Maclaurin series for cos x and prove it represents cos x

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Suresh Chand · Accepted Answer

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

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

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

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