Taylorpolynom/xsin y- e^xy/Nullpunkt/Grad 4/Aufgabe/Lösung

Der Funktionswert ist

${}f(0,0)=-1\,.$

Die relevanten Ableitungen sind

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

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

${}{\frac {\partial ^{2}f}{\partial ^{2}x}}=-y^{2}e^{xy}\,,$

${}{\frac {\partial ^{2}f}{\partial ^{2}y}}=-x\sin y-x^{2}e^{xy}\,,$

${}{\frac {\partial }{\partial y}}{\frac {\partial f}{\partial x}}=\cos y-e^{xy}-xye^{xy}\,.$

${}{\frac {\partial ^{3}f}{\partial ^{3}x}}=-y^{3}e^{xy}\,,$

${}{\frac {\partial ^{3}f}{\partial ^{3}y}}=-x\cos y-x^{3}e^{xy}\,,$

${}{\frac {\partial }{\partial y}}{\frac {\partial }{\partial y}}{\frac {\partial f}{\partial x}}=-\sin y-2xe^{xy}-x^{2}ye^{xy}\,,$

${}{\frac {\partial }{\partial y}}{\frac {\partial }{\partial x}}{\frac {\partial f}{\partial x}}=-2ye^{xy}-xy^{2}e^{xy}\,.$

${}{\frac {\partial ^{4}f}{\partial ^{4}x}}=-y^{4}e^{xy}\,,$

${}{\frac {\partial ^{4}f}{\partial ^{4}y}}=-x\cos y-x^{3}e^{xy}\,,$

${}{\frac {\partial }{\partial y}}{\frac {\partial }{\partial y}}{\frac {\partial }{\partial y}}{\frac {\partial f}{\partial x}}=-\cos y-2x^{2}e^{xy}-x^{2}e^{xy}-x^{3}ye^{xy}\,,$

${}{\frac {\partial }{\partial y}}{\frac {\partial }{\partial y}}{\frac {\partial }{\partial x}}{\frac {\partial f}{\partial x}}=-2e^{xy}-2xye^{xy}-2xye^{xy}-x^{2}y^{2}e^{xy}\,,$

${}{\frac {\partial }{\partial y}}{\frac {\partial }{\partial x}}{\frac {\partial }{\partial x}}{\frac {\partial f}{\partial x}}=-3y^{4}e^{xy}-xy^{3}e^{xy}\,.$

Wenn man den Punkt ${}(0,0)$ einsetzt, so ergibt sich überall ${}0$, außer beim Funktionswert, bei

${}{\frac {\partial }{\partial y}}{\frac {\partial }{\partial y}}{\frac {\partial }{\partial y}}{\frac {\partial f}{\partial x}}(0,0)=-1\,$

und bei

${}{\frac {\partial }{\partial y}}{\frac {\partial }{\partial y}}{\frac {\partial }{\partial x}}{\frac {\partial f}{\partial x}}(0,0)=-2\,.$

Daher ist das Taylor-Polynom der Ordnung vier gleich

$-1-{\frac {1}{6}}xy^{3}-{\frac {1}{2}}x^{2}y^{2}.$

Zur gelösten Aufgabe