taylor x/(1+2x^2)

Lösung

taylor

\frac{x}{1 + 2 x ^{2}}

x - 2 x^{3} + 4 x^{5} - 8 x^{7} + 16 x^{9} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 0 + \frac{1}{1 !} x + \frac{0}{2 !} x^{2} + \frac{- 12}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{480}{5 !} x^{5} + \frac{0}{6 !} x^{6} + \frac{- 40320}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{5806080}{9 !} x^{9} + \dots

Fasse zusammen

= x - 2 x^{3} + 4 x^{5} - 8 x^{7} + 16 x^{9} + \dots

\int \frac{7 x ^{2} - 4 x + 4}{( x + 1 ) ( x ^{2} - x + 1 )} d x

d er i v a t i v e (3 + csc (2 t))^{- 6}

t an g e n t f (x) = - 2 - cos (x), at x = - \frac{π}{4}

in v erse l a pl a ce \frac{1}{( s - 1 )}

t an g e n t f (x) = x^{2} - 7, (2, - 3)