Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

Complexity Science

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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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Complexity Science

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.

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

Complexity Science

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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: 10−6m - 10−4m Plants: 10−3m - 10m (perhaps 102) Clusters of trees: 102 m Forests: 103m - 104m c) Cell body of neuron: 10−6 - 10−4 The axon 𝑚𝑁 Brain regions linear size: 10−2m 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.

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

Complexity Science

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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.

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

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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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Complexity Science

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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Complexity Science

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

𝐴𝑠−𝜏 = 1

𝑠=1

𝐴−1 =

∞ ÿ −𝜏

𝑠 = 𝜁 (𝜏)

𝑠=1

which is Riemann’s Zeta function. To ensure 𝐴 ă 8 we need 𝜏 ą 1 c) Moments ă 𝑆 ą=

∞ ÿ

𝑠𝑃(𝑠) = 𝐴

∞

𝑠=1

=𝐴

∞ ÿ

ÿ

𝑠(𝑎−𝜏)

𝑠=1

𝑆−(𝜏−𝑎) = 𝐴𝜁 (𝜏 ´ 𝑎)

𝑠=1

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

Exercise 2 a)

𝑔𝑠𝑁 (𝑠) =

∞

ÿ

𝑃𝑠𝑁 (𝑞)𝑠 𝑞

𝑞=0

where

𝑃𝑠𝑁 (𝑞) = 𝑝𝑟𝑜𝑏t𝑠𝑁 =

𝑁 ÿ

𝑋𝑡 = 𝑞u

𝑡 =1

∞

ÿ 𝑛

𝑛=0

𝑡 1

= ÿ 𝑃𝑟𝑜𝑏t𝑁 = 𝑛u𝑃𝑟𝑜𝑏t ∞ ÿ

=

𝑋𝑡 = 𝑞u

=

𝑃𝑁 (𝑛)

𝑛=0

ÿ

𝑞1

ÿ ...

𝑃𝑋(𝑞1)...𝑃𝑋(𝑞𝑛)

𝑞𝑛

𝑞1 ` 𝑞2 ` ... ` 𝑞𝑛 = 𝑞 ∞

𝑔𝑠𝑁 (𝑠) = = 𝑛=0

∞ ÿ

ÿ

∞ ÿ

𝑞=0 𝑛=0

𝑃𝑁 (𝑛) 𝑞=0

ÿ

ÿ

𝑁 𝑃 (𝑛) 𝑞 ...𝑞 1 𝑛

𝑃𝑋(𝑞)...𝑃𝑋(𝑞𝑛)𝑠

∞ ÿ ÿ ÿ ( 𝑃𝑋(𝑞1)𝑠𝑞1 )...( 𝑃𝑋(𝑞𝑛)𝑠

𝑞1

𝑞

) 𝑞𝑛

𝑞𝑛

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

Complexity Science

∞ ÿ

∞ ÿ

𝑃𝑁 (𝑛)(

∞

ÿ

𝑃𝑋 (𝑞)𝑠 )𝑞 𝑛

𝑛=0

𝑞=0

𝑃𝑁 (𝑛)(𝑔𝑋

(𝑠)) 𝑛 = 𝑔𝑁 (𝑔𝑋 (𝑠))

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

𝑍𝑛 =

𝑍1 ÿ

𝑍𝑛(𝑡)

𝑡 =1

According to the theorem in (i) we have,

𝑔𝑍𝑛 (𝑠) = 𝑔𝑍1 (𝑔𝑍𝑛 ( 𝑡 )(𝑠)) Next note that the generator for 𝑍1 is identical to the generator for the branching probabilities ∞ ÿ

𝑔( 𝑠 ) =

𝑃 𝑡𝑠𝑡

𝑡 =0

with 𝑃𝑡 = proban individual gets offspring.

The generator for 𝑍𝑛(𝑡) is the same as the generator for the total size of the generation after 𝑛 ´ 1 generations i.e. 𝑔𝑍𝑛 (𝑡 )(𝑠) = 𝑔(𝑔𝑍𝑛—1 (𝑠)). c)

𝑔𝑍𝑛 (𝑠) = 𝑔(𝑔𝑍𝑛—1 (𝑠)) 𝑔𝑍𝑛 (𝑠) = 𝑔(𝑔𝑍𝑛—1 (𝑠)) = 𝛼 ` 𝛽𝑔𝑍𝑛—1 (𝑠) = 𝛼 ` 𝛽(𝛼 ` 𝛽𝑔𝑍𝑛—2 (𝑠)) = 𝛼 ` 𝛼𝛽 ` 𝛽 2 𝑔𝑍𝑛—2 (𝑠) = 𝛼 ` 𝛼𝛽 ` 𝛽2 (𝛼 ` 𝛽𝑔𝑍𝑛—3 (𝑠))

. = 𝛼(1 ` 𝛽 ` 𝛽2 ` ... ` 𝛽𝑛−1) ` 𝛽𝑛𝑠 1 ´ 𝛽𝑛 𝑛 =𝛼 `𝛽 𝑠 1´ 𝛽

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Complexity Science From normalisation,

𝑔(1) = 𝛼 ` 𝛽 = 1 → — 𝛼 = 1´ 𝛽 we obtain,

𝑔𝑍 𝑛 (𝑠) = 1 ´ 𝛽𝑛 ` 𝛽𝑛𝑠 d) Recall the definition of the generator function for 𝑔𝑍𝑛 (𝑠): ∞

ÿ

𝑔𝑍𝑛 (𝑠) =

𝑃𝑟𝑜𝑏t𝑍𝑛

= 𝑘u𝑠 𝑘

𝑘=0

From the result in (iii) we conclude,

𝑃𝑟𝑜𝑏t𝑍𝑛 = 0u = 1 ´ 𝛽𝑛 𝑃𝑟𝑜𝑏t𝑍𝑛 = 1u = 𝛽𝑛 𝑃𝑟𝑜𝑏t𝑍𝑛 = 𝑘u = 0 𝑓 𝑜𝑟𝑘 ě 2 Exercise 3 a) Use the expression for a geometric series ∞ ÿ

1 1´𝑥

𝑎𝑥𝑘 = 𝑎

𝑘=0

and

𝑔(𝑠) =

∞ ÿ

𝑘=0

ÿ 1´𝑝 𝑃𝑘 𝑠 = 1 ´ 𝑝𝑠 = (1 ´ 𝑝) (𝑝𝑠)𝑘 ∞

𝑘

𝑘=0

∞

= ÿ r(1 ´ 𝑝)𝑝𝑘 s𝑠 𝑘 𝑘=0

to conclude that 𝑃𝑘 = (1 ´ 𝑝)𝑝𝑘 b)

𝜇 = 𝑔2(𝑠)|

𝑠=1

=

𝑝 (1 ´ 𝑝 ) | 𝑠=1 (1 ´ 𝑝𝑠)2

c) Critical when

𝜇=1→ —

𝑝 1´ 𝑝

=

𝑝 1´𝑝

=1

— 𝑝=1´ 𝑝 → → — 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 , 𝑔(𝑠), for 𝑍𝑘 and the generator for 𝑌𝑛 are related according to

𝑔𝑌𝑛 (𝑠) = 𝑔(𝑔𝑌𝑛—1 (𝑠)) (See Eq. 5.30) So we have,

𝑔 𝑌𝑛(𝑠) =

1´ 𝑝 1 ´ 𝑝𝑔𝑌𝑛—1

1´ 𝑝 1 ´ 𝑝𝑔(𝑔𝑌𝑛—2 (𝑠)

(𝑠)

=

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 all members originate from member 𝑘 at 1st generation. Let 𝑍(𝑘) denote the number of elements in 𝑛 this group. We now have, 𝑍1

𝑍𝑛 =

ÿ

𝑘

𝑍 (𝑛)

𝑘=1

See Fig. 5.2.

In general, we have for,

𝑆𝑁 =

𝑁 ÿ

𝑋𝑛

𝑛−1

where 𝑁 = 0, 1, 2, ... is a stochastic variable and 𝑋𝑘 P 0,1,2,... are i.i.d that the generators are related according to 𝑔𝑆𝑁 (𝑠 ) = 𝑔𝑁 (𝑔𝑋 𝑠( )) . 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. 𝑔(𝑠) and 𝑋𝑘 = 𝑍𝑛( 𝑘). The generator for 𝑍(𝑘𝑛 ) must, due to the self-similarity, be the same as the generator forthe whole tree offer 𝑛 ´ 1 generation i.e. 𝑔𝑍 (𝑘) (𝑠) = 𝑔𝑍𝑛—1 (𝑠) and we conclude from above that, 𝑛

𝑔𝑍𝑛 (𝑠) = 𝑔(𝑔𝑍𝑛—1 (𝑠)) b) The relation 𝑋𝑛 = 𝑔(𝑋𝑛−1) follows from (a) by noticing that

𝑥𝑛 = 𝑃𝑟𝑜𝑏t𝑍𝑛 = 0u = 𝑔𝑍𝑛 (0) c) This is most easily seen geometrically by iterations the map 𝑥𝑛 = 𝑔(𝑍𝑛−1). Notice 𝑋0 = 𝑝0 andthat 𝑔 is convex since 𝑔„(𝑠) ą 0 and the slope at 𝑠 = 1, 𝑔2(1) = 𝜇

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d) The average branching ratio

𝜇=

∞

ÿ

= 2𝑝

𝑘𝑃 𝑘

𝑘=0

The process is critical for 𝜇 = 1 i.e. for 𝑝 = 𝑝𝑐 = 1 . 2 e) ă 𝑍𝑛 ą= 𝑔2𝑍𝑛 ( 1) = = r𝑔2(𝑔 𝑍𝑛—1 ( 𝑆 ))

𝑑𝑔(𝑔𝑍𝑛—1 (𝑠) 𝑑𝑠

𝑑𝑔 ( 𝑔 𝑍 𝑛— 1 (𝑠) 𝑑𝑠

2 2 s𝑠=1 = r𝑔 (𝑔 𝑍𝑛—1 (𝑠))𝑔 (𝑔

|𝑠=1

𝑍𝑛—2 ( 𝑠 ))

𝑑𝑔(𝑔𝑍𝑛—2 (𝑠) 𝑑𝑠

s𝑠=1

... = r𝑔2(𝑔𝑍𝑛—1 (𝑠))𝑔2(𝑔𝑍𝑛—2 (𝑠))𝑔2(𝑔𝑍0 (𝑠))s𝑠=1 Next, we have,

𝑔𝑍𝑛 (1) = 𝑔(𝑔𝑍𝑛—1 (1)) = .. = 𝑔(𝑔(...(𝑔(1))...))

Furthermore 𝑔(1) = 1 and 𝑔2(1) = 𝜇 = 2𝑝 ă 𝑍𝑛 ą= r𝑔2(1)s𝑛 = (2𝑝)𝑛 f) The probability that the process goes extinct is the smallest root, 𝑥 of the equation,

𝑠 = 𝑔(𝑠) i.e.

a 1 ˘ 1 ´ 4𝑝𝑞 𝑝𝑠𝑠 ` 𝑞 = 𝑠 ùñ 𝑠 = 2𝑝

the smallest root is for

𝑝ă

1

:𝑥=1 2 1 1 𝑞 𝑝ą : ´1= 2 𝑝 𝑝

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

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

𝑔(𝑠) =

𝑞

𝑞(𝑝𝑠)𝑘 =

1 ´ 𝑝𝑠

𝑘=0

Normalising, we get, 1 = 𝑔(1) =

𝑞

ùñ

1´ 𝑝

1

𝑞

ùñ 𝑞 ` 𝑝 = 1

1

=

1´ 𝑝

Average branching ratio is

𝑞𝑝

𝜇 = 𝑔2(1) =

|

(1 ´ 𝑝𝑠)2

𝑞𝑝

= 𝑝

(1 ´ 𝑝)2

𝑞

=

𝑠=1

and therefore, 1 1 1 ´ 𝑝𝑠 𝑞 ´ = ´ 1 ´ 𝑔(𝑠) 𝜇(1 ´ 𝑠) 𝑝(1 ´ 𝑠) 𝑝(1 ´ 𝑠) 1 ´(1 ´ 𝑝) ´ 𝑝𝑠 1 ´ 𝑞 ´ 𝑝𝑠 = =1 𝑝(1 ´ 𝑠) 𝑝(1 ´ 𝑠) QED h) For simplicity use the notation 𝑔𝑛(𝑠) = 𝑔𝑍𝑛 (𝑠) and use the expression 𝑔𝑛(𝑠) = 𝑔(𝑔𝑛−1(𝑠)) from (a). In the relation from (g), substitute 𝑠 by 𝑔𝑛−1(𝑠) : r𝑞 ´ 𝑔(𝑔𝑛−1(𝑠))s−1 ´r 𝜇 (1 ´ 𝑔𝑛−1(𝑠))s−1 = 1 1 1 ´ 𝑔𝑛(𝑠)

=1`

1

𝜇(1 ´ 𝑔𝑛 −1 (𝑠))

=1`

1

𝜇

r1 `

1

𝜇(1 ´ 𝑔 𝑛−2(𝑠))

s

which is a repeat of the relation we see in (a).

1 1 ´𝑛 𝑔 (𝑠)

1 1 = 1 ` ` 2 ` ... `

𝜇

𝜇

1

𝜇(𝑛 ´ 1)

`

1

1

𝜇𝑛 1 ´ 𝑔0(𝑠)

we have 𝑔0(𝑠) = 𝑔𝑍0 (𝑠) = 𝑠 since the population at generation0 is of size 1 with probability 1.

1 1 ´ 𝑔 𝑛 (𝑠 )

=

1 ´ 𝑔𝑛 ( 𝑠) =

1 ´ ( 1𝜇 )𝑛 1

1´ 𝜇

`(

1 𝑛 )

𝜇

1 1´𝑠

𝜇𝑛 (𝜇 ´ 1)( 1 ´ 𝑠) 𝜇(𝜇𝑛 ´ 1))(1 ´ 𝑠) ` 𝜇 ´ 1

The extinction probability 𝑋𝑛 = 𝑔𝑛(0), thus, 1 ´ 𝑥𝑛 =

𝜇𝑛(𝜇 ´ 1) 𝜇𝑛+1 ´ 1

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Complexity Science

i) Assume 𝜇 ą 1 : 1 ´ 𝑥𝑛 =

𝜇𝑛+1 ´ 𝜇𝑛 𝜇𝑛+1 ´ 1

1´ 𝜇 1

=

1´

1

→ — 1´

𝜇𝑛+1

1

𝜇

for 𝑛 → — 8

Assume 𝜇 ă 1 : 1 ´ 𝑥𝑛 =

𝜇𝑛 (𝜇 ´ 1 )

𝑛 (𝜇 ´ 1)

— 0 for 𝑛 → → — 8 ´1 𝜇𝑛+1 ´ 1 j) We proved in the chapter that 𝑝(𝑇 )𝛼 1 for 𝜇 = 1. See figure in the next page as the simulations =𝜇

𝑇2

— 𝑝𝑐 = 2 1. approach this behaviour as 𝑝 →

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

Complexity Science

17

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Complexity Science

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

ÿ

ÿ

𝑍 = 𝑒−𝛽𝐸𝑛 =

𝑒 𝐽𝛽𝑆 1 𝑆2

𝑆1=−1,1 𝑆2=−1,1

𝑛

=𝑒

ÿ

𝑥(−1)(−1)

`𝑒

𝑥1(−1)

`𝑒

𝑥(−1).1

`𝑒

𝑥.1.1

= 2(𝑒𝑥 ` 𝑒−𝑥 ) = 4 cosh(𝑥) b) According to Eq. (6.10), ă 𝐸 ą= ´

B ln 𝑧

=´

B𝑥 B

B𝛽

B 𝛽 B𝑥

ln(4 cosh(𝑥))

sinh(𝑥) ´𝐽 = ´𝐽 tanh(𝑥) cosh(𝑥) c)

𝑇→ — 0 ùñ 𝛽 → — 8 ùñ 𝑥 → — 8 and since lim tanh(𝑥) = 1

𝑥→ —∞

we have than ă 𝐸 ą= ´𝐽

𝑇→ — 8 ùñ 𝛽 → — 0 ùñ 𝑥 → — 0 and since lim tanh( 𝑥) = 0

𝑥→ —0

we have lim ă 𝐸 ą= 0

𝑇→ —∞

Exercise 2 𝑥 —𝑥 —𝑥 𝑥 The probabilities are 𝑝1 = 𝑒 , 𝑝1 = 𝑒 , 𝑝1 = 𝑒 , 𝑝1 = 𝑒 .

