Complexity Science The Study Of Emergence By Henrik Jeldtoft Jensen SOLUTIONS MANUAL

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Chapter 1 Exercise 1 For example, atoms forming molecules, molecules forming proteins etc.

Exercise 2 For example, both stress how essential complex phenomenon are and that they can be classified.

Anderson even argues that understanding emergence is a conceptually fundamental activity at least as important as is particle physics.

Exercise 3 For example, an ontology consisting of hierarchies of processes seems to contain emergence as a defining aspect of the world. It appears that the distinction between weak and strong emergence evaporates and is replaced by an investigation of relationships between processes at different levels of the hierarchy.

Project 1 a) Epistemic: Phenomena that emerge because of the way we represent the world or obtain knowledge about the world.

Ontological: genuinely new entities at higher levels such as objects, properties or causes. b) Perhaps it would be interesting to discuss this in relation to optical phenomena such as the colour of an object and the lightning or gestalt images such as the duck-rabbit drawing. c) A short answer could be that we know the world epistemologically. d) Simon seems to think in terms of epistemic complexity when he says "... not in an ultimate, metaphysical sense, but in the important pragmatic sense...". His discussions of evolution and the advantage of hierarchies is ontological. The summary blends epistemic and ontological emergence: Simon mentions that the level of complexity depends on the description but the processes he mentions appear to have an ontological status. Anderson’s description of emergence through symmetry breaking is about ontology. e) Seems that if at the deepest ontological level we have processes rather than things and if we understand the world as processes the distinction between ontology and epistemology may fade away.

Project 2 a) Biology for example in cells, organs, organisms,..., flocks etc. Also, society for example with individuals, families, organisations etc. b) The answer may include a discussion about hierarchies of selective pressure, the relationship between reproductive unit, the relation between "genotype" and "phenotype", and the difference between adaptive change and evolutionary change. c) Both ecosystems and society is organised hierarchically and even if ideal definitions of the building blocks of these hierarchies are at present lacking it appears to be impossible to describe the functionality and structure of, say, an ecosystem or, say, a political strucutre without some notion of building blocks consisting of collections (a species for example)of individuals sharing some relevant relations.

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d) First to answer this, think about how to define biological species and social classes. One would then probably conclude that we (at present) only can use these concepts at an epistemological level.

Project 3 a) Carefully go through Section 1.1 b) For example, when using a bicycle pump it feels like the air inside the pump is an elastic medium.

That the pressure of the gas gives rise to elasticity. c) No, Anderson will not consider pressure as an emergent property belonging to the "more is different" category. d) If we go with Wheeler, any accumulated property seems to acquire "emergent property" status. e) No, I cannot think of any.

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Chapter 2 Exercise 1 a) Wolf packs, ant colonies, beehives, human tribes, companies, resting state neural network. Perhaps neurosis, depression. b) Punctuated equilibrium in macro-evolution. Social history is interspersed with political, cultural, and technological revolutions. Neuronal bursts (or avalanches) In psychology, attention and distraction are examples as well.

Exercise 2 Biology: Fireflies, crickets etc.

Physics: Coupled pendula, electronic circuits, etc.

Economics: Herding behaviour, exchange rate variation, etc.

Neuroscience: Epileptic seizures, circadian clock, brain waves as measured by EEG, etc.

Exercise 3 Social class as given by income level is an emergent structure in society. Self-perception emerges as a consequence of our personal experience and history. Manstead demonstrates how the two are linked.

Exercise 4 Simon discusses typically shared properties amongst complex systems. He focuses on hierarchical structure and presents arguments for why hierarchies naturally arise in heterogeneous many component systems. The key term is "near decomposability". Simon concentrates on epistemology, in the conclusion he writes: " How complex or simple a structure is dependent critically upon the way in which we describe it." Anderson focuses on the ontological differences between the whole and the components. He refers to symmetry breaking to describe the new reality that can emerge when many (in principle infinitely many) interacting (and possibly identical) components form coherent structures.

Exercise 5 Below are some snapshots of possible considerations: a) The individual scale: 1 The family scale: 2, ..., 10 The town or city: 103 - 106 Nations: 106 - 109 b) The biological cell: 10´6 m - 10´4 m Plants: 10´3 m - 10m (perhaps 102 ) Clusters of trees: 102 m Forests: 103 m - 104 m c) Cell body of neuron: 10´6 - 10´4 𝑁 The axon 𝑚 Brain regions linear size: 10´2 m Entire brain linear size 0.15m d) The answer will depend on whether we think of the size of structures in terms of money or the number of employees. One also needs to decide what to include in the structures of the financial system. For example, one could focus on banks and find data on bank-size in Italy and China, it would be of interest to look at the distribution of bank sizes. Perhaps look up the paper: Rank-size

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Complexity Science distributions for banks: A cross-country analysis by R. Cequeti, C. Lupi, F, Peirtrovito, and A.F. Pozzolo. Physica A 585 126336 (2022)

Exercise 6 a) At least some degree of robustness: - the family - the tribe - the nation-state - institutions like hospitals, schools, companies, etc. b) How about: - individual plants, animals, ...

- species or types - perhaps sub-networks of food webs c) Various brain regions have some functional robustness. For example, the visual cortex, the olfactory bulb, the auditory cortex etc. Perhaps if we think of the mind moods like happy, sad, anxiety, etc. can be considered "robust" emergent structures. d) For eg. financial institutions such as banks, building societies, pension funds and regulatory bodies such as central banks, and government institutions.

Exercise 7 Below are some suggestions as to how the disappearance or breaking of symmetry may be related to these changes. a) For example change in the symmetry of the distribution of productivity per worker. Pre-industrial revolution: One narrow, presumably peaked formed distribution. Post-industrial revolution: Bimodal distribution with one peak at low value for non-industrial workers and another peak at higher value for industry workers. b) For example change from autocracy to democracy and change in the distribution of possible political influence. c) For example the change in the distribution of accepted moral behaviours when a change occurs from a narrow, strict (perhaps religiously motivated) set of values to a liberal tolerant "anything goes", individual-focused values. d) For example distribution of accepted female dress code. The change from dress and skirt only to a broader range including trousers and trouser suits.

Exercise 8 The question is intended to inspire discussion.

Project 1 Read de Haan’s paper carefully.

Project 2 a) Matter of scale and aggregation. The intermittent dynamics occur at a systems level. For ecology that is the abrupt changes in taxonomic composition. For example mass extinction of species. For the economy, the intermittency is similarly seen in the type of extant companies. Suddenly nearly all cartwrights disappear for example.

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b) The arrival of a new type which is disruptive to the stability of the current configuration c) Better understanding of dynamics may enable us to identify potential dangerous developments and act to avoid or minimise consequences. For example stress testing of banks followed by prudential action or firefighters to minimize damage by forest fires. d) No sample solution needed.

Project 3 At present, a detailed answer to (a) and (b) is not known. But it is presumably related to the time averaging of spatial structures.

Project 4 No sample solution needed Project 5 No sample solution needed

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Chapter 3 Exercise 1 No sample solution needed Exercise 2 a) For example, the cell life of red blood cells (RBCs) is approximately 4 months, white blood cells (WBCs) is about a year, and skin cells are just a few weeks. Heartbeats are about 1 second apart.

Breath at rest is at a rate of 15 per minute, and digestion takes about 36 hours total. The lifespan for the entire body is about 70 years. b) Travel speed of signal along a motor neuron is approximately 430 km/hr. The average neuron fires about 0.16 times per second. The up-down cycle of the neuronal action potential is approximately 1 msec. The reaction time to stimuli is 200 msec and thoughts can vary from seconds to minutes, and to hours.

Exercise 3 Some examples: Ice to water: Since ice and water can co-exist the length scale characterising the transition remains finite.

Magnetic to non-magnetic: The magnetic ordering spans across the materials, i.e. the length scale equals the size of the system. Therefore, the characteristic length is infinite compared with the microscopic scale, say the separation between the atoms.

When we look at the transitions in the biological or social systems (say a collapse of an ecosystem or a political change), we can check if the change penetrates across the entire system.

Exercise 4 a) Despite critical fluctuations that have been suggested by Scheffer et al. to often occurring before a transition, it is difficult by a web search for "critical fluctuations as precursor" to find established examples. However, Olthof et al. suggest "critical fluctuations as an early-warning signal for sudden gains and losses in patients receiving psychotherapy for mood disorders" [Clinical Psychological Science 8(1) 2020. First published 24.09.2019] that the mechanism may be clinically relevant. b) There seems to be no indication that, for example, earthquakes are preceded by fluctuations. Nor does the appearance of a new infectious virus with the ensuing pandemic seem to be foreseeable from fluctuations.

Exercise 5 No sample solution needed Project 1 No sample solution needed Project 2 a) No solution needed b) No solution needed c) Generically, the strength of the approach is a possible identification of essential mechanisms. The weakness is often that it can be difficult to go beyond the Ising approach in an incremental way.

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Project 3 No sample solution needed.

Project 4 a) Considerations may include the observation that one will need to distinguish between prey and predator links: if a species is a prey of many other species, i.e. high prey-degree it has a harder time avoiding being eaten. If a predator has a small predator-degree, meaning it only preys on a few species, it may be difficult to find food. b) Keystone species do not affect the stability of an ecosystem through tropic interactions only, as the example of the bee indicates. The bees are of crucial importance as pollinators, and hence for reproduction in the ecosystem, but are not eaten by many species. If all species could be considered to be of equal pairwise importance, i.e. all links between nodes representing species of equal weight, keystone species would presumably, e.g. be represented by hub nodes - these are nodes of the unusually high number of links.

In reality, some links between species may be more important than others. In this case, a key species is probably connected to nodes (species) with important links. c) In a food-web structure a keystone species might sit at the bottom of a pyramid structured food-web with very many different predators dependent on it, i.e. with high prey-degree. Or the keystone could sit as a bottleneck between different communities in the network. It could possibly also be a top predatory, the removal of which will allow prey species populations to grow unchecked and thereby destablise the ecosystem. This could be difficult to see from the network structure alone and will only become clear when population dynamics on the network is accounted for. d) Foodwebs have very broad degree distributions. Such networks are stable against random removal of nodes (since a typical node is of low degree and in this respect of little consequence) fragile against target removal of high degree nodes. e) See last paragraph of ref. [220]. f) That a focus on pairwise interaction between species misses important aspects of higher order interactions and that non-local global effects may not be captured by the pairwise network representation. g) No sample solution needed.

Project 5 No sample solution needed.

Project 6 a) No sample solution needed. b) The result of the body’s temperature regulation is similar to the result of the Gaia mechanism, namely homeostasis. But the Gaia hypothesis involves evolutionary dynamics and selection pressures. The temperature regulation involves a negative feedback loop controlled centrally from the brain’s thermal regulation centre in the hypothalamus. c) See for example John S. Torday. "Homeostasis as the mechanism of evolution", Biology (Basel).

2015 Sep:4(3):575-590 For examples of sociological feedback loops see J.J. Fernandez and A.M. Jamie-Castillo. " Positive or negative policy feedback? Explaining popular attitudes towards pragmatic pension policy reforms. European Sociological Review. 29(4) (2013) 803-815. d) No sample solution needed.

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e) Entropic hierarchy: Sequences of systemic conditions allowing for an increasing number of configurations corresponding to an increasing entropy of the system. f) One may argue that given the intricacy of the phenomena under consideration, namely how the self-regulation of Gaia arises spontaneously, the model in [28] does very well on all three: the model extracts the most essential (P1). Though the model is simple compared to reality it tries not to over-simplify (P2) by using an individual-based stochastic model. Finally, even P3 seems to be well fulfilled given that the Gaia hypothesis may be at work in many different situations.

Chapter 4 Exercise 1 This is obviously an open-ended question intended for discussion. Presumably one would formulate a model which includes components like borrowers, investors, and bank agents. The paper by Viegas et al. "Ecosystems perspectives on Financial Networks: Diagnostic Tools". COMPLEXITY, 18(6) 3448(2013) found that an element of evolutionary dynamics is essential for crashes to occur.

Exercise 2 One approach could be to start from simple and then add features to the model until the desired phenomenology appears. Another is to start from a description (model) which reproduces the phenomenology, but which appears to have irrelevant details. One may remove details to study which are essential.

Exercise 3 a) Probably not much. The Hodgkin-Huxley model already describes well the dynamics of a single neuron. The biggest challenge for an understanding of the human brain appears to consist in understanding the emergent collective behaviour of the huge number of interacting neurons in the brain. The computer neuronal network will present the same problem. But of course, it will be easier to carry out an experiment on the computer representation of the brain. b) That is at the moment a very open question. Perhaps the artificial "brain" can help to understand the formation of information propagation and the formation of functional networks. However, it is highly unclear if the computer brain will possess anything similar to the human mind or consciousness. c) Not necessarily. It is possible. For example, the mind and consciousness are dependent on the entire history of experience back from the formation of the first brain cells in the foetus or other unknown dependencies. d) Maybe, if the computers were to develop through a "social environment" like a human does more or less all the way back to germination.

Exercise 4 Very difficult term to handle rigorously. Even the so-called "fitness" used in evolutionary biology as genotype fitness or phenotype fitness needs to be defined for a given environment. Hence, it is not an intrinsic property of the organism.

Project 1 a) Depends. Consider a primary school. The group of stakeholders will include: the teachers, parents, pupils, city council etc. b) Discussion question. No sample solution needed

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c) It will depend on the time scale considered. For example, the union organising a group of employees could well be considered a robust structure for the duration of the existence of the firm. d) This is definitely an open question and will depend on context. Few stakeholders (e.g. individual parents or interest groups) can be considered as robust over time scales of the society (say decades and centuries). And if a certain stakeholder can be identified, say the Agriculture and Horticulture Development Board Potatoes or AHDB Potatoes (former Potato Council), they most likely keep changing their strategies and aims over the years. This will make it problematic to consider them as well-defined society building blocks in their own right. e) Perhaps on shorter time scales and for some specific subset of phenomena. A way to go about this question could be to ask for a model of, say, an university.

Project 2 a) Yes. Nice, clear, and schematic model. b) Perhaps not. The paper has been useful to much further research, as indicated by 200 citations, and for example mentioned in Mahadevan et al. "Evaluating the sensitivity of functional connectivity measures to motion artifact in resting-state fMRI data". NeuroImage 241, 118408 (2021).

But it has also been pointed out that the model approach by Vertes et al. needs to be supplemented by information theory. See e.g. Srivastava et al. "Models of communication and control for brain networks: distinctions, convergence, and future outlook". Network Neuroscience. 4(4), 1122-1159 (2020). c) Perhaps some kind of network of information processing and exchanging nodes. d) Yes. Probably some "aspects" as we already see rudimentary indications of it in current AI systems such as speaking and listening robots.

Project 3 a) Most importantly, the waiting time distribution for ants leaving the nest follows not an exponential (i.e. a Poisson process) but behaves like a log-Poisson. The probability that ant leaves at time 𝜔 is proportional to 𝜔1 . b) N different types of agents (ants) on a periodic lattice of size 𝐿 ˆ 𝐿. Agents move according to a utility function representing the effect of interaction between ants. For a low "degree of stochasticity" (DS) moves are dominated by interaction. For high DS, ants move at random and independently. For low-DS, its log-Poisson. For high-DS, its Poisson. c) Difficult to answer since we do not know exactly how ants make decisions. But the model framework makes suggestions that in principle can be tested: dependence on density, behaviour independent of location of exit. etc. d) This is an open question, which can be addressed by trying out simplifications. e) The model predicts that the log-Poisson behaviour should very generally apply to agents moving around and could be applied to different types of ants, perhaps other insects, and maybe even humans moving in, say, a public space.

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Chapter 5 Exercise 1 a) See 7th paragraph of Section 3.2 b) We need 8 ÿ

𝐴𝑠´ 𝜏 “ 1

𝑠 “1

𝐴´1 “

8 ÿ

𝑠´ 𝜏 “ 𝜁p𝜏q

𝑠 “1

which is Riemann’s Zeta function. To ensure 𝐴 ă 8 we need 𝜏 ą 1 c) Moments 8 8 ÿ ÿ ă 𝑆 ą“ 𝑠𝑃p𝑠q “ 𝐴 𝑠 p 𝑎´ 𝜏 q 𝑠 “1 8 ÿ

“𝐴

𝑠 “1

𝑆 ´p 𝜏 ´𝑎q “ 𝐴𝜁p𝜏 ´ 𝑎q

𝑠 “1

To ensure ă 𝑆 𝑎 ąă 8 we need, 𝜏 ´ 𝑎 ą 1 or 𝑎 ă 𝜏 ´ 1.

Exercise 2 a) 8 ÿ

𝑃𝑠𝑁 p𝑞q𝑠 𝑞

𝑔𝑠𝑁 p𝑠q “ 𝑞 “0

where 𝑁 ÿ

𝑃𝑠𝑁 p𝑞q “ 𝑝𝑟𝑜𝑏t𝑠 𝑁 “

𝑋𝑡 “ 𝑞u 𝑡 “1

8 ÿ

“

𝑛 ÿ

𝑃𝑟𝑜𝑏t𝑁 “ 𝑛u𝑃𝑟𝑜𝑏t 𝑛“0 8 ÿ

“

𝑋𝑡 “ 𝑞u 𝑡 “1

ÿ 𝑃 𝑁 p𝑛q

𝑛 “0

ÿ 𝑃 𝑋 p𝑞 1 q...𝑃 𝑋 p𝑞 𝑛 q

... 𝑞1

𝑞𝑛

𝑞 1 ` 𝑞 2 ` ... ` 𝑞 𝑛 “ 𝑞 8 ÿ 8 ÿ

𝑔𝑠𝑁 p𝑠q “

ÿ 𝑃 𝑁 p𝑛q

𝑞 “0 𝑛“0 8 ÿ

“

𝑃 𝑁 p𝑛q 𝑛“0

ÿ ...

𝑞1

𝑃 𝑋 p𝑞q...𝑃 𝑋 p𝑞 𝑛 q𝑠 𝑞

𝑞𝑛

8 ÿ ÿ ÿ p 𝑃 𝑋 p𝑞 1 q𝑠 𝑞1 q...p 𝑃 𝑋 p𝑞 𝑛 q𝑠 𝑞𝑛 q

𝑞 “0 𝑞1

𝑞𝑛

Complexity Science

8 ÿ

8 ÿ

𝑃 𝑁 p𝑛qp 𝑛 “0 8 ÿ

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𝑃 𝑋 p𝑞q𝑠 𝑞 q𝑛

𝑞 “0

𝑃 𝑁 p𝑛qp𝑔 𝑋 p𝑠qq𝑛 “ 𝑔 𝑁 p𝑔 𝑋 p𝑠qq

𝑛“0

b) Group the individuals at generation 𝑛 into 𝑍1 groups according to which of the 𝑍1 individuals at generation 1 they originate from. See figure 5.2. We have 𝑍1 ÿ

𝑍 𝑛 p𝑡q

𝑍𝑛 “ 𝑡 “1

According to the theorem in (i) we have, 𝑔 𝑍𝑛 p𝑠q “ 𝑔 𝑍1 p𝑔 𝑍𝑛 p 𝑡 q p𝑠qq Next note that the generator for 𝑍1 is identical to the generator for the branching probabilities 8 ÿ

𝑔p𝑠q “

𝑃𝑡 𝑠 𝑡

𝑡 “0

with 𝑃𝑡 = proban individual gets offspring.

