Chaitanya's Random Pages

August 19, 2011

Biological systems in various animal groups

Filed under: nature — ckrao @ 1:41 pm

I was curious to find out something about the inner workings of animals. For example do invertebrates have blood? (Answer: some, such as earthworms) Which animals have a brain? (Answer coming up.) Below is a list of the biological systems of humans. We shall see the extent to which these are present in other animals.

System Organs/Cell types Function
Circulatory heart, blood, blood vessels circulating blood to and from body and lungs
Digestive tongue, salivary glands, esophagus, stomach, liver, gall bladder, pancreas, small intestine, large intestine, appendix, rectum, anus processing food
Endocrine hypothalamus, pituitary gland, pineal gland, thyroid, parathyroids, adrenal glands, Islets of Langerhans (in pancreas) release of hormones to send messages to other parts of body
Integumentary skin, hair, fat, nails protects body from external damage, insulation, detection of pain/pressure/temperature, vitamin D synthesis
Lymphatic (inc. immune) lymph, lymph vessels, lymphoid tissue, lymph nodes, leukocytes, bone marrow, tonsils, adenoids, thymus, spleen removal of interstitial fluid from tissue, transport of white blood cells into bones, absorption of fatty acids from circulatory system
Muscular muscles, tendons, fasciae movement
Nervous brain, spinal cord, peripheral nerves, nerves, includes sensory systems of ear, eye etc. sensing and processing information
Reproductive ovaries, fallopian tubes, uterus, placenta, vulva, vagina, mammary glands, testes, prostate, epididymis, vas deferens, seminal vesicles, prostate, penis reproduction
Respiratory nose, pharynx, larynx, trachea, bronchi, lungs, diaphragm breathing
Skeletal bones, cartilage, ligaments, tendons structural support
Urinary kidneys, ureters, bladder, urethra fluid balance, electrolyte balance, excretion

Next are some notes on the presence of these systems in some of the main animal groups. In the following an “open circulatory system” means there is no difference between blood and tissue fluid, while in a closed circulatory system blood (in vessels) is separate from tissue fluid.

System

Vertebrates

(e.g. fish, amphibians, reptiles, birds, mammals)

Echinoderms

(starfish, sea urchins, sea cucumbers etc.)

Molluscs

(snails, slugs, squid, clams, etc.)

Arthropods

 (arachnids, insects, crustaceans, centipedes etc.)

Circulatory closed system, platelets unique to mammals open circulatory system, no heart blood and coelomic fluid circulate in main body cavity, heart present in cephalapods (squids/octopus etc.) open circulatory system, hexapods have tracheae and no gases in their blood
Digestive simple digestive gut most graze on algae, possess stomach and gut spiders etc. have chelicerae at mouth opening
Endocrine
Integumentary most fish/reptiles have scales, birds have feathers, mammals have hair some have a shell jointed limbs
Lymphatic (inc immune) insects have hemolymph
Muscular muscular foot attached to inside of exoskeleton
Nervous brain and spinal cord simple radial nervous system, no brain some have brains, most have two nerve cords; octopus and squids have eyes, snails have antennae paired ganglia in each segment; sensory organs (e.g. compound eyes)
Reproductive larval stage in bony fish, amphibians; most reptiles, birds, amphibians, birds, monotremes lay eggs while marsupials and placental mammals give birth to live young; in birds and playpuses the right ovary never matures; mammary glands unique to mammals gonads may occupy entire cavity terrestrial species have external fertilisation, metamorphosis in some insects, no larval stage in arachnids
Respiratory gills in most fish, otherwise lungs; gas exchange with water/air also occurs through skin in amphibians; syrinx in birds responsible for singing varying methods of gas exchange most have one pair of gills; snails/slugs have lungs varying, e.g. booklungs in spiders, gills in crustaceans, tracheae in insects
Skeletal sharks, rays etc. have cartilaginous skeleton; otherwise bony skeleton contain spines emerging from endoskeleton calcareous endoskeleton exoskeleton made of cuticle
Urinary waste diffuses, no specialised excretory organs nephridia play role of kidneys, they also have role in reproduction malpighian tubules in insects, green glands in crustaceans
 System

Annelids

(segmented worms)

Nematodes

(roundworms)

Platyhelminthes

(flatworms)

Cnidarians

(corals, sea anemones, jellyfish, hydra etc.)

Poriferans

(sponges)

Circulatory closed circulatory system, some with blood vessels along length circulation of gases and nutrients through pseudocoelom fluid. no specialised organs no organs no organs
Digestive full gut from mouth to anus no stomach specialised system no specialised system no specialised system
Endocrine
Integumentary segments covered by cuticle  no symmetry
Lymphatic (inc immune)
Muscular muscular, movement by peristalsis muscular pharynx hydras have muscle no muscle
Nervous bundle of nerves serve as brain four nerves run its length, dense nerve ring serves as brain first forms in which we see brain-like structure no distinct brain no neurons in adults
Reproductive some reproduce asexually, no larval stage in earthworms ovaries and testes present sexual or asexual reproduction sperm cells released into water
Respiratory variable, via skin or through gills no specialised organs, rely on diffusion no specialised organs
Skeletal mesohyl acts as endoskeleton
Urinary waste excreted through body wall primitive, made of flame cells, excretory ducts and excretory pores. no specialised system no specialised system

Edit: I later found just the thing I was looking for here.

