$ \newcommand{\qPochhammer}[3][\infty]{\left( #2;#3 \right)_{#1}} \newcommand{\setQ}{\mathbb{Q}} $ Via jupytext this file can be shown as a jupyter notebook.

This notebook find and checks relations for the Rogers-Ramanujan functions $G$, $H$, and the Rogers-Ramanujan continued fraction $R$.

)cd ..
)read input/jfricas-test-support.input )quiet

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All )browse facility databases have been cleared.
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)set output algebra on
)set output formatted off

Init

We start with a few functions that we will need further below.

-------------------------------------------------------------------
--setup
-------------------------------------------------------------------

C ==> QQ
)set mess type off

lauex ==> laurentExpansion $ QEQSI C
idxSPEC idx ==> if #idx=3 then eqSPEC[idx] else eqSPEC[concat(idx,1)]
rf spec ==> rationalFunction(spec :: SPEX(C))

For REL1(id) we assume that multiplier(id)=1.

REL1(id) ==> (_
  if fpol = 'fpol then fpol := 1;_
  idpol := identityPolynomial(id);_
  rpol := 'F - (inv(coefficient(idpol,'F, 1))::C) * idpol;_
  tracePrint("rpol", rpol);_
  assertEquals(rpol,fpol);_
  rspec := definingSpecification id;_
  sspec := coSpecification id;_
  mspecs := monoidSpecifications id;_
  msyms := indexedSymbols("M", #mspecs);_
  f := lauex(sspec*rspec);_
  agens := [lauex mspec for mspec in mspecs];_
  zser := eval(idpol, cons('F,msyms), cons(f,agens))_
            $ PolynomialEvaluation(C, A1 C);_
  assertTrue(zero? zser);_
  rf rspec = rf(inv sspec))

commonFactor id ==> (_
  pol := inv coefficient id * identityPolynomial id - 'F;_
  cs := [leadingCoefficient x for x in monomials pol];_
  den := lcm [denom x for x in cs];_
  num := gcd [numer(den*x) for x in cs];_
  num/den)

The following command determines how many terms of a series will be shown on output. You can change it if you want to see more.

)set stream calc 4

-------------------------------------------------------------------
--endsetup
-------------------------------------------------------------------

-------------------------------------------------------------------
--test:Rogers-Ramanujan-1-5
-------------------------------------------------------------------

Rogers-Ramanujan functions $G$, $H$, $R$

We define the Rogers-Ramanujan functions as in. \cite{FryeGarvan_AutomaticProofOfThetaFunctionIdentities_2019}

\begin{align} G(q) &= \sum_{n=0}^\infty \frac{q^{n^2}}{(q;q)_n} = \prod_{n=0}^\infty \frac{1}{(1-q^{5n+1})(1-q^{5n+4})} = \frac{1}{\qPochhammer{q,q^4}{q^5}} \\ H(q) &= \sum_{n=0}^\infty \frac{q^{n(n+1)}}{(q;q)_n} = \prod_{n=0}^\infty \frac{1}{(1-q^{5n+2})(1-q^{5n+3})} = \frac{1}{\qPochhammer{q^2,q^3}{q^5}} \end{align}

In QEta we represent $G(q)$ and $H(q)$ by two macros. Here qP stands for the (infinite) qPochhammer product.

gg(q) ==> inv(qP[q,q^4,q^5])   :: SPEX(C)
hh(q) ==> inv(qP[q^2,q^3,q^5]) :: SPEX(C)
qq ==> qPower 1

Verify Ramanujan identity

Let us first check the following identity..

\begin{align} H(q) G(q)^{11}-q^2 G(q) H(q)^{11} &= 1 + 11 q G(q)^6 H(q)^6 %\tag{1.2} \end{align}

\begin{align} \frac{1}{\eta_{5,1}(\tau)^{11} \eta_{5,2}(\tau)} - \frac{1}{\eta_{5,1}(\tau) \eta_{5,2}(\tau)^{11}} &= 1 + 11 \frac{\eta(5\tau)^6}{\eta(\tau)^6} %\tag{1.5} \end{align}

lspex := hh(q) * gg(q)^11 - qq^2 * gg(q) * hh(q)^11
rspex := 1 + 11 * qq * gg(q)^6 * hh(q)^6
spex := lspex - rspex;

\[ \frac{1}{{{\left({q}; {{q}}^{5}\right)}_{\infty }}^{11}\, {\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }\, {\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{11}}-{{q}}^{2}\, \frac{1}{{\left({q}; {{q}}^{5}\right)}_{\infty }\, {{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{11}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{11}\, {\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }} \]

\[ 11\, {q}\, \frac{1}{{{\left({q}; {{q}}^{5}\right)}_{\infty }}^{6}\, {{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{6}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{6}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{6}}+1 \]

Expressed in eta-quotients with $q=\exp(2\pi i\tau)$, it looks like this.

etaExpression spex

\[ -\frac{1}{{\eta}_{5, 1}\left(\tau\right)\, {{\eta}_{5, 2}\left(\tau\right)}^{11}}-11\, \frac{1}{{{\eta}_{5, 1}\left(\tau\right)}^{6}\, {{\eta}_{5, 2}\left(\tau\right)}^{6}}+\frac{1}{{{\eta}_{5, 1}\left(\tau\right)}^{11}\, {\eta}_{5, 2}\left(\tau\right)}-1 \]

Each of the terms and therefore the whole expression is a modular function for $\Gamma_1(5)$.

MG ==> MGAMMA1 5
assertTrue(qModular?(C,MG) spex)

\[ \mathtt{true} \]

For verifying that the above modular function is zero, it is enough to expand it at all cusps of $\Gamma_1(5)$ and show that at each cusp the respective expansion has no pole and a constant zero term.

We start with the cusps (or corresponding transformation matrices).

trfs := cuspMatrices() $ MG
spitzen := map(cusp, trfs)

\[ \left[\begin{bmatrix}0&-1\\1&0\end{bmatrix}, \begin{bmatrix}2&-1\\5&-2\end{bmatrix}, \begin{bmatrix}1&0\\2&1\end{bmatrix}, \begin{bmatrix}1&0\\0&1\end{bmatrix}\right] \]

\[ \left[0, \frac{2}{5}, \frac{1}{2}, \infty \right] \]

The expansion at cusps other than $\infty$ may require an extension of the coefficient domain. The following code determines the extension CX of the domain $C$ (rational numbers) by a certain root of unity.

xiord := minRootOfUnity(C,MG) spex
EXTENDEDCOEFFICIENTRING(C, xiord, CX, xi);

\[ 10 \]

We can determine the pole orders (qetaGrade) of the terms of spex and realize that there are poles at the cusps $0$ and $\frac{1}{2}$.

qGrades(C,MG) spex

\[ \left[\left[1, -2, 1, 0\right], \left[1, 0, 1, -2\right], \left[1, -1, 1, -1\right], \left[0, 0, 0, 0\right]\right] \]

Since the terms have poles not only at infinity, we must look at all cusps.

a := spexMAn(trfs,CX,MG)(spex)
assertTrue(zero? a)

\[ \left[O\left({{q}}^{9}\right), O\left({{q}}^{9}\right), O\left({{q}}^{9}\right), O\left({{q}}^{9}\right)\right] \]

\[ \mathtt{true} \]

If you want to see the expansions of the terms, you can achieve it like this.

mons := monomials etaExpression spex

\[ \left[-\frac{1}{{\eta}_{5, 1}\left(\tau\right)\, {{\eta}_{5, 2}\left(\tau\right)}^{11}}, -11\, \frac{1}{{{\eta}_{5, 1}\left(\tau\right)}^{6}\, {{\eta}_{5, 2}\left(\tau\right)}^{6}}, \frac{1}{{{\eta}_{5, 1}\left(\tau\right)}^{11}\, {\eta}_{5, 2}\left(\tau\right)}, -1\right] \]

as := [spexMAn(trfs,CX,MG)(m) for m in mons]; #as

\[ 4 \]

as.1
as.2
as.3
as.4

\[ \left[\left(\frac{1}{25}\, {{ξ}}^{3}-\frac{1}{25}\, {{ξ}}^{2}+\frac{3}{125}\right)\, {{q}}^{-1}+\frac{1}{5}\, {{ξ}}^{3}-\frac{1}{5}\, {{ξ}}^{2}+\frac{17}{125}+\left(\frac{11}{25}\, {{ξ}}^{3}-\frac{11}{25}\, {{ξ}}^{2}+\frac{22}{125}\right)\, {q}+\frac{11}{25}\, {{q}}^{2}-\frac{33}{25}\, {{q}}^{3}+O\left({{q}}^{4}\right), 1+11\, {q}+67\, {{q}}^{2}+298\, {{q}}^{3}+1090\, {{q}}^{4}+O\left({{q}}^{5}\right), \left(\frac{8}{125}\, {{ξ}}^{3}-\frac{3}{125}\, {{ξ}}^{2}+\frac{3}{125}\, {ξ}-\frac{8}{125}\right)\, {{q}}^{-1}-\frac{1}{5}\, {{ξ}}^{3}+\frac{1}{5}\, {{ξ}}^{2}+\frac{42}{125}+\left(-\frac{11}{25}\, {{ξ}}^{2}-\frac{22}{125}\, {ξ}-\frac{11}{25}\right)\, {q}+\frac{11}{25}\, {{ξ}}^{2}\, {{q}}^{2}+\frac{33}{25}\, {{ξ}}^{3}\, {{q}}^{3}+O\left({{q}}^{4}\right), -{{q}}^{2}-{{q}}^{3}-12\, {{q}}^{4}-23\, {{q}}^{5}-90\, {{q}}^{6}+O\left({{q}}^{7}\right)\right] \]

\[ \left[-\frac{11}{125}\, {{q}}^{-1}+\frac{66}{125}-\frac{99}{125}\, {q}-\frac{22}{25}\, {{q}}^{2}+\frac{66}{25}\, {{q}}^{3}+O\left({{q}}^{4}\right), -11\, {q}-66\, {{q}}^{2}-297\, {{q}}^{3}-1078\, {{q}}^{4}-3465\, {{q}}^{5}+O\left({{q}}^{6}\right), \left(-\frac{11}{125}\, {{ξ}}^{3}+\frac{11}{125}\, {{ξ}}^{2}-\frac{11}{125}\, {ξ}+\frac{11}{125}\right)\, {{q}}^{-1}+\frac{66}{125}+\frac{99}{125}\, {ξ}\, {q}-\frac{22}{25}\, {{ξ}}^{2}\, {{q}}^{2}-\frac{66}{25}\, {{ξ}}^{3}\, {{q}}^{3}+O\left({{q}}^{4}\right), -11\, {q}-66\, {{q}}^{2}-297\, {{q}}^{3}-1078\, {{q}}^{4}-3465\, {{q}}^{5}+O\left({{q}}^{6}\right)\right] \]

\[ \left[\left(-\frac{1}{25}\, {{ξ}}^{3}+\frac{1}{25}\, {{ξ}}^{2}+\frac{8}{125}\right)\, {{q}}^{-1}-\frac{1}{5}\, {{ξ}}^{3}+\frac{1}{5}\, {{ξ}}^{2}+\frac{42}{125}+\left(-\frac{11}{25}\, {{ξ}}^{3}+\frac{11}{25}\, {{ξ}}^{2}+\frac{77}{125}\right)\, {q}+\frac{11}{25}\, {{q}}^{2}-\frac{33}{25}\, {{q}}^{3}+O\left({{q}}^{4}\right), -{{q}}^{2}-{{q}}^{3}-12\, {{q}}^{4}-23\, {{q}}^{5}-90\, {{q}}^{6}+O\left({{q}}^{7}\right), \left(\frac{3}{125}\, {{ξ}}^{3}-\frac{8}{125}\, {{ξ}}^{2}+\frac{8}{125}\, {ξ}-\frac{3}{125}\right)\, {{q}}^{-1}+\frac{1}{5}\, {{ξ}}^{3}-\frac{1}{5}\, {{ξ}}^{2}+\frac{17}{125}+\left(\frac{11}{25}\, {{ξ}}^{2}-\frac{77}{125}\, {ξ}+\frac{11}{25}\right)\, {q}+\frac{11}{25}\, {{ξ}}^{2}\, {{q}}^{2}+\frac{33}{25}\, {{ξ}}^{3}\, {{q}}^{3}+O\left({{q}}^{4}\right), 1+11\, {q}+67\, {{q}}^{2}+298\, {{q}}^{3}+1090\, {{q}}^{4}+O\left({{q}}^{5}\right)\right] \]

\[ \left[-1+O\left({{q}}^{5}\right), -1, -1+O\left({{q}}^{5}\right), -1\right] \]

Relations between $G$ and $H$

Let us determine a (Gröbner) basis of all relations between $G$ and $H$. In other words, find an ideal $I$ in $K[x,y]$ such that $p(g,h)=0$ for \begin{align*} g(\tau) &= \frac{1}{\eta_{5,1}(\tau)} = \frac{1}{q^{\frac{1}{60}}\qPochhammer{q,q^4}{q^5}}, \\ h(\tau) &= \frac{1}{\eta_{5,2}(\tau)} = \frac{q^{\frac{11}{60}}}{\qPochhammer{q^2,q^3}{q^5}} \end{align*} where $q=\exp(2\pi i \tau)$.

