taylor sqrt((1+x)/(1-x))-1

Lösung

taylor

\frac{1 + x}{1 - x} - 1

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

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 0 + \frac{1}{1 !} x + \frac{1}{2 !} x^{2} + \frac{3}{3 !} x^{3} + \frac{9}{4 !} x^{4} + \frac{45}{5 !} x^{5} + \dots

Fasse zusammen

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

d er i v a t i v e f (x) = sec (x^{2})

\frac{d}{d x} (9 + 3 x + sin (3 x))

t an g e n t f (x) = 2 x^{- 1}, at x = 5

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

\frac{d y}{d x} + 2 x y = 0