nLab logarithm

Logarithms

Logarithms

Idea

Classically, a logarithm is a partially-defined smooth homomorphism from a multiplicative group of numbers to an additive group of numbers. As such, it is a local section of an exponential map. As exponential maps can be generalised to Lie groups, so can logarithms.

Definitions

Logarithms of real numbers

Consider the field of real numbers; these numbers form a Lie group under addition (which we will call simply \mathbb{R}), while the nonzero numbers form a Lie group under multiplication (which we will call *\mathbb{R}^*). The multiplicative group has two connected components; we will focus attention on the identity component (which we will call +\mathbb{R}^+), consisting of the positive numbers.

The Lie groups \mathbb{R} and +\mathbb{R}^+ are in fact isomorphic. In fact, there is one isomorphism for each positive real number bb other than 11; this number bb is called the base. Fixing a base, the map from +\mathbb{R}^+ to \mathbb{R} is called the real logarithm with base bb, written xlog bxx \mapsto \log_b x; the map from \mathbb{R} to +\mathbb{R}^+ is the real exponential map with base bb, written xb xx \mapsto b^x.

The real logarithms are handily defined using the Riemann integral of the reciprocal as follows:

(1)lnx 1 x1tdt; log bx lnxlnb. \array { \ln x & \coloneqq \int_1^x \frac{1}{t} \,\mathrm{d}t ;\\ \log_b x & \coloneqq \frac{\ln x}{\ln b} .\\ }

Note that ln\ln is itself a logarithm, the natural logarithm, whose base is e=2.71828182845\mathrm{e} = 2.71828182845\ldots. (The exponential map may similarly be defined as an infinite series, but I'll leave that for its own article.)

Logarithms of complex numbers

Now consider the field of complex numbers; these also form a Lie group under addition (which we call \mathbb{C}), while the nonzero numbers form a Lie group under multiplication (which we call *\mathbb{C}^*). Now the multiplicative group is connected, so we would like to use all of it.

However, \mathbb{C} and *\mathbb{C}^* are not isomorphic. Indeed, the multiplication map

*×S 1 *\mathbb{R}^* \times S^1 \to \mathbb{C}^*

exhibits *\mathbb{C}^* as a biproduct (in Ab) of *\mathbb{R}^* and the circle group S 1S^1, so that homomorphisms *\mathbb{C}^* \to \mathbb{C} are given by pairs of homomorphisms f: *f \colon \mathbb{R}^* \to \mathbb{C}, g:S 1g \colon S^1 \to \mathbb{C}. But every homomorphisms g:S 1g \colon S^1 \to \mathbb{C} is trivial: the restriction of gg to the torsion subgroup of S 1S^1 is trivial since \mathbb{C} is torsionfree, and since the torsion subgroup is dense in S 1S^1, any Lie group homomorphism S 1S^1 \to \mathbb{C} must also be trivial. Therefore, every homomorphism h: *h \colon \mathbb{C}^* \to \mathbb{C} factors through the projection * *\mathbb{C}^* \to \mathbb{R}^*. It quickly follows that no such hh can be injective, nor can such hh be surjective.

Taking advantage of biproduct representations \mathbb{C} \cong \mathbb{R} \oplus \mathbb{R} and * *S 1\mathbb{C}^* \cong \mathbb{R}^* \oplus S^1, we can classify homomorphisms from \mathbb{C} to *\mathbb{C}^*. Each is given by a 4-tuple of real numbers (a,b,c,d)(a, b, c, d):

ϕ a,b,c,d(x+iy)=e axe ibxe cye idy.\phi_{a, b, c, d}(x + i y) = e^{a x} e^{i b x} e^{c y} e^{i d y}.

The cases where a=da = d, b=cb = -c correspond to those homomorphisms that are holomorphic functions (i.e., that satisfy the Cauchy-Riemann equations). Putting w=a+biw = a + b i, we have

ϕ a,b,b,a(z)=e wz\phi_{a, b, -b, a}(z) = e^{w z}

with one such homomorphism for each complex number ww, and these homomorphisms are surjections whenever w0w \ne 0. (N.B.: these homomorphisms are not uniquely determined by their values at z=1z = 1, since we have e w=e we^w = e^{w'} whenever www - w' is an integer multiple of 2πi2 \pi i, and yet the homomorphisms ze wzz \mapsto e^{w z} and ze wzz \mapsto e^{w' z} will be different unless w=ww = w'.)

So we have these surjections (the complex exponential map ze wzz \mapsto e^{w z}, for w0w \ne 0), which are regular epimorphisms but not split epimorphisms. However, while they have no sections (being not split), they have quite a few local sections, and the domains of the maximal local sections are precisely the connected simply connected open dense subspaces RR of *\mathbb{C}^*. A complex logarithm with exponential base ww on RR is this RR-defined section of the complex exponential map ze wzz \mapsto e^{w z}. Supposing RR given, we denote this by log [w]\log_{[w]} (but please note that in the context of real logarithms, this would ordinarily be denoted log b\log_b where b=e wb = e^w).