We can easily find all relations between the eta-functions $\eta_{5,1}(\tau)$ and $\eta_{5,2}(\tau)$.

er := etaRelations([[5,1],[5,2]]$List(List ZZ))$QEtaIdealHemmecke(MG)

-- numOfGaps:=[0, -1]
-- numOfGaps:=[0, -1]
-- TIME:=0.0

\[ \left[{{e5\_1}}^{11}\, {{e5\_2}}^{11}-{{e5\_2}}^{10}+11\, {{e5\_1}}^{5}\, {{e5\_2}}^{5}+{{e5\_1}}^{10}\right] \]

There is only one and we can recover the Ramanujan identity from above by dividing this generator appropriately and noting that \begin{align*} \frac{1}{\eta_{5,1}(\tau)^6 \eta_{5,2}(\tau)^6} &= \frac{\eta(5\tau)^6}{\eta(\tau)^6}. \end{align*}

erel := eqSPEX(ZZ)(first(er)/(e5_1*e5_2)^11)

\[ \frac{1}{{\eta}_{5, 1}\left(\tau\right)\, {{\eta}_{5, 2}\left(\tau\right)}^{11}}+11\, \frac{1}{{{\eta}_{5, 1}\left(\tau\right)}^{6}\, {{\eta}_{5, 2}\left(\tau\right)}^{6}}-\frac{1}{{{\eta}_{5, 1}\left(\tau\right)}^{11}\, {\eta}_{5, 2}\left(\tau\right)}+1 \]

We can also do the steps in such a way that we directly get a polynomial for $g$ and $h$.

eligs := etaLaurentIdealGenerators([[5,1],[5,2]])$QEtaIdealHemmecke(MG)

-- numOfGaps:=[0, -1]
-- numOfGaps:=[0, -1]

\[ \left[{e5\_1}\, {y5\_1}-1, {e5\_2}\, {y5\_2}-1, {{e5\_2}}^{10}\, {{y5\_1}}^{10}-11\, {{e5\_2}}^{5}\, {{y5\_1}}^{5}-{e5\_1}\, {{e5\_2}}^{11}-1\right] \]

In the polynomial above y5_1 and y5_2 correspond to the inverses of e5_1 and e5_2. We replace them by g and h and eliminate the e variables by a Gröbner basis computation.

ghxideal := [eval(p,[y5_1=g,y5_2=h]) for p in eligs]

\[ \left[{e5\_1}\, {g}-1, {e5\_2}\, {h}-1, {{e5\_2}}^{10}\, {{g}}^{10}-11\, {{e5\_2}}^{5}\, {{g}}^{5}-{e5\_1}\, {{e5\_2}}^{11}-1\right] \]

ghideal:=groebnerEliminate(ghxideal, [e5_1,e5_2], [g,h])$QEtaGroebnerBasisTools

-- TIME:=0.0

\[ \left[-{g}\, {{h}}^{11}-11\, {{g}}^{6}\, {{h}}^{6}+{{g}}^{11}\, {h}-1\right] \]

Again, the ideal of all relations of $g$ and $h$ is generated by the above relation.

We can include the eta-functions of level 5 in this computation.

alleligs := etaLaurentIdealGenerators([[1],[5],[5,1],[5,2]])$QEtaIdealHemmecke(MG)

-- numOfGaps:=[0, -1]
-- numOfGaps:=[0, -1]

\[ \left[{e1}\, {y1}-1, {e5}\, {y5}-1, {e5\_1}\, {y5\_1}-1, {e5\_2}\, {y5\_2}-1, {e1}\, {y5}\, {y5\_1}\, {y5\_2}-1, {e5}\, {e5\_1}\, {e5\_2}\, {y1}-1, {{e1}}^{10}\, {{y5}}^{10}\, {{y5\_1}}^{20}+\left(-{{e1}}^{11}\, {{y5}}^{11}-11\, {{e1}}^{5}\, {{y5}}^{5}\right)\, {{y5\_1}}^{10}-1\right] \]

allghxideal := [eval(p,[y5_1=g,y5_2=h]) for p in alleligs]

\[ \left[{e1}\, {y1}-1, {e5}\, {y5}-1, {e5\_1}\, {g}-1, {e5\_2}\, {h}-1, {e1}\, {g}\, {h}\, {y5}-1, {e5}\, {e5\_1}\, {e5\_2}\, {y1}-1, -{{e1}}^{11}\, {{g}}^{10}\, {{y5}}^{11}+{{e1}}^{10}\, {{g}}^{20}\, {{y5}}^{10}-11\, {{e1}}^{5}\, {{g}}^{10}\, {{y5}}^{5}-1\right] \]

allghideal:=groebnerEliminate(allghxideal, [y1,y5,e5_1,e5_2], [e1,e5,g,h])$QEtaGroebnerBasisTools

-- TIME:=0.0

\[ \left[{e1}\, {e5}\, {{h}}^{11}+11\, {{e5}}^{2}\, {{g}}^{4}\, {{h}}^{5}+{{e1}}^{2}\, {h}-{{e5}}^{2}\, {{g}}^{9}, -{g}\, {{h}}^{11}-11\, {{g}}^{6}\, {{h}}^{6}+{{g}}^{11}\, {h}-1, -{e5}\, {{h}}^{10}-11\, {e5}\, {{g}}^{5}\, {{h}}^{5}+{e5}\, {{g}}^{10}-{e1}, {e1}\, {g}\, {h}-{e5}\right] \]

Above we have all relations between $g(\tau)$, $h(\tau)$, $\eta(\tau)$, $\eta(5\tau)$ and rediscover as the last entry in the list a relation that we have used above.

last allghideal

\[ {e1}\, {g}\, {h}-{e5} \]

Congruences for $G$ and $H$

Divisibility property of $H(q) G(q)^{11} = \frac{1}{\eta_{5;1}(\tau)^{11} \eta_{5;2}(\tau)}$

We can prove by the following witness identity that $a(5n+4) \equiv 0 \pmod{5}$ for \begin{gather*} A_1(q) = \sum_{n=0}^\infty a_1(n) q^n = H(q)G(q)^{11}. \end{gather*}

We enter the data as follows.

a1 := ("a"::Symbol)[1];
gf1 := generatingFunction((hh(q)*gg(q)^11)::QSPEC, a1)
dissect1 := choose(5,4) gf1

\[ \sum_{{n}=0}^{\infty }{{{a}}_{1}\left({n}\right)\, {{q}}^{{n}}} \]

\[ \sum_{{n}=0}^{\infty }{{{a}}_{1}\left(5\, {n}+4\right)\, {{q}}^{{n}}} \]

We easily find an identity in terms of generalized eta-functions $\eta_{5,1}(\tau)$ and $\eta_{5,2}(\tau)$.

idxs := [[5,1],[5,2]]
id1 := findId(C,MG)(dissect1,idxs);
eq1 := etaEquation id1

\[ \left[\left[5, 1\right], \left[5, 2\right]\right] \]

-- == z:=[zinhom=[[4, 2]], zhom=[], zfree=[]]
-- numOfGaps:=[0, 0]

\[ {{q}}^{\frac{4}{5}}\, {{\eta}_{5, 1}\left(\tau\right)}^{11}\, {{\eta}_{5, 2}\left(\tau\right)}^{49}\, \left(\sum_{{n}=0}^{\infty }{{{a}}_{1}\left(5\, {n}+4\right)\, {{q}}^{{n}}}\right)=90-2415\, \frac{{{\eta}_{5, 2}\left(\tau\right)}^{5}}{{{\eta}_{5, 1}\left(\tau\right)}^{5}}+25480\, \frac{{{\eta}_{5, 2}\left(\tau\right)}^{10}}{{{\eta}_{5, 1}\left(\tau\right)}^{10}}+108030\, \frac{{{\eta}_{5, 2}\left(\tau\right)}^{15}}{{{\eta}_{5, 1}\left(\tau\right)}^{15}}+685050\, \frac{{{\eta}_{5, 2}\left(\tau\right)}^{20}}{{{\eta}_{5, 1}\left(\tau\right)}^{20}}+1797095\, \frac{{{\eta}_{5, 2}\left(\tau\right)}^{25}}{{{\eta}_{5, 1}\left(\tau\right)}^{25}}+1198980\, \frac{{{\eta}_{5, 2}\left(\tau\right)}^{30}}{{{\eta}_{5, 1}\left(\tau\right)}^{30}}+110040\, \frac{{{\eta}_{5, 2}\left(\tau\right)}^{35}}{{{\eta}_{5, 1}\left(\tau\right)}^{35}}+1090\, \frac{{{\eta}_{5, 2}\left(\tau\right)}^{40}}{{{\eta}_{5, 1}\left(\tau\right)}^{40}} \]

Factoring the right-hand side, reveals divisibility by 5.

factor((numer rationalFunction rhs eq1)::Pol(ZZ))

\[ 5\, \left(218\, {{e5\_2}}^{40}+22008\, {{e5\_1}}^{5}\, {{e5\_2}}^{35}+239796\, {{e5\_1}}^{10}\, {{e5\_2}}^{30}+359419\, {{e5\_1}}^{15}\, {{e5\_2}}^{25}+137010\, {{e5\_1}}^{20}\, {{e5\_2}}^{20}+21606\, {{e5\_1}}^{25}\, {{e5\_2}}^{15}+5096\, {{e5\_1}}^{30}\, {{e5\_2}}^{10}-483\, {{e5\_1}}^{35}\, {{e5\_2}}^{5}+18\, {{e5\_1}}^{40}\right) \]

The same thing is extracted directly by the function commonFactor that is defined above.

assertEquals(commonFactor id1, 5)

\[ \mathtt{true} \]

Divisibility property of $q^2 G(q) H(q)^{11} = \frac{1}{\eta_{5;1}(\tau) \eta_{5;2}(\tau)^{11}}$

We can prove by the following witness identity that $a(5n+4) \equiv 0 \pmod{5}$ for \begin{gather*} A_2(q) = \sum_{n=0}^\infty a_2(n) q^n = G(q)H(q)^{11}. \end{gather*}

a2 := ("a"::Symbol)[2]
gf2 := generatingFunction((gg(q)*hh(q)^11)::QSPEC,a2)
dissect2 := choose(5,4) gf2

\[ {{a}}_{2} \]

\[ \sum_{{n}=0}^{\infty }{{{a}}_{2}\left({n}\right)\, {{q}}^{{n}}} \]

\[ \sum_{{n}=0}^{\infty }{{{a}}_{2}\left(5\, {n}+4\right)\, {{q}}^{{n}}} \]

id2 := findId(C,MG)(dissect2,idxs);
eq2 := etaEquation id2

-- == z:=[zinhom=[[4, 2]], zhom=[], zfree=[]]
-- numOfGaps:=[0, 0]

\[ {{q}}^{\frac{6}{5}}\, {{\eta}_{5, 1}\left(\tau\right)}^{9}\, {{\eta}_{5, 2}\left(\tau\right)}^{51}\, \left(\sum_{{n}=0}^{\infty }{{{a}}_{2}\left(5\, {n}+4\right)\, {{q}}^{{n}}}\right)=1090-110040\, \frac{{{\eta}_{5, 2}\left(\tau\right)}^{5}}{{{\eta}_{5, 1}\left(\tau\right)}^{5}}+1198980\, \frac{{{\eta}_{5, 2}\left(\tau\right)}^{10}}{{{\eta}_{5, 1}\left(\tau\right)}^{10}}-1797095\, \frac{{{\eta}_{5, 2}\left(\tau\right)}^{15}}{{{\eta}_{5, 1}\left(\tau\right)}^{15}}+685050\, \frac{{{\eta}_{5, 2}\left(\tau\right)}^{20}}{{{\eta}_{5, 1}\left(\tau\right)}^{20}}-108030\, \frac{{{\eta}_{5, 2}\left(\tau\right)}^{25}}{{{\eta}_{5, 1}\left(\tau\right)}^{25}}+25480\, \frac{{{\eta}_{5, 2}\left(\tau\right)}^{30}}{{{\eta}_{5, 1}\left(\tau\right)}^{30}}+2415\, \frac{{{\eta}_{5, 2}\left(\tau\right)}^{35}}{{{\eta}_{5, 1}\left(\tau\right)}^{35}}+90\, \frac{{{\eta}_{5, 2}\left(\tau\right)}^{40}}{{{\eta}_{5, 1}\left(\tau\right)}^{40}} \]

assertEquals(commonFactor id2, 5)

\[ \mathtt{true} \]

Curious divisibility by $5^3$

Since there is an equation involving $G(q)H(q)^{11}$ and $H(q)G(q)^{11}$, we can try the difference of the representations that we got above.

equ := (eq1 - eq2) * inv(eqSPEX(C)(e5_1^10 * e5_2^50));
qequ := map(qExpression, qPower(-1) * equ)

\[ -\frac{{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }\, {\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}{{\left({q}; {{q}}^{5}\right)}_{\infty }\, {\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}\, \left(\sum_{{n}=0}^{\infty }{{{a}}_{2}\left(5\, {n}+4\right)\, {{q}}^{{n}}}\right)+\frac{{\left({q}; {{q}}^{5}\right)}_{\infty }\, {\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}{{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }\, {\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}\, \left(\sum_{{n}=0}^{\infty }{{{a}}_{1}\left(5\, {n}+4\right)\, {{q}}^{{n}}}\right)=1000\, \frac{1}{{{\left({q}; {{q}}^{5}\right)}_{\infty }}^{50}\, {{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{10}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{10}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{50}}-1000\, {{q}}^{8}\, \frac{1}{{{\left({q}; {{q}}^{5}\right)}_{\infty }}^{10}\, {{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{50}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{50}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{10}}+107625\, {{q}}^{7}\, \frac{1}{{{\left({q}; {{q}}^{5}\right)}_{\infty }}^{15}\, {{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{45}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{45}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{15}}-1173500\, {{q}}^{6}\, \frac{1}{{{\left({q}; {{q}}^{5}\right)}_{\infty }}^{20}\, {{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{40}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{40}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{20}}+1905125\, {{q}}^{5}\, \frac{1}{{{\left({q}; {{q}}^{5}\right)}_{\infty }}^{25}\, {{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{35}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{35}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{25}}+1905125\, {{q}}^{3}\, \frac{1}{{{\left({q}; {{q}}^{5}\right)}_{\infty }}^{35}\, {{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{25}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{25}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{35}}+1173500\, {{q}}^{2}\, \frac{1}{{{\left({q}; {{q}}^{5}\right)}_{\infty }}^{40}\, {{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{20}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{20}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{40}}+107625\, {q}\, \frac{1}{{{\left({q}; {{q}}^{5}\right)}_{\infty }}^{45}\, {{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{15}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{15}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{45}} \]

