Question

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

Answer 1

Answer on Question #64245 – Math – Differential Equations

Question

Taylor's expansion of $f(x) = \sin x$ at $x = 0$.

Solution

By the definition, Taylor expansion is given by

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

where $a$ is a point where the expansion is evaluated,

$n! = 1 \cdot 2 \cdot \cdots \cdot n$ is the factorial of $n$,

$f^{(n)}(a)$ is the $n$th derivative of $f$ evaluated at the point $a$.

In this problem

$$f(x) = \sin(x), \quad f^{(n)}(a) = \sin\left(a + \frac{n\pi}{2}\right), \quad a = 0,$$ $$f^{(n)}(0) = \sin\left(\frac{n\pi}{2}\right) = \begin{cases} 1, & \text{if } n = 4m + 1, \text{ in is integer}, \\ 0, & \text{if } n = 2m, \quad \text{m is integer}, \\ -1, & \text{if } n = 4m + 3, \text{m is integer}. \end{cases}$$

In other words, the derivatives of even degree are zero.

General formula of expansion of $f(x) = \sin x$ at $x = 0$:

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

Answer: $\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$

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