maclaurin f(x)=x^4sin(4x)

Lösung

maclaurin

f (x) = x^{4} sin (4 x)

4 x^{5} - \frac{32}{3} x^{7} + \frac{128}{15} x^{9} + \dots

Schritte zur Lösung

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 0 + \frac{0}{1 !} x + \frac{0}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{480}{5 !} x^{5} + \frac{0}{6 !} x^{6} + \frac{- 53760}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{3096576}{9 !} x^{9} + \frac{0}{10 !} x^{10} + \dots

Fasse zusammen

= 4 x^{5} - \frac{32}{3} x^{7} + \frac{128}{15} x^{9} + \dots

y^{^{''}} + 9 y = cos (3 x) + sin (3 x)

n = 3 \sum \infty \frac{3 ^{n}}{1 0 ^{n}}

\frac{\partial}{\partial x} (\frac{1}{ln ( \frac{2}{x} )})

y^{^{''}} + 6 y^{^{'}} + 8 y = e^{2 x}

\int e^{2 - 5 x} d x