@HDN46 There are only two ways to get out of the loop for n from 0 to 19: Either n goes above 19, or the break statement (after the then in the loop) is executed. The break is executed when we have a non-solution: either a non-integer was encountered (r<>0) or two consecutive zeroes were found. Otherwise, we have a solution, and n goes above 19, so it is 20. 

Does that answer your question?

@marc005 What trouble did you have trying to make it into a procedure?

@marc005 What trouble did you have trying to make it into a procedure?

Capital O is a protected name. I am skeptical about your attempt to use it as a variable.

@jenniferchloe You asked:

1. Here, what do a and b represent?

They are the positions of the first and second letters of a digraph in the "long form" of the key. For example, if the key is the example that I used before,

"PLAYFIREXMBCDGHKNOQSTUVWZ"

and the digraph is "HI", then a = 15, b = 6.

You asked:

2. When you write the command (r1,c1):=to_row_col(a), are you substituting the value of 'a' into 'n' in the to_row_col expression above?

Yes.

You asked:

3. If I had (r1,c1)=(2,3) [2nd row,3rd column] then how would I find to_row_col(a)?

If r=2 and c=3, then a = (2-1)*5+3 = 8. In other words, row 2 column 3 is the 8th position in the long form key. To go the other way

8-1 = 7.

Divide 7 by 5 with remainder, obtaining quotient q=1 and remainder m=2. Then the row is q+1 = 2 and the column is m+1 = 3.

@jenniferchloe You asked:

1. Here, what do a and b represent?

They are the positions of the first and second letters of a digraph in the "long form" of the key. For example, if the key is the example that I used before,

"PLAYFIREXMBCDGHKNOQSTUVWZ"

and the digraph is "HI", then a = 15, b = 6.

You asked:

2. When you write the command (r1,c1):=to_row_col(a), are you substituting the value of 'a' into 'n' in the to_row_col expression above?

Yes.

You asked:

3. If I had (r1,c1)=(2,3) [2nd row,3rd column] then how would I find to_row_col(a)?

If r=2 and c=3, then a = (2-1)*5+3 = 8. In other words, row 2 column 3 is the 8th position in the long form key. To go the other way

8-1 = 7.

Divide 7 by 5 with remainder, obtaining quotient q=1 and remainder m=2. Then the row is q+1 = 2 and the column is m+1 = 3.

I understand where b and d come from, that's just

X minus {indices}(T, 'nolist');

But where do you get {3,4} and 4 from?

@mutaz Okay, then I guess you should ignore what I said about x2. But I was only basing that on what you said in the original question: You included x2 and called it a PDE.

No need to reply.

@mutaz Okay, then I guess you should ignore what I said about x2. But I was only basing that on what you said in the original question: You included x2 and called it a PDE.

No need to reply.

@Joe Riel Joe, this is a gem of code. ListTools:-Classify is also a gem, and it would be well worth any Maple programmer's time to contemplate it. It's essentially only three lines; the other two are only to handle a degenerate case. It uses a table of tables. So simple, like a haiku or Zen koan.

Markiyan wrote:

The Levi-Lindeberg theorem implies that  X has the approximately normal distribution with parameters 49 and 8.

I think that the parameters are 49 and 8/sqrt(81).

Markiyan wrote:

The Levi-Lindeberg theorem implies that  X has the approximately normal distribution with parameters 49 and 8.

I think that the parameters are 49 and 8/sqrt(81).

askstathelp wrote:

I am trying to get Maple to calculate this for me: Let X₁,X₂,...X₈₁ be a random sample from a distsribution (not necessarily normal) with mean µ=49 and standard deviation σ =8. Let ¯X = (1/81)* ∑_i=1⁸¹ * X_{i  I need to find the mean, the standard deviation and the approximate distribition of ‾X,}

This is not a calculation or computation problem, other than the trivial computation of taking the square root of 81; so Maple will not be much help. The purpose of this problem is to get you to think about the Central Limit Theorem.

@marc005 Clicking on that link, it now asks me for a username and password.

Is the file still there (is your "working link" still working)? I get the message "Cloud not found" when I click the link.

Question to the general audience: How feasible would it be to implement HTTPS in high-level Maple code?