Method of Least Square
Consider the points (x1, y1), (x2, y2), (x3, y3), ………., (xn, yn) shown on the
graph.
x1
x2
x3
The blue line has the equation
y=mx+b
We will find m and b so that the error is least.
E = ( m * x1 + b - y1 ) 2 + ( m * x2 + b - y2 ) 2 + ( m * x3 + b - y3 ) 2 + …….
We must have:
dE
 0
dm at cons tan t b
dE
 0
db at cons tan t m
2(m*x1 + b -y1)*x1 + 2(m*x2 + b -y2)*x2 + 2(m*x3 + b -y3)*x3 + ….. = 0
and
2(m*x1 + b -y1) + 2(m*x2 + b -y2) + 2(m*x3 + b -y3) + ….. = 0
where we used the chain rule. The equations can be written as:
m ( x1*x1 + x2*x2 + x3*x3 + ……) + b ( x1 + x2 + x3 + ….) =
x1*y1 + x2*y2 + x3*y3 + …..
m ( x1 + x2 + x3 + …)
+ b ( 1 + 1 + 1 + …) = y1 + y2 + y3 +…..
Using the summation notation we can write:
m
  nk 1
m
x
xk2
 b
 b 
(n
x )    nk 1 xk yk
y
Solving for m we get:
1 n
 k 1 xk yk )  x y
m  n
1
(  nk 1 xk 2 )  x 2
n
(

If you can show that this is the same equation as the one in the
book you will get 10 points added to your score of 1000.