How clear is this post?

## How Behavior Spreads: The Science of Complex Contagions, by Damon Centola, a review

NB. This book was sent to me as a review copy.

The idea of this book is relatively simple, but the consequences are huge, and in fact some of the ideas are far more subtle and complex than they may first appear.

Essentially this book is based on a series of experiments which Damon Centola has run, which are all related to changes in behaviour which can be tracked, and made to occur, through a social network (in the broadest sense of the word). This is the study diffusion in a network.

The fundamentals of the research lie on two distinctions: One in the complexity of a contagion/behaviour, meaning how many connections with others who have the contagion/behaviour do you need until you adopt it, and the other in the topology of the social network, meaning loosely, how much like a street where each person only talks to their neighbours, versus a small world-network where there are a lot of disparate connections does the network look like.…

## Why did we choose that range for theta when doing trig substitutions?

Remember when we are doing a trig substitution, for instance for an integral with:

$\sqrt{a^2-x^2}$

We said that we should choose $x=a\sin\theta$, which seemed reasonable, but we also said that $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$. Where did this last bit come from?

Well, we want a couple of things to hold true. The first is that any substitution that we make, we have to be able to undo. That is, we will substitute $x$ for a function of $\theta$ but in the end we need to convert back to $x$ and so to do that we have to be able to write the inverse function of, in this case $x=a\sin\theta$. The $\sin$ function is itself not invertible because it’s not one to one, so we have to choose a range over which it is one to one. We could choose $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$ or we could choose $\frac{\pi}{2}\le\theta\le\frac{3\pi}{2}$ (amongst an infinite set of possibilities). That would also be invertible. However, remember that we are going to end up with a term of the form:

$\sqrt{1-\sin^2\theta}=\sqrt{\cos^2\theta}$

So if we want this to simplify, we had better choose our range of $\theta$ such that $\cos\theta$ is positive, so that we can write $\sqrt{\cos^2\theta}=\cos\theta$.…