Reference

Zoology Review – http://www.biology-questions-and-answers.com/zoology-review.html

August 14, 2011

Big wins in the VFL/AFL

Filed under: sport — ckrao @ 4:03 am

In recent times there have been some big winning margins in AFL football. This table shows that five of the biggest wins in the 115-year history of the VFL/AFL have come this season.

# Marg Team 1 Score Team 2 Score Gnd Rd Date Year
1 190 Fitzroy 36.22 (238) Melbourne 6.12 (48) Wav 17 Sat 28th Jul 1979
2 186 Geelong 37.11 (233) Melbourne 7.5 (47) SS 19 Sat 30th Jul 2011
3 178 Collingwood 31.21 (207) St Kilda 3.11 (29) VP 4 Sat 28th Apr 1979
4 171 Sth Melb 29.15 (189) St Kilda 2.6 (18) AP 12 Sat 26th Jul 1919
5 168 Richmond 30.19 (199) Nth Melb 4.7 (31) PR 2 Sat 9th May 1931
6 165 Hawthorn 31.11 (197) Port Adel 5.2 (32) MCG 21 Sat 13th Aug 2011
6 165 Essendon 28.16 (184) Sth Melb 2.7 (19) WH 18 Sat 22nd Aug 1964
8 164 Geelong 37.17 (239) Bris Bears 11.9 (75) Car 7 Sun 3rd May 1992
9 163 Sydney 36.20 (236) Essendon 11.7 (73) SCG 17 Sun 26th Jul 1987
10 162 Bris Bears 33.21 (219) Sydney 8.9 (57) Gab 8 Sun 16th May 1993
11 161 Geelong 23.24 (162) St Kilda 0.1 (1) CO F3B Sat 9th Sep 1899
12 160 Hawthorn 32.24 (216) Essendon 8.8 (56) MCG 20 Sat 1st Aug 1992
13 157 Geelong 35.12 (222) Richmond 9.11 (65) Eti 6 Sun 6th May 2007
13 157 Hawthorn 36.15 (231) Fitzroy 11.8 (74) NHO 6 Sun 28th Apr 1991
15 152 Richmond 34.18 (222) St Kilda 11.4 (70) SCG 16 Sun 20th Jul 1980
16 151 Richmond 28.19 (187) Fitzroy 5.6 (36) MCG 21 Sun 25th Aug 1996
17 150 Geelong 29.14 (188) Gold Coast 6.2 (38) SS 20 Sat 6th Aug 2011
17 150 Fitzroy 34.16 (220) Nth Melb 10.10 (70) JO 13 Sat 18th Jun 1983
19 147 Collingwood 30.20 (200) Essendon 7.11 (53) VP 14 Sat 3rd Jul 1971
20 146 Essendon 32.16 (208) Footscray 9.8 (62) WO 22 Sat 28th Aug 1982
21 143 Hawthorn 28.26 (194) Footscray 7.9 (51) Wav 3 Sat 10th Apr 1982
22 142 Essendon 25.10 (160) West Coast 1.12 (18) WH 15 Sat 15th Jul 1989
23 141 Bris Lions 29.15 (189) Adelaide 6.12 (48) Gab 17 Sat 24th Jul 2004
23 141 Nth Melb 32.17 (209) Richmond 9.14 (68) MCG 2 Fri 6th Apr 1990
23 141 Melbourne 21.28 (154) Hawthorn 1.7 (13) MCG 9 Sat 26th Jun 1926
26 140 Essendon 29.20 (194) Bris Bears 8.6 (54) MCG 9 Fri 27th May 1988
26 140 Carlton 25.19 (169) St Kilda 2.17 (29) Mor 2 Sat 6th Apr 1985
26 140 Nth Melb 27.23 (185) Sydney 5.15 (45) SCG 3 Sun 10th Apr 1983
29 139 Essendon 31.11 (197) Gold Coast 8.10 (58) Eti 6 Sun 1st May 2011
29 139 St Kilda 28.18 (186) Bris Lions 7.5 (47) Eti 22 Sat 27th Aug 2005
29 139 Adelaide 26.15 (171) Richmond 4.8 (32) AAMI 16 Fri 16th Jul 1993
32 138 Collingwood 23.21 (159) Port Adel 3.3 (21) AAMI 20 Sat 6th Aug 2011

The Collingwood-Port Adelaide game was the first time a team only had 6 scoring shots (3 goals + 3 behinds) in a game since 1971. Only 68 times has a team had 5 or fewer scoring shots, only 4 of which were after 1915. In the following week against Hawthorn, Port Adelaide only had 7 scoring shots. As a result their percentage went down from 69.7% to 61.2% in two rounds. Geelong’s percentage went up from 138.5% to 160.4% from their two huge wins. Both are massive changes this late in the season.

