taylor e^x,-1

Lösung

taylor

e^{x}, - 1

\frac{1}{e} + \frac{1}{e} (x + 1) + \frac{1}{2 e} (x + 1)^{2} + \frac{1}{6 e} (x + 1)^{3} + \frac{1}{24 e} (x + 1)^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

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

Fasse zusammen

= \frac{1}{e} + \frac{1}{e} (x + 1) + \frac{1}{2 e} (x + 1)^{2} + \frac{1}{6 e} (x + 1)^{3} + \frac{1}{24 e} (x + 1)^{4} + \dots

d er i v a t i v e f (x) = x^{2} (x + 1)

d er i v a t i v e f (x) = \frac{2 x ^{2}}{x ^{2} - 2 x - 8}

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

d er i v a t i v e (\frac{t ^{3}}{t ^{8} + 1})^{2}

y^{^{''}} + 4 y = cos (t)