2

2

2

2

According to Eq (6.11),

𝑆r𝑝1, 𝑝2, 𝑝3, 𝑝4s = ´

4 ÿ

𝑝𝑖 ln 𝑝𝑖

𝑖=1

´2 𝑥 𝑒−𝑥 𝑒𝑥 −𝑥 ln( )) = 𝑍 (𝑒 ln ( ) ` 𝑒 𝑍 𝑍

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

Complexity Science

=

´2

𝑍

=

(𝑥𝑒𝑥 ´ 𝑒𝑥 ln 𝑍 ´ 𝑥𝑒−𝑥 ´ 𝑒−𝑥 ln 𝑍)

´2

𝑍

𝑥(𝑒𝑥 ´ 𝑒−𝑥 ) `

2 ln 𝑍

𝑍

( 𝑒 𝑥 ` 𝑒 −𝑥 )

4 sinh 𝑥 4 cosh 𝑥 = ´𝑥 `ln 𝑍 = ln(4 cosh 𝑥) ´ 𝑥 tanh 𝑥 4 cosh 𝑥 4 cosh 𝑥 Exercise 3 In the microcanonical ensemble, 𝑝𝑖 = 1 Ωfor all 𝑖 =1,2,...,Ω

𝑆= ´

ÿ Ω 1 Ω 𝑖=1

ln

1

= ´Ω

Ω

1 Ω

1

ln

= ln Ω

Ω

Exercise 4 a) 𝑇 → — 0

𝛽→ — 8→ — 𝑥→ — 8

— →

𝑃−1,−1 = 𝑃1,1 =

𝑒𝑥 4 cosh 𝑥

=

𝑒𝑥 2(𝑒𝑥 ` 𝑒−𝑥 )

— →

1 2

for 𝑇 → — 0

𝑃−1,1 = 𝑃1,−1 =

𝑒−𝑥 → — 4 sinh 𝑥

0

for 𝑇 → — 0 Entropy in the limit 𝑇 → — 0 1

1

1

1

2

2

2

2

𝑆r , 0, 0, s = ´2 ln

= ´ln

1 2

= ln 2

b) 𝑇 → — 8→ — 𝛽→ — 0→ — 𝑥→ — 0

𝑃−1,−1 = 𝑃1,1 =

𝑒𝑥 2(𝑒𝑥 ` 𝑒−𝑥 )

→ —

1 4

for 𝑇 → — 8

𝑃−1,1 = 𝑃1,−1 =

𝑒−𝑥 — → 2(𝑒𝑥 ` 𝑒−𝑥 ) 4

for 𝑇 → — 8 Entropy in the limit 𝑇 → — 8 1 1 1 1

1

1

4 4 4 4

4

4

𝑆r , , , s = ´4 ln

= ln 4

1

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Complexity Science

c) From Eq. (6.121)

𝑆r𝑝1, 𝑝2, 𝑝3, 𝑝4s = ln(4 cosh 𝑥) ´ 𝑥 tanh 𝑥 𝑇→ — 0→ — 𝑥→ — 8 𝑒𝑥 ` 𝑒−𝑥

𝑆 = ln(4 𝑥

= ln(2𝑒 ) ´ 𝑥

)´𝑥 𝑥

𝑒𝑥 ´ 𝑒−𝑥

𝑒 ` 𝑒−𝑥

2

1 ´ 𝑒−2𝑥 1`𝑒

−2𝑥

= ln 2 ` 𝑥 ´ 𝑥(1 ´ 𝑒 −2𝑥

−2𝑥

)(1 ´ 𝑒

)

ln 2 ` 𝑥 ´ 𝑥(1 ´ 2𝑒−2𝑥 ` 𝑒−4𝑥 ) → — ln 2

𝑇→ — 8→ — 𝑥→ — 0 𝑆→ — ln(4 cosh(0)) ´ 0 tanh(0) = ln 4 Exercise 5 a) Since each position is assumed to be occupied by the same probability, we have in box 𝑖 that 𝑝𝑖 =

1

𝐿2𝑖

b) Same as above 𝑃(𝑖)𝑒𝑠𝑐𝑎𝑝𝑒 = 𝑝 𝑖 = 𝐿 𝑖2 c) Particle must be at the site next to the exit, and since we assume equal probability for each site, this occurs with prob 𝐿12 . To move to box 𝑖 ` 1, the particle needs to move to the right which is one 1

𝑖

specific direction out of three, so occurs with probability 13. Hence,

𝑃𝑖→ — 𝑖+1 =

1 1 3 𝐿2 𝑖

d) Consideration similar to above gives:

𝑃𝑖→ — 𝑖+1 =

1 1 3 𝐿2

𝑖 +1

e) The current = 𝑃 𝑖 → — 𝑖+1 ´ 𝑃 𝑖 + 1 → —𝑖 1 1 1 1 1 1 = r 2𝑖 ´ 2𝑖+2 s = 2𝑖 r1 ´ s 3 2 2 32 4 1 4(22𝑖 )

=

1 22(𝑖+1)

Exercise 6

𝑒𝐽˜+𝐵˜ ´ 𝜆

𝑒−𝐽˜ 𝑒𝐽˜−𝐵˜ ´ 𝜆

𝑒−𝐽˜ ˜

˜

˜

˜

˜

=0 ˜

𝑒2𝐽 ´ 𝜆(𝑒𝐽 +𝐵 ` 𝑒𝐽 −𝐵 ) ` 𝜆2 ´ 𝑒−2𝐽 = 0

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

Complexity Science

21

˜

𝜆2 ´ 2𝑒𝐽 cosh 𝐵˜ ` 2 sinh 2𝐽˜ = 0 𝜆

1 2𝐽˜ 2 ˜ 𝐽˜ ˜ 1 ˜ = (2𝑒 cosh 𝐵 ˘ r4𝑒 cosh 𝐵 ´ 4 sinh 2𝐽s 2 ) 2

Using cosh2 𝑥 ´sinh2 𝑥 = 1, 𝐽˜

˜

𝜆 = 𝑒 cosh 𝐵˜ ˘ r𝑒2𝐽 (1 `sinh2 𝐵˜) ´ 2 sinh 2𝐽˜s ˜ ˜ ˜ ˜ ˜ = 𝑒𝐽 cosh 𝐵˜ ˘ r𝑒2𝐽 sinh2 𝐵˜ ` 𝑒2𝐽 ´ 𝑒2𝐽 ` 𝑒−2𝐽 s 1

˜

˜

˜

𝜆 = 𝑒𝐽 cosh 𝐵˜ ˘ r𝑒2𝐽 sinh2 𝐵˜ ` 𝑒−2𝐽 s 1

1 2

2

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 𝐿2. The energy of the vortex increases as log 𝐿. The energy of the configuration ÒÓ Ó ... ÓÒ Ò... 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 B𝐻 from B𝜙

𝐻r𝜙 `B 𝜙 s ´ 𝐻r𝜙s =

𝐽 𝑎

ż∞ −∞

and conclude

2

𝑑𝑥 r´ 𝑎 B𝐻 B𝜙

Hence,

B𝐻 B𝜙

ℎ 2 ` 𝛿𝜙 sin 𝜙s ` 𝑂(𝛿𝜙 )...

2

𝑑2𝜙 𝑑𝑥

= ´𝐽𝑎

=0→ —

𝐽

𝑑2𝜙 𝑑𝑥

` 2

𝑑2𝜙

ℎ

𝑑𝑥

𝐽𝑎2

= 2

2𝜙

b) We demonstrate that 𝜙(𝑥) = 4 tan−1(𝑒𝜇𝑥 ) satisfies 𝑑

𝑎

ℎ sin 𝜙

2

sin 𝜙 = 𝜇 sin 𝜙

= 𝜇 sin 𝜙 by use of the following identities

𝑑𝑥2

𝑑 1 tan−1 𝑥 = 𝑑𝑥 1 ` 𝑥2 tan−1 𝑥 =

´𝑖 2

ln

1 ` 𝑖𝑥 1 ´ 𝑖𝑥

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Complexity Science

𝑒𝑖𝑥 ´ 𝑒−𝑖𝑥 2𝑖

sin 𝑥 = First,

𝑑𝜙

4𝜇𝑒𝜇𝑥 4𝜇 2𝜇 = = = 𝑑𝑥 1 ` 𝑒2𝜇𝑥 𝑒−𝜇𝑥 ` 𝑒 𝜇𝑥 cosh 𝜇𝑥

𝑑2𝜙 = 𝑑𝑥2

´2𝜇2 sinh 𝜇𝑥 cosh 2 𝜇𝑥

( 𝑦1´ 𝑦) ( = 4𝜇 2 1 ` 𝑦)2 𝑦

with 𝑦 Ξ 𝑒𝜇𝑥 c) For 𝑣 ă 𝑁2𝐸, 1 𝛽increases with increasing energy. Similar to what we usually expect the temperature does. The singularity at 𝑣 = 𝐸 𝑁2and negative values for 𝑣 ą 𝐸 𝑁 do not correspond to 2 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 𝑅ÿ ÿ𝑅 ÿ 𝑅 B𝐽 B = t𝑁 ln 𝑁 ´ 𝑁 ´ (𝑛 ln 𝑛 ´ 𝑛 ) ` 𝛼(𝑁 ´ 𝑛 ) ` 𝛽(𝑣 ´ 𝜖 𝑛 ) 𝑗 𝑗 𝑗 𝑗 𝑗 𝑗 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 bedifferentiated with respect to 𝑛𝑖.

We now get, B𝐽 =0→ — 𝑛𝑖 = 𝑁 exp(𝛼 ` 𝛽𝜖𝑖) B𝑛𝑖 ř ř b) From 𝑁 = 𝑖 𝑛𝑖 = 𝑁 𝑖 exp(𝛼 ` 𝛽𝜖𝑖) ÿ → — exp 𝛼 = ( 𝑒𝛽𝜖𝑖 )−1 Ξ 𝑧−1 𝑖

c)

𝑈=

ÿ

𝜖𝑖𝑛𝑖 =

𝑖

d)

𝑁𝑒𝛽𝜖

ÿ 𝜖𝑖𝑒𝛽𝜖𝑖 𝑖

𝑧

𝛽𝜖

𝑛1 = 𝑁 ,𝛽𝜖𝑛2 = 𝑈 = 0𝑛1 ` 𝜖𝑛2 = 𝜖𝑁 𝑒 𝛽𝜖 1+𝑒𝑣 1+1𝑒𝛽𝜖 1 1+𝑒 Let 𝑥 = and 𝑦 = then we get 𝑦 = 𝑥 𝜖𝑁

𝛽𝜖

ln 1—𝑥

And 𝑦 ą 0 for 1 𝑥− 𝑥ą 1 → — 𝑥 ą 1 𝑦 ă 0 for 1 𝑥 ă 1 → — 𝑥ă 1 2 −𝑥 2 1 1 We notice 𝑦 for . We are free to change the sign of the Lagrange multiplier in = 𝛽𝜖 𝑥 = 𝜖𝑁 ă 2 𝛽 the expression for 𝑦(𝑛). When we do this, we get 𝛽 → — ´𝛽 ùñ 𝑦 → — ´𝑦 and we conclude:

𝑣 𝜖𝑁

ă

1 2

→ — 𝛽ą0

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

Complexity Science 𝑣

23

approaching 1 from below gives us

𝜖𝑁

2

𝑦=

1

𝛽𝜖

— `8 →

And for 𝑣 just above 1 we have, 𝜖𝑁

2

𝑦=

1

𝛽𝜖

— ´8 →

Finally, when

e) Project 2 a)

𝐸1 = 𝐻rone kinks ´ 𝐻rno kinks 1) ` l oo m 1 ons ´ 𝐽(´𝑁) = 2𝐽

𝐽 rĺoo(o𝑁 om´oooon parallel pairs

kink

b) Each of the 𝑁 bonds can contain no kink, energy 𝐸0 = 0, or a kink, energy 𝐸1 = 2𝐽. Partition function

𝑍=

ÿ ...

(1)

𝐸𝑖 =0,2𝐽

=r

ÿ

ÿ 𝑁) 𝐸𝑖( =0,2𝐽

𝑁

exp(´𝛽

ÿ

𝑘

𝐸( 𝑖 ))

𝑘=1

exp(´𝛽𝐸𝑖)s𝑁 = (1 ` 𝑒−𝛽𝐸1 )𝑁

𝐸= 𝑖 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 = 𝑛𝐸1 → —ă 𝐸 ą= 𝐸1 ă 𝑛 ą and it follows from Eq (6.10) that ă 𝑛 ą= 𝐸11 ă 𝐸 ą= ´ 𝐸11 B𝛽 Bln 𝑍. d) 1 B ln(1 ` 𝑒−𝛽𝐸1 )𝑁 ă 𝑛 ą= ´ 𝐸1 B 𝛽

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Complexity Science

𝑁 ´𝐸1𝑒−𝛽𝐸1

´

𝐸 1 1 ` 𝑒−𝛽𝐸1

=

𝑁 1 ` 𝑒𝛽𝐸1

Therefore,

𝑁

ă 𝑙 ą= = 1 ` 𝑒𝛽𝐸1 = ă𝑛ą

"

8 , for 𝑇 → — 0 ùñ 𝛽 → — 8 2 , for 𝑇 → — 8 ùñ 𝛽 → —0

Project 3 No sample solution needed

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25

Chapter 7 Exercise 1 a) Define Δ = 𝜃1 ´ 𝜃2 and 𝛼 = 𝜔1 ´ 𝜔2.

𝑑Δ 𝑑𝑡

= 𝛼 ` 𝜖 (𝜃2 ´ 𝜃1) ´ 𝜖 (𝜃1 ´ 𝜃2) = 𝛼 ´ 𝜖 Δ2

𝑑Δ 𝑑𝑡

` 2𝜖 Δ = 𝛼

𝑑Δ 𝐼 ` 2𝜖 Δ = 𝛼 𝑑𝑡 𝑑Δ 𝐼 `2 𝜖 𝐼Δ = 𝐼𝛼 𝑑𝑡

ş Which with 𝐼 = exp( 2𝑒𝑝𝑠𝑖𝑙𝑜𝑛𝑑𝑡) = 𝑒2𝜖 𝑡) is equivalent to

𝑑(𝐼Δ) 𝑑𝑡

= 2𝜖 𝐼Δ ùñ 𝑒2𝜖 𝑡 Δ =

𝛼 2𝜖 𝑡 𝑒 =𝐴 2𝜖

where 𝐴 is a constant determined by the initial condition. We conclude Δ = 𝐴𝑒−2𝜖 𝑡 ` 𝛼 . 2𝜖

𝜃1 = 𝜃2 ` Δ = 𝜃2 ` 𝐴−2𝜖 𝑡 ` — 𝜃2 ´ →

𝜔2 ´ 𝜔2 2𝜖

𝜔1 ´ 𝜔2 2𝜖

for 𝑡 → — 8

b) Let again Δ = 𝜃1 ´ 𝜃2

𝑑Δ

= 𝜔Δ ` 𝜖 r𝜃2 ´ 𝜃1 ´ 𝜃1 ` 𝜃2s 𝑑𝑡 = 𝜔Δ ´ 2𝜖 Δ = (𝜔 ´ 2𝜖 )ΔΔ(𝑡) = Δ(0)𝑒(𝜔−2𝜖 )𝑡

Therefore, we have, For 𝜔 ą 2𝜖 → — Δ(𝑡) → — 8 with 𝑡 → — 8 i.e. no synchronisation. For 𝜔 ă 2𝜖 → — Δ(𝑡) → — 0 with 𝑡 → — 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>𝐵, the function 𝜓 will increase monotonously with time. In terms of Eq. (7.6) this corresponds to rotors for which |𝜔𝑘 𝜖´ 𝜖𝐾 | ą. For these rotors, 𝜓𝑘 will keep increasing with time, and therefore the corresponding phase 𝜃𝑘 𝑡 in( )Eq. (7.5) will not lock to 𝜔¯ 𝑡. So, these rotors will not synchronize.

