Question

Taylors expansion of f(x) = Sin x at x = 0.

Answer 1

Taylors expansion of $f(x) = \sin x$ at $x = 0$.

Solution

$$f(x) = \sin x \Rightarrow f(x) = f(0) + \frac{f'(0)}{1!} x + \frac{f''(0)}{2!} x^2 + \cdots + \frac{f^{(n)}(x)}{n!} x^n + \frac{f^{(n+1)}(\theta x)}{(n+1)!} x^{n+1},$$

where $\theta \in (0,1)$;

Calculate

$$f(0) = \sin 0 = 0;$$ $$f'(0) = \cos 0 = 1;$$ $$f''(0) = -\sin 0 = 0;$$ $$f'''(0) = -\cos 0 = -1;$$ $$f^{(4)}(0) = \sin 0 = 0;$$ $$f^{(5)}(0) = \cos 0 = 1;$$ $$f^{(6)}(x) = -\sin x;$$

Hence:

$$\sin x = \frac{1}{1!} x + \frac{-1}{3!} x^3 + \frac{1}{5!} x^5 + \frac{-\sin(\theta x)}{6!} x^6 = x - \frac{x^3}{6} + \frac{x^5}{120} - \frac{\sin \theta x}{720} x^6.$$