Question

Calcular el valor de la función f(x)=ex en el punto x=0.5, utilizando la representación seno como serie de McLaurin.​

Answer 1

Respuesta:

1,648...

Explicación paso a paso:

${e}^{x} = 1 + x + \frac{ {x}^{2} }{2!} + \frac{ {x}^{3} }{3!} + \frac{ {x}^{4} }{4!}\cdot{\cdot{\cdot{}}}$

${e}^{0.5} = 1 + 0.5 + \frac{ {0.5}^{2} }{2!} + \frac{ {0.5}^{3} }{3!} + \frac{ {0.5}^{4} }{4!}\cdot{\cdot{\cdot{}}}$

${e}^{0.5} = 1 + 0.5 + \frac{ {0.25}}{1 \times 2} + \frac{ {0.125} }{1 \times 2 \times 3} + \frac{ {0.0625} }{1 \times 2 \times 3 \times 4}\cdot{\cdot{\cdot{}}}$

${e}^{0.5} = 1 + 0.5 + \frac{ {0.25}}{2} + \frac{ {0.125} }{6} + \frac{ {0.0625} }{24}\cdot{\cdot{\cdot{}}}$

${e}^{0.5} = 1 + 0.5 + 0.125+ 0.0208333... + 0.0026041666...\cdot{\cdot{\cdot{}}}$

${e}^{0.5} = 1.648\cdot{}\cdot{}\cdot{}$