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February 23, 2014

Australian marsupial genera

Filed under: nature — ckrao @ 2:09 am

Native Australian mammals (those that predate human times) include monotremes (platypus and echidna), marsupials, bats, rodents (all mouse-like) and sea mammals (seals, whales and dolphins, dugongs). The marsupials of Australia extend far beyond the kangaroo/possum/koala/wombat (diprotodont) variety to the marsupial moles, carnivorous marsupials such as quolls, dunnarts, numbat and Tasmanian devil, and the omnivorous bandicoots and bilbies. (The remaining marsupial orders are confined to the Americas and comprise of the opossums, shrew opossums and Monito del Monte.) Here is a list of extant marsupial genera from those orders that are native to Australia, made simply to learn more about their diversity.

Order Suborder/Family Subfamily/Tribe Genus Genus meaning # Australian species # world species notes
Notoryctemorphia (Marsupial moles) Notoryctidae Notoryctes southern digger 2 2 marsupial moles
Dasyuromorphia   (marsupial carnivores) Dasyuridae Dasyurinae – Dasyurini Dasycercus hairy tail 2 2 mulgaras
Dasykaluta hairy kaluta 1 1 little red kaluta
Dasyuroides resembling Dasyurus 1 1 Kowari
Dasyurus hairy tail 4 6 quolls
Myoictis mouse weasel 0 4 dasyures – New Guinea
Neophascogale new Phascogale 0 1 Speckled dasyure  – New   Guinea
Parantechinus near Antechinus 1 1 Dibbler
Phascolosorex pouched shrew 0 2 Marsupial shrews – New Guinea
Pseudantechinus false Antechinus 6 6 False antechinuses
Sarcophilus flesh lover 1 1 Tasmanian devil
Dasyurinae –   Phascogalini Antechinus hedgehog equivalent 10 10
Micromurexia small Murexia 0 1 Habbema dasyure – rocky areas of New Guinea
Murexechinus hedgehog mouse 0 1 Black-tailed dasyure – tropical dry forests of New Guinea
Murexia marsupial mouse (significance unknown) 0 1 Short-furred dasyure – New Guinea
Paramurexia near Murexia 0 1 Broad-striped dasyure – South east Papua New Guinea
Phascomurexia pouched Murexia 0 1 Long-nosed dasyure – tropical dry forests of New Guinea
Phascogale pouched weasel 2 2
Sminthopsinae –   Sminthopsini Antechinomys antechinus-mouse 1 1 Kultarr
Ningaui Aboriginal mythical creature 3 3
Sminthopsis mouse appearance 21 21 dunnarts
Sminthopsinae – Planigalini Planigale flat weasel 4 5
Myrmecobiidae Myrmecobius ant-living 1 1 numbat
Peramelemorphia   (Bilbies and bandicoots) Peramelidae Echymiperinae Echymipera pouched hedgehog 1 1 long-nosed spiny bandicoot
Microperoryctes small Peroryctes 0 5 striped bandicoots – New Guinea
Rhynchomeles beaked badger 0 1 Seram bandicoot – existence only recorded in 1920
Peramelinae Isoodon equal tooth 3 3 short-nosed bandicoots
Perameles pouched badger 3 3 long-nosed bandicoots
Peroryctidae Peroryctes pouched digger 0 2 New Guinean long-nosed bandicoots
Thylacomyidae Macrotis big-ear 1 1 bilby
Diprotodontia   (“two front teeth” – kangaroos and relatives) Vombatiformes – Phascolarctidae Phascolarctos pouched bear 1 1 koala
Vombatiformes –   Vombatidae Vombatus wombat 1 1
Lasiorhinus hairy-nose 2 2
Phalangeriformes   (possums and gliders) – Phalangeridae Ailurops cat-like 0 2 bear cuscuses – NE Indonesia inc Sulawesi
Phalanger notable digits 1 13 cuscus
Spilocuscus spotted cuscus 1 5
Strigocuscus thin cuscus 0 2
Trichosurus hairy tail 5 5 bushtail possums
Wyulda brush-tail possum (mistakenly assigned) 1 1 scaly-tailed possum
Phalangeriformes –   Burramyidae (pygmy possums) Burramys stony-place mouse 1 1 Mountain Pygmy Possum
Cercartetus possibly tail-in-air 4 4
Phalangeriformes – Tarsipedidae Tarsipes tarsier-foot 1 1 Honey possum
Phalangeriformes –   Petauridae Dactylopsila naked finger 1 4 striped possums
Gymnobelideus naked Belideus 1 1 Leadbeater’s possum
Petaurus rope dancer 4 6 gliders
Phalangeriformes –   Pseudocheiridae (ringtailed possums and relatives) Hemibelideus half Belideus (fluffy-tailed glider) 1 1 Lemur-like ringtail possum
Petauroides Petaurus-like 1 1 Greater Glider
Petropseudes rock-Pseudocheirus 1 1 Rock-haunting ringtail possum
Pseudocheirus false hand 2 1 Common ringtail possum
Pseudochirops Pseudocheirus-like 1 5
Pseudochirulus little Pseudocheirus 2 8
Phalangeriformes –   Acrobatidae Acrobates acrobat 1 1 Feathertail glider
Distoechurus tail in two rows 0 1 Feather-tailed possum – New Guinea
Macropodiformes –   Macropodidae Lagostrophus turning hare 1 1 Banded hare-wallaby
Dendrolagus tree hare 2 13 tree-kangaroos
Lagorchestes dancing hare 2 2
Macropus long foot 13 13 kangaroos and wallaroos
Onychogalea nailed weasel 2 2 nail-tail wallabies
Petrogale rock weasel 16 16 rock-wallabies
Setonix bristle-claw 1 1 Quokka
Thylogale pouched weasel 3 7 pademelons
Wallabia wallaby 1 1 Swamp wallaby
Macropodiformes –   Potoroidae Aepyprymnus high rump 1 1 Rufous rat-kangaroo
Bettongia bettong 4 4
Potorous potoroo (Aboriginal) 3 3
Macropodiformes – Hypsiprymnodontidae Hypsiprymnodon high rump tooth 1 1 Musky rat-kangaroo
Totals 151 224


