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

Wir möchten für die Funktion

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

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

${\displaystyle {}f'(x)=\sin x+x\cos x\,,}$
${\displaystyle {}f^{\prime \prime }(x)=\cos x+\cos x-x\sin x=2\cos x-x\sin x\,,}$
${\displaystyle {}f^{\prime \prime \prime }(x)=-2\sin x-\sin x-x\cos x=-3\sin x-x\cos x\,,}$
${\displaystyle {}f^{\prime \prime \prime \prime }(x)=-3\cos x-\cos x+x\sin x=-4\cos x+x\sin x\,.}$

Unter Verwendung von

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

ist somit

${\displaystyle {}f{\left({\frac {\pi }{4}}\right)}={\frac {\pi }{4}}{\frac {1}{\sqrt {2}}}={\frac {\pi }{4{\sqrt {2}}}}\,,}$
${\displaystyle {}f'{\left({\frac {\pi }{4}}\right)}={\frac {1}{\sqrt {2}}}{\left(1+{\frac {\pi }{4}}\right)}={\frac {4+\pi }{4{\sqrt {2}}}}\,,}$
${\displaystyle {}f^{\prime \prime }{\left({\frac {\pi }{4}}\right)}={\frac {1}{\sqrt {2}}}{\left(2-{\frac {\pi }{4}}\right)}={\frac {8-\pi }{4{\sqrt {2}}}}\,,}$
${\displaystyle {}f^{\prime \prime \prime }{\left({\frac {\pi }{4}}\right)}={\frac {1}{\sqrt {2}}}{\left(-3-{\frac {\pi }{4}}\right)}={\frac {-12-\pi }{4{\sqrt {2}}}}\,,}$
${\displaystyle {}f^{\prime \prime \prime \prime }{\left({\frac {\pi }{4}}\right)}={\frac {1}{\sqrt {2}}}{\left(-4+{\frac {\pi }{4}}\right)}={\frac {-16+\pi }{4{\sqrt {2}}}}\,.}$

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

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