In the translation of the right-hand side into a polynomial, the variables g5_i stand for $\frac{1}{\qPochhammer{q^i}{q^5}}$.

polynomial rhs qequ

\[ -1000\, {{g5\_1}}^{10}\, {{g5\_2}}^{50}\, {{g5\_3}}^{50}\, {{g5\_4}}^{10}\, {{q}}^{8}+107625\, {{g5\_1}}^{15}\, {{g5\_2}}^{45}\, {{g5\_3}}^{45}\, {{g5\_4}}^{15}\, {{q}}^{7}-1173500\, {{g5\_1}}^{20}\, {{g5\_2}}^{40}\, {{g5\_3}}^{40}\, {{g5\_4}}^{20}\, {{q}}^{6}+1905125\, {{g5\_1}}^{25}\, {{g5\_2}}^{35}\, {{g5\_3}}^{35}\, {{g5\_4}}^{25}\, {{q}}^{5}+1905125\, {{g5\_1}}^{35}\, {{g5\_2}}^{25}\, {{g5\_3}}^{25}\, {{g5\_4}}^{35}\, {{q}}^{3}+1173500\, {{g5\_1}}^{40}\, {{g5\_2}}^{20}\, {{g5\_3}}^{20}\, {{g5\_4}}^{40}\, {{q}}^{2}+107625\, {{g5\_1}}^{45}\, {{g5\_2}}^{15}\, {{g5\_3}}^{15}\, {{g5\_4}}^{45}\, {q}+1000\, {{g5\_1}}^{50}\, {{g5\_2}}^{10}\, {{g5\_3}}^{10}\, {{g5\_4}}^{50} \]

assertEquals(content((polynomial rhs qequ)::Pol(ZZ)), 5^3)

\[ \mathtt{true} \]

Written in a more symmetric form this gives \begin{align*} \frac{H(q)}{G(q)} \sum_{n=0}^\infty a_1(5n+4) q^n - \frac{G(q)}{H(q)} \sum_{n=0}^\infty a_2(5n+4) q^n &\equiv 0 \pmod{5^3} \\ \frac{\qPochhammer{q,q^4}{q^5}}{\qPochhammer{q^2,q^3}{q^5}} \sum_{n=0}^\infty a_1(5n+4) q^n - \frac{\qPochhammer{q^2,q^2}{q^5}}{\qPochhammer{q,q^4}{q^5}} \sum_{n=0}^\infty a_2(5n+4) q^n &\equiv 0 \pmod{5^3}. \end{align*}

Rogers-Ramanujan R function

Rogers-Ramanujan continued fraction. \begin{gather*} R(q) = q^{\frac15} \dfrac{1}{1+\dfrac{q}{1+\dfrac{q^2}{1+\dfrac{q^3}{1+\ddots}}}} = q^{\frac15} \frac{\qPochhammer{q,q^4}{q^5}} {\qPochhammer{q^2,q^3}{q^5}} = \frac{\eta_{5,1}(\tau)}{\eta_{5,2}(\tau)} \end{gather*}

\begin{align} \frac{1}{R(q)^5} - R(q)^5 &= 11 + \frac{\eta(\tau)^6}{\eta(5\tau)^6} %\tag{1.2} \end{align}

\begin{align} \frac{\eta_{5;2}(\tau)^5}{\eta_{5;1}(\tau)^5} - \frac{\eta_{5;1}(\tau)^5}{\eta_{5;2}(\tau)^5} &= 11 + \frac{\eta(\tau)^6}{\eta(5\tau)^6} %\tag{1.5} \end{align}

Compare with \begin{align} \frac{1}{\eta_{5;1}(\tau)^{11} \eta_{5;2}(\tau)} - \frac{1}{\eta_{5;1}(\tau) \eta_{5;2}(\tau)^{11}} &= 1 + 11 \frac{\eta(5\tau)^6}{\eta(\tau)^6} %\tag{1.5} \end{align} and note that we have \begin{align} \eta_{5;1}(\tau) \, \eta_{5;2}(\tau) &= \frac{\eta(\tau)}{\eta(5\tau)}. \end{align}

idxs := [[1],[5],[5,1]];
id := findId(C,MGAMMA1 5)(idxs, idxSPEC [5,2,-1], 1, 0, idxs);
REL1 id

-- numOfGaps:=[0, 0]
-- rpol:=1

\[ \frac{1}{{e5\_2}}=\frac{{e5}\, {e5\_1}}{{e1}} \]

Divisibility property of $\frac{\eta(5\tau)^6}{q\cdot\eta(\tau)^{6}}=\frac{\qPochhammer{q^5}{q^5}}{\qPochhammer{q}{q}}$

rqspec3 := qP(q^5)/qP(q)
s1 := lauex rqspec3

\[ \frac{{\left({{q}}^{5}\right)}_{\infty }}{{\left({q}\right)}_{\infty }} \]

\[ 1+{q}+2\, {{q}}^{2}+3\, {{q}}^{3}+5\, {{q}}^{4}+O\left({{q}}^{5}\right) \]

idxs := [[5,1],[5,2]]
id3 := findId(C,MG)(lift rqspec3,5,4,idxs);
qEquation id3

\[ \left[\left[5, 1\right], \left[5, 2\right]\right] \]

-- numOfGaps:=[0, 0]

\[ {{\left({q}; {{q}}^{5}\right)}_{\infty }}^{5}\, {{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{5}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{5}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{5}\, \left(\sum_{{n}=0}^{\infty }{a\left(5\, {n}+4\right)\, {{q}}^{{n}}}\right)=5 \]

assertEquals(commonFactor(id3), 5)

\[ \mathtt{true} \]

Divisibility for $(5n+t)$-disection ($t=1,2,3,4$) of $R(q)^5 = q \frac{\qPochhammer{q,q^4}{q^5}^5}{\qPochhammer{q^2,q^3}{q^5}^5}$ and $\frac{1}{R(q)^5}$

It is perhaps not so surprising that these disections give rise to a divisibility by 5.

t := 1
idxs := [[5,1],[5,2]]
rrspec := eqSPEC [[5,1,1],[5,2,-1]];
r5spec := rrspec^5;
gfv1 := generatingFunction(    r5spec, ("b"::Symbol)[1]);
gfv2 := generatingFunction(inv r5spec, ("b"::Symbol)[2]);

\[ 1 \]

\[ \left[\left[5, 1\right], \left[5, 2\right]\right] \]

e1 := qEquation  findId(C,MG)(choose(5,t) gfv1,idxs)

-- == z:=[zinhom=[[1, 0]], zhom=[], zfree=[]]
-- numOfGaps:=[0, 0]

\[ {{q}}^{-1}\, {{\left({q}; {{q}}^{5}\right)}_{\infty }}^{4}\, {{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{8}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{8}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{4}\, \left(\sum_{{n}=0}^{\infty }{{{b}}_{1}\left(5\, {n}+1\right)\, {{q}}^{{n}}}\right)=-5\, {{q}}^{-1}\, \frac{{{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{5}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{5}}{{{\left({q}; {{q}}^{5}\right)}_{\infty }}^{5}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{5}}+15 \]

e2 := qEquation  findId(C,MG)(choose(5,t) gfv2,idxs)

-- == z:=[zinhom=[[1, 0]], zhom=[], zfree=[]]
-- numOfGaps:=[0, 0]

\[ {{q}}^{-1}\, {{\left({q}; {{q}}^{5}\right)}_{\infty }}^{6}\, {{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{6}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{6}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{6}\, \left(\sum_{{n}=0}^{\infty }{{{b}}_{2}\left(5\, {n}+1\right)\, {{q}}^{{n}}}\right)=5\, {{q}}^{-1}\, \frac{{{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{5}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{5}}{{{\left({q}; {{q}}^{5}\right)}_{\infty }}^{5}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{5}}-40 \]

However, the sum of both is divisible by 25.

e1+e2

\[ {{q}}^{-1}\, {{\left({q}; {{q}}^{5}\right)}_{\infty }}^{4}\, {{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{8}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{8}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{4}\, \left(\sum_{{n}=0}^{\infty }{{{b}}_{1}\left(5\, {n}+1\right)\, {{q}}^{{n}}}\right)+{{q}}^{-1}\, {{\left({q}; {{q}}^{5}\right)}_{\infty }}^{6}\, {{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{6}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{6}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{6}\, \left(\sum_{{n}=0}^{\infty }{{{b}}_{2}\left(5\, {n}+1\right)\, {{q}}^{{n}}}\right)=-25 \]

assertEquals(gcd coefficients(rhs(e1+e2)::SPEX(ZZ)), 25)

\[ \mathtt{true} \]

qeqs := [_
  qEquation findId(C,MG)(choose(5,tt) gfv1,idxs) _
  + qEquation findId(C,MG)(choose(5,tt) gfv2,idxs) _
  for tt in 1..4];

-- == z:=[zinhom=[[1, 0]], zhom=[], zfree=[]]
-- numOfGaps:=[0, 0]
-- == z:=[zinhom=[[1, 0]], zhom=[], zfree=[]]
-- numOfGaps:=[0, 0]
-- == z:=[zinhom=[[1, 0]], zhom=[], zfree=[]]
-- numOfGaps:=[0, 0]
-- == z:=[zinhom=[[1, 0]], zhom=[], zfree=[]]
-- numOfGaps:=[0, 0]
-- == z:=[zinhom=[[1, 0]], zhom=[], zfree=[]]
-- numOfGaps:=[0, 0]
-- == z:=[zinhom=[[1, 0]], zhom=[], zfree=[]]
-- numOfGaps:=[0, 0]
-- == z:=[zinhom=[[1, 0]], zhom=[], zfree=[]]
-- numOfGaps:=[0, 0]
-- == z:=[zinhom=[[1, 0]], zhom=[], zfree=[]]
-- numOfGaps:=[0, 0]

assertEquals([gcd coefficients(rhs(x)::SPEX(ZZ)) for x in qeqs],_
            [25,25,25,25])

\[ \mathtt{true} \]

Check relation $e1+e2$ for $5n+1$

We can check the relation from above by expanding it at all cusps of $\Gamma_1(5)$. For that we need to extend the the rational numbers by a primitive 10-th root of unity.

t := 1
qeq1 := qEquation findId(C,MG)(choose(5,t) gfv1,idxs);
qeq2 := qEquation findId(C,MG)(choose(5,t) gfv2,idxs);
qeq :=  qeq1 + qeq2

\[ 1 \]

-- == z:=[zinhom=[[1, 0]], zhom=[], zfree=[]]
-- numOfGaps:=[0, 0]
-- == z:=[zinhom=[[1, 0]], zhom=[], zfree=[]]
-- numOfGaps:=[0, 0]

\[ {{q}}^{-1}\, {{\left({q}; {{q}}^{5}\right)}_{\infty }}^{4}\, {{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{8}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{8}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{4}\, \left(\sum_{{n}=0}^{\infty }{{{b}}_{1}\left(5\, {n}+1\right)\, {{q}}^{{n}}}\right)+{{q}}^{-1}\, {{\left({q}; {{q}}^{5}\right)}_{\infty }}^{6}\, {{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{6}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{6}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{6}\, \left(\sum_{{n}=0}^{\infty }{{{b}}_{2}\left(5\, {n}+1\right)\, {{q}}^{{n}}}\right)=-25 \]

xiord := minRootOfUnity(ZZ,MG) concat(support lhs qeq, support rhs qeq)
EXTENDEDCOEFFICIENTRING(C, xiord, CX, xi);
trfs := cuspMatrices() $ MG
)set stream calc 2

\[ 10 \]

\[ \left[\begin{bmatrix}0&-1\\1&0\end{bmatrix}, \begin{bmatrix}2&-1\\5&-2\end{bmatrix}, \begin{bmatrix}1&0\\2&1\end{bmatrix}, \begin{bmatrix}1&0\\0&1\end{bmatrix}\right] \]

al := spexMAn(trfs,CX,MG) lhs qeq
ar := spexMAn(trfs,CX,MG) rhs qeq
qetaGrades al, qetaGrades ar

\[ \left[-25+O\left({{q}}^{3}\right), -25+O\left({{q}}^{3}\right), -25+O\left({{q}}^{3}\right), -25+O\left({{q}}^{3}\right)\right] \]

\[ \left[-25+O\left({{q}}^{3}\right), -25, -25+O\left({{q}}^{3}\right), -25\right] \]

\[ \left[\left[0, 0, 0, 0\right], \left[0, 0, 0, 0\right]\right] \]

assertTrue(zero?(al + 25 * 1$An(trfs)(CX)))
assertTrue(zero?(ar + 25 * 1$An(trfs)(CX)))

\[ \mathtt{true} \]

\[ \mathtt{true} \]