Dictionary of Australian and New Guinean Mammals, edited by Ronald Strahan & Pamela Conder, CSIRO Publishing, 2007.

February 9, 2014

The validity of the index laws for complex numbers

Filed under: mathematics — ckrao @ 11:47 am

In an earlier post we saw that some of the index laws fail when the base is negative or zero. Now we shall see what occurs when the allowed values of base and exponent are extended to be complex numbers.

For z_1 \in \mathbb{C}\backslash \{ 0\} we can proceed analogously to the real case and define

\displaystyle z_1^{z_2} = \exp(z_2 \log (z_1)).\quad \quad(1)

However what are the exponential and logarithm of a complex number? Last time we defined the logarithm before the exponential and we can do the same here. For complex z we can define \log(z) as \int_1^z \frac{1}{t}\ \text{d}t as for the real case, but note that this time it is a contour integral along any path from 1 to z not including the origin. By allowing the path to wind around the origin any number of times, we find the function is no longer single-valued. For example if we choose the contour to be the unit circle from +1 to itself in an anticlockwise direction, then the contour may be parametrised by t = \cos \theta + i \sin \theta for \theta \in [0, 2\pi] (\text{d}t = (-\sin \theta + i \cos \theta)\text{d}\theta) and we have

\begin{aligned} \log(1) &= \int_1^1 \frac{1}{t}\ \text{d}t\\ &= \int_0^{2\pi} (\cos \theta - i \sin \theta)(-\sin \theta + i \cos \theta)\text{d}\theta\\ &= \int_0^{2\pi} i( \cos^2 \theta + \sin^2 \theta )\ \text{d}\theta \\ &= 2\pi i. \end{aligned}

More generally if w is a logarithm of z, so is w + 2\pi i n where n is an integer. Defining \exp(w) for complex w to be z where w is a logarithm of z, we have

z = \exp(w + 2\pi i n) = \exp(w),\quad n \in \mathbb{Z}.

(Note that \displaystyle \exp(\log (z)) = z while \displaystyle \log(\exp (z)) = z + 2\pi i k.)

