Two points:

Make sure the trendline formula is viewed with sufficient significant
figures. Apply a number format of 0.000000000000000E00.

A poly fit may be useful for making a line that kind of follows the points,
but in most cases there is no physical basis for such a high order equation.

- Jon
-------
Jon Peltier, Microsoft Excel MVP
Tutorials and Custom Solutions
Peltier Technical Services, Inc. - http://PeltierTech.com
_______


"Aaron Davies" wrote in message
...
I think I figured out the problem: sixth-order polynomials are just
wrong. A sixth-order trend looks fine on my sample data below, but the
equation is nonsense. With my real data, a fifth-order equation works
fine.

On Feb 5, 12:51 pm, Aaron Davies wrote:
Sorry, that's what I meant by "the integer of the date". 1/1/2001 is
36892, and using 36892 for x in the equation given by Excel yields
1.361956e+09, not 1.

On Feb 5, 12:13 pm, "Jon Peltier"
wrote:

The x values are in days, but 1/1/2001 is not day one. Day one is
1/1/1900,
and 1900 was treated as a leap year. If you want to make 1/1/2001 day
one,
use the date minus 12/31/2007 for your x values.

- Jon
-------
Jon Peltier, Microsoft Excel MVP
Tutorials and Custom Solutions
Peltier Technical Services, Inc. -http://PeltierTech.com
_______

"Aaron Davies" wrote in message

...

What are the units of x in a date-based trend equation? If I take a
simple integer-integer dataset,

1 1
2 4
3 9
4 16

and fit a second-order polynomial to it, I get y = 1x^2 +
5.6843418860808e-14, which is reasonably accurate at the known
points.
If, however, I have dates in my x axis, I can't figure out the units
of the equation.

1/1/2001 1
1/2/2001 4
1/3/2001 9
1/4/2001 16
1/5/2001 25
1/6/2001 36
1/7/2001 49
1/8/2001 64
1/9/2001 81
1/10/2001 100

yields y = x^2 + 7.3782x + 1.360945881, but what values for x yield
the y's I already know? It's not the integer of the date--that series
starts 1.361292e+09, 1.361366e+09, 1.361439e+09.