Find the Taylor series for f(x) centered at the given value of a. [Assume that f

Question

Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that R_n(x) rightarrow 0.] f(x) = x^4 - 7x^2 + 9, a = 1 sigma^infinity_n = 0 f^n(1)/n! (x - 1)^n = -3 - 10(x - 1) - 1(x - 1)^2 + 4(x - 1)^3 + (x - 1)^4 sigma^infinity_n = 0 f^n(1)/n! (x - 1)^n = 10 + 3(x - 1) - 4(x - 1)^2 - 1(x - 1)^3 + (x - 1)^4 sigma^infinity_n = 0 f^n(1)/n! (x - 1)^n = 10 + 3(x - 1) + 4(x - 1)^2 - 1(x - 1)^3 + (x - 1)^4 sigma^infinity_n = 0 f^n(1)/n! (x - 1)^n = 3 - 10(x - 1) + 4(x - 1)^2 - 1(x - 1)^3 + (x - 1)^4 sigma^infinity_n = 0 f^n(1)/n! (x - 1)^n = 3 - 10(x - 1) - 1(x - 1)^2 + 4(x - 1)^3 + (x - 1)^4

Explanation / Answer

This is actually a straight forward problem. The Taylor Series of f(x) centered at the point x = a is given by:

f(x) = f(a) + f ‘(a)*(x-a)/ 1! + f ‘’(a)*(x-a)^2/ 2! + f ‘’’(a)*(x-a)^3 /3! + ....

Using this formula

.f(a) = X4-7X2+9 = 14-7(1)4+9 = 3

F’(a) = 4x3-14x = 4(1)3-14(1) = -10

F’’ (a) = 12x2-14 = 12(1)2-14 = -2

F’’’(a) = 24x = 24(1) = 24

F’’’’(a) = 24

Substituting this values in above equation

F(x) = 3-10(x-1)-2(x-1)2/2! + 24(x-1)3/3! + 24(x-1)4/4!

F(x) = 3-10(x-1)–1(x-1)2+4(x-1)+(x-1)4

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