taylor e^x+0+5

Lösung

taylor

e^{x} + 0 + 5

6 + x + \frac{1}{2} x^{2} + \frac{1}{6} x^{3} + \frac{1}{24} x^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

= 6 + \frac{\frac{d}{d x} ( e ^{x} + 0 + 5 ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( e ^{x} + 0 + 5 ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( e ^{x} + 0 + 5 ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 6 + \frac{1}{1 !} x + \frac{1}{2 !} x^{2} + \frac{1}{3 !} x^{3} + \frac{1}{4 !} x^{4} + \dots

Fasse zusammen

= 6 + x + \frac{1}{2} x^{2} + \frac{1}{6} x^{3} + \frac{1}{24} x^{4} + \dots

x \to 0 lim ((1 + 4 x)^{\frac{3}{x}})

\int (\frac{tan ( u )}{4} - \frac{4}{7} u^{6} - \frac{2}{u}) d u

x \to a lim (\frac{1}{2})

d er i v a t i v e f (x) = (5 x + 4)^{3}

\int_{0}^{5} \frac{1}{u ^{2} - 25} d u