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Complexity Science

Project 1 a) Fourier transformation:

ż 𝑖𝑛 𝑓 𝑡 𝑦 𝑑𝑘

𝜃(𝑥, 𝑡) =

2𝜋

∞

ż∞

B2𝜖

=

B𝑥2

𝑑𝑘 (´𝑘 2𝜋

∞

−𝑖𝑘𝑥

𝜃ˆ(𝑘, 𝑡)𝑒

and

𝜃)𝑒 −𝑖𝑘𝑥

2 ˆ

The F.T of the equation becomes B𝑡 𝜃ˆ(𝑘, 𝑡) = Ω̂(𝑘) ´ 𝛾𝑘2𝜃ˆ(𝑘, 𝑡) B𝑡 𝜃ˆ `

𝛾𝑘2𝜃ˆ = Ω̂(𝑘) ş 2 Integration factor 𝐼 = expr 𝑑𝑡𝛾𝑘 2 s = 𝑒𝛾𝑘 𝑡 . Multiply eq. by IF and get, B 𝛾𝑘2 𝑡 ˆ 𝛾𝑘 2 𝑡 ( 𝑒 𝜃 ) = 𝑒 Ω̂ B𝑡

𝑒𝛾𝑘 𝑡 Ω̂ = 2

𝜃ˆ(𝑘, 𝑡) =

2

𝑒 𝛾𝑘 𝑡 ˆ Ω ` 𝐴(𝑘) 𝛾𝑘 2

Ω̂

` 𝐴(𝑘)𝑒=𝛾𝑘 𝑡 2

𝛾𝑘 2

where 𝐴(𝑘) is independent of t. Use that for 𝑡 = 0, we have 𝜃ˆ(𝑘, 𝑡) = 𝜙ˆ(𝑘) to eliminate 𝐴(𝑘).

𝜃ˆ(𝑘, 𝑡) =

Ω̂

𝛾𝑘

b) Δ=

=

ż

∞

2`(ˆ´

ż

𝜙

∞

Ω̂

𝛾𝑘 2

) 𝑒 − 𝛾𝑘 𝑡 2

𝑑𝑥|B𝑥 𝜃(𝑥, 𝑡)|2

−∞

𝑑𝑥|B𝑥ˆ𝜃(𝑥, 𝑡)|2 Ð — According to Plancherel theorem

−∞

ż =Δ

∞

𝑑𝑘| ´ 𝑖𝑘 𝜃ˆ(𝑘, 𝑡)|2 =

−∞

=

ż ∞ 𝑑𝑘

lim Δ = 𝑡→ —∞

𝑑𝑘|𝑖𝜃ˆ(𝑘, 𝑡)|2

−∞

Ω̄ 𝑘(𝜙ˆ ´ Ωˆ )𝑒−𝛾𝑘 𝑡2 |2 | `

𝜋 𝛾𝑘

−∞ 2

ż∞

ż∞

𝛾𝑘2

𝑑𝑘 Ω̂ 2 | | −∞ 2𝜋 𝛾𝑘

c) Synchronisation either complete (i.e. 𝜃𝑛(𝑡) = 𝜃𝑛+1(𝑡) for all 𝑛) or phase locking (i.e. 𝜃𝑛(𝑡) = 𝜃𝑛+1(𝑡) ` 𝛼𝑛+1) corresponds to Δ(𝑡) becoming independent of 𝑡 as 𝑡 → — 8.

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

Complexity Science

27

d) We have used continuous 𝑥 as an approximation. In reality, the rotors 𝜃𝑛 sit at discrete positions 𝑥𝑛 = 𝜎𝑛, where 𝜎 is the separation between rotors. To respect2that the shortest period of spatial variation is given by 𝜎 we need the Fourier 𝑘 to satisfy |𝑘| ă 𝜋 . Let us also assume the chain is of 𝜎

finite length 𝐿. The longest period of variation must be smaller than 𝐿 and therefore |𝑘| ą 2𝐿𝜋 Ξ 𝜖 . ˆ To ensure that the l i m 𝑡 → — 8 and therefore 𝜖 → — 0 we need | Ω |2 to — ∞ Δ remains finite as we let 𝐿 → 𝑘 be integrable near 𝑘 = 0. Assume |Ω(ˆ𝑘)| » 𝑘𝑎 for 𝑘 → — 0. This means we need 𝑘2𝑎−2 to be ş 2𝑎—1 1 integrable for 𝑘→ — 0. Since, the integral to exist the root. limit 0. So, i.e. ˆ 𝑘2𝑎−2𝑑𝑘 = 𝑘 2𝑎−1we need 2𝑎 ´ 1 ą 0 0for more slowly than the in square 𝜖→ — 𝑎ą 2 Ω(𝑘) must not approach zero as 𝑘 → —

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

ÿ

— 𝑙𝛼𝑗 = 1 ` 3 ˆ 2 ` 3 ` 3 ˆ 2 = 14 →

𝑗

ùñ 𝐶𝛼 =

11 14

1 ÿ

𝑁

𝑗

» 0.79

Node 𝛾: ÿ

𝑙𝛾𝑗 = 3 ` 2 ` 2 ˆ 3 ` 3 ` 3 ˆ 4 = 26

𝑗

1 ÿ

𝑁

𝑙 𝛾𝑗 =

𝑗

Exercise 2 No sample solution Exercise 3 We move a node from clusters 𝐴 to cluster 𝐵

26 11

— 𝑐𝛾 = →

11 » 0.42 26

14

𝑙 𝛼𝑗 = 11

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Complexity Science

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

𝑄r𝐶s =

1 2𝐿

ÿ

(𝑎

𝑖, 𝑗P 𝐴−𝑎

𝑘𝑖 𝑘 𝑗

1 ´ )` 𝑖𝑗 2𝐿 2𝐿

ÿ

(𝑎

´ 𝑘𝑖 𝑘𝑎 )` 1 𝑖𝑎 2𝐿 2𝐿

𝑖P 𝐴−𝑎

ÿ

𝑘𝑎 𝑘 𝑗

(𝑎

1 ´ )` 𝑎𝑗 2𝐿 2𝐿

𝑗 P 𝐴−𝑎

ÿ

(𝑎 𝑖𝑗

𝑖, 𝑗P𝐵

´ 𝑘𝑖 𝑘 𝑗 ) 2𝐿

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

𝑄r 𝐶 2s =

1

ÿ

2𝐿 𝑖, 𝑗P 𝐴−𝑎

(𝑎

𝑘𝑘 1 ´ 𝑖 𝑗 )` 𝑖𝑗 2𝐿 2𝐿

ÿ

(𝑎

´ 𝑖𝑗

𝑘𝑖 𝑘 𝑗

)`

2𝐿

𝑖, 𝑗P𝐵

1 ÿ (𝑎 2𝐿 𝑖P𝐵

´ 𝑖𝑎

𝑘𝑖 𝑘𝑎

)`

2𝐿

1 ÿ (𝑎 2𝐿 𝑗P𝐵

𝑎𝑗

´ 𝑘𝑎 𝑘 𝑗 ) 2𝐿

1 ÿ 𝑘𝑖 𝑘𝑎 𝑘𝑖 𝑘𝑎 ÿ (𝑎 𝑖𝑎 ´ s= r ´ )s 𝑖 𝑎 2𝐿 ) ´ (𝑎 𝐿 𝑖 P𝐴 −𝑎 2𝐿 𝑖P𝐵

The change Δ𝑄 = 𝑄r𝐶s ´ 𝑄r𝐶2

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

𝑄r𝐶s =

1 ÿ

𝐿 𝑞=1 𝐿𝐶 𝑞

´

1 (2𝐿)2

𝐾 ÿ ÿ ( 𝑘2) 𝑞=1 𝑖P𝐶𝑞

𝑖

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

Partition (a)

𝑞 = 1 : 𝐿𝐶1 = 4,

ÿ

𝑘2𝑖 = 1 ` 52 ` 22 ` 32 = 39

𝑖P𝐶1

𝑞 = 2 : 𝐿𝐶2 = 3,

ÿ

𝑘𝑖 2 = 1 ` 42 ` 22 ` 32 ` 22 = 34

𝑖P𝐶2

𝑞 = 3 : 𝐿𝐶3 = 5,

ÿ

𝑘𝑖2 = 42 ` 22 ` 32 ` 22 = 33

𝑖P𝐶3

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

Complexity Science

𝑄r𝐶𝑎s =

1 15

(4 ` 3 ` 5) ´

1 (30)2

29

(39 ` 34 ` 33) » 0.682 In

the second partition, 𝐾 = 2,

𝑞 = 1 : 𝐿𝐶1 = 10,

ÿ

𝑘𝑖 2 = 1 ` 52 ` 22 ` 32 ` 22 ` 52 ` 22 ` 1 = 73

𝑖P𝐶1

𝑞 = 2 : 𝐿𝐶2 = 5,

ÿ

𝑘𝑖2 = 42 ` 22 ` 32 ` 22 = 33

𝑖P𝐶2

1

𝑄r𝐶𝑏s =

15

(10 ` 5) ´

1 (2𝐿)2

(73 ` 33) = 0.882

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

𝑃𝑘 = (1 ´ 𝑒−𝜆)𝑒−𝜆𝑘 Eq. (8.59) is given by

𝑔r𝑠 s =

∞ ÿ

∞ ÿ

−𝜆

𝑘

𝑃𝑘 𝑠

= (1 ´ 𝑒

𝑘=0

)

1 ´ 𝑒−𝜆

−𝜆𝑘 𝑘

𝑒

𝑠 =

𝑘=0

1 ´ 𝑠𝑒−𝜆

=

𝑒𝜆 ´ 1 𝑒𝜆 ´ 𝑠

Note

𝑔2(𝑠) =

∞ ÿ

𝑘=0

𝑔”(𝑠) =

∞

ÿ

𝑘(𝑘 ´ 1)𝑃𝑘

𝑘𝑃 𝑘 𝑠𝑘−1 ùñ 𝑔2(1) =

∞

ÿ

𝑘𝑃 𝑘

=ă 𝑘 ą

𝑘=0

𝑠𝑘−2 ùñ 𝑔”(1) =ă 𝑘(𝑘 ´ 1) ą=ă 𝑘2 ą ´ ă 𝑘 ą

𝑘=2

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

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Complexity Science

1

𝑔2(𝑠) = (𝑒𝜆 ´ 1)

ùñ 𝑔2(1) =

(𝑒𝜆 ´ 𝑠)2 2

𝑔”(𝑠) = (𝑒𝜆 ´ 1)

(𝑒𝜆 ´ 𝑠)3

ùñ 𝑔”(1) =

1

𝑒𝜆 ´ 1 2

(𝑒𝜆 ´ 1)2

So, the M-R condition becomes given by: 2

2

ă 𝑘 ą ´2 ă 𝑘 ą=ă 𝑘 ą ´ ă 𝑘 ą ´ ă 𝑘 ą= 𝑔”(1)´𝑔 (1) =

2

2

1

´ = 𝜆 (𝑒𝜆 ´ 1)2 𝑒 𝜆 ´ 1 (𝑒 ´ 1)

3 ´ 𝑒𝜆 ą0 2

ùñ 𝑒𝜆 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=

1 (𝑒ln 3 −1)

=1 2

Exercise 9 Each time the walker reaches a new node, it finds with probability: ˆ

𝑃𝑘 =

˙

𝑁´2 𝑘 𝑝 (1 ´ 𝑝)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

𝑔(𝑠) =

𝑁ÿ −2 ˆ 𝑘=0

˙

𝑁 ´ 2 𝑝𝑘 (1 ´ 𝑝)𝑁 −2−𝑘 𝑠𝑘 = (𝑝𝑠 ` 1 ´ 𝑝)𝑁 −2 𝑘

. The average branching ratio is

𝜇 = 𝑔2(𝑠)|𝑠=1 = (𝑁 ´ 2)𝑝 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 𝑝𝑐 = 1 » 1 . Note: strictly speaking, we should exclude all (𝑁 −2)

𝑁

previously visited nodes to avoid loops. But in the limit, 𝑁 → — 8, we can ignore this.

The Mean Field Equation

𝑃∞ is given by the probability that an occupied site belongs to the spanning cluster. Hence, 1 ´ 𝑃∞ = Prob. that no neighbour site belongs to the spanning cluster = (1𝑝 𝑃∞)𝑞 The critical exponent 𝛽 describes how 𝑃∞ becomes non-zero, 𝑃∞9(𝑝 ´𝑝𝑐)𝛽 i.e. we seek solutions to 1 ´ 𝑥 = 𝑓 (𝑥) for small x. Hence, using the Taylor expansion, we expand 𝑓 (𝑥) about 𝑥 = 0.

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

Complexity Science

31

𝑓 (0) = 1, 𝑓 2(0) = ´𝑝𝑞(1 ´ 𝑝𝑥)(𝑞−1)|𝑥=0 = ´𝑝𝑞 𝑓 ”(0) = 𝑝2𝑞(𝑞 ´ 1)(1 ´ 𝑝𝑥)(𝑞−2)|𝑥=0 = 𝑝2𝑞(𝑞 ´ 1) So, this becomes: 1 1 ´ 𝑥 = 1 ´ 𝑝𝑞𝑥 ` 𝑝2𝑞(𝑞 ´ 1)𝑥2 ùñ 2

𝑝𝑞 ´ 1

ùñ 𝑥 = 2

1 2 𝑝 𝑞(𝑞 ´ 1)

Now substitute 𝑝 = 𝑝𝑐 ` Δ = 1 `Δ 𝑞 1

𝑥 » 2𝑞Δr1 ´ ` 𝜃(Δ)s−1 𝑞 𝑃∞ »

2𝑞2 (𝑝 ´ 𝑝𝑐) 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 𝑝 (1 ´ 𝑝)𝑁 −2 𝑃𝑘 =

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.

𝑃𝑘 =

𝜆𝑘 𝑒−𝜆 𝑘!

The connectivity, see Eq(8.1), 𝐶 = 𝑁 (2𝑁𝐿 −1) is the average number of links 𝑁 ÿ 2𝐿 =ă

𝑘𝑖 ą= 𝑁 ă 𝑘𝑖 ą= 𝑁𝜆 = 𝑁2 𝑝

𝑖=1

Therefore,

𝐶=

𝑁2 𝑝 𝑁(𝑁 ´ 1)

»𝑃

b) According to Eq(8.99), the return time for a node 𝑖 of degree 𝑘𝑖 is given by 𝑟𝑖 = 2𝐿 . 𝑘𝑖 ř To average over degree 𝑘𝑖 ą 0, we need to renormalise 𝑃𝑘 = 𝛼𝑃˜𝑘 such that ∞ 1 𝑃˜𝑘 = 1. We have 𝑘 ř∞ = 𝑃 𝑘 =1 𝑃𝑘 = 1 ´ 𝑝0, so 𝑃˜𝑘 = 𝑘1−for 𝑝0 𝑘 ě 1.

32

Complexity Science We neglect degree correlations and have and have for the joint probability 𝑃˜ (𝑘 𝑞 , 𝑘𝑖 ) = 𝑃˜𝑘𝑞𝑃˜𝑘𝑖.

This gives, ř ă 𝑟 ą=ă 2𝐿 ą=ă 𝑖

𝑞= 1 𝑘𝑞

𝑁

𝑘𝑖

𝑘𝑖

𝑞‰𝑖

∞ ∞ 𝑘𝑞 ÿ ÿ ÿ ˜

𝑘𝑖

` ą 1

𝑃 𝑘

=1`

𝑘𝑖

𝑞‰𝑖 𝑘𝑞 =1 𝑘𝑖 =1

1`

ÿ 𝑘𝑞

ą=ă

( 𝑞 , 𝑘𝑖)

∞ ÿ ∞ ÿ ÿ 𝑘𝑞 𝜆𝑘𝑞

𝑞 𝑖𝑘

1𝑘 ‰ 𝑞 = 𝑖=

𝑒−2𝜆 𝑘 𝑘𝑖! (1 ´ 𝑒−𝜆)2 1 𝑖

∞ ÿ𝜆 ÿ

1 ` 1 (𝑒𝜆 ´ 1)2

𝑞‰𝑖

=1`

𝜆𝑘𝑞 −1

ÿ 8 𝜆𝑘𝑖

𝑞

𝑘𝑖 =1

(𝑘 ´ 1)!

