Does anyone know a closed form expression for the Taylor series of the function $f(x) = \log(x)$ where $\log(x)$ denotes the natural logarithm function?

  • 1
    $\begingroup$ wolframalpha.com/input/?i=taylor+log%28x%29 $\endgroup$
    – Amzoti
    Commented Nov 29, 2013 at 0:40
  • 1
    $\begingroup$ It is easy to find a closed-form expression for $f^{(n)}(a)$ for any $a>0$ you wish, then let $c_n = f^{(n)}(a)/n!$, and $f(x)=\sum_{n=0}^\infty c_n (x-a)^n$ for $|x-a|<a$. $\endgroup$ Commented Nov 29, 2013 at 0:46
  • $\begingroup$ the Taylor series for ln(x) is relatively simple : 1/x , -1/x^2, 1/x^3, -1/x^4, and so on iirc. log(x) = ln(x)/ln(10) via the change-of-base rule, thus the Taylor series for log(x) is just the Taylor series for ln(x) divided by ln(10). $\endgroup$ Commented Mar 18 at 14:35

4 Answers 4


Abromowitz & Stegun gives a number of forms.

Among these, the bilinear expansion is known for its used in digital filter theory:

$$\log(z) = 2\left[\left({z-1\over z+1}\right)+ {1\over3} \left({z-1\over z+1}\right)^3 + {1\over5} \left({z-1\over z+1}\right)^5+ \cdots\right],$$

for $\Re z > 0.$

Series of log(x)

  • $\begingroup$ +1 but the LHS should be $\log(z)$ rather than $\log(x)$ $\endgroup$
    – greg
    Commented Jun 28, 2021 at 14:48
  • 1
    $\begingroup$ Any details about the applicatins in digital filter theory? $\endgroup$
    – MathArt
    Commented Apr 24 at 19:00

The Taylor series centered at $1$ can be easily derived with the geometric series

$$\sum_{n=0}^\infty x^n=\frac{1}{1-x}$$

We start with the derivative of $\ln(x)$, which is given by $1/x$ for every $x>0$. This can be rewritten as


so if $|-(x-1)|<1$, i.e. $|x-1|<1$, this can be expanded as a geometric series:

\begin{align} \frac{1}{1-[-(x-1)]} &= \sum_{n=0}^\infty [-(x-1)]^n\\ &=\sum_{n=0}^\infty (-1)^n(x-1)^n \end{align}

It follows that $(\ln)'(t)=\sum_{n=0}^\infty (-1)^n(t-1)^n$ holds whenever $|t-1|<1$. We can then get a series expression for $\ln(x)$ by integrating this identity from $1$ to $x$:

\begin{align} \ln(x)-\ln(1) &= \sum_{n=0}^\infty (-1)^n\frac{\left(x-1\right)^{n+1}}{n+1}-\sum_{n=0}^\infty (-1)^n\frac{(1-1)^{n+1}}{n+1}\\ &= \sum_{n=1}^\infty (-1)^{n-1}\frac{\left(x-1\right)^n}{n}-\sum_{n=1}^\infty (-1)^{n-1}\frac{0^{n}}{n}\\ &= \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\left(x-1\right)^n\\ &= \left(x-1\right)-\frac{\left(x-1\right)^2}{2}+\frac{\left(x-1\right)^3}{3}-\frac{\left(x-1\right)^4}{4}+\cdots \end{align}

What if we want a Taylor series centered at a point other than $1$, say at $a>0$? You can always do this directly by computing the derivatives of $\ln(x)$ at $a$, but an easier method is to leverage the series we just derived and the identity


To see this, evaluate $\ln$ at $x/a$, where $x$ is any positive real number. If $|x-a|<a$, we will have that $|x/a-1|<1$, so the Taylor series centered at $1$ will converge to $\ln(x/a)$. We can then write

\begin{align} \ln\left(\frac{x}{a}\right) &= \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\left(\frac{x}{a}-1\right)^n\\ &= \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\left(\frac{x-a}{a}\right)^n\\ &= \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\frac{(x-a)^n}{a^n}\\&= \sum_{n=1}^\infty \frac{(-1)^{n-1}}{na^n}(x-a)^n\\ \end{align}

From $\ln(x/a)=\ln(x)-\ln(a)$, we immediately get

\begin{align} \ln(x) &= \ln(a)+\sum_{n=1}^\infty \frac{(-1)^{n-1}}{na^n}(x-a)^n\\ &= \ln(a)+\frac{1}{a}(x-a)-\frac{1}{2a^2}(x-a)^2+\frac{1}{3a^3}(x-a)^3-\frac{1}{4a^4}(x-a)^4+\cdots \end{align}

  • 1
    $\begingroup$ Your writing style is excellent! $\endgroup$
    – MathArt
    Commented Apr 24 at 19:12
  • $\begingroup$ @MathArt thank you! $\endgroup$ Commented Apr 27 at 7:47

For $x \in \mathbb{R}$ satisfying $0 < x < 2$,

$$f(x) = \ln(x) = \left(x-1\right)-\frac{1}{2}\left(x-1\right)^2 + \frac{1}{3} \left(x-1\right)^3-\frac{1}{4} \left(x-1\right)^4 + \cdots$$ $$ f(x) = \displaystyle\sum\limits_{n=1}^{\infty} \left[\frac{\left(-1\right)^{n+1}}{n}\left(x-1\right) ^n\right] $$


$$-\log(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots \qquad (|x|<1)$$

There is no expansion around $x=1$ because the log is singular at $0$.

  • 2
    $\begingroup$ is the sign supposed to alternate? $\endgroup$
    – Kurtoid
    Commented Feb 11, 2023 at 23:21

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .