taylor z(z+1)e^z

Lösung

taylor

z (z + 1) e^{z}

z + 2 z^{2} + \frac{3}{2} z^{3} + \frac{2}{3} z^{4} + \frac{5}{24} z^{5} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 0 + \frac{1}{1 !} z + \frac{4}{2 !} z^{2} + \frac{9}{3 !} z^{3} + \frac{16}{4 !} z^{4} + \frac{25}{5 !} z^{5} + \dots

Fasse zusammen

= z + 2 z^{2} + \frac{3}{2} z^{3} + \frac{2}{3} z^{4} + \frac{5}{24} z^{5} + \dots

\frac{\partial}{\partial x} ((2 x y - 1) e^{- x y})

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

t an g e n t y = 8 x + 12 - 15 e^{x}, (0, - 3)

\int \frac{3 + x}{1 + x ^{2}} d x

a re a x^{2} + 10, 9 x, 0