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

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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

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