Question

Find the Taylors expansion of f(x) = Sin x at x = 0

Answer 1

Answer on Question #37226 – Math - Differential Calculus

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

Solution.

Taylor's theorem.

Suppose $f(x)$ has derivatives of all orders at $x = b$. The Taylor series of $f(x)$ expanded about $x = b$ is

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

Let's find the Taylor series of $f(x) = \sin x$ expanded about $x = 0$.

First, compute the derivatives and look for a pattern.

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

So we see

$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \cdots +$$

The $n^{th}$ term is $\frac{(-1)^{n-1} x^{2n-1}}{(2n-1)!}$.

To find the interval of convergence, observe that for all $x$, and for all $x$

$$\left| f^{(n+1)}(x) \right| \leq 1$$

So in Taylor's inequality

$$|R_n(x)| \leq \frac{M}{(n+1)!} |x - b|^{n+1}$$

we can take $M = 1$ and $C = \infty$, for each $x$ obtaining $|R_n(x)| \leq \frac{|x|^{n+1}}{(n+1)!}$

Now apply the Ration Test

$$\lim_{n \to \infty} \frac{|x|^{n+2} / (n+2)!}{|x|^{n+1} / (n+1)!} = \lim_{n \to \infty} \frac{|x|}{(n+2)!} = 0$$

for all $x$ and we see the radius of convergence is infinite.