# Confused on Linear Least Square Fits very basic.

1. Mar 29, 2010

### bobbo7410

Quite confused. I've read the book/online definitions yet I suppose I may need a simpler explanation.

Lets say I have a table of x y values.

x 1 2 3 4 5
y 6 7 8 9 10

how would I carry out the linear least squares fit of the data to determine the slope and y-intercept?

2. Mar 29, 2010

### Staff: Mentor

How about you tell us a definition that you've seen (or give us a link to it), and tell us where you get stuck in trying to apply it?

3. Mar 30, 2010

### HallsofIvy

Staff Emeritus
As jtbell says, there are a number of equivalent ways to do that. Not necessarily the simplest, but the most direct is this: Let the line be y= mx+ b where "m" is the slope and "b" is the y-intercept. For x= 1, the "calculated value" would be y= m+ h while the true value is 6. The "square error" is $(m+ b- 6)^2$. Similarly, for x= 2 the "calculated value" would be y= 2m+ b while the true value is 7. The "square error" is $(2m+b- 7)^2$. Repeating that for all given values, the "total square error" would be [itex](m+ b- 6)^2+ (2m+ b- 7)^2+ (3m+ b- 8)^2+ (4m+ b- 9)^2+ (5m+ b- 10)^2[/math]. To minimize that, take the partial derivatives with respect to m and b and set them equal to 0.

Of course, for this particular example, it is obvious that the line y= x+ 5 goes exactly through every point so that is what you would get.