Divisibility for $(5n)$-disection of $R(q)^5 = \frac{\qPochhammer{q,q^4}{q^5}^5}{\qPochhammer{q^2,q^3}{q^5}^5}$ and $\frac{1}{R(q)^5}$

We have an exception when $t=0$. There is no divisibility by 5. However, the 5-disections of $\frac{R(q)^5}{q}$ and $\frac{q}{R(q)^5}$ are equal.

t := 0;
e1 := qEquation(id1 := findId(C,MG)(choose(5,t) gfv1,idxs))
e2 := qEquation(id2 := findId(C,MG)(choose(5,t) gfv2,idxs))

-- == z:=[zinhom=[[1, 1]], zhom=[], zfree=[]]
-- numOfGaps:=[0, 0]

\[ {{q}}^{-2}\, {{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{12}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{12}\, \left(\sum_{{n}=0}^{\infty }{{{b}}_{1}\left(5\, {n}\right)\, {{q}}^{{n}}}\right)=-36\, {{q}}^{-1}\, \frac{{{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{5}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{5}}{{{\left({q}; {{q}}^{5}\right)}_{\infty }}^{5}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{5}}+{{q}}^{-2}\, \frac{{{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{10}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{10}}{{{\left({q}; {{q}}^{5}\right)}_{\infty }}^{10}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{10}}-1 \]

-- == z:=[zinhom=[[1, 1]], zhom=[], zfree=[]]
-- numOfGaps:=[0, 0]

\[ {{q}}^{-2}\, {{\left({q}; {{q}}^{5}\right)}_{\infty }}^{2}\, {{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{10}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{10}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{2}\, \left(\sum_{{n}=0}^{\infty }{{{b}}_{2}\left(5\, {n}\right)\, {{q}}^{{n}}}\right)=-36\, {{q}}^{-1}\, \frac{{{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{5}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{5}}{{{\left({q}; {{q}}^{5}\right)}_{\infty }}^{5}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{5}}+{{q}}^{-2}\, \frac{{{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{10}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{10}}{{{\left({q}; {{q}}^{5}\right)}_{\infty }}^{10}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{10}}-1 \]

assertEquals(rhs e1, rhs e2)

\[ \mathtt{true} \]

Adding both equations reveals divisibility by 2.

eq := e1+e2
reqspex := rhs eq
etaExpression reqspex

\[ {{q}}^{-2}\, {{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{12}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{12}\, \left(\sum_{{n}=0}^{\infty }{{{b}}_{1}\left(5\, {n}\right)\, {{q}}^{{n}}}\right)+{{q}}^{-2}\, {{\left({q}; {{q}}^{5}\right)}_{\infty }}^{2}\, {{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{10}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{10}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{2}\, \left(\sum_{{n}=0}^{\infty }{{{b}}_{2}\left(5\, {n}\right)\, {{q}}^{{n}}}\right)=-72\, {{q}}^{-1}\, \frac{{{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{5}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{5}}{{{\left({q}; {{q}}^{5}\right)}_{\infty }}^{5}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{5}}+2\, {{q}}^{-2}\, \frac{{{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{10}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{10}}{{{\left({q}; {{q}}^{5}\right)}_{\infty }}^{10}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{10}}-2 \]

\[ -72\, {{q}}^{-1}\, \frac{{{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{5}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{5}}{{{\left({q}; {{q}}^{5}\right)}_{\infty }}^{5}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{5}}+2\, {{q}}^{-2}\, \frac{{{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{10}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{10}}{{{\left({q}; {{q}}^{5}\right)}_{\infty }}^{10}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{10}}-2 \]

\[ -2-72\, \frac{{{\eta}_{5, 2}\left(\tau\right)}^{5}}{{{\eta}_{5, 1}\left(\tau\right)}^{5}}+2\, \frac{{{\eta}_{5, 2}\left(\tau\right)}^{10}}{{{\eta}_{5, 1}\left(\tau\right)}^{10}} \]

The above congruence modulo 2 follows trivially from the equality of e1 and e2 and their integer coefficients.

In fact, after multiplication by appropriat cofactors, for the difference of the identities for the dissection of $R(q)^5$ and $1/R(q)^5$, we get

(e1 - e2) * qq^2 * gg(q) * hh(q)^11

\[ \frac{{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }\, {\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}{{\left({q}; {{q}}^{5}\right)}_{\infty }\, {\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}\, \left(\sum_{{n}=0}^{\infty }{{{b}}_{1}\left(5\, {n}\right)\, {{q}}^{{n}}}\right)-\frac{{\left({q}; {{q}}^{5}\right)}_{\infty }\, {\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}{{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }\, {\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}\, \left(\sum_{{n}=0}^{\infty }{{{b}}_{2}\left(5\, {n}\right)\, {{q}}^{{n}}}\right)=0 \]

where

gfv1::OF = eulerProduct definingSpecification id1
gfv2::OF = eulerProduct definingSpecification id2

\[ \sum_{{n}=0}^{\infty }{{{b}}_{1}\left({n}\right)\, {{q}}^{{n}}}=\frac{{{\left({q}; {{q}}^{5}\right)}_{\infty }}^{5}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{5}}{{{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{5}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{5}} \]

\[ \sum_{{n}=0}^{\infty }{{{b}}_{2}\left({n}\right)\, {{q}}^{{n}}}=\frac{{{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{5}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{5}}{{{\left({q}; {{q}}^{5}\right)}_{\infty }}^{5}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{5}} \]

It says for \begin{gather*} r(q) = q^{-1/5} R(q) = \frac{H(q)}{G(q)} = \frac{\qPochhammer{q,q^4}{q^5}}{\qPochhammer{q^2,q^3}{q^5}} \end{gather*} that \begin{align*} \frac{1}{r(q)} \, U_5(r(q)^5) &= r(q) \, U_5\!\left(\frac{1}{r(q)^5}\right) \end{align*} or \begin{align*} \frac{G(q)}{H(q)} \, U_5\!\left(\frac{H(q)^5}{G(q)^5}\right) &= \frac{H(q)}{G(q)} \, U_5\!\left(\frac{G(q)^5}{H(q)^5}\right) \end{align*} or \begin{align*} U_5\!\left(\frac{R(q)^5}{R(q^5)}\right) &= U_5\!\left(\frac{R(q^5)}{R(q)^5}\right) \end{align*}

where $U_5$ is the operator that acts as follows: \begin{gather*} U_5\left(\sum_{n=k}^\infty a(n)q^n\right) = \sum_{n=\lceil k/5 \rceil}^\infty a(5n)q^n \end{gather*}

Check relation for $5n$

Let us extract the co-specifications from the identities above. Note that these co-factors have been chosen in such a way that

spex := lhs(e1 - e2)

\[ {{q}}^{-2}\, {{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{12}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{12}\, \left(\sum_{{n}=0}^{\infty }{{{b}}_{1}\left(5\, {n}\right)\, {{q}}^{{n}}}\right)-{{q}}^{-2}\, {{\left({q}; {{q}}^{5}\right)}_{\infty }}^{2}\, {{\left({{q}}^{2}; {{q}}^{5}\right)}_{\infty }}^{10}\, {{\left({{q}}^{3}; {{q}}^{5}\right)}_{\infty }}^{10}\, {{\left({{q}}^{4}; {{q}}^{5}\right)}_{\infty }}^{2}\, \left(\sum_{{n}=0}^{\infty }{{{b}}_{2}\left(5\, {n}\right)\, {{q}}^{{n}}}\right) \]

is an identity of modular functions for $\Gamma_1(5)$.

qModular?(C,MG) spex

\[ \mathtt{true} \]

We have to extend our coefficient domain (the rational numbers) such that the expansions at all cusps can be represented in this extended field.

xiord := minRootOfUnity(ZZ,MG) support spex
EXTENDEDCOEFFICIENTRING(C, xiord, CX, xi);

\[ 10 \]

It turns out that adjoining a primitive 10-th root of unity is enough.

aid := spexMAn(trfs,CX,MG)(spex)
assertTrue(zero? aid)

\[ \left[O\left({{q}}^{5}\right), O\left({{q}}^{5}\right), O\left({{q}}^{5}\right), O\left({{q}}^{3}\right)\right] \]

\[ \mathtt{true} \]

Expansion at infinity gives 0 up to high order

Although, the above is already enough to prove equality, we here show that the coefficients of the series for the difference vanishes up to $q^{200}$.

)set stream calc 10
rrser := eulerExpansion specEQI(C)(rrspec)
r5ser := eulerExpansion specEQI(C)(r5spec)
r55ser := choose(5,0,r5ser)

\[ 1-{q}+{{q}}^{2}-{{q}}^{4}+{{q}}^{5}-{{q}}^{6}+{{q}}^{7}-{{q}}^{9}+2\, {{q}}^{10}+O\left({{q}}^{11}\right) \]

\[ 1-5\, {q}+15\, {{q}}^{2}-30\, {{q}}^{3}+40\, {{q}}^{4}-26\, {{q}}^{5}-30\, {{q}}^{6}+125\, {{q}}^{7}-220\, {{q}}^{8}+245\, {{q}}^{9}-124\, {{q}}^{10}+O\left({{q}}^{11}\right) \]

\[ 1-26\, {q}-124\, {{q}}^{2}-550\, {{q}}^{3}-1926\, {{q}}^{4}-6074\, {{q}}^{5}-17251\, {{q}}^{6}-45924\, {{q}}^{7}-114750\, {{q}}^{8}-273626\, {{q}}^{9}-624848\, {{q}}^{10}+O\left({{q}}^{11}\right) \]

irrser := inv rrser
ir5ser := inv r5ser
ir55ser := choose(5,0,ir5ser)

\[ 1+{q}-{{q}}^{3}+{{q}}^{5}+{{q}}^{6}-{{q}}^{7}-2\, {{q}}^{8}+2\, {{q}}^{10}+O\left({{q}}^{11}\right) \]

\[ 1+5\, {q}+10\, {{q}}^{2}+5\, {{q}}^{3}-15\, {{q}}^{4}-24\, {{q}}^{5}+15\, {{q}}^{6}+70\, {{q}}^{7}+30\, {{q}}^{8}-125\, {{q}}^{9}-175\, {{q}}^{10}+O\left({{q}}^{11}\right) \]

\[ 1-24\, {q}-175\, {{q}}^{2}-826\, {{q}}^{3}-3100\, {{q}}^{4}-10174\, {{q}}^{5}-30024\, {{q}}^{6}-81926\, {{q}}^{7}-209577\, {{q}}^{8}-508800\, {{q}}^{9}-1181123\, {{q}}^{10}+O\left({{q}}^{11}\right) \]

ser := irrser*r55ser - rrser*ir55ser
cs := removeDuplicates [ser.i for i in 0..200]
assertEquals(cs,[0])

\[ O\left({{q}}^{21}\right) \]

\[ \left[0\right] \]

\[ \mathtt{true} \]

Check series expansion at all cusps

Here we check that the $q$-expansion at all cusps agree in the principal part for the following equality. \begin{gather*} U_5\left(\frac{R(q)^5}{R(q^5)}\right) = U_5\left(\frac{R(q^5)}{R(q)^5}\right) \end{gather*}

We first compute the arguments of the $U_5$ operator and check that they are modular functions for $\Gamma_1(25)$.

rrspec   := eqSPEC [[ 5,1,1],[ 5, 2,-1]]
d5rrspec := eqSPEC [[25,5,1],[25,10,-1]]
xspec := rrspec^5 / d5rrspec
ixspec := inv xspec
assertTrue(qModular?(C,MGAMMA1 25) xspec)
assertTrue(qModular?(C,MGAMMA1 25) ixspec)

\[ \frac{{\eta}_{5, 1}\left(\tau\right)}{{\eta}_{5, 2}\left(\tau\right)} \]

\[ \frac{{\eta}_{25, 5}\left(\tau\right)}{{\eta}_{25, 10}\left(\tau\right)} \]

\[ \frac{{{\eta}_{5, 1}\left(\tau\right)}^{5}\, {\eta}_{25, 10}\left(\tau\right)}{{{\eta}_{5, 2}\left(\tau\right)}^{5}\, {\eta}_{25, 5}\left(\tau\right)} \]

\[ \frac{{{\eta}_{5, 2}\left(\tau\right)}^{5}\, {\eta}_{25, 5}\left(\tau\right)}{{{\eta}_{5, 1}\left(\tau\right)}^{5}\, {\eta}_{25, 10}\left(\tau\right)} \]

\[ \mathtt{true} \]

\[ \mathtt{true} \]

We do not need a cofactor to make the 5-disection modular in $\Gamma_1(25)$.

MG25 ==> MGAMMA1 25
v  := (etaCofactorSpace(idxs, xspec,5,0)$MG25).particular
iv := (etaCofactorSpace(idxs,ixspec,5,0)$MG25).particular
assertEquals(v ::Vector(ZZ), vector [0,0])
assertEquals(iv::Vector(ZZ), vector [0,0])

\[ \left[0, 0\right] \]

\[ \left[0, 0\right] \]

\[ \mathtt{true} \]

\[ \mathtt{true} \]

Thus, we can choose 1 as a cofactor for both dissections, but the $U_5$ operator brings us from $\Gamma_1(5)$ to $\Gamma_1(25)$.

