@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?

@mutaz  the domain is 0≤x1≤2pi.

I was afraid that you would say that that's the domain.... The solution we have so far is only good for 0 ≤ x1 ≤ 0.066, as Adri explained. To find solutions on the rest of the domain we need more initial conditions. Indeed, the initial condition we have for the current solution is just one that Adri made up, V(0)=0. We can do it for an arbitrary intial condition like this

dsolve({DE, V(x0)=v0}, V(x1));

and you'll get a solution only very slightly more complicated than what we already have. You will not be able to numerically evaluate this until you supply numeric values for x0 and v0. You may be able get some more information from the original PDE problem.

We also need to address the issue of the significance of x2 in your original problem. Whatever solution we get for V(x1), we need to add to it an arbitrary function of x2 but not x1: V(x1,x2) = V1(x1) + F(x2), where the V1 represents the solution from dsolve. Note that the F(x2) disappears when you take the derivative with respect to x1.

@mutaz  the domain is 0≤x1≤2pi.

I was afraid that you would say that that's the domain.... The solution we have so far is only good for 0 ≤ x1 ≤ 0.066, as Adri explained. To find solutions on the rest of the domain we need more initial conditions. Indeed, the initial condition we have for the current solution is just one that Adri made up, V(0)=0. We can do it for an arbitrary intial condition like this

dsolve({DE, V(x0)=v0}, V(x1));

and you'll get a solution only very slightly more complicated than what we already have. You will not be able to numerically evaluate this until you supply numeric values for x0 and v0. You may be able get some more information from the original PDE problem.

We also need to address the issue of the significance of x2 in your original problem. Whatever solution we get for V(x1), we need to add to it an arbitrary function of x2 but not x1: V(x1,x2) = V1(x1) + F(x2), where the V1 represents the solution from dsolve. Note that the F(x2) disappears when you take the derivative with respect to x1.

@mutaz The integral can only be evaluated numerically, for specific numeric values of x1. What is the domain for x1?

@mutaz The integral can only be evaluated numerically, for specific numeric values of x1. What is the domain for x1?

@adel-00 I don't understand what you mean. The plot command that I gave below does it all without any loops.