𝑘𝑞 =1

𝜆𝑒𝜆 (𝑒𝜆 ´ 1)2

𝑘𝑘! 𝑖 𝑖

𝑁𝑆(𝜆)

Here we defined,

𝑆(𝜆) =

∞ ÿ 𝜆𝑘 . 𝑘𝑘 ! 𝑘=1

Note 𝑆(0) = 0

we have, ∞ 𝑘−1 ÿ 𝜆

2

𝑆 (𝜆 ) =

𝑘!

𝑘=1

𝑆(𝜆) =

ż

=

1 ÿ 𝜆𝑘 ∞

𝜆

𝑘=1

𝑘!

=

𝑒𝜆 ´ 1 𝜆

𝜆

𝑒 𝑑𝜆 ´ln 𝜆 ` 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝜆

The first term is the exponential integral ż

𝐸𝑖(𝑥) =

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 𝐸 (𝑥 ) 𝛾» ln` 𝑥 for 𝑥 0+ → where 𝛾 is the Euler-Mascheroni constant. So, we — conclude that the constant in expression above is given by 𝛾 and have

𝑆(𝜆) = 𝐸𝑖(𝜆) ´ln(𝜆) ` 𝛾 In summary, ă 𝑟𝑖 ą= 1`

𝜆𝑒𝜆 (𝑒𝜆 ´ 1)2

𝑁r𝐸𝑖(𝜆) ´

ln 𝜆 ` 𝛾s

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

Complexity Science c) We estimate ă 𝑟𝑖 ą by replacing 𝐿 → — ă 𝐿 ą and 1 𝑘→ — 𝑖 ă 𝑟𝑖 ą» ă 2𝐿 ą = ă ă 𝑘𝑖 ą

ř

1 ă 𝑘𝑖 ą

𝑖 𝑘𝑖 ą

𝑁 ă 𝑘𝑖 ą

𝜆𝑒𝜆

∞ ÿ 𝜆𝑘

= ă 𝑘𝑖 ą

=𝑁 ă 𝑘𝑖 ą

We found above ă 𝑟𝑖 ą= 1 `

(𝑒𝜆 ´ 1)

𝑁 2

𝑘𝑘 !

𝑘=1

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

𝜆(1 ` 𝜆) (𝜆

𝜆2

` 2)

2

(𝜆 `

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

𝜆2 4

)»𝑁

33

34

Complexity Science

Chapter 9 Exercise 1

𝑋=1

with probability 𝑝

𝑋 = ´1

with probability 1 ´ 𝑝

𝐻r𝑃𝑋s = ´(𝑝 ln 𝑝 `(1 ´ 𝑝) ln(1 ´ 𝑝)) 𝑑𝐻 =0→ — ´(ln 𝑝 ` 1 ´ 𝑝 ln(1 ´ 𝑝)) ´ 1 = 0 𝑑𝑝

Maxima:

ùñ ln(

𝑝

ùñ

1´ 𝑝

𝑝

1´ 𝑝

)=0

= 1 ùñ 𝑝 =

1 2

Exercise 2 The probabilities for 𝑌 = 𝑋2: Case (1):

"

𝑃𝑋(𝑥) = "

𝑃 𝑌 ( 𝑦) = Case (2):

"

𝑃𝑋(𝑥) =

𝑁 1 , for 𝑥 P t0, 1, ..., 𝑁u 1

+

0,

otherwise

𝑁 1 , for 𝑦 P t0, 1, 4..., 𝑁 u 1

2

+

0,

otherwise

2

1 , for 𝑥 P t´𝑁, ´(𝑁 ´ 1), ...0, 1, ..., 𝑁u

0,

otherwise

1

𝑁+

$ 1 for 𝑦 = 0 & 2 1, 𝑁+ 𝑃𝑌 (𝑦) = 2 2 1 , for 𝑦 P t1, 4, 9, ..., 𝑁2u % 𝑁+ otherwise 0, Joint probability distributions

𝑃𝑋,𝑌 (𝑥, 𝑦) = Prob t𝑋 = 𝑥 and 𝑌 = 𝑦 = 𝑥2u = 𝑃𝑋(𝑥)𝛿𝑦, 𝑥2 Case (1)

"

𝑃𝑋,𝑌 (𝑥, 𝑦) =

𝑁 1 , (𝑥, 𝑦) P t(0, 0), (1, 1), (2, 4), ..., (𝑁, 𝑁 )u Ξ 𝑆1 1

2

+

0

otherwise

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

Complexity Science Case(2)

"

1 2

𝑃𝑋,𝑌 (𝑥, 𝑦)=

2 2 2 1 , (𝑥, 𝑦) P t(´𝑁, 𝑁 ), (´(𝑁 ´ 1), (𝑁 ´ 1) ),.......... , (𝑁, 𝑁 )u Ξ 𝑆2

𝑁+

otherwise

0

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

ă 𝑋𝑌 ą=

𝑥𝑦𝑃𝑋𝑌 (𝑥, 𝑦)

𝑥=0 𝑦=0

ÿ𝑁 ÿ

𝑦 = 0𝑁2 𝑥𝑦𝑃 (𝑥)𝛿

𝑁 ÿ

=

2

2 𝑌 (𝑥 ) =

𝑥=1

𝑁(𝑁 ` 1) 2 s

𝑋 ( 𝑥) =

𝑁(𝑁 ` 1) 1 ÿ𝑁 1 𝑁 𝑥= = 𝑁 ` 1 𝑥=0 𝑁`1 2 2

𝑁`1 𝑥=0

𝑁`1

2

𝑁(𝑁 ` 1)(2𝑁 ` 1) 𝑁(2𝑁 ` 1) 1 ÿ𝑁 1 𝑥= = 𝑁 ` 1 𝑥=1 𝑁`1 6 6

𝑁2(𝑁 ` 1) 𝐶𝑋𝑌 9

r

𝑦, 𝑥

𝑥=0 𝑁 ă 𝑌 ą= ÿ 𝑥2𝑃

1

=

𝑋 𝑥=0 𝑁 ă 𝑋 ą= ÿ 𝑥𝑃

𝑥3

𝑁 𝑁(2𝑁 ` 1)

´

4

2

6

Case (2)

ÿ𝑁 ă 𝑋𝑌 ą=

𝑥

𝑥𝑥

𝑁

1

2

2𝑁 ` 1

=−

ÿ𝑁 ă 𝑋 ą= ă 𝑌 ą=

𝑥

1

𝑁

=−

ÿ𝑁

=

𝑥

2𝑁 ` 1 2

𝑥2

𝑁2(𝑁 ` 2) 12

‰0

=0

=0 ‰0

2𝑁 ` 1

𝑥=0

So, 𝐶𝑋,𝑌 = 0 b) Mutual information

𝐼 (𝑋, 𝑌 ) = Case (1)

ÿ

So,

𝐼(𝑋, 𝑌 ) =

) 𝑥, 𝑦 ln

𝑥𝑦

𝑃𝑋,𝑌 (𝑥, 𝑦=) 𝑃 (𝑥)𝑃 (𝑦) 𝑋

𝑃𝑋,𝑌 (𝑥, 𝑦) 𝑃𝑋( 𝑥)𝑃𝑌 (𝑦)

𝑃 𝑋,𝑌 (

𝑌

ÿ

1

= 𝑁 `1

𝑁 +1 𝑁 +1

1

𝑁 `1 (𝑥, 𝑦)P𝑆 1

𝑁 1+1 1

ln(𝑁 ` 1) = ln(𝑁 ` 1) ‰ 0

35

36

Complexity Science Case (2)

"

𝑃𝑋,𝑌 (𝑥, 𝑦)

1 1 −1 = 2𝑁 ` 1, (𝑥, 𝑦) = (0, 0) 1r2 12 1s 𝑁+ 𝑁+ 𝑁+ 2𝑁 +1 1 r 1 2 s−1 = 2 , (𝑥, 𝑦) P 𝑆2 2𝑁 +1 2𝑁 +1 2𝑁 +1 1

=

𝑃𝑋 (𝑥)𝑃𝑌 (𝑦)

*

2

1

𝐼(𝑋, 𝑌 ) =

1

ln(2𝑁 ` 1) ` 2𝑁

2𝑁 ` 1

ln

2𝑁 ` 1

2𝑁 ` 1

ln(2𝑁 ` 1) ´

2

2𝑁 log 2 ‰ 0 2𝑁 ` 1

Exercise 3 𝛽𝑆1 𝑆2

Two Ising spins: 𝐻 = ´𝑆1𝑆2 with Boltzmann weights: 𝑃(𝑆1, 𝑆2) = 𝑒

𝑍=

ÿÿ

𝑆1

where 𝑍

−𝛽 −𝛽 𝛽 𝑒𝛽𝑆1𝑆2 = loo𝑒m𝛽 on ` lo𝑒o m on ` lo𝑒o m on ` loo𝑒m on

𝑆2

(−1,−1)

(−1,1)

(1,−1)

(1,1)

= 2(𝑒𝛽 ` 𝑒−𝛽) = 4 cosh(𝛽) Marginal distributions

𝑃 (𝑠) = 𝑆1

1 ÿ 𝛽𝑠2

𝑒

=

𝑍 𝑠2 cosh 𝛽𝑠

=

2 cosh 𝛽

1

𝑍

t

𝑒−𝛽𝑠 ` 𝑒𝛽𝑠 u

= 𝑃𝑆2 (𝑠)

Correlation coefficient

𝐶𝑆1 𝑆2 =ă 𝑆1𝑆2 ą ´ ă 𝑆1 ąă 𝑆2 ą Here, ă 𝑆1𝑆2 ą=

𝑍

1 𝛽 r𝑒 ´ 𝑒

−𝛽

ă 𝑆1 ą=ă 𝑆2 ą= ´

𝛽

𝛽

´𝑒 `𝑒 s=

1 2

tanh 𝛽

cosh(´𝛽) cosh 𝛽 ` =0 2 cosh 𝛽 2 cosh 𝛽

" 1 2, for 𝑇 → — 0 ùñ 𝛽 → — 8 1 — 8 ùñ 𝛽 → — 0 2 𝛽, for 𝑇 →

1

𝐶𝑆 1 𝑆2 = tanh 𝛽 = 2

Mutual Information

𝐼(𝑆 1 ; 𝑆2 ) =

ÿÿ

𝑆1

ÿ 𝑒𝛽𝑠1 𝑠2

= 𝑠1 𝑠2

4 cosh 𝛽

𝑃(𝑆1, 𝑆2) ln

𝑃(𝑆1)𝑃(𝑆2)

𝑆2

ln

𝑃(𝑆1, 𝑆2)

𝑒 𝛽𝑠 1 𝑠2

2 cosh 𝛽 2 cosh 𝛽 4 cosh 𝛽 cosh 𝛽𝑠1 cosh 𝛽𝑠2

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

Complexity Science

=

ÿ 1 𝑒 𝛽𝑠 1 𝑠2 𝑒 𝛽𝑠 1𝑠 2 ln 4 cosh 𝛽 𝑠1 𝑠2 cosh 𝛽

where we used cosh 𝛽𝑠1 = cosh 𝛽𝑠2 = cosh 𝛽 for 𝑠1, 𝑠2 P t´1, 1u. Hence, 1 (𝛽𝑒𝛽 ´ 𝑒𝛽 log cosh 𝛽 ´ 𝛽𝑒−𝛽 ´ 𝑒𝛽 ln cosh 𝛽) = 𝛽 ´ln cosh 𝛽 2 cosh 𝛽

𝐼(𝑆1; 𝑆2) =

As 𝑇 → — 0 ùñ 𝛽 → — 𝛽8 ùñ cosh 𝛽 » 1 𝑒2𝛽.

So, 𝐼(𝑆1; 𝑆2) = 𝛽 ´ln 𝑒 = log 2 2 As 𝑇 𝛽 cosh 𝛽 » ` 21 2 1 𝛽. → — 0 ùñ 1 2 1 — 8 ùñ So, → 1 2 ùñ 𝛽 = 𝑘𝐵𝑇 𝐼(𝑆1; 𝑆2) = 𝛽 ´ln(1 ` 2 𝛽 ) » 𝛽 ´ 2 𝛽 Exercise 4 We have per definition

𝐼(𝑋; 𝑌 ) =

ÿÿ

𝑃

𝑥

𝑋𝑌

(

𝑃𝑋𝑌 (𝑥, 𝑦) 𝑃𝑋( 𝑥)𝑃𝑌 (𝑦)

) 𝑥, 𝑦 log

𝑦

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

𝐻(𝑋) = ´ 𝑃𝑋(𝑥) log 𝑃𝑋(𝑥) 𝑥

ÿ

𝐻 (𝑋, 𝑌 ) = ´

𝑃 𝑋𝑌 (𝑥, 𝑦 )log 𝑥, 𝑦

𝑃𝑋𝑌 (𝑥, 𝑦) 𝑃𝑌 (𝑦)

So, we have,

𝐻(𝑋) ` 𝐻(𝑌 ) ´ 𝐻(𝑋𝑌 ) ÿ ÿ ÿÿ = ´ 𝑃𝑋(𝑥) log 𝑃𝑋(𝑥) ´ 𝑃𝑌 (𝑦) log 𝑃𝑌 (𝑦) ` 𝑃𝑋𝑌 (𝑥, 𝑦) log 𝑃𝑋𝑌 (𝑥, 𝑦) 𝑥

𝑦

𝑥

𝑦

And from definitions of the mutual information:

𝐼(𝑋; 𝑌 ) = 𝑥

ÿÿ

𝑃𝑋𝑌 (𝑥, 𝑦) log 𝑃𝑋𝑌 (𝑥, 𝑦)

𝑦

ÿÿ ÿ ÿ ´ ( 𝑃𝑋𝑌 (𝑥, 𝑦) log 𝑃𝑋(𝑥)) ´ ( 𝑃𝑋𝑌 (𝑥, 𝑦) log 𝑃𝑌 (𝑦)) 𝑥

ÿÿ 𝑥

𝑦

Further, we have,

𝑦

𝑦

ÿ

𝑥

ÿ

𝑃𝑋𝑌 (𝑥, 𝑦) log 𝑃𝑋𝑌 (𝑥, 𝑦) ´ 𝑃𝑋(𝑥) log 𝑃𝑋(𝑥) ´ 𝑃𝑌 (𝑦) log 𝑃𝑌 (𝑦) 𝑥

𝑦

37

38

Complexity Science

𝐻( 𝑋, 𝑌 ) ´ 𝐻( 𝑋|𝑌 ) ´ 𝐻( 𝑌 |𝑋) = ´

ÿÿ

𝑦𝑃𝑋𝑌 (𝑥, 𝑦) log 𝑃𝑋𝑌 (𝑥, 𝑦)

𝑥

´r´

ÿÿ 𝑥

´r´

ÿÿ

ÿÿ

𝑥

𝑥

` 𝑥

=

ÿÿ 𝑥

s

𝑃𝑋𝑌 ( 𝑥, 𝑦 )log

𝑃𝑋𝑌 (𝑥, 𝑦) 𝑃𝑋(𝑥)

s

𝑦

=´ ÿÿ

𝑃𝑋𝑌 (𝑥, 𝑦) 𝑃𝑌 (𝑦)

𝑦

𝑥

`

𝑃𝑋𝑌 ( 𝑥, 𝑦 )log

𝑃𝑋𝑌 (𝑥, 𝑦) log 𝑃𝑋𝑌 (𝑥, 𝑦)

𝑦

ÿ

𝑃𝑋𝑌 (𝑥, 𝑦) log 𝑃𝑋𝑌 (𝑥, 𝑦) ´ 𝑃𝑌 (𝑦) log 𝑃𝑌 (𝑦)

𝑦

𝑦

ÿÿ

ÿ

𝑃𝑋𝑌 (𝑥, 𝑦) log 𝑃𝑋𝑌 (𝑥, 𝑦) ´ 𝑃𝑋(𝑥) log 𝑃𝑋(𝑥)

𝑦

𝑥

ÿ

ÿ

𝑃𝑋𝑌 (𝑥, 𝑦) ´ 𝑃𝑋(𝑥) log 𝑃𝑋(𝑥) ´ 𝑃𝑌 (𝑦) log 𝑃𝑌 (𝑦) 𝑦

𝑥

ÿ

𝐻(𝑋) ´ 𝐻(𝑋|𝑌 ) = ´ 𝑃 𝑋

(‹)

𝑦

(𝑥) log 𝑃𝑋 (𝑥) `

𝑥

ÿÿ 𝑥

𝑃 𝑋𝑌 (𝑥, 𝑦 )log

𝑦

𝑃𝑋𝑌 (𝑥, 𝑦) 𝑃(𝑦)

ÿ ÿÿ ÿ = ´ 𝑃𝑋(𝑥) log 𝑃𝑋(𝑥) ` 𝑃𝑋𝑌 (𝑥, 𝑦) log 𝑃𝑋𝑌 (𝑥, 𝑦) ´ 𝑃𝑌 (𝑦) log 𝑃𝑌 (𝑦) 𝑥

𝑥

𝑦

𝑦

And similar for 𝐻(𝑌 ) ´ 𝐻(𝑌 |𝑋).

