taylor f(x)=x

Lösung

taylor

f (x) = x

x

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 0 + \frac{1}{1 !} x + \frac{0}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{0}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{0}{9 !} x^{9} + \frac{0}{10 !} x^{10}

Fasse zusammen

= x

x \to 0 lim (cos (π x))

\int e^{- 2 x} x^{2} d x

s l o p e f (x) = 0.01 (2)^{x} 11 a^{15}

\int_{0}^{\frac{π}{2}} cos^{3} (31 x) d x

t an g e n t f (x) = ln (6 - x^{2} + 2 x^{4}), at x = 1