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I'm using the definition $$[n]=\frac{A^{2n}-A^{-2n}}{A^2-A^{-2}},$$ although the usual definition is recovered by substituting $q=A^2$.

In trying to prove various identities (and also reading the literature) I've had a lot of problems passing between the LHS and RHS of the top equation. I know that there are a lot of identities out there (I'm not sure where either) but I'm moreso looking for a program/ paper that collects this information.

For example, one can use a computer algebra system (I used sage) that can "prove" an identity, but what is returned is some mess in the RHS notation, rather than something meaningful with the quantum integers. I'm unfamiliar with manipulating them.

A successful program in my mind, would be able to translate $$\frac{A^{-1/4}}{A^{1/4}+A^{-1/4}}$$ into a reasonable closed form in quantum integer notation, since I struggle even to get these types of expressions going.

Andres Mejia
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  • What are the operations you are allowing exactly ? Addition, multiplication, division of expressions $C \ f_n(A)$ where $c \in \mathbb{C}$ and $f_n(A) = \frac{A^{2n}-A^{-2n}}{A^2-A^{-2}},$ ? $n$ is integer or rational ? – reuns Jul 17 '17 at 02:50
  • $n$ is an integer. I'm okay with quotients of the form $[m]/[n]$ and products of a similar form. Addition and subtraction is allowed as well, although my ultimate goal is to translate a sum into a product. algebraically I am mostly interested in seeing when some expression in $A$ has a "simple" closed form in quantum integer notation. – Andres Mejia Jul 17 '17 at 02:58

2 Answers2

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The product/quotient of quantum integers can be factored into homogeneous cyclotomic polynomials.

For example: $$\Phi_3(q)=[3],\; \Phi_4(q)=\frac{[4]}{[2]},\; \Phi_5(q)=[5],\; \Phi_6(q)=\frac{[6][1]}{[2][3]},\; \Phi_7(q)=[7],\;\Phi_8(q)= \frac{[8]}{[4]}.$$ A computer algebra system should be able to factor a product/quotient of quantum integers into cyclotomic polynomial factors.

As for list of identities, there is one source I know of. My list of Simple Algebraic Identities has many identities that are satisfied by trigonometric sine (indicated by $\texttt{[TS]}$ tag). For example: $$ \texttt{id2_3_1_2a = +a*a -b*b -(a-b)*(a+b)}.$$ This corresponds to the trigonometric identity $$ 0 = \sin(a)\sin(a) - \sin(b)\sin(b) - \sin(a-b)\sin(a+b),$$ but also the equivalent quantum integer identity $$ 0 = [n][n] - [m][m] - [n-m][n+m].$$

Somos
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  • Oh, this is a nice point of view. How, for example would someone check the last identity I gave with this method? What about sums and differences of quantum integers? – Andres Mejia Jul 17 '17 at 02:25
  • Yes, thank you. For your identity you have $\frac{A^{-1/4}}{A^{1/4}+A^{-1/4}}$, but you rewrite it as $\frac{A^{-2}}{A^2+A^{-2}}$ which is $1/[2]$. If you give a more realistic exmaple I may show how to do it. Sums are harder just like trig identities. – Somos Jul 17 '17 at 02:35
  • Why would it be $1/[2]$? $1/[2]$ needs another summand in the numerator, if I'm not mistaken. I think this is a fairly realistic example, it shows up in kuperberg's paper – Andres Mejia Jul 17 '17 at 02:39
  • Yes, tanks for catching that. Stray factors of $A$ can cause problems. I should have seen it since the numerator is not balanced but the denominator is. In general you have to be careful about this. – Somos Jul 17 '17 at 02:45
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With your CAS you obtain a rational function in $A^{2/m}$ $$f(A^{2/m})=f(A^{-2/m}),\qquad f(x) = \frac{g(x)}{h(x)}, \qquad g,h \in \mathbb{C}[x]$$

$$f(x) = \frac{g(x)h(x^{-1})+g(x^{-1})h(x)}{2h(x)h(x^{-1})} = \frac{\sum_{n=0}^N b_n (x^n+x^{-n})}{\sum_{n=0}^M c_n (x^n+x^{-n})}$$ $$f(A^{2/m}) =\frac{\sum_{n=0}^N b_n \frac{[2n/m]}{[n/m]}}{\sum_{n=0}^M c_n \frac{[2n/m]}{[n/m]}}, \qquad [n/m] = \frac{A^{2n/m}-A^{-2n/m}}{A^{2}-A^{-2}}$$

That you can try to simplify with some other rules.

reuns
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