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Trendline factor extraction for 6th order
I am attempting to extract polonomial factors for a data set. At 6th order,
the process falls apart. My data is X=0,100,200,300,400,500,600,700,800,900,100 Y=0,0.1,-3.1,-12,-29.5,59.2,-106.4,177.6,280.2,421.3,606.2 At 6th order the C6,b and R-aquared values are crazy. My formula a C6=INDEX(LINEST(y,x^{1,2,3,4,5,6}),1) b=INDEX(LINEST(y,x^{1,2,3,4,5,6}),1,7) R-squared=INDEX(LINEST(y,x^{1,2,3,4,5,6},TRUE,TRUE), 3) Any suggestions? In addition to this, I would like to determine if using these functions, can the trendline factor calculation be forced thru the origin 0,0? |
Trendline factor extraction for 6th order
On Oct 9, 4:40 pm, Kmartin wrote:
I am attempting to extract polonomial factors for a data set. At 6th order, the process falls apart. My data is X=0,100,200,300,400,500,600,700,800,900,100 Y=0,0.1,-3.1,-12,-29.5,59.2,-106.4,177.6,280.2,421.3,606.2 At 6th order the C6,b and R-aquared values are crazy. The worksheet functions LINEST etc are only marginally stable for difficult cubic polynomials. They are dead in the water at 6th order. The polynomial fit in the chart of XL2002 and earlier is much better numerically and can sometimes solve much harder problems accurately. It has been discussed here previously. You need to display all siginifcant digits in the equation to get sensible results recreating the fitted line in a spreadsheet. A quick eyeball of your data suggests that not even a 6th order polynomial fit will make sense of that set of points. You really need to fit a physical model to your data rather than the highest order polynomial available. My formula a C6=INDEX(LINEST(y,x^{1,2,3,4,5,6}),1) b=INDEX(LINEST(y,x^{1,2,3,4,5,6}),1,7) R-squared=INDEX(LINEST(y,x^{1,2,3,4,5,6},TRUE,TRUE), 3) Any suggestions? In addition to this, I would like to determine if using these functions, can the trendline factor calculation be forced thru the origin 0,0? The chart algorithm can. Although it will make the fit worse. Normalising the x values so that they are in the range [-1,1] will improve the condition number of the problem and at least give LINEST a sporting chance of solving it numerically. That is define a new column X' = (X - 500)/500 and use that as the new dependent variable. Regards, Martin Brown |
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