Your f@p is not periodic; try it for -8..8

@Christian Wolinski

@Bendesarts

You cannot have an infinity of branches in piecewise. And you don't need that.

If f is defined in [-1,1) instead of [0,2) then use:

f := x -> exp(-x);   #  -1 <= x  <  1
f_persym:= x -> f( x - 2*floor(x/2+1/2) );

# for a period T>0, use   f( x - T*floor(x/T+1/2) )  

Probably "product" means "componentwise product", *~  in Maple.

f is probably assumed to be linear on the span of the vectors; but the vectors not being assumed linearly independent, the definition of f could be incorrect.

For example all the vectors could be 0; or all <1,1,...,1>.

Conclusion: problem not doable as it is.

@Carl Love 

I am not a native English speaker (probaby this is obvious). In my native language (Romanian, btw) numeral=number.

So, is it incorrect to say that that "the number 123 has three distinct digits"? Is it mandatory to use numeral?

@nMaple 

Carl's solutions works; probably you have an older version.

In Rouben's solution you will have to consider the cases t <= -2 and t>= 10 which are not covered.

@H-R 

It's not so complicated; it is in this way that Maple and maths works.

You can't ignore the complex numbers even if you are working in R. E.g. you will not be able to completely understand why the power series of 1/(1+x^2) converges only in (-1,1).

RealDomain is an acceptable (somehow) compromise.

@Christopher2222 

1 - 9^3/10^3;

    271/1000

@H-R 

You must think mathematically and anticipate the difficulties that may arise.
Here is a short list of problems.

- The entire CAS works over the complex field because this is the best option.
- It is well known that choosing the convenient branch of a multivalued complex function could be a hard task, Maple is not working with Riemann surfaces.
As a consequence, several definite integrals could be wrongly computed or indefinite integrals could be discontinuous. It would be almost imposible to avoid all such problems; there are scientific papers addressing them.
- It is not unusual to have intermediate complex expressions even if the final result is real.
- The assume facility is far from perfect and RealDomains relies on it.

Example:

Why not simply:

Heaviside(0):=0: Heaviside(0.0):=0.0:

?

@Axel Vogt 

I have also obtained a symbolic result using some conversions such as
Expand(%); Change(%,xp=t^(1/3));
subsindets(%,'specfunc(anything,int)', Split, [infinity]);

It's a pitty that one has to do such acrobatics instead of a simple convert(%,GAMMA).

use e.g.

'f()' $i=1..5;

@acer 

@H-R 

A shorter formatted proof:

by definition, for any branch.

@Markiyan Hirnyk 

I answer with great pleasure when the person who asked the question does not understand the solution. And I did it each time. Now I do not know what to think.

Is it not clear that

-RootOf(_Z^2*s-s+1)*s/(s-1)

equals

+-  sqrt(s*(s-1))/(s-1)

?

Must I elaborate? Must I prove that Maple is correct? Are you serious?

@Markiyan Hirnyk 

Why mine and not yours?

But more important, why must we waste our time for such a simple problem?

Just apply allvalues as you did in your post.
Or even simpler. use the definition of RootOf.

@Markiyan Hirnyk