Question

Use the definition of Taylor series to find the Taylor series, centered at c, for the function. f(x) = sin x, c = 3π/4

Answer 1

`\sin(x) = \sum\limit^(\infty)_(n = 0) (1)/(\sqrt 2)((-1)^(n(n+1)/2))/(n!)(x - (3\pi)/(4))^n`

Explanation:

Given

`f(x) = \sin x\\`

`c = (3\pi)/(4)`

Required

Find the Taylor series

The Taylor series of a function is defines as:

`f(x) = f(c) + f'(c)(x -c) + (f`

We have:

`c = (3\pi)/(4)`

`f(x) = \sin x\\`

`f(c) = \sin(c)`

`f(c) = \sin((3\pi)/(4))`

This gives:

`f(c) = (1)/(\sqrt 2)`

We have:

`f(c) = \sin((3\pi)/(4))`

Differentiate

`f'(c) = \cos((3\pi)/(4))`

This gives:

`f'(c) = -(1)/(\sqrt 2)`

We have:

`f'(c) = \cos((3\pi)/(4))`

Differentiate

`f`

This gives:

`f`

We have:

`f`

Differentiate

`f`

This gives:

`f`

`f`

So, we have:

`f(c) = (1)/(\sqrt 2)`

`f'(c) = -(1)/(\sqrt 2)`

`f`

`f`

`f(x) = f(c) + f'(c)(x -c) + (f`

becomes

`f(x) = (1)/(\sqrt 2) - (1)/(\sqrt 2)(x - (3\pi)/(4)) -(1/\sqrt 2)/(2!)(x - (3\pi)/(4))^2 +(1/\sqrt 2)/(3!)(x - (3\pi)/(4))^3 + ... +(f^n(c))/(n!)(x - (3\pi)/(4))^n`

Rewrite as:

`f(x) = (1)/(\sqrt 2) + ((-1))/(\sqrt 2)(x - (3\pi)/(4)) +((-1)/\sqrt 2)/(2!)(x - (3\pi)/(4))^2 +((-1)^2/\sqrt 2)/(3!)(x - (3\pi)/(4))^3 + ... +(f^n(c))/(n!)(x - (3\pi)/(4))^n`

Generally, the expression becomes

`f(x) = \sum\limit^(\infty)_(n = 0) (1)/(\sqrt 2)((-1)^(n(n+1)/2))/(n!)(x - (3\pi)/(4))^n`

Hence:

`\sin(x) = \sum\limit^(\infty)_(n = 0) (1)/(\sqrt 2)((-1)^(n(n+1)/2))/(n!)(x - (3\pi)/(4))^n`