References

Fewest scoring shots from afl.allthestats.com

Greatest margins from afl.allthestats.com

AFL Tables – 2011 Season scores

August 9, 2011

A Collection of Infinite Products – II

Filed under: mathematics — ckrao @ 11:57 am

Following on from my previous post on infinite products, here are some more to enjoy! Like last time, proof outlines are at the end.

(1) Seidel: \displaystyle \prod_{n=1}^{\infty} \frac{2}{1 + x^{1/2^{n}}} = \frac{\log x}{x-1} = \frac{2}{1 + \sqrt{x}}\cdot \frac{2}{1 + \sqrt{\sqrt{x}}}\cdot \frac{2}{1 + \sqrt{\sqrt{\sqrt{x}}}} \cdot \ldots

In particular x=2 gives von Seidel‘s product:

\displaystyle \log 2 = \prod_{n=1}^{\infty} \frac{2}{1 + 2^{1/2^{n}}} = \frac{2}{1 + \sqrt{2}}\cdot \frac{2}{1 + \sqrt{\sqrt{2}}}\cdot \frac{2}{1 + \sqrt{\sqrt{\sqrt{2}}}} \cdot \ldots

(2) \displaystyle \frac{\sin x}{x} = \cos \frac{x}{2} \cos \frac{x}{4} \cos \frac{x}{8} \ldots = \prod_{n=1}^{\infty} \cos \frac{x}{2^n}

(3) Viète: \displaystyle \frac{2}{\pi} = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2 + \sqrt{2}}}{2}\cdot \frac{\sqrt{2 + \sqrt{2+\sqrt{2}}}}{2}\cdots

(4) \displaystyle x = x^{1/2}x^{1/4}x^{1/8}\ldots = \prod_{n=1}^{\infty} x^{1/2^n}

(5) \displaystyle 1-x = \left(\frac{1-x}{1+x}\right)^{1/2}\left(\frac{1-x^2}{1+x^2}\right)^{1/4}\left(\frac{1-x^4}{1+x^4}\right)^{1/8}\ldots

(6) \displaystyle\prod_{k=0}^{\infty} \left(1 + x^{2^k}\right) = \frac{1}{1-x} for |x| < 1

(7) \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^s} = \prod_{p \text{ prime}} \frac{p^s}{p^s-1}

(8) \displaystyle \prod_{p \text{ prime}} \frac{p^2}{p^2 - 1} = \frac{\pi^2}{6}

(9) \displaystyle \prod_{p \text{ prime}} \frac{p^2}{p^2 +1} = \frac{\pi^2}{15}

(10) \displaystyle \prod_{p \text{ prime}} \frac{p^2+1}{p^2 -1} = \frac{5}{2}

(11) \displaystyle \frac{\pi}{4} = \frac{3}{4} \cdot \frac{5}{4} \cdot \frac{7}{8} \cdot \frac{11}{12} \cdot \frac{13}{12} \cdot \frac{17}{16} \cdot \frac{19}{20} \cdot \frac{23}{24} \cdot \ldots =\prod_{p \text{ odd prime}} \frac{p}{p - (-1)^{(p-1)/2}}
(in each fraction the numerator is an odd prime, the denominator is the nearest multiple of 4 to the numerator)

(12) \displaystyle \frac{\pi}{2} = \frac{3}{2} \cdot \frac{5}{6} \cdot \frac{7}{6} \cdot \frac{11}{10} \cdot \frac{13}{14} \cdot \frac{17}{18} \cdot \frac{19}{18} \cdot \frac{23}{22} \cdot \ldots = \prod_{p \text{ odd prime}} \frac{p}{p + (-1)^{(p-1)/2}}
(in each fraction the numerator is an odd prime, the denominator is the nearest even non-multiple of 4 to the numerator)

(13) \displaystyle 2 = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{9} \cdot \frac{10}{9} \cdot \frac{12}{11} \cdot \ldots
(in each fraction the numerator and denominator differ by 1, sum to the odd primes, and numerator is even)

(14) \displaystyle \frac{\pi}{4} = \frac{4}{5} \cdot \frac{8}{7} \cdot \frac{10}{11} \cdot \frac{12}{13} \cdot \frac{14}{13} \cdot \frac{16}{17} \cdot \frac{17}{18} \cdot \frac{20}{19} \cdot \ldots
(in each fraction the numerator and denominator differ by 1, sum to odd numbers that are not prime, and numerator is even)

(15) \displaystyle e = \frac{2}{1} \cdot \frac{4}{3} \cdot \frac{15}{14} \cdot \frac{64}{63} \cdot \ldots = \prod_{n=1}^{\infty} \frac{e_n+1}{e_n},

where e_1 = 1, e_{n+1} = (n+1)(e_n + 1)

(16) Pippenger: \displaystyle \frac{e}{2} = \left(\frac{2}{1}\right)^{1/2}\left(\frac{2}{3} \cdot \frac{4}{3}\right)^{1/4}\left(\frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{7}\right)^{1/8} \ldots

