Question

Use an appropriate series to find Taylor series of the given function centered at the indicated value of a. Write your answer in summation notation.

sinx, a= 2π

Answer 1

The Taylor series is
`$$\sum_(n=0)^(\infty) [((-1)^n)/((2n +1)!) (x)^(2n+1)]`

Explanation:

From the question we are told that

The function is
`f(x) = sin (x)`

This is centered at

`a = 2 \pi`

Now the next step is to represent the function sin (x) in it Maclaurin series form which is

`sin (x) = (x^3)/(3! ) + (x^5)/(5!) - (x^7)/(7 !) +***`

=>
`sin (x) = $$\sum_(n=0)^(\infty) [((-1)^n)/((2n +1)!) (x)^(2n+1)]`

Now since the function is centered at
`a = 2 \pi`

We have that

`sin (x - 2 \pi ) = (x-2 \pi ) - ((x - 2 \pi)^3 )/(3 \ !) + ((x - 2 \pi)^5 )/(5 \ !) - ((x - 2 \pi)^7 )/(7 \ !) + ***`

This above equation is generated because the function is not centered at the origin but at
`a = 2 \pi`

`sin (x-2 \pi ) = $$\sum_(n=0)^(\infty) [((-1)^n)/((2n +1)!) (x - 2 \pi)^(2n+1)]`

Now due to the fact that
`sin (x- 2 \pi) = sin (x)`

This because
`2 \pi` is a constant

Then it implies that the Taylor series of the function centered at
`a = 2 \pi` is

`$$\sum_(n=0)^(\infty) [((-1)^n)/((2n +1)!) (x)^(2n+1)]`