taylor 1/(x-1),5

Lösung

taylor

\frac{1}{x - 1}, 5

\frac{1}{4} - \frac{1}{16} (x - 5) + \frac{1}{64} (x - 5)^{2} - \frac{1}{256} (x - 5)^{3} + \frac{1}{1024} (x - 5)^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

= \frac{1}{4} + \frac{\frac{d}{d x} ( \frac{1}{x - 1} ) ( 5 )}{1 !} (x - 5) + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{1}{x - 1} ) ( 5 )}{2 !} (x - 5)^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{1}{x - 1} ) ( 5 )}{3 !} (x - 5)^{3} + \dots

Evaluate Derivatives

= \frac{1}{4} + \frac{- \frac{1}{16}}{1 !} (x - 5) + \frac{\frac{1}{32}}{2 !} (x - 5)^{2} + \frac{- \frac{3}{128}}{3 !} (x - 5)^{3} + \frac{\frac{3}{128}}{4 !} (x - 5)^{4} + \dots

Fasse zusammen

= \frac{1}{4} - \frac{1}{16} (x - 5) + \frac{1}{64} (x - 5)^{2} - \frac{1}{256} (x - 5)^{3} + \frac{1}{1024} (x - 5)^{4} + \dots

t an g e n t y = 8 x - x^{2}

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

\frac{d y}{d x} = (2 x)

x \to \infty lim (\frac{4 x ^{2}}{e ^{4 x}})

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