@Art Kalb Ah, I think that I understand your problem: I think that you have the mistaken impression that the behaviour of f(g(x)) can be controlled by a procedure named `f/g` for arbitrary f and g. The reality is that that mechanism only works for a few select system routines f, for example, evalf, print, type, value, diff.

@Art Kalb What do you expect the code above to do? Show the command that you give and the output that you expect. What is name of the help file that you referred to, the one that you based this on?

@Art Kalb What do you expect the code above to do? Show the command that you give and the output that you expect. What is name of the help file that you referred to, the one that you based this on?

@Art Kalb You misinterpretted what I wrote. It is only in the places that you currently have unevaluation quotes that you need to change them to be both types of quotes. In summary

`f/g` is fine, it's just a name.

'f/g' is nonsense, unless you mean unevaluated division.

'`f/g`' is the unevaluated form of `f/g`.

`'f/g'` is just a name, albeit one likely to cause confusion.

I am absolutely 100% sure about this.

@Art Kalb You misinterpretted what I wrote. It is only in the places that you currently have unevaluation quotes that you need to change them to be both types of quotes. In summary

`f/g` is fine, it's just a name.

'f/g' is nonsense, unless you mean unevaluated division.

'`f/g`' is the unevaluated form of `f/g`.

`'f/g'` is just a name, albeit one likely to cause confusion.

I am absolutely 100% sure about this.

@Kitonum The default time command is showing the sum of the times on all processors. Parallel processing always takes more time, when time is measured that way, because there is significant overhead involved in dividing up the tasks among processors. The benefit of parallel processing, if any, is only evident when you look at the real time via time[real].

time[real](PP({$1..20}));
                             6.113
time[real](PowerSet({$1..20}));
                             4.302

You can also simply replace the seq in PP with Threads:-Seq. It turns out that there is not much benefit to the parallel processing here. But the benefit should increase as computers get more and more processors (I have 8). To make a better comparison, compare procedures where the only change is from seq to Threads:-Seq.

@Kitonum The default time command is showing the sum of the times on all processors. Parallel processing always takes more time, when time is measured that way, because there is significant overhead involved in dividing up the tasks among processors. The benefit of parallel processing, if any, is only evident when you look at the real time via time[real].

time[real](PP({$1..20}));
                             6.113
time[real](PowerSet({$1..20}));
                             4.302

You can also simply replace the seq in PP with Threads:-Seq. It turns out that there is not much benefit to the parallel processing here. But the benefit should increase as computers get more and more processors (I have 8). To make a better comparison, compare procedures where the only change is from seq to Threads:-Seq.

You say "d will be x(t) + a(t)". What is a(t)? Did you mean "d will be x(t) + a[1](t)"?

This brute force technique is far from optimal, especially considering that the sum is converted to a continuous increasing function of N.

This brute force technique is far from optimal, especially considering that the sum is converted to a continuous increasing function of N.

It is important for you to note that Markiyan changed your Sum to sum (with lowercase s). It is easy for a reader to miss that detail. The fsolve will not work without that. With the evaluated sum, it converts the summation to a continuous increasing function of N.

Also, he changed your pi to Pi. The latter is the well-known constant, while the former is just another variable.

It is important for you to note that Markiyan changed your Sum to sum (with lowercase s). It is easy for a reader to miss that detail. The fsolve will not work without that. With the evaluated sum, it converts the summation to a continuous increasing function of N.

Also, he changed your pi to Pi. The latter is the well-known constant, while the former is just another variable.

@Przemek I can't give any more advice until you post the equations.

You did not enter the procedure exactly as I typed it. It's Statistics :- CDF, not Statistics - CDF. Cut-and-paste the boldface code below. The non-boldface part is Maple's responses.

normalcdf:= proc(x1,x2,mu,sigma)
local t,dist:= Statistics:-CDF(Normal(mu, sigma), t);
     eval(dist, t= x2) - eval(dist, t= x1)
end proc:
normalcdf(2100,2200,2150,50);
                            /1  (1/2)\
                         erf|- 2     |
                            \2       /
evalf(%);
                       0.682689492137088

You did not enter the procedure exactly as I typed it. It's Statistics :- CDF, not Statistics - CDF. Cut-and-paste the boldface code below. The non-boldface part is Maple's responses.

normalcdf:= proc(x1,x2,mu,sigma)
local t,dist:= Statistics:-CDF(Normal(mu, sigma), t);
     eval(dist, t= x2) - eval(dist, t= x1)
end proc:
normalcdf(2100,2200,2150,50);
                            /1  (1/2)\
                         erf|- 2     |
                            \2       /
evalf(%);
                       0.682689492137088