RE: How do I adjust a curve ?

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Jerry:

Thank´s a lot.
I do the change in the format on the displayed equation and you have reason.
The data serie is the followin:

542,195 -7,59
542,404 -7,36
542,65 -7,09
542,827 -6,88
543,034 -6,62
543,212 -6,40
543,388 -6,17
543,602 -5,89
543,819 -5,59
543,997 -5,33
544,212 -5,01
544,39 -4,73
544,568 -4,44
544,747 -4,15
544,926 -3,84
545,136 -3,47
545,315 -3,14
545,532 -2,75
545,748 -2,33
545,965 -1,91
546,146 -1,55
546,364 -1,11
546,585 -0,65
546,8 -0,19
547,056 0,37
547,312 0,93
547,607 1,59
547,973 2,43
548,346 3,28
549,004 4,82
549,485 5,93
549,814 6,69
549,995 7,10
550,247 7,67
550,466 8,15
550,649 8,54
550,831 8,92
551,018 9,29
551,199 9,65
551,383 10,00
551,563 10,30
551,708 10,60
551,889 10,90
552,029 11,20
552,209 11,50
552,354 11,70
552,461 11,90
552,566 12,00
552,709 12,20
552,814 12,40
552,921 12,50
553,063 12,70
553,204 12,80
553,344 13,00
553,484 13,10
553,588 13,20
553,728 13,40
553,866 13,50
553,969 13,50
554,072 13,60
554,177 13,70
554,313 13,80
554,449 13,80
554,586 13,90
554,721 13,90
554,857 14,00
554,958 14,00
555,091 14,00
555,226 14,00
555,36 14,00
555,492 14,00
555,622 14,00
555,755 13,90
555,887 13,90
556,054 13,80
556,187 13,80
556,319 13,70
556,447 13,60
556,61 13,50
556,772 13,40
556,935 13,20
557,064 13,10
557,224 13,00
557,385 12,80
557,546 12,60
557,707 12,50
557,868 12,30
558,059 12,00
558,252 11,80
558,442 11,50
558,631 11,20
558,885 10,90
559,171 10,50
559,461 10,00
559,897 9,32
560,496 8,35
560,964 7,59
561,345 6,98
561,753 6,34
562,167 5,70
562,511 5,18
562,894 4,62
563,236 4,12
563,583 3,63
563,933 3,16
564,185 2,82
564,439 2,49
564,758 2,09
565,074 1,70
565,426 1,29
565,714 0,96
566,032 0,60
566,352 0,25
566,673 -0,09
566,964 -0,39
567,257 -0,68
567,384 -0,81
567,608 -1,02
567,832 -1,23
568,123 -1,49
568,477 -1,80
568,863 -2,12
569,283 -2,45
569,607 -2,69
569,993 -2,97
570,411 -3,26
570,738 -3,47
571,192 -3,77
571,678 -4,07
572,199 -4,37
572,655 -4,63
573,046 -4,83
573,673 -5,15
574,292 -5,44
575,276 -5,88
575,828 -6,12


what do you think ? Exist a better solution than
y = -1,56667460056046E-07x6 + 5,02755756786137E-04x5 -
6,70625116968844E-01x4 + 4,75836028914398E+02x3 - 1,89364117750962E+05x2 +
4,00628517416978E+07x - 3,51900855534185E+09 ?

How can I obtain the area under the curve for each (xi) ? in other words how
can I obtained a integral with a high precision using execl ?

"Jerry W. Lewis" wrote:

If your equation is from the chart display, then by default the coefficient
values will be too heavily rounded to use. Right click on the dilpayed
equation and format in scientific notation with 14 decimal places. Also make
sure that your chart is an "XY (Scatter)" chart and not a "Line" chart.

You may not have a broad enough range of x-values to permit accurate
numerical calculation of the coefficients (is the data set small enough to
include in the body of a reply?). This can be especially problematic if your
polynomial coefficients are from LINEST() in Excel versions prior to 2003.
In 2003's LINEST, coeffients of exactly zero are also suspect.

Jerry

"Mauro" wrote:

I have a serie of experimental data (x1,...xn) and (y1, ..., yn). I plote the
points on an XY chart and I try to fit the trendline. The equation of this
trendline is like the following: f(x)=a*x^6 + b*x^5+c*x^4+....+h. But when i
evalute the function f(x) in the each value of x, f(xi) I obtained a linear
function that is different from the data tendency. What i am doing worng or
how can i get the correct function of the plot.

The plot of the experimental data is like the nromal distribution, and I
need to obtain the area under the peak ? Is there any function on the excel
that permits this ?
.

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