Question

How do you find the taylor series for `lnx` in powers of `x-1`?

Answer 1

`lnx = sum_(n=1)^oo (-1)^(n-1) (x-1)^n/n`

Explanation:

The general formula for the Taylor series of a function `f(x)` around `x=1` is:

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

we can immediately note that:

`f^((0))(1) = lnx |_(x=1) = 0`

so the constant term is null. For the following terms, we have to calculate the derivatives of `f(x) = lnx` for all orders:

`d/(dx)lnx = 1/x = x^(-1)`

`(d^2)/(dx^2) lnx = -x^(-2)`

`(d^3)/(dx^3) lnx = 2x^(-3)`

and we can easily see that in general:

`(d^n)/(dx^n) lnx = (-1)^(n-1)(n-1)!x^(-n)`

and for `x=1`

`f^((n))(1) =[(d^n)/(dx^n) lnx]_(x=1) = (-1)^(n-1)(n-1)!`

The Taylor series is then:

`lnx = sum_(n=1)^oo (-1^(n-1)(n-1)!)/(n!)(x-1)^n = sum_(n=1)^oo (-1)^(n-1) (x-1)^n/n`

As a bonus, we can note that this expansion can be used to calculate the sum of the alternating harmonic series:

`1-1/2+1/3-1/4+... = sum_(n=1)^oo (-1)^(n-1)/n`

In fact if we use the Taylor series above for `x=2` we have:

`ln2 = sum_(n=1)^oo (-1)^(n-1) (2-1)^n/n = sum_(n=1)^oo (-1)^(n-1)/n`

Answer 2

`ln x=(x-1)-(x-1)^2/2-(x-1)^3/3-,,,.-1<x-1<=1` that gives `-2<x<=2`

Explanation:

Using `X = x-1`, the The Maclaurin series

`ln(1+X)=X-X^2/2+X^3/3-..., -1<X<1`

Is the Taylor series for

`ln x = ln(1+(x-1))=(x-1)-(x-1)^2/2-(x-1)^3/3-,,,.-1<x-1<=1` that gives `-2<x<=2`