taylor f(x)=6sin(2x)

Lösung

taylor

f (x) = 6 sin (2 x)

12 x - 8 x^{3} + \frac{8}{5} x^{5} - \frac{16}{105} x^{7} + \frac{8}{945} x^{9} + \dots

Schritte zur Lösung

Apply the Taylor Formula

= 0 + \frac{\frac{d}{d x} ( 6 sin ( 2 x ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( 6 sin ( 2 x ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( 6 sin ( 2 x ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 0 + \frac{12}{1 !} x + \frac{0}{2 !} x^{2} + \frac{- 48}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{192}{5 !} x^{5} + \frac{0}{6 !} x^{6} + \frac{- 768}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{3072}{9 !} x^{9} + \dots

Fasse zusammen

= 12 x - 8 x^{3} + \frac{8}{5} x^{5} - \frac{16}{105} x^{7} + \frac{8}{945} x^{9} + \dots

y - 2 y^{^{'}} - 24 y = 0

\frac{\partial}{\partial x} (- 3 y cos (9 x))

\int_{0}^{6} 1 + x^{4} + 2 x^{2} d x

\frac{d}{d y} (\frac{A}{[ - 4.2 ]} y \frac{B}{( - 1.5 )})

k = 1 \sum \infty k e^{- k^{2}}