Again, we only need a primitive 10-th root of unity in the coefficient domain. But there are a lot more cusps in $\Gamma_1(25)$ than in $\Gamma_1(5)$.

spex1 := choose(5,0) generatingFunction xspec
spex2 := choose(5,0) generatingFunction ixspec
xiord := minRootOfUnity(C,MG25)([spex1,spex2])
EXTENDEDCOEFFICIENTRING(C, xiord, CX, xi);
trfs := cuspMatrices() $ MG25
spitzen := map(cusp, trfs)

\[ \sum_{{n}=0}^{\infty }{a\left(5\, {n}\right)\, {{q}}^{{n}}} \]

\[ \sum_{{n}=0}^{\infty }{a\left(5\, {n}\right)\, {{q}}^{{n}}} \]

\[ 10 \]

\[ \left[\begin{bmatrix}0&-1\\1&0\end{bmatrix}, \begin{bmatrix}2&-1\\25&-12\end{bmatrix}, \begin{bmatrix}1&0\\12&1\end{bmatrix}, \begin{bmatrix}1&0\\11&1\end{bmatrix}, \begin{bmatrix}1&0\\10&1\end{bmatrix}, \begin{bmatrix}1&0\\9&1\end{bmatrix}, \begin{bmatrix}3&-1\\25&-8\end{bmatrix}, \begin{bmatrix}1&0\\8&1\end{bmatrix}, \begin{bmatrix}1&0\\7&1\end{bmatrix}, \begin{bmatrix}4&-1\\25&-6\end{bmatrix}, \begin{bmatrix}1&0\\6&1\end{bmatrix}, \begin{bmatrix}1&0\\5&1\end{bmatrix}, \begin{bmatrix}6&-1\\25&-4\end{bmatrix}, \begin{bmatrix}1&0\\4&1\end{bmatrix}, \begin{bmatrix}7&-2\\25&-7\end{bmatrix}, \begin{bmatrix}3&-1\\10&-3\end{bmatrix}, \begin{bmatrix}8&-1\\25&-3\end{bmatrix}, \begin{bmatrix}1&0\\3&1\end{bmatrix}, \begin{bmatrix}9&-4\\25&-11\end{bmatrix}, \begin{bmatrix}2&-1\\5&-2\end{bmatrix}, \begin{bmatrix}11&-4\\25&-9\end{bmatrix}, \begin{bmatrix}12&-1\\25&-2\end{bmatrix}, \begin{bmatrix}1&0\\2&1\end{bmatrix}, \begin{bmatrix}3&1\\5&2\end{bmatrix}, \begin{bmatrix}7&2\\10&3\end{bmatrix}, \begin{bmatrix}4&-1\\5&-1\end{bmatrix}, \begin{bmatrix}9&-1\\10&-1\end{bmatrix}, \begin{bmatrix}1&0\\0&1\end{bmatrix}\right] \]

\[ \left[0, \frac{2}{25}, \frac{1}{12}, \frac{1}{11}, \frac{1}{10}, \frac{1}{9}, \frac{3}{25}, \frac{1}{8}, \frac{1}{7}, \frac{4}{25}, \frac{1}{6}, \frac{1}{5}, \frac{6}{25}, \frac{1}{4}, \frac{7}{25}, \frac{3}{10}, \frac{8}{25}, \frac{1}{3}, \frac{9}{25}, \frac{2}{5}, \frac{11}{25}, \frac{12}{25}, \frac{1}{2}, \frac{3}{5}, \frac{7}{10}, \frac{4}{5}, \frac{9}{10}, \infty \right] \]

We compute the expansions for each of $U_5\left(\frac{R(q)^5}{R(q^5)}\right)$ and $U_5\left(\frac{R(q^5)}{R(q)^5}\right)$ and see their pole orders at the above cusps below.

ax  := spexMAn(trfs,CX,MG25)(spex1);
aix := spexMAn(trfs,CX,MG25)(spex2);
[qetaGrades ax, qetaGrades aix]

\[ \left[\left[5, 0, 5, 5, 0, 5, 0, 5, 5, 0, 5, 0, 0, 5, 0, 0, 0, 5, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0\right], \left[5, 0, 5, 5, 0, 5, 0, 5, 5, 0, 5, 0, 0, 5, 0, 0, 0, 5, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0\right]\right] \]

Eventually, we check that the difference is zero at each cusp.

assertTrue(zero?(ax - aix))

\[ \mathtt{true} \]

Check Relation for $R(q)$ and $R(q^5)$

In Entry 3.2.14 of Andrews and Berndt: Ramanujan's Lost Notebook, Part I, Springer, 2005. there is a relation \begin{align*} R(q)^5 &= R(q^5)\cdot \frac{1 - 2 R(q^5) + 4 R(q^5)^2 - 3 R(q^5)^3 + R(q^5)^4} {1 + 3 R(q^5) + 4 R(q^5)^2 + 2 R(q^5)^3 + R(q^5)^4} \end{align*} which we are going to verify by multiplying by the denominator and then checking that every term is a modular function for $\Gamma_1(25)$. Then it is enough to check that the expansion at every cusp has no pole and a zero constant term. We repeat the definition: \begin{gather*} R(q) = \frac{\eta_{5,1}(\tau)}{\eta_{5,2}(\tau)} \end{gather*}

r := eqSPEC(e5_1/e5_2)::SPEX(C)
s := dilate(r, 5)
rhsnum := 1 - 2*s + 4*s^2 - 3*s^3 + s^4;
rhsden := 1 + 3*s + 4*s^2 + 2*s^3 + s^4;
spex := r^5 * rhsden - s * rhsnum;

\[ \frac{{\eta}_{5, 1}\left(\tau\right)}{{\eta}_{5, 2}\left(\tau\right)} \]

\[ \frac{{\eta}_{25, 5}\left(\tau\right)}{{\eta}_{25, 10}\left(\tau\right)} \]

Each of the terms is a modular function for $\Gamma_1(25)$.

MG ==> MGAMMA1 25
assertTrue(qModular?(C,MG)(spex))

\[ \mathtt{true} \]

For verifying that the above modular function is zero, it is enough to expand it at all cusps of $\Gamma_1(25)$ and show that each expansion has no pole and a constant zero term for each cusp.

We start with the cusps (or corresponding transformation matrices).

trfs := cuspMatrices() $ MG

\[ \left[\begin{bmatrix}0&-1\\1&0\end{bmatrix}, \begin{bmatrix}2&-1\\25&-12\end{bmatrix}, \begin{bmatrix}1&0\\12&1\end{bmatrix}, \begin{bmatrix}1&0\\11&1\end{bmatrix}, \begin{bmatrix}1&0\\10&1\end{bmatrix}, \begin{bmatrix}1&0\\9&1\end{bmatrix}, \begin{bmatrix}3&-1\\25&-8\end{bmatrix}, \begin{bmatrix}1&0\\8&1\end{bmatrix}, \begin{bmatrix}1&0\\7&1\end{bmatrix}, \begin{bmatrix}4&-1\\25&-6\end{bmatrix}, \begin{bmatrix}1&0\\6&1\end{bmatrix}, \begin{bmatrix}1&0\\5&1\end{bmatrix}, \begin{bmatrix}6&-1\\25&-4\end{bmatrix}, \begin{bmatrix}1&0\\4&1\end{bmatrix}, \begin{bmatrix}7&-2\\25&-7\end{bmatrix}, \begin{bmatrix}3&-1\\10&-3\end{bmatrix}, \begin{bmatrix}8&-1\\25&-3\end{bmatrix}, \begin{bmatrix}1&0\\3&1\end{bmatrix}, \begin{bmatrix}9&-4\\25&-11\end{bmatrix}, \begin{bmatrix}2&-1\\5&-2\end{bmatrix}, \begin{bmatrix}11&-4\\25&-9\end{bmatrix}, \begin{bmatrix}12&-1\\25&-2\end{bmatrix}, \begin{bmatrix}1&0\\2&1\end{bmatrix}, \begin{bmatrix}3&1\\5&2\end{bmatrix}, \begin{bmatrix}7&2\\10&3\end{bmatrix}, \begin{bmatrix}4&-1\\5&-1\end{bmatrix}, \begin{bmatrix}9&-1\\10&-1\end{bmatrix}, \begin{bmatrix}1&0\\0&1\end{bmatrix}\right] \]

The expansion at cusps other than $\infty$ may require an extension of the coefficient domain. The following code determines the extension of the domain $C$ (rational numbers) by a certain root of unity.

xiord := minRootOfUnity(C,MG) support spex
EXTENDEDCOEFFICIENTRING(C, xiord, CX, xi);

\[ 50 \]

We can determine the pole orders (qetaGrade) of the terms of spex and realize that there are poles at 9 of the 28 cusps.

grdss := [qGrades(C,MG) x for x in support spex]
poles := [[c for c in cusps()$MG for g in grds|g>0] for grds in grdss]
joinedpoles := removeDuplicates concat poles
[#joinedpoles, #spitzen]

\[ \left[\left[0, 5, 0, 0, 0, 0, 5, 0, 0, -5, 0, 0, -5, 0, 5, 0, 5, 0, -5, 0, -5, 5, 0, 0, 0, 0, 0, -5\right], \left[0, 5, 0, 0, -5, 0, 5, 0, 0, -5, 0, -5, -5, 0, 5, 5, 5, 0, -5, 5, -5, 5, 0, 5, 5, -5, -5, -5\right], \left[0, 4, 0, 0, 0, 0, 4, 0, 0, -4, 0, 0, -4, 0, 4, 0, 4, 0, -4, 0, -4, 4, 0, 0, 0, 0, 0, -4\right], \left[0, 4, 0, 0, -5, 0, 4, 0, 0, -4, 0, -5, -4, 0, 4, 5, 4, 0, -4, 5, -4, 4, 0, 5, 5, -5, -5, -4\right], \left[0, 3, 0, 0, 0, 0, 3, 0, 0, -3, 0, 0, -3, 0, 3, 0, 3, 0, -3, 0, -3, 3, 0, 0, 0, 0, 0, -3\right], \left[0, 3, 0, 0, -5, 0, 3, 0, 0, -3, 0, -5, -3, 0, 3, 5, 3, 0, -3, 5, -3, 3, 0, 5, 5, -5, -5, -3\right], \left[0, 2, 0, 0, 0, 0, 2, 0, 0, -2, 0, 0, -2, 0, 2, 0, 2, 0, -2, 0, -2, 2, 0, 0, 0, 0, 0, -2\right], \left[0, 2, 0, 0, -5, 0, 2, 0, 0, -2, 0, -5, -2, 0, 2, 5, 2, 0, -2, 5, -2, 2, 0, 5, 5, -5, -5, -2\right], \left[0, 1, 0, 0, 0, 0, 1, 0, 0, -1, 0, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 1, 0, 0, 0, 0, 0, -1\right], \left[0, 1, 0, 0, -5, 0, 1, 0, 0, -1, 0, -5, -1, 0, 1, 5, 1, 0, -1, 5, -1, 1, 0, 5, 5, -5, -5, -1\right]\right] \]

\[ \left[\left[\frac{2}{25}, \frac{3}{25}, \frac{7}{25}, \frac{8}{25}, \frac{12}{25}\right], \left[\frac{2}{25}, \frac{3}{25}, \frac{7}{25}, \frac{3}{10}, \frac{8}{25}, \frac{2}{5}, \frac{12}{25}, \frac{3}{5}, \frac{7}{10}\right], \left[\frac{2}{25}, \frac{3}{25}, \frac{7}{25}, \frac{8}{25}, \frac{12}{25}\right], \left[\frac{2}{25}, \frac{3}{25}, \frac{7}{25}, \frac{3}{10}, \frac{8}{25}, \frac{2}{5}, \frac{12}{25}, \frac{3}{5}, \frac{7}{10}\right], \left[\frac{2}{25}, \frac{3}{25}, \frac{7}{25}, \frac{8}{25}, \frac{12}{25}\right], \left[\frac{2}{25}, \frac{3}{25}, \frac{7}{25}, \frac{3}{10}, \frac{8}{25}, \frac{2}{5}, \frac{12}{25}, \frac{3}{5}, \frac{7}{10}\right], \left[\frac{2}{25}, \frac{3}{25}, \frac{7}{25}, \frac{8}{25}, \frac{12}{25}\right], \left[\frac{2}{25}, \frac{3}{25}, \frac{7}{25}, \frac{3}{10}, \frac{8}{25}, \frac{2}{5}, \frac{12}{25}, \frac{3}{5}, \frac{7}{10}\right], \left[\frac{2}{25}, \frac{3}{25}, \frac{7}{25}, \frac{8}{25}, \frac{12}{25}\right], \left[\frac{2}{25}, \frac{3}{25}, \frac{7}{25}, \frac{3}{10}, \frac{8}{25}, \frac{2}{5}, \frac{12}{25}, \frac{3}{5}, \frac{7}{10}\right]\right] \]

\[ \left[\frac{2}{25}, \frac{3}{25}, \frac{7}{25}, \frac{8}{25}, \frac{12}{25}, \frac{3}{10}, \frac{2}{5}, \frac{3}{5}, \frac{7}{10}\right] \]

\[ \left[9, 28\right] \]

Since the terms have poles not only at infinity, we must look at all cusps.