(17) Catalan: \displaystyle e = \frac{2}{1} \left( \frac{4}{3}\right)^{1/2} \left( \frac{6}{5} \cdot \frac{8}{7} \right)^{1/4} \left( \frac{10}{9} \cdot \frac{12}{11} \cdot \frac{14}{13} \cdot \frac{16}{15}\right)^{1/8}\ldots

(18) Pentagonal number theorem: \displaystyle \prod_{n=1}^{\infty} (1-x^n) = \sum_{k=1}^{\infty} (-1)^k x^{k(3k-1)/2}

Proof Outlines

A. Products based on geometric series

The following proof of (1) is based on [1]. We start with repeated application of the difference of perfect squares identity:

\begin{array}{lcl} x-y &=& (x^{1/2} - y^{1/2})(x^{1/2} + y^{1/2})\\&=& (x^{1/4} - y^{1/4})(x^{1/4} + y^{1/4})(x^{1/2} + y^{1/2})\\&=& (x^{1/8} - y^{1/8})(x^{1/8} + y^{1/8})(x^{1/4} + y^{1/4})(x^{1/2} + y^{1/2})\\&=&\ldots\\&=& (x^{1/2^N} - y^{1/2^N})\prod_{n=1}^N (x^{1/2^n} + y^{1/2^n})\\&=& 2^N (x^{1/2^N} - y^{1/2^N}) \prod_{n=1}^N \left(\frac{x^{1/2^n} + y^{1/2^n}}{2}\right) \quad \quad (*) \end{array}

In this expression let y=1. After rearrangement this gives

\displaystyle 2^N \left( x^{1/2^N} - 1 \right) = (x-1) \prod_{n=1}^N \left( \frac{2}{x^{1/2^n}+ 1} \right).

Next we take the limit of both sides as N\rightarrow \infty. The left side becomes \lim_{\epsilon \rightarrow 0+} \frac{x^{\epsilon}-1}{\epsilon} = \log x and so we have

\displaystyle \prod_{n=1}^{\infty} \frac{2}{1 + x^{1/2^{k}}} = \frac{\log x}{x-1} = \frac{2}{1 + \sqrt{x}}\cdot \frac{2}{1 + \sqrt{\sqrt{x}}}\cdot \frac{2}{1 + \sqrt{\sqrt{\sqrt{x}}}} \cdot \ldots

As noted in [1] one can also set x = e^{i\theta}, y = e^{-i\theta} in (*) and arrive at

\displaystyle \prod_{n=1}^{N} \cos \left(\frac{x}{2^n}\right) = \frac{\sin x}{2^N \sin \frac{x}{2^n}} so letting N\rightarrow \infty gives us (2). It need not be arrived at via the double angle formula!

Setting x = \pi/2 in (2) automatically gives (3), which also can be derived by considering regular (2^n)-gons being approximated by a circle in the limit.

Next (4) is based on the infinite series 1 = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots while (5) is apparent from the similar equality

\displaystyle a_0 = \left(\frac{a_0}{a_1}\right)^{1/2}\left(\frac{a_0 a_1}{a_2}\right)^{1/4}\left(\frac{a_0 a_1 a_2}{a_3}\right)^{1/8}\ldots

with \displaystyle a_0 = 1-x, a_1 = 1+x, a_2 = 1+x^2, a_3 = 1+x^4, \ldots [2].

Finally, (6) is based on the fact that every positive integer has a unique representation in binary, so every term x^n is represented in exactly one product on the left side (e.g. x^{13} = x^8.x^4.x^1), and so

\displaystyle\prod_{k=0}^{\infty} \left(1 + x^{2^k}\right) = 1 + x + x^2 + x^3 + x^4 + \ldots,

which is summed as a geometric series for |x| < 1.

B. Euler products

The next products over the primes [(7)-(14)] are due to Euler and come from his expansion of the Riemann zeta function in (7):

\displaystyle \zeta(s) := \sum_{n=1}^{\infty} \frac{1}{n^s} = \prod_{p \text{ prime}} \left(\sum_{k=0}^{\infty} \frac{1}{p^{ks}}\right) = \prod_{p \text{ prime}} \frac{p^s}{p^s-1}

This is a essentially a statement of unique factorisation, that every positive integer n can be written uniquely as a product of primes.

Setting s=2 gives (8). Setting s=4 gives

\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^4} = \prod_{p \text{prime}} \frac{p^4}{p^4-1} = \prod_{p \text{prime}} \frac{p^2}{p^2-1} \cdot \frac{p^2}{p^2+1}. \quad \quad (**)

The left side is equal to \frac{\pi^4}{90}.

One way of seeing this is by taking the x^5 term in the infinite product expansion \sin \pi x = \pi x \prod_{n=1}^{\infty} (1-\frac{x^2}{n^2}) (seen in my previous post). This gives

\displaystyle \frac{(\pi x)^5}{5!} = \pi x \sum_{m>n} \frac{1}{m^2n^2}.

