Question

Repeat problem 16 for each of the following set of data points.

(a)$\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{}\mathbf{-}\mathbf{}\mathbf{1}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{}\mathbf{-}\mathbf{}\mathbf{8}\mathbf{)}$

(b)$\mathbf{(}\mathbf{}\mathbf{-}\mathbf{}\mathbf{2}\mathbf{,}\mathbf{}\mathbf{-}\mathbf{}\mathbf{6}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{}\mathbf{-}\mathbf{}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{-}\mathbf{}\mathbf{3}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{9}\mathbf{/}\mathbf{2}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{7}\mathbf{)}$

(c)$\mathbf{(}\mathbf{}\mathbf{-}\mathbf{}\mathbf{2}\mathbf{,}\mathbf{4}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{}\mathbf{-}\mathbf{}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{-}\mathbf{}\mathbf{1}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{-}\mathbf{}\mathbf{8}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{}\mathbf{-}\mathbf{}\mathbf{10}\mathbf{)}$

To find the best straight line fit to a set of data points$\left(x_n,y_n\right)$in the "least squares" sense means the following: Assume that the equation of the line is$\mathit{y}\mathbf{}\mathbf{=}\mathbf{}\mathit{m}\mathit{x}\mathbf{}\mathbf{+}\mathbf{}\mathit{b}$and verify that the vertical deviation of the line from the point$\left(x_n,y_n\right)$is${\mathit{y}}_{\mathbf{n}}\mathbf{-}\left(m{x}_n+b\right)$Write$\mathit{S}\mathbf{}\mathbf{=}\mathbf{}$sum of the squares of the deviations, substitute the given values of${\mathit{x}}_{\mathbf{n}}\mathbf{,}{\mathit{y}}_{\mathbf{n}}$to give$\mathit{S}$as a function of$\mathit{m}$and$\mathit{b}$, and then find$\mathit{m}$and$\mathit{b}$to minimize S. Carry through this routine for the set of points:. $\mathbf{(}\mathbf{}\mathbf{-}\mathbf{}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{-}\mathbf{}\mathbf{2}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{3}\mathbf{)}$. Check your results by computer, and also computer plot (on the same axes) the given points and the approximating line.

Short answer

(a) The resultant answer is $y=-\; 4x+\; 5$.

(b) The resultant answer is $y=3.35x+\; 0.5$.

(c) The resultant answer is $y=-\; 3.6x-\; 3$.

Step-by-step solution

Given data

A set of data points.

Concept of Partial differential equation

A partial differential equation consists of two or more independent variables, an unknown function (depending on those variables), and partial derivatives of the unknown function with respect to the independent variables.

Simplify the expression

(a)

Objective is to determine the best straight line fit to a given set of the data points

$(1,0),(2,-1),(3,-8)$

Also,this value is plotted the points and the approximating line

Let the equation be: $y=mx+b$

With the data point $\left(x_n,y_n\right)$to localid="1664349153354" $y=mx+b$will belocalid="1664349156757" $y_n-y$atlocalid="1664349160381" $x_n$

The value oflocalid="1664349164823" $y$some pointlocalid="1664349168224" $x_n$islocalid="1664349171176" $m{x}_n+b$

Using this value and get the vertical deviation as:localid="1664349175060" $y_n-\left(m{x}_n+b\right)$

Where S is the sum of the squares of the deviations, substitute the given values of localid="1664349179780" $\left(x_n,y_n\right)$.

To give as a function of localid="1664349195465" $m$and localid="1664349199335" $b$, and then m and localid="1664349184037" $b$to minimize

And this way for the set of points localid="1664349189625" $(1,0),(2,-1),(3,-8)$.

Simplify the expression

(b)

Objective is to determine the best straight line fit to a given set of the data poin$(-2,-6),(-1,-3),(0,0),(1,9/2),(2,7)$$(-2,-6),(-1,-3),(0,0),(1,9/2),(2,7)$

Also, this value is plotted the points and the approximating line

Let the equation be:localid="1664349780756" $y=mx+b$

With the data pointlocalid="1664349785314" $\left(x_n,y_n\right)$tolocalid="1664349789388" $y=mx+b$will belocalid="1664349793210" $y_n-y$atlocalid="1664349796829" $x_n$

The value oflocalid="1664349800646" $y$some pointlocalid="1664349804142" $x_n$islocalid="1664349807475" $m{x}_n+b$

Using this value and get the vertical deviation as:localid="1664349812280" $y_n-\left(m{x}_n+b\right)$

Where S is the sum of the squares of the deviations, substitute the given values of localid="1664349817181" $\left(x_n,y_n\right)$.

To give as a function of localid="1664349820841" $m$and localid="1664349824180" $b$, and then localid="1664349828022" $m$and localid="1664349831638" $b$to minimize

And this way for the set of points localid="1664349835755" $(-2,-6),(-1,-3),(0,0),(1,9/2),(2,7)$.

Simplify the expression

(c)

Objective is to determine the best straight line fit to a given set of the data points

$(-2,4),(-1,0),(0,-1),(1,-8),(2,-10)$

Also, this value is plotted the points and the approximating line

Let the equation be:$y=mx+b$

With the data point$\left(x_n,y_n\right)$to$y=mx+b$will be$y_n-y$at$x_n$

The value of $y$some point$x_n$is$m{x}_n+b$

Using this value and get the vertical deviation as:$y_n-\left(m{x}_n+b\right)$

Where S is the sum of the squares of the deviations, substitute the given values of $\left(x_n,y_n\right)$.

To give as a function of $m$and $b$, and then $m$and $b$to minimize

And this way for the set of points $(-2,4),(-1,0),(0,-1),(1,-8),(2,-10)$.