How do you find the Taylor series of f(x)=ln(x) ?

Sep 10, 2014

Any Taylor series of a function $f \left(x\right)$ can be found by calculating
sum_(n=0)^oo(f^(n)(a)*(x-a)^n)/(n!)
where $a$ is the point where you need to approximate the function.

Let's say you need to approximate $\ln \left(x\right)$ around the point $x = 1$. So:

• The Taylor series of degree 0 is simply $f \left(1\right) = \ln \left(1\right) = 0$
• The Taylor series of degree 1 is the Taylor series of degree 0, plus (f'(1)(x-1)^1)/(1!)
We know that $f ' \left(x\right) = \frac{1}{x}$ for $x > 0$, that 1! =1, and that ${\left(x - 1\right)}^{1} = \left(x - 1\right)$
So the Taylor series of degree 1 is
$0 + \frac{1 \left(x - 1\right)}{1} = \left(x - 1\right)$.
• The Taylor series of degree 2 is the Taylor series of degree 1, plus (f''(1)(x-1)^2)/(2!)
We know that $f ' ' \left(x\right) = - \frac{1}{x} ^ 2$ for $x > 0$, and that 2! =2
So the Taylor series of degree 2 is $\left(x - 1\right) + \frac{- 1 {\left(x - 1\right)}^{2}}{2}$
$= \left(x - 1\right) - {\left(x - 1\right)}^{2} / 2$.
• The Taylor series of degree 3 is the Taylor series of degree 2, plus (f'''(1)(x-1)^3)/(3!)
We know that $f ' ' ' \left(x\right) = \frac{2}{x} ^ 3$ for $x > 0$, and that 3! =6
So the Taylor series of degree 3 is $\left(x - 1\right) - {\left(x - 1\right)}^{2} / 2 + \frac{\frac{2}{1} {\left(x - 1\right)}^{3}}{6}$
$= \left(x - 1\right) - {\left(x - 1\right)}^{2} / 2 + {\left(x - 1\right)}^{3} / 3$.

Keep on working in this vein until you reach the degree that has been asked for. Usually during high school and early university, you won't need it beyond about the sixth degree as it gets quite time-consuming to calculate by hand.