Hence

\begin{array}{lcl} \sum_{n=1}^{\infty} \frac{1}{n^4} &=& \left( \sum_{n=1}^{\infty} \frac{1}{n^2} \right)^2 -2 \sum_{m\neq n}^{\infty} \frac{1}{m^2n^2}\\&=& \left( \frac{\pi^2}{6} \right)^2 - 2\frac{\pi^4}{5!}\\ &=& \pi^4 \left( \frac{1}{36} - \frac{2}{120} \right) \\ &=& \frac{\pi^4}{90}. \end{array}

Finally one obtains (9) by dividing (**) by (8). (10) simply arises from dividing (8) by (9).

To prove (11) we proceed in a similar manner to proving (7), this time using unique factorisation of the odd natural numbers, and that an odd number that is 3 modulo 4 must have all prime factors that are 3 modulo 4 occurring an odd number of times:

\begin{array}{lcl} \sum_{n=0}^{\infty} \frac{(-1)^ n}{(2n+1)^s} &=& \left(1 - \frac{1}{3^s} + \frac{1}{3^{2s}} - \ldots\right)\left(1 + \frac{1}{5^s} + \frac{1}{5^{2s}} + \ldots\right)\left(1 - \frac{1}{7^s} + \frac{1}{7^{2s}} - \ldots\right) \ldots \\ &=&\prod_{p \text{ odd prime}} \sum_{k=0}^{\infty} \left(\frac{(-1)^{(p-1)/2}}{p^s}\right)^{k}\\ &=& \prod_{p \text{ odd prime}} \frac{p^s}{p^s - (-1)^{(p-1)/2}}\end{array}

To be more precise, if m is the largest prime less than N, one can show [3] that the difference

\displaystyle \left| \sum_{n=0}^{\infty} \frac{(-1)^ n}{(2n+1)^s} - \prod_{p \text{ odd prime}}^m \frac{p^s}{p^s - (-1)^{(p-1)/2}}\right| \leq \sum_{n=N}^{\infty} \left|\frac{(-1)^ n}{(2n+1)^{\text{Re }s}} \right|\rightarrow 0

as n \rightarrow \infty.

Setting s to 1 and using the sum \displaystyle \sum_{n=0}^{\infty} \frac{(-1)^ n}{2n+1} = \frac{\pi}{4} gives (11). We obtain (12) by dividing 3/4 times (8) by (11).

Dividing 3/4 times (8) by the square of (11) leads to (13) after some cancellation.

The next formula uses result (9) (the Wallis product) of my previous post:

\displaystyle \frac{\pi}{2} = \prod_{n=1}^{\infty} \left(1- \frac{1}{(2n+1)^2}\right) = \frac{2}{3}\cdot \frac{4}{3}\cdot \frac{4}{5}\cdot \frac{6}{5}\cdot \frac{6}{7}\cdot\frac{8}{7}\cdot \ldots \quad \quad (***)

Dividing this by (13) leads to (14).

C. Formulas involving e

We obtain (14) by using the following infinite sum form of e:

\displaystyle e = \sum_{n=1}^{\infty} \frac{1}{n!}

We can see by the definition of e_n that the product \prod_{n=1}^N \frac{e_n + 1}{e_n} = \frac{e_{N+1}}{(N+1)!}. Then it can be seen that e_N is the numerator of the partial sum \sum_{n=0}^{N} \frac{1}{n!}. The result follows by taking the limit N \rightarrow \infty.

To obtain (15) we use Stirling’s formula for the asymptotic form of the factorial function:

\displaystyle n! \sim \left(\frac{n}{e}\right)^n \sqrt{2 \pi n} \Rightarrow e = \lim_{n \rightarrow \infty} \frac{n}{(n!)^{1/n}}

Next we observe that each term in the right side can be written in terms of factorials and powers of 2. For example,

\begin{array}{lcl} \frac{8}{9} \cdot \frac{10}{9} \cdot \frac{10}{11} \cdot \ldots \cdot \frac{14}{15} \cdot \frac{16}{15} &=&\frac{2^8.(4.5.5.6.6.7.8)}{(9.11.13.15)^2}\\ &=& 2^8\frac{(8!)^2}{(4!)^2}\frac{4}{8} \frac{(8!)^2.(10.12.14.16)^2}{(16!)^2}\\ &=& 2^{15}\frac{(8!)^2}{(4!)^2}\frac{(8!)^2 (8!)^2}{(16!)^2(4!)^2}\\ &=& \frac{2^{15}.(8!)^6}{(4!)^4.(16!)^2} \end{array}

The general term is of the form

\displaystyle \left[\frac{2^{2^n-1} (2^{n-1}!)^6}{(2^{n-2}!)^4(2^n!)^2} \right]^{\frac{1}{2^n}}.