Exercise 5 We derive, Eq. (9.27), 𝐷𝑇𝐸𝑥𝑖 → — 𝑥𝑗 = 𝑇 𝐸 X → — 𝑋𝑗 ´ 𝑇𝐸X(—𝑖 ) → — 𝑋𝑗

𝐷𝑇𝐸 𝑋𝑖 → — 𝑋𝑗

− Probt 𝑋 + 𝑗 |X u ą =ă log Probt𝑋+|X− u

𝑗

ÿ

(−𝑖)

− − − − + + + rProbt 𝑋 + 𝑗 , X u log Probt 𝑋 𝑗 |X u ´ Probt 𝑋 𝑗 , X u log Probt 𝑋 𝑗 |X (−𝑖)us

− + − = ´𝐻 r𝑋 + 𝑗 |X s ` 𝐻 r 𝑋 𝑗 |X (−i)s

+ − + + − = 𝐻r𝑋 + s 𝑗 s ´ 𝐻 r 𝑋 𝑗 |X s ´ 𝐻 ( 𝑋 𝑗 ) ´ 𝐻 r 𝑋 𝑗 |X (−𝑖)

From Eq. (9.20), − + − = 𝐼r𝑋+ 𝑗 ; X s ´ 𝐼 r 𝑋 𝑗 ; X (−𝑖)s

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

Complexity Science

39

From Eq. (9.21),

𝑇 𝐸𝑌 → — 𝑋 = 𝐼 ( 𝑋 + ; (𝑋−, 𝑌 −)) ´ 𝐼 ( 𝑋 + ; 𝑋−) − − + − 𝑇 𝐸 X→ — 𝑋𝑗 = 𝐼 ( 𝑋 + 𝑗 ; ( 𝑋 𝑗 , X )) ´ 𝐼 ( 𝑋 𝑗 ; 𝑋 𝑗 )

+ − = 𝐼(𝑋+ 𝑗 ; X) ´ 𝐼 ( 𝑋 𝑗 ; 𝑋 𝑗 )

(A)

− − + − 𝑇𝐸X(—𝑖 ) → — 𝑋𝑗 = 𝐼 ( 𝑋 + 𝑗 ; ( 𝑋 𝑗 , X (−𝑖))) ´ 𝐼 ( 𝑋 𝑗 ; 𝑋 𝑗 ) +

−

= ( 𝐼 𝑗𝑋 ; X(−𝑖) ) ´ 𝐼 ( 𝑋 𝑗 ; 𝑋 𝑗 ) Since, 𝑋− is contained in X−

(−𝑖)

𝑗

+

−

(B)

. And (9.27) follows since subtracting (B) from (A) gives (‹)

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 𝑋 → — 𝑍 and

𝐷𝑇 𝐸 𝑋 → —𝑍 Exercise 7

1´

𝑆𝑞r𝑝s =

Eq.(9.36) Tsallis entropy

ř𝑊

𝑝𝑞

𝑖=1 𝑖

𝑞´1 ÿ

𝐻r𝑝s = ´ 𝑝𝑖 ln 𝑝𝑖

Eq.(9.31) Shannon entropy

𝑖 1

exp𝑞(𝑥) = r1 ´(1 ´ 𝑞)𝑥)s 1—𝑞

Eq.(9.38) q-exponential Recall

𝑥

lim (1 ` ) 𝑘 = exp(𝑥) 𝑘 𝑘→ —∞ To determine the limit lim 𝑆𝑞

𝑞→ —1

we make use of l’Hopital’s rule:

𝑓 2(𝑥) lim 𝑓 (𝑥) = lim 𝑥→ — 𝑥0 𝑔(𝑥) 𝑥→ — 𝑥0 𝑔2(𝑥) for 𝑓 (𝑥) = 𝑔(𝑥) = 0. So, we have 𝑑

ř (1 ´

𝑖 lim 𝑆𝑞 lim 𝑑𝑞 = 𝑑 ( 𝑞 ´ 1) 𝑞→ —1 𝑞→ —1 𝑑𝑞

ř ´ 𝑝𝑞 ln 𝑝

𝑝𝑞) 𝑖

=

lim 𝑞→ —1

𝑖

𝑖

1

𝑖

40

Complexity Science ÿ = ´ 𝑝𝑖 ln 𝑝𝑖 = Eq. (9.31)) 𝑖

For the q-exponential, we have r1 ´ (1 ´ 𝑞)𝑥s 1

= r1 ´ 𝑥s𝜅 with 𝜅 =

1 —𝑞

𝜅

1

1 1−𝑞

and since 𝜅 → — 8 for

𝑞→ — 1, we conclude that r1 ´(1 ´ 𝑞)𝑥s 1—𝑞 → — exp(´𝑥) when 𝑞 → — 1.

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

Eq(9.41) 𝑆 r 𝑝 s =

ÿ

𝑖=1

1

𝑝 𝐺𝑖 (ln )𝑝𝑖 ř

and the non-trace form

𝐺(ln( 𝑊 𝑖=1

Eq (9.42) 𝑆r𝑝s =

𝑝𝛼) 𝑖

1´𝛼

and make use of the extensivity requirement.

1 Eq. (9.44) 𝑆( ) = 𝜆𝑁 for 𝑁 → — 8

𝑊

Trace form

𝑆(

1

𝑊

𝑊

ÿ 1 𝐺 ln 𝑊 𝜆𝑁 )= 𝑊 (

— ) = 𝜆𝑁 →

𝑖= 1

ùñ 𝐺(ln 𝑊) = 𝜆𝑁 We find the functional form for 𝐺(𝑡) by inversion. Let 𝑡 = ln 𝑊 → — −1 𝑡 𝜆𝑊 (𝑒 ). To ensure 𝐺(0) = 0, we subtract and have

𝑁 = 𝑊−1(𝑒𝑡 ), so 𝐺(𝑡) =

𝐺(𝑡) = 𝜆r𝑊−1(𝑒𝑡 ) ´ 𝑊−1(1)s The three cases:

𝑊(𝑁) = 𝑁𝑎 = 𝑡 → — 𝑊−1(𝑡) = 𝑁 = 𝑡 which gives 𝑡

1

𝑎

𝑡

1

𝐺(𝑡) = 𝜆r(𝑒 ) 𝑎 ´ 1s = 𝜆r𝑒 𝑎 ´ 1s and

𝑆r𝑝s =

ÿ 𝑊

𝑝 𝐺(ln 𝑖

𝑖=1

1

)=

𝑝𝑖

ÿ𝑊

𝑝𝜆

𝑖=1

1 𝑖 r(

1

)𝑎 ´

1s i.e. Eq.(9.47)

𝑝𝑖

𝑊(𝑁) = 𝑘𝑁 = 𝑡 → — 𝑁 ln 𝑘 = ln 𝑡

ùñ 𝑊−1(𝑡) =

ln 𝑡 ln 𝑘

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

Complexity Science So,

ln 𝑒𝑡 𝑡 𝐺(𝑡) = 𝜆r ´ 0s = 𝜆 ln 𝑘 ln 𝑘

and

𝑊

ÿ

𝜆

𝑆r𝑝s = ln 𝑘 𝑊(𝑁) = 𝑁

𝛾𝑁

𝑖=1

1

𝑝 𝑖 ln 𝑝𝑖

𝑁

=𝑡→ — 𝑁

Eq.(9.49) 1

1

= 𝑡 ùñ 𝑁 ln 𝑁 = ln 𝑡 𝛾 𝛾

Now recall the Lambert function 𝐿(𝑡) defined according to 𝐿(𝑡)𝑒𝐿(𝑡) = 𝑡 and let ln 𝑋 = 𝐿(𝑡). Then 𝑥 ln 𝑥 = 𝐿−1(ln 𝑥) or 𝐿(𝑥 ln 𝑥) = ln 𝑥. So from, we have 𝑡 = 𝐿−1(ln 𝑥) and ln 𝑥𝑒ln 𝑥 = 𝐿−1(ln 𝑋) → — 1

𝑁 ln 𝑁 = ln 𝑡 𝛾 we have 1

𝐿(𝑁 ln 𝑁) = 𝐿(ln 𝑡 ) 𝛾

1

ln 𝑁 = 𝐿(ln 𝑡 𝛾 ) 1

𝑁 = exp(𝐿(ln 𝑡 𝛾 )) and conclude that

𝑊−1(𝑡) = expr𝐿(

1

𝛾

ln 𝑡)s

So, 1

𝐺(𝑡) = 𝜆texpr𝐿( 𝑡)s ´ 1u 𝛾 and ÿ

1

𝑆r𝑝s = 𝜆 𝑝 rexpr 𝐿( ln 𝑖 𝛾 𝑖

1

) ´ 1ss

Eq. (9.50)

𝑝𝑖

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

𝑊(𝑁) = 𝑁𝑎 → — 𝑊−1(𝑡) = 𝑡 𝑎 and 1

𝑆r𝑝s = 𝜆rexp(

ln

ř 𝑖

𝑝𝛼 𝑖

𝑎(1 ´ 𝛼)

) ´ 1s

which is Eq.(9.51).

𝑊(𝑁) = 𝑘𝑁 → — 𝑊−1(𝑡) = So,

𝑆r𝑝s =

𝜆 ln 𝑘(1 ´ 𝛼)

ln 𝑡 ln 𝑘

ÿ exprln 𝑊 𝑝 𝛼 s 𝑖 𝑖=1

41

42

Complexity Science = 𝜆 ř𝑊 𝑝 𝛼 𝑖=1 𝑖 ln 𝑘 1´𝛼 1

𝑊(𝑁) = 𝑁𝛾𝑁 → — 𝑊−1(𝑡) = expr𝐿( ln 𝑡)s 𝛾 ùñ 𝑊−1(1) = 1 So, 1 𝑡 𝐺(𝑡) = 𝜆(1 ´ 𝛼)texpr𝐿( ln(exp ))s ´ 1u 𝛾 1´𝛼 = 𝜆(1 ´ 𝛼)texpr𝐿(

𝑡 )s ´ 1u 𝛾(1 ´ 𝛼)

and

𝑆r 𝑝 s =

1

ÿ

𝐺(ln 𝑊 𝑝 𝛼) 𝑖

1´𝛼 𝑖=1 ř𝑊 ln 𝑝𝛼 𝑖=1 𝑖 = 𝜆texpr𝐿( )s ´ 1u 𝛾(1 ´ 𝛼) 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

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

Complexity Science

43

Project 2

a)

𝑋1

𝑋2

𝑋3

𝑋2𝑛𝑎𝑛𝑑𝑋3

𝑃(𝑋3

𝑋1𝑛𝑜𝑟𝑋2

𝑃(𝑋3)

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 (𝑥 ´ 𝑥0)2

1 Eq. (10.57) 𝑃(𝑥, 𝑡) = , (expr´ 4𝜋𝐷𝑡 1

𝐴

ż =2

4𝐷𝑡

s)

2

0

𝑃(𝑥, 𝑡)𝑑𝑧 ż∞

(𝑥 ´ 𝑥0)2

= , ( expr´ 4𝜋𝐷𝑡 0

4𝐷𝑡

𝑒−𝑧 𝑑𝑧

0

ż∞

(𝑥 ` 𝑥0)2

𝑥0

.

ż∞

1

4𝐷𝑡

s ´expr´

s𝑑𝑥 ´

4𝐷𝑡

(𝑥 ` 𝑥0)2

expr´ 0

4𝐷𝑡

s𝑑𝑥)

𝑥−𝑥0 𝑥0 In first integral use substitution 𝑧 = ?4𝐷𝑡 in the second 𝑧 = ?4𝐷𝑡𝑥+. Then,

ż∞

𝑃(𝑥 , 𝑡 ) 𝑑𝑥 = ,

4𝜋𝐷𝑡

0

´

ż∞ ( 𝑥

1

ż∞ 𝑥 . 0 4𝐷𝑡

𝑒𝑧

− .40𝐷𝑡

2

=

𝑥

− .4𝐷𝑡0

ż

, 4𝐷𝑡𝑑𝑧

𝑥0

.

4𝐷𝑡

𝑒 𝑑𝑧 =2 −𝑧

2

, 4𝐷𝑡𝑑𝑧)

ż. 𝑥0 4𝐷𝑡

𝑒 −𝑧

2

𝑒−𝑧 𝑑𝑧 2

0

Exercise 2 a)

1

1

𝑛(𝑥, 𝑡 ` 1) = 𝑛(𝑥 ´ 1, 𝑡) ` 𝑛(𝑥 ` 1, 𝑡) 2

2

1

𝑛(𝑥, 𝑡 ` 1) ´ 𝑛(𝑥, 𝑡) = (𝑛(𝑥 ´ 1, 𝑡) ` 𝑛(𝑥 ` 1, 𝑡) ´ 2𝑛(𝑥, 𝑡)) 2

In the continuous limit, B𝑡 𝑛(𝑥, 𝑡) =

2

1 2 B𝑥 𝑛(𝑥, 𝑡)

(‹)

b) Include source term 1 2 B𝑡 𝑛 = B𝑥 𝑛 ` 𝑔(𝑥, 𝑡) 2 Fourier transform of Eq.(‹), 1

𝑖𝜔𝑛ˆ = ´ 𝑘2𝑛ˆ `𝑔ˆ 2

ˆ

𝑛 = 𝑔ˆ(𝑘1, 𝜔) 𝑖𝜔 ` 𝑘2 2

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

Complexity Science ż∞

𝑛(𝑥, 𝑡) = c)

= ż∞

ż∞ −∞

𝑑𝜔2 ∞ 2𝜋

ż∞

ż∞

𝑑𝑡

𝑑𝜔2 = ∞ 2𝜋

𝑑𝑘 𝑔ˆ(𝑘, 𝜔) 𝑖𝑘𝑡 +𝑖𝜔 𝑒 1 2 −∞ 2𝜋 𝑖𝜔 ` 𝑘2

ż∞

𝑁ˆ (𝜔) =

=

ż∞

𝑑𝜔 −∞ 2𝜋

−∞

ż∞ −∞

ż∞

𝑑𝑡𝑁(𝑡)𝑒−𝑖𝜔𝑡

𝑛(𝑥0, 𝑡)𝑒−𝑖𝜔𝑡 𝑑𝑘 𝑔ˆ(𝑘, 𝜔)

−∞ 2𝜋 𝑖𝜔

𝑑𝑘 𝑔ˆ(𝑘, 𝜔)

−∞ 2𝜋 𝑖𝜔2 `

𝑁ˆ (𝜔 ) =

2

` 1 𝑘22

𝑒𝑖𝑘𝑥0

ż∞

𝑒𝑖𝑘𝑥0 +𝑖𝑤 𝑡 𝑒−𝑖𝜔𝑡 1

𝑑𝑡𝑒𝑖(𝜔 −𝜔)𝑡 1

1 2 𝑘2

∞

ż∞

𝑑𝑘 𝑔ˆ(𝑘, 𝜔) 𝑖𝑘𝑥0 𝑒 1 2 ∞ 2𝜋 𝑖𝜔 ` 2𝑘

Use 𝑔(𝑥, 𝑡) = 𝛿(𝑡)𝜒(𝑡) ùñ 𝑔ˆ(𝑘, 𝜔) = 𝜒ˆ(𝜔) Since, ş∞ 𝑑𝑡𝛿(𝑡)𝑒−𝑖𝑘 𝜔 =1. −∞ Therefore,

𝑁ˆ (𝜔) =

ż∞

𝑑𝑘 𝜒ˆ(𝜔 1 2 2 −∞ 𝜋 𝑖𝜔 ` 𝑘2

d) Compute the integral over 𝑘

𝑁ˆ (𝜔 ) =

𝑒𝑖𝑘𝑥0 𝜒ˆ(𝜔) ż ∞ 𝑑𝑘 𝜋 ∞ (𝑘 ´ 𝑘 +)(𝑘 ´ 𝑘−)

where 𝑘+ and 𝑘− are the roots of 1 2 𝑘 ` 𝑖𝜔 = 0 ùñ 𝑘 2

˘= ˘

, 2𝜔

1 𝑖 (´ , ` , ) 2 2

Since, 𝑥0 ą 0 we can close the contour in the upper half plane and pick up the residue , exp(𝑖 2𝜔 𝑖?− 1 𝑥0) 𝑖𝑘 + 𝑥0 𝑒 2 Res(𝑘+) = 𝑘 + ´ 𝑘− = 2,𝜔(𝑖 ´ 1) Hence,

𝑁ˆ ( 𝜔 ) =

𝜒ˆ(𝜔) 2𝜋𝑖Res(𝑘 +) 𝜋

45

46

Complexity Science and the power spectrum, |𝜒ˆ(𝜔)|2

𝑁ˆ (𝜔 ) =

Since, 𝑒

?

