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

Lösung

taylor

\frac{e ^{x^{2}} - 1}{x}

e - 1 + (e + 1) (x - 1) + (2 e - 1) (x - 1)^{2} + \frac{4 e + 3}{3} (x - 1)^{3} + \frac{11 e - 6}{6} (x - 1)^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= e - 1 + \frac{e + 1}{1 !} (x - 1) + \frac{2 ( 2 e - 1 )}{2 !} (x - 1)^{2} + \frac{2 ( 4 e + 3 )}{3 !} (x - 1)^{3} + \frac{4 ( 11 e - 6 )}{4 !} (x - 1)^{4} + \dots

Fasse zusammen

= e - 1 + (e + 1) (x - 1) + (2 e - 1) (x - 1)^{2} + \frac{4 e + 3}{3} (x - 1)^{3} + \frac{11 e - 6}{6} (x - 1)^{4} + \dots

n \to \infty lim (- n^{2})

\frac{d}{d x} (2 x^{3} (1 - x^{2})^{4})

x y^{^{'}} - 2 y + 8 = 0

\int e^{4 x + 5} d x

l a pl a ce t r an s f or m e^{- 8 s}