Taylorpolynom/x sin x/x ist pi viertel/Ordnung 4/Beispiel

Wir möchten für die Funktion

$f\colon {\mathbb {C} }\longrightarrow {\mathbb {C} },\,x\longmapsto x\sin x,$

das Taylor-Polynom der Ordnung ${}4$ im Entwicklungspunkt ${}a={\frac {\pi }{4}}$ bestimmen. Es ist

${}f'(x)=\sin x+x\cos x\,,$

${}f^{\prime \prime }(x)=\cos x+\cos x-x\sin x=2\cos x-x\sin x\,,$

${}f^{\prime \prime \prime }(x)=-2\sin x-\sin x-x\cos x=-3\sin x-x\cos x\,,$

${}f^{\prime \prime \prime \prime }(x)=-3\cos x-\cos x+x\sin x=-4\cos x+x\sin x\,.$

Unter Verwendung von

${}\sin {\frac {\pi }{4}}=\cos {\frac {\pi }{4}}={\frac {1}{\sqrt {2}}}\,$

ist somit

${}f{\left({\frac {\pi }{4}}\right)}={\frac {\pi }{4}}{\frac {1}{\sqrt {2}}}={\frac {\pi }{4{\sqrt {2}}}}\,,$

${}f'{\left({\frac {\pi }{4}}\right)}={\frac {1}{\sqrt {2}}}{\left(1+{\frac {\pi }{4}}\right)}={\frac {4+\pi }{4{\sqrt {2}}}}\,,$

${}f^{\prime \prime }{\left({\frac {\pi }{4}}\right)}={\frac {1}{\sqrt {2}}}{\left(2-{\frac {\pi }{4}}\right)}={\frac {8-\pi }{4{\sqrt {2}}}}\,,$

${}f^{\prime \prime \prime }{\left({\frac {\pi }{4}}\right)}={\frac {1}{\sqrt {2}}}{\left(-3-{\frac {\pi }{4}}\right)}={\frac {-12-\pi }{4{\sqrt {2}}}}\,,$

${}f^{\prime \prime \prime \prime }{\left({\frac {\pi }{4}}\right)}={\frac {1}{\sqrt {2}}}{\left(-4+{\frac {\pi }{4}}\right)}={\frac {-16+\pi }{4{\sqrt {2}}}}\,.$

Das Taylor-Polynom vom Grad ${}4$ ist daher

${\frac {\pi }{4{\sqrt {2}}}}+{\frac {4+\pi }{4{\sqrt {2}}}}{\left(x-{\frac {\pi }{4}}\right)}+{\frac {8-\pi }{8{\sqrt {2}}}}{\left(x-{\frac {\pi }{4}}\right)}^{2}+{\frac {-12-\pi }{24{\sqrt {2}}}}{\left(x-{\frac {\pi }{4}}\right)}^{3}+{\frac {-16+\pi }{96{\sqrt {2}}}}{\left(x-{\frac {\pi }{4}}\right)}^{4}.$