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Ron Rosenfeld[_2_] Ron Rosenfeld[_2_] is offline
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Default Formula Help on Interpolation

On Wed, 29 Aug 2012 03:13:00 +0000, YuriTulchinbork wrote:

Thank you very much Sir, using the trend has helped out but it still
isn't coming out with a smooth transition between two points. I need to
be able to grab the cells and know what they would look like together
smoothed out I guess between the points I select. I don't know if I'm
asking the question right.

If I grab 3 cells, and dont care what the cell #2 says, rather I want
the three points to average out to something smooth, for example If cell
one has a value of 0.00, two has a value of 0.56, and cell three has a
value of 1.00, I know that cell two should be .50, not .56 which would
create a much smoother graph between the two points that have been
selected... I guess once I figure that part out it should be easy to do
up to 10 cells at once.

Can someone help me out here? Or just tell me im in over my head lol.


You are not describing clearly enough what you want. I don't know how you "know" that cell two should be 0.50 and not some other value.

In particular, if you have a group of points and want to draw a straight line through those points, TREND will do that using the least squares method. If your data is better fitted by an exponential or polynomial type of curve fitting, there are ways of doing that also.

But in your above description, you seem to be wanting to completely exclude the value of 0.56, but include the values 0.00 and 1.00. Without knowing the criteria that leads you to include the latter, and exclude the former, it is difficult to supply you with an answer.

The TREND function can account for either forcing or not forcing the intercept to be zero.

As an example, if your known Y's are 0.00, 0.56 and 1.00; and your known X's are 0, 1, and 2, a straight line, computed using the least squares method, through that data returns: 0.02, 0.52 and 1.02. If you force the intercept to be zero, then the straight line for those points would pass through 0.00, 0.512 and 1.024.

So, if curve-fitting using the least squares method is not what you want, I think you need to think more about exactly what you want, and describe it more accurately here.