taylor sqrt(1+x),0

Lösung

taylor

1 + x, 0

1 + \frac{1}{2} x - \frac{1}{8} x^{2} + \frac{1}{16} x^{3} - \frac{5}{128} x^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{\frac{1}{2}}{1 !} x + \frac{- \frac{1}{4}}{2 !} x^{2} + \frac{\frac{3}{8}}{3 !} x^{3} + \frac{- \frac{15}{16}}{4 !} x^{4} + \dots

Fasse zusammen

= 1 + \frac{1}{2} x - \frac{1}{8} x^{2} + \frac{1}{16} x^{3} - \frac{5}{128} x^{4} + \dots

x \to 1 lim (e^{6 x^{2} - 6 x})

d er i v a t i v e \frac{120 x ^{2}}{x ^{2} + 4}

\int 62 + 6 x d x

y^{^{''}} - y = 9 e^{5 t} + 6 t e^{5 t} + 3 t^{2} e^{5 t}

x \to 0 lim (3 3 x)