Replace both lines with the single command

remove(type, identify(d), float);

Matrix multiplication is associative, so grouping in different ways with parentheses is worthless.

If B is invertible, then it is trivial to prove that your expression equals A. So choose a B that is not square but such that B.B' is invertible. (Make B m x n with m < n.)

What your purpose in doing the exercise? That is, Is the goal to perform an exercise of writing the most robust numeric code possible? Or is the goal simply to get the answers? What is the form of the coefficients? Are they exact rational numbers? floating-point numbers? or something more complicated?

If the goal is simply to get the answers, the best option is to use the command solve; it will handle any type of coefficient. Example:

solve(2*x^2 + 3*x + 4 = 0, x);

If you know that you want floating-point answers, then use fsolve:

fsolve(2*x^2 + 3*x + 4 = 0, x);

If your goal is the robust-code-writing exercise, then you need to write a procedure, the skeleton of which looks like this:

SolveQuadratic:= proc(A::float, B::float, C::float)
# Solve A*x^2 + B*x + C = 0 numerically
local ...  #Any other variables in the procedure go here

... #Body of procedure

end proc;

Feel free to ask for more details. All of the above will work in Maple 5.

Here's a start. I have a feeling that this can be improved. I can't test without having your procedure elem.

restart:
Elem:= proc(i,j)
option remember;
     Elem(j,i):= elem(i,j)
end proc:

N:= 2^10:
A:= Matrix(N, N, shape= symmetric);
for i to N do
     A[i,..]:= < Threads:-Seq(Elem(i,j), j= 1..N) >
end do:

Motivated by Axel's analysis, I looked more closely for solutions for the situations that RootFinding:-NextZero rejected. I used a "microscope", searching very close to the origin. By that I mean that I used semilogplot and applied fsolve to numer(f(10^d)) rather than to f(d). I plotted in all cases, and the sequence of plots really shows what's happening.

The first point remains a mystery because the limit(f(d), d=0, right) = -infinity (as Axel noted) where one would expect 0. The limit for the other points is in fact 0, and the plots seem like a continuous progression.

Download erf_solve_2.mw

Download triangle.mw

In order to get a numeric result, you need two numeric limits of integration. Use Int instead of Intat, and then apply evalf.

RootFinding:-NextZero is not as strict as fsolve is with the accuracy. It obtains a solution for several of your (x, N) pairs. For the pairs for which no solution is found, a plot indicates that there is truly no solution.

Download erf_solve.mw

Use Statistics:-NonlinearFit. Your data is the listlist L, and your parameter is d. Your model function is NN+N (so that it is in solved-for-N form).

You should change the title and tags of this Question. The fact that the data came from Excel, or that they were imported from anywhere, is totally irrelevant.

All that you need is

str1:= convert(Xor(f(number10), f(number8)), hex);

Note that there are no quotation marks in this command!

The command convert(..., bytes) has nothing to do with this. It deals with ASCII character values. However, if you still want to know the difference between the two forms of convert(..., bytes), it is this: They are inverses of each other. To convert the string expr to a list of ASCII decimal character values, use L:= convert(expr, bytes). To convert L back to the string expr, use convert(L, bytes). Note that quotation marks are only used to enclose literal string values: see ?symbol .

You can automate the conversion from/to hex so that it will work for any of the operators in Bits like this,

Hex:= Op-> convert(Bits[Op]((x-> convert(x, decimal, hex))~([_rest])[]), hex):

And then call it like this:

Hex(Xor, number8, number10);

It is not currently possible to access programmatically the directory name of the currently active worksheet.

Download Sqrfree.mw

There is some information pertaining to your question on the help page ?dsolve,numeric,BVP .

I meant quitting the iterator programmatically. To do this, the iterator needs to be the index in a for loop.

Download Iterator_forloop.mw

Download sinPi16.mw