taylor ze^z

Lösung

taylor

z e^{z}

z + z^{2} + \frac{1}{2} z^{3} + \frac{1}{6} z^{4} + \frac{1}{24} z^{5} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

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

Fasse zusammen

= z + z^{2} + \frac{1}{2} z^{3} + \frac{1}{6} z^{4} + \frac{1}{24} z^{5} + \dots

x \to 2 lim (3 x)

n = 0 \sum \infty (\frac{n}{n + 3})^{n}

x \to 1 lim (4 x - 2)

d er i v a t i v e 2 sin (x) cos (x)

\frac{d y}{d x} = y, y (7) = 8