If 1R1 \in R, then a complex natural logarithm on RR may be defined using the contour integral with the same formula (1) as for the real natural logarithm. We merely insist that the integral be done along a contour within the region RR. (Since RR is connected, there is such a contour; since RR is simply connected and t1/tt \mapsto 1/t is holomorphic, the result is unique.) Note that if x +Rx \in \mathbb{R}^+ \subseteq R, then the real and complex natural logarithms of xx will be equal.

The natural exponential map is periodic (with period 2πi2 \pi \mathrm{i}), and it is possible to add any multiple of this period to the natural logarithm of any x1x \ne 1 by suitably changing the region RR. We then obtain the most general notion of maximally-defined complex logarithm with any base by using the formulas

lnx C+ e C x1tdt, log [w]x lnxw. \array { \ln x & \coloneqq C + \int_{\mathrm{e}^C}^x \frac{1}{t} \,\mathrm{d}t,\\ \log_{[w]} x & \coloneqq \frac{\ln x}{w} .\\ }

Logarithms and Lie groups

In the classical examples, the multiplicative groups +\mathbb{R}^+ and *\mathbb{C}^* are both Lie groups. The additive groups \mathbb{R} and \mathbb{C} are also Lie groups, but they are more than this: they are Lie algebras. (The additive group of a Lie algebra is always a Lie group. Actually, since these are abelian Lie algebras, their Lie-algebra structure is easy to miss, but of course they are vector spaces.) And what's more, each additive group is the Lie algebra of the corresponding Lie group.

This generalises. Given any Lie group GG, let 𝔤\mathfrak{g} be its Lie algebra. Then we have an exponential map exp:𝔤G\exp\colon \mathfrak{g} \to G, which is surjective under certain conditions (most famously when GG is connected and compact, but also in the classical cases, even though GG is not compact). More generally, given any automorphism ϕ\phi of 𝔤\mathfrak{g}, we have a map xexp(ϕ(x))x \mapsto \exp(\phi(x)), which is a homomorphism of Lie groups. Any local section of this map may be called a logarithm base ϕ\phi on GG (denoted log [ϕ]\log_{[\phi]} with the bracket as in the previous section); any local section of exp\exp itself may be called a natural logarithm on GG.

Properties

Proposition

(Mercator series)
The Taylor series of the natural logarithm around 11 \in \mathbb{R} is the following series:

(2)n=01n!(d ndx nln(1+x)) |x=0x n =n=1(1) n+1nx n =x12x 2+13x 314x 4+. \begin{aligned} \underoverset{n = 0}{\infty}{\sum} \tfrac{1}{n!} \left( \frac{d^n}{ d x^n} ln(1 + x) \right)_{\vert x = 0} x^n & \;\; = \;\; \underoverset{n = 1}{\infty}{\sum} \frac {(-1)^{n+1}} {n} x^n \\ & \;\; = \;\; x - \tfrac{1}{2} x^2 + \tfrac{1}{3} x^3 - \tfrac{1}{4} x^4 + \cdots \,. \end{aligned}

Proof

For the first two terms notice that

ln(1+x)x0ln(1)=0 ln(1 + x) \;\xrightarrow{x \to 0}\; ln(1) \,=\, 0

and that the derivative of the natural logarithm is:

ddxln(1+x)=11+xx01. \frac{d}{d x} \ln(1 + x) \;=\; \tfrac{1}{1+x} \;\xrightarrow{ x \to 0 }\; 1 \,.

From here on, noticing for k +k \in \mathbb{N}_+ that:

ddx(1(1+x) k)=k1(1+x) k+1x0k \frac{d}{d x} \left( \frac{1}{(1 + x)^k} \right) \;=\; - k \frac{1}{(1 + x)^{k+1}} \;\xrightarrow{x \to 0}\; - k

we get, for n +n \in \mathbb{N}_+:

d ndx nln(1+x) =d n1dx n1(11+x) =(n1)!(1) n11(1+x) n1 x0(n1)!(1) n+1. \begin{aligned} \frac{d^n}{d x^n} ln(1 + x) & \;=\; \frac{d^{n-1}}{d x^{n-1}} \left( \frac{1}{1 + x} \right) \\ & \;=\; (n-1)! \cdot (-1)^{n-1} \frac{1}{(1 + x)^{n-1}} \\ & \;\xrightarrow{ x \to 0 }\; (n-1)! \cdot (-1)^{n+1} \end{aligned} \,.

Plugging this into the defining equation on the left of (2) and using

(n1)!n!=1n \frac{(n-1)!}{n!} = \frac{1}{n}

yields the claim.

References

Historical textbooks:

See also:

Discussion in point-free topology:

Last revised on December 14, 2023 at 00:56:36. See the history of this page for a list of all contributions to it.