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Additional resources for Calculus 1c-6, Examples of Taylor's Formula and Limit Processes
We shall then prove that this assumption will lead to a contradiction, hence that the assumption is false. 3. Assume that (4). Then by (2) for x = 1, m 1 1 1 = 1 + + + ··· + + Rn+1 (1). n 1! 2! n! First prove that 0 < n! Rn+1 (1) < 1. Multiply the equation by n! and then derive the contradiction. A. Taylor expansion. There are given some guidelines. D. Follow the guidelines. I. 1) Since f (n) (x) = ex > 0, we get Rn+1 (x) = eξ xn+1 > 0, (n + 1)! for x > 0, hence ex = 1 + x2 xn x xn + + ··· + + Rn+1 (x) > , 1!
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 .
From these results follows that a general term of the Taylor expansion is Q(j) (0) j 8! x = a8−j xj = j! (8 − j)! 8 j a8−j xj , j = 0, 1, . . , 8. Then we ﬁnd the Taylor polynomial 8 8 j P8 (x) = j=0 a8−j xj . 2) Since Q(9) (x) ≡ 0, it follows from Taylor’s formula that Q(x) = (x + a)8 = P8 (x) + 1 (9) Q (ξ) x9 = P8 (x). 9! 3) When we use the same method as above we obtain the general binomial formula n (x + a)n = j=0 n j an−j xj . 3 Prove that for positive x, 1+ 1 3 1 6 x − x < 2 8 1 + x3 < 1 + 1 3 1 6 1 9 x − x + x , 2 8 16 and ﬁnd the corresponding bounds for 1 2 1 + x3 dx.