maclaurin x/((1+x^2)^2)

Lösung

maclaurin

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

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

Schritte zur Lösung

Apply the Maclaurin Formula

= 0 + \frac{\frac{d}{d x} ( \frac{x}{( 1 + x ^{2} ) ^{2}} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{x}{( 1 + x ^{2} ) ^{2}} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{x}{( 1 + x ^{2} ) ^{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{360}{5 !} x^{5} + \frac{0}{6 !} x^{6} + \frac{- 20160}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{1814400}{9 !} x^{9} + \dots

Fasse zusammen

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

f (x) = 2 x e^{x^{2}}

n = 0 \sum \infty (1 - 1 0^{42})^{n}

\frac{d}{d x} (ln (x) sin (x))

\frac{d ^{2} y}{d x ^{2}} = 3 e^{- x}, y (0) = 1, y^{^{'}} (0) = 4

\int 1 \frac{1}{x ^{2}} d x