taylor e^{3x},0

Lösung

taylor

e^{3 x}, 0

1 + 3 x + \frac{9}{2} x^{2} + \frac{9}{2} x^{3} + \frac{27}{8} x^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{3}{1 !} x + \frac{9}{2 !} x^{2} + \frac{27}{3 !} x^{3} + \frac{81}{4 !} x^{4} + \dots

Fasse zusammen

= 1 + 3 x + \frac{9}{2} x^{2} + \frac{9}{2} x^{3} + \frac{27}{8} x^{4} + \dots

\int x^{2} (5 x^{2} + x)^{2} d x

\frac{\partial}{\partial h} (\frac{1}{2} (a + b) h)

\frac{d}{d x} (\frac{2}{x ^{\frac{1}{3}}} - 2)

y^{^{''}} - 2 y^{^{'}} + 5 y = 5, y (0) = 2, y^{^{'}} (0) = 1

t an g e n t f (x) = (\frac{x - 3}{x + 1})^{2}, at x = 0