an := spexMAn(trfs,CX,MG)(spex)
assertTrue(zero? an)

\[ \left[O\left({{q}}^{21}\right), O\left({{q}}^{20}\right), O\left({{q}}^{21}\right), O\left({{q}}^{21}\right), O\left({{q}}^{22}\right), O\left({{q}}^{21}\right), O\left({{q}}^{20}\right), O\left({{q}}^{21}\right), O\left({{q}}^{21}\right), O\left({{q}}^{22}\right), O\left({{q}}^{21}\right), O\left({{q}}^{22}\right), O\left({{q}}^{22}\right), O\left({{q}}^{21}\right), O\left({{q}}^{20}\right), O\left({{q}}^{16}\right), O\left({{q}}^{20}\right), O\left({{q}}^{21}\right), O\left({{q}}^{22}\right), O\left({{q}}^{16}\right), O\left({{q}}^{22}\right), O\left({{q}}^{20}\right), O\left({{q}}^{21}\right), O\left({{q}}^{16}\right), O\left({{q}}^{16}\right), O\left({{q}}^{22}\right), O\left({{q}}^{22}\right), O\left({{q}}^{22}\right)\right] \]

\[ \mathtt{true} \]

Find Relation for $R(q)^5$ and $R(q^5)$

In Entry 3.2.14 of Andrews and Berndt: Ramanujan's Lost Notebook, Part I, Springer, 2005. there is a relation \begin{align*} R(q)^5 &= R(q^5)\cdot \frac{1 - 2 R(q^5) + 4 R(q^5)^2 - 3 R(q^5)^3 + R(q^5)^4} {1 + 3 R(q^5) + 4 R(q^5)^2 + 2 R(q^5)^3 + R(q^5)^4}, \end{align*} where $R(q)$ is defined by \begin{gather*} R(q) = \frac{\eta_{5,1}(\tau)}{\eta_{5,2}(\tau)}. \end{gather*} This relation involves only $R(q)^5$ and $R(q^5)$, which are both modular functions for $\Gamma_1(25)$. Let us compute the modular polynomial between the two.

r0 := eqSPEC(e5_1/e5_2)::SPEX(C);
r := r0^5
s := dilate(r0, 5)

\[ \frac{{{\eta}_{5, 1}\left(\tau\right)}^{5}}{{{\eta}_{5, 2}\left(\tau\right)}^{5}} \]

\[ \frac{{\eta}_{25, 5}\left(\tau\right)}{{\eta}_{25, 10}\left(\tau\right)} \]

qGrades(C,MG) r

\[ \left[\left[0, 1, 0, 0, -5, 0, 1, 0, 0, -1, 0, -5, -1, 0, 1, 5, 1, 0, -1, 5, -1, 1, 0, 5, 5, -5, -5, -1\right]\right] \]

gammas := cuspMatrices() $ MG;
grdss := [first qGrades(C,MG)(r), first qGrades(C,MG)(s)];

Cusps that correspond to non-poles in the series expansions for both $R(q)^5$ and $R(q^5)$ can be excluded for the computation of the modular polynomial between the two.

trfs := [gammas.i for i in 1..#grdss.1 | grdss.1.i > 0 or grdss.2.i > 0]
spitzen := map(cusp, trfs)

\[ \left[\begin{bmatrix}2&-1\\25&-12\end{bmatrix}, \begin{bmatrix}3&-1\\25&-8\end{bmatrix}, \begin{bmatrix}7&-2\\25&-7\end{bmatrix}, \begin{bmatrix}3&-1\\10&-3\end{bmatrix}, \begin{bmatrix}8&-1\\25&-3\end{bmatrix}, \begin{bmatrix}2&-1\\5&-2\end{bmatrix}, \begin{bmatrix}12&-1\\25&-2\end{bmatrix}, \begin{bmatrix}3&1\\5&2\end{bmatrix}, \begin{bmatrix}7&2\\10&3\end{bmatrix}\right] \]

\[ \left[\frac{2}{25}, \frac{3}{25}, \frac{7}{25}, \frac{3}{10}, \frac{8}{25}, \frac{2}{5}, \frac{12}{25}, \frac{3}{5}, \frac{7}{10}\right] \]

For the expansions at the remaining cusps, the coefficient ring must be extended. From the data we have, we conclude that an extension by a primitive 10-th root of unity (we call it $\xi$) is enough.

xiord := minRootOfUnity(C,MG)(r+s, trfs)
EXTENDEDCOEFFICIENTRING(C, xiord, CX, xi);

\[ 10 \]

ans := [spexMAn(trfs,CX,MG)(x) for x in [r,s]];
a1 := ans.1
a2 := ans.2

\[ \left[-{{q}}^{-1}-5-10\, {q}-5\, {{q}}^{2}+15\, {{q}}^{3}+24\, {{q}}^{4}-15\, {{q}}^{5}-70\, {{q}}^{6}-30\, {{q}}^{7}+125\, {{q}}^{8}+175\, {{q}}^{9}+O\left({{q}}^{10}\right), -{{q}}^{-1}-5-10\, {q}-5\, {{q}}^{2}+15\, {{q}}^{3}+24\, {{q}}^{4}-15\, {{q}}^{5}-70\, {{q}}^{6}-30\, {{q}}^{7}+125\, {{q}}^{8}+175\, {{q}}^{9}+O\left({{q}}^{10}\right), -{{q}}^{-1}-5-10\, {q}-5\, {{q}}^{2}+15\, {{q}}^{3}+24\, {{q}}^{4}-15\, {{q}}^{5}-70\, {{q}}^{6}-30\, {{q}}^{7}+125\, {{q}}^{8}+175\, {{q}}^{9}+O\left({{q}}^{10}\right), -{{q}}^{-5}-5-10\, {{q}}^{5}+O\left({{q}}^{6}\right), -{{q}}^{-1}-5-10\, {q}-5\, {{q}}^{2}+15\, {{q}}^{3}+24\, {{q}}^{4}-15\, {{q}}^{5}-70\, {{q}}^{6}-30\, {{q}}^{7}+125\, {{q}}^{8}+175\, {{q}}^{9}+O\left({{q}}^{10}\right), -{{q}}^{-5}-5-10\, {{q}}^{5}+O\left({{q}}^{6}\right), -{{q}}^{-1}-5-10\, {q}-5\, {{q}}^{2}+15\, {{q}}^{3}+24\, {{q}}^{4}-15\, {{q}}^{5}-70\, {{q}}^{6}-30\, {{q}}^{7}+125\, {{q}}^{8}+175\, {{q}}^{9}+O\left({{q}}^{10}\right), -{{q}}^{-5}-5-10\, {{q}}^{5}+O\left({{q}}^{6}\right), -{{q}}^{-5}-5-10\, {{q}}^{5}+O\left({{q}}^{6}\right)\right] \]

\[ \left[-{{q}}^{-1}-{{q}}^{4}+O\left({{q}}^{10}\right), -{{q}}^{-1}-{{q}}^{4}+O\left({{q}}^{10}\right), -{{q}}^{-1}-{{q}}^{4}+O\left({{q}}^{10}\right), {{ξ}}^{3}+{ξ}-1+\left({{ξ}}^{3}-{{ξ}}^{2}-3\right)\, {{q}}^{5}+\left(-3\, {{ξ}}^{3}-{{ξ}}^{2}-3\, {ξ}\right)\, {{q}}^{10}+O\left({{q}}^{11}\right), -{{q}}^{-1}-{{q}}^{4}+O\left({{q}}^{10}\right), {{ξ}}^{2}-{ξ}+\left(-{{ξ}}^{3}+{{ξ}}^{2}-2\right)\, {{q}}^{5}+\left(-3\, {{ξ}}^{2}+4\, {ξ}-3\right)\, {{q}}^{10}+O\left({{q}}^{11}\right), -{{q}}^{-1}-{{q}}^{4}+O\left({{q}}^{10}\right), -{{ξ}}^{2}+{ξ}-1+\left(-{{ξ}}^{3}+{{ξ}}^{2}-2\right)\, {{q}}^{5}+\left(-{{ξ}}^{3}+4\, {{ξ}}^{2}-4\, {ξ}+1\right)\, {{q}}^{10}+O\left({{q}}^{11}\right), -{{ξ}}^{3}-{ξ}+\left({{ξ}}^{3}-{{ξ}}^{2}-3\right)\, {{q}}^{5}+\left(4\, {{ξ}}^{3}+3\, {ξ}-3\right)\, {{q}}^{10}+O\left({{q}}^{11}\right)\right] \]

modPolynomial ==> modularPolynomial$QEtaModularEquation(CX, An(trfs)(CX))

The modular polynomial between $x=R(q)^5$ and $y=R(q^5)$ can easily be computed and agrees with the equation given above.

mp := modPolynomial([a1,a2],[x,y],[])
c0 := coefficient(mp,x,0); c1 := -coefficient(mp,x,1);
x = c0/c1
assertEquals(c0/c1, y * ((1-2*y+4*y^2-3*y^3+y^4)) / (1+3*y+4*y^2+2*y^3+y^4))

\[ {{y}}^{5}+\left(-{x}-3\right)\, {{y}}^{4}+\left(-2\, {x}+4\right)\, {{y}}^{3}+\left(-4\, {x}-2\right)\, {{y}}^{2}+\left(-3\, {x}+1\right)\, {y}-{x} \]

\[ {x}=\frac{{{y}}^{5}-3\, {{y}}^{4}+4\, {{y}}^{3}-2\, {{y}}^{2}+{y}}{{{y}}^{4}+2\, {{y}}^{3}+4\, {{y}}^{2}+3\, {y}+1} \]

\[ \mathtt{true} \]

Note that $\xi$ is a primitive 10-th root of unity.

factor c0
factor c1

\[ {y}\, \left({y}+{ξ}-1\right)\, \left({y}+{{ξ}}^{3}-1\right)\, \left({y}-{{ξ}}^{2}-1\right)\, \left({y}-{{ξ}}^{3}+{{ξ}}^{2}-{ξ}\right) \]

\[ \left({y}+{{ξ}}^{3}+{ξ}\right)\, \left({y}-{{ξ}}^{2}+{ξ}\right)\, \left({y}+{{ξ}}^{2}-{ξ}+1\right)\, \left({y}-{{ξ}}^{3}-{ξ}+1\right) \]

-------------------------------------------------------------------
--endtest
-------------------------------------------------------------------

Find Modular Equations for $R(q)$ and $R(q^n)$

Modular Equations for $R(q)$ and $R(q^{n})$, $n\not\equiv0\pmod5$

-------------------------------------------------------------------
--test:RogersRamanujanModularPolynomials
-------------------------------------------------------------------

In this section we compute the modular polynomial between $R(q)$ and $R(q^n)$ for $n$ not congruent to 0 modulo 5.

For this demonstration we choose $n=2$ and find the relation \begin{align*} \frac{R(q^2) - R(q)^2}{R(q^2) + R(q)^2} &= R(q) R(q^2)^2 \end{align*} that was already given by Ramanujan in his Lost Notebook (p. 365). See Entry 3.2.10 of Andrews and Berndt: Ramanujan's Lost Notebook, Part I, Springer, 2005.

n := 2;
r1spec := eqSPEC(e5_1/e5_2);
r2spec := dilate(r1spec, n);

Unfortunately, these eta-quotients are not modular functions for $\Gamma_1(10)$.

[qModular?(C,MGAMMA1 10) x for x in [r1spec,r2spec]]

-- Chen: Theorem 10.1(3) sspec:=1
-- Chen: Theorem 10.1(3) rspec:=subscript(eta, [5, 1])(tau)/subscript(eta, [5, 2])(tau)
-- Chen: Theorem 10.1(3) alpha:=1/5
-- Chen: Theorem 10.1(3) sspec:=1
-- Chen: Theorem 10.1(3) rspec:=subscript(eta, [10, 2])(tau)/subscript(eta, [10, 4])(tau)
-- Chen: Theorem 10.1(3) alpha:=2/5

\[ \left[\mathtt{false}, \mathtt{false}\right] \]

Thus we look for a modular equation of $R(q^5)$ and $R(q^{5n})$.

rspecs := [dilate(x,5) for x in [r1spec, r2spec]]

\[ \left[\frac{{\eta}_{25, 5}\left(\tau\right)}{{\eta}_{25, 10}\left(\tau\right)}, \frac{{\eta}_{50, 10}\left(\tau\right)}{{\eta}_{50, 20}\left(\tau\right)}\right] \]

Those are modular functions for $\Gamma_1(50)$.

