taylor ln(1+x),0

Solución

taylor

ln (1 + x), 0

n = 1 \sum \infty (- 1)^{n + 1} \frac{x ^{n}}{n}

Pasos de solución

Aplicar la fórmula de Taylor

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

Evaluar derivadas

= 0 + \frac{1}{1 !} x + \frac{- 1}{2 !} x^{2} + \frac{2}{3 !} x^{3} + \frac{- 6}{4 !} x^{4} + \frac{24}{5 !} x^{5} + \dots

Simplificar

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

La sumatoria que representa la serie

= n = 1 \sum \infty (- 1)^{n + 1} \frac{x ^{n}}{n}

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\frac{d}{d y} (x e^{- y})

\int sin^{4} (x co) s^{5} x d x