# Taylor Series for $\log(x)$

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?

• wolframalpha.com/input/?i=taylor+log%28x%29 Commented Nov 29, 2013 at 0:40
• 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 $|x-a|<a$. Commented Nov 29, 2013 at 0:46
• 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). Commented Mar 18 at 14:35

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)

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

$$\frac{1}{1-[-(x-1)]}$$

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

$$\ln\left(\frac{x}{y}\right)=\ln(x)-\ln(y)$$

To see this, evaluate $$\ln$$ at $$x/a$$, where $$x$$ is any positive real number. If $$|x-a|, we will have that $$|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}

• Your writing style is excellent! Commented Apr 24 at 19:12
• @MathArt thank you! 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 $$0$$.

• is the sign supposed to alternate? Commented Feb 11, 2023 at 23:21