I am not sure how many times I have to say it. The results are because you
have not set the formula to show enough precision. Doing anything but
setting the formula to display more precision is a waste of time.

=1.5555555*a1^6

will give a lot different answer than

=2*a1^6

With 01/03/05 in A1, the difference is like 1.4 x 10 to 27th power.

This is the type of problem you are having.

You need to increase the precision (number of digits) displayed in the
formula.

--
Regards,
Tom Ogilvy



"David" wrote in message
...
Hi Tom,
I did get the same results after format change, but this is not at all
what
I am trying to accomplish. The graph produced has underlying data points
associated with it for each date/data point. That is what I am trying to
find. Maybe I am back to the old stats book and trying to figure that all
out
and creating a table, mean, std dev, R squared and all of that. A single
answer that is a negative number is not even close to what I am trying to
find. Thanks for your help.

"Tom Ogilvy" wrote:

format the cell as number. (it defaults to date - that causes the
#######;
a negative date)

produces:
-445236297867245000

As I said, you didn't set the formula with enough precision.

There are no cell references in the formula. the formula is like an
algebraic formula you would write using the variable x. Many of the
numbers
are in scientific/exponential notation.

--
Regards,
Tom Ogilvy


"David" wrote in message
...
Maybe I can start over. I have simplified the data as much as possible
and
will be literal with what I have come with on the chart and the
equations.
Sample data:

Date Adj. Close*
01/03/05 10729.43
12/27/04 10776.13
12/20/04 10661.6
12/13/04 10638.32
12/06/04 10547.06
....
01/04/99 9643.32

Equation showing on graph:
y = -6E-11x6 + 1E-05x5 - 1.2819x4 + 65247x3 - 2E+09x2 + 3E+13x - 2E+17

Conversion:
y = -6E-11^6 + 1E-05^5 - 1.2819^4 + 65247^3 - 2E+09^2 + 3E+13x - 2E+17

Tried to make it more literal at this point, but several areas of
confusion:
=.000011^6 + .5^5 - 1.2819^4+65247^3 - 9^2 + 3E(confused)
+13*A2(confused,
x
is a date, ref to cell?)- 2E(?)+17

The 13x appears to be the only ref to a cell? The 3E & 2E I can not
tell
what numbers they are in ref to?

When I copied out the equation you provided I end up with
########.....,
which indicated the number might be REALLY big. The actual trend line
on
the
graph indicates the number is fairly close to 10729, which is the last
close
for the date 1/3/05. Maybe it is slightly larger.

"Tom Ogilvy" wrote:

Assume you values are in A2

Then you would modify what you copied to

= -6E-11*A2^6 + 1E-05*A2^5 - 1.2819*A2^4 + 65247*A2^3 - 2E+09*A2^2 +
3E+13*A2 - 2E+17

I paste that in the formula bar

and get a result. However, you need the to select the trendline
formula
and
format it to display more precision.

--
Regards,
Tom Ogilvy
..


"David" wrote in message
...
Hi Again,

This is the literal equation:
y = -6E-11x6 + 1E-05x5 - 1.2819x4 + 65247x3 - 2E+09x2 + 3E+13x -
2E+17

What I copied out and pasted was:
=-6E-11x6 + 1E-05x5 - 1.2819x4 + 65247x3 - 2E+09x2 + 3E+13x -
2E+17

The equation was corrected by Excel to:
=-E6-11*6+E1-5*5-1.2819*4+65247*3-E2+9*2+E3+13*-E2+17

This gives me a Value Error, not recognizing the "E"? I also ws
expecting
references to ranges and stuff and hoping I could copy a formula
down
my
sheet to arrive at specific values accross from Dates and data
values.

Thanks again.


"Tom Ogilvy" wrote:

one of the options on the graph is to display the equation of
the
trendline.
You need to format the equation to show many decimal places of
precision.
Once you have the constants/coefficients associated with the
terms
of
the
equation, you can calculate the predicted points.



--
Regards,
Tom Ogilvy

"David" wrote in message
...
Hi Group,
This is a little difficult to explain, but the underlying data
is
simple,
so
please bear with me. I have Dates and Closings for the Dow
Jones
Industrial.
Similar to below:
Date Adj. Close*
01/03/05 10729.43
12/27/04 10776.13
12/20/04 10661.6
12/13/04 10638.32
The above is easy to graph and I have automated the process,
which
includes
a 6th Order Polynomial Trend line added to the graph. What I
am
trying
to
do
is find the data points associated with the 6th Order
Polynomial
Trend
line.
It has simply been just too long since I have done this type
of
statistics. I
believe it may be necessary to create a new table to find
these
data
points,
which I am willing to do. I can calculate the sample mean,
number
of
sample
variables, sample variance, sample standard deviation, etc.,
but
it
has
just
been too many years to bring the necessary statistical
expertise
to
arrive
at
the data points. I am trying to get a table that looks
something
like
this:
Date Adj. Close* Trend
01/03/05 10729.43 10730.25
12/27/04 10776.13 10750.31
12/20/04 10661.6 10765.03
12/13/04 10638.32 10750.00
I have tried using some of the built in functions, but they do
not
yield
the
same data points that have been graphed by the 6th Order
Polynomial
Trend
line. I have tried Trend and Forecast. I created a table many
years
ago,
which I think calculated the data points, but it has simply
been
to
many
years and I have lost the statistical expertise. Any help
would be
greatly
appreciated.
--
David