There is a constant Pr in the second equation. I assumed that this was a typo for Pi. Using that and attempting to solve the system, I get the dreaded error "Newton iteration is not converging".

@Murilo V The multiplication operator was likely caused by typing an extra space between evalf and its opening parenthesis. This only affects 2D input.

@kamrul126 Does the file exist already? If it does, then that is the problem. Do you have write permission for the directory/folder specified by currentdir()? If you are not sure, try using a fully qualified file name such as "C:/Users/Carl/Desktop/file1.mpl"

@DJJerome1976 asked:

If I change the problem to 2*sin(2*x+Pi/4), and the desired form is sqrt(2)*sin(2*x)+sqrt(2)*cos(2*x), expand(,trig) no longer works for me. How might this one be tackled? Clearly, one may just replace "x" with "2x," but is this the only way?

combine(expand(%));

@Markiyan Hirnyk My comment was directed to the OP.

@Markiyan Hirnyk Isn't that close enough to the desired output to count? Is the problem that you want the sqrt(2) in the front?

This output can be obtained with a simple expand; the trig modifier is unnecessary.

Perhaps you're missing a semicolon on the first line.

@Markiyan Hirnyk Hence the use of a midpoint method, either bvp[midrich] or bvp[middefer].

After correcting for "initial Newton iteration is not converging" by using a continuation parameter (I used A[1]:=1/C, continuation= C), I am left with the much more difficult-to-fix error "Newton iteration is not converging". See ?dsolve,numeric_bvp,advanced .

@Ratch You use option implicit when you do not want the answer to be in a form solved for y(x).

@Axel Vogt What he wants is

seq(fsolve(eval(NN, [x,N]=~ L[k])), k= 1..nops(L));

But all of the fsolves return unevaluated. I guess it's because either the accuracy cannot be achieved, or there is truly no solution. Further explorations with RootFinding:-NextZero and plot (see my other Answer) show that both situations ensue.

@J4James I misinterpretted your question. I thought that you wanted the one value of d that is the best fit for all the data. If you want a separate d for each (x, N) pair, you can use fsolve without NonlinearFit.

@sakhan Converting the result back into hex is required. It is a nescessary step to getting back to plaintext. You cannot take an arbitrary integer and apply convert([...], bytes) to it. The integers in the list need to be in the range 0-255. Here's a little procedure to convert hex to plaintext:

HexToPlaintext:= H-> convert(sscanf(H, cat("%2x" $ length(H))), bytes);

However, if I apply this to your hex strings, I just get garbage characters rather than English text. So I wonder how you got your number1 thru number11.

What is the result when you call rtable_eval(A) ?

@John Fredsted It is just a coincidence that the final result is also equal B . < 3, 0, 1 >. If you change any entry in A, this is no longer true.

For shorthand, note that Vector([a,b,c]) is equivalent to < a, b, c >, and that Transpose(A) is equivalent to A^%T.

@Joe Riel How is it possible that the whole procedure takes only 48 seconds when simply the call to Iterator takes over 200 seconds for me? I am using a compiled iterator. That is, the line

AllP:= [seq(P, P= Iterator:-SetPartitions({S[]}, [[4,4]]))]:

takes over 200 seconds for me. Could there be that much difference between our compilers?