# Can you express $\ln(x)$ as a series with negative or fractional powers of $x$?

I read in my book that $\ln(x)$ cannot be written as a power series. That is a series whose terms contain only non-negative integer powers of $x$. But can $\ln(x)$ be written as an infinite series whose terms contain negative or fractional powers of $x$? Then there is the bigger question. Can all functions be expressed with such terms?

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## 3 Answers

The "problem" with Taylor series is, that they (have to!) converge in discs around the point of expansion. Laurent series have the same "problem" (rings around the point of expansion). In both cases the multi-value property of the ln function cannot be represented. To have any chance of convergence we have to restrict ourselves to a cut plan.

An example of a series with convergence in a cut plane is a series of Padé approximants (converges in complex plane without the negative real axis). And indeed, the Padé approximants converge to the main branch of the logarithm.

Such approximants consist of a polynomial devided by a polynomial. Eg $$P_6^6(x) = \frac{7 \left(-7-132 x-375 x^2+375 x^4+132 x^5+7 x^6\right)}{10 \left(1+36 x+225 x^2+400 x^3+225 x^4+36 x^5+x^6\right)} .$$ This is the (6,6) Padé approximant around the expansion point 1. If you plot it, you will see, that it already resembles the logarithm farely well for small phases of the complex argument.

Note that the main Padés are equivalent to continued fractions. So if you don't like the term Padé approximant you might also say, that the logarithm can be represented as a continued fraction.

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Wow, never thought that Padé approximation is so powerful! – Ruslan Nov 21 '14 at 13:41

I think you're probably referring to an expansion about the value $x=0$, correct? If this were possible, then we would have what we call a Laurent series for $\log z$, where $z\in\mathbb{C}$. Laurent series are essentially power series expanded about singular points.

The problem with $\log z$ in the complex plane is that it is a "multi-valued function", so we must specify what range of values we are considering the function to have. Because we must make this choice, the function fails to be continuous in any (punctured) disk about $z=0$, and thus is not (complex) differentiable in any neighborhood about the point. However, we have that any convergent Laurent series is (complex) differentiable in some annulus about the pole at which it is centered.

So in conclusion, the answer is "no", because the $\log$ function is badly behaved in the complex plane. I hope this helps answer your question; if not, I suggest reading up a little on complex analysis.

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You can have $\log(x+1)$ expressed as series with negative powers of $x$, when you expand it at $x=\infty$: $$\log(x+1)=\log(x)+\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}\cdots$$ See here.

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+1 for daring to be different, like your style! – Arjang Mar 26 '12 at 9:59
Hmm, but this in fact requires computation of $\log(x)$ to get $\log(x+1)$... seems kind of pointless, and not really expansion in powers of $x$. – Ruslan Nov 21 '14 at 13:44
@Ruslan I agree on the problem of using $\log(x)$, but the OP asked "an infinite series whose terms contain negative ... powers of $x$"... – draks ... Nov 30 '14 at 19:36