The generator for 𝑍 𝑛 p𝑡q is the same as the generator for the total size of the generation after 𝑛 ´ 1 generations i.e. 𝑔 𝑍𝑛 p𝑡 q p𝑠q “ 𝑔p𝑔 𝑍𝑛´1 p𝑠qq. c) 𝑔 𝑍𝑛 p𝑠q “ 𝑔p𝑔 𝑍𝑛´1 p𝑠qq

𝑔 𝑍𝑛 p𝑠q “ 𝑔p𝑔 𝑍𝑛´1 p𝑠qq “ 𝛼 ` 𝛽𝑔 𝑍𝑛´1 p𝑠q

“ 𝛼 ` 𝛽p𝛼 ` 𝛽𝑔 𝑍𝑛´2 p𝑠qq “ 𝛼 ` 𝛼𝛽 ` 𝛽2 𝑔 𝑍𝑛´2 p𝑠q “ 𝛼 ` 𝛼𝛽 ` 𝛽2 p𝛼 ` 𝛽𝑔 𝑍𝑛´3 p𝑠qq

.. . “ 𝛼p1 ` 𝛽 ` 𝛽2 ` ... ` 𝛽 𝑛´1 q ` 𝛽 𝑛 𝑠

“𝛼

1 ´ 𝛽𝑛 ` 𝛽𝑛 𝑠 1´𝛽

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Complexity Science From normalisation, 𝑔p1q “ 𝛼 ` 𝛽 “ 1 Ñ Ý 𝛼 “1´𝛽 we obtain, 𝑔 𝑍𝑛 p𝑠q “ 1 ´ 𝛽 𝑛 ` 𝛽 𝑛 𝑠

d) Recall the definition of the generator function for 𝑔 𝑍𝑛 p𝑠q: 8 ÿ

𝑔 𝑍𝑛 p𝑠q “

𝑃𝑟𝑜𝑏t𝑍 𝑛 “ 𝑘u𝑠 𝑘

𝑘 “0

From the result in (iii) we conclude, 𝑃𝑟𝑜𝑏t𝑍 𝑛 “ 0u “ 1 ´ 𝛽 𝑛 𝑃𝑟𝑜𝑏t𝑍 𝑛 “ 1u “ 𝛽 𝑛 𝑃𝑟𝑜𝑏t𝑍 𝑛 “ 𝑘u “ 0 𝑓 𝑜𝑟 𝑘 ě 2 Exercise 3 a) Use the expression for a geometric series 8 ÿ

𝑎𝑥 𝑘 “ 𝑎

𝑘 “0

1 1´𝑥

and 8 ÿ

𝑔p𝑠q “

𝑃𝑘 𝑠 𝑘 “

𝑘 “0 8 ÿ

“

8 ÿ 1´ 𝑝 p𝑝𝑠q 𝑘 “ p1 ´ 𝑝q 1 ´ 𝑝𝑠 𝑘 “0

rp1 ´ 𝑝q𝑝 𝑘 s𝑠 𝑘

𝑘 “0

to conclude that 𝑃 𝑘 “ p1 ´ 𝑝q𝑝 𝑘 b) 𝜇 “ 𝑔 1 p𝑠q|𝑠“1 “

𝑝p1 ´ 𝑝q 𝑝 |𝑠“1 “ 2 1´ 𝑝 p1 ´ 𝑝𝑠q

c) Critical when 𝜇“1Ñ Ý

𝑝 “1 1´ 𝑝

Ñ Ý 𝑝 “1´ 𝑝 Ñ Ý 𝑝“

1 2

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d) The size of the progeny of n generations, 𝑌𝑛 , is given by 𝑛 ÿ

𝑌𝑛 “

𝑍𝑘 𝑘 “1

The generator , 𝑔p𝑠q, for 𝑍 𝑘 and the generator for 𝑌𝑛 are related according to 𝑔𝑌𝑛 p𝑠q “ 𝑔p𝑔𝑌𝑛´1 p𝑠qq (See Eq. 5.30) So we have, 𝑔𝑌𝑛 p𝑠q “

1´ 𝑝 1´ 𝑝 “ 1 ´ 𝑝𝑔𝑌𝑛´1 p𝑠q 1 ´ 𝑝𝑔p𝑔𝑌𝑛´2 p𝑠q

The expression in the question follows by iteration.

Project 1 a) The relation is obtained by use of the self-similar structure of the branching process. Let the first generation have 𝑍1 members. Divide the population at generation 𝑛 into 1 groups. In the 𝑘th group p𝑘q all members originate from member 𝑘 at 1st generation. Let 𝑍 𝑛 denote the number of elements in this group. We now have, 𝑍1 ÿ

𝑍𝑛 “

p𝑘q

𝑍𝑛 𝑘 “1

See Fig. 5.2.

In general, we have for, 𝑁 ÿ

𝑆𝑁 “

𝑋𝑛 𝑛´1

where 𝑁 “ 0, 1, 2, ... is a stochastic variable and 𝑋 𝑘 P 0,1,2,... are i.i.d that the generators are related according to 𝑔𝑆 𝑁 p𝑠q “ 𝑔 𝑁 p𝑔 𝑋 p𝑠qq. See e.g. Feller "An Introduction to Probability theory" V1, Chapter 12 section 1.

In our case, 𝑁 “ 𝑍1 which is generated by the generator for the branching process, i.e. 𝑔p𝑠q and p𝑘q p 𝑋 𝑘 “ 𝑍 𝑛 𝑘q. The generator for 𝑍 𝑛 must, due to the self-similarity, be the same as the generator for the whole tree offer 𝑛 ´ 1 generation i.e. 𝑔 𝑍 p𝑘q p𝑠q “ 𝑔 𝑍𝑛´1 p𝑠q and we conclude from above that, 𝑛

𝑔 𝑍𝑛 p𝑠q “ 𝑔p𝑔 𝑍𝑛´1 p𝑠qq b) The relation 𝑋𝑛 “ 𝑔p𝑋𝑛´1 q follows from (a) by noticing that 𝑥 𝑛 “ 𝑃𝑟𝑜𝑏t𝑍 𝑛 “ 0u “ 𝑔 𝑍𝑛 p0q c) This is most easily seen geometrically by iterations the map 𝑥 𝑛 “ 𝑔p𝑍 𝑛´1 q. Notice 𝑋0 “ 𝑝 0 and that 𝑔 is convex since 𝑔 2 p𝑠q ą 0 and the slope at 𝑠 “ 1, 𝑔 1 p1q “ 𝜇

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d) The average branching ratio 8 ÿ

𝜇“

𝑘 𝑃 𝑘 “ 2𝑝

𝑘 “0

The process is critical for 𝜇 “ 1 i.e. for 𝑝 “ 𝑝 𝑐 “ 21 . e) ă 𝑍 𝑛 ą“ 𝑔 1𝑍𝑛 p1q “

“ r𝑔 1 p𝑔 𝑍𝑛´1 p𝑆qq

𝑑𝑔p𝑔 𝑍𝑛´1 p𝑠qq |𝑠“1 𝑑𝑠

𝑑𝑔p𝑔 𝑍𝑛´1 p𝑠qq 𝑑𝑔p𝑔 𝑍𝑛´2 p𝑠qq s𝑠“1 “ r𝑔 1 p𝑔 𝑍𝑛´1 p𝑠qq𝑔 1 p𝑔 𝑍𝑛´2 p𝑠qq s𝑠“1 𝑑𝑠 𝑑𝑠 ... “ r𝑔 1 p𝑔 𝑍𝑛´1 p𝑠qq𝑔 1 p𝑔 𝑍𝑛´2 p𝑠qq𝑔 1 p𝑔 𝑍0 p𝑠qqs𝑠“1

Next, we have, 𝑔 𝑍𝑛 p1q “ 𝑔p𝑔 𝑍𝑛´1 p1qq “ .. “ 𝑔p𝑔p...p𝑔p1qq...qq Furthermore 𝑔p1q “ 1 and 𝑔 1 p1q “ 𝜇 “ 2𝑝 ă 𝑍 𝑛 ą“ r𝑔 1 p1qs𝑛 “ p2𝑝q𝑛 f) The probability that the process goes extinct is the smallest root, 𝑥 of the equation, 𝑠 “ 𝑔p𝑠q i.e. 𝑠

𝑝𝑠 ` 𝑞 “ 𝑠 ùñ 𝑠 “

1˘

a 1 ´ 4𝑝𝑞 2𝑝

the smallest root is for 𝑝ă

𝑝ą

1 :𝑥“1 2

1 1 𝑞 : ´1“ 2 𝑝 𝑝

Complexity Science g) The generator for the geometric process 8 ÿ

𝑔p𝑠q “

𝑞p𝑝𝑠q 𝑘 “

𝑘 “0

𝑞 1 ´ 𝑝𝑠

Normalising, we get, 1 “ 𝑔p1q “

1 1 𝑞 ùñ “ ùñ 𝑞 ` 𝑝 “ 1 1´ 𝑝 𝑞 1´ 𝑝

Average branching ratio is 𝜇 “ 𝑔 1 p1q “

𝑞𝑝 𝑞𝑝 𝑝 |𝑠“1 “ “ 𝑞 p1 ´ 𝑝𝑠q2 p1 ´ 𝑝q2

and therefore, 1 1 1 ´ 𝑝𝑠 𝑞 ´ “ ´ 1 ´ 𝑔p𝑠q 𝜇p1 ´ 𝑠q 𝑝p1 ´ 𝑠q 𝑝p1 ´ 𝑠q 1 ´ p1 ´ 𝑝q ´ 𝑝𝑠 1 ´ 𝑞 ´ 𝑝𝑠 “ “1 𝑝p1 ´ 𝑠q 𝑝p1 ´ 𝑠q QED h) For simplicity use the notation 𝑔𝑛 p𝑠q “ 𝑔 𝑍𝑛 p𝑠q and use the expression 𝑔𝑛 p𝑠q “ 𝑔p𝑔𝑛´1 p𝑠qq from (a). In the relation from (g), substitute 𝑠 by 𝑔𝑛´1 p𝑠q : r𝑞 ´ 𝑔p𝑔𝑛´1 p𝑠qqs´1 ´ r𝜇p1 ´ 𝑔𝑛´1 p𝑠qqs´1 “ 1 1 1 1 1 “1` “ 1 ` r1 ` s 𝜇 1 ´ 𝑔𝑛 p𝑠q 𝜇p1 ´ 𝑔𝑛´1 p𝑠qq 𝜇p1 ´ 𝑔𝑛´2 p𝑠qq which is a repeat of the relation we see in (a).

1 1 1 1 1 1 ` 𝑛 “ 1 ` ` 2 ` ... ` p 𝜇 𝜇 1 ´ 𝑔0 p𝑠q 1 ´ 𝑔𝑛 p𝑠q 𝜇 𝜇 𝑛 ´ 1q we have 𝑔0 p𝑠q “ 𝑔 𝑍0 p𝑠q “ 𝑠 since the population at generation0 is of size 1 with probability 1.

1 ´ p 𝜇1 q𝑛 1 1 1 “ ` p q𝑛 𝜇 1´𝑠 1 ´ 𝑔𝑛 p𝑠q 1 ´ 𝜇1

1 ´ 𝑔𝑛 p𝑠q “

𝜇 𝑛 p𝜇 ´ 1qp1 ´ 𝑠q 𝜇p𝜇 𝑛 ´ 1qqp1 ´ 𝑠q ` 𝜇 ´ 1

The extinction probability 𝑋𝑛 “ 𝑔𝑛 p0q, thus, 1 ´ 𝑥𝑛 “

𝜇 𝑛 p𝜇 ´ 1q 𝜇 𝑛`1 ´ 1

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Complexity Science

i) Assume 𝜇 ą 1 : 1 ´ 𝜇1 𝜇 𝑛`1 ´ 𝜇 𝑛 1 1 ´ 𝑥 𝑛 “ 𝑛`1 “ Ý 8 Ñ Ý 1 ´ for 𝑛 Ñ 1 𝜇 𝜇 ´1 1 ´ 𝑛`1 𝜇

Assume 𝜇 ă 1 : 1 ´ 𝑥𝑛 “

p𝜇 ´ 1q 𝜇 𝑛 p𝜇 ´ 1q “ 𝜇𝑛 Ñ Ý 0 for 𝑛 Ñ Ý 8 𝑛 ` 1 ´1 𝜇 ´1

j) We proved in the chapter that 𝑝p𝑇q𝛼 𝑇12 for 𝜇 “ 1. See figure in the next page as the simulations approach this behaviour as 𝑝 Ñ Ý 𝑝 𝑐 “ 21 .

Complexity Science

17

18

Complexity Science

Chapter 6 Exercise 1 a) According to Eq. (6.9) ÿ 𝑍“

ÿ

𝑒 ´𝛽𝐸𝑛 “

ÿ

𝑒 𝐽 𝛽𝑆1 𝑆2

𝑆1 “´1,1 𝑆2 “´1,1

𝑛

“ 𝑒 𝑥 p´1qp´1q ` 𝑒 𝑥1p´1q ` 𝑒 𝑥 p´1q.1 ` 𝑒 𝑥.1.1

“ 2p𝑒 𝑥 ` 𝑒 ´ 𝑥 q “ 4 coshp𝑥q b) According to Eq. (6.10), ă 𝐸 ą“ ´

´𝐽

B ln 𝑧 B𝑥 B “´ lnp4 coshp𝑥qq B𝛽 B𝛽 B𝑥 sinhp𝑥q “ ´𝐽 tanhp𝑥q coshp𝑥q

c) 𝑇Ñ Ý 0 ùñ 𝛽 Ñ Ý 8 ùñ 𝑥 Ñ Ý 8 and since lim tanhp𝑥q “ 1

𝑥Ñ Ý8

we have than ă 𝐸 ą“ ´𝐽 𝑇Ñ Ý 8 ùñ 𝛽 Ñ Ý 0 ùñ 𝑥 Ñ Ý 0 and since lim tanhp𝑥q “ 0

𝑥Ñ Ý0

we have lim ă 𝐸 ą“ 0

𝑇Ñ Ý8

Exercise 2 ´𝑥

𝑥

´𝑥

𝑥

The probabilities are 𝑝 1 “ 𝑒2 , 𝑝 1 “ 𝑒 2 , 𝑝 1 “ 𝑒 2 , 𝑝 1 “ 𝑒2 .

According to Eq (6.11), 4 ÿ

𝑆r𝑝 1 , 𝑝 2 , 𝑝 3 , 𝑝 4 s “ ´

𝑝 𝑖 ln 𝑝 𝑖

𝑖 “1

“

´2 𝑥 𝑒𝑥 𝑒´𝑥 p𝑒 lnp q ` 𝑒 ´ 𝑥 lnp qq 𝑍 𝑍 𝑍

Complexity Science

“ ´𝑥

“

´2 p𝑥𝑒 𝑥 ´ 𝑒 𝑥 ln 𝑍 ´ 𝑥𝑒 ´ 𝑥 ´ 𝑒 ´ 𝑥 ln 𝑍q 𝑍

“

´2 2 ln 𝑍 𝑥 𝑥p𝑒 𝑥 ´ 𝑒 ´ 𝑥 q ` p𝑒 ` 𝑒 ´ 𝑥 q 𝑍 𝑍

4 cosh 𝑥 4 sinh 𝑥 ` ln 𝑍 “ lnp4 cosh 𝑥q ´ 𝑥 tanh 𝑥 4 cosh 𝑥 4 cosh 𝑥

Exercise 3 In the microcanonical ensemble, 𝑝 𝑖 “ Ω1 for all 𝑖 “1,2,...,Ω

𝑆“´

Ω ÿ 1

Ω 𝑖 “1

ln

1 1 1 “ ´Ω ln “ ln Ω Ω Ω Ω

Exercise 4 a) 𝑇 Ñ Ý 0Ñ Ý 𝛽Ñ Ý 8Ñ Ý 𝑥Ñ Ý 8 𝑃´1,´1 “ 𝑃1,1 “

𝑒𝑥 1 𝑒𝑥 Ñ Ý “ 𝑥 4 cosh 𝑥 2 2p𝑒 ` 𝑒 ´ 𝑥 q

for 𝑇 Ñ Ý 0 𝑃´1,1 “ 𝑃1,´1 “

𝑒´𝑥 Ñ Ý 0 4 sinh 𝑥

for 𝑇 Ñ Ý 0 Entropy in the limit 𝑇 Ñ Ý 0 1 1 1 1 1 𝑆r , 0, 0, s “ ´2 ln “ ´ ln “ ln 2 2 2 2 2 2 b) 𝑇 Ñ Ý 8Ñ Ý 𝛽Ñ Ý 0Ñ Ý 𝑥Ñ Ý 0 𝑃´1,´1 “ 𝑃1,1 “

𝑒𝑥 1 Ñ Ý 4 2p𝑒 𝑥 ` 𝑒 ´ 𝑥 q

𝑃´1,1 “ 𝑃1,´1 “

𝑒´𝑥 1 Ñ Ý 4 2p𝑒 𝑥 ` 𝑒 ´ 𝑥 q

for 𝑇 Ñ Ý 8

for 𝑇 Ñ Ý 8 Entropy in the limit 𝑇 Ñ Ý 8 1 1 1 1 1 1 𝑆r , , , s “ ´4 ln “ ln 4 4 4 4 4 4 4

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Complexity Science

c) From Eq. (6.121) 𝑆r𝑝 1 , 𝑝 2 , 𝑝 3 , 𝑝 4 s “ lnp4 cosh 𝑥q ´ 𝑥 tanh 𝑥 𝑇Ñ Ý 0Ñ Ý 𝑥Ñ Ý 8 𝑆 “ lnp4

“ lnp2𝑒 𝑥 q ´ 𝑥

𝑒 𝑥 ` 𝑒´𝑥 𝑒 𝑥 ´ 𝑒´𝑥 q´𝑥 𝑥 2 𝑒 ` 𝑒´𝑥

1 ´ 𝑒 ´2𝑥 “ ln 2 ` 𝑥 ´ 𝑥p1 ´ 𝑒 ´2𝑥 qp1 ´ 𝑒 ´2𝑥 q 1 ` 𝑒 ´2𝑥

ln 2 ` 𝑥 ´ 𝑥p1 ´ 2𝑒 ´2𝑥 ` 𝑒 ´4𝑥 q Ñ Ý ln 2 𝑇Ñ Ý 8Ñ Ý 𝑥Ñ Ý 0 𝑆Ñ Ý lnp4 coshp0qq ´ 0 tanhp0q “ ln 4 Exercise 5 a) Since each position is assumed to be occupied by the same probability, we have in box 𝑖 that 𝑝 𝑖 “ 𝐿12 𝑖

p𝑖 q

b) Same as above 𝑃𝑒𝑠𝑐𝑎 𝑝𝑒 “ 𝑝 𝑖 “ 𝐿12 𝑖 c) Particle must be at the site next to the exit, and since we assume equal probability for each site, this 1 occurs with prob 𝐿 2 . To move to box 𝑖 ` 1, the particle needs to move to the right which is one 𝑖

specific direction out of three, so occurs with probability 31 . Hence, 𝑃𝑖 Ñ Ý 𝑖`1 “

1 1 3 𝐿 𝑖2

d) Consideration similar to above gives: 𝑃𝑖 Ñ Ý 𝑖`1 “

1 1 3 𝐿 𝑖2`1

e) The current = 𝑃𝑖Ñ Ý 𝑖`1 ´ 𝑃𝑖`1Ñ Ý𝑖 “

1 1 1 1 1 1 r 2𝑖 ´ 2𝑖`2 s “ r1 ´ s 2𝑖 3 2 32 4 2 1 1 “ 2p𝑖`1q 2𝑖 4p2 q 2

Exercise 6 ˜

˜

˜

˜

˜

𝑒 𝐽 `𝐵 ´ 𝜆 𝑒´𝐽 “0 ˜ ˜ ˜ 𝑒´𝐽 𝑒 𝐽 ´𝐵 ´ 𝜆 ˜

˜

˜

˜

𝑒 2 𝐽 ´ 𝜆p𝑒 𝐽 ` 𝐵 ` 𝑒 𝐽 ´ 𝐵 q ` 𝜆2 ´ 𝑒 ´2 𝐽 “ 0

Complexity Science

21

˜

𝜆2 ´ 2𝑒 𝐽 cosh 𝐵˜ ` 2 sinh 2𝐽˜ “ 0 1 ˜ ˜ ˜ 12 q p2𝑒 𝐽 cosh 𝐵˜ ˘ r4𝑒 2 𝐽 cosh2 𝐵˜ ´ 4 sinh 2𝐽s 2

𝜆“

Using cosh2 𝑥 ´ sinh2 𝑥 “ 1, ˜ ˜ ˜ ´ 2 sinh 2𝐽s ˜ 12 𝜆 “ 𝑒 𝐽 cosh 𝐵˜ ˘ r𝑒 2 𝐽 p1 ` sinh2 𝐵q

˜

˜

˜

˜

˜

1

“ 𝑒 𝐽 cosh 𝐵˜ ˘ r𝑒 2 𝐽 sinh2 𝐵˜ ` 𝑒 2 𝐽 ´ 𝑒 2 𝐽 ` 𝑒 ´2 𝐽 s 2 ˜ ˜ ˜ 1 𝜆 “ 𝑒 𝐽 cosh 𝐵˜ ˘ r𝑒 2 𝐽 sinh2 𝐵˜ ` 𝑒 ´2 𝐽 s 2

Exercise 7 We can think of the location of the anti-parallel pair Ö or Œ as similar to the location of the centre of a vortex or an anti-vortex. Since the Ö configuration only changes the energy of the two anti-parallel and leaves the other spins unperturbed, the energy of a single Ö does not depend on the system size. This is in contrast to a single vortex in a system of size 𝐿 2 . The energy of the vortex increases as log 𝐿. The energy of the configuration ÒÓÓ ... ÓÒÒ ... does not depend on the separation between Ö and Œ. This is different of a vortex-anti-vortex pair of separation 𝑅. The energy of the pair grows like log 𝑅.

