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A little twist in the problem of a configuration model for an exponential distribution of degrees. Imagine that the probability pk of a node of degree k in this network is proportional to kak (obviously with the constant a being a number smaller than one, otherwise that expression would diverge for large k). (1) Calculate the normalization of the degrees. In other words obtain an expression for the Pk- (2) Calculate the average degree of the network. (3) Calculate the average number of second neighbors. (4) Find out for which values of a does this neighbor have a giant component. These three expressions might be useful kak k=0 a (1-a)² Σk² ak k=0 = a + a² (1 - a)³ Σk²³ ak= ª +4a² + a² (1-a)4 k=0

A little twist in the problem of a configuration model for an exponential distribution of degrees. Imagine that the probability pk of a node of degree k in this network is proportional to kak (obviously with the constant a being a number smaller than one, otherwise that expression would diverge for large k). (1) Calculate the normalization of the degrees. In other words obtain an expression for the Pk- (2) Calculate the average degree of the network. (3) Calculate the average number of second neighbors. (4) Find out for which values of a does this neighbor have a giant component. These three expressions might be useful kak k=0 a (1-a)² Σk² ak k=0 = a + a² (1 - a)³ Σk²³ ak= ª +4a² + a² (1-a)4 k=0

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter1: Functions
Section1.EA: Extended Application Using Extrapolation To Predict Life Expectancy
Problem 6EA
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A little twist in the problem of a configuration model for an exponential distribution of degrees. Imagine that the probability pk of a node of degree k in this network is proportional to kak (obviously with the constant a being a number smaller than one, otherwise that expression would diverge for large k). (1) Calculate the normalization of the degrees. In other words obtain an expression for the Pk- (2) Calculate the average degree of the network. (3) Calculate the average number of second neighbors. (4) Find out for which values of a does this neighbor have a giant component. These three expressions might be useful kak k=0 a (1-a)² Σk² ak k=0 = a + a² (1 - a)³ Σk²³ ak= ª +4a² + a² (1-a)4 k=0
Transcribed Image Text:A little twist in the problem of a configuration model for an exponential distribution of degrees. Imagine that the probability pk of a node of degree k in this network is proportional to kak (obviously with the constant a being a number smaller than one, otherwise that expression would diverge for large k). (1) Calculate the normalization of the degrees. In other words obtain an expression for the Pk- (2) Calculate the average degree of the network. (3) Calculate the average number of second neighbors. (4) Find out for which values of a does this neighbor have a giant component. These three expressions might be useful kak k=0 a (1-a)² Σk² ak k=0 = a + a² (1 - a)³ Σk²³ ak= ª +4a² + a² (1-a)4 k=0
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