Question

Write the Taylor Series for f(x) = sin(x)center

Answer 1

Answer:

Taylor series of sin(x) centered at x = 0.

Step-by-step explanation:

Taylor series expansion:

$\sum_{n=0}^N f^n(a)\displaystyle\frac{(x-a)^n}{n!}$

Here, f(x) = sin (x) and a = 0

$f(x) = \sin x, f(0) = 0\\f'(x) = \cos x, f'(0) = 1\\f''(x) = -\sin x, f''(0) = 0\\f'''(x) = -\cos x, f'''(0) = -1\\f^4(x) = \sin x, f^4(0) = 0\\f^5(x) = \cos x, f^5(0) = 0$

Putting all the values and expanding, we get,

$f(x) = f(a) + \displaystyle\frac{f'(a)(x-a)}{1!} + \displaystyle\frac{f''(a)(x-a)^2}{2!} + \displaystyle\frac{f'''(a)(x-a)^3}{3!} + \displaystyle\frac{f^4(a)(x-a)^4}{4!} + \displaystyle\frac{f^5(a)(x-a)^5}{5!} + ...\\\\= \sin 0 + \displaystyle\frac{x}{1!} + \displaystyle\frac{(0)x^2}{2!} + \displaystyle\frac{(-1)x^3}{3!} + \displaystyle\frac{(0)x^4}{4!} + \displaystyle\frac{(1)x^5}{5!} + ...$

Solving, we get

$\sin x = x - \displaystyle\frac{x^3}{3!} + \displaystyle\frac{x^5}{5!} - \displaystyle\frac{x^7}{7!} + ...\\\\\sin x = x - \displaystyle\frac{x^3}{6} + \displaystyle\frac{x^5}{120} - \displaystyle\frac{x^7}{5040} + ...$