How do you find the Taylor series for ln(x) about the value x=1?

May 20, 2015

firstly we look at the formula for the Taylor series, which is:

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

which equals:

f(a) + f'(a)(x-a) + (f''(a)(x-a)^2)/(2!) + (f'''(a)(x-a)^3)/(3!) + ...

So you would like to solve for $f \left(x\right) = \ln \left(x\right)$ at $x = 1$ which I assume mean centered at $1$ of which you would make $a = 1$

To solve:

$f \left(x\right) = \ln \left(x\right)$ and $f \left(1\right) = \ln \left(1\right) = 0$

$f ' \left(x\right) = \frac{1}{x}$ and $f ' \left(1\right) = \frac{1}{1} = 1$

$f ' ' \left(x\right) = - \frac{1}{x} ^ 2$ and $f ' ' \left(1\right) = - \frac{1}{1} ^ 2 = - 1$

${f}^{\left(3\right)} \left(x\right) = \frac{2}{x} ^ 3$ and ${f}^{\left(3\right)} \left(1\right) = \frac{2}{1} ^ 3 = 2$

${f}^{\left(4\right)} \left(x\right) = - \frac{\left(2\right) \left(3\right)}{x} ^ 4$ and ${f}^{\left(4\right)} \left(1\right) = - \frac{\left(2\right) \left(3\right)}{1} ^ 4 = - \left(2\right) \left(3\right)$

Where now we can already start to see a pattern forming, so we starting using our formula(2):

0 + 1(x-1) - (1(x-1)^2)/(2!) + (2(x-1)^3)/(3!) - ((2)(3)(x-1)^4)/(4!) .....

and now try try see how we can write this as a series, which we get: (we start will n=1 as our first term is 0)

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

Which can then simplify to:

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