Question

What is the Taylor series for `cosx` centered around `x = pi`?

Answer 1

The general formula for a Taylor series is:

`f(x) = sum_(n=0)^(oo) (f^((n))(a))/(n!)(x-a)^n`

where `f^((n))(a)` is the `n`th derivative of `f` defined at `x = a`. Thus, it makes sense to take some derivatives when finding the Taylor series.

`f^((0))(x) = f(x) = cosx`

`f'(x) = -sinx`

`f''(x) = -cosx`

`f'''(x) = sinx`

`f''''(x) = cosx`

etc.

Therefore, the Taylor series centered at `x = a` is:

`= (f(a))/(0!)(x-a)^0 + (f'(a))/(1!)(x-a)^1 + (f''(a))/(2!)(x-a)^2 + (f'''(a))/(3!)(x-a)^3 + (f''''(a))/(4!)(x-a)^4 + ...`

But `sin(pi) = 0`, so for all odd derivatives, the term vanishes. Thus, we get:

`= cospi + cancel((f'(a))/(1!)(x-a)^1)^(0) + (-cospi)/(2!)(x-pi)^2 + cancel((f'''(a))/(3!)(x-a)^3)^(0) + (cospi)/(4!)(x-pi)^4 + ...`

`= color(blue)(-1 + 1/2(x-pi)^2 - 1/24(x-pi)^4 + . . . )`

Generalizing this...

• We have that the series alternates sign, so we include a `(-1)^n` term somewhere...
• We also have the `(x-pi)^n` term, obviously, but we specify even `n` to ignore the odd terms (which are zero), so we can use `2n`.
• We have a negative sign out front, since the first term in a Taylor series tends to be `1`.
• We would have to specify even `n` for the factorial terms, so the factorial terms go as `(2n)!`, i.e. `0!`, `2!`, `4!`, etc.

So, our instinct gives us a series representation as:

`sum_(n=0)^(oo) -((-1)^(n)(x-pi)^(2n))/((2n)!)`

`= color(blue)(sum_(n=0)^(oo) ((-1)^(n+1)(x-pi)^(2n))/((2n)!))`

`= ((-1)^(0+1)(x-pi)^(2*0))/((2*0)!) + ((-1)^(1+1)(x-pi)^(2*1))/((2*1)!) + ((-1)^(2+1)(x-pi)^(2*2))/((2*2)!) + ((-1)^(3+1)(x-pi)^(2*3))/((2*3)!) + . . .`

`= color(blue)(-1 + 1/2(x-pi)^2 - 1/24(x-pi)^4 + 1/720(x-pi)^6 - . . . )`