taylor 1/(3-x)

Lösung

taylor

\frac{1}{3 - x}

\frac{1}{3} + \frac{1}{9} x + \frac{1}{27} x^{2} + \frac{1}{81} x^{3} + \frac{1}{243} x^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

= \frac{1}{3} + \frac{\frac{d}{d x} ( \frac{1}{3 - x} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{1}{3 - x} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{1}{3 - x} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= \frac{1}{3} + \frac{\frac{1}{9}}{1 !} x + \frac{\frac{2}{27}}{2 !} x^{2} + \frac{\frac{2}{27}}{3 !} x^{3} + \frac{\frac{8}{81}}{4 !} x^{4} + \dots

Fasse zusammen

= \frac{1}{3} + \frac{1}{9} x + \frac{1}{27} x^{2} + \frac{1}{81} x^{3} + \frac{1}{243} x^{4} + \dots

\int \frac{- 1}{2} cos (2 x) d x

\frac{d}{d x} (\frac{x}{z ( x )})

\frac{\partial}{\partial x} (14 x y e^{- x^{2} - y^{2}})

t an g e n t f (x) = x + 3, at x = 6

y^{^{''}} + 16 y = 6 tan (4 x)