MG ==> MGAMMA1 50
[qModular?(C,MG) x for x in rspecs]

\[ \left[\mathtt{true}, \mathtt{true}\right] \]

gammas := cuspMatrices() $ MG;
grdss := [qGrades(C,MG)(x) for x in rspecs];

Cusps that correspond to non-poles in the series expansions for both $R(q^5)$ and $R(q^{5n})$ can be excluded for the computation of the modular polynomial between the two.

trfs0 := [gammas.i for i in 1..#grdss.1 | grdss.1.i > 0 or grdss.2.i > 0]
#trfs0

\[ \left[\begin{bmatrix}1&0\\25&1\end{bmatrix}, \begin{bmatrix}3&1\\50&17\end{bmatrix}, \begin{bmatrix}2&-1\\25&-12\end{bmatrix}, \begin{bmatrix}3&-1\\25&-8\end{bmatrix}, \begin{bmatrix}7&-1\\50&-7\end{bmatrix}, \begin{bmatrix}4&-1\\25&-6\end{bmatrix}, \begin{bmatrix}6&-1\\25&-4\end{bmatrix}, \begin{bmatrix}13&-6\\50&-23\end{bmatrix}, \begin{bmatrix}7&-2\\25&-7\end{bmatrix}, \begin{bmatrix}8&-1\\25&-3\end{bmatrix}, \begin{bmatrix}17&1\\50&3\end{bmatrix}, \begin{bmatrix}9&-4\\25&-11\end{bmatrix}, \begin{bmatrix}11&-4\\25&-9\end{bmatrix}, \begin{bmatrix}23&-6\\50&-13\end{bmatrix}, \begin{bmatrix}12&-1\\25&-2\end{bmatrix}\right] \]

\[ 15 \]

Unfortunately, if we look at the pole orders of the remaining cusps, neither $R(q^5)$ nor $R(q^{5n})$ has positive pole order at all these cusps.

matrix [qGrades(C,MG)(x,trfs0) for x in rspecs]

\[ \begin{bmatrix}-2&1&2&2&1&-2&-2&1&2&2&1&-2&-2&1&2\\1&2&-1&-1&2&1&1&2&-1&-1&2&1&1&2&-1\end{bmatrix} \]

We realize, however, that if we start with $R(q^5)R(q^{25n})^n$ and $R(q^{25n})$, we can remove more cusps and end with an element that has proper poles at all the remaining cusps.

rrspecs := [dilate(x,5) for x in [r1spec*r2spec^n, r2spec]]

\[ \left[\frac{{\eta}_{25, 5}\left(\tau\right)\, {{\eta}_{50, 10}\left(\tau\right)}^{2}}{{\eta}_{25, 10}\left(\tau\right)\, {{\eta}_{50, 20}\left(\tau\right)}^{2}}, \frac{{\eta}_{50, 10}\left(\tau\right)}{{\eta}_{50, 20}\left(\tau\right)}\right] \]

qgrdss := [qGrades(C,MG) x for x in rrspecs];
trfs := [gammas.i for i in 1..#qgrdss.1 | qgrdss.1.i > 0 or qgrdss.2.i > 0];
#trfs

\[ 10 \]

For the expansions at the remaining cusps, the coefficient ring must be extended. From the data we have, we conclude that an extension by a 10-th root of unity (we call it $\xi$) is enough.

xiord := minRootOfUnity(C,MG)(rrspecs,trfs)
EXTENDEDCOEFFICIENTRING(C, xiord, CX, xi);
ans := [specMAn(trfs,CX,MG)(x) for x in rrspecs];
matrix [qetaGrades x for x in ans]

\[ 10 \]

\[ \begin{bmatrix}0&5&5&0&0&5&5&0&0&5\\1&2&2&1&1&2&2&1&1&2\end{bmatrix} \]

The modular polynomial between $x=R(q^5)R(q^{5n})^n$ and $y=R(q^{5n})$ can easily be computed.

modPolynomial ==> modularPolynomial$QEtaModularEquation(CX, An(trfs)(CX))
mp := modPolynomial(reverse ans,[y,x],[])

\[ \left({x}-1\right)\, {{y}}^{5}+{{x}}^{3}+{{x}}^{2} \]

Let us express that polynomial in the original functions $u=R(q^5)$ and $v=R(q^{5n})$. The first factor is a power of $y=v$ and can be ignored.

)clear prop v
fl := factorList factor eval(mp,[x=u*v^n,y=v]);
fl.1
modpol := fl.2.factor

\[ \left[{flag}=\mathtt{\texttt{"}prime\texttt{"}}, {factor}={v}, {exponent}=4\right] \]

\[ {u}\, {{v}}^{3}+{{u}}^{3}\, {{v}}^{2}-{v}+{{u}}^{2} \]

The second factor agrees with the equation of Ramanujan given above.

if n=2 then assertEquals(u*v^2*(v+u^2) - (v-u^2), modpol)

\[ \mathtt{true} \]

For $n=3$ the following should be true.

if n=3 then assertEquals((v-u^3)*(1+u*v^3)-3*u^2*v^2, modpol)

-------------------------------------------------------------------
--endtest
-------------------------------------------------------------------

Modular Equations for $R(q)$ and $R(q^{5n})$

-------------------------------------------------------------------
--test:RogersRamanujanModularPolynomials-5
-------------------------------------------------------------------

In this section we compute the modular polynomial between $R(q)$ and $R(q^{5n})$. In a previous section we found the modular polynomial between $R(q)^5$ and $R(q^5)$. This time we do not actually pretend that the modular polynomial is in $R(q)^5$ instead of $R(q)$. However, the computation delivers that $R(q)$ always comes in powers of 5.

For demonstration let us set $n=2$. We have to find a modular polynomial between the following two eta-quotients.

n := 2
r1spec := eqSPEC(e5_1/e5_2);
r2spec := dilate(r1spec, 5*n);
rspecs := [r1spec, r2spec]

\[ 2 \]

\[ \left[\frac{{\eta}_{5, 1}\left(\tau\right)}{{\eta}_{5, 2}\left(\tau\right)}, \frac{{\eta}_{50, 10}\left(\tau\right)}{{\eta}_{50, 20}\left(\tau\right)}\right] \]

However, the first quotient, i.e. $R(q)$ is not a modular function for $\Gamma_1(50)$.

assertEquals([qModular?(C,MGAMMA1 50) x for x in rspecs],[false,true])

-- Chen: Theorem 10.1(3) sspec:=1
-- Chen: Theorem 10.1(3) rspec:=subscript(eta, [5, 1])(tau)/subscript(eta, [5, 2])(tau)
-- Chen: Theorem 10.1(3) alpha:=1/5

\[ \mathtt{true} \]

So instead, we consider $R(q^5)$ and $R(q^{25n})$ as modular functions for $\Gamma_1(250)$.

rrspecs0 := [dilate(x,5) for x in rspecs]
nn := lcm [level x for x in rrspecs0]
MG ==> MGAMMA1 nn
assertEquals([qModular?(C,MG) x for x in rrspecs0],[true,true])

\[ \left[\frac{{\eta}_{25, 5}\left(\tau\right)}{{\eta}_{25, 10}\left(\tau\right)}, \frac{{\eta}_{250, 50}\left(\tau\right)}{{\eta}_{250, 100}\left(\tau\right)}\right] \]

\[ 250 \]

\[ \mathtt{true} \]

Let us compute the pole orders (qetaGrades) of the respective $q$-expansions at the cusps without computing the actual series.

gammas := cuspMatrices() $ MG;
qgrdss0 := [qGrades(C,MG) x for x in rrspecs0]

\[ \left[\left[0, -2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, -10, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 10, 0, 0, 0, 1, 0, 0, 5, 0, 0, 0, 0, -2, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, -10, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, -2, 0, 1, 0, -10, 0, 0, 2, 0, 0, 0, 5, 0, 0, 0, 2, 0, 1, 5, 0, -2, 0, 0, -1, 0, 0, 0, 10, 0, 0, -1, 0, 0, -2, 0, -5, 0, 1, 10, 2, 0, 2, 10, 1, 0, -5, 0, -2, 0, 0, -1, 0, 10, -1, 0, -2, 0, 5, 1, 0, 2, 5, 0, 2, 0, -10, 1, 0, -2, -1, 0, -10, -1, 0, -2, 5, 1, 10, 0, 2, 0, -5, 2, -10, 1, -5, -2, -1, 0, -1, -2, -5, 0, 1, -10, 2, 0, -5, 2, 0, 10, 1, 5, -2, 0, -1, -10, 0, -1, -2, 0, 1, -10, 2, 5, 2, 0, 1, 5, -2, -1, 10, -1, -2, -5, 1, 10, 2, 0, 2, 10, 1, -5, -2, -1, 0, 10, -1, -2, 5, 1, 0, 2, 5, 2, 1, 0, -2, -1, -10, -1, -2, 5, 1, 10, 2, -5, 2, 1, -5, -2, -1, 0, -1, -2, -5, 1, -10, 2, -5, 2, 1, 5, -2, 0, -1, -10, -1, -2, 0, 1, -10, 2, 5, 2, 1, 5, -2, -1, 10, -1, -2, -5, 1, 10, 2, 0, 10, 0, 5, -10, 10, -5, 0, -10, -5, 0, -10, -10, 5, 10, 0, 10, 5, -10, -5, 0, -5, -10, 5, 10, 0, 10, 5, -10, -5, -1\right], \left[0, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 0, 0, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 0, -10, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -10, 0, 0, 0, 0, 0, 5, 0, 10, 0, 0, 0, 0, -5, 0, 0, 0, 0, 0, 0, 0, -5, 0, 10, 0, 0, 5, 0, 0, -10, 0, 0, 0, 0, 0, 0, -10, 0, 0, 5, 0, 0, 0, 10, 0, -5, 0, -5, 0, 10, 0, 0, 0, 5, 0, 0, -10, 0, 0, -10, 0, 5, 0, 0, 10, 0, -5, 0, 0, -5, 0, 0, 10, 0, 5, -10, 0, 0, -10, 0, 5, 0, 10, 0, 0, -5, 0, 0, -5, 0, 10, 0, 5, -10, 0, -10, 5, 0, 0, 10, 0, -5, 0, 0, -5, 0, 0, 10, 0, 5, 0, -10, 0, 0, -10, 5, 0, 10, 0, -5, 0, -5, 0, 10, 0, 5, -10, 0, -10, 5, 0, 10, 0, -5, 0, -5, 0, 10, 0, 5, -10, 0, 0, -10, 5, 0, 10, 0, -5, 0, -5, 10, 0, 5, -10, 0, -10, 5, 0, 10, 0, -5, 0, -5, 10, 0, 5, -10, 0, -10, 5, 0, 10, 0, -5, 0, -5, 10, 0, 5, 0, -10, 0, -10, 5, 0, 10, 0, -5, 0, -5, 10, 0, 5, -10, 0, -10, 5, 0, 10, 0, -5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -10\right]\right] \]

Cusps that correspond to non-poles in the series expansions for both $R(q^5)$ and $R(q^{25k})$ can be excluded for the computation of the modular polynomial between the two.

trfs0 := [gammas.i for i in 1..#qgrdss0.1 | qgrdss0.1.i > 0 or qgrdss0.2.i > 0]
#trfs0

