Table adjusting ?

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Table adjusting ?

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Answer 1 · 2016-08-31T17:10:27

$y = - 0.489421 x + 25.3113$ $y = -0.489421 x + 25.3113$

Explanation:

Given a list of values ${x_{k}, y_{k}}, k = 1, 2, \dots, n$ ${x_k,y_k}, k=1,2,cdots,n$

a line can be adjusted such that the accumulated deviation error is minimum.

Let the line be given by

$y = a x + b$ $y = a x + b$

then the error at point $x_{k}$ $x_k$ is $e_{k} = y_{k} - (a x_{k} + b)$ $e_k = y_k - (a x_k + b)$

The accumulated quadratic error will be given by

$E (a, b) = n \sum k = 1 e_{k}^{2} = n \sum k = 1 {(y_{k} - (a x_{k} + b))}_{k}^{2}$ $E(a,b) = sum_(k=1)^n e_k^2 =sum_(k=1)^n( y_k - (a x_k + b))^2$

$E (a, b)$ $E(a,b)$ have a minimum for $a_{0}, b_{0}$ $a_0,b_0$ such that

$\frac{\partial E}{\partial a} (a_{0}, b_{0}) = 0 \to b_{0} \sum x_{k} + a_{0} \sum x_{k}^{2} = \sum x_{k} y_{k}$ $(partial E)/(partial a)(a_0,b_0) = 0-> b_0 sum x_k+a_0 sumx_k^2=sum x_ky_k$
$\frac{\partial E}{\partial b} (a_{0}, b_{0}) = 0 \to n b_{0} + a_{0} \sum x_{k} = \sum y_{k}$ $(partial E)/(partial b)(a_0,b_0) = 0 -> n b_0 + a_0 sum x_k =sum y_k$

solving for $a_{0}, b_{0}$ $a_0,b_0$

$a_{0} = \frac{(\sum x_{k}) (\sum y_{k}) - n \sum x_{k} y_{k}}{{(\sum x_{k})}_{k}^{2} - n \sum x_{k}^{2}}$ $a_0 = ( (sum x_k)( sum y_k)-n sum x_ky_k)/((sum x_k)^2-nsum x_k^2)$
$b_{0} = \frac{(\sum x_{k}) (\sum x_{k} y_{k}) - (\sum x_{k}^{2}) (\sum y_{k})}{{(\sum x_{k})}_{k}^{2} - n \sum x_{k}^{2}}$ $b_0 = ((sum x_k)(sum x_ky_k)-(sum x_k^2)(sum y_k))/((sum x_k)^2-nsum x_k^2$

appliying it to the table

${{10, 15}, {10, 25}, {15, 17}, {17, 20}, {25, 11.5}, {30, 11}, {35, 5}, {35, 10}}$ ${{10,15}, {10,25}, {15,17}, {17,20}, {25,11.5}, {30,11},{35,5},{35,10}}$

we obtain:

$a_{0} = - 0.489421, b_{0} = 25.3113$ $a_0 = -0.489421, b_0 = 25.3113$

so the adjusting line is

$y = - 0.489421 x + 25.3113$ $y = -0.489421 x + 25.3113$

Attached a plot

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Table adjusting ?

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