taylor (1-x^2)^{-3/2}

Lösung

taylor

(1 - x^{2})^{- \frac{3}{2}}

1 + \frac{3}{2} x^{2} + \frac{15}{8} x^{4} + \frac{35}{16} x^{6} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{0}{1 !} x + \frac{3}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{45}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{1575}{6 !} x^{6} + \dots

Fasse zusammen

= 1 + \frac{3}{2} x^{2} + \frac{15}{8} x^{4} + \frac{35}{16} x^{6} + \dots

x^{2} y^{^{'}} - x = 1 + y + x y

\int_{- 1}^{3} (5 - \frac{1}{4} x^{3}) d x

\int \frac{7 x ^{2}}{1 + x ^{6}} d x

(x^{2} y^{2} - x^{4}) y^{^{'}} = x^{2} y^{2} - y^{4}

x \to - 1 lim (ln (1 + x))