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To: Jan Robert Wolansky who wrote (5201)	7/5/1998 11:30:00 PM
From: TechTrader42	Read Replies (1) \| Respond to of 11149

It isn't because the formula is already using a for loop. Lots of scans have nested loops. Someone will probably come up with the correct coding soon. I probably made errors in the way I nested the loops, among other things. The darn scan drove me loopy. I kept getting the wrong values for slope when I looked back 10 days with a for loop. Brooke

To: Jan Robert Wolansky who wrote (5201)	7/6/1998 2:33:00 PM
From: Roy Yorgensen	Read Replies (3) \| Respond to of 11149

Jan & Brook, I'm not certain, but I think this will do what you are looking for. I don't know what the results mean, but the logic seems correct. //Slope, Linear Regression Formula, by Bob Jagow output = "linregslope.lst"; issuetype common; Daystoload = 50; //needed for R2 bug!!! integer i,j, S, Sx, Sxx; float b,b1, Sxy, Sy; b1:=0; // only initialized first time for j = 0 to -10 step -1 DO // reset all variables at stat of next loop S := 14; Sx := 0; Sxx := 0; Sxy := 0; Sy := 0; for i = j+(1 - S) to j do Sx := Sx + i; Sy := Sy + close(i); //kills println wo d2ld Sxx := Sxx + ii; Sxy := Sxy + iclose(i); //kills println wo d2ld //Sxy := Sxy + iclose(0); next i; b := (SSxy - SxSy)/(SSxx - Sx*Sx); if j<0 and // eliminate first pass thru loop b< b1 then println symbol,",", date(i),",",close(0):8:2,",",b:8:4; b1:=b; // save current for next test else b1:=b; // must alway set ot last day's val endif; next j;