial y}}\left(0,\,{\frac {\pi }{2}}\right)=1\,,$

${}{\frac {\partial ^{2}f}{\partial ^{2}x}}\left(0,\,{\frac {\pi }{2}}\right)=1\,,$

${}{\frac {\partial ^{2}f}{\partial ^{2}y}}\left(0,\,{\frac {\pi }{2}}\right)=1\,,$

${}{\frac {\partial }{\partial y}}{\frac {\partial f}{\partial x}}\left(0,\,{\frac {\pi }{2}}\right)=1\,.$

Daher ist das Taylor-Polynom der Ordnung zwei gleich

$1+x+{\left(y-{\frac {\pi }{2}}\right)}+{\frac {1}{2}}x^{2}+x{\left(y-{\frac {\pi }{2}}\right)}+{\frac {1}{2}}{\left(y-{\frac {\pi }{2}}\right)}^{2}.$

Zur gelösten Aufgabe