−2 𝜔𝑥0

?

𝑒 −2 𝜔𝑥0

2𝜔 , → — 1 for 2 𝜔𝑥0 ăă 1 we obtain a 1 behaviour when 𝜔 ăă

1 2

.

(2𝑥0 )

𝑓

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,

𝑃(0, 𝑡 ` 1) = 𝑃(0, 𝑡) ` 𝑝−𝑃(1, 𝑡) and

𝑃𝑅(0, 𝑡 ` 1) = 𝑃(0, 𝑡 ` 1) ´ 𝑃(0, 𝑡) = 𝑝−𝑃(1, 𝑡) b) Master Equation:

𝑃(𝑥, 𝑡 ` 1) = 𝑃(𝑥, 𝑡) ` 𝑝+𝑃(𝑥 ´ 1, 𝑡) ` 𝑝−𝑃(𝑥 ` 1, 𝑡) ´ 𝑃(𝑥, 𝑡) and since, 𝑝+ = 𝑝− = 1 ,2we have 1

𝑃(𝑥, 𝑡 ` 1) = (𝑃(𝑥 ´ 1, 𝑡) ` 𝑃(𝑥 ` 1, 𝑡)) 2

c) Simply use Eq. (10.61) with 𝑥0 = 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 ż∞ 2 𝑆 (𝜔) = | 𝑓 (ˆ𝜔 )| =

−∞

𝑑𝑡𝐶 𝑡𝑒( )

𝜏=

−𝑖𝜔𝑡

1 =

(𝜏𝐴)2

4 1 ` ( 𝜏𝜔2 )2

1

𝜈

c) From (a) we have,

𝑆(𝜔) → —

(𝜏𝐴)2 4

= constant for 𝜔 → — 0

(𝜏𝐴)2 2 2 𝐴2 𝑆(𝜔) » ( ) = ( ) for 𝜔 ąą 1 4 𝜏𝜔 𝜔

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

Complexity Science

47

Project 1 a) Write a code to simulate the walk. Determine the time dependence of the average ă () 2(𝑡 ) ą and use log-log plots to find the Hurst exponent from Eq. (10.85). b) 𝑃(𝑥, 𝑡 ` 𝛿) = 𝑃(𝑥, 𝑡) ` 𝑝+𝑃(𝑥 ´ 1, 𝑡) ` 𝑝−𝑃(𝑥 ` 1, 𝑡) ´(1 ´ 𝑝0)𝑃(𝑥, 𝑡) 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 "

1, for 𝑥 = 0 0, otherwise

𝑃(𝑥, 0) =

) And continue to iterate until 𝑃(´𝐿, 𝑡) and 𝑃(𝐿, 𝑡) become non-zero. d) For 𝑝+ = 𝑝− Ξ 𝑝, we have from (b)

𝑃(𝑥, 𝑡 ` 𝛿) ´ 𝑃(𝑥, 𝑡) = 𝑝r𝑃(𝑥 ´ 1, 𝑡) ` 𝑃(𝑥 ` 1, 𝑡) ´ 2𝑃(𝑥, 𝑡)s `r2 𝑝 ´(1 ´ 𝑝0)s𝑃(𝑥, 𝑡) Here the last term vanishes since 1 = 𝑝+ ` 𝑝− ` 𝑝0 = 2𝑝 ` 𝑝0 We get then in the limit 𝑎 → — 0 and → — 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 𝑋2 (𝑡) ą= 2𝐷𝑡 From Eq. (10.85), we get 𝐻 = 12 Project 2 No sample solution needed.

Project 3 a) Fourier transform

𝑛(𝑥, 𝑡) =

ż∞

𝑑𝑘 −∞ 2𝜋

ż∞

𝑑𝜔 𝑛ˆ(𝑘, 𝜔)𝑒𝑖(𝑘𝑥+𝜔𝑡 ) −∞ 2𝜋

Substitute into equation and get,

𝑖𝜔𝑛ˆ = 𝛾𝑘2𝑛ˆ ` 𝛽𝑖𝑘𝑛ˆ ` 𝑔ˆ ùñ 𝑛ˆ =

𝑔ˆ 𝑖𝜔 ` 𝛾𝑘2 ´ 𝛽𝑖𝑘

48

b)

Complexity Science Substitute into expression (‹) above tożobtain 𝑛(𝑥, 𝑡) ∞ ż 𝑑𝑘 ∞ 𝑑𝜔 𝑔ˆ(𝑘{𝑤) 𝑛(𝑥, 𝑡) = −∞ 2

Note 𝑘𝑥 ` 𝜔𝑡 = 𝑘 ` ` 𝜔𝑡 = 𝑍1 𝑥

𝑡

𝑥 `𝑍

𝜋 −∞ 2𝜋 𝑖𝜔 ´ 𝛽𝑖𝑘

𝑒𝑖(𝑘𝑥+𝜔𝑡 )

2 with

𝑡

𝑍1 = 𝑘𝑡 → — 𝑘=

𝑑𝑍1 𝑍1 — 𝑑𝑘 = → 𝑡 𝑡

𝑍2 = 𝑘𝑡 → — 𝑘=

𝑑𝑍2 𝑍2 — 𝑑𝑘 = → 𝑡 𝑡

1 1 ż ∞ 𝑑𝑍1 (2𝜋)2 𝑖 −∞ 𝑡

ż∞

Therefore,

𝑛(𝑥, 𝑡) =

= 1 2𝜋1 2 𝑖1 𝑡 ( )

ż∞

𝑍1 𝑑𝑍 2 𝑔ˆ( 𝑍2) 𝑖(𝑍1 𝑥 +𝑡 𝑍2) 𝑒 𝑍2 𝑍1 −∞ 𝑡 𝑡 ´ 𝛽 𝑡

ż∞

𝑑𝑍

1 ∞

−∞

𝑑𝑍

2

𝑔ˆ( 𝑍𝑍1 ) 2 𝑖(𝑍1 𝑥 +𝑡𝑍2) 𝑍 ´ 𝛽𝑍 𝑒 2

1

1 𝑥 = 𝐹( )

𝑡

c) 𝛾 ‰ 0, 𝛽 = 0

𝑛(𝑥, 𝑡) =

𝑁(𝑡) =

ż∞

ż∞ −∞ 2

𝑑𝑥𝑛(𝑥 , 𝑡 ) =

−∞

ż∞

𝜋

𝑑𝑘

) 𝑑𝜔 𝑔ˆ(𝑘, 𝜔

∞ 2𝜋 𝑖𝜔 ` 𝛾𝑘 2

ż∞

∞ 2𝜋

−∞

Recall ş∞

ż∞

𝑑𝑘

𝑡

𝑒 𝑖 ( 𝑘𝑥 + 𝜔𝑡 )

𝑑𝜔 𝑔ˆ(𝑘, 𝜔) 2 −∞ 2𝜋 𝑖𝜔 ` 𝛾𝑘

r

ż∞

𝑑𝑥𝑒 𝑖 𝑘 𝑥 s𝑒𝑖𝜔𝑡

−∞

𝑑𝑥𝑒𝑖𝑘𝑥 = 2𝜋𝛿(𝑥) and perform the integration over 𝑘: 𝑁(𝑡) =

ż∞

−∞

𝑑𝜔 𝑔ˆ(0, 𝜔) 𝑖𝜔𝑡 𝑒 2𝜋 𝑖𝜔

d) The fourier transform of 𝐶(𝑡) is the power spectrum of 𝑁(𝑡) i.e.

𝐶ˆ(𝜔) = 𝑆(𝜔) = |𝑁ˆ(𝜔)| = | e)

𝑔ˆ(0, 𝜔) 2 | 𝑖𝜔

The fourier transform of 𝛿(𝑡) is a constant independent of frequency 𝜔. So, we need 𝑔ˆ(0, 𝜔)9𝜔. ż

𝐶( 𝑡 ) = ż∞ = Substitute 𝑍 = 𝜔𝑡 ùñ 𝜔 = 𝑍 𝑡

∞

𝑑𝜔 𝑔ˆ(0, 𝜔) 2 𝑖𝜔𝑡 | |𝑒 𝑖𝜔 −∞ 2𝜋

𝑑𝜔 |𝜔| 𝛼

−∞ 2𝜋

𝜔

𝑒 2

𝑖𝜔𝑡

ż∞ =

−∞

𝑑𝜔 2𝜋

|𝜔|

𝛼−2

𝑒

𝑖𝑡

1 ùñ 𝑑𝑤 = 𝑡 𝑑𝑍

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

Complexity Science

𝐶 ( 𝑡) =

1 2𝜋

ż∞

𝑑𝑍 −∞

𝑍 𝛼−2 𝑖𝑍 1−𝛼 𝑒 9𝑡

| |

𝑡 𝑡

49

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Chapter 11 Exercise 1 a) The probability that a randomly chosen person doesn’t belong to infinite 𝐵 cluster= 1 ´ 𝑃∞ Ξ 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 (1 ´ 𝑝𝑃∞)𝑞𝑐 Ξ 𝐼 𝐼 I = II 1 ´ 𝑃∞ = (1 ´ 𝑝𝑃∞)𝑞𝑐

No zero solution for 𝑃∞ appears when 𝑝 ą 𝑝𝑐, where 𝑝𝑐 is determined by the following criterion:

𝑓 (𝑥) = (1 ´ 𝑝𝑥)𝑞𝑐 , 𝑓 2(0) = ´1

𝑓 2(𝑥) = ´𝑞𝑐 𝑝(1 ´ 𝑝𝑥)𝑞𝑐 −1

𝑓 2(0) = ´𝑞𝑐 𝑝 Hence, 𝑓 2(0) = ´1 ùñ 𝑝𝑐 = 1 𝑞𝑐 b) Expand 𝑓 (𝑥) = (1 ´ 𝑝𝑥)𝑞𝑐 for small x

𝑓 ”(𝑥) = 𝑝2𝑞𝑐(𝑞𝑐 ´ 1)(1 ´ 𝑝𝑥)𝑞𝑐 −2

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

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51

ùñ 𝑓 ”(0) = 𝑝2 𝑞𝑐(𝑞𝑐 ´ 1) The Eq. from (a) now becomes 1 2 𝑝 𝑞𝑐(𝑞𝑐 ´ 1)𝑥2 2

1 ´ 𝑥 = 1 ´ 𝑝𝑞𝑐𝑥 `

ùñ 𝑥 = 2

𝑝𝑞𝑐 ´ 1 𝑝2 𝑞𝑐(𝑞𝑐 ´ 1)

1

Substitute 𝑝 = 𝑝𝑐 ` Δ = 1 𝑞𝑐`Δ

𝑥»

2𝑞𝑐Δ 𝑐 2 ´Δ 1 ` 𝑞𝑐 2Δ2 ` 𝑞𝑐 Δ ´ 𝑞1𝑐 ´ 𝑞2Δ

» 2𝑞𝑐Δr1 ´

𝑃∞ »

2𝑞2 𝑐

1

−1

` 𝑂(Δ)s

𝑞𝑐

2𝑞2 𝑐

Δ=

𝑞𝑐 ´ 1

(𝑝 ´ 𝑝𝑐)𝛽

𝑞𝑐 ´ 1 𝛽=1

c)

𝑃(𝑠) = 𝑠(1 ´ 𝑝)𝑝𝑠(1 ´ 𝑝) 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=

∞

ÿ

𝑠𝑃(𝑠)

𝑠=1

= (1 ´ 𝑝)2

∞ ÿ 2 𝑠

𝑠𝑝

𝑠=1

= (1 ´ 𝑝) 𝑝 2

∞ 𝑑 𝑑 ÿ 𝑝 𝑝𝑠 𝑑𝑝 𝑑𝑝 𝑠=1

= (1 ´ 𝑝) 2𝑝 𝑑 𝑝 𝑑

𝑝

𝑑𝑝 𝑑𝑝 1 ´ 𝑝

= 𝑝1 ` 𝑝 1´ 𝑝

e) In one dimension, a system spanning infinite clusters apprears as 𝑝 → — 1 = 𝑝𝑐. Hence from (d), ă 𝑠 ą 9(𝑝𝑐 ´ 𝑝)−1 , therefore, 𝛾 = 1

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Exercise 2 Merger is initiated with probability 𝑝𝑚𝑒𝑟 = 𝑝𝑚𝑒𝑟𝑔𝑒𝑟 = 𝑝( 1 ` 𝑛𝐴 )𝛼 where 𝑛𝐴 is the ancestry of the initiating cell. We will think of 𝑛𝐴 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. 𝑃(𝑛𝐴, 𝑡) = probability that a randomly chosen company (bin) has ancestry 𝑛𝐴 (contains 𝑛𝐴 balls) Consider 𝑛𝐴 = 0 𝑡 ÿ

𝑃(0, 𝑡 ` 1) = 𝑃(0, 𝑡) `

𝑝𝑚𝑒𝑟 (𝑛)𝑃(𝑛, 𝑡)(1 ´ 𝑃(0, 𝑡))

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 𝑃(𝑛, 𝑡) and triggers the merger with probability 𝑝𝑚𝑒𝑟 (𝑛). 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 ´ 𝑃(0, 𝑡).

Consider 𝑛𝐴 = 1

𝑃(1, 𝑡 ` 1) = 𝑃(1, 𝑡) ´ 𝑃(1, 𝑡)𝑝𝑚𝑒𝑟 (1)

𝑡 ÿ

𝑃(𝑘, 𝑡)

𝑘=1

´𝑃(1, 𝑡)

𝑡 ÿ

𝑝𝑚𝑒𝑟 (𝑛)𝑃(𝑛, 𝑡)

𝑛=1

Explanation of 2nd term on the righthand side 𝑃(1, 𝑡) is the probability that bin with one ball is chosen to initiate merger and therefore cease to be a 𝑛 = 1 bin. 𝑝𝑚𝑒𝑟 (1) is probability that merger goes ahead. 𝑝 𝑚 𝑒 𝑟 (1) = probability that merger goes ahead. ř 𝑡 𝑘=1 𝑃(𝑘, 𝑡) = 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 𝑛=1 𝑝𝑚𝑒𝑟 (𝑛)𝑃(𝑛, 𝑡) is the total probability that the merger is initiated by a bin with 𝑛 = 1, 2, ..., 𝑡 balls. 𝑃(1, 𝑡) = probability that site chosen for merger with initiator has 𝑛𝐴 = 1 ball and therefore after the merger cease to be a 𝑛𝐴 = 1 bin.

