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) = 4 cos(x), a = 7π

Answer 1

Answer:

The Taylor series of f(x) around the point a, can be written as:

$f(x) = f(a) + \frac{df}{dx}(a)*(x -a) + (1/2!)\frac{d^2f}{dx^2}(a)*(x - a)^2 + .....$

Here we have:

f(x) = 4*cos(x)

a = 7*pi

then, let's calculate each part:

f(a) = 4*cos(7*pi) = -4

df/dx = -4*sin(x)

(df/dx)(a) = -4*sin(7*pi) = 0

(d^2f)/(dx^2) = -4*cos(x)

(d^2f)/(dx^2)(a) = -4*cos(7*pi) = 4

Here we already can see two things:

the odd derivatives will have a sin(x) function that is zero when evaluated in x = 7*pi, and we also can see that the sign will alternate between consecutive terms.

so we only will work with the even powers of the series:

f(x) = -4 + (1/2!)*4*(x - 7*pi)^2 - (1/4!)*4*(x - 7*pi)^4 + ....

So we can write it as:

f(x) = ∑fₙ

Such that the n-th term can written as:

$fn = (-1)^{2n + 1}*4*(x - 7*pi)^{2n}$