Question

Use an appropriate Taylor series to find the first four nonzero terms of an infinite series that is equal to sin(5/2).

Answer 1

Answer:

2.5, -2.61, 0.81, -0.12

Step-by-step explanation:

The taylor series of the function sin(x) around zero is given by

$sin(x)=\sum_{n=0}^{\infty}\dfrac{(-1)^k}{(2k-1)!}x^{2k+1}=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}$

Therefore,

$\sin(\frac{5}{2})=\dfrac{5}{2}-\dfrac{[\frac{5}{2}]^3}{3!}+\dfrac{[\frac{5}{2}]^5}{5!}-\dfrac{[\frac{5}{2}]^7}{7!}+...$

hence the first four nonzero terms of the series are

$\dfrac{5}{2}=2.5\\\\-\dfrac{[\frac{5}{2}]^3}{3!} \approx -2.61\\\\\dfrac{[\frac{5}{2}]^5}{5!} \approx 0.81\\\\-\dfrac{[\frac{5}{2}]^7}{7!} \approx -0.12$