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Thank you!

Is it possible to make sage recognize that a certain rational function is a Laurent polynomial, and treat it as such?

A simple example:

sage: R = LaurentPolynomialRing(ZZ, ['q1'])
sage: R.inject_variables(verbose=False)
sage: f = (1/(1-q1) + 1/(1-q1^-1))
sage: parent(f)
Fraction Field of Univariate Polynomial Ring in q1 over Integer Ring

Here, f lives correctly in the fraction field. However, it is in particular a polynomial, to which I'd like to apply the polynomial methods, e.g. get its monomial list.

Is this possible to achieve?

the error it gives is: fraction must have unit denominator.

Thanks, that's also what I thought, but why is it not working e.g. here? R.<q1,q2> = LaurentPolynomialRing(ZZ,2)

recognize a rational function is a polynomial Is it possible to make sage recognize that a certain rational function is

recognize a rational function is a polynomial Is it possible to make sage recognize that a certain rational function as

Thanks, although I thought symbolic computations would be slower here, right?

Thanks a lot! it seems to work well.

I'm looking for an efficient (i.e. fast enough for $k=4$ and higher) way in Sage to build large rational functions (ratio of polynomials with integer coefficients), and then take residues at simple poles (which I know by construction), by multiplication and taking limit (via subs).

I can do that in the symbolic ring, which I believe uses maxima or sympy, but at some point it becomes slow in factorization (say compared to mathematica), so I tried to work with polynomial rings, hoping that singular would be faster.

However, at present I have issues:

1. since by default it expands whatever factorized polynomial I define, then it has a hard time factorizing it back (see how long it takes even just to return $\chi$).

2. Moreover, I cannot operate with factorization objects, e.g. I need to multiply them, but it seems this is not allowed.

This is my current attempt:

def br(x):
    return 1-1/x

def measNew():
    k = 3
    R = PolynomialRing(ZZ, ['x%d' %p for p in range(k)]+['q1', 'q2', 'q3', 'q4', 'm'], order='invlex')
    R.inject_variables()
    K = 1 + q2 + q2^2 + q3 + q4
    X = R.gens()[:k]
    rho = [p.substitute({q4:(q1*q2*q3)^-1}) for p in K.monomials()]
    rho.reverse()
    chix = prod([ br(X[j]) for j in range(k)])
    chim = prod([ br(X[j]/m) for j in range(k)])
    chiup = prod([ prod([ br(q1*q2*X[i]/X[j])*br(q1*q3*X[i]/X[j])*br(q2*q3*X[i]/X[j])*(br(X[i]/X[j]))^2 for i in range(k) if i > j]) for j in range(k)])
    chiups = prod([ prod([ br(X[i]/(X[j]*q1*q2))*br(X[i]/(X[j]*q1*q3))*br(X[i]/(X[j]*q2*q3)) for i in range(k) if i > j]) for j in range(k)])
    chido = prod([ prod([ br(q1*X[i]/X[j])*br(q2*X[i]/X[j])*br(q3*X[i]/X[j])*br(q1*q2*q3*X[i]/X[j]) for i in range(k) if i > j]) for j in range (k)])
    chidos = prod([ prod([ br(X[i]/(X[j]*q1))*br(X[i]/(X[j]*q2))*br(X[i]/(X[j]*q3))*br(X[i]/(X[j]*q1*q2*q3)) for i in range(k) if i > j]) for j in range (k)])
    dx = prod([ X[j] for j in range(k)])
    chinum = (chim*chiup*chiups)
    chiden = (chix*chido*chidos*dx)
    chi = (chinum/chiden)
    # return chi # just to check the intermediate step
    for xi,rhoi in zip(X,rho):
        chi = (chi*(xi-rhoi)).factor().subs({xi: rhoi})
    return chi.factor()

Any help/suggestions?

Thanks.

Yes, this assumption is correct. Let me run a few more checks. Note that k=1 is still broken as I pointed out.

Thanks for helping me. Two comments: it seems that R = PolynomialRing(QQ, is better here, e.g. k=1 does not complain. Wh

Sorry, but this does not look right: cancellations between numerator and denominator of chi will happen along the way (i

Simultaneous substitution is not allowed here: we always compute an iterated residue, for which order matters.

I've also reversed rho, to get correct ordering.

factorization and multiplication in polynomial ring I'm looking for an efficient (i.e. fast enough for $k=4$ and higher)

You're right, that's not correct.

factorization and multiplication in polynomial ring I'm looking for an efficient (i.e. fast enough for $k=4$ and higher)

in case I did not make it clear: the order in which residues are taken is crucial

Thanks, let me study your suggestions. Is there a reason why you changed the order from invlex? the actual ordering shou

You're right. I never use X[p] with p \geq k, I just didn't know how to code it correctly. Thanks.

I wrote a new question, trying to explain what I want and what goes wrong https://ask.sagemath.org/question/61151/factor

factorization and multiplication in polynomial ring I'm looking for an efficient (i.e. fast enough for $k=4$ and higher)

the change is that what I called chi, is then used to compute a residue (see the last 3 lines), so one needs to be able

do you think using the dictionary could work for the above?

But see the edited question, as there will be cancellations.

I added a full explanation, sorry for being not complete. I'm not sure using a dictionary is the fastest way here?

fast factorization of ratios of polynomials I consider polynomials of the form $P=\prod_i (1-1/y_i)$, where each $y_i$ i

fast factorization of ratios of polynomials I consider polynomials of the form $P=\prod_i (1-1/y_i)$, where each $y_i$ i

fast factorization of ratios of polynomials I consider polynomials of the form $P=\prod_i (1-1/y_i)$, where each $y_i$ i

Sure, but let me be more precise. What I asked is only part of the story, and I'd like to understand if using Singular i

@tmonteil I added a code example, does this clarify my question?

fast factorization of ratios of polynomials I consider polynomials of the form $P=\prod_i (1-1/y_i)$, where each $y_i$ i

Well, numerator and denominator are factored separately, but I want to cancel common factors when taking the ratio, name

fast factorization of ratios of polynomials I consider polynomials of the form $P=\prod_i (1-1/y_i)$, where each $y_i$ i

fast factorization of ratios of polynomials I consider polynomials of the form $P=\prod_i (1-1/y_i)$, where each $y_i$ i

Thanks, I did see that, although in my case it just prints the sums and does not simplify them to zero.

Let $n$ be a positive integer and $m = (m_1, \ldots, m_n)$ an $n$-dimensional vector of real numbers. Let $g$ be a real number.

I want to prove, for any $n$ and $m$, an equality of the form $$ \sum_{i=1}^n f_i (m,g) = 0 $$ where the function $f_i$ is a rational function of $m$ and $g$.

Of course it's easy to check this by substituting finite values of $n$, but is there a way in Sage to prove it for any integer?