Question

How do you find the taylor series series for `f(x)=lnx` at a=2?

Answer 1

`ln(2)+1/2(x-2)-1/8(x-2)^2+1/24(x-2)^3-1/64(x-2)^4+\cdots`

Explanation:

Use the following expression for the Taylor series of an infinitely differentiable function at `x=a`:

`f(a)+f'(a)(x-a)+(f''(a))/(2!)(x-a)^2+(f'''(a))/(3!)(x-a)^3+(f''''(a))/(4!)(x-a)^4\cdots`

Since `f(x)=ln(x)`, we get `f'(x)=1/x=x^{-1}`, `f''(x)=-x^{-2}`, `f'''(x)=2x^{-3}`, `f''''(x)=-6x^{-4}`, etc...

Since `a=2`, we calculate `f(2)=ln(2)`, `f'(2)=1/2`, `f''(2)=-1/4`, `f'''(2)=2/8=1/4`, `f''''(2)=-6/16=-3/8`, etc...

Therefore, we can write the answer as

`ln(2)+1/2(x-2)-1/8(x-2)^2+1/24(x-2)^3-1/64(x-2)^4+\cdots`

This series happens to equal `ln(x)` for `0 < x < 4` (the "radius of convergence" is 2 and it equals the function for these values as well).

Answer 2

First, we can start with the general definition of the Taylor series expansion, which is:

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

where `f^((n))(a)` is the `n`th derivative of `f(x)` evaluated at `x -> a`, `n` varies, `a` does not, and `n!` is `1xx2xx3xxcdotsxx(n-1)xxn`. Note that `x->a` does not apply to `(x-a)^n` (yes, I've seen that happen).

Thus you have to take the derivative up to some `n`th order term. Let's say `n = 4`. Then you have:

`f^((0))(x) = f(x) = lnx`
`f'(x) = 1/x`
`f''(x) = -1/x^2`
`f'''(x) = 2/(x^3)`
`f''''(x) = -6/(x^4)`

Now you have the final result to simplify:

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

`= (ln2) + (1/2)(x-2) + (-1/4)/(2)(x-2)^2 + (2/8)/(6)(x-2)^3 + (-6/16)/(24)(x-2)^4 + ...`

`= color(blue)(ln2 + 1/2(x-2) - 1/8(x-2)^2 + 1/24(x-2)^3 - 1/64(x-2)^4 + ...)`

Just remember the following:
- `x->a` only for `f(a)`, not `(x-a)^n`
- `n` varies, but `a` does not
- It's probably wise to take the `n`th derivatives first so you aren't doing them as you are writing out the next step from the point you write out the general series formula