Based on these definitions we can show that \log(z_1z_2) = \log(z_1) + \log(z_2) and \exp(z_1 + z_2) = \exp(z_1)\exp(z_2) in an analogous way to the real case, except that the first equation is to be interpreted as an equality of sets of values rather than individual values. Note that for this reason we have to be careful when adding or subtracting logarithms: for example for complex numbers,

\log(z) - \log(z) = \log(1) = 2\pi i n \neq 0


\begin{aligned} \log(z^2) &= \log (z) + \log (z)\\ &= (\log (z) + 2\pi i m) + (\log (z) + 2\pi i n)\\ &= 2\log (z) + 2\pi i k, \quad k \in \mathbb{Z}\\ \text{while }\quad 2 \log (z) &= 2 (\log (z) + 2\pi i n)\\ &= 2\log (z) + 4 \pi i n, \quad n \in \mathbb{Z}. \end{aligned}

Hence we cannot write a\log (z) + b \log (z) = (a+b)\log (z) without paying special attention to the values of a,b,z. If we want to know when \log (z^c) = c \log (z), we can verify that \log (z^c) = c \log (z) + 2\pi i k, k \in \mathbb{Z} which is only equal to c \log (z) = c (\log (z) + 2\pi i m) when cm covers all integers, i.e. if c = 1/n for some non-zero integer n:

\displaystyle \log\left(z^{1/n}\right)= \frac{1}{n} \log(z), \quad\quad n \in \mathbb{Z}.

Now in general,

\begin{aligned} z_1^{z_2} &= \exp(z_2 \log (z_1))\\ &= \exp(z_2 (\log (z_1) + 2\pi i n) )\\ &= \exp(z_2 \log (z_1))\exp(2\pi i n z_2).\quad\quad (2)\end{aligned}

This will be multi-valued if nz_2 takes on non-integer values, as n varies over the integers. It will only be single-valued if z_2 is an integer. For example treated as a complex power, 2^{1/2} will have two values: \sqrt{2} and -\sqrt{2} while 2^{1/3} will take three values. The number 2^{\sqrt 2} will have infinitely many complex values \exp(\sqrt{2} \log (2))\exp(2\pi i n \sqrt{2}), n \in \mathbb{Z} although only one of them is real-valued. Note that through (2) we can work out quantities such as:

  • (-1)^{\sqrt{2}} = \exp(i \pi \sqrt{2})\exp(2\sqrt{2} \pi i n), n \in \mathbb{Z} (infinitely many non-real values!)
  • i^i = \exp(-\pi/2)\exp(-2\pi n), n\in \mathbb{Z} (infinitely many real values!).

Also note from (2) that

\log(z_1^{z_2}) = z_2 \log (z_1) + 2\pi i k = z_2 \textrm{Log} (z_1) + z_2 2\pi i k_1 + 2\pi i k_2.

One can define the principal value of the logarithm \textrm{Log}(z) to be that with imaginary part in the interval (-\pi, \pi]. Similarly one can define the principal value of the power function as

\displaystyle z_1^{z_2} := \exp(z_2 \textrm{Log} (z_1)).\quad \quad (3)

This gives single-valued results but they may not be as expected. For example, since \textrm{Log}(-1) = i \pi, (-1)^{1/3} = \exp(i \pi/3) rather than the real-valued root -1. However we can now say a \textrm{Log}(z) + b \textrm{Log}(z) = (a+b)\textrm{Log}(z), being single-valued.

We would like to know which of the index laws hold. In the remainder of the post we verify the identities summarised in the following table. The real number case was already treated in this post.

Real numbers a,b Complex numbers z,z_1,z_2,a,b
Positive real x,y

Multiple-valued power

z_1^{z_2} = \exp(z_2 \log z_1)

Single-valued power

z_1^{z_2} = \exp(z_2 \textrm{Log} z_1)

