Answer:
The Taylor series of f(x) around the point a, can be written as:
[tex]f(x) = f(a) + \frac{df}{dx}(a)*(x -a) + (1/2!)\frac{d^2f}{dx^2}(a)*(x - a)^2 + .....[/tex]
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:
[tex]fn = (-1)^{2n + 1}*4*(x - 7*pi)^{2n}[/tex]
In this exercise we must calculate the Taylor series for the given function in this way;
[tex]f_n= (-1)^{2n+1}(4)(x-7\pi)^{2n}[/tex]
The Taylor series of f(x) around the point a, can be written as:
[tex]f(x) = f(a) + f'(a)(x-a)+\frac{1}{2!} f''(a)(x-a)^2+....[/tex]
Here we have:
[tex]f(x) = 4cos(x)\\a = 7\pi[/tex]
Then, let's calculate each part:
[tex]f(a) = 4cos(7\pi) = -4\\df/dx = -4sin(x)\\(df/dx)(a) = -4sin(7\pi) = 0\\(d^2f)/(dx^2) = -4cos(x)\\(d^2f)/(dx^2)(a) = -4cos(7\pi) = 4[/tex]
Here we already can see two things:
1) The odd derivatives will have a sin(x) function that is zero when evaluated in [tex]x=7\pi[/tex].
2) We also can see that the sign will alternate between consecutive terms.
So we only will work with the even powers of the series:
[tex]f(x) = -4 + (1/2!)*4*(x - 7\pi)^2 - (1/4!)*4*(x - 7\pi)^4 + ....[/tex]
So we can write it as:
[tex]f(x)=\sum f_n[/tex]
Such that the n-th term can written as:
[tex]f_n= (-1)^{2n+1}(4)(x-7\pi)^{2n}[/tex]
See more abour Taylor series at: brainly.com/question/6953942
