taylor sin(4x)cos(2x)

Lösung

taylor

sin (4 x) cos (2 x)

4 x - \frac{56}{3} x^{3} + \frac{488}{15} x^{5} - \frac{8752}{315} x^{7} + \frac{5624}{405} x^{9} + \dots

Schritte zur Lösung

Apply the Taylor Formula

= 0 + \frac{\frac{d}{d x} ( sin ( 4 x ) cos ( 2 x ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( sin ( 4 x ) cos ( 2 x ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( sin ( 4 x ) cos ( 2 x ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 0 + \frac{4}{1 !} x + \frac{0}{2 !} x^{2} + \frac{- 112}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{3904}{5 !} x^{5} + \frac{0}{6 !} x^{6} + \frac{- 140032}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{5039104}{9 !} x^{9} + \dots

Fasse zusammen

= 4 x - \frac{56}{3} x^{3} + \frac{488}{15} x^{5} - \frac{8752}{315} x^{7} + \frac{5624}{405} x^{9} + \dots

\int \frac{1.2 x}{x ^{2} x ^{4} - 9} d x

(\frac{sin ( x )}{1 + x})^{^{'}}

y^{^{''}} + 2 y^{^{'}} + 2 y = 0, y (0) = - 2, y^{^{'}} (0) = - 4

(- x + 2) \cdot y^{^{'}} - y = e^{- 2 x}

\frac{d y}{d x} = a (y^{2} + 1)