taylor (1+1/x)^{-2}

Lösung

taylor

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

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

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= \frac{1}{4} + \frac{\frac{1}{4}}{1 !} (x - 1) + \frac{- \frac{1}{8}}{2 !} (x - 1)^{2} + \frac{0}{3 !} (x - 1)^{3} + \frac{\frac{3}{8}}{4 !} (x - 1)^{4} + \frac{- \frac{15}{8}}{5 !} (x - 1)^{5} + \dots

Fasse zusammen

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

t an g e n t f (x) = sin (x), at x = 0

\frac{d}{d x} (\frac{- 1}{x + c})

x y^{^{'}} + y = y^{2} ln (x)

y^{^{''}} - 2 y^{^{'}} + y = 2 cos (t) + 4 sin (t), y (0) = 3, y^{^{'}} (0) = 2

\frac{d}{d x} (x^{6})