Exercise 8 a) We determine BB𝐻𝜙 from 𝐻r𝜙 ` B𝜙s ´ 𝐻r𝜙s “

𝐽 𝑎

ż8 ´8

𝑑𝑥r´𝑎 2

𝑑2 𝜙 ℎ ` 𝛿𝜙 sin 𝜙s ` 𝑂p𝛿𝜙2 q... 𝐽 𝑑𝑥 2

and conclude 𝑑2 𝜙 B𝐻 ℎ “ ´𝐽𝑎 2 ` sin 𝜙 B𝜙 𝑎 𝑑𝑥 Hence, B𝐻 𝑑2 𝜙 ℎ “0Ñ Ý “ 2 sin 𝜙 “ 𝜇2 sin 𝜙 B𝜙 𝑑𝑥 2 𝐽𝑎 2

b) We demonstrate that 𝜙p𝑥q “ 4 tan´1 p𝑒 𝜇𝑥 q satisfies 𝑑𝑑 𝑥𝜙2 “ 𝜇 sin 𝜙 by use of the following identities 𝑑 1 tan´1 𝑥 “ 𝑑𝑥 1 ` 𝑥2 tan´1 𝑥 “

´𝑖 1 ` 𝑖𝑥 ln 2 1 ´ 𝑖𝑥

22

Complexity Science

sin 𝑥 “

𝑒 𝑖 𝑥 ´ 𝑒 ´𝑖𝑥 2𝑖

First, 4𝜇 𝑑𝜙 4𝜇𝑒 𝜇𝑥 2𝜇 “ ´ 𝜇𝑥 “ “ 𝑑𝑥 𝑒 ` 𝑒 𝜇𝑥 cosh 𝜇𝑥 1 ` 𝑒 2𝜇𝑥 p 1 ´ 𝑦q ´2𝜇2 sinh 𝜇𝑥 𝑑2 𝜙 2 𝑦 “ “ 4𝜇 1 𝑑𝑥 2 p 𝑦 ` 𝑦q2 cosh2 𝜇𝑥 with 𝑦 ” 𝑒 𝜇𝑥 c) For 𝑣 ă 𝑁2 𝐸, 𝛽1 increases with increasing energy. Similar to what we usually expect the temperature does. The singularity at 𝑣 “ 𝐸 𝑁2 and negative values for 𝑣 ą 𝐸 𝑁2 do not correspond to our usual expectations of how the (absolute) temperature should behave. However, see (g). d) Negative temperature can occur when the allowed number of states decreases with increasing energy.

Project 1 a) We assume 𝑁 and all 𝑛𝑖 large and use Stirling’s approximation ln 𝑛! » 𝑛 ln 𝑛 ´ 𝑛 𝑅 𝑅 𝑅 ÿ ÿ ÿ B𝐽 B “ t𝑁 ln 𝑁 ´ 𝑁 ´ p𝑛 𝑗 ln 𝑛 𝑗 ´ 𝑛 𝑗 q ` 𝛼p𝑁 ´ 𝑛 𝑗 q ` 𝛽p𝑣 ´ 𝜖 𝑗𝑛𝑗q B𝑛𝑖 B𝑛𝑖 𝑗 “1 𝑗 “1 𝑗 “1

“ ln 𝑁 ´ ln 𝑛𝑖 ´ 𝛼 ´ 𝛽𝜖𝑖 A word of caution- The 𝑁 and the 𝑣 in the 𝛼 and 𝛽 terms are the constraints and should not be differentiated with respect to 𝑛𝑖 .

We now get, B𝐽 “0Ñ Ý 𝑛𝑖 “ 𝑁 expp𝛼 ` 𝛽𝜖𝑖 q B𝑛𝑖 ř ř b) From 𝑁 “ 𝑖 𝑛𝑖 “ 𝑁 𝑖 expp𝛼 ` 𝛽𝜖𝑖 q ÿ Ñ Ý exp 𝛼 “ p 𝑒 𝛽 𝜖𝑖 q´1 ” 𝑧 ´1 𝑖

c) ÿ 𝑈“

𝜖 𝑖 𝑛𝑖 “

ÿ 𝜖 𝑖 𝑒 𝛽 𝜖𝑖

𝑖

𝑧 𝑖

𝛽𝜖

𝛽𝜖

d) 𝑛1 “ 1`𝑁𝑒 𝛽 𝜖 , 𝑛2 “ 1𝑁`𝑒𝑒 𝛽 𝜖 𝑈 “ 0𝑛1 ` 𝜖𝑛2 “ 𝜖 𝑁 1`𝑒 𝑒 𝛽 𝜖 Let 𝑥 “ 𝜖𝑣𝑁 and 𝑦 “ 𝛽1𝜖 then we get 𝑦 “ ln 1𝑥 1´ 𝑥

And 𝑦 ą 0 for 1´𝑥 𝑥 ą 1 Ñ Ý 𝑥 ą 12 𝑦 ă 0 for 1´𝑥 𝑥 ă 1 Ñ Ý 𝑥 ă 12 We notice 𝑦 “ 𝛽1𝜖 for 𝑥 “ 𝜖 𝑁 ă 12 . We are free to change the sign of the Lagrange multiplier 𝛽 in the expression for 𝑦p𝑛q. When we do this, we get 𝛽 Ñ Ý ´𝛽 ùñ 𝑦 Ñ Ý ´𝑦 and we conclude: 𝑣 1 ă Ñ Ý 𝛽ą0 𝜖𝑁 2

Complexity Science

23

𝑣 1 𝜖 𝑁 approaching 2 from below gives us

𝑦“

1 Ñ Ý `8 𝛽𝜖

𝑦“

1 Ñ Ý ´8 𝛽𝜖

And for 𝜖𝑣𝑁 just above 12 we have,

Ý 𝑦“0 Finally, when 𝜖𝑣𝑁 “ 1 Ñ

e) Project 2 a) 𝐸 1 “ 𝐻rone kinks ´ 𝐻rno kinks 𝐽r´p𝑁 ´ 1q ` loomo 1 ons ´ 𝐽p´𝑁q “ 2𝐽 loooomoooon kink

parallel pairs

b) Each of the 𝑁 bonds can contain no kink, energy 𝐸 0 “ 0, or a kink, energy 𝐸 1 “ 2𝐽. Partition function ÿ

ÿ

𝑍“

... p1q

𝐸𝑖 “0,2𝐽

ÿ “r

p𝑁 q

𝐸𝑖

“0,2𝐽

expp´𝛽

𝑁 ÿ

p𝑘q

𝐸𝑖 q 𝑘 “1

expp´𝛽𝐸 𝑖 qs 𝑁 “ p1 ` 𝑒 ´𝛽𝐸1 q 𝑁

𝐸𝑖 “0,2𝐽

c) The simplest way to derive the result is by noting that the energy of a state is given by the number of kinks 𝐸 “ 𝑛𝐸 1 Ñ Ý ă 𝐸 ą“ 𝐸 1 ă 𝑛 ą and it follows from Eq (6.10) that 1 B ă 𝑛 ą“ 𝐸11 ă 𝐸 ą“ ´ 𝐸1 B 𝛽 ln 𝑍. d) 1 B ă 𝑛 ą“ ´ lnp1 ` 𝑒 ´𝛽𝐸1 q 𝑁 𝐸 1 B𝛽

24

Complexity Science

´

𝑁 ´𝐸 1 𝑒 ´𝛽𝐸1 𝑁 “ 𝐸 1 1 ` 𝑒 ´𝛽𝐸1 1 ` 𝑒 𝛽𝐸1

Therefore, ă 𝑙 ą“

Project 3 No sample solution needed

𝑁 “ 1 ` 𝑒 𝛽𝐸1 “ ă𝑛ą

"

8, for 𝑇 Ñ Ý 0 ùñ 𝛽 Ñ Ý 8 2, for 𝑇 Ñ Ý 8 ùñ 𝛽 Ñ Ý 0

Complexity Science

25

Chapter 7 Exercise 1 a) Define Δ “ 𝜃 1 ´ 𝜃 2 and 𝛼 “ 𝜔1 ´ 𝜔2 . 𝑑Δ “ 𝛼 ` 𝜖p𝜃 2 ´ 𝜃 1 q ´ 𝜖p𝜃 1 ´ 𝜃 2 q “ 𝛼 ´ 2𝜖Δ 𝑑𝑡 𝑑Δ ` 2𝜖Δ “ 𝛼 𝑑𝑡 𝐼

𝐼

𝑑Δ ` 2𝜖Δ “ 𝛼 𝑑𝑡

𝑑Δ ` 2𝜖 𝐼Δ “ 𝐼𝛼 𝑑𝑡

ş Which with 𝐼 “ expp 2𝑒 𝑝𝑠𝑖𝑙𝑜𝑛𝑑𝑡q “ 𝑒 2𝜖 𝑡 q is equivalent to 𝑑p𝐼Δq 𝛼 2𝜖 𝑡 “ 2𝜖 𝐼Δ ùñ 𝑒 2𝜖 𝑡 Δ “ 𝑒 “𝐴 𝑑𝑡 2𝜖 𝛼 . where 𝐴 is a constant determined by the initial condition. We conclude Δ “ 𝐴𝑒 ´2𝜖 𝑡 ` 2𝜖

𝜃 1 “ 𝜃 2 ` Δ “ 𝜃 2 ` 𝐴´2𝜖 𝑡 `

Ñ Ý 𝜃2 ´

𝜔1 ´ 𝜔2 2𝜖

𝜔2 ´ 𝜔2 for 𝑡 Ñ Ý 8 2𝜖

b) Let again Δ “ 𝜃 1 ´ 𝜃 2 𝑑Δ “ 𝜔Δ ` 𝜖r𝜃 2 ´ 𝜃 1 ´ 𝜃 1 ` 𝜃 2 s 𝑑𝑡 “ 𝜔Δ ´ 2𝜖Δ “ p𝜔 ´ 2𝜖qΔ Δp𝑡q “ Δp0q𝑒 p 𝜔´2𝜖 q𝑡 Therefore, we have, For 𝜔 ą 2𝜖 Ñ Ý Δp𝑡q Ñ Ý 8 with 𝑡 Ñ Ý 8 i.e. no synchronisation.

For 𝜔 ă 2𝜖 Ñ Ý Δp𝑡q Ñ Ý 0 with 𝑡 Ñ Ý 0 i.e. synchronisation.

Exercise 2 Go through equations (7.1) to (7.19) step by step.

Exercise 3 Fig. 7.1 shows that if 𝐴 ą 𝐵, the function 𝜓 will increase monotonously with time. In terms of Eq. (7.6) this corresponds to rotors for which |𝜔 𝑘 ´ 𝜖| ą 𝜖 𝐾. For these rotors, 𝜓 𝑘 will keep increasing with time, and therefore the corresponding phase 𝜃 𝑘 p𝑡q in Eq. (7.5) will not lock to 𝜔𝑡. ¯ So, these rotors will not synchronize.

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Complexity Science

Project 1 a) Fourier transformation: ż 𝑖𝑛 𝑓 𝑡 𝑦 𝜃p𝑥, 𝑡q “ 8

B2 𝜖 “ B𝑥 2

𝑑𝑘 ˆ 𝜃p𝑘, 𝑡q𝑒 ´𝑖𝑘 𝑥 and 2𝜋

ż8

𝑑𝑘 ˆ ´𝑖𝑘 𝑥 p´𝑘 2 𝜃q𝑒 2𝜋 8

The F.T of the equation becomes ˆ 𝑡q “ Ω̂p𝑘q ´ 𝛾𝑘 2 𝜃p𝑘, ˆ 𝑡q B𝑡 𝜃p𝑘, B𝑡 𝜃ˆ ` 𝛾𝑘 2 𝜃ˆ “ Ω̂p𝑘q ş 2 Integration factor 𝐼 “ expr 𝑑𝑡𝛾𝑘 2 s “ 𝑒 𝛾𝑘 𝑡 . Multiply eq. by IF and get, 2 B 𝛾𝑘 2 𝑡 ˆ p𝑒 𝜃q “ 𝑒 𝛾𝑘 𝑡 Ω̂ B𝑡

2

2

𝑒 𝛾𝑘 𝑡 Ω̂ “

𝑒 𝛾𝑘 𝑡 Ω̂ ` 𝐴p𝑘q 𝛾𝑘 2

ˆ 𝑡q “ Ω̂ ` 𝐴p𝑘q𝑒 “𝛾𝑘 2 𝑡 𝜃p𝑘, 𝛾𝑘 2 ˆ 𝑡q “ 𝜙p𝑘q ˆ where 𝐴p𝑘q is independent of t. Use that for 𝑡 “ 0, we have 𝜃p𝑘, to eliminate 𝐴p𝑘q. ˆ 𝑡q “ Ω̂ ` p𝜙ˆ ´ Ω̂ q𝑒 ´𝛾𝑘 2 𝑡 𝜃p𝑘, 𝛾𝑘 2 𝛾𝑘 2 b) ż8 Δ“

𝑑𝑥|B 𝑥 𝜃p𝑥, 𝑡q|2

´8

ż8 “

𝑑𝑥|B 𝑥ˆ 𝜃p𝑥, 𝑡q|2 Ð Ý According to Plancherel theorem

´8

ż8 Δ“

ˆ 𝑡q|2 “ 𝑑𝑘| ´ 𝑖𝑘 𝜃p𝑘,

´8

ż8

ˆ 𝑡q|2 𝑑𝑘|𝑖 𝜃p𝑘,

´8

ż8 “

2 𝑑𝑘 Ω̄ Ω̂ | ` 𝑘p𝜙ˆ ´ 2 q𝑒 ´𝛾𝑘 𝑡 |2 2𝜋 𝛾𝑘 𝛾𝑘 ´8

ż8 lim Δ “

𝑡Ñ Ý8

𝑑𝑘 Ω̂ 2 | | 2𝜋 𝛾𝑘 ´8

c) Synchronisation either complete (i.e. 𝜃 𝑛 p𝑡q “ 𝜃 𝑛`1 p𝑡q for all 𝑛) or phase locking (i.e. 𝜃 𝑛 p𝑡q “ 𝜃 𝑛`1 p𝑡q ` 𝛼𝑛`1 q corresponds to Δp𝑡q becoming independent of 𝑡 as 𝑡 Ñ Ý 8.

Complexity Science

27

d) We have used continuous 𝑥 as an approximation. In reality, the rotors 𝜃 𝑛 sit at discrete positions 𝑥 𝑛 “ 𝜎𝑛, where 𝜎 is the separation between rotors. To respect that the shortest period of spatial variation is given by 𝜎 we need the Fourier 𝑘 to satisfy |𝑘| ă 2𝜎𝜋 . Let us also assume the chain is of finite length 𝐿. The longest period of variation must be smaller than 𝐿 and therefore |𝑘| ą 2𝐿𝜋 ” 𝜖.

To ensure that the lim𝑡 Ñ Ý 8 and therefore 𝜖 Ñ Ý 0 we need | Ω̂𝑘 |2 to Ý 8 Δ remains finite as we let 𝐿 Ñ ˆ » 𝑘 𝑎 for 𝑘 Ñ be integrable near 𝑘 “ 0. Assume |Ωp𝑘q| Ý 0. This means we need 𝑘 2𝑎´2 to be ş 2𝑎´2 𝑘 2𝑎´1 integrable for 𝑘 Ñ Ý 0. Since, 𝑘 𝑑𝑘 “ 2𝑎´1 we need 2𝑎 ´ 1 ą 0 for the integral to exist in the 1 limit 𝜖 Ñ Ý 0. So, 𝑎 ą 2 i.e. Ω̂p𝑘q must not approach zero as 𝑘 Ñ Ý 0 more slowly than the square root.

Chapter 8 Exercise 1 Size of the connected component 𝑁 “ 11. Node 𝛼:

ÿ

𝑙 𝛼 𝑗 “ 1 ` 3 ˆ 2 ` 3 ` 3 ˆ 2 “ 14 Ñ Ý

𝑗

ùñ 𝐶 𝛼 “

1 ÿ 14 𝑙 𝛼𝑗 “ 𝑁 𝑗 11

11 » 0.79 14

Node 𝛾: ÿ

𝑙 𝛾 𝑗 “ 3 ` 2 ` 2 ˆ 3 ` 3 ` 3 ˆ 4 “ 26

𝑗

1 ÿ 26 11 Ñ Ý 𝑐𝛾 “ » 0.42 𝑙𝛾𝑗 “ 𝑁 𝑗 11 26 Exercise 2 No sample solution Exercise 3 We move a node from clusters 𝐴 to cluster 𝐵

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Complexity Science

We need only to consider the two clusters 𝐴 and 𝐵. Before:

𝑄r𝐶s “

ÿ 𝑘𝑖 𝑘 𝑗 𝑘𝑎𝑘 𝑗 𝑘𝑖 𝑘 𝑗 1 1 ÿ 𝑘𝑖 𝑘 𝑎 1 ÿ 1 ÿ p𝑎 𝑖 𝑗 ´ q` p𝑎 𝑖𝑎 ´ q` p𝑎 𝑎 𝑗 ´ q` q p𝑎 𝑖 𝑗 ´ 2𝐿 𝑖, 𝑗 P 𝐴´𝑎 2𝐿 2𝐿 𝑖P 𝐴´𝑎 2𝐿 2𝐿 𝑗 P 𝐴´𝑎 2𝐿 2𝐿 𝑖, 𝑗 P 𝐵 2𝐿

where 𝐴 ´ 𝑎 denotes all nodes in 𝐴 different from node 𝑎. After,

𝑄r𝐶 1 s “

ÿ 𝑘𝑖 𝑘 𝑗 𝑘𝑖 𝑘 𝑗 𝑘𝑎𝑘 𝑗 1 1 ÿ 1 ÿ 1 ÿ 𝑘𝑖 𝑘 𝑎 q` q` q` q p𝑎 𝑖 𝑗 ´ p𝑎 𝑖 𝑗 ´ p𝑎 𝑖𝑎 ´ p𝑎 𝑎 𝑗 ´ 2𝐿 𝑖, 𝑗 P 𝐴´𝑎 2𝐿 2𝐿 𝑖, 𝑗 P 𝐵 2𝐿 2𝐿 𝑖P 𝐵 2𝐿 2𝐿 𝑗 P 𝐵 2𝐿 The change Δ𝑄 “ 𝑄r𝐶s ´ 𝑄r𝐶 1 s “

ÿ 1 ÿ 𝑘𝑖 𝑘 𝑎 𝑘𝑖 𝑘 𝑎 r p𝑎 𝑖𝑎 ´ q ´ p𝑎 𝑖𝑎 ´ qs 𝐿 𝑖P 𝐴´𝑎 2𝐿 2𝐿 𝑖P𝐵

Exercise 4 We use the expression for 𝑄r𝐶s in Eq.(8.34)

𝑄r𝐶s “

𝐾 𝐾 1 ÿ 1 ÿ ÿ 2 𝐿 𝐶𝑞 ´ p 𝑘 q 𝐿 𝑞 “1 p2𝐿q2 𝑞“1 𝑖P𝐶 𝑖 𝑞

We have the total number of nodes 𝑁 “ 12 and the total number of links 𝐿 “ 15.