When these terms are multiplied, the product telescopes and we end up with

\begin{array}{lcl} \left(\frac{2}{1}\right)^{1/2}\left(\frac{2}{3} \cdot \frac{4}{3}\right)^{1/4}\left(\frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{7}\right)^{1/8} \ldots &=& 2\sqrt{2} \lim_{N \rightarrow \infty} \prod_{n=2}^N \left[\frac{2^{2^n-1} (2^{n-1}!)^6}{(2^{n-2}!)^4(2^n!)^2} \right]^{\frac{1}{2^n}}\\ &=& 2\sqrt{2} \lim_{N \rightarrow \infty} 2^{N-1} \frac{1}{2^{1/4 + 1/8 + \ldots + 1/2^N}}.2.\frac{\left[ 2^{N-1}! \right]^{6/2^N - 2/2^{N-1}}}{\left[ 2^N !\right]^{2/2^N}}\\&=& 2\sqrt{2} \lim_{N \rightarrow \infty} 2^N.2^{-\left(1/2 - 1/2^N \right)}. \left[\frac{2^{N-1}!}{2^N!} \right]^{\frac{1}{2^{N-1}}}\\ &=& 2 \lim_{N \rightarrow \infty} 2^{1/2^N}.\left(\frac{2^N}{\left(2^N! \right)^{1/2^N}}\right)^2.\left(\frac{\left(2^{N-1}! \right)^{1/2^{N-1}}}{2^{N-1}}\right).\frac{1}{2}\\&=& 2.1.e^2.\frac{1}{e}.\frac{1}{2}\\&=& e\end{array}

We remark that the product converges rapidly to e, since even if we use the more precise form \displaystyle e \sim \left(\frac{n^{n+1/2}}{n!}\sqrt{2 \pi} \right)^{1/n} and replace \pi with the Wallis product expansion, the left and right sides of (15) can be shown to be similar for N large.

It is amazing to see that (15) becomes Wallis’s product (***) for \displaystyle \frac{\pi}{2} when the exponents are set to 1 (while keeping the same fractions)!

(16) can be easily obtained from (15) by multiplying both sides by

\displaystyle 2 = 2^{1/2}2^{1/4}2^{1/8}\ldots

(although Catalan’s result, obtained through more involved means, long preceded that of Pippenger).

More similar products are given in [4].

D. Pentagonal number theorem

This result, again due to Euler, shows how much cancellation there is when the product (1-x)(1-x^2)(1-x^3)\ldots is expanded. The coefficient of x^n is the number of partitions of n into an even number of parts minus the number of partitions of n into an odd number of parts. The theorem says that this number is zero unless n is a pentagonal number (i.e. of the form k(3k-1)/2), in which case it is 1 or -1. Refer to this Wikipedia entry of this theorem for a combinatorial proof.

References

[1] T.J. Osler, Interesting finite and infinite products from simple algebraic identities, The Mathematical Gazette, 90(2006), pp. 90-93. Available here.

[2] K. Brown, Infinite Products and a Tangent Fan. Mathpages link

[3] P. Loya, Amazing and Aesthetic Aspects of Analysis: On the incredible infinite, available at http://www.math.binghamton.edu/dennis/478.f07/EleAna.pdf

[4] J. Sondow and H. Yi, New Wallis- and Catalan-Type Infinite Products for π , e, and sqrt(2+ sqrt(2)), available at http://arxiv.org/ftp/arxiv/papers/1005/1005.2712.pdf

August 5, 2011

A Collection of Infinite Products – I

Filed under: mathematics — ckrao @ 12:25 pm

This post and the next one show some cool infinite products (mostly taken from [1] and [2]) with an attempt to demystify most of them. A few ideas are mentioned, but they are not necessarily the approaches with which the formulas were discovered in the first place! If you wish to continue to be captivated by the mystery, admire the formulas as they are and no need to proceed to the proof section. 🙂

(1) \displaystyle \prod_{n=2}^{\infty} \left(1 - \frac{1}{n^2}\right) = \frac{1}{2}

(2) \displaystyle \prod_{n=3}^{\infty} \left(1 - \frac{4}{n^2}\right) = \frac{1}{6}

(3) In general \displaystyle \prod_{n=k+1}^{\infty} \left(1 - \frac{k^2}{n^2}\right) = \frac{1}{\binom{2k}{k}}

(4) \displaystyle \prod_{n=2}^{\infty} \frac{n^3 -1}{n^3 + 1} = \frac{2}{3}

(5) \displaystyle \frac{1}{\Gamma(z)} = \lim_{n \rightarrow \infty} \frac{z(z+1)\ldots (z+n)}{n! n^z}= ze^{\gamma z} \prod_{n=1}^{\infty} \left[\left(1 + \frac{z}{n}\right) e^{-z/n}\right] = z\prod_{n=1}^{\infty} \frac{1 + z/n}{\left(1 + 1/n\right)^z}

Here \displaystyle \Gamma (z) = \int_0^{\infty} u^{z-1}e^{-u}\ du generalises the factorial function, so that \Gamma (n+1) = n! for n a non-negative integer. Also \displaystyle \gamma := \lim_{n \rightarrow \infty} \left( \sum_{k=1}^n \frac{1}{k} - \log n\right) \approx 0.577 is the Euler-Mascheroni constant. The first equality in (5) is due to Gauss, the second to Weierstrass, the third to Euler.

