Download Calculus 1c-6, Examples of Taylor's Formula and Limit by Mejlbro L. PDF

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By Mejlbro L.

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I. 1) From the series of the exponential we get 2 e−x = 1 − x2 + 1 4 1 6 1 8 x − x + x + x8 ε(x). 2 6 24 Furthermore, cos x = 1 − 1 2 1 4 1 6 1 x + x − x + x8 + x8 ε(x), 2 24 720 40 320 hence f (x) 2 = e−x − cos x 1 1 1 = − x2 + − x4 − 2 2 24 11 4 119 6 1 x − x + = − x2 + 2 24 720 1 1 1 1 − − x6 + 6 6! 24 8! 1679 8 x + x8 ε(x). 40 320 x8 + x8 ε(x) 2) Since 2 f (x) = sin x − 2xe−x = 2 d e−x − cos x , dx one may wrongly conclude that by differentiation of the result of (1) should obtain 2 f (x) = sin x − 2xe−x 11 3 119 5 1679 7 x − x + = −x + x + x7 ε(x).

I. 1) If f (x) = 2x = ex ln 2 , then f (k) (x) = (ln 2)k · 2x , so (ln 2)k f (k) (0) = , k! k! and the Taylor expansion is n f (x) = 2x = k=0 1 (ln 2)k xk + xn ε(x). k! (x + 2)−k−1 , so f (k) (0) (−1)k = k+1 , k! 2 and the Taylor expansion is f (x) = 1 1 = 2+x 2 n (−1)k k=0 x 2 k + xn ε(x), x into the development which we of course also could have obtained directly by putting y = 2 1 1 . 12 Find the Taylor expansion with the point of expansion x0 = 0 and of any order n for the functions (1) f (x) = sin x + cos x, (2) f (x) = 1 + x2 − 1 − x2 .

And the Taylor expansion is n f (x) = 2x = k=0 1 (ln 2)k xk + xn ε(x). k! (x + 2)−k−1 , so f (k) (0) (−1)k = k+1 , k! 2 and the Taylor expansion is f (x) = 1 1 = 2+x 2 n (−1)k k=0 x 2 k + xn ε(x), x into the development which we of course also could have obtained directly by putting y = 2 1 1 . 12 Find the Taylor expansion with the point of expansion x0 = 0 and of any order n for the functions (1) f (x) = sin x + cos x, (2) f (x) = 1 + x2 − 1 − x2 . A. Taylor expansion for any n. D. Substitute into known developments.

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