taylor x^{-2},3

Lösung

taylor

x^{- 2}, 3

\frac{1}{9} - \frac{2}{27} (x - 3) + \frac{1}{27} (x - 3)^{2} - \frac{4}{243} (x - 3)^{3} + \frac{5}{729} (x - 3)^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

= \frac{1}{9} + \frac{\frac{d}{d x} ( x ^{- 2} ) ( 3 )}{1 !} (x - 3) + \frac{\frac{d ^{2}}{d x ^{2}} ( x ^{- 2} ) ( 3 )}{2 !} (x - 3)^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( x ^{- 2} ) ( 3 )}{3 !} (x - 3)^{3} + \dots

Evaluate Derivatives

= \frac{1}{9} + \frac{- \frac{2}{27}}{1 !} (x - 3) + \frac{\frac{2}{27}}{2 !} (x - 3)^{2} + \frac{- \frac{8}{81}}{3 !} (x - 3)^{3} + \frac{\frac{40}{243}}{4 !} (x - 3)^{4} + \dots

Fasse zusammen

= \frac{1}{9} - \frac{2}{27} (x - 3) + \frac{1}{27} (x - 3)^{2} - \frac{4}{243} (x - 3)^{3} + \frac{5}{729} (x - 3)^{4} + \dots

\frac{d}{d s} (\frac{k}{s ^{2} + k ^{2}})

\int (x - 1) d x

t^{2} y^{^{''}} + 5 t y^{^{'}} - 12 y = 0

ma c l a u r in \frac{1}{( 1 - x ^{2} ) ^{2}}

10 x - 6 y x^{2} + 1 \frac{d y}{d x} = 0