Partition (a) 𝑞 “ 1 : 𝐿 𝐶1 “ 4,

ÿ

𝑘 𝑖2 “ 1 ` 52 ` 22 ` 32 “ 39

𝑖 P𝐶1

𝑞 “ 2 : 𝐿 𝐶2 “ 3,

ÿ

𝑘 𝑖2 “ 1 ` 42 ` 22 ` 32 ` 22 “ 34

𝑖 P𝐶2

𝑞 “ 3 : 𝐿 𝐶3 “ 5,

ÿ 𝑖 P𝐶3

𝑘 𝑖2 “ 42 ` 22 ` 32 ` 22 “ 33

Complexity Science

𝑄r𝐶𝑎 s “

29

1 1 p39 ` 34 ` 33q » 0.682 p4 ` 3 ` 5q ´ 15 p30q2

In the second partition, 𝐾 “ 2, 𝑞 “ 1 : 𝐿 𝐶1 “ 10,

ÿ

𝑘 𝑖2 “ 1 ` 52 ` 22 ` 32 ` 22 ` 52 ` 22 ` 1 “ 73

𝑖 P𝐶1

𝑞 “ 2 : 𝐿 𝐶2 “ 5,

ÿ

𝑘 𝑖2 “ 42 ` 22 ` 32 ` 22 “ 33

𝑖 P𝐶2

𝑄r𝐶𝑏 s “

1 1 p73 ` 33q “ 0.882 p10 ` 5q ´ 15 p2𝐿q2

Exercise 5 Probably most easily done by writing a general algorithm.

Exercise 6 Develop the algorithm from Exercise 5 so it can perform steps [1] and [2] of the Louvain procedure.

Exercise 7 a) The more modern view represented by the Leiden algorithm avoids, e.g., disconnected communities. b) The Leiden algorithm is presumably for speed and better partitioning. c) One would probably like to compare with some social structures such as year groups, departments, faculties, campus, student societies, etc.

Exercise 8 The generator for the degree distribution 𝑃 𝑘 “ p1 ´ 𝑒 ´𝜆 q𝑒 ´𝜆𝑘 Eq. (8.59) is given by 8 ÿ

𝑔r𝑠s “

𝑃 𝑘 𝑠 𝑘 “ p1 ´ 𝑒 ´𝜆 q

8 ÿ

𝑒 ´𝜆𝑘 𝑠 𝑘 “

𝑘 “0

𝑘 “0

1 ´ 𝑒 ´𝜆 𝑒𝜆 ´ 1 “ 1 ´ 𝑠𝑒 ´𝜆 𝑒𝜆 ´ 𝑠

Note 𝑔 1 p𝑠q “

8 ÿ

𝑘 “0 8 ÿ

𝑔”p𝑠q “

𝑘 𝑃 𝑘 𝑠 𝑘 ´1 ùñ 𝑔 1 p1q “

8 ÿ

𝑘 𝑃 𝑘 “ă 𝑘 ą 𝑘 “0

𝑘p𝑘 ´ 1q𝑃 𝑘 𝑠 𝑘 ´2 ùñ 𝑔”p1q “ă 𝑘p𝑘 ´ 1q ą“ă 𝑘 2 ą ´ ă 𝑘 ą

𝑘 “2

For the generator belonging to the exponential distribution, we have,

30

Complexity Science

𝑔 1 p𝑠q “ p𝑒 𝜆 ´ 1q

𝑔”p𝑠q “ p𝑒 𝜆 ´ 1q

1 1 ùñ 𝑔 1 p1q “ 𝜆 𝑒 ´1 p𝑒 𝜆 ´ 𝑠q2

2 2 ùñ 𝑔”p1q “ 𝜆 p𝑒 𝜆 ´ 𝑠q3 p𝑒 ´ 1q2

So, the M-R condition becomes given by:

ă 𝑘 2 ą ´2 ă 𝑘 ą“ă 𝑘 2 ą ´ ă 𝑘 ą ´ ă 𝑘 ą“ 𝑔”p1q´𝑔 1 p1q “

2 1 3 ´ 𝑒𝜆 ´ “ ą0 p𝑒 𝜆 ´ 1q2 𝑒 𝜆 ´ 1 p𝑒 𝜆 ´ 1q2

ùñ 𝑒 𝜆 ă 3 ùñ 𝜆 ă ln 3 . which is the same conclusion as reached after Eq. (8.64). For 𝜆 “ ln 3 ùñ ă 𝑘 ą“ p𝑒ln 31´1q “ 21 Exercise 9 Each time the walker reaches a new node, it finds with probability: ˆ 𝑃𝑘 “

˙ 𝑁 ´2 𝑘 𝑝 p1 ´ 𝑝q 𝑁 ´2´ 𝑘 𝑘

k new edges to proceed along. (There are only 𝑁 ´ 2 nodes to connect with, since going back is not allowed and self-loops are illegal). The generator for this branching process is 𝑁ÿ ´2 ˆ

𝑔p𝑠q “ 𝑘 “0

˙ 𝑁 ´2 𝑘 𝑝 p1 ´ 𝑝q 𝑁 ´2´ 𝑘 𝑠 𝑘 “ p𝑝𝑠 ` 1 ´ 𝑝q 𝑁 ´2 𝑘

. The average branching ratio is 𝜇 “ 𝑔 1 p𝑠q|𝑠“1 “ p𝑁 ´ 2q𝑝 when 𝜇 ă 1 the probability for extinction is 1. For 𝜇 ě 1, the walk (branching process) may continue forever i.e. the sought value for p is 𝑝 𝑐 “ p 𝑁 1´2q » 𝑁1 . Note: strictly speaking, we should exclude all previously visited nodes to avoid loops. But in the limit, 𝑁 Ñ Ý 8, we can ignore this.

The Mean Field Equation 𝑃8 is given by the probability that an occupied site belongs to the spanning cluster. Hence, 1 ´ 𝑃8 “ Prob. that no neighbour site belongs to the spanning cluster “ p1 𝑝 𝑃8 q𝑞 The critical exponent 𝛽 describes how 𝑃8 becomes non-zero, 𝑃8 9p𝑝 ´ 𝑝 𝑐 q𝛽 i.e. we seek solutions to 1 ´ 𝑥 “ 𝑓 p𝑥q for small x. Hence, using the Taylor expansion, we expand 𝑓 p𝑥q about 𝑥 “ 0.

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31

𝑓 p0q “ 1, 𝑓 1 p0q “ ´𝑝𝑞p1 ´ 𝑝𝑥qp𝑞´1q | 𝑥 “0 “ ´𝑝𝑞 𝑓 ”p0q “ 𝑝 2 𝑞p𝑞 ´ 1qp1 ´ 𝑝𝑥qp𝑞´2q | 𝑥 “0 “ 𝑝 2 𝑞p𝑞 ´ 1q So, this becomes: 1 ´ 𝑥 “ 1 ´ 𝑝𝑞𝑥 `

1 2 𝑝 𝑞p𝑞 ´ 1q𝑥 2 ùñ 2

ùñ 𝑥 “ 1 2

𝑝𝑞 ´ 1 𝑝 2 𝑞p𝑞 ´ 1q

Now substitute 𝑝 “ 𝑝 𝑐 ` Δ “ 𝑞1 ` Δ 𝑥 » 2𝑞Δr1 ´

𝑃8 »

1 ` 𝜃pΔqs´1 𝑞

2𝑞 2 p𝑝 ´ 𝑝 𝑐 q ùñ 𝛽 “ 1 𝑞´1

Project 2 No sample solution needed.

Project 3 a) The degree distribution is given by the binomial distribution ˆ ˙ 𝑁 ´1 𝑘 𝑃𝑘 “ 𝑝 p1 ´ 𝑝q 𝑁 ´2 𝑘 For large 𝑁, we have 𝑁 ´ 1 » 𝑁 and for fixed 𝑝𝑁 “ 𝜆 the binomial is well approximated by the Poisson distribution i.e. 𝜆 𝑘 𝑒 ´𝜆 𝑃𝑘 “ 𝑘!

The connectivity, see Eq(8.1), 𝐶 “ 𝑁 p 2𝐿 𝑁 ´1q is the average number of links 2𝐿 “ă

𝑁 ÿ

𝑘 𝑖 ą“ 𝑁 ă 𝑘 𝑖 ą“ 𝑁𝜆 “ 𝑁 2 𝑝

𝑖 “1

Therefore, 𝐶“

𝑁2 𝑝 »𝑃 𝑁p𝑁 ´ 1q

b) According to Eq(8.99), the return time for a node 𝑖 of degree 𝑘 𝑖 is given by 𝑟 𝑖 “ 2𝐿 𝑘𝑖 .

To average over degree 𝑘 𝑖 ą 0, we need to renormalise 𝑃 𝑘 “ 𝛼 𝑃˜ 𝑘 such that ř8 𝑃𝑘 ˜ 𝑘 “1 𝑃 𝑘 “ 1 ´ 𝑝 0 , so 𝑃 𝑘 “ 1´ 𝑝0 for 𝑘 ě 1.

ř8

˜

𝑘 “1 𝑃 𝑘 “ 1. We have

32

Complexity Science ˜ 𝑞 , 𝑘 𝑖 q “ 𝑃˜ 𝑘𝑞 𝑃˜ 𝑘𝑖 .

We neglect degree correlations and have and have for the joint probability 𝑃p𝑘 This gives, ř𝑁

2𝐿 ă 𝑟 𝑖 ą“ă ą“ă 𝑘𝑖

“1`

𝑞 “1 𝑘 𝑞

𝑘𝑖

𝑘𝑖

`1ą

𝑖

8 ÿ 8 ÿ ÿ 𝑘 𝑞 𝜆 𝑘𝑞

𝑒 ´2𝜆 𝑘 𝑘 ! p1 ´ 𝑒 ´𝜆 q2 𝑞 ‰𝑖 𝑘 “1 𝑘 “1 𝑖 𝑖 𝑞

1`

𝑞 ‰𝑖

8 ÿ 8 ÿ ÿ 𝑘𝑞 ˜ 𝑞, 𝑘𝑖q 𝑃p𝑘 𝑘 𝑞 ‰𝑖 𝑘 “1 𝑘 “1 𝑖 𝑞

1`

ÿ 𝑘𝑞 ą“ă

𝑖

8 ÿ ÿ 1 𝜆 𝑘𝑞 ´1 ÿ 𝜆 𝑘𝑖 𝜆 8 p𝑒 𝜆 ´ 1q2 𝑞‰𝑖 𝑘 “1 p𝑘 𝑞 ´ 1q! 𝑘 “1 𝑘 𝑖 𝑘 𝑖 ! 𝑞

“1`

𝑖

𝜆𝑒 𝜆 p𝑒 𝜆 ´ 1q2

𝑁 𝑆p𝜆q

Here we defined, 8 ÿ 𝜆𝑘

𝑆p𝜆q “

Note 𝑆p0q “ 0

. 𝑘 “1

𝑘 𝑘!

we have, 𝑆 1 p𝜆q “

8 ÿ 𝜆 𝑘 ´1

𝑘!

𝑘 “1

ż 𝑆p𝜆q “

8

“

1 ÿ 𝜆𝑘 𝑒𝜆 ´ 1 “ 𝜆 𝑘 “1 𝑘! 𝜆

𝑒𝜆 𝑑𝜆 ´ ln 𝜆 ` 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝜆

The first term is the exponential integral ż 𝐸 𝑖 p𝑥q “

𝑒𝜆 𝑑𝜆 𝜆

See e.g. Gradshteyn and Ryzik, "Table of Integrals, series, and products" formula 2.325. According to formula 8.213.2 we have 𝐸p𝑥q » 𝛾 ` ln 𝑥 for 𝑥 Ñ Ý 0` where 𝛾 is the Euler-Mascheroni constant.

So, we conclude that the constant in expression above is given by 𝛾 and have 𝑆p𝜆q “ 𝐸 𝑖 p𝜆q ´ lnp𝜆q ` 𝛾 In summary, ă 𝑟 𝑖 ą“ 1 `

𝜆𝑒 𝜆 𝑁r𝐸 𝑖 p𝜆q ´ ln 𝜆 ` 𝛾s p𝑒 𝜆 ´ 1q2

Complexity Science c) We estimate ă 𝑟 𝑖 ą by replacing 𝐿 Ñ Ý ă 𝐿 ą and 𝑘1𝑖 Ñ Ý ă 𝑘1𝑖 ą ř ă 𝑖 𝑘𝑖 ą 𝑁 ă 𝑘𝑖 ą ă 2𝐿 ą “ “ “𝑁 ă 𝑟 𝑖 ą» ă 𝑘𝑖 ą ă 𝑘𝑖 ą ă 𝑘𝑖 ą We found above ă 𝑟 𝑖 ą“ 1 `

8 ÿ 𝜆𝑒 𝜆 𝜆𝑘 𝑁 𝜆 2 𝑘 𝑘! p𝑒 ´ 1q 𝑘 “1

Consider the limit 𝜆 ăă 1 and 𝑁 ąą 1, ă 𝑟 𝑖 ą» 1 ` 𝑁

𝜆p1 ` 𝜆q 𝜆2

p𝜆 ` 2 q2

p𝜆 `

So, in this limit, we recover the expression derived in (b).

𝜆2 q»𝑁 4

33

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Complexity Science

Chapter 9 Exercise 1 𝑋“1

with probability 𝑝 with probability 1 ´ 𝑝

𝑋 “ ´1

𝐻r𝑃 𝑋 s “ ´p𝑝 ln 𝑝 ` p1 ´ 𝑝q lnp1 ´ 𝑝qq

Maxima:

𝑑𝐻 “0Ñ Ý ´pln 𝑝 ` 1 ´ 𝑝 lnp1 ´ 𝑝qq ´ 1 “ 0 𝑑𝑝 ùñ lnp

𝑝 q“0 1´ 𝑝

𝑝 1 “ 1 ùñ 𝑝 “ 1´ 𝑝 2

ùñ

Exercise 2 The probabilities for 𝑌 “ 𝑋 2 : Case (1): " 𝑃 𝑋 p𝑥q “ " 𝑃𝑌 p𝑦q “

1 𝑁 `1 , for 𝑥 P t0, 1, ..., 𝑁u

0,

otherwise

1 2 𝑁 `1 , for 𝑦 P t0, 1, 4..., 𝑁 u

0,

otherwise

Case (2): " 𝑃 𝑋 p𝑥q “

1 2𝑁 `1 , for 𝑥 P t´𝑁, ´p𝑁 ´ 1q, ...0, 1, ..., 𝑁u

0,

otherwise

$ 1 for 𝑦 “ 0 & 2𝑁 `1 , 2 𝑃𝑌 p𝑦q “ 2𝑁 `1 , for 𝑦 P t1, 4, 9, ..., 𝑁 2 u % 0, otherwise Joint probability distributions 𝑃 𝑋,𝑌 p𝑥, 𝑦q “ Prob t𝑋 “ 𝑥 and 𝑌 “ 𝑦 “ 𝑥 2 u

“ 𝑃 𝑋 p𝑥q𝛿 𝑦,𝑥 2 Case (1) " 𝑃 𝑋,𝑌 p𝑥, 𝑦q “

1 2 𝑁 `1 , p𝑥, 𝑦q P tp0, 0q, p1, 1q, p2, 4q, ..., p𝑁, 𝑁 qu ” 𝑆 1

0

otherwise

Complexity Science Case(2) " 𝑃 𝑋,𝑌 p𝑥, 𝑦q “

1 2 2 2 2𝑁 `1 , p𝑥, 𝑦q P tp´𝑁, 𝑁 q, p´p𝑁 ´ 1q, p𝑁 ´ 1q q, ...., p𝑁, 𝑁 qu ” 𝑆 2

0

otherwise

a) Correlation coefficient 𝐶𝑋,𝑌 9 ă 𝑋𝑌 ą ´ ă 𝑋 ąă 𝑌 ą Case (1) 2

𝑁 ÿ 𝑁 ÿ

ă 𝑋𝑌 ą“

𝑥𝑦𝑃 𝑋𝑌 p𝑥, 𝑦q 𝑥 “0 𝑦 “0

𝑁 ÿ ÿ

2

𝑦 “ 0 𝑁 𝑥𝑦𝑃 𝑋 p𝑥q𝛿 𝑦, 𝑥 2 “

𝑥 “0 𝑁 ÿ

ă 𝑋 ą“

𝑥𝑃 𝑋 p𝑥q “ 𝑥 “0

𝑁 ÿ

ă 𝑌 ą“

𝑥 2 𝑃𝑌 p𝑥 2 q “

𝑥 “1

𝐶𝑋𝑌 9

𝑁p𝑁 ` 1q 2 1 𝑥3 “ r s 𝑁 ` 1 𝑁 ` 1 2 𝑥 “0 𝑁 ÿ

𝑁 1 ÿ 1 𝑁p𝑁 ` 1q 𝑁 “ 𝑥“ 𝑁 ` 1 𝑥 “0 𝑁 `1 2 2

𝑁 𝑁p2𝑁 ` 1q 1 ÿ 1 𝑁p𝑁 ` 1qp2𝑁 ` 1q 𝑥“ “ 𝑁 ` 1 𝑥 “1 𝑁 `1 6 6

𝑁 2 p𝑁 ` 1q 𝑁 2 p𝑁 ` 2q 𝑁 𝑁p2𝑁 ` 1q ´ “ ‰0 4 2 6 12

Case (2) 𝑁 ÿ

𝑥𝑥 2

ă 𝑋𝑌 ą“ 𝑥 “´ 𝑁 𝑁 ÿ

ă 𝑋 ą“

𝑥 𝑥 “´ 𝑁 𝑁 ÿ

ă 𝑌 ą“ 𝑥 “0

𝑥2

1 “0 2𝑁 ` 1

1 “0 2𝑁 ` 1

2 ‰0 2𝑁 ` 1

So, 𝐶𝑋,𝑌 “ 0 b) Mutual information ÿ 𝐼p𝑋, 𝑌 q “

𝑃 𝑋,𝑌 p𝑥, 𝑦q ln

𝑥𝑦

𝑃 𝑋,𝑌 p𝑥, 𝑦q 𝑃 𝑋 p𝑥q𝑃𝑌 p𝑦q

Case (1) 1

𝑃 𝑋,𝑌 p𝑥, 𝑦q 𝑁 `1 “ 1 “ 𝑁 `1 1 𝑃 𝑋 p𝑥q𝑃𝑌 p𝑦q 𝑁 `1 𝑁 `1 So, ÿ 𝐼p𝑋, 𝑌 q “ p 𝑥,𝑦 qP𝑆1

1 lnp𝑁 ` 1q “ lnp𝑁 ` 1q ‰ 0 𝑁 `1

35

36

Complexity Science Case (2) 𝑃 𝑋,𝑌 p𝑥, 𝑦q “ 𝑃 𝑋 p𝑥q𝑃𝑌 p 𝑦 q

"

𝐼p𝑋, 𝑌 q “

* 1 1 1 ´1 “ 2𝑁 ` 1, p𝑥, 𝑦q “ p0, 0q 2𝑁 `1 r 2𝑁 `1 2𝑁 `1 s 1 1 2 ´1 “ 2𝑁2`1 , p𝑥, 𝑦q P 𝑆2 2𝑁 `1 r 2𝑁 `1 2𝑁 `1 s

1 1 2𝑁 ` 1 lnp2𝑁 ` 1q ` 2𝑁 ln 2𝑁 ` 1 2𝑁 ` 1 2 lnp2𝑁 ` 1q ´

2𝑁 log 2 ‰ 0 2𝑁 ` 1

Exercise 3 𝛽𝑆 𝑆

Two Ising spins: 𝐻 “ ´𝑆1 𝑆2 with Boltzmann weights: 𝑃p𝑆1 , 𝑆2 q “ 𝑒 𝑍1 2 where ÿÿ 𝑍“

𝛽 ´𝛽 ´𝛽 𝛽 𝑒 𝛽𝑆1 𝑆2 “ loo𝑒mo on ` lo𝑒omoon ` lo𝑒omoon ` loo𝑒moon

𝑆1 𝑆2

p´1, ´1q

p´1,1q

p1, ´1q

p1,1q

“ 2p𝑒 𝛽 ` 𝑒 ´𝛽 q “ 4 coshp𝛽q Marginal distributions 1 1 ÿ 𝛽𝑠𝑠2 𝑒 “ t𝑒 ´𝛽𝑠 ` 𝑒 𝛽𝑠 u 𝑍 𝑠 𝑍

𝑃𝑆1 p𝑠q “

2

“

cosh 𝛽𝑠 “ 𝑃𝑆2 p𝑠q 2 cosh 𝛽

Correlation coefficient 𝐶𝑆1 𝑆2 “ă 𝑆1 𝑆2 ą ´ ă 𝑆1 ąă 𝑆2 ą Here, ă 𝑆1 𝑆2 ą“

1 𝛽 1 r𝑒 ´ 𝑒 ´𝛽 ´ 𝑒 𝛽 ` 𝑒 𝛽 s “ tanh 𝛽 𝑍 2

ă 𝑆1 ą“ă 𝑆2 ą“ ´ 1 𝐶𝑆1 𝑆2 “ tanh 𝛽 “ 2

coshp´𝛽q cosh 𝛽 ` “0 2 cosh 𝛽 2 cosh 𝛽

"1

Ý 0 ùñ 𝛽 Ñ Ý 8 2 , for 𝑇 Ñ 1 𝛽, for 𝑇 Ñ Ý 8 ùñ 𝛽 Ñ Ý 0 2

Mutual Information 𝐼p𝑆1 ; 𝑆2 q “

ÿÿ 𝑆1 𝑆2

“

𝑃p𝑆1 , 𝑆2 q ln

𝑃p𝑆1 , 𝑆2 q 𝑃p𝑆1 q𝑃p𝑆2 q

ÿ 𝑒 𝛽𝑠1 𝑠2 𝑒 𝛽𝑠1 𝑠2 2 cosh 𝛽 2 cosh 𝛽 ln 4 cosh 𝛽 4 cosh 𝛽 cosh 𝛽𝑠1 cosh 𝛽𝑠2 𝑠 𝑠 1 2

Complexity Science

“

ÿ 1 𝑒 𝛽𝑠1 𝑠2 𝑒 𝛽𝑠1 𝑠2 ln 4 cosh 𝛽 𝑠 𝑠 cosh 𝛽 1 2

where we used cosh 𝛽𝑠1 “ cosh 𝛽𝑠2 “ cosh 𝛽 for 𝑠1 , 𝑠2 P t´1, 1u. Hence, 𝐼p𝑆1 ; 𝑆2 q “

1 p𝛽𝑒 𝛽 ´ 𝑒 𝛽 log cosh 𝛽 ´ 𝛽𝑒 ´𝛽 ´ 𝑒 𝛽 ln cosh 𝛽q “ 𝛽 ´ ln cosh 𝛽 2 cosh 𝛽

As 𝑇 Ñ Ý 0 ùñ 𝛽 Ñ Ý 8 ùñ cosh 𝛽 » 21 𝑒 𝛽 . 𝛽 So, 𝐼p𝑆1 ; 𝑆2 q “ 𝛽 ´ ln 𝑒2 “ log 2 As 𝑇 Ñ Ý 8 ùñ 𝛽 Ñ Ý 0 ùñ cosh 𝛽 » 1 ` 21 𝛽2 .

