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10 Nov 2019
Suppose that an unspecified function f(x) has the degree 7 Taylor polynomial at x = 0 p_7 (x) = -2x^7 - 18x^6 - 8x^5 - 17x^4. Because p_7 (x) = sigma_k = 0^7 f^(k) (0) x^k/k! syou can read off the value of the derivatives of f(x) at x = 0 by looking at the coefficients of x^k/k! as follows: the coefficient of x^1/1! In p_7 (x) is f^(1) (0) = the coefficient of x^2/2! In p_7 (x) is f^(2) (0) = the coefficient of x^3/3! In p_7 (x) is f^(3) (0) = the coefficient of x^4/4! In p_7 (x) is f^(4) (0) = And so without differentiating, we see that the smallest positive integer n > 1 for which f^(n)(0) notequalto 0 is
Suppose that an unspecified function f(x) has the degree 7 Taylor polynomial at x = 0 p_7 (x) = -2x^7 - 18x^6 - 8x^5 - 17x^4. Because p_7 (x) = sigma_k = 0^7 f^(k) (0) x^k/k! syou can read off the value of the derivatives of f(x) at x = 0 by looking at the coefficients of x^k/k! as follows: the coefficient of x^1/1! In p_7 (x) is f^(1) (0) = the coefficient of x^2/2! In p_7 (x) is f^(2) (0) = the coefficient of x^3/3! In p_7 (x) is f^(3) (0) = the coefficient of x^4/4! In p_7 (x) is f^(4) (0) = And so without differentiating, we see that the smallest positive integer n > 1 for which f^(n)(0) notequalto 0 is
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Collen VonLv2
12 Jan 2019