taylor e^x,-4

Lösung

taylor

e^{x}, - 4

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

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

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

Fasse zusammen

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

\frac{\partial}{\partial y} ((2 x + 3 y)^{10})

t an g e n t y = \frac{2}{x}, (4, \frac{1}{2})

t an g e n t f (x) = e^{- x} ln (x), (1, 0)

\int \frac{1}{x ( x ^{2} + 4 ) ^{2}} d x

\frac{d}{d θ} (2 sin (5 θ))