So, 𝐼p𝑆1 ; 𝑆2 q “ 𝛽 ´ lnp1 ` 12 𝛽2 q » 𝛽 ´ 12 𝛽2 ùñ 𝛽 “ 𝑘𝐵1𝑇 Exercise 4 We have per definition ÿÿ 𝐼p𝑋; 𝑌 q “ 𝑥

𝑃 𝑋𝑌 p𝑥, 𝑦q log

𝑦

𝑃 𝑋𝑌 p𝑥, 𝑦q 𝑃 𝑋 p𝑥q𝑃𝑌 p𝑦q

To derive the expressions in Eq. (9.20), we make use of ÿ 𝐻p𝑋q “ ´

𝑃 𝑋 p𝑥q log 𝑃 𝑋 p𝑥q

𝑥

ÿ 𝐻p𝑋, 𝑌 q “ ´

𝑃 𝑋𝑌 p𝑥, 𝑦q log

𝑥,𝑦

𝑃 𝑋𝑌 p𝑥, 𝑦q 𝑃𝑌 p𝑦q

So, we have, 𝐻p𝑋q ` 𝐻p𝑌 q ´ 𝐻p𝑋𝑌 q

ÿ “´

𝑃 𝑋 p𝑥q log 𝑃 𝑋 p𝑥q ´

𝑥

ÿ

𝑃𝑌 p𝑦q log 𝑃𝑌 p𝑦q `

𝑦

ÿÿ 𝑥

𝑃 𝑋𝑌 p𝑥, 𝑦q log 𝑃 𝑋𝑌 p𝑥, 𝑦q

𝑦

And from definitions of the mutual information: ÿÿ 𝐼p𝑋; 𝑌 q “ 𝑥

´

ÿÿ ÿÿ p 𝑃 𝑋𝑌 p𝑥, 𝑦q log 𝑃 𝑋 p𝑥qq ´ p 𝑃 𝑋𝑌 p𝑥, 𝑦q log 𝑃𝑌 p𝑦qq 𝑥

ÿÿ 𝑥

𝑦

𝑃 𝑋𝑌 p𝑥, 𝑦q log 𝑃 𝑋𝑌 p𝑥, 𝑦q ´

𝑦

Further, we have,

𝑃 𝑋𝑌 p𝑥, 𝑦q log 𝑃 𝑋𝑌 p𝑥, 𝑦q

𝑦

𝑦

ÿ 𝑥

𝑥

𝑃 𝑋 p𝑥q log 𝑃 𝑋 p𝑥q ´

ÿ 𝑦

𝑃𝑌 p𝑦q log 𝑃𝑌 p𝑦q

37

38

Complexity Science

ÿÿ 𝐻p𝑋, 𝑌 q ´ 𝐻p𝑋|𝑌 q ´ 𝐻p𝑌 |𝑋q “ ´

𝑦𝑃 𝑋𝑌 p𝑥, 𝑦q log 𝑃 𝑋𝑌 p𝑥, 𝑦q

𝑥

ÿÿ ´r´ 𝑥

𝑃 𝑋𝑌 p𝑥, 𝑦q s 𝑃𝑌 p𝑦q

𝑃 𝑋𝑌 p𝑥, 𝑦q log

𝑃 𝑋𝑌 p𝑥, 𝑦q s 𝑃 𝑋 p𝑥q

𝑦

ÿÿ ´r´ 𝑥

𝑃 𝑋𝑌 p𝑥, 𝑦q log

𝑦

ÿÿ “´ 𝑥

ÿÿ ` 𝑥

𝑃 𝑋𝑌 p𝑥, 𝑦q log 𝑃 𝑋𝑌 p𝑥, 𝑦q ´

𝑦

ÿÿ ` 𝑥

𝑃 𝑋𝑌 p𝑥, 𝑦q log 𝑃 𝑋𝑌 p𝑥, 𝑦q ´

𝑦

ÿ

ÿ

𝑃 𝑋 p𝑥q log 𝑃 𝑋 p𝑥q

𝑃 𝑋 p𝑥q log 𝑃 𝑋 p𝑥q ´

ÿ

𝑃 𝑋 p𝑥q log 𝑃 𝑋 p𝑥q `

𝑃 𝑋 p𝑥q log 𝑃 𝑋 p𝑥q `

𝑥

ÿÿ 𝑥

ÿÿ 𝑥

𝑃𝑌 p𝑦q log 𝑃𝑌 p𝑦q

𝑦

𝑥

ÿ

ÿ

𝑥

𝑦

𝐻p𝑋q ´ 𝐻p𝑋|𝑌 q “ ´

“´

𝑃𝑌 p𝑦q log 𝑃𝑌 p𝑦q

𝑥

𝑃 𝑋𝑌 p𝑥, 𝑦q ´ 𝑥

ÿ 𝑦

ÿÿ “

𝑃 𝑋𝑌 p𝑥, 𝑦q log 𝑃 𝑋𝑌 p𝑥, 𝑦q

𝑦

𝑃 𝑋𝑌 p𝑥, 𝑦q log

𝑃 𝑋𝑌 p𝑥, 𝑦q 𝑃p𝑦q

𝑦

𝑃 𝑋𝑌 p𝑥, 𝑦q log 𝑃 𝑋𝑌 p𝑥, 𝑦q ´

𝑦

ÿ

𝑃𝑌 p𝑦q log 𝑃𝑌 p𝑦q

𝑦

And similar for 𝐻p𝑌 q ´ 𝐻p𝑌 |𝑋q.

Exercise 5 We derive, Eq. (9.27), 𝐷𝑇 𝐸 𝑥𝑖 Ñ Ý 𝑥 𝑗 “ 𝑇 𝐸 XÑ Ý 𝑋 𝑗 ´ 𝑇 𝐸 Xp´𝑖q Ñ Ý 𝑋𝑗 𝐷𝑇 𝐸 𝑋𝑖 Ñ Ý 𝑋 𝑗 “ă log ÿ

´ Probt𝑋 ` 𝑗 |X u ´ Probt𝑋 ` 𝑗 |Xp´𝑖 q u

ą

` ´ ` ` ´ ´ ´ rProbt𝑋 ` 𝑗 , X u log Probt𝑋 𝑗 |X u ´ Probt𝑋 𝑗 , X u log Probt𝑋 𝑗 |Xp´𝑖 q us

` ´ ´ “ ´𝐻r𝑋 ` 𝑗 |X s ` 𝐻r𝑋 𝑗 |Xp´iq s

` ´ ` ´ ` “ 𝐻r𝑋 ` 𝑗 s ´ 𝐻r𝑋 𝑗 |X s ´ 𝐻p𝑋 𝑗 q ´ 𝐻r𝑋 𝑗 |Xp´𝑖 q s

From Eq. (9.20), ` ´ ´ “ 𝐼r𝑋 ` 𝑗 ; X s ´ 𝐼r𝑋 𝑗 ; Xp´𝑖 q s

(‹q

Complexity Science

39

From Eq. (9.21), ` ´ ´ ` ´ 𝑇 𝐸𝑌 Ñ Ý 𝑋 “ 𝐼p𝑋 ; p𝑋 , 𝑌 qq ´ 𝐼p𝑋 ; 𝑋 q

` ´ ` ´ ´ 𝑇 𝐸 XÑ Ý 𝑋 𝑗 “ 𝐼p𝑋 𝑗 ; p𝑋 𝑗 , X qq ´ 𝐼p𝑋 𝑗 ; 𝑋 𝑗 q

` ´ “ 𝐼p𝑋 ` 𝑗 ; Xq ´ 𝐼p𝑋 𝑗 ; 𝑋 𝑗 q

(A)

´ ` ´ ´ ` 𝑇 𝐸 Xp´𝑖q Ñ Ý 𝑋 𝑗 “ 𝐼p𝑋 𝑗 ; p𝑋 𝑗 , Xp´𝑖q qq ´ 𝐼p𝑋 𝑗 ; 𝑋 𝑗 q

´ ` ´ “ 𝐼p𝑋 ` 𝑗 ; Xp´𝑖 q q ´ 𝐼p𝑋 𝑗 ; 𝑋 𝑗 q

(B)

´ Since, 𝑋 ´ 𝑗 is contained in Xp´𝑖 q . And (9.27) follows since subtracting (B) from (A) gives p‹q

Exercise 6 Write a code to simulate the process in Eq. (9.91). From the simulation of the process estimate the probabilities in Eq.(9.24) and Eq.(9.26) from histograms and use these to estimate 𝑇 𝐸 𝑋Ñ Ý 𝑍 and 𝐷𝑇 𝐸 𝑋Ñ Ý𝑍 Exercise 7 Eq.(9.36) Tsallis entropy

ř 𝑞 1´ 𝑊 𝑖 “1 𝑝 𝑖 𝑆 𝑞 r𝑝s “ 𝑞´1

Eq.(9.31) Shannon entropy

ÿ 𝐻r𝑝s “ ´

𝑝 𝑖 ln 𝑝 𝑖

𝑖 1

Eq.(9.38) q-exponential

exp𝑞 p𝑥q “ r1 ´ p1 ´ 𝑞q𝑥qs 1´𝑞

Recall lim p1 `

𝑘Ñ Ý8

𝑥 𝑘 q “ expp𝑥q 𝑘

To determine the limit lim 𝑆 𝑞

𝑞Ñ Ý1

we make use of l’Hopital’s rule: 𝑓 p𝑥q 𝑓 1 p𝑥q lim “ lim 1 𝑥Ñ Ý 𝑥0 𝑔p𝑥q 𝑥Ñ Ý 𝑥0 𝑔 p𝑥q for 𝑓 p𝑥q “ 𝑔p𝑥q “ 0. So, we have ř 𝑞 ř 𝑑 ´ 𝑖 𝑝 𝑖𝑞 ln 𝑝 𝑖 𝑖 𝑝𝑖 q 𝑑𝑞 p1 ´ lim 𝑆 𝑞 “ lim “ lim 𝑑 𝑞Ñ Ý1 𝑞Ñ Ý1 𝑞Ñ Ý1 1 𝑑𝑞 p𝑞 ´ 1q

40

Complexity Science ÿ “´

𝑝 𝑖 ln 𝑝 𝑖 “ Eq. (9.31))

𝑖 1

For the q-exponential, we have r1 ´ p1 ´ 𝑞q𝑥s 1´𝑞 “ r1 ´ 𝑥𝜅 s 𝜅 with 𝜅 “ 1´1 𝑞 and since 𝜅 Ñ Ý 8 for 1

𝑞Ñ Ý 1, we conclude that r1 ´ p1 ´ 𝑞q𝑥s 1´𝑞 Ñ Ý expp´𝑥q when 𝑞 Ñ Ý 1.

Exercise 8 Derivation of expressions for 𝑆r𝑝s is on page 242. We start from the trace form

Eq(9.41) 𝑆r𝑝s “

𝑊 ÿ

𝑝 𝑖 𝐺pln 𝑖 “1

1 q 𝑝𝑖

and the non-trace form ř 𝛼 𝐺plnp 𝑊 𝑖 “1 𝑝 𝑖 q Eq (9.42) 𝑆r𝑝s “ 1´𝛼 and make use of the extensivity requirement.

Eq. (9.44) 𝑆p

1 q “ 𝜆𝑁 for 𝑁 Ñ Ý 8 𝑊

Trace form 𝑆p

𝑊 ÿ 1 1 q “ 𝜆𝑁 Ñ Ý 𝐺pln 𝑊q “ 𝜆𝑁 𝑊 𝑊 𝑖 “1

ùñ 𝐺pln 𝑊q “ 𝜆𝑁 We find the functional form for 𝐺p𝑡q by inversion. Let 𝑡 “ ln 𝑊 Ñ Ý 𝑁 “ 𝑊 ´1 p𝑒 𝑡 q, so 𝐺p𝑡q “ ´1 𝑡 𝜆𝑊 p𝑒 q. To ensure 𝐺p0q “ 0, we subtract and have 𝐺p𝑡q “ 𝜆r𝑊 ´1 p𝑒 𝑡 q ´ 𝑊 ´1 p1qs The three cases: 1

𝑊p𝑁q “ 𝑁 𝑎 “ 𝑡 Ñ Ý 𝑊 ´1 p𝑡q “ 𝑁 “ 𝑡 𝑎 which gives 1

𝑡

𝐺p𝑡q “ 𝜆rp𝑒 𝑡 q 𝑎 ´ 1s “ 𝜆r𝑒 𝑎 ´ 1s and 𝑊 ÿ

𝑆r𝑝s “

𝑝 𝑖 𝐺pln 𝑖 “1

𝑊 ÿ 1 1 1 q“ 𝑝 𝑖 𝜆rp q 𝑎 ´ 1s i.e. Eq.(9.47) 𝑝𝑖 𝑝 𝑖 𝑖 “1

𝑊p𝑁q “ 𝑘 𝑁 “ 𝑡 Ñ Ý 𝑁 ln 𝑘 “ ln 𝑡

ùñ 𝑊 ´1 p𝑡q “

ln 𝑡 ln 𝑘

Complexity Science

41

So, 𝐺p𝑡q “ 𝜆r

ln 𝑒 𝑡 𝑡 ´ 0s “ 𝜆 ln 𝑘 ln 𝑘

and 𝑊

𝑆r𝑝s “

1 𝜆 ÿ 𝑝 𝑖 ln ln 𝑘 𝑖“1 𝑝𝑖

Eq.(9.49) 1

1

𝑊p𝑁q “ 𝑁 𝛾 𝑁 “ 𝑡 Ñ Ý 𝑁 𝑁 “ 𝑡 𝛾 ùñ 𝑁 ln 𝑁 “ ln 𝑡 𝛾 Now recall the Lambert function 𝐿p𝑡q defined according to 𝐿p𝑡q𝑒 𝐿 p𝑡 q “ 𝑡 and let ln 𝑋 “ 𝐿p𝑡q. Then we have 𝑡 “ 𝐿 ´1 pln 𝑥q and ln 𝑥𝑒 ln 𝑥 “ 𝐿 ´1 pln 𝑋q Ñ Ý 𝑥 ln 𝑥 “ 𝐿 ´1 pln 𝑥q or 𝐿p𝑥 ln 𝑥q “ ln 𝑥. So from, 1 𝑁 ln 𝑁 “ ln 𝑡 𝛾 we have 1

𝐿p𝑁 ln 𝑁q “ 𝐿pln 𝑡 𝛾 q 1

ln 𝑁 “ 𝐿pln 𝑡 𝛾 q 1

𝑁 “ expp𝐿pln 𝑡 𝛾 qq and conclude that 1 𝑊 ´1 p𝑡q “ expr𝐿p ln 𝑡qs 𝛾 So, 1 𝐺p𝑡q “ 𝜆texpr𝐿p 𝑡qs ´ 1u 𝛾 and 1 1 𝑝 𝑖 rexpr𝐿p ln q ´ 1ss 𝛾 𝑝 𝑖 𝑖

ÿ 𝑆r𝑝s “ 𝜆

Eq. (9.50)

The trace form in Eq. (9.42) together with Eq. (9.46) gives us 1

𝑊p𝑁q “ 𝑁 𝑎 Ñ Ý 𝑊 ´1 p𝑡q “ 𝑡 𝑎 and ř ln 𝑖 𝑝 𝑖𝛼 𝑆r𝑝s “ 𝜆rexpp q ´ 1s 𝑎p1 ´ 𝛼q which is Eq.(9.51). 𝑊p𝑁q “ 𝑘 𝑁 Ñ Ý 𝑊 ´1 p𝑡q “

ln 𝑡 ln 𝑘

So, 𝑆r𝑝s “

𝑊 ÿ 𝜆 exprln 𝑝 𝑖𝛼 s ln 𝑘p1 ´ 𝛼q 𝑖 “1

42

Complexity Science

ř𝑊 𝛼 𝜆 𝑖 “1 𝑝 𝑖 “ ln 𝑘 1 ´ 𝛼 1 𝑊p𝑁q “ 𝑁 𝛾 𝑁 Ñ Ý 𝑊 ´1 p𝑡q “ expr𝐿p ln 𝑡qs 𝛾 ùñ 𝑊 ´1 p1q “ 1 So, 1 𝑡 𝐺p𝑡q “ 𝜆p1 ´ 𝛼qtexpr𝐿p lnpexp qqs ´ 1u 𝛾 1´𝛼 “ 𝜆p1 ´ 𝛼qtexpr𝐿p

𝑡 qs ´ 1u 𝛾p1 ´ 𝛼q

and

𝑆r𝑝s “

𝑊 ÿ 1 𝐺pln 𝑝 𝑖𝛼 q 1´𝛼 𝑖 “1

ř 𝛼 ln 𝑊 𝑖 “1 𝑝 𝑖 “ 𝜆texpr𝐿p qs ´ 1u 𝛾p1 ´ 𝛼q Exercise 9 a) and (b) This is a difficult question. Somehow experiments persistently find that the more aware brain produces electrical signals with more diversity as indicated by the Lempel-Ziv measure.

Perhaps, this is reasonable at an intuitive level since we may have more "complicated" and diverse thoughts processes when in a state of higher awareness.

Project 1 The computational details are given in Section 7 of Dr. Fatimah Abdul Razak’s PhD thesis which canbe downloaded from the spiral depository of Imperial College London, accessible via this link: https://spiral.imperial.ac.uk/bitstream/10044/1/11579/1/Abdul%20Razak-F-2013-PhD-Thesis.PDF

Complexity Science

43

Project 2

a)

𝑋1

𝑋2

𝑋3

𝑋2 𝑛𝑎𝑛𝑑𝑋3

𝑃p𝑋3

𝑋1 𝑛𝑜𝑟 𝑋2

𝑃p𝑋3 q

0 0 0 1 1 1 0 1

0 0 1 0 1 0 1 1

0 1 0 0 0 1 1 1

NP1 1 NP NP 0 1 1 NP

0 1/4 0 0 1/4 1/4 1/4 0

NP 1 0 NP 0 0 NP NP

0 1/4 1/4 0 1/4 1/4 0 0

b) to (f) straightforward numerical work plotting the liner combinations similar to Eq.(9.67) and (9.68).

