taylor 1/(x^2),5

Lösung

taylor

\frac{1}{x ^{2}}, 5

\frac{1}{25} - \frac{2}{125} (x - 5) + \frac{3}{625} (x - 5)^{2} - \frac{4}{3125} (x - 5)^{3} + \frac{1}{3125} (x - 5)^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

= \frac{1}{25} + \frac{\frac{d}{d x} ( \frac{1}{x ^{2}} ) ( 5 )}{1 !} (x - 5) + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{1}{x ^{2}} ) ( 5 )}{2 !} (x - 5)^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{1}{x ^{2}} ) ( 5 )}{3 !} (x - 5)^{3} + \dots

Evaluate Derivatives

= \frac{1}{25} + \frac{- \frac{2}{125}}{1 !} (x - 5) + \frac{\frac{6}{625}}{2 !} (x - 5)^{2} + \frac{- \frac{24}{3125}}{3 !} (x - 5)^{3} + \frac{\frac{24}{3125}}{4 !} (x - 5)^{4} + \dots

Fasse zusammen

= \frac{1}{25} - \frac{2}{125} (x - 5) + \frac{3}{625} (x - 5)^{2} - \frac{4}{3125} (x - 5)^{3} + \frac{1}{3125} (x - 5)^{4} + \dots

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

t an g e n t f (x) = \frac{1}{x}, at x = 0.204

\int \frac{u ^{2} - 1}{u ^{2} + 1} d u

t an g e n t f (x) = 5 e^{x}, at x = 3

\frac{d}{d x} (\frac{x ^{3}}{( x + 2 ) ^{2}})