@acer Good idea. But selectremove works just as well with functions as lists, and the Vector form makes the shuffling easier. So I reduced the code to this:

rearrangeCurves:= proc(
     v_items::specfunc(anything, PLOT),
     v_reorder::list([integer,integer]):= []
)
local p, curves, rest;
     (curves,rest):= selectremove(type, v_items, specfunc(anything, CURVES));
     curves:= < op(curves) >:         
     for p in v_reorder do
          curves[p]:= curves[p[[2,1]]]
     end do;
     PLOT(convert(curves,list)[], op(rest))
end proc:

@taro Note that op(a) and a[] are the same only for lists and sets.

@mehdi jafari Sorry, I meant imaginaryunit, not complexunit. I have corrected the original.

Regarding method= float: If you have a large Matrix of exact values (say, Gaussian integers as in this example), then it will be much faster to use method= float; for a 2x2 Matrix, it makes no significant difference.

@MuriloLobo Yes, I understand that extra values may need to be added after the trimming, if trimming is used.

I don't know how to account for the changing mean and standard deviation. Perhaps there is a standard distribution that is like a Normal but has finite bounds. I don't know enough about distributions to answer that. If you check ?Statistics,Distributions , you might find something.

Note that the mean and standard deviation of a random sample are unlikely to be exactly equal to the mean and standard deviation of the distribution from which the sample is drawn.

@MuriloLobo Your question is now well formed to me. Thank you.

To answer your question, would it be acceptable to generate a normal Normal sample with given mean and sd and then trim off the values that lie outside the 95% bounds? Do you want it trimmed on bottom end also?

I executed your worksheet twice using the menu command Edit -> Execute -> Worksheet and I cannot reproduce your problem. I get fully displayed matrices.

@Christopher2222 Are you referring to

kernelopts(limitjvmheap= true);
kernelopts(jvmheaplimit= 2000*Unit(mebibyte));

?

@Joe Riel Joe, see the Wikipedia article "Longest common subsequence problem". The command StringTools:-LongestCommonSubsequnce would return the characterwise longest common subsequence. What Markiyan wants is the wordwise longest common subsequence.

I am adapting the algorithm described in the Wikipedia article. The first step is to break the strings into Arrays of words.

Definitions: For a positive integer n, let [n] denote {1, 2, ..., n}. Define a (finite) sequence as a function f: [n]-> A for some n and some set A. Define the length of f to be n. Let n, A, and sequence f be given. Let S be a subset of [n]. Let O: [|S|]->  S be the unique increasing bijection determined by S. Then g:= f@O is the subsequence of f determined by S. Note that g is itself a sequence as defined above. Note that g: [|S|]-> A and thus, when viewed as a sequence, g does not depend on S, merely on |S|.

Let f1 and f2 be two sequences, and let g1 and g2 be respective subsequences. If g1 = g2 (as sequences), then g1 is a common subsequence of f1 and f2.

@Mac Dude To my mind, the original question and the followup both make no sense. So I can't really say that the Question has been Answered, although the Replies do contain some useful information for the OP. I am waiting for a reformulation of the original question.

@MTata94 Change ln to log10 in my code. I already updated the Answer. If I do this, then I get 1860.633190 for evalf(Fi(30)).

@MTata94 Use evalf(Fi(30)) instead of Fi(30).

You should show your erroneous code. I am sure that we could correct it.

@shingy Shingy, you need to begin your session with the command

with(PDEtools);

otherwise Maple cannot find the DeterminePDE command.

An alternative is to type the fully qualified command name (i.e., including the module prefix PDEtools:- ) at each invocation, which is what Edgardo did:

PDEtools:-DeterminePDE(equ,  integrabilityconditions= false);

@mehdi jafari But the OP needs a programmatic way to do it---explicitly with Maple commands. This is easy with plotsetup and plotoptions, as in the Answer by itsme.

@Muhammad Usman 

The algorithm is fine, but its coding needs to be retrofitted for your older Maple 13, which doesn't have the numelems command. Try this:

B:= proc(X::{Vector,list})
local n;
     n:= `if`(X::list, nops(X), op(1,X));
     LinearAlgebra:-Determinant(
          Matrix(
               n,
               (i,j)-> `if`(j < i, -1, binomial(n-i,j-i)*X[j-i+1]),
               shape= Hessenberg[upper]
          )
      )
end proc: