# Find the Maclaurin series for cos x and prove that it represents cos x for all x.

Solution:

Let f(x) = cos x

By Maclaurin expansion we have,

f(x) = f(0) + x f'(o) + $$\frac{x^2}{2!}$$f(0) + $$\frac{x^3}{3!}$$f”’(0) + …

Here,

f(x) = cos x

Differentiating we get,

f(x) = -sin x

f(x) = -cos x

f”’(x) = sin x

fiv(x) = cos x

Therefore,  Maclaurin Series for cos x is

$$cosx = cos0 – sin0x – \frac{cos0}{2!}x^2 – \frac{cos0}{3!}x^3 + \frac{sin0}{4!}x^4 + ….$$

$$= 1 – \frac{x^2}{2!} + \frac{x^4}{4!} – \frac{x^6}{6!} + …$$

Since the derivative and cosine function have absolute value less than or equal to 1. So, by Taylor’s inequality

$$|R_{2n}(x)| \leq \frac{|x|^{2n + 1}}{(2n + 1)!}$$

Now

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

i.e

$$\lim_{x \to ∞} R_{2n}(x) = 0$$

This implies that series is converges to cos x for all value of x