Project 3 No sample solution is needed.

44

Complexity Science

Chapter 10 Exercise 1 Eq. (10.57) 𝑃p𝑥, 𝑡q “ ?

1 4𝜋𝐷𝑡

pexpr´

p𝑥 ´ 𝑥0 q2 p𝑥 ` 𝑥 0 q2 s ´ expr´ sq 4𝐷𝑡 4𝐷𝑡

ż ?𝑥0 4𝐷𝑡 2 1 𝑒 ´𝑧 𝑑𝑧 “2 𝐴 0 ż8 𝑃p𝑥, 𝑡q𝑑𝑧 0

“?

ż8 ż8 p𝑥 ´ 𝑥0 q2 p𝑥 ` 𝑥0 q2 p expr´ s𝑑𝑥 ´ s𝑑𝑥q expr´ 4𝐷𝑡 4𝐷𝑡 4𝜋𝐷𝑡 0 0 1

𝑥 ` 𝑥0 𝑥 ´ 𝑥0 in the second 𝑧 “ ? . Then, In first integral use substitution 𝑧 “ ?

4𝐷𝑡 4𝐷𝑡

ż8 𝑃p𝑥, 𝑡q𝑑𝑥 “ ?

0

1 4𝜋𝐷𝑡

ż8 𝑒𝑧

´

2

ż8 p 𝑥

𝑒´𝑧

2

?

4𝐷𝑡𝑑𝑧

´? 0

4𝐷𝑡

?

𝑥 4𝐷𝑡

4𝐷𝑡𝑑𝑧q

? 0

ż ?𝑥0

4𝐷𝑡

“

𝑒

´𝑧2

𝑥 ´? 0 4𝐷𝑡

ż ?𝑥0

4𝐷𝑡

𝑑𝑧 “ 2

2

𝑒 ´𝑧 𝑑𝑧

0

Exercise 2 a) 𝑛p𝑥, 𝑡 ` 1q “ 𝑛p𝑥, 𝑡 ` 1q ´ 𝑛p𝑥, 𝑡q “

1 1 𝑛p𝑥 ´ 1, 𝑡q ` 𝑛p𝑥 ` 1, 𝑡q 2 2 1 p𝑛p𝑥 ´ 1, 𝑡q ` 𝑛p𝑥 ` 1, 𝑡q ´ 2𝑛p𝑥, 𝑡qq 2

In the continuous limit, B𝑡 𝑛p𝑥, 𝑡q “

1 2 B 𝑛p𝑥, 𝑡q 2 𝑥

b) Include source term B𝑡 𝑛 “

1 2 B 𝑛 ` 𝑔p𝑥, 𝑡q 2 𝑥

Fourier transform of Eq.(‹), 1 𝑖𝜔𝑛ˆ “ ´ 𝑘 2 𝑛ˆ ` 𝑔ˆ 2 𝑛ˆ “

𝑔p𝑘, ˆ 𝜔q 𝑖𝜔 ` 12 𝑘 2

(‹q

Complexity Science

ż8

𝑑𝜔 ´8 2𝜋

𝑛p𝑥, 𝑡q “

ż8

ˆ 𝜔q 𝑖𝑘𝑡 `𝑖 𝜔 𝑑𝑘 𝑔p𝑘, 𝑒 1 2 ´8 2𝜋 𝑖𝜔 ` 2 𝑘

c) ż8 ˆ 𝑁p𝜔q “

𝑑𝑡𝑁p𝑡q𝑒 ´𝑖 𝜔𝑡 ´8

ż8 𝑛p𝑥0 , 𝑡q𝑒 ´𝑖 𝜔𝑡

“ ´8

ż8

ż8

´8

𝑑𝜔1 8 2𝜋

ż8

ż8

“

𝑑𝑡

“

𝑑𝜔1 8 2𝜋

ż8

ˆ 𝜔q 𝑑𝑘 𝑔p𝑘,

´8 2𝜋 𝑖𝜔

1

1 ` 1 𝑘2 2

ˆ 𝜔q 𝑖𝑘 𝑥0 𝑑𝑘 𝑔p𝑘, 𝑒 2𝜋 𝑖𝜔1 ` 21 𝑘 2 ´8

𝑒 𝑖𝑘 𝑥0 `𝑖𝑤 𝑡 𝑒 ´𝑖 𝜔𝑡

ż8

1

𝑑𝑡𝑒 𝑖p 𝜔 ´ 𝜔q𝑡 8

ż8 ˆ 𝑁p𝜔q “

ˆ 𝜔q 𝑖𝑘 𝑥0 𝑑𝑘 𝑔p𝑘, 𝑒 2𝜋 𝑖𝜔 ` 12 𝑘 2 8

Use 𝑔p𝑥, 𝑡q “ 𝛿p𝑡q𝜒p𝑡q ùñ 𝑔p𝑘, ˆ 𝜔q “ 𝜒p𝜔q ˆ ş8 Since, ´8 𝑑𝑡𝛿p𝑡q𝑒 ´𝑖𝑘 𝜔 “ 1.

Therefore, ż8 ˆ 𝑁p𝜔q “

ˆ 𝑑𝑘 𝜒p𝜔 2𝜋 𝑖𝜔 ` 21 𝑘 2 ´8

d) Compute the integral over 𝑘 𝜒p𝜔q ˆ ˆ 𝑁p𝜔q “ 𝜋

ż8 𝑑𝑘 8

𝑒 𝑖𝑘 𝑥0 p𝑘 ´ 𝑘 ` qp𝑘 ´ 𝑘 ´ q

where 𝑘 ` and 𝑘 ´ are the roots of ?

1 𝑖 1 2 𝑘 ` 𝑖𝜔 “ 0 ùñ 𝑘 ˘ “ ˘ 2𝜔p´ ? ` ? q 2 2 2 Since, 𝑥0 ą 0 we can close the contour in the upper half plane and pick up the residue ? ´1 expp𝑖 2𝜔 𝑖? 𝑥 q 𝑒 𝑖𝑘` 𝑥0 2 0 ?

Resp𝑘 ` q “ “ 𝑘` ´ 𝑘´ 2 𝜔p𝑖 ´ 1q Hence, 𝜒p𝜔q ˆ ˆ 𝑁p𝜔q “ 2𝜋𝑖Resp𝑘 ` q 𝜋

45

46

Complexity Science and the power spectrum, 2

| 𝜒p𝜔q| ˆ ˆ 𝑁p𝜔q “ 𝑒 ´2 2𝜔 ?

?

Since, 𝑒 ´2

𝜔 𝑥0 Ñ Ý 1 for 2

?

𝜔 𝑥0

𝜔𝑥0 ăă 1 we obtain a 1𝑓 behaviour when 𝜔 ăă p2𝑥1 q2 .

0

Exercise 3 a) When the walker steps from position 𝑥 “ 1 to 𝑥 “ 0, it arrices for the first time at 𝑥 “ 0 due to the absorbing boundary condition i.e. no step is possible from 𝑥 “ 0 to 𝑥 “ 1. Accordingly, 𝑃p0, 𝑡 ` 1q “ 𝑃p0, 𝑡q ` 𝑝 ´ 𝑃p1, 𝑡q and 𝑃 𝑅 p0, 𝑡 ` 1q “ 𝑃p0, 𝑡 ` 1q ´ 𝑃p0, 𝑡q “ 𝑝 ´ 𝑃p1, 𝑡q b) Master Equation: 𝑃p𝑥, 𝑡 ` 1q “ 𝑃p𝑥, 𝑡q ` 𝑝 ` 𝑃p𝑥 ´ 1, 𝑡q ` 𝑝 ´ 𝑃p𝑥 ` 1, 𝑡q ´ 𝑃p𝑥, 𝑡q and since, 𝑝 ` “ 𝑝 ´ “ 12 , we have 𝑃p𝑥, 𝑡 ` 1q “

1 p𝑃p𝑥 ´ 1, 𝑡q ` 𝑃p𝑥 ` 1, 𝑡qq 2

c) Simply use Eq. (10.61) with 𝑥 0 “ 𝜖 “ 1 Exercise 4 a) The details are given in Sec.6.6.2 for 𝐴 “ 1. See Eq. (6.92), (6.94) to (6.97) and again considered in Eq. (10.80). b) The power spectrum is computed in detail in Eq. (10.81) to be 𝑆p𝜔q “ | 𝑓ˆp𝜔q|2 “

ż8 𝑑𝑡𝐶p𝑡q𝑒 ´𝑖 𝜔𝑡 “ ´8

𝜏“

1 p𝜏 𝐴q2 4 1 ` p 𝜏2𝜔 q2

1 𝜈

c) From (a) we have,

𝑆p𝜔q Ñ Ý

𝑆p𝜔q »

p𝜏 𝐴q2 “ constant for 𝜔 Ñ Ý 0 4

p𝜏 𝐴q2 2 2 𝐴 p q “ p q2 for 𝜔 ąą 1 4 𝜏𝜔 𝜔

Complexity Science

47

Project 1 a) Write a code to simulate the walk. Determine the time dependence of the average ă pq2 p𝑡q ą and use log-log plots to find the Hurst exponent from Eq. (10.85). b) 𝑃p𝑥, 𝑡 ` 𝛿q “ 𝑃p𝑥, 𝑡q ` 𝑝 ` 𝑃p𝑥 ´ 1, 𝑡q ` 𝑝 ´ 𝑃p𝑥 ` 1, 𝑡q ´ p1 ´ 𝑝 0 q𝑃p𝑥, 𝑡q c) Iterate the equation in (b) e.g. on the integers from ´𝐿 to 𝐿 for some reasonable 𝐿 ą 0. Use 𝑎 “ 𝛿 “ 1. Use initial condition " 𝑃p𝑥, 0q “

1, for 𝑥 “ 0 0, otherwise

) And continue to iterate until 𝑃p´𝐿, 𝑡q and 𝑃p𝐿, 𝑡q become non-zero. d) For 𝑝 ` “ 𝑝 ´ ” 𝑝, we have from (b) 𝑃p𝑥, 𝑡 ` 𝛿q ´ 𝑃p𝑥, 𝑡q “ 𝑝r𝑃p𝑥 ´ 1, 𝑡q ` 𝑃p𝑥 ` 1, 𝑡q ´ 2𝑃p𝑥, 𝑡qs ` r2𝑝 ´ p1 ´ 𝑝 0 qs𝑃p𝑥, 𝑡q Here the last term vanishes since 1 “ 𝑝 ` ` 𝑝 ´ ` 𝑝 0 “ 2𝑝 ` 𝑝 0 We get then in the limit 𝑎 Ñ Ý 0 and Ñ Ý 0 B𝑃 B2 𝑃 “𝐷 2 B𝑡 B𝑥 which corresponds to Eq (10.11) with 𝛾 “ 0 and we can use the solution in Eq. (10.13). e) From the solution in (d) follows according to Eq. (10.15) that ă 𝑋 2 p𝑡q ą“ 2𝐷𝑡 From Eq. (10.85), we get 𝐻 “ 21 Project 2 No sample solution needed.

Project 3 a) Fourier transform ż8 𝑛p𝑥, 𝑡q “

𝑑𝑘 ´8 2𝜋

ż8

𝑑𝜔

´8 2𝜋

𝑛p𝑘, ˆ 𝜔q𝑒 𝑖p 𝑘 𝑥 ` 𝜔𝑡 q

Substitute into equation and get, 𝑖𝜔𝑛ˆ “ 𝛾𝑘 2 𝑛ˆ ` 𝛽𝑖𝑘 𝑛ˆ ` 𝑔ˆ

ùñ 𝑛ˆ “

𝑔ˆ 𝑖𝜔 ` 𝛾𝑘 2 ´ 𝛽𝑖𝑘

48

Complexity Science Substitute into expression p‹q above to obtain 𝑛p𝑥, 𝑡q

b) ż8

𝑑𝑘 ´8 2𝜋

𝑛p𝑥, 𝑡q “

ż8

ˆ 𝑑𝜔 𝑔p𝑘{𝑤q

´8 2𝜋 𝑖𝜔 ´ 𝛽𝑖𝑘

𝑒 𝑖p 𝑘 𝑥 ` 𝜔𝑡 q

Note 𝑘𝑥 ` 𝜔𝑡 “ 𝑘 ` 𝑥𝑡 ` 𝜔𝑡 “ 𝑍1 𝑥𝑡 ` 𝑍2 with 𝑍1 “ 𝑘𝑡 Ñ Ý 𝑘“

𝑑𝑍1 𝑍1 Ñ Ý 𝑑𝑘 “ 𝑡 𝑡

𝑍2 “ 𝑘𝑡 Ñ Ý 𝑘“

𝑍2 𝑑𝑍2 Ñ Ý 𝑑𝑘 “ 𝑡 𝑡

Therefore, 1 1 𝑛p𝑥, 𝑡q “ p2𝜋q2 𝑖

“

1 1 1 𝑡 p2𝜋q2 𝑖

ż8

𝑑𝑍1 ´8 𝑡

ż8

ˆ 𝑍𝑍21 q 𝑖p 𝑍 𝑥 ` 𝑍 q 𝑑𝑍2 𝑔p 𝑒 1𝑡 2 𝑍2 ´ 𝛽 𝑍1 ´8 𝑡

ż8

𝑡

𝑔p ˆ 𝑍𝑍12 q

ż8 𝑑𝑍1

𝑑𝑍2

´8

8

“

𝑡

𝑥

𝑒 𝑖p 𝑍1 𝑡 ` 𝑍2 q

𝑍2 ´ 𝛽𝑍1

1 𝑥 𝐹p q 𝑡 𝑡

c) 𝛾 ‰ 0, 𝛽 “ 0 ż8 𝑛p𝑥, 𝑡q “

𝑑𝑘 ´8 2𝜋

ż8 𝑁p𝑡q “

ż8

ˆ 𝜔q 𝑖p 𝑘 𝑥 ` 𝜔𝑡 q 𝑑𝜔 𝑔p𝑘, 𝑒 2 8 2𝜋 𝑖𝜔 ` 𝛾𝑘

ż8 𝑑𝑥𝑛p𝑥, 𝑡q “

´8

𝑑𝑘 8 2𝜋

ż8

ż8 ˆ 𝜔q 𝑑𝜔 𝑔p𝑘, r 𝑑𝑥𝑒 𝑖𝑘 𝑥 s𝑒 𝑖 𝜔𝑡 2 ´8 2𝜋 𝑖𝜔 ` 𝛾𝑘 ´8

ş8

Recall ´8 𝑑𝑥𝑒 𝑖𝑘 𝑥 “ 2𝜋𝛿p𝑥q and perform the integration over 𝑘: ż8 𝑁p𝑡q “

ˆ 𝜔q 𝑖 𝜔𝑡 𝑑𝜔 𝑔p0, 𝑒 𝑖𝜔 ´8 2𝜋

d) The fourier transform of 𝐶p𝑡q is the power spectrum of 𝑁p𝑡q i.e. 𝑔p0, ˆ 𝜔q 2 ˆ ˆ 𝐶p𝜔q “ 𝑆p𝜔q “ | 𝑁p𝜔q| “| | 𝑖𝜔 The fourier transform of 𝛿p𝑡q is a constant independent of frequency 𝜔. So, we need 𝑔p0, ˆ 𝜔q9𝜔. e) ż8 𝐶p𝑡q “

ˆ 𝜔q 2 𝑖 𝜔𝑡 𝑑𝜔 𝑔p0, | | 𝑒 𝑖𝜔 ´8 2𝜋

ż8 “

𝑑𝜔 |𝜔| 𝛼 𝑖 𝜔𝑡 𝑒 “ 2 ´8 2𝜋 𝜔

Substitute 𝑍 “ 𝜔𝑡 ùñ 𝜔 “ 𝑍𝑡 ùñ 𝑑𝑤 “ 1𝑡 𝑑𝑍

ż8

𝑑𝜔 𝛼´2 𝑖𝑡 |𝜔| 𝑒 ´8 2𝜋

Complexity Science

1 𝐶p𝑡q “ 2𝜋

ż8

𝑑𝑍 𝑍 𝛼´2 𝑖𝑍 1´ 𝛼 | | 𝑒 9𝑡 𝑡 ´8 𝑡

49

50

Complexity Science

Chapter 11 Exercise 1 a) The probability that a randomly chosen person doesn’t belong to infinite 𝐵 cluster “ 1 ´ 𝑃8 ” 𝐼 For a person not to belong to the infinite cluster none of the person’s neighbours can belong to the infinite cluster. This happens with probability p1 ´ 𝑝𝑃8 q𝑞𝑐 ” 𝐼 𝐼 I = II 1 ´ 𝑃8 “ p1 ´ 𝑝𝑃8 q𝑞𝑐

No zero solution for 𝑃8 appears when 𝑝 ą 𝑝 𝑐 , where 𝑝 𝑐 is determined by the following criterion: 𝑓 p𝑥q “ p1 ´ 𝑝𝑥q𝑞𝑐 , 𝑓 1 p0q “ ´1

𝑓 1 p𝑥q “ ´𝑞 𝑐 𝑝p1 ´ 𝑝𝑥q𝑞𝑐 ´1

𝑓 1 p0q “ ´𝑞 𝑐 𝑝 Hence, 𝑓 1 p0q “ ´1 ùñ 𝑝 𝑐 “ 𝑞1𝑐 b) Expand 𝑓 p𝑥q “ p1 ´ 𝑝𝑥q𝑞𝑐 for small x 𝑓 ”p𝑥q “ 𝑝 2 𝑞 𝑐 p𝑞 𝑐 ´ 1qp1 ´ 𝑝𝑥q𝑞𝑐 ´2

Complexity Science

ùñ 𝑓 ”p0q “ 𝑝 2 𝑞 𝑐 p𝑞 𝑐 ´ 1q The Eq. from (a) now becomes 1 ´ 𝑥 “ 1 ´ 𝑝𝑞 𝑐 𝑥 `

1 2 𝑝 𝑞 𝑐 p𝑞 𝑐 ´ 1q𝑥 2 2

𝑝𝑞 𝑐 ´ 1 2 2 𝑝 𝑞 𝑐 p𝑞 𝑐 ´ 1q

ùñ 𝑥 “ 1 Substitute 𝑝 “ 𝑝 𝑐 ` Δ “ 𝑞1𝑐 ` Δ 𝑥»

2𝑞 𝑐 Δ 2 2 1 ` 𝑞 𝑐 Δ ` 𝑞 𝑐 Δ ´ 𝑞1𝑐 ´ 𝑞 2𝑐 Δ2 ´ Δ » 2𝑞 𝑐 Δr1 ´

𝑃8 »

1 ` 𝑂pΔqs´1 𝑞𝑐

2𝑞 2𝑐 2𝑞 2𝑐 Δ“ p𝑝 ´ 𝑝 𝑐 q𝛽 𝑞𝑐 ´ 1 𝑞𝑐 ´ 1 𝛽“1

c) 𝑃p𝑠q “ 𝑠p1 ´ 𝑝q𝑝 𝑠 p1 ´ 𝑝q where 𝑠 is number of ways to choose B-people from size 𝑠 clusters and rest of the terms give the probability that 𝑠 sites are B-sites and that a A-site is situated at each end. d) Average ă 𝑠 ą“

8 ÿ

𝑠𝑃p𝑠q 𝑠 “1

“ p1 ´ 𝑝q2

8 ÿ

𝑠2 𝑝 𝑠

𝑠 “1

“ p1 ´ 𝑝q2 𝑝

“ p1 ´ 𝑝q2 𝑝

8 𝑑 𝑑 ÿ 𝑠 𝑝 𝑝 𝑑𝑝 𝑑𝑝 𝑠“1

𝑑 𝑑 𝑝 1` 𝑝 𝑝 “𝑝 𝑑𝑝 𝑑𝑝 1 ´ 𝑝 1´ 𝑝

e) In one dimension, a system spanning infinite clusters apprears as 𝑝 Ñ Ý 1 “ 𝑝 𝑐 . Hence from (d), ă 𝑠 ą 9p𝑝 𝑐 ´ 𝑝q´1 , therefore, 𝛾 “ 1

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Exercise 2 Merger is initiated with probability 𝑝 𝑚𝑒𝑟 “ 𝑝 𝑚𝑒𝑟 𝑔𝑒𝑟 “ 𝑝p1 ` 𝑛 𝐴q 𝛼 where 𝑛 𝐴 is the ancestry of the initiating cell. We will think of 𝑛 𝐴 as the number of balls in the bin. a) Write simple code and simulate b) To derive the mean-field equations, we just need to establish the mean field master equation describing how the balls are moved around. 𝑃p𝑛 𝐴, 𝑡q “ probability that a randomly chosen company (bin) has ancestry 𝑛 𝐴 (contains 𝑛 𝐴 balls) Consider 𝑛 𝐴 “ 0 𝑡 ÿ

𝑃p0, 𝑡 ` 1q “ 𝑃p0, 𝑡q `

𝑝 𝑚𝑒𝑟 p𝑛q𝑃p𝑛, 𝑡qp1 ´ 𝑃p0, 𝑡qq 𝑛“1

The probability that a bin contains zero balls will increase every time balls are moved from a bin chosen to be merged with a bin that initiate the merger. At time 𝑡, bins can contain 1, 2, ..., 𝑡 balls, so we need to sum over 𝑛 “ 1, ..., 𝑡 where 𝑛 is the number of balls of the bin that initiate the merger.

