Chapter 4.5, Problem 1E

The Gauss-Newton Method can be applied to find the point $\bar{x},\bar{y}$ for which the sum of the squared distances to the three circles is minimized. Using initial vector $\left(x_0,y_0\right)=(0,0)$, carry out the first step to find $\left(x_1,y_1\right)$

(a) centers (0, 1), (1, 1), (0, 1) and all radii 1 (b) centers

(-1, 0), (1, 1), (1, -1) and all radii 1. (Computer Problem 1 asks for $\left(\bar{x},\bar{y}\right)$

).)

To determine

To find out the point $\stackrel{-}{x},\stackrel{-}{y}$ by gauss-newton method.

Answer to Problem 1E

  $v=(2-\surd 2,0),and x^{{}_1}=x^{{}_0}+v=(2-\surd 2,0)$

Explanation of Solution

Given information:

The given initial vector $\left(x_0,y_0\right)=\left(0,0\right)\text{ and center }(0,1),\left(1,1\right),(0,-1),\text{all radii 1}$ .

Calculation:

 Here, we find the point  $\stackrel{-}{x},\stackrel{-}{y}$ that minimizes the sum of the squared distances to the three circles. The distance to the i^th circle is  $r_i=\sqrt{{(x-x_i)}^2+{(y-y_i)}^2}-R_i$

where  $\left(x_i,y_i\right)$ is the center and R_i the radius.

Therefore

  $Dr=\left[\begin{array}{l}\frac{{\text{x-x}}_{\text{1}}}{{\text{S}}_{\text{1}}}\frac{{\text{y-y}}_{\text{1}}}{{\text{S}}_1}\\ \frac{{\text{x-x}}_2}{{\text{S}}_2}\frac{{\text{y-y}}_2}{{\text{S}}_2}\\ \frac{{\text{x-x}}_3}{{\text{S}}_3}\frac{{\text{y-y}}_3}{{\text{S}}_3}\end{array}\right]=\left[\begin{array}{l}\text{0 -1}\\ \frac{\text{-1}}{\sqrt{\text{2}}}\frac{\text{-1}}{\sqrt{\text{2}}}\\ \text{0 1}\end{array}\right]$

where  $S_i=\sqrt{{(x-x_i)}^2+{(y-y_i)}^2}$ . To take a Gauss-Newton step, start with  $x^0=\left[0,0\right]$ and compute

  $D{r}^{T}Dr=\left[\begin{array}{l}\frac{1}{2}\frac{1}{2}\\ \frac{1}{2}\frac{5}{2}\end{array}\right]$ and  $D{r}^{T}r=\left[\begin{array}{l}\frac{1}{\sqrt{2}}\text{-1}\\ \frac{1}{\sqrt{2}}\text{-1}\end{array}\right]$ Solving  $D{r}^{T}Drv=-D{r}^{T}r$ yields  $v=(2-\surd 2,0),and x^{{}_1}=x^{{}_0}+v=(2-\surd 2,0)$

To determine

To find out the point $\stackrel{-}{x},\stackrel{-}{y}$ by gauss-newton method.

Answer to Problem 1E

  $v=\left(1-\frac{\surd 2}{2},0\right)and x^{{}_1}=x^{{}_0}+v=\left(1-\frac{\surd 2}{2},0\right)$ .

Explanation of Solution

Given information:

The given initial vector $\left(x_0,y_0\right)=\left(0,0\right)\text{ and center }(-1,0),\left(1,1\right),(1,-1),\text{all radii 1}$ .

Calculation:

 Here, we find the point  $\stackrel{-}{x},\stackrel{-}{y}$ that minimizes the sum of the squared distances to the three circles. The distance to the i^th circle is  $r_i=\sqrt{{(x-x_i)}^2+{(y-y_i)}^2}-R_i$

where  $\left(x_i,y_i\right)$ is the center and R_i the radius.

Therefore

  $Dr=\left[\begin{array}{l}\frac{{\text{x-x}}_{\text{1}}}{{\text{S}}_{\text{1}}}\frac{{\text{y-y}}_{\text{1}}}{{\text{S}}_1}\\ \frac{{\text{x-x}}_2}{{\text{S}}_2}\frac{{\text{y-y}}_2}{{\text{S}}_2}\\ \frac{{\text{x-x}}_3}{{\text{S}}_3}\frac{{\text{y-y}}_3}{{\text{S}}_3}\end{array}\right]=\left[\begin{array}{l}\text{0 -1}\\ \frac{\text{-1}}{\sqrt{\text{2}}}\frac{\text{-1}}{\sqrt{\text{2}}}\\ \frac{\text{-1}}{\sqrt{\text{2}}}\frac{\text{-1}}{\sqrt{\text{2}}}\end{array}\right]$

where  $S_i=\sqrt{{(x-x_i)}^2+{(y-y_i)}^2}$ . To take a Gauss-Newton step, start with  $x^0=\left[0,0\right]$ and compute

  $D{r}^{T}Dr=\left[\begin{array}{l}20\\ 01\end{array}\right]$ and  $D{r}^{T}r=\left[\begin{array}{l}\sqrt{2}\text{-1}\\ \text{ 0}\end{array}\right]$ Solving  $D{r}^{T}Drv=-D{r}^{T}r$ yields  $v=(1-\frac{\surd 2}{2},0),and x^{{}_1}=x^{{}_0}+v=(1-\frac{\surd 2}{2},0)$