Artes

@user64494 Yes, I don't since you incorrectly calculate the integral.

Residue hasn't been updated since version 2.0 and so you should not expect evaluating Residue[(n/z/(z^n - 1)), {z, 0}] for abstract n. If you prescirbe n e.g. with Table or With it returns correct values. See in your comment inappropriate z -> 1/2*Exp[I*t])*I*Exp[I*t].

@user64494 Since when Residue has the Assumptions option? I'm using version 13.0.1.

Question has been changed a few times since I've read it. Residue at zero is obvious, see e.g. Table[Residue[n/(z (z^n - 1)), {z, 0}], {n, 10}]

What do you mean by residue of $-n$? There is no singularity in $-n$ for $n>1$.

@halirutan @halirutan It's not the first time when this user exhibits several psychological or even psychiatric problems. I suggest that any dialog will not be constructive in this case. Most likely this are aspects of the low self-esteem problem.

@Szabolcs Indeed, I wrote first, than I read the comments.

@Szabolcs Do you know when WRI launches version 10.4? I find they made quite a lot of bugs in 10.3, and I believe they should take a closer look at clarification existing functionality, rather than adding new trivial functions.

@JasonB It's all right

@JasonB No, it doesnt matter. Sometimes I get a bit angry when some MSE users provide wrong answers to simple questions, and then they get upvotes. I was in the chat quite a long time ago. Now I'm not sure how comunicate to other MSE users.

@rm-rf Sometimes I used to ping someone this way like here...

@MichaelE2 Thanks for your response. I think that duplicates to earlier posts make this site disordered. Of course there is some kind of freedom of deciding whether a post is a duplicate or not however in this case it seems to me quite clear it's a duplicate. Your method seems to be an overkill but I haven't known it, so +1.

@rm-rf I mean I asked them here [Sum with Levi-Civita ](mathematica.stackexchange.com/questions/56063/…). They voted to leave that post open but as far as I understand it was an exact duplicate of another one.

@RunnyKine @michaele2 @bobthechemist Could you explain how this post is not a duplicate of this one Contracting with Levi-Civita (totally antisymmetric) tensor? I've asked you in the comments to that post but got no response. Have I overlooked anything or just don't understand a "deeper level"?

Read this answer How do I work with Root objects? for more

You can if you put numeric values for A and L, when they are symbolic then solutions are in terms of Root objects and there is nothing more to add

If you expect some clarification edit your question and make it precise

and it does't matter

I said I don't know what is W

however they are exact solutions

It is represented in terms of Root objects

and it can be factored: Factor[p[x, 0, 2]] returns 32 (-2 + x)^2 x^4 (4 + 3 x^2), so you can see there are two solutions 2, four solutions 0 and two complex solutions to 4 + 3 x^2

For example Solve[p[x, 0, 2] == 0 && x > 0, x] returns a double solution: {{x -> 2}, {x -> 2}}, then the polynomial is quite simple p[x, 0, 2] == 512 x^4 - 512 x^5 + 512 x^6 - 384 x^7 + 96 x^8

Having such involved coefficients it is unlikely to expect that you will find a general representation in terms of radicals.

there are solutions in terms of radicals only for 4-th order polynomials

than Root object

When A and L are symbolic you cannot expect more

If you expect any help you have to ask a precise question. I don't know what is W and nobody knows it. It doesn't matter at all what it is, but overall I don't know what you are looking for

I have done it in the answer, Root objects represetns solutions in terms of A and L, if you apply N you will find numerical approximation.

Solve[p[x, 2, 8.753275] == 0 && x > 0, x, Reals] returns two solutions {{x -> 2.01469}, {x -> 3.00001}}.

What is $W$?. Edit your question and make it clear since I'm not sure what you really want if my answer does not satisfy you needs.

@J.D'Almbert I updated my answer. Root's are symbolic representations of exact solutions, under specific conditions one might apply ToRadicals to Root objects finding solutions in terms of radicals but in general you cannot expect such solutions because it is mathematically impossible to find solutions in terms of radicals for polynomials of order higher than $4$.

Bye

Therefore I couldn't answer too many questions

If you have another questions I will help when I have time, but recently I was quite a busy

But did you agree with my arguments?

Remamber that Mathematica works on complex numbers by default

I don't know if you understand the problem correctly

Of course you can

What is it Mapleprimes?

I have one question: why do you mean by "our views are different"?

Nevertheless the complete set of solutions includes also negative numbers.

thus for the sake of simplicity you have to postulate that x is positive

It is another story, in school children don't know about numbers Sqrt[x] for negative x

However it is a negative number thus Sqrt[x] and Sqrt[1 - x^2] are complex

I demonstrated in my answer with FillSimplify that -1 - Sqrt[2] is also a solution to Sqrt[x] + Sqrt[1 - x^2] == Sqrt[2 - 3 x - 4 x^2]

they are not a consequence of the equation Sqrt[x] + Sqrt[1 - x^2] == Sqrt[2 - 3 x - 4 x^2]