The site randomly chosen to initiate the merger contains 𝑛 balls with probability 𝑃p𝑛, 𝑡q and triggers the merger with probability 𝑝 𝑚𝑒𝑟 p𝑛q. The bin randomly chosen for merger with the initiating site will become an empty bin if it contains balls. This bin is non-empty with probability 1 ´ 𝑃p0, 𝑡q.

Consider 𝑛 𝐴 “ 1 𝑃p1, 𝑡 ` 1q “ 𝑃p1, 𝑡q ´ 𝑃p1, 𝑡q𝑝 𝑚𝑒𝑟 p1q

𝑡 ÿ

𝑃p𝑘, 𝑡q 𝑘 “1

𝑡 ÿ

´𝑃p1, 𝑡q

𝑝 𝑚𝑒𝑟 p𝑛q𝑃p𝑛, 𝑡q 𝑛“1

Explanation of 2nd term on the righthand side 𝑃p1, 𝑡q is the probability that bin with one ball is chosen to initiate merger and therefore cease to be a 𝑛 “ 1 bin. 𝑝 𝑚𝑒𝑟 p1q is probability that merger goes ahead. 𝑝 p1q “ probability that merger goes ahead. ř𝑚𝑒𝑟 𝑡 𝑘 “1 𝑃p𝑘, 𝑡q “ the sum over the probabilities that the bin chosen to merge with initiator contains 𝑘 balls with 𝑘 “ 1, 2, ..., 𝑡 Explanation of the 3rd term ř𝑡 𝑝 p𝑛q𝑃p𝑛, 𝑡q is the total probability that the merger is initiated by a bin with 𝑛 “ 1, 2, ..., 𝑡 𝑛“1 𝑚𝑒𝑟 balls. 𝑃p1, 𝑡q “ probability that site chosen for merger with initiator has 𝑛 𝐴 “ 1 ball and therefore after the merger cease to be a 𝑛 𝐴 “ 1 bin.

Consider 2 ă 𝑛 𝐴 ď 𝑡 𝑃p𝑛, 𝑡 ` 1q “ 𝑃p𝑛, 𝑡q `

𝑛ÿ ´1

𝑝 𝑚𝑒𝑟 p𝑘q𝑃p𝑘, 𝑡q𝑃p𝑛 ´ 𝑘, 𝑡q 𝑘 “1 𝑡 ÿ

´𝑝 𝑚𝑒𝑟 p𝑛q𝑃p𝑛, 𝑡q

𝑃p𝑘, 𝑡q 𝑘 “1

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𝑡 ÿ

´

𝑝 𝑚𝑒𝑟 p𝑘q𝑃p𝑘, 𝑡q𝑃p𝑛, 𝑡q 𝑘 “1

Explanation of 2nd term on the righthand side: 𝑃p𝑘, 𝑡q “ probability bin with 𝑘 balls chosen to initiate merger. 𝑝 𝑚𝑒𝑟 p𝑘q “ probability merger goes ahead. 𝑃p𝑛 ´ 𝑘, 𝑡q “ probability selected bin to be merged contains 𝑛 ´ 𝑘 balls so initiating bin will contain 𝑛 balls after merger.

Explanation of the 3rd term: 𝑃p𝑛, 𝑡q “ Probability bin chosen to initiate merger contains 𝑛 balls. When merged with bin containing 𝑘 “ 1, ..., 𝑡 balls, the bin will contain 𝑛 ` 𝑘 balls and hence probability weight has to be subtracted from 𝑃p𝑛, 𝑡q. 𝑝 𝑚𝑒𝑟 p𝑛q “ Probability merger goes ahead. ř 𝑡 𝑘 “1 𝑝p𝑘, 𝑡q “ total probability that bin to be merged contains 𝑘 “ 1, ..., 𝑡 balls.

Explanation of 4th term 𝑃p𝑘, 𝑡q “ probability bin chosen to initiate merger contains 𝑘 balls. 𝑝 𝑚𝑒𝑟 p𝑘q “ probability merger goes ahead. 𝑃p𝑛, 𝑡q “ probability bin to be merged contains 𝑛 balls. Since, after merger the bin will be empty, the merger will remove probability weight from 𝑃p𝑛, 𝑡q. c) No sample solution needed.

Exercise 3 See Section 11.2. One just need to repeat the steps from Eq. (11.28) to Eq. (11.35) with this new set of parameters.

Exercise 4 A full comprehensive study of the model is obviously a major undertaking. But simply, "playing" around with a few simulations of a few parameter choices will indicate that even a small proportion of agents with 𝜔𝑖 close to zero has a strong effect on the asymptotic state (after long time) of the system.

Project 1 a) ă 𝑥 𝑖 ą“ 𝑓 p𝛽𝑥˜𝑖 q ´ 𝑓 p´𝛽𝑥˜𝑖 q “

1 1 1 p ´ q 2 1 ` 𝑒 ´𝛽 𝑥˜𝑖 1 ` 𝑒 𝛽 𝑥˜𝑖

1 sinhp𝛽𝑥˜𝑖 q 1 “ tanhp𝛽𝑥˜𝑖 q 2 1 ` coshp𝛽𝑥˜𝑖 q 2 b) In the above expression replace 𝑥˜𝑖 by 𝑞 ă 𝑥 ą and ă 𝑥𝑖 ą by ă 𝑥 ą

ă 𝑥 ą“ c)

1 sinhp𝛽𝑞 ă 𝑥 ąq ” 𝐹pă 𝑥 ąq 2 1 ` coshp𝛽𝑞 ă 𝑥 ąq

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d) Slope of 𝐹pă 𝑥 ąq at ă 𝑥 ą“ 0 sinhp𝑢q 𝑑𝐹 1 𝑑 “ 𝛽𝑞 𝑑ă𝑥ą 2 𝑑𝑢 1 ` coshp𝑢q “

1 ` coshp𝑢q 1 𝛽𝑞 2 p1 ` coshp𝑢qq2

𝛽𝑞 𝑑𝐹 For 𝑢 “ 0 Ñ Ý 𝑑ă 𝑥 ą |0 “ 4 Critical value: 𝛽𝑐4𝑞 “ 1 Ñ Ý 𝛽𝑐 “ 𝑞4

e) Expand about ă 𝑥 ą“ 0 at 𝛽 “ 𝛽𝑐 𝐹 2 p0q “

p𝑝𝑞q2 p𝛽𝑞q2 𝑑 1 sinh 𝑢 |𝑢 “0 “ |𝑢“0 “ 0 2 𝑑𝑢 1 ` cosh 𝑢 2 p1 ` cosh 𝑢q2 𝐹 3 p0q “

“´

p𝛽𝑞q2 𝑑 sinh 𝑢 |𝑢“0 2 𝑑𝑢 p1 ` cosh 𝑢q2

p𝛽𝑞q3 coshp1 ` cosh 𝑢q2 ´ sinh 𝑢p1 ` cosh 𝑢q2 sinh 𝑢 q|𝑢“0 p 2 p1 ` cosh 𝑢q4 “

´1 p𝛽𝑞q3 8

The ME Equation becomes 3rd order: ă 𝑥 ą“

𝛽𝑞 1 1 ă𝑥ą´ p𝛽𝑞q3 ă 𝑥 ą3 4 3! 8

use 𝛽𝑐 “ 𝑞4 : 𝛽 𝛽𝑞 3 𝛽𝑐 ´ 1 4 ´1 ă 𝑥 ą “ 3!8 “ 4 p 𝛽 q3 p𝛽𝑞q3 𝛽𝑐 2

3 Δ ùñ ă 𝑥 ą» 4𝛽𝑐

a 3𝑞Δ 4

Therefore, in the MF the order parameter exponent is 12 . f) Master Equation: ÿ ÿ 𝑑 𝑃p𝑥, 𝑡q “ 𝑊p𝑥, 𝑥 1 q𝑃p𝑥, 𝑡q ´ 𝑊p𝑥 1 , 𝑥q𝑃p𝑥, 𝑡q 𝑑𝑡 𝑥1 𝑥1 The only non-zero transition coefficients are: 𝑊p0, ´1q “ 𝛾, 𝑊p0, 1q “ 𝜂, 𝑊p´1, 0q “ 𝜔, 𝑊p1, 0q “ 𝜔

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55

Therefore, p𝐸1q

𝐴9 “ 𝜔𝐶 ´ 𝛾 𝐴 ” 𝐻

p𝐸2q

𝐵9 “ 𝜔𝐶 ´ 𝜂𝐵 ” 𝐺

p𝐸3q

𝐶9 “ 𝛾 𝐴 ` 𝜂𝐵 ´ 2𝜔𝐶

And, 𝐴 ` 𝐵 ` 𝐶 “ 𝑁 ùñ 𝐶 “ 𝑁 ´ 𝐴 ´ 𝐵 g) Fixed point 𝐻 “ 𝐺 “ 0 𝐻“0Ñ Ý 𝜔𝑁 ´ p𝜔 ` 𝛾q𝐴 ´ 𝜔𝐵 “ 0 𝐺“0Ñ Ý 𝜔𝑁 ´ 𝜔𝐴 ´ p𝜔 ` 𝜂q𝐵 “ 0 Note 𝐸 1 ´ 𝐸 2 Ñ Ý 𝐴 “ 𝛾𝜂 𝐵 and solve to get: 𝐴“

𝜔𝜂 𝑁 𝜔p𝛾 ` 𝜂q ` 𝛾𝛿

𝐵“

𝜔𝛾 𝑁 𝜔p𝛾 ` 𝜂q ` 𝛾

𝐶“

𝛾𝜂 𝑁 𝜔p𝛾 ` 𝜂q ` 𝛾𝜂

h) Linear stability analysis: B𝐻 B𝐻 “ ´p𝜔 ` 𝛾q; “ ´𝜔 B𝐴 B𝐵 B𝐺 B𝐺 “ ´𝜔; “ ´p𝜔 ` 𝜂q B𝐴 B𝐵 Eigenvalues: ´p𝜔 ` 𝛾q ´ 𝜆 ´𝜔 “0 ´𝜔 ´p𝜔 ` 𝜂q ´ 𝜆 𝜆˘ “

1 t´p2𝜔 ` 𝛾 ` 𝜂q ˘ rp2𝜔 ` 𝛾 ` 𝜂q2 ´ 4p𝜔𝜂 ` 𝛾𝜔 ` 𝛾𝜂qs1{2 u 2

Clearly, 𝜆˘ ă 0 Ñ Ý attractive fixed point. Further reduction leads to 𝜆˘ “

1 p´p2𝜔 ` 𝛾 ` 𝜂q ˘ r4𝜔2 ` p𝛾 ´ 𝜂q2 s1{2 q 2

The relaxation towards the fixed point is controlled by the factors 𝑒 𝜆` 𝑡 and 𝑒 𝜆´ 𝑡 , so the time scales are given by 𝜏` “

1 1 and 𝜏´ “ |𝜆` | |𝜏´ |

The longest time scale 𝜏` will dominate the approach to the asymptotic fixed point.

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i) From (f) we now have 𝑊p0, ´1q “ 𝛾𝐶 and 𝑊p0.1q “ 𝜂𝐶. This gives us the following set of equations 𝐴9 “ 𝜔𝐶 ´ 𝛾 𝐴𝐶 ” 𝐻 𝐵9 “ 𝜔𝐶 ´ 𝜂𝐵𝐶 ” 𝐺

𝐶 “ 𝑁´𝐴´𝐵 j) We have 𝐴9 “ 𝐻 “ r𝜔 ´ 𝛾 𝐴sr𝑁 ´ 𝐴 ´ 𝐵s 𝐵9 “ 𝐺 “ r𝜔 ´ 𝜂𝐵sr𝑁 ´ 𝐴 ´ 𝐵s 𝜔 ˚ Fixed point at 𝐴˚ “ 𝜔 𝛾 and 𝐵 “ 𝜂 . Note also that evolution stops if 𝐴 ` 𝐵 “ 𝑁.

Linear stability matrix:

B𝐻 “ ´𝛾𝑁 ` 2𝛾 𝐴 “ 𝛾𝐵 ´ 𝜔 B𝐴 B𝐻 “ ´r𝜔 ´ 𝛾 𝐴s B𝐵 B𝐺 “ ´r𝜔 ´ 𝜂𝐵s B𝐴 B𝐺 “ ´𝜂𝑁 ` 𝛿 𝐴 ` 2𝜂𝐵 ´ 𝜔 B𝐵 At fixed point, B𝐻 𝛾𝜔 B𝐻 “ ´𝛾𝑁 ` 𝜔 ` , “0 B𝐴 𝜂 B𝐵 B𝐺 B𝐻 𝜂𝜔 “ 0, “ ´𝜂𝑁 ` 𝜔 ` B𝐵 B𝐵 𝛾 Eigenvalues: 0 ´p𝛾𝑁 ` 𝜔 ` 𝛾𝜂𝜔 q ´ 𝜆 “0 0 ´p𝜂𝑁 ` 𝜔 ` 𝜂𝛾𝜔 q ´ 𝜆 ùñ 𝜆 𝐴 “ 𝜔p1 `

𝜆 𝐵 “ 𝜔p1 `

𝛾 q ´ 𝛾𝑁 𝜂

𝜂 q ´ 𝜂𝑁 𝛾

Note: 𝜆 𝐴 ě 0 ùñ 𝑁 ď

𝜔 𝛾 p1 ` q “ 𝐴˚ ` 𝐵˚ ” 𝑁 𝑐 𝛾 𝜂

Complexity Science

𝜆 𝐵 ě 0 ùñ 𝑁 ď

𝜔 𝜂 p1 ` q “ 𝐴˚ ` 𝐵˚ ” 𝑁 𝑐 𝜂 𝛾

We have, For 𝑁 ă 𝑁 𝑐 fixed point repulsive For 𝑁 ą 𝑁 𝑐 fixed point attractive For 𝑁 ą 𝑁 𝑐 , 𝐴 Ñ Ý 𝐴˚ and 𝐵 Ñ Ý 𝐵˚ and evolution asymptotically stop with lim𝑡 Ñ Ý 8 𝐶p𝑡q ą 0 For 𝑁 ă 𝑁 𝑐 , we return to, and recall 𝐴 ` 𝐵 ă 𝑁 𝑐 A. “ 𝛾p𝐴˚ ´ 𝐴qr𝑁 ´ p𝐴 ` 𝐵qs B. “ 𝜂p𝐵˚ ´ 𝐵qr𝑁 ´ p𝐴 ` 𝐵qs where 𝑁 ´ p𝐴 ` 𝐵q is positive and if 𝐴p0q ą 𝐴˚ , 𝐴p𝑡q will decrease towards 𝐴˚ . If 𝐴p0q ă 𝐴˚ , 𝐴p𝑡q will increase towards 𝐴˚ .

The evolution of 𝐴p𝑡q and 𝐵p𝑡q stops when 𝐴p𝑡q ` 𝐵p𝑡q “ 𝑁 with lim𝑡 Ñ Ý 8 𝐶p𝑡q “ 0

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Project 2 a) In the voter model, each agent updates its state (opinion) by randomly copying the opinion of one of its neighbours. b) Axelrod model. A randomly chosen agent selects another agent at random. The two agents interact with a probability given by the two agent’s "cultural similarity". c) Agents on a directed network exchange opinions like in the voter model if they are sufficiently similar like in the Axelrod model. d) When the network contains loops that can act as echochambers. e) Intermittent changes in the distribution of worldviews occuer when influential agents change their opinion on issues of importance. This can trigger a cascade of opinion change. f) The Tangled Axelrod Model consists of agents copying "strategies" from each other. So, this is similar to Axelrod and voter. But, in contrast to Axelrod, the agents are placed on a network and strategies are copied accoprding to their frequencies rather than the similarity between two agents. g) Intermittency in the tangled Axelrod model is due to the appearence of disruptive innovation from new agents which can destabilise and trigger the change. h) Both models attempt to extract few imnportant features. They do well on P1. They are very simple relative to the considered phenomena. Fine on P2. Their generaility is not yet fully clear. Uncertain on P3.

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Chapter 12 Exercise 1 Solar flares also show deviation from pure power law behaviour and the driven OFC model seems to exhibit correlations similar to those seen in solar flares. See also the paper: "Universality in Solar Flares and Earthquake Occurance", L.de Arcangelis, C.Godano, E. Lippiello, and M. Nicodemi. Phys. Rev. Lett. 96, 051102 (2006).

Exercise 2 No sample solution needed Exercise 3 To derive Eq. (12.23) 𝑃p𝑡q “ 𝑡 `1 1 , just follow Eqns. (12.16) to (12.23) step by step. All details are given in the text. Eq.(12.27) 𝑃𝑛 p𝑡q “

1 pln 𝑡qp𝑛´1q 𝑡 p𝑛 ´ 1q!

First consider the case 𝑛 “ 3 i.e. two records in addition to the one at 𝑡 “ 0. In thi case, Eq. (12.25) becomes 𝑃3 p𝑡; 0, 𝑡1 , 𝑡2 q “

1 𝑡𝑡2 𝑡 1

.

We sum over 𝑡 1 and 𝑡2 , 𝑡

𝑃3 p𝑡q “

𝑡 ´1

2 1 ÿ ÿ 1 1 » 𝑡 𝑡 “2 𝑡 “1 𝑡2 𝑡1 𝑡 2

1

ż𝑡

ż 𝑡2 𝑑𝑡 2

1

𝑑𝑡 1 1

1 𝑡1 𝑡2

Note the integrand is symmetric under reflection in the identity line 𝑦 “ 𝑥. So, we can write 𝑃3 p𝑡q »

1 “ 2𝑡

ż ż 𝑡1 ż𝑡 ż 𝑡2 11 𝑡 1 1 t 𝑑𝑡 1 𝑑𝑡 2 ` 𝑑𝑡 2 𝑑𝑡 1 u 𝑡2 1 𝑡2 𝑡1 𝑡2 𝑡1 1 1 1

ż𝑡 𝑑𝑡 1 1

ż pln 𝑡q2 1 1 𝑡 1 2 𝑑𝑡 2 “ r 𝑑𝑧 s “ 𝑡1 𝑡2 2𝑡 1 𝑧 2𝑡 1

ż𝑡

which is the form giveb in Eq. (12.27) for 𝑛 “ 3. We now do the induction step i.e. we assume Eq(12.27) is correct of 𝑛 and show the case 𝑛 ` 1 follows. We write:

𝑃𝑛 p𝑡q “

𝜇𝑡𝑛´1 1 pln 𝑡q𝑛´1 “ 𝑒 ´ 𝜇𝑡 𝑡 p𝑛 ´ 1q! p𝑛 ´ 1q!

where we introduced "logarithmic time" 𝜇𝑡 “ ln 𝑡.

Next, we repeat the argument used in Eq. (6.96) ÿ 𝑃𝑛`1 p𝑡q “

𝑃𝑛 p𝜇𝑡 ˚ q𝑃1 p𝜇𝑡 ´ 𝜇𝑡 ˚ q 𝜇𝑡 ˚

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i.e. 𝑛 records upto 𝜇𝑡 ˚ and exactly one record during 𝜇𝑡 ˚ to 𝜇𝑡 . ÿ 𝑃𝑛`1 p𝑡q “

𝑒 ´ 𝜇𝑡 ˚

𝜇𝑡 ˚

“

𝜇𝑡 ˚ 𝑒 ´p 𝜇𝑡 ´ 𝜇𝑡 ˚ q p𝑛 ´ 1q!

