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 rn(x) → 0. ] f(x) = 8/x, a = −4

Answer 1

The taylor series for the f(x)=8/x centered at the given value of a=-4 is -2+2(x+4)/1!-24/16 $(x+4)^{2}$/2!+...........

Given a function f(x)=9/x,a=-4.

We are required to find the taylor series for the function f(x)=8/x centered at the given value of a and a=-4.

The taylor series of a function f(x)=$f(a)+f^{1}(a)(x-a)/1!+ f^{11}(a)(x-a)^{2} /2! +f^{111}(a)(x-a)a^{3}/3!+..........$

Where the terms in f prime $f^{1}$(a) represent the derivatives of x valued at a.

For the given function.f(x)=8/x and a=-4.

So,f(a)=f(-4)=8/(-4)=-2.

$f^{1}$(a)=$f^{1}$(-4)=-8/($-4)^{2}$

=-8/16

=-1/2

The series of f(x) is as under:

f(x)=f(-4)+$f^{1}(-4)(x+4)/1!+ f^{11}(-4)(x+4)^{2}/2!.............$

$=8/(-4)-8/(-4)^{2} (-4)(x+4)/1!+ 24/(-4)^{3} (-4)(x+4)^{2}/2!.............$

=-2+2(x+4)/1!-24/16 $(x+4)^{2}$/2!+...........

Hence the taylor series for the f(x)=8/x centered at the given value of a=-4 is -2+2(x+4)/1!-24/16 $(x+4)^{2}$/2!+...........

Learn more about taylor series at brainly.com/question/23334489

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