Consider 2 ă 𝑛𝐴 ď 𝑡

𝑃 (𝑛, 𝑡 ` 1) = 𝑃(𝑛, 𝑡 )`

𝑛−1

ÿ

𝑝𝑚𝑒𝑟 (𝑘)𝑃(𝑘, 𝑡)𝑃(𝑛 ´ 𝑘, 𝑡)

𝑘=1

´𝑝𝑚𝑒𝑟 (𝑛)𝑃(𝑛, 𝑡)

𝑡 ÿ

𝑃(𝑘, 𝑡)

𝑘=1

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

Complexity Science 𝑡 ÿ ´

53

𝑝𝑚𝑒𝑟 (𝑘)𝑃(𝑘 , 𝑡)𝑃(𝑛, 𝑡)

𝑘=1

Explanation of 2nd term on the righthand side: 𝑃 (𝑘, 𝑡 ) = probability bin with 𝑘 balls chosen to initiate merger. 𝑝𝑚𝑒𝑟 𝑘( ) = probability merger goes ahead. 𝑃( 𝑛 ´ 𝑘, 𝑡) = probability selected bin to be merged contains 𝑛 ´ 𝑘 balls so initiating bin will contain 𝑛 balls after merger.

Explanation of the 3rd term: 𝑃(𝑛, 𝑡) = 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 𝑃(𝑛, 𝑡). 𝑝 𝑚 𝑒 𝑟 (𝑛) = Probability merger goes ahead. ř 𝑡 𝑘=1 𝑝(𝑘, 𝑡) = total probability that bin to be merged contains 𝑘 = 1, ..., 𝑡 balls.

Explanation of 4th term 𝑃 (𝑘, 𝑡 ) = probability bin chosen to initiate merger contains 𝑘 balls. 𝑝𝑚𝑒𝑟 𝑘( ) = probability merger goes ahead. 𝑃( 𝑛, 𝑡) = probability bin to be merged contains 𝑛 balls. Since, after merger the bin will be empty, the merger will remove probability weight from 𝑃 (𝑛, 𝑡 ). 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) 1 1 1 ă 𝑥𝑖 ą= 𝑓 (𝛽𝑥˜𝑖) ´ 𝑓 (´𝛽𝑥˜𝑖) = ( ´ ) 2 1 ` 𝑒−𝛽 𝑥˜𝑖 1 ` 𝑒𝛽 𝑥˜𝑖 1 sinh(𝛽𝑥˜𝑖) 1 = tanh(𝛽𝑥˜ ) 𝑖 2 1 ` cosh( 𝛽𝑥˜𝑖 ) 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 sinh(𝛽𝑞 ă 𝑥 ą) Ξ 𝐹(ă 𝑥 ą) 2 1 `cosh(𝛽𝑞 ă 𝑥 ą)

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d) Slope of 𝐹(ă 𝑥 ą) at ă 𝑥 ą= 0 sinh(𝑢) = 1 𝛽𝑞 𝑑 𝑑ă𝑥ą 2 𝑑𝑢 1 `cosh(𝑢)

𝑑𝐹

1 `cosh(𝑢) 1 = 𝛽𝑞 2 (1 `cosh(𝑢))2 For 𝑢 = 0 → — Critical value:

𝑑𝐹 𝑑ă𝑥ą 𝛽𝑐𝑞

𝛽𝑞

| = 4 — 𝛽𝑐 = 𝑞 4= 1 → 0

4

e) Expand about ă 𝑥 ą= 0 at 𝛽 = 𝛽𝑐 „

𝐹 (0) =

(𝛽𝑞)2 𝑑 2

1

𝑑𝑢 1 `cosh 𝑢 3

𝐹 (0) =

|𝑢=0 =

sinh 𝑢

2

(1 `cosh 𝑢)2

(𝛽𝑞)2 𝑑

sinh 𝑢

2

𝑑𝑢 (1 `cosh 𝑢)2 𝑢=0

|𝑢=0 = 0

|

( 𝛽𝑞) cosh(1 `cosh 𝑢) ´sinh 𝑢(1 `cosh 𝑢)2 sinh 𝑢 ( 2 (1 `cosh 𝑢)4 3

=´

(𝑝𝑞)2

2

=

)|𝑢=0

´1 (𝛽𝑞)3 8

The ME Equation becomes 3rd order:

𝛽𝑞

ă 𝑥 ą=

4

ă𝑥ą´

11 (𝛽𝑞) 3! 8

3

ă𝑥ą

3

use 𝛽𝑐 = 𝑞: 4

𝛽𝑞

ă 𝑥 ą2 = 3!8

4

´1

𝛽

=

(𝛽𝑞)3 3 Δ ùñ ă 𝑥 ą» 4𝛽𝑐

3 𝛽𝑐 ´ 1 4 ( 𝛽 )3 𝛽𝑐

a 3𝑞Δ 4

Therefore, in the MF the order parameter exponent is 12. f) Master Equation: ÿ ÿ 𝑑 𝑃 (𝑥, 𝑡) = 𝑊 ( 𝑥 , 𝑥 2 ) 𝑃( 𝑥, 𝑡) ´ 𝑊(𝑥2, 𝑥)𝑃(𝑥, 𝑡) 𝑑𝑡 1 1 𝑥

𝑥

The only non-zero transition coefficients are:

𝑊(0, ´1) = 𝛾, 𝑊(0, 1) = 𝜂, 𝑊(´1, 0) = 𝜔, 𝑊(1, 0) = 𝜔

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

Complexity Science Therefore, (𝐸1)

𝐴9 = 𝜔𝐶 ´ 𝛾𝐴 Ξ 𝐻

(𝐸2)

𝐵9 = 𝜔𝐶 ´ 𝜂𝐵 Ξ 𝐺

(𝐸3)

𝐶9 = 𝛾𝐴 ` 𝜂𝐵 ´ 2𝜔𝐶

And,

𝐴 ` 𝐵 ` 𝐶 = 𝑁 ùñ 𝐶 = 𝑁 ´ 𝐴 ´ 𝐵 g) Fixed point 𝐻 = 𝐺 = 0

𝐻=0→ — 𝜔𝑁 ´(𝜔 ` 𝛾)𝐴 ´ 𝜔𝐵 = 0 𝐺=0→ — 𝜔𝑁 ´ 𝜔𝐴 ´(𝜔 ` 𝜂)𝐵 = 0 Note 𝐸1 ´ 𝐸2 → — 𝐴 = 𝐵𝛾 and solve to get: 𝜂

𝐴= 𝐵= 𝐶=

𝜔𝜂 𝜔(𝛾 ` 𝜂) ` 𝛾𝛿 𝜔𝛾

𝑁

𝜔 ( 𝛾 ` 𝜂) ` 𝛾 𝛾𝜂

𝑁

𝜔(𝛾 ` 𝜂) ` 𝛾𝜂

𝑁

h) Linear stability analysis: B𝐻 B𝐴 B𝐺 B𝐴

= ´(𝜔 ` 𝛾);

= ´𝜔 ;

B𝐺 B𝐵

B𝐻

= ´𝜔 B𝐵

= ´(𝜔 ` 𝜂)

Eigenvalues: ´(𝜔 ` 𝛾) ´ 𝜆 ´𝜔 ´𝜔 ´(𝜔 ` 𝜂) ´ 𝜆

𝜆

1 ˘ =

2

=0

2 2 t´( 𝜔 ` 𝛾 ` 𝜂) ˘ r(2𝜔 ` 𝛾 ` 𝜂) ´ 4(𝜔𝜂 ` 𝛾𝜔 ` 𝛾𝜂 )s

1{2

u

Clearly, 𝜆˘ ă 0 → — attractive fixed point. Further reduction leads to 1 𝜆 2 2 2 1{2 ˘ = (´( 𝜔 ` 𝛾 ` 𝜂) ˘ r4𝜔 `( 𝛾 ´ 𝜂) s ) 2 The relaxation towards the fixed point is controlled by the factors 𝑒𝜆+ 𝑡 and 𝑒𝜆— 𝑡 , so the time scales are given by

𝜏+ =

1 |𝜆+ |

and

𝜏 −=

1 |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 𝑊( 0,´ 1) = 𝛾𝐶 and 𝑊 (0.1 ) = 𝜂𝐶. This gives us the following set of equations

𝐴9 = 𝜔𝐶 ´ 𝛾𝐴𝐶 Ξ 𝐻 𝐵9 = 𝜔𝐶 ´ 𝜂𝐵𝐶 Ξ 𝐺 𝐶=𝑁´ 𝐴´ 𝐵 j) We have

𝐴9 = 𝐻 = r𝜔 ´ 𝛾𝐴sr𝑁 ´ 𝐴 ´ 𝐵s 𝐵9 = 𝐺 = r𝜔 ´ 𝜂𝐵sr𝑁 ´ 𝐴 ´ 𝐵s Fixed point at 𝐴x = 𝜔 and 𝐵x = 𝜔 . Note also that evolution stops if 𝐴 ` 𝐵 = 𝑁. 𝛾

𝜂

Linear stability matrix: B𝐻 B𝐴

= ´𝛾𝑁 ` 2𝛾𝐴 = 𝛾𝐵 ´ 𝜔 B𝐻 B𝐺 B𝐵 = ´r𝜔 ´ 𝛾𝐴s B𝐴

B𝐺 B𝐵 At fixed point,

B𝐻 B𝐴

Eigenvalues:

= ´𝜂𝑁 ` 𝛿𝐴 ` 2𝜂𝐵 ´ 𝜔

= ´𝛾𝑁 ` 𝜔 `

𝛾𝜔 B𝐻 ,

𝜂

B𝐺 B𝐵

= ´r𝜔 ´ 𝜂𝐵s

= 0,

B𝐻 B𝐵

=0 B𝐵

= ´𝜂𝑁 ` 𝜔 `

𝜂𝜔 𝛾

𝛾𝜔

´(𝛾𝑁 ` 𝜔 ` 𝜂 ) ´ 𝜆 0

0 ´(𝜂𝑁 ` 𝜔 ` 𝜂 𝜔𝛾 ) ´ 𝜆

=0

𝛾

ùñ 𝜆 𝐴 = 𝜔( 1` ) ´ 𝛾𝑁

𝜂

𝜂 𝜆𝐵 = 𝜔( 1` ) ´ 𝜂𝑁 𝛾 Note:

𝜔 𝛾 𝜆 𝐴 ě 0 ùñ 𝑁 ď ( 1 ` ) = 𝐴x ` 𝐵x Ξ 𝑁 𝑐 𝛾 𝜂

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

Complexity Science

𝜆 ě 0 ùñ 𝑁 ď 𝐵

𝜔

𝜂

(1 ` ) = 𝐴x ` 𝐵x Ξ 𝑁

𝜂

𝛾

𝑐

We have, For 𝑁 ă 𝑁𝑐fixed point repulsive For 𝑁 ą 𝑁𝑐fixed point attractive For 𝑁 ą 𝑁𝑐, 𝐴 → — 𝐴x and 𝐵 → — 𝐵x and evolution asymptotically stop with l i m 𝑡 → — ∞ 𝐶(𝑡) ą 0 For 𝑁 ă 𝑁𝑐, we return to, and recall 𝐴 ` 𝐵 ă 𝑁𝑐 A. = 𝛾(𝐴x ´ 𝐴)r𝑁 ´(𝐴 ` 𝐵)s B. = 𝜂(𝐵x ´ 𝐵)r𝑁 ´(𝐴 ` 𝐵)s where 𝑁 ´( 𝐴 ` 𝐵) is positive and if 𝐴(0) ą 𝐴x, 𝐴(𝑡) will decrease towards 𝐴x. If 𝐴(0) ă 𝐴x, 𝐴(𝑡) will increase towards 𝐴x.

The evolution of 𝐴(𝑡) and 𝐵(𝑡) stops when 𝐴(𝑡) ` 𝐵(𝑡) = 𝑁 with l i m 𝑡 → — ∞ 𝐶(𝑡) = 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.

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

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59

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) 𝑃(𝑡) = 𝑡 1 1 , just follow Eqns. (12.16) to (12.23) step by step. All details are given + in the text. Eq.(12.27) 1 (ln 𝑡)(𝑛−1) 𝑃𝑛(𝑡) = 𝑡 (𝑛 ´ 1)!

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 1

𝑃3(𝑡; 0, 𝑡1, 𝑡2) =

𝑡𝑡2𝑡1

.

We sum over 𝑡1 and 𝑡2, 2ÿ 𝑡 1 ÿ

𝑡 −1

𝑃 3( 𝑡 ) =

𝑡 𝑡2=2 𝑡1 =1

1

𝑡 2𝑡 1

»

1

ż𝑡

ż 𝑡2

𝑑𝑡2

𝑡

1

𝑑𝑡1 1

1

𝑡1𝑡2

Note the integrand is symmetric under reflection in the identity line 𝑦 = 𝑥. So, we can write ż ż 𝑡1 ż𝑡 ż 𝑡2 11 𝑡 1 1 𝑑𝑡 2 𝑃3(𝑡) » t 𝑑𝑡1 𝑑𝑡2 ` 𝑑𝑡1 u 𝑡2 1 𝑡2 𝑡1 𝑡2𝑡1 1 1 1 =

1 2𝑡

ż𝑡

ż𝑡

𝑑𝑡1 1

𝑑𝑡2 1

𝑡 1𝑡 2

1 =

2𝑡

r

1 1

ż𝑡

𝑑𝑧 s = 𝑧

1 2

(ln 𝑡)2 2𝑡

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(𝑡) =

1 (ln 𝑡)𝑛−1 𝑡 (𝑛 ´ 1)!

= 𝑒−

𝜇𝑡

𝜇 𝑡𝑛− 1 (𝑛 ´ 1)!

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

Next, we repeat the argument used in Eq. (6.96)

𝑃𝑛+1(𝑡) =

ÿ 𝜇𝑡+

𝑃𝑛(𝜇𝑡+ )𝑃1(𝜇𝑡 ´ 𝜇𝑡+ )

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Complexity Science

i.e. 𝑛 records upto 𝜇𝑡+ and exactly one record during 𝜇𝑡+ to 𝜇𝑡 . ÿ

𝜇𝑡 + (𝑛 ´ 1)!

𝑃𝑛+1(𝑡) = 𝑒−𝜇𝑡+ 𝜇𝑡+

ÿ

𝑒−𝜇𝑡 =

−𝜇

𝑒−(𝜇𝑡 −𝜇𝑡+ ) 𝑛−1

𝜇𝑡 +

𝑒

𝑡+

𝑡ÿ −1

(ln 𝑡x)𝑛−1

𝑡 +ě𝑛

𝑡x

(𝑛 ´ 1)! 𝜇+ 𝑡

For the sum, we have ÿ

𝑒

=

−𝜇𝑡+ 𝑛 1 −

𝜇

𝑡+

𝜇𝑡+

»

ż 𝑡 −1 ln 𝑡x 𝑛−1 ( )

=

𝑑𝑡x =

𝑡x

𝑛

(ln(𝑡 ´ 1))𝑛

𝑧𝑛−1 𝑑𝑧

ln 𝑛

(ln 𝑛)𝑛−1

=

𝑛

ż ln(𝑡 −1)

»

𝑛

(ln 𝑡)𝑛

𝑛

So, in summary,

𝑃𝑛+1(𝑡) =

1

(ln 𝑡)𝑛

1

𝑡 (𝑛 ´ 1)!

𝑛

(ln 𝑡)𝑛 =

𝑡𝑛!

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

Exercise 4 Logistic map 𝑓 (𝑥) = 𝑟𝑥(1 ´ 𝑥) fixed point: 𝑥x = 𝑓 (𝑥x) i.e. 𝑥x = 𝑟𝑥x(1 ´ 𝑥x) → — 𝑥x = 1−𝑟 2 2 Tangent point 𝑥𝑡𝑎𝑛 : 𝑓 (𝑥𝑡𝑎𝑛) = 1, 𝑓 (𝑥) = 𝑟 ´ 2𝑟𝑥 i.e. 𝑟 ´ 2𝑟𝑥𝑡𝑎𝑛 = 1 → — 𝑥𝑡𝑎𝑛 = 𝑟 −1

𝑟 2𝑟

𝜋 𝑇 = , 𝜖𝛽

Eq. (12.40) Here we need to use |𝛽| since our 𝛽 ă 0.

