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

Jan 13, 2017

$\ln x = {\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n - 1} {\left(x - 1\right)}^{n} / n$

Explanation:

The general formula for the Taylor series of a function $f \left(x\right)$ around $x = 1$ is:

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

we can immediately note that:

${f}^{\left(0\right)} \left(1\right) = \ln x {|}_{x = 1} = 0$

so the constant term is null. For the following terms, we have to calculate the derivatives of $f \left(x\right) = \ln x$ for all orders:

$\frac{d}{\mathrm{dx}} \ln x = \frac{1}{x} = {x}^{- 1}$

$\frac{{d}^{2}}{{\mathrm{dx}}^{2}} \ln x = - {x}^{- 2}$

$\frac{{d}^{3}}{{\mathrm{dx}}^{3}} \ln x = 2 {x}^{- 3}$

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:

$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots = {\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n - 1} / n$

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

$\ln 2 = {\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n - 1} {\left(2 - 1\right)}^{n} / n = {\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n - 1} / n$

Jan 13, 2017

$\ln x = \left(x - 1\right) - {\left(x - 1\right)}^{2} / 2 - {\left(x - 1\right)}^{3} / 3 - , , , . - 1 < x - 1 \le 1$ that gives $- 2 < x \le 2$

Explanation:

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

$\ln \left(1 + X\right) = X - {X}^{2} / 2 + {X}^{3} / 3 - \ldots , - 1 < X < 1$

Is the Taylor series for

$\ln x = \ln \left(1 + \left(x - 1\right)\right) = \left(x - 1\right) - {\left(x - 1\right)}^{2} / 2 - {\left(x - 1\right)}^{3} / 3 - , , , . - 1 < x - 1 \le 1$ that gives $- 2 < x \le 2$