𝑒 ´ 𝜇𝑡 ÿ ´ 𝜇𝑡 ˚ 𝑛´1 𝑒 𝜇𝑡 ˚ p𝑛 ´ 1q! 𝜇 ˚ 𝑡

For the sum, we have ÿ

𝑒 ´ 𝜇𝑡 ˚ 𝜇𝑡𝑛˚´1 “

𝑡 ˚ ě𝑛

𝜇𝑡 ˚

ż 𝑡 ´1 » 𝑛

“

𝑡ÿ ´1

pln 𝑡 ˚ q𝑛´1 ˚ 𝑑𝑡 “ 𝑡˚

pln 𝑡 ˚ q𝑛´1 𝑡˚

ż lnp𝑡 ´1q

𝑧 𝑛´1 𝑑𝑧

ln 𝑛

plnp𝑡 ´ 1qq𝑛 pln 𝑛q𝑛´1 pln 𝑡q𝑛 “ » 𝑛 𝑛 𝑛

So, in summary, 𝑃𝑛`1 p𝑡q “

pln 𝑡q𝑛 pln 𝑡q𝑛 1 1 “ 𝑡 p𝑛 ´ 1q! 𝑛 𝑡𝑛!

which is of the form (12.27) for 𝑛 ` 1 records during 𝑡.

Exercise 4 Logistic map 𝑓 p𝑥q “ 𝑟𝑥p1 ´ 𝑥q fixed point: 𝑥 ˚ “ 𝑓 p𝑥 ˚ q i.e. 𝑥 ˚ “ 𝑟𝑥 ˚ p1 ´ 𝑥 ˚ q Ñ Ý 𝑥 ˚ “ 1´𝑟 𝑟 ´1 1 1 Tangent point 𝑥 𝑡 𝑎𝑛 : 𝑓 p𝑥 𝑡 𝑎𝑛 q “ 1, 𝑓 p𝑥q “ 𝑟 ´ 2𝑟𝑥 i.e. 𝑟 ´ 2𝑟𝑥 𝑡 𝑎𝑛 “ 1 Ñ Ý 𝑥 𝑡 𝑎𝑛 “ 𝑟2𝑟 Eq. (12.40)

𝜋 𝑇“? 𝜖𝛽

Here we need to use |𝛽| since our 𝛽 ă 0. 𝛽 “ 𝑓 2 p𝑥 𝑡 𝑎𝑛 q 𝜖 “ 𝑓 p𝑥 𝑡 𝑎𝑛 q ´ 𝑥 𝑡 𝑎𝑛 Since, 𝑓 2 p𝑥q “ ´2𝑟, we have 𝛽 “ ´2𝑟 and further: 𝜖 “ 𝑓p

𝑟 ´1 𝑟 ´1 1 3 q´ “ 1 ´ p ` 𝑟q 2𝑟 2𝑟 4 𝑟

with 𝑟 “ 1 ` Δ and 0 ă Δ ăă 1, we can write, 1 3 1 𝜖 “1´ p ` 1 ` Δq » 1 ´ r3p1 ´ Δ ` Δ2 q ` 1 ` Δs 4 1`Δ 4

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1 Δ “ 1 ´ r4 ´ 2Δ ´ 3Δ2 s » 4 2 𝜋 𝑇»b

Δ 2 2p1 ` Δq

𝜋 »? Δ

Exercise 5 No sample solution needed.

Project 1 a) 𝑝 0 “ 1 ´ 𝑝 , 𝑝 1 “ 0, 𝑝 2 “ 𝑝, 𝑝 𝑘 “ 0, 𝑘 ą 𝑧 b) 8 ÿ 𝑔p𝑠q “ 𝑝 𝑘 𝑠 𝑘 “ p1 ´ 𝑝q ` 𝑝𝑠2 𝑘 “0

Process critical when average branch 𝜇 “ 1. 𝜇 “ 𝑔 1 p1q “ 2𝑝 so 𝜇 “ 1 ùñ 𝑝 “ 𝑝 𝑐 “

1 2

c) From lectures 𝑃p𝜏 ą 𝑛q9𝜇 𝑛 “ expp𝑛 ln 𝜇q. Define 𝑛0 “ ln´1𝜇 P R` and we have 𝑛 𝑃p𝜏 ă 𝑛q “ expp´ 𝑛0 q. d) 1 1 ´1 » 𝑛0 “ ´ “ ln 𝜇 Δ lnp1 ´ Δq 𝜇 “ 1 ´ Δ ùñ 𝑎 “ 1 e) When an energy unit is added, one of the 𝑁 “ 2p𝑛`1q ´ 1 sites receives a unit with probability 𝑁1 .

As a result, the site moves towards the threshold for relaxation, which is represented by an increase in the overall relaxation probability 𝑝p𝑡q “ 𝑝2, namely 𝑝p𝑡q Ñ Ý 𝑝p𝑡q ` 𝑁1 . Similarly, relaxation at the boundary lowers the amount of energy on the system and the relaxation probability is lowered by an amount equal to the average loss per site: 𝑝p𝑡q Ñ Ý 𝑝p𝑡q ´

𝜎𝑛 p𝑡q 𝑁

This is summarised by the mean-field eq. 𝑝p𝑡 ` 1q “ 𝑝p𝑡q `

𝜎𝑛 p𝑡q 1 ´ 𝑁 𝑁

So, in continuous time, 1 ´ 𝜎𝑛 p𝑡q 𝑑𝑃 “ 𝑑𝑡 𝑁 If two units are lost, each boundary relaxation we have: 1 ´ 2𝜎𝑛 p𝑡q 𝑑𝑃 “ 𝑑𝑡 𝑁

62

Complexity Science On average (as claimed in the the paper), ă 𝜎𝑛 ą“ p2𝑝q𝑛 , so the fixed point value of 𝑝 will be given by 1 ´ 2p2𝑝q𝑛 “ 0 ùñ 𝑝 “

1 ln 2 1 expp´ qÑ Ý as 𝑛 Ñ Ý 8 2 𝑛 2

f) No sample solution needed. g) From Eq. (3), neglecting the noise term, we have 𝑑𝑝 1 “ 𝑑𝑡 𝑁 A linear estimate for 𝑝p𝑡q is then 𝑝p𝑡q “

1 𝑡 ` 𝑝0 𝑁

𝑝p𝑡 ˚ q “

ùñ

1 2

1 ˚ 1 𝑡 ` 𝑝0 “ 𝑁 2

1 𝑡 ˚ “ 𝑁p ´ 𝑝 0 q 2 1 “ p ´ 𝑝 0 qp2𝑛`1 ´ 1q92𝑛`1 for large n 2 h) No sample solution needed i) No sample solution needed. j) We know that ă 𝜎𝑛 ą“

𝑑 𝑑 𝑔 𝜎 p𝑠q|𝑠“1 “ 𝑔p𝑔 𝜎𝑛´1 p𝑠qq|𝑠“1 𝑑𝑆 𝑛 𝑑𝑠 “ r𝑔 1 p1qs𝑛 “ 𝜇 𝑛

and from (b) we have 𝜇 “ 2𝑝 Ñ Ý ă 𝜎𝑛 ą“ 𝜇 𝑛 .

Total progeny, 8 8 8 ÿ ÿ ÿ ă 𝑌8 ą“ă 𝜎𝑛 ą“ ă 𝜎𝑛 ą“ 𝜇𝑛 “ 𝑛“0

𝑛 “0

𝑛“0

1 1´𝜇

From the lecture notes, we have 𝑔𝑌8 p𝑠q “ 𝑠𝑔p𝑔𝑌8 p𝑠qq and ă 𝑌8 ą“ 𝑔𝑌1 8 p1q By implicit differentiation, we have, 𝑔𝑌1 8 p𝑠q “ 𝑔p𝑔𝑌8 p𝑠qq ` 𝑠𝑔 1 p𝑔𝑌8 p𝑠qq𝑔𝑌1 8 p𝑠q

𝑔𝑌1 8 p𝑠q “

𝑔𝑌8 p𝑠q 1 ´ 𝑠𝑔 1 p𝑔𝑌8 p𝑠q

Complexity Science For 𝜇 ă 1, we have 𝑔 1 p𝑔𝑌8 p1qq “ 𝑔 1 p1q “ 𝜇 ùñ ă 𝑌8 ą“

1 1´𝜇

k) Let 𝑔 𝑋 p𝑠q be the generator function for the stochastic variable 𝑋 we then have ă 𝑋 ą“ 𝑔 1𝑋 p1q. 𝑔”𝑋 p1q “ă 𝑋 2 ą ´ ă 𝑋 ą and therefore, Var𝑋 “ă 𝑋 2 ą ´ ă 𝑋 ą2 “ 𝑔”𝑋 p1q ` 𝑔 1𝑋 p1q ´ p𝑔 1𝑋 p1qq2 From the lecture notes, we have, 𝑔 𝜎𝑛 p𝑠q “ 𝑔p𝑔 𝜎𝑛´1 p𝑠qq From (b), we know, 𝑔p𝑠q “ 1 ´ 𝑝 ` 𝑝𝑠2 𝑔 1 p𝑠q “ 2𝑝𝑠 “ 𝜇𝑠 𝑔 2 p𝑠q “ 2𝑝 “ 𝜇 Notation: 𝑔𝑛 p𝑠q ” 𝑔 𝜎𝑛 p𝑠q

𝑑 𝑑 𝑑𝑔𝑛 “ 𝑔p𝑔𝑛´1 p𝑠qq “ p1 ´ 𝑝 ` 𝑝p𝑔𝑛´1 p𝑠qq2 q 𝑑𝑠 𝑑𝑠 𝑑𝑠 “ 2𝑝𝑔𝑛´1 p𝑠q𝑔𝑛1 ´1 p𝑠q “ 𝜇𝑔𝑛´1 p1q𝑔𝑛1 ´1 p𝑠q 𝑑 2 𝑔𝑛 “ 𝜇pp𝑔𝑛1 ´1 p𝑠qq2 ` 𝑔𝑛´1 p𝑠q𝑔𝑛´1 p𝑠q𝑔𝑛2´1 p𝑠qq 𝑑𝑠2 Recall that 𝑔𝑛´1 p1q “ 1 and 𝑔𝑛1 ´1 p1q “ 𝜇 𝑛´1 𝑑 2 𝑔𝑛 |𝑠 “ 1 “ 𝜇r𝜇2p𝑛´1q ` 𝑔𝑛2´1 p𝑠qs 𝑑𝑠2 Notation: 𝑥 𝑛 ” 𝑔𝑛2 p1q and 𝑥 𝑛 “ 𝜇p𝑥 𝑛´1 ` 𝜇2p𝑛´1q q Recall 𝑋1 “ 𝑔12 p1q “ 𝜇 Hence, 𝑥 𝑛 “ 𝜇p𝑥 𝑛´1 ` 𝜇2p𝑛´1q q “ 𝜇r𝜇p𝑥 𝑛´2 ` 𝜇2p𝑛´2q q ` 𝜇2p𝑛´1q s

63

64

Complexity Science “ 𝜇2 𝑥 𝑛´2 ` 𝜇2𝑛´2 ` 𝜇2𝑛´1

.. .

“ 𝜇 𝑛´1 𝑥1 ` 𝜇2𝑛´p𝑛´1q ` 𝜇2𝑛´p𝑛´2q ` ... ` 𝜇2𝑛´1

“ 𝜇 𝑛 r1 ` 𝜇 ` 𝜇2 ` ... ` 𝜇 𝑛´1 s “ 𝜇 𝑛

1 ´ 𝜇2 1´𝜇

𝑛

So, 𝑔𝑛2 p1q “ 𝜇 𝑛 11´´𝜇𝜇 and, Var 𝜎𝑛 “ 𝑔𝑛2 p1q ` 𝑔𝑛1 p1q ´ r𝑔𝑛1 p1qs2 “ 𝜇𝑛 p

1 ´ 𝜇𝑛 q ` 𝜇 𝑛 ´ 𝜇2𝑛 1´𝜇 𝑛

Ý 𝑛 as 𝜇 Ñ Ý 1.

When 𝑝 Ñ Ý 𝑝𝑐 , 𝜇 Ñ Ý 1, the limit 𝜇 Ñ Ý 1´ . L’Hospital rule 11´´𝜇𝜇 Ñ So, Var 𝜎𝑛 Ñ Ý 𝑛 as 𝜇 Ñ Ý 1´ Project 2 a) Infection occurs when 𝜂 is drawn as a record. We are asked to complete the probability 𝑃p𝑡1 , 𝑡2 q that a record occurs at 𝑡1 and 𝑡 2 . ż𝜂 𝐹p𝜂q “

𝑓 p𝜂q𝑑𝜂 0

ż1 𝑃p𝑡1 , 𝑡2 q “

ż1 𝑑𝜂0 𝑓 p𝜂0 q𝐹

𝑡1 ´1

p𝜂0 q

0

ż1 “ 0

ż1 𝑑𝜂1 𝑓 p𝜂1 q𝐹

𝜂0

𝑑𝑍0 𝑍0𝑡1 ´1

ż1

𝑡2 ´𝑡1 ´2

p𝜂1 q

𝑑𝜂2 𝑑p𝜂2 q 𝜂1

𝑑𝑍1 𝑍1𝑡2 ´𝑡1 ´2 p1 ´ 𝑍1 q with 𝑍𝑖 “ 𝐹p𝜂𝑖 q

𝑍0

“

1 𝑡 1 p𝑡2 ´ 1q𝑡 2

b) We note that 𝑄 is equal to the number of records during the years 𝑡1 to 𝑡 2 . We know that a record process starting at time 0 running to time 𝑡 to a good approximation is a Poisson process 𝑃p𝑋 “ 𝑛q “

𝜆 𝑛´1 𝑒 ´𝜆 p𝑛 ´ 1q!

for 𝑛 ´ 1 events. The ´1 occurs because the event at 𝑡 “ 0 always is a record with average number of events 𝜆 replaced by ln 𝑡. So, we have,

Complexity Science

𝑃𝑟p𝑄 “ 𝑞q “

65

pln 𝑡q𝑞´1 𝑒 ´ ln 𝑡 pln 𝑡q𝑞´1 “ p𝑞 ´ 1q! 𝑡p𝑞 ´ 1q!

Since, we now look at number of events during the years 𝑡1 to 𝑡 2 we replace ln 𝑡 by ln 𝑡 2 ´ ln 𝑡 1 “ ln 𝑡𝑡12 and arrive at rln 𝑡𝑡2 s𝑞´1 𝑃𝑟p𝑄 “ 𝑞q “ 𝑡2 1 𝑡1 p𝑞 ´ 1q! c) The generator function for 𝑃 𝑘 is 8 ÿ

𝑔p𝑠q “

𝑃 𝑘 𝑠 𝑘 “ p1 ´ 𝑒 ´ 𝛼 q

𝑘 “0

8 ÿ

𝑒𝛼 ´ 1 𝑠 p 𝛼 q𝑘 “ 𝛼 𝑒 𝑒 ´𝑠 𝑘 “0

The average 𝜇 “ 𝑔 1 p1q “

𝑒𝛼 ´ 1 1 |𝑠“1 “ 𝛼 𝛼 2 𝑒 ´1 p𝑒 ´ 𝑠q

d) ă 𝑍 𝑛 ą“ 𝑔 1𝑍𝑛 p1q “ “ r𝑔 1 p𝑔 𝑍𝑛´1 p𝑠qq

𝑑 𝑔p𝑔 𝑍𝑛´1 p𝑠qq|𝑠“1 𝑑𝑠 𝑑 𝑔 𝑍 p𝑠qs𝑠“1 𝑑𝑠 𝑛´1

.. . “ r𝑔 1 p𝑔 𝑍𝑛´1 p𝑠qq𝑔 1 p𝑔 𝑍𝑛´2 p𝑠qq...𝑔 1 p𝑔 𝑍0 p𝑠qq𝑔 1 p𝑠qs𝑠“1 Use 𝑔 𝑍𝑛 p1q “ 𝑔p𝑔 𝑍𝑛´1 p1qq “ ... ´ 𝑔p𝑔...𝑔p𝑠qq|𝑠“1 and 𝑔p1q “ 1 to obtain ă 𝑍 𝑛 ą“ r𝑔 1 p1qs𝑛 “ 𝜇 𝑛 and we have 𝑍𝑛 “ 𝜇 𝑛´ 𝑚 𝑍𝑚 e) We know that for a branching process the time to extinction 𝜏 satisfies 𝜇 ă 1 : Probp𝜏 ą 𝑛q9𝜇 𝑛 for 𝑛 ąą 1 1 𝜇 “ 1 : Probp𝜏 ą 𝑛q9 for 𝑛 ąą 1 𝑛 (assuming that the variance of the branching probabilities is finite) We therefore have: 𝜇ă1Ñ Ý

𝑃𝑟p𝑇1 q “ 𝜇𝑇1 ´𝑇2 𝑃𝑟p𝑇2 q

𝜇“1Ñ Ý

𝑃𝑟p𝑇1 q 𝑇2 “ 𝑇1 𝑃𝑟p𝑇2 q

So to have the ratio to only depend on 𝑇𝑇21 we need 𝜇 “ 1 or 𝑒 𝛼1´1 “ 1 ùñ 𝛼 “ ln 2

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Complexity Science

f) Consider 𝑥 𝑛`1 “ 𝑓 p𝑥 𝑛 q Assume that 𝑥 𝑐 exist such that 𝑓 1 p𝑥 𝑐 q “ 1 and let 𝜖 “ 𝑓 p𝑥 𝑐 q ´ 𝑥 𝑐 and Taylor expand

𝑓 p𝑥q “ 𝑓 p𝑥 𝑐 q ` 𝑓 1 p𝑥 𝑐 qp𝑥 ´ 𝑥 𝑐 q `

“ 𝑥 𝑐 ` 𝜖 ` p𝑥 ´ 𝑥 𝑐 q `

1 2 𝑓 p𝑥 𝑐 qp𝑥 ´ 𝑥 𝑐 q2 2

1 𝑓 ”p𝑥 𝑐 qp𝑥 ´ 𝑥 𝑐 q2 2

Define Δ𝑛 “ 𝑥 𝑛 ´ 𝑥 𝑐 and 𝛽 “ 21 𝑓 ”p𝑥 𝑐 q. In this approximation we have from 𝑥 𝑛`1 “ 𝑓 p𝑥 𝑛 q.

Δ𝑛`1 “ 𝜖 ` Δ𝑛 ` 𝛽Δ2𝑛 ùñ Δ𝑛`1 ´ Δ𝑛 “ 𝜖 ` 𝛽Δ2𝑛

assume |𝜖| ăă 1 and |𝛽| ăă 1 and use a continuous approximation

Δ𝑛`1 ´ Δ𝑛 Ñ Ý

𝑑Δ , Δ𝑛 Ñ Ý Δp𝑡q 𝑑𝑡

𝑑Δ 𝑑Δ “ 𝜖 ` 𝛽Δ2 Ñ Ý “ 𝑑𝑡 𝑑𝑡 𝜖 ` 𝛽Δ2

c ùñ Δ “

𝜖 a 𝑡𝑔r 𝜖 𝛽p𝑡 ` 𝑡 0 qs 𝛽

Since, 𝑡 𝑔 goes from ´8 to 8 when the argument varies from ´2𝜋 to 𝜋2 , we estimate the number of iterations needed to pass through the bottleneck at 𝑥 𝑐 as 𝑇 “ ?𝜋 . 𝜖𝛽

g) See figure below

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67

h) In (4), we derived, 𝑇 “ ?𝜋 . We have 𝛽 “ 12 𝑓 ”p𝑥 𝑐 q. In the specific case 𝑓 ”p𝑠q “ 1 for all 𝑥, so 𝜖𝛽

𝛽 “ 1. And we have, 𝜖

𝑇 “ ?𝜋𝜖 ” 𝑇p𝜖q

10´2 10´3 10´4

31.4 99.3 314.2 height

The analytic estimate of 𝑇 becomes better as 𝜖 decreases which is reasonable since we replaced the discrete iterative map by a differential eq. in continous time. As 𝜖 Ñ Ý 0 the number of iterations needed to get through the bottleneck increases and the behaviour is close to the continuous description.

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Complexity Science

Chapter 13 Exercise 1 It is not at all easy to forecast the transition from the fluctuations.

Exercise 2 𝑄p𝑡q is more robust i.e. fewer false positives.

Exercise 3 Identifying the logarithmic time dependence can be difficult.

Project 1 No sample solution needed.