𝛽 = 𝑓 „(𝑥𝑡𝑎𝑛) 𝜖 = 𝑓 (𝑥𝑡𝑎𝑛) ´ 𝑥𝑡𝑎𝑛 Since, 𝑓 „(𝑥) = ´2𝑟, we have 𝛽 = ´2𝑟 and further:

𝜖 = 𝑓(

𝑟´1

)´

𝑟´1

2𝑟

=1

2𝑟

1 3 𝑟) ´ ( ` 4 𝑟

with 𝑟 = 1 ` Δ and 0 ă Δ ăă 1, we can write,

𝜖=1

´

1

(

3

4 1 `Δ

` 1 `Δ) » 1

´

1 3(1 ´ Δ `Δ 2 ) ` 1 `Δs r 4

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

Complexity Science

61

1 Δ 2 = 1 ´ r4 ´ 2Δ ´ 3Δ s » 4 2

𝑇» b

𝜋

» ,

Δ 22(1 `Δ)

𝜋 Δ

Exercise 5 No sample solution needed.

Project 1 a) 𝑝0 = 1 ´ 𝑝 , 𝑝1 = 0, 𝑝2 = 𝑝, 𝑝𝑘 = 0, 𝑘 ą 𝑧 b)

𝑔( 𝑠) =

∞ ÿ

𝑝𝑘 𝑠𝑘 = (1 ´ 𝑝) ` 𝑝𝑠2

𝑘=0

Process critical when average branch 𝜇 = 1.

𝜇 = 𝑔2(1) = 2𝑝 so 𝜇 = 1 ùñ 𝑝 = 𝑝𝑐 =

1 2

c) From lectures 𝑃(𝜏 ą 𝑛)9𝜇𝑛 = exp(𝑛 ln 𝜇). Define 𝑛0 = l −n 1 P R𝜇 + and we have 𝑃(𝜏 ă 𝑛) = exp(´ 𝑛 ).𝑛0 d) ´1 𝑛 =´ 1 = » 1 0

ln(1 ´Δ)

ln 𝜇

Δ

𝜇 = 1 ´ Δ ùñ 𝑎 = 1 e) When an energy unit is added, one of the 𝑁 = 2(𝑛+1) ´ 1 sites receives a unit with probability 1 . 𝑁 As a result, the site moves towards the threshold for relaxation, which is represented by an increase in the overall relaxation probability 𝑝(𝑡) = 𝑝2, namely 𝑝(𝑡) → — 𝑝(𝑡) ` 1 . Similarly, relaxation at 𝑁 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:

𝜎𝑛(𝑡)

𝑝(𝑡) → — 𝑝(𝑡) ´

𝑁

This is summarised by the mean-field eq.

𝑝(𝑡 ` 1) = 𝑝(𝑡

)`

So, in continuous time,

𝑑𝑃 𝑑𝑡

=

1

𝑁

´

1 ´ 𝜎𝑛(𝑡)

𝑁

If two units are lost, each boundary relaxation we have:

𝑑𝑃 𝑑𝑡

=

1 ´ 2𝜎𝑛(𝑡)

𝑁

𝜎𝑛(𝑡) 𝑁

62

Complexity Science On average (as claimed in the the paper), ă 𝜎𝑛 ą= ( 2𝑝 )𝑛, so the fixed point value of 𝑝 will be given by 1 1 1 ´ 2(2𝑝)𝑛 = 0 ùñ 𝑝 = exp as 𝑛 → — 8 ln 2 (´ )→ — 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 𝑝(𝑡) is then

𝑝(𝑡) =

1

𝑁

𝑡 ` 𝑝0

𝑝(𝑡x) = ùñ

1 2

1 x

1

𝑁

2

𝑡 ` 𝑝0 = 1

𝑡x = 𝑁 ( ´ 𝑝0) 2

1 𝑛+1 ´ 1)92 𝑛+1 for large n = ( ´ 𝑝0)(2 2 h) No sample solution needed i) No sample solution needed. j) We know that

𝑑

ă 𝜎𝑛 ą=

𝑑𝑆

𝑔𝜎𝑛 (𝑠)|𝑠=1 =

𝑑 𝑔(𝑔𝜎𝑛—1 (𝑠))|𝑠=1 𝑑𝑠

= r𝑔2(1)s𝑛 = 𝜇𝑛 and from (b) we have 𝜇 = 2𝑝 → — ă 𝜎𝑛 ą= 𝜇𝑛.

Total progeny, ă 𝑌∞ ą=ă

∞

ÿ

𝜎𝑛 ą=

𝑛=0

∞

ÿ

𝑛=0

ă 𝜎𝑛 ą=

∞

ÿ

𝑛=0

𝑛

1

𝜇 = 1´ 𝜇

From the lecture notes, we have

𝑔𝑌œ (𝑠) = 𝑠𝑔(𝑔𝑌œ (𝑠)) and ă 𝑌∞ ą= 𝑔𝑌2 œ (1) By implicit differentiation, we have,

𝑔𝑌2 œ (𝑠) = 𝑔(𝑔𝑌œ (𝑠)) ` 𝑠𝑔2(𝑔𝑌œ (𝑠))𝑔𝑌2 œ (𝑠) 𝑔𝑌2 œ (𝑠) =

𝑔𝑌œ (𝑠) 1 ´ 𝑠𝑔2(𝑔𝑌œ (𝑠)

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

Complexity Science For 𝜇 ă 1, we have 𝑔2(𝑔𝑌œ (1)) = 𝑔2(1) = 𝜇 ùñ ă 𝑌∞ ą=

1 1´ 𝜇

k) Let 𝑔𝑋(𝑠) be the generator function for the stochastic variable 𝑋 we then have ă 𝑋 ą= 𝑔2𝑋 (1).

𝑔”𝑋(1) =ă 𝑋2 ą ´ ă 𝑋 ą and therefore, Var𝑋 =ă 𝑋2 ą ´ ă 𝑋 ą2= 𝑔”𝑋(1) ` 𝑔2𝑋 (1) ´(𝑔2𝑋 (1))2 From the lecture notes, we have,

𝑔𝜎𝑛 (𝑠) = 𝑔(𝑔𝜎𝑛—1 (𝑠)) From (b), we know,

𝑔(𝑠) = 1 ´ 𝑝 ` 𝑝𝑠2 𝑔2(𝑠) = 2𝑝𝑠 = 𝜇𝑠 𝑔„(𝑠) = 2𝑝 = 𝜇 Notation:

𝑔𝑛(𝑠) Ξ 𝑔𝜎𝑛 (𝑠) 𝑑𝑔𝑛

𝑑 𝑑𝑠

=

𝑑𝑠

𝑔(𝑔𝑛−1(𝑠)) =

𝑑 𝑑𝑠

(1 ´ 𝑝 ` 𝑝(𝑔𝑛−1(𝑠)) )

= 2𝑝𝑔𝑛−1(𝑠)𝑔𝑛2 −1(𝑠) = 𝜇𝑔𝑛−1(1)𝑔𝑛2 −1(𝑠)

𝑑2𝑔𝑛

2

„

2

= 𝜇((𝑔𝑛−1(𝑠)) ` 𝑔𝑛−1(𝑠)𝑔𝑛−1(𝑠)𝑔𝑛−1(𝑠))

𝑑𝑠2

𝑛−1 Recall that 𝑔𝑛−1(1) = 1 and 𝑔𝑛2 − 1(1) = 𝜇

𝑑2𝑔𝑛 𝑑𝑠2

|𝑠 = 1 = 𝜇r𝜇

2(𝑛−1)

` 𝑔𝑛„−1(𝑠)s

2(𝑛−1) Notation: 𝑥𝑛 Ξ 𝑔𝑛„( 1) and 𝑥𝑛 = 𝜇 (𝑥𝑛 − 1 ` 𝜇 ) „ Recall 𝑋1 = 𝑔1 (1 ) = 𝜇 Hence,

𝑥𝑛 = 𝜇(𝑥𝑛−1 ` 𝜇2(𝑛−1)) = 𝜇r𝜇(𝑥𝑛−2 ` 𝜇2(𝑛−2)) ` 𝜇2(𝑛−1)s

2

63

64

Complexity Science = 𝜇2𝑥𝑛−2 ` 𝜇2𝑛−2 ` 𝜇2𝑛−1

.

= 𝜇𝑛−1𝑥1 ` 𝜇2𝑛−(𝑛−1) ` 𝜇2𝑛−(𝑛−2) ` ... ` 𝜇2𝑛−1

= 𝜇 𝑛 r1 ` 𝜇 ` 𝜇2 ` ... ` 𝜇 𝑛−1 s = 𝜇 „

𝑛

𝑛

1−𝜇

So, 𝑔 (1) = 𝜇𝑛 1−𝜇

𝑛1´ 𝜇

2

1´𝜇

and, Var𝜎𝑛 = 𝑔𝑛„(1) ` 𝑔𝑛2 (1) ´ r𝑔𝑛2 (1)s2

1 𝜇𝑛 𝑛 = 𝜇𝑛 ( ´ ) ` 𝜇 ´ 𝜇 2𝑛 1´ 𝜇 𝑛 When 𝑝 → — 𝑝 ,𝜇→ — 1, the limit 𝜇 → — 1−. L’Hospital rule 1−𝜇 → — 𝑐

𝑛 as 𝜇 → — 1.

1−𝜇

So,

Var𝜎𝑛 → — 𝑛 as 𝜇 → — 1− Project 2 a) Infection occurs when 𝜂 is drawn as a record. We are asked to complete the probability 𝑃 𝑡1, 𝑡(2 that)a record occurs at 𝑡1 and 𝑡2.

𝐹( 𝜂 ) =

𝑑𝜂0 𝑓 (𝜂0)𝐹 1

= 0

ż1

𝑡 −1

0

ż1

𝑓 (𝜂)𝑑𝜂

0

ż1

𝑃(𝑡1, 𝑡2) =

ż 𝜂

𝑡1 −1

𝑑𝑍0𝑍0

(𝜂0)

𝑑𝜂1 𝑓 (𝜂1)𝐹 2 1 𝜂0

ż1

ż1

𝑡 −𝑡 −2

(𝜂1)

𝑑𝜂2𝑑(𝜂2) 𝜂1

𝑡2 −𝑡1 −2

𝑍0

(1 ´ 𝑍1) with 𝑍𝑖 = 𝐹(𝜂𝑖)

𝑑𝑍1𝑍1

=

1

𝑡1(𝑡2 ´ 1)𝑡2

b) We note that 𝑄 is equal to the number of records during the years 𝑡1 to 𝑡2. We know that a record process starting at time 0 running to time 𝑡 to a good approximation is a Poisson process

𝑃( 𝑋 = 𝑛 ) =

𝜆𝑛−1 𝑒−𝜆 (𝑛 ´ 1)!

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,

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

Complexity Science

𝑃𝑟(𝑄 = 𝑞) =

(ln 𝑡)𝑞−1 𝑒− ln 𝑡

(ln 𝑡)𝑞−1

= (𝑞 ´ 1)! 𝑡(𝑞 ´ 1)!

Since, we now look at number of events during the years 𝑡1 to 𝑡2 we replace ln 𝑡 by ln 𝑡2 ´ln 𝑡1 = ln 𝑡2 and𝑡1arrive at rln 𝑡𝑡21 s𝑞−1

𝑃𝑟 (𝑄 = 𝑞 ) =

𝑡2 𝑡1 (𝑞 ´ 1)!

c) The generator function for 𝑃𝑘 is

𝑔(𝑠) =

∞ ÿ

𝑃𝑘 𝑘𝑠

= (1 ´ 𝑒 − 𝛼 )

∞ 𝑒𝛼 ´ 1 𝑠 𝑘 ÿ ( 𝛼) = 𝑒𝛼 ´ 𝑠 𝑘=0 𝑒

2

𝑒𝛼 ´ 1

1

𝑘=0

The average

𝜇 = 𝑔 (1) =

(𝑒𝛼 ´ 𝑠)

d)

| = 2 𝑠=1

ă 𝑍𝑛 ą= 𝑔2𝑍𝑛 ( 1) =

𝑑 𝑔(𝑔 𝑑𝑠

= r𝑔2(𝑔 𝑍𝑛—1 (𝑠))

𝑑 𝑔

𝑑𝑠

𝑒𝛼 ´ 1

𝑍𝑛—1 (𝑠))| 𝑠=1

𝑍𝑛—1 (𝑠)s 𝑠=1

. = r 𝑔2(𝑔𝑍𝑛—1 (𝑠))𝑔2(𝑔𝑍𝑛—2 (𝑠))...𝑔2(𝑔𝑍0 (𝑠))𝑔2(𝑠)s𝑠=1 Use 𝑔𝑍𝑛 1( ) = 𝑔 (𝑔𝑍 𝑛—1 (1)) = ... ´ 𝑔(𝑔...𝑔(𝑠))|𝑠=1 and 𝑔(1) = 1 to obtain ă 𝑍𝑛 ą= r𝑔2(1)s𝑛 = 𝜇𝑛 and we have

𝑍𝑛 = 𝜇𝑛−𝑚 𝑍𝑚 e) We know that for a branching process the time to extinction 𝜏 satisfies

𝜇 ă 1 : Prob(𝜏 ą 𝑛)9𝜇𝑛for 𝑛 ąą 1 1

𝜇 = 1 : Prob(𝜏 ą 𝑛)9 for 𝑛 ąą 1 𝑛 (assuming that the variance of the branching probabilities is finite) We therefore have:

𝜇ă 1→ —

𝑃𝑟(𝑇1) 𝑇1 −𝑇2 =𝜇 𝑃𝑟(𝑇2)

𝜇=1→ —

𝑃𝑟(𝑇1)

=

𝑇2

𝑃𝑟(𝑇2) 𝑇1 So to have the ratio to only depend on 𝑇1 we need 𝜇 = 1 or 1 𝑇2

𝑒𝛼 −1

= 1 ùñ 𝛼 = ln 2

65

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Complexity Science

f) Consider 𝑥𝑛+1 = 𝑓 (𝑥𝑛) Assume that 𝑥𝑐 exist such that 𝑓 2(𝑥𝑐) = 1 and let 𝜖 = 𝑓 (𝑥𝑐) ´ 𝑥𝑐 and Taylor expand

1 2

𝑓 (𝑥) = 𝑓 (𝑥𝑐) ` 𝑓 2(𝑥𝑐)(𝑥 ´ 𝑥𝑐) ` 𝑓 „(𝑥𝑐)(𝑥 ´ 𝑥𝑐)2

= 𝑥𝑐 ` 𝜖 `( 𝑥 ´ 𝑥𝑐) `

1 2

𝑓 ”(𝑥𝑐)(𝑥 ´ 𝑥𝑐)

2

Define Δ𝑛 = 𝑥𝑛 ´ 𝑥𝑐 and 𝛽 = 1 𝑓 2”(𝑥𝑐). In this approximation we have from 𝑥𝑛+1 = 𝑓 (𝑥𝑛).

Δ𝑛+1 = 𝜖 `Δ 𝑛 ` 𝛽Δ2

𝑛

ùñ Δ𝑛+1 ´Δ 𝑛 = 𝜖 ` 𝛽Δ𝑛 2

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

𝑑Δ

Δ𝑛+1 ´Δ 𝑛

𝑑𝑡

— →

𝑑Δ 𝑑𝑡

2

= 𝜖 ` 𝛽Δ → —

c ùñ Δ =

, Δ𝑛 → — Δ(𝑡)

𝑑Δ 𝜖 ` 𝛽Δ2

= 𝑑𝑡

𝜖 a 𝑡𝑔r 𝜖 𝛽(𝑡 ` 𝑡0)s 𝛽

Since, 𝑡𝑔 goes from ´8 to 8 when the argument varies from −𝜋 to 𝜋 , we estimate the number of iterations needed to pass through the bottleneck at 𝑥𝑐 as 𝑇

2

= ,𝜋 .

2

𝜖𝛽

g) See figure below

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen

Complexity Science

h) In (4), we derived, 𝑇 = ,𝜋 . We have 𝛽 = 1 2𝑓 ”(𝑥𝑐). In the specific case 𝑓 ”(𝑠) = 1 for all 𝑥, so 𝜖𝛽

𝛽 = 1. And we have, 𝜖

𝑇 = ?𝜋 Ξ 𝑇 (𝜖 )

10−2 10−3 10−4

31.4 99.3 314.2 height

𝜖

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 → — 0 the number of iterations needed to get through the bottleneck increases and the behaviour is close to the continuous description.

67

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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

𝑄(𝑡) is more robust i.e. fewer false positives.

Exercise 3 Identifying the logarithmic time dependence can be difficult.

Project 1 No sample solution needed.

Solutions Manual for Complexity Science The Study of Emergence by Henrik Jeldtoft Jensen