## 3 Analysis
### Convergence Analysis
#### 3.1.1 Convergence in One-Dimensional Setup  
We first show the algorithm convergence property in the simple one-dimensional (1-D) setup. We shall use the concept of spring-relaxation for the discussion, which is equivalent to our algorithm. We assume that there is a single location, which we called a _stopping point_, where all forces from beacons cancel out. This assumption is often valid for common configurations. Consider \(n\) beacons where \(n\geq 2\). Each beacon is located at \(x_{i},i=1,2,...,n\). All distance variables are normalized to the length of the spring, _i.e._, the length of the spring is of one unit long. The net force applied on a node located at \(x\) is  
\[F(x)=\sum_{i=1}^{n}F_{x_{i}}(x)\]  
where  
\[F_{x_{i}}(x)=\left\{ \begin{array}{ll}-x+x_{i}-1,&x<x_{i}\\ 0,&x=x_{i}\\ -x+x_{i}+1,&x>x_{i}\end{array} \right. \tag{4}\]  
In the following, we shall prove that in 1-D setup, if there exist a unique stopping point where the net force is zero, at any other location, the net force is nonzero, and the force points towards the stopping point. In (4), we define that the leftward (resp. rightward) force carries a negative (resp. positive) sign. **Proposition 1**: _Consider \(n\) beacons where \(n\geq 2\). Each beacon is located at \(x_{i},i=1,2,...,n\). The net force applied on a node located at \(x\) is \(F(x)=\sum_{i=1}^{n}F_{x_{i}}(x)\). If \(F(x)=0\) produces a single unique solution, and let the solution be \(\hat{x}\), then \(\forall x<\hat{x},F(x)>0\) and \(\forall x>\hat{x},F(x)<0\)._  
**Proof** We first show that \(\forall x<\hat{x},F(x)>0\). By (4), \(F_{x_{i}}(x)\) is a monotone decreasing function except at \(x=x_{i}\) where the curve steps up by two units. Thus, \(F(x)\) is also a piecewise monotone decreasing function with \(n\) upward steps occurring at \(x_{i},i=1,2,...,n\). In other words, at the discontinuity point \(x_{i}\), if \(F(x_{i}^{-})>0\), \(F(x_{i})>0\) and \(F(x_{i}^{+})>0\). Given that as \(x\rightarrow-\infty\), \(F_{x_{i}}(x)\rightarrow\infty\) which implies \(\lim_{x\rightarrow-\infty}F(x)>0\). Since \(F(x)\) is a piecewise monotone decreasing function where at any of its discontinuity points \(x_{i}\), if \(F(x_{i}^{-})>0\), then \(F(x_{i})>0\) and \(F(x_{i}^{+})>0\), and \(\hat{x}\) is the only point that the \(F(x)\) crosses the x-axis, with condition \(\lim_{x\rightarrow-\infty}F(x)>0\), in the range where \(x\in(-\infty,\hat{x})\), we must have \(F(x)>0\). We now show that \(\forall x>\hat{x},F(x)<0\), and we shall prove this by using contradiction. Given \(F(\hat{x})=0\) as the unique solution, there exists a point \(x>\hat{x}\) where \(F(x)>0\). Recall that \(F(x)\) is also a piecewise monotone decreasing function with \(n\) upward steps occurring at \(x_{i},i=1,2,...,n\) where at any of its discontinuity points \(x_{i}\), if \(F(x_{i}^{-})>0\), then \(F(x_{i})>0\) and \(F(x_{i}^{+})>0\). As \(x\rightarrow\infty\), by (4) \(F_{x_{i}}(x)\rightarrow\infty\) which implies \(\lim_{x\rightarrow\infty}F(x)<0\). Given that there is a point \(x>\hat{x}\) where \(F(x)>0\), since \(F(x)\) is a piecewise monotone decreasing function, and at its discontinuity point \(x_{i}\), if \(F(x_{i}^{-})>0\), then \(F(x_{i})>0\) and \(F(x_{i}^{+})>0\), with \(F(x)>0\) and \(\lim_{x\rightarrow\infty}F(x)<0\), there must exist a point \(u\in(x,\infty)\) such that \(F(u)=0\) which contradict with our condition that \(F(\hat{x})=0\) is the only solution. Thus, if \(x>\hat{x}\) we must have \(F(x)<0\). In the above discussion, we make no description on the step size of each movement of the estimated location. The analysis is valid for infinitesimal step size, or equivalently \(\delta_{1}\) is very small. In fact, \(\delta_{1}<1\) represents the upper bound condition for the setting of \(\delta_{1}\) as \(\delta_{1}=1\) corresponds to applying \(100\%\) of the net force to the movement of the estimated location. Moreover, it is possible that a system setup gives more than one stopping points. In practice, such a setup should be avoided as it creates ambiguity in the localization.