Answer:
The Taylor series is [tex]$\sum_{n=0}^{\infty} [\frac{(-1)^n}{(2n +1)!} (x)^{2n+1}][/tex]
Step-by-step explanation:
From the question we are told that
The function is [tex]f(x) = sin (x)[/tex]
This is centered at
[tex]a = 2 \pi[/tex]
Now the next step is to represent the function sin (x) in it Maclaurin series form which is
[tex]sin (x) = \frac{x^3}{3! } + \frac{x^5}{5!} - \frac{x^7}{7 !} +***[/tex]
=> [tex]sin (x) = $\sum_{n=0}^{\infty} [\frac{(-1)^n}{(2n +1)!} (x)^{2n+1}][/tex]
Now since the function is centered at [tex]a = 2 \pi[/tex]
We have that
[tex]sin (x - 2 \pi ) = (x-2 \pi ) - \frac{(x - 2 \pi)^3 }{3 \ !} + \frac{(x - 2 \pi)^5 }{5 \ !} - \frac{(x - 2 \pi)^7 }{7 \ !} + ***[/tex]
This above equation is generated because the function is not centered at the origin but at [tex]a = 2 \pi[/tex]
[tex]sin (x-2 \pi ) = $\sum_{n=0}^{\infty} [\frac{(-1)^n}{(2n +1)!} (x - 2 \pi)^{2n+1}][/tex]
Now due to the fact that [tex]sin (x- 2 \pi) = sin (x)[/tex]
This because [tex]2 \pi[/tex] is a constant
Then it implies that the Taylor series of the function centered at [tex]a = 2 \pi[/tex] is
[tex]$\sum_{n=0}^{\infty} [\frac{(-1)^n}{(2n +1)!} (x)^{2n+1}][/tex]