(6) \displaystyle \prod_{n=0}^{\infty} \frac{(n+a_1)(n+a_2)}{(n+b_1)(n+b_2)} = \frac{\Gamma (b_1) \Gamma (b_2)}{\Gamma (a_1) \Gamma (a_2)}, where a_1 + a_2 = b_1 + b_2.

(7) Wallis (1655): \displaystyle \prod_{n=1}^{\infty} \frac{4n^2}{4n^2-1} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \ldots = \frac{\pi}{2}

(8) \displaystyle \prod_{n=1}^{\infty} \left(1 - \frac{1}{(2n)^2}\right) = \frac{2}{\pi}

(9) \displaystyle \prod_{n=1}^{\infty} \left(1 - \frac{1}{(2n+1)^2}\right) = \frac{\pi}{4}

(10) \displaystyle \sin \pi z = \pi z \prod_{n=0}^{\infty}\left(1 - \frac{z^2}{n^2}\right)

(11) \displaystyle \cos \pi z = \prod_{n=-\infty}^{\infty}\left(1 + \frac{z}{n - 1/2}\right) = \prod_{n=0}^{\infty}\left(1 - \frac{4z^2}{(2n+1)^2}\right)

(12) \begin{array}{lcl} \prod_{n=0}^{\infty} \left(1 + \frac{(-1)^n}{2n+1} \right) &=& \left(1 + \frac{1}{1}\right) \left(1 - \frac{1}{3}\right) \left(1 + \frac{1}{5}\right) \left(1 - \frac{1}{7}\right) \ldots\\ &=& \frac{2}{1}\cdot \frac{2}{3}\cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \ldots\\ &=& \sqrt{2} \end{array}
(From Wallis’s product (7) remove fractions containing multiples of 4 and we go from \pi/2 to \sqrt{2}!)

(13) \displaystyle \prod_{n=-\infty}^{\infty} \frac{n-z_1}{n-z_2} = \frac{\sin \pi z_1}{\sin \pi z_2}

(14) \displaystyle \prod_{n=1}^{\infty} \left(1 +\frac{1}{n^2}\right) = \frac{1}{\pi} \sinh \pi

(15) \displaystyle \prod_{n=1}^{\infty} \left(1 +\frac{1}{n^3}\right) = \frac{1}{\pi} \cosh \left(\frac{\pi \sqrt{3}}{2}\right)

(16) \displaystyle \prod_{n=2}^{\infty} \left(1 -\frac{1}{n^3}\right) = \frac{1}{3\pi} \cosh \left(\frac{\pi \sqrt{3}}{2}\right)

Proofs:

A. Telescoping products

The idea is that terms cancel and only early and late terms remain, then take the limit as n goes to infinity. For example, to prove (1) we have

\begin{array}{lcl} \prod_{n=2}^{\infty} \left(1 - \frac{1}{n^2}\right) &=& \lim_{N \rightarrow \infty} \prod_{n=2}^N \frac{n-1}{n} \prod_{n=2}^N \frac{n+1}{n}\\ & = & \lim_{N \rightarrow \infty} \left(\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \ldots \cdot \frac{N-1}{N} \right) \left(\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \ldots \cdot \frac{N+1}{N} \right) \\ & =& \lim_{N \rightarrow \infty} \frac{1}{N} \frac{N+1}{2}\\ &=& \frac{1}{2}. \end{array}

Equations (2) and (3) are proved similarly. To prove (4) we have

\begin{array}{lcl} \prod_{n=2}^{\infty} \frac{n^3-1}{n^3+1} &=& \lim_{N \rightarrow \infty} \prod_{n=2}^{N}\frac{(n-1)(n^2+n + 1)}{(n+1)(n^2-n + 1)}\\ &=& \lim_{N \rightarrow \infty} \prod_{n=2}^{\infty} \frac{n-1}{n+1} \prod_{n=2}^{\infty} \frac{n(n+1) + 1}{(n-1)n + 1}\\&=& \lim_{N \rightarrow \infty} \left(\frac{1}{3} \cdot \frac{2}{4} \cdot \frac{3}{5} \cdot \ldots \cdot \frac{N+1}{N+1} \right)\left(\frac{2\times 3+1}{1\times 2+1} \cdot \frac{3\times 4+1}{2\times 3+1} \cdot \frac{5\times 4+1}{3\times 4+1} \cdot \ldots \cdot \frac{N(N+1)+1}{(N-1)N+1} \right)\\&=& \lim_{N \rightarrow \infty} \frac{1 \times 2}{N(N+1)} \cdot \frac{N(N+1)+1}{1\times 2+1}\\&=& \frac{2}{3}.\end{array}

B. Use of the Gamma function:

By induction we may prove \displaystyle \int_0^1 t^{z-1} (1-t)^n\ dt = \frac{n!}{z(z+1)\ldots (z+n)}. Then setting t=u/n and letting n tend to infinity we arrive at the first equality in (5):

\displaystyle \Gamma (z) = \int_0^{\infty} u^{z-1}e^{-u}\ du = \lim_{n \rightarrow \infty} \frac{n^{z-1}n!}{z(z+1)\ldots (z+n-1)}

To prove the second equality in (5) refer to this post from the Abstract Nonsense blog. The third equality follows similarly from the definition of the exponential and \gamma.