(1)  x^a x^b = x^{a+b}  z^{a+b} a subset of  z^a z^b  z^a z^b = z^{a+b}
(2)  (x^a)^b = x^{ab}  z^{ab} a subset of (z^a)^b  (z^a)^b = z^{ab}\exp(2\pi i b n_0)
(3)  (xy)^a = x^a y^a  (z_1 z_2)^a = z_1^a z_2^a  (z_1 z_2)^a = z_1^a z_2^a \exp(2 \pi i a n_{+})
 (4)  x^0 = 1  z^0 = 1   z^0 = 1
 (5)  1^a = 1  1^a = \exp(a 2\pi i k)   1^a = 1
 (6)  x^{-a} = 1/x^a  z^{-a} = 1/z^a but z^{-a}z^a = \exp(2\pi i k)  z^{-a} = 1/z^a and  z^{-a}z^a = 1
 (7)  x^a / x^b = x^{a-b}  z^{a-b} a subset of z^{a}/z^{b}   z^a / z^b = z^{a-b}
 (8)  (x/y)^a = x^a/y^a  (z_1/z_2)^a = z_1^a/z_2^a  (z_1/z_2)^a = \frac{z_1^a}{z_2^a}\exp(2\pi i a n_{-})

Note that in the table, n_0, n_{+},n_{-} are particular integers chosen to enable equality.

For verifying identity (1) in the multi-valued power case we have

\begin{aligned} z^{a+b} &= \exp((a+b)\log (z))\\ &= \exp\left((a+b)(\textrm{Log} (z) + 2\pi i k)\right)\\ &= \exp\left((a+b)\textrm{Log} (z) \right) \exp\left( 2\pi i k (a+b)\right) \end{aligned}


\begin{aligned}z^a z^b &= \exp(a \log(z)) \exp(b \log(z))\\&= \exp\left(a (\textrm{Log} z + 2\pi ik) \right)\exp\left(b (\textrm{Log} z + 2\pi in) \right)\\ &= \exp\left((a+b)\textrm{Log} (z) \right) \exp\left(2\pi i (ka + nb) \right).\end{aligned}

This shows that the set of values of z^{a+b} is a subset of the set of values of z^a z^b. In the single-valued case,

\begin{aligned} z^az^b &= \exp(a \textrm{Log} (z)) \exp(b \textrm{Log} (z))\\ &= \exp(a \textrm{Log} (z) + b \textrm{Log} (z))\\&= \exp((a+b) \textrm{Log} (z))\\ &= z^{a+b}.\end{aligned}

For identity (2) in the multi-valued power case we have

\begin{aligned} (z^a)^b &=(\exp(a \log (z)))^b\\ &= \exp(b \log(\exp(a \log( z))))\\ &= \exp(b(a \log (z)+2 \pi ik))\\ &= \exp(ba \log (z)) \exp(2 \pi ibk)\\ &= z^{ab}\exp(2\pi ibk). \end{aligned}

This shows that the set of values of z^{ab} is a subset of the values of (z^a)^b. We have the equality (z^a)^b = z^{ab} if \exp(2\pi ibk) =1, or bk \in \mathbb{Z} for all k, which is true if b \in \mathbb{Z}. In the single-valued case, the integer k is chosen so that a \textrm{Log}(z)+2 \pi ik has imaginary part in the interval (-\pi, \pi].

For identity (3) in the multi-valued power case we have

\begin{aligned} (z_1 z_2)^a &= \exp(a \log(z_1 z_2))\\ &= \exp(a(\log (z_1)+\log (z_2)))\\ &= \exp(a \log (z_1))\exp(a \log (z_2))\\ &= z_1^a z_2^a.\end{aligned}

In the single-valued power case,

\begin{aligned} (z_1 z_2)^a &= \exp(a \textrm{Log}(z_1 z_2))\\ &= \exp(a(\textrm{Log} (z_1)+\textrm{Log} (z_2) + 2\pi i a n_{+}))\\ &= \exp(a \textrm{Log} (z_1))\exp(a \textrm{Log} (z_2))\\ &= z_1^a z_2^a \exp(2 \pi i an_{+}). \end{aligned}

Here the value n_{+} is chosen so that \textrm{Log} (z_1)+\textrm{Log} (z_2) + 2\pi i a n_{+} has imaginary part in the interval (-\pi, \pi].

Identity (4) comes from setting b=0 in identity (1). Identity (5) results from setting y or z_2 to 1 in identity (3). Identity (6) results from setting b = -a in identity (1) and using identity (4). Finally identities (7) and (8) follow from identities (1) and (3).

The moral of all this is that care is to be taken when applying the index laws to complex numbers (or indeed even when adding logarithms) by virtue of the multi-valued nature of the complex logarithm.


H. Haber, The complex logarithm, exponential and power functions, UC Santa Cruz Physics 116A notes (2011) available at

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