@Rouben Rostamian Thanks, I had just come to the point of guessing that.

@vv Perhaps I don't understand what "concurrent" means. I'd guess that it means that They are along the same line. But I can only imagine pairs, not quadruples, of points on an ellipse having that property, and they'd be diametrically opposed. Does it instead mean that they intersect at a common point?

Considering the specific operation that you ask about timing---modular exponentiation---and that you emphasized a desire for its exact or precise time, I wonder if your interest is exploring the possibility of a so-called side-channel attack on RSA? Specifically, depending on the algorithm used for modular exponentiation, with some experimental timing data and a knowledge of the precise timing of C&^d mod n, one can figure out the total number of bits and the number of one bits of the decryption key d. One may be able to figure out something else about d also----of that I'm not sure. 

@ntan113 I still see no attachment. What is the name of the file that you're trying to attach? Sometimes weird names are the problem. Please also state what Maple version you're using.

@syhue 

Tom is correct that &^ (like many functions) is only defined when it's used in conjunction with mod. Anyway, there'd be no point in using it for a trivial one-word computation like 2^50.

Regarding selecting a prime factor at random: I assume that your w will always be even and that you never want to select 2. In that case, do

combinat:-randcomb(ifactors(w)[2][2..], 1)[][1];

Regarding selecting a prime at random from an interval: For better crypto-security, you probably want to select it uniformly at random. The technique that you used is commonly used, but it is biased towards primes that are at larger distances from their predecessors. To make a uniform (i.e., unbiased) random selection, use

UniformRandPrime:= proc(R::range(And(realcons, positive)))
local RI:= rand(ceil(lhs(R)) .. floor(rhs(R)));
   proc() local r; do r:= RI() until isprime(r) end proc
end proc
:
URP:= UniformRandPrime(2^50..2^51):
(p,q):= 'URP()'$2;

I assume that you meant to attach some code or a Maple worksheet. Nothing got attached. Please edit your Question and try that again. 

Is your actual problem that you want to find the value of the model parameter a such that the level surface UC(...) = 1 best approximates the experimental data?

A Google search for "Joaminsthal" returns 0 hits. So, this name doesn't appear in a single web page anywhere in the world?

I assume that your Question is essentially Does this Maple code and its accompanying comments fulfill the requirements of this assignment, i.e., does it verify the theorem? Without a clear reference for the theorem, I can't Answer that.

I have a complete solution to your problem done, including producing the graph that you showed. I'm happy to help, but you need to show some effort. If all you can get is a worksheet full of error messages, that's fine, show that.

Here's my plot:

@Angelo Melino 

Here's why Acer's trick works: If you try to apply evalhf to Acer's procedures Hre or Him (e.g., evalhf(Hre(1))), you'll get an error message because evalhf can't handle lists (such as Acer's empty list []) in any form. The numeric integrator (`evalf/int` and its subprocedures) is prepared for this possibility: It traps that error and switches to evalf instead of evalhf.

This doesn't explain how he knew that evalhf was the cause of the problem; rather, it explains how he knew how to construct a workaround once he knew that.

@maple_user_2017 If you'd prefer not explaining a "workaround" to your students, I'll note again that solve followed by evalf gets you your results without issue. Your initial problem was that you were thrown off course by RootOf. Those can usually be resolved with evalf if there are no symbolic parameters.

@maple_user_2017 You're absolutely right that it's difficult to understand why this trick works. I only said that it was obvious to me (as 20-year hard-core Maple user), not that it was obvious in general.

Here is my insight---just some heuristics, not an algorithm. This is also just me guessing---I haven't investigated in this case. The command fsolve is unfortunately very sensitive to precision issues. It tries to guarantee that its results are accurate to the last decimal place. That's good, but what's unfortunate is that there's no way for the user to control how many extra digits it uses internally. The result is that fsolve often gives up and returns unevaluated even though it has actually found a result accurate to almost the last decimal place. The variable a is essentially part of an exponent. Tiny changes in an exponent can lead to much larger overall changes. With my substitution, the first two equations are simple polynomials (no variables in exponents) that can be solved independently of the third equation.

Note that your solve command also reduced the problem to a polynomial. If you had pressed forward by applying evalf to solve's result, you would've obtained the same result that we just got from fsolve. You can usually do this with RootOf results from solve.

Do you mean What aspects of a convergent transition matrix^[*1l affect the rate of convergence of its powers?

[*1] A transition matrix is a square matrix of probabilities each row of which sums to 1. It's convergent if its powers approach a limit.

@tomleslie The function is piecewise differentiable on the same intervals on which the derivative is piecewise continuous. That's enough. The derivative of the function is the stated derivative; so it is a solution.

In Maple 2018, and perhaps earlier, sum(binomial(k+j, k), j= 0..n-k) immediately evaluates to binomial(n+1, k+1), even without the assumptions.