From the first equality of (5) we arrive at (6):

\displaystyle \prod_{n=0}^{\infty} \frac{(n+a_1)(n+a_2)}{(n+b_1)(n+b_2)} = \frac{\Gamma (b_1) \Gamma (b_2)}{\Gamma (a_1) \Gamma (a_2)},

where a_1 + a_2 = b_1 + b_2.

In general$a_1, a_2, b_1, b_2$ can be any algebraic expressions and (7) easily generalises to a product of more than two terms.

Setting a_1, a_2, b_1, b_2 to 1, 3, 2 and 2 respectively, gives equation (1). Setting a_1, a_2, b_1, b_2 to 1, 1, 1/2, 3/2 respectively gives Wallis’s formula (7):

\displaystyle \prod_{n=1}^{\infty} \frac{4n^2}{4n^2-1} = \prod_{n=0}^{\infty} \frac{(n+1)^2}{(n+1/2)(n+3/2)} = \frac{\Gamma (1/2) \Gamma(3/2)}{\Gamma (1) \Gamma (1)} = \frac{\sqrt{\pi}\times \sqrt{\pi}/2}{1\times 1} = \frac{\pi}{2}

C. Trigonometric/Hyperbolic results

Formula (7) can also be proved via the infinite product expansion for \sin \pi z (10) shown in my previous mathematical blog post on Tannery’s theorem.

\displaystyle \sin \pi z = \pi z \prod_{n=0}^{\infty}\left(1 - \frac{z^2}{n^2}\right)

By a similar approach the infinite product expansion for \cos \pi z (11) can be derived. These two formulas and the expansion of the gamma function (5) can also be derived by application of the Weierstrass factorization theorem.

Setting z to 1/2 in (10) leads to (7). Formula (8) is simply the reciprocal of (7), while (9) is obtained via (6).  It is interesting to compare (1), (8) and (9) to see how different their values can be.

Setting z to 1/4 in the sine formula (10) and then taking reciprocals yields

\begin{array}{lcl} \sqrt{2} &=& \prod_{n=0}^{\infty}\left(\frac{(2n+1)^2}{(2n+1)^2 - 1/4}\right)\\ &=& \prod_{n=0}^{\infty}\left(\frac{2(2n+1).2(2n+1)}{(2(2n+1)-1).(2(2n+1) +1}\right)\\&=& \frac{2}{1}\cdot \frac{2}{3}\cdot \frac{6}{5}\cdot \frac{6}{7}\cdot \frac{10}{9}\cdot \frac{10}{11}\cdot\ldots \end{array}

leading to (12).

Formula (13) results from direct application of (10) (a ratio of sines). The hyperbolic results (14)-(16) result partly from (10) and (11) (via \cosh z = \cos iz, \sinh z = -i \sin iz) and partly from Wallis’s formula (7).  To prove (15) ((16) is proved similarly) we firstly note that

\displaystyle \cosh \left( \frac{\sqrt{3} \pi}{2} \right) = \prod_{n=1}^{\infty} \left(1 + \frac{3}{(2n-1)^2}\right).

Then

\begin{array}{lcl} \prod_{n=1}^{\infty} \left(1 + \frac{1}{n^3} \right) &=& \prod_{n=1}^{\infty} \frac{(n+1)(n^2-n+1)}{n^3}\\&=& \prod_{n=1}^{\infty} \frac{n+1}{n} \cdot \frac{4(n^2 - n + 1)}{4n^2}\\&=& \prod_{n=1}^{\infty} \frac{n+1}{n} \cdot \frac{(2n-1)^2(2n+1)}{(2n+1)4n^2} \cdot \frac{4(n^2 - n + 1)}{(2n-1)^2}\\&=&\prod_{n=1}^{\infty} \frac{n+1}{n} \cdot \frac{2n-1}{2n+1} \cdot \frac{4n^2-1}{4n^2} \cdot \frac{4(n^2 - n + 1)}{(2n-1)^2}\\&=&\lim_{N \rightarrow \infty}\prod_{n=1}^{N} \frac{n+1}{n} \cdot \frac{2n-1}{2n+1} \cdot \frac{2}{\pi} \cdot \prod_{n=1}^{\infty} \left(1 + \frac{3}{(2n-1)^2}\right)\\&=& \lim_{N \rightarrow \infty} \frac{N+1}{2N+1} \cdot \frac{2}{\pi} \cdot \cosh \left( \frac{\sqrt{3} \pi}{2} \right)\\ &= & \frac{1}{\pi}\cosh \left( \frac{\sqrt{3} \pi}{2}\right).\end{array}

This last formula shows how several ideas mentioned previously can be combined to form new products.

References

[1] Weisstein, Eric W. “Infinite Product.” From MathWorld–A Wolfram Web Resource. http://mathworld.wolfram.com/InfiniteProduct.html

[2] A, Dieckmann, Collection of Infinite Products and Series

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