| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
Here's one for you lunchtime mathematicians. And a solution is
urgently needed. I am without any reference books and working on
customer site.
I need a subroutine that will give me a best-fit linear equation
through 3 calibration points.
by best-fit, i mean that I desire the linear equation that minimizes
the errors^2 for each of the three points supplied.
C
C
C
CALL BESTFIT( X1,Y1 , X2,Y2 , X3,Y3 , M_COEF, B_COEF)
C
C
C
How does the math work??
dj brown
(I can guarantee that no two of the calib pts have the same X value)
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1552.1 | CLT::KOBAL::GILBERT | Ownership Obligates | Thu Jan 30 1992 12:50 | 27 | |
Problem: Given x[i] and y[i] for i=1..n, find m and b that minimize: n f = sum (m�x[i]+b - y[i])� i=1 Solution: Using partial derivatives we get these two equations: m�Sx + b�n - Sy = 0 m�Sxx + b�Sx - Sxy = 0 where: n n Sx = sum x[i], Sy = sum y[i], i=1 i=1 n n Sxx = sum x[i]�, Sxy = sum x[i]�y[i] i=1 i=1 So finally, m = (Sxy�n - Sx�Sy)/(Sxx�n - (Sx)�) b = (Sxx�Sy - Sx�Sxy)/(Sxx�n - (Sx)�) | |||||
| 1552.2 | solution found ! | PTOVAX::DJBROWN | To feel we are able, is to be so | Thu Jan 30 1992 14:24 | 9 |
This is wonderful ! Thanks a million.
You know, its resources like this (having a nation-wide NOTE-ing
network) and the fact that we have a lot of top-notch people working
for this company that *REALLY* impress our customers !
I have my solution tested and implimented, and works great.
Again, Thanks ::GILBERT !
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