Question

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

Answer 1

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

Explanation:

Taylor series expansion:

`\sum_(n=0)^N f^n(a)\displaystyle((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(f'(a)(x-a))/(1!) + \displaystyle(f''(a)(x-a)^2)/(2!) + \displaystyle(f'''(a)(x-a)^3)/(3!) + \displaystyle(f^4(a)(x-a)^4)/(4!) + \displaystyle(f^5(a)(x-a)^5)/(5!) + ...\\\\= \sin 0 + \displaystyle(x)/(1!) +  \displaystyle((0)x^2)/(2!) + \displaystyle((-1)x^3)/(3!) + \displaystyle((0)x^4)/(4!) + \displaystyle((1)x^5)/(5!) + ...`

Solving, we get

`\sin x = x - \displaystyle(x^3)/(3!) + \displaystyle(x^5)/(5!) - \displaystyle(x^7)/(7!) + ...\\\\\sin x = x - \displaystyle(x^3)/(6) + \displaystyle(x^5)/(120) - \displaystyle(x^7)/(5040) + ...`