@lemelinm I notice that the first value in your dataset is 100404, which is an extreme outlier, more than 60 standard deviations from the mean of the others. My guess is that this was transcribed wrong, and it should be 100484. If I make that change, then the standard deviation is 1.289, which may have been incorrectly rounded to 1.28. So, please check whether you entered the first datum correctly.

Note that there is a difference between RMS and RMS deviation. You have used both terms in this post. I am certain that you mean RMS deviation.

@JEBland When dsolve({sys1, inits}) gives NULL output, it usually means that it was unable to find a symbolic solution; it is not usually an indication of user error.

@lemelinm I did not try the worksheet.

To cut and paste: Select the column of data from the post with the mouse. Hit Control-C (that's the "cut"). Place your cursor in a Maple (Standard) worksheet. Hit Control-V. That's all there is to it.

@Thomas Richard Yes, that's what I was refering to.

Please don't change a Question after an Answer has been given! Make a new Question instead. I have an answer to your question about the error plot, but I won't give it until you put back the original Question.

@lemelinm According to the Wikipedia page "Root-mean-square deviation", for a dataset it is the same as the standard deviation. But the standard deviation of your dataset is 10.906, not 1.28. So what is your formula for RMS deviation?

Note that you can cut and paste your dataset directly into a Maple worksheet, and it thus becomes a 55x1 Matrix. There's no need to use file I/O for such a small dataset.

@casperyc Yes, a seq is usually a little faster than the corresponding for loop.

@mehdi jafari Since the stats package is deprecated, you might as well use Statistics:-QuadraticMean.

What do you mean by exporting? Do you want the data written a file in two columns: the first coordinate value and the second coordinate value? Note that your plot data exists in tabular form before you plot it, so you don't need to "extract" it from the plot. The data is simply

[seq([k[i],cos_phi2[i]],i=0..number_of_points)]

@mehdi jafari evalr is surely the correct thing to use. Indeed, the vote up is mine. If you use lprint to look at the internal structure of the objects generated by the Tolerances package, you will see that they are all of the of the form INTERVAL(........), which are precisely the things that evalr is for. I just find it amazing that there is no example of this on the page ?Tolerances and no mention of the use of INTERVAL.

For example, using the code in the original Question, take a look at lprint(Phi). Then look at lprint(evalr(Phi)).

Does anyone here know if this need to use evalr is documented anywhere? I don't see it mentioned at ?Tolerances . That page merely has a link to ?evalr at the bottom.

@mounir Matrix is spelled with a capital M. You used a lowercase m.

Also, there's no point in including the command M:= Matrix(convert(A, list)) inside the loop. It's just wasted operations. Put it after the loop.

@love maths I followed the link, and I see that Afn means Average(F, f||n), where Average is the procedure with the corrections made by Preben Alsholm. It seems that f||infinity is a Matrix of all ones for the example f0 given, although it is not immediately obvious to me how to prove that. Given that f||n converges to a Matrix with all entries equal, it is obvious to me that "triangle"(f||n) converges to the 0 Matrix.

Your code works for me in Maple 17; s contains all four constants of integration. What does your s contain?

@mounir 

M:= Matrix(convert(m, list));