\[ \left[\begin{bmatrix}1&0\\125&1\end{bmatrix}, \begin{bmatrix}3&-1\\250&-83\end{bmatrix}, \begin{bmatrix}2&-1\\125&-62\end{bmatrix}, \begin{bmatrix}3&1\\125&42\end{bmatrix}, \begin{bmatrix}2&-1\\75&-37\end{bmatrix}, \begin{bmatrix}7&-3\\250&-107\end{bmatrix}, \begin{bmatrix}3&-1\\100&-33\end{bmatrix}, \begin{bmatrix}4&-1\\125&-31\end{bmatrix}, \begin{bmatrix}6&1\\125&21\end{bmatrix}, \begin{bmatrix}13&4\\250&77\end{bmatrix}, \begin{bmatrix}7&1\\125&18\end{bmatrix}, \begin{bmatrix}3&1\\50&17\end{bmatrix}, \begin{bmatrix}8&3\\125&47\end{bmatrix}, \begin{bmatrix}17&7\\250&103\end{bmatrix}, \begin{bmatrix}7&3\\100&43\end{bmatrix}, \begin{bmatrix}9&1\\125&14\end{bmatrix}, \begin{bmatrix}2&-1\\25&-12\end{bmatrix}, \begin{bmatrix}11&-3\\125&-34\end{bmatrix}, \begin{bmatrix}23&8\\250&87\end{bmatrix}, \begin{bmatrix}7&-3\\75&-32\end{bmatrix}, \begin{bmatrix}12&-5\\125&-52\end{bmatrix}, \begin{bmatrix}13&-5\\125&-48\end{bmatrix}, \begin{bmatrix}8&-3\\75&-28\end{bmatrix}, \begin{bmatrix}27&-4\\250&-37\end{bmatrix}, \begin{bmatrix}14&1\\125&9\end{bmatrix}, \begin{bmatrix}3&-1\\25&-8\end{bmatrix}, \begin{bmatrix}16&-5\\125&-39\end{bmatrix}, \begin{bmatrix}13&-3\\100&-23\end{bmatrix}, \begin{bmatrix}33&-7\\250&-53\end{bmatrix}, \begin{bmatrix}17&-3\\125&-22\end{bmatrix}, \begin{bmatrix}7&-1\\50&-7\end{bmatrix}, \begin{bmatrix}18&1\\125&7\end{bmatrix}, \begin{bmatrix}37&-4\\250&-27\end{bmatrix}, \begin{bmatrix}19&-7\\125&-46\end{bmatrix}, \begin{bmatrix}21&1\\125&6\end{bmatrix}, \begin{bmatrix}17&-8\\100&-47\end{bmatrix}, \begin{bmatrix}43&-16\\250&-93\end{bmatrix}, \begin{bmatrix}13&-4\\75&-23\end{bmatrix}, \begin{bmatrix}22&-3\\125&-17\end{bmatrix}, \begin{bmatrix}23&-7\\125&-38\end{bmatrix}, \begin{bmatrix}47&-22\\250&-117\end{bmatrix}, \begin{bmatrix}24&-5\\125&-26\end{bmatrix}, \begin{bmatrix}26&-5\\125&-24\end{bmatrix}, \begin{bmatrix}53&-7\\250&-33\end{bmatrix}, \begin{bmatrix}27&-8\\125&-37\end{bmatrix}, \begin{bmatrix}28&-13\\125&-58\end{bmatrix}, \begin{bmatrix}17&-5\\75&-22\end{bmatrix}, \begin{bmatrix}57&-13\\250&-57\end{bmatrix}, \begin{bmatrix}23&-3\\100&-13\end{bmatrix}, \begin{bmatrix}29&-13\\125&-56\end{bmatrix}, \begin{bmatrix}31&-1\\125&-4\end{bmatrix}, \begin{bmatrix}63&-31\\250&-123\end{bmatrix}, \begin{bmatrix}32&11\\125&43\end{bmatrix}, \begin{bmatrix}13&-6\\50&-23\end{bmatrix}, \begin{bmatrix}33&-14\\125&-53\end{bmatrix}, \begin{bmatrix}67&-26\\250&-97\end{bmatrix}, \begin{bmatrix}27&-10\\100&-37\end{bmatrix}, \begin{bmatrix}34&-3\\125&-11\end{bmatrix}, \begin{bmatrix}7&-2\\25&-7\end{bmatrix}, \begin{bmatrix}36&-17\\125&-59\end{bmatrix}, \begin{bmatrix}73&-33\\250&-113\end{bmatrix}, \begin{bmatrix}22&-5\\75&-17\end{bmatrix}, \begin{bmatrix}37&-8\\125&-27\end{bmatrix}, \begin{bmatrix}38&-7\\125&-23\end{bmatrix}, \begin{bmatrix}23&-4\\75&-13\end{bmatrix}, \begin{bmatrix}77&4\\250&13\end{bmatrix}, \begin{bmatrix}39&-5\\125&-16\end{bmatrix}, \begin{bmatrix}8&-1\\25&-3\end{bmatrix}, \begin{bmatrix}41&20\\125&61\end{bmatrix}, \begin{bmatrix}33&-1\\100&-3\end{bmatrix}, \begin{bmatrix}83&-1\\250&-3\end{bmatrix}, \begin{bmatrix}42&1\\125&3\end{bmatrix}, \begin{bmatrix}17&1\\50&3\end{bmatrix}, \begin{bmatrix}43&11\\125&32\end{bmatrix}, \begin{bmatrix}87&8\\250&23\end{bmatrix}, \begin{bmatrix}44&19\\125&54\end{bmatrix}, \begin{bmatrix}46&-7\\125&-19\end{bmatrix}, \begin{bmatrix}37&-10\\100&-27\end{bmatrix}, \begin{bmatrix}93&-16\\250&-43\end{bmatrix}, \begin{bmatrix}28&-3\\75&-8\end{bmatrix}, \begin{bmatrix}47&3\\125&8\end{bmatrix}, \begin{bmatrix}48&-5\\125&-13\end{bmatrix}, \begin{bmatrix}97&-26\\250&-67\end{bmatrix}, \begin{bmatrix}49&-20\\125&-51\end{bmatrix}, \begin{bmatrix}51&-20\\125&-49\end{bmatrix}, \begin{bmatrix}103&7\\250&17\end{bmatrix}, \begin{bmatrix}52&-5\\125&-12\end{bmatrix}, \begin{bmatrix}53&-14\\125&-33\end{bmatrix}, \begin{bmatrix}107&-3\\250&-7\end{bmatrix}, \begin{bmatrix}43&3\\100&7\end{bmatrix}, \begin{bmatrix}54&19\\125&44\end{bmatrix}, \begin{bmatrix}56&-13\\125&-29\end{bmatrix}, \begin{bmatrix}113&-33\\250&-73\end{bmatrix}, \begin{bmatrix}57&-26\\125&-57\end{bmatrix}, \begin{bmatrix}23&-6\\50&-13\end{bmatrix}, \begin{bmatrix}58&-13\\125&-28\end{bmatrix}, \begin{bmatrix}117&-22\\250&-47\end{bmatrix}, \begin{bmatrix}47&-8\\100&-17\end{bmatrix}, \begin{bmatrix}59&-17\\125&-36\end{bmatrix}, \begin{bmatrix}12&-1\\25&-2\end{bmatrix}, \begin{bmatrix}61&20\\125&41\end{bmatrix}, \begin{bmatrix}123&-31\\250&-63\end{bmatrix}, \begin{bmatrix}37&-1\\75&-2\end{bmatrix}, \begin{bmatrix}62&-1\\125&-2\end{bmatrix}, \begin{bmatrix}13&1\\25&2\end{bmatrix}, \begin{bmatrix}27&7\\50&13\end{bmatrix}, \begin{bmatrix}43&4\\75&7\end{bmatrix}, \begin{bmatrix}33&-2\\50&-3\end{bmatrix}, \begin{bmatrix}17&2\\25&3\end{bmatrix}, \begin{bmatrix}18&5\\25&7\end{bmatrix}, \begin{bmatrix}37&17\\50&23\end{bmatrix}, \begin{bmatrix}43&6\\50&7\end{bmatrix}, \begin{bmatrix}22&7\\25&8\end{bmatrix}, \begin{bmatrix}23&11\\25&12\end{bmatrix}, \begin{bmatrix}47&-16\\50&-17\end{bmatrix}\right] \]

\[ 115 \]

Unfortunately, if we look at the pole orders of the remaining cusps, neither $R(q^5)$ nor $R(q^{25n})$ has positive pole order at all these cusps.

rrspecs0

\[ \left[\frac{{\eta}_{25, 5}\left(\tau\right)}{{\eta}_{25, 10}\left(\tau\right)}, \frac{{\eta}_{250, 50}\left(\tau\right)}{{\eta}_{250, 100}\left(\tau\right)}\right] \]

qgrdss0 := [qGrades(C,MG)(x,trfs0) for x in rrspecs0]

\[ \left[\left[-2, 1, 2, 2, 10, 1, 5, -2, -2, 1, 2, 5, 2, 1, 5, -2, 10, -2, 1, 10, 2, 2, 10, 1, -2, 10, -2, 5, 1, 2, 5, 2, 1, -2, -2, 5, 1, 10, 2, 2, 1, -2, -2, 1, 2, 2, 10, 1, 5, -2, -2, 1, 2, 5, 2, 1, 5, -2, 10, -2, 1, 10, 2, 2, 10, 1, -2, 10, -2, 5, 1, 2, 5, 2, 1, -2, -2, 5, 1, 10, 2, 2, 1, -2, -2, 1, 2, 2, 1, 5, -2, -2, 1, 2, 5, 2, 1, 5, -2, 10, -2, 1, 10, 2, 10, 5, 10, 5, 10, 10, 5, 5, 10, 10, 5\right], \left[5, 10, -5, -5, 0, 10, 0, 5, 5, 10, -5, 0, -5, 10, 0, 5, 0, 5, 10, 0, -5, -5, 0, 10, 5, 0, 5, 0, 10, -5, 0, -5, 10, 5, 5, 0, 10, 0, -5, -5, 10, 5, 5, 10, -5, -5, 0, 10, 0, 5, 5, 10, -5, 0, -5, 10, 0, 5, 0, 5, 10, 0, -5, -5, 0, 10, 5, 0, 5, 0, 10, -5, 0, -5, 10, 5, 5, 0, 10, 0, -5, -5, 10, 5, 5, 10, -5, -5, 10, 0, 5, 5, 10, -5, 0, -5, 10, 0, 5, 0, 5, 10, 0, -5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0\right]\right] \]

We realize, however, that if we start with $R(q^5)R(q^{25n})$ and $R(q^{25n})$, we can remove more cusps and end with an element that has proper poles at all the remaining cusps.

rrspecs := [rrspecs0.1*rrspecs0.2, rrspecs0.2]

\[ \left[\frac{{\eta}_{25, 5}\left(\tau\right)\, {\eta}_{250, 50}\left(\tau\right)}{{\eta}_{25, 10}\left(\tau\right)\, {\eta}_{250, 100}\left(\tau\right)}, \frac{{\eta}_{250, 50}\left(\tau\right)}{{\eta}_{250, 100}\left(\tau\right)}\right] \]

qgrdss := [qGrades(C,MG)(x) for x in rrspecs]
trfs := [gammas.i for i in 1..#qgrdss.1 | qgrdss.1.i > 0 or qgrdss.2.i > 0]
#trfs
qgrdss := [qGrades(C,MG)(x,trfs) for x in rrspecs]

\[ \left[\left[0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 11, 0, 0, 0, 0, 0, 0, 0, 0, -10, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -3, 0, 0, 0, 0, 0, 10, 0, 0, 0, 11, 0, 0, 5, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, -11, 0, 0, 0, 0, 0, -10, 0, 0, 0, 0, -11, 0, 0, 0, 0, 0, 3, 0, 11, 0, -10, 0, 0, -3, 0, 0, 0, 5, 0, 0, 0, -3, 0, 11, 5, 0, 3, 0, 0, -11, 0, 0, 0, 10, 0, 0, -11, 0, 0, 3, 0, -5, 0, 11, 10, -3, 0, -3, 10, 11, 0, -5, 0, 3, 0, 0, -11, 0, 10, -11, 0, 3, 0, 5, 11, 0, -3, 5, 0, -3, 0, -10, 11, 0, 3, -11, 0, -10, -11, 0, 3, 5, 11, 10, 0, -3, 0, -5, -3, -10, 11, -5, 3, -11, 0, -11, 3, -5, 0, 11, -10, -3, 0, -5, -3, 0, 10, 11, 5, 3, 0, -11, -10, 0, -11, 3, 0, 11, -10, -3, 5, -3, 0, 11, 5, 3, -11, 10, -11, 3, -5, 11, 10, -3, 0, -3, 10, 11, -5, 3, -11, 0, 10, -11, 3, 5, 11, 0, -3, 5, -3, 11, 0, 3, -11, -10, -11, 3, 5, 11, 10, -3, -5, -3, 11, -5, 3, -11, 0, -11, 3, -5, 11, -10, -3, -5, -3, 11, 5, 3, 0, -11, -10, -11, 3, 0, 11, -10, -3, 5, -3, 11, 5, 3, -11, 10, -11, 3, -5, 11, 10, -3, 0, 10, 0, 5, -10, 10, -5, 0, -10, -5, 0, -10, -10, 5, 10, 0, 10, 5, -10, -5, 0, -5, -10, 5, 10, 0, 10, 5, -10, -5, -11\right], \left[0, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 0, 0, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 0, -10, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -10, 0, 0, 0, 0, 0, 5, 0, 10, 0, 0, 0, 0, -5, 0, 0, 0, 0, 0, 0, 0, -5, 0, 10, 0, 0, 5, 0, 0, -10, 0, 0, 0, 0, 0, 0, -10, 0, 0, 5, 0, 0, 0, 10, 0, -5, 0, -5, 0, 10, 0, 0, 0, 5, 0, 0, -10, 0, 0, -10, 0, 5, 0, 0, 10, 0, -5, 0, 0, -5, 0, 0, 10, 0, 5, -10, 0, 0, -10, 0, 5, 0, 10, 0, 0, -5, 0, 0, -5, 0, 10, 0, 5, -10, 0, -10, 5, 0, 0, 10, 0, -5, 0, 0, -5, 0, 0, 10, 0, 5, 0, -10, 0, 0, -10, 5, 0, 10, 0, -5, 0, -5, 0, 10, 0, 5, -10, 0, -10, 5, 0, 10, 0, -5, 0, -5, 0, 10, 0, 5, -10, 0, 0, -10, 5, 0, 10, 0, -5, 0, -5, 10, 0, 5, -10, 0, -10, 5, 0, 10, 0, -5, 0, -5, 10, 0, 5, -10, 0, -10, 5, 0, 10, 0, -5, 0, -5, 10, 0, 5, 0, -10, 0, -10, 5, 0, 10, 0, -5, 0, -5, 10, 0, 5, -10, 0, -10, 5, 0, 10, 0, -5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -10\right]\right] \]

\[ \left[\begin{bmatrix}1&0\\125&1\end{bmatrix}, \begin{bmatrix}3&-1\\250&-83\end{bmatrix}, \begin{bmatrix}2&-1\\75&-37\end{bmatrix}, \begin{bmatrix}7&-3\\250&-107\end{bmatrix}, \begin{bmatrix}3&-1\\100&-33\end{bmatrix}, \begin{bmatrix}4&-1\\125&-31\end{bmatrix}, \begin{bmatrix}6&1\\125&21\end{bmatrix}, \begin{bmatrix}13&4\\250&77\end{bmatrix}, \begin{bmatrix}3&1\\50&17\end{bmatrix}, \begin{bmatrix}17&7\\250&103\end{bmatrix}, \begin{bmatrix}7&3\\100&43\end{bmatrix}, \begin{bmatrix}9&1\\125&14\end{bmatrix}, \begin{bmatrix}2&-1\\25&-12\end{bmatrix}, \begin{bmatrix}11&-3\\125&-34\end{bmatrix}, \begin{bmatrix}23&8\\250&87\end{bmatrix}, \begin{bmatrix}7&-3\\75&-32\end{bmatrix}, \begin{bmatrix}8&-3\\75&-28\end{bmatrix}, \begin{bmatrix}27&-4\\250&-37\end{bmatrix}, \begin{bmatrix}14&1\\125&9\end{bmatrix}, \begin{bmatrix}3&-1\\25&-8\end{bmatrix}, \begin{bmatrix}16&-5\\125&-39\end{bmatrix}, \begin{bmatrix}13&-3\\100&-23\end{bmatrix}, \begin{bmatrix}33&-7\\250&-53\end{bmatrix}, \begin{bmatrix}7&-1\\50&-7\end{bmatrix}, \begin{bmatrix}37&-4\\250&-27\end{bmatrix}, \begin{bmatrix}19&-7\\125&-46\end{bmatrix}, \begin{bmatrix}21&1\\125&6\end{bmatrix}, \begin{bmatrix}17&-8\\100&-47\end{bmatrix}, \begin{bmatrix}43&-16\\250&-93\end{bmatrix}, \begin{bmatrix}13&-4\\75&-23\end{bmatrix}, \begin{bmatrix}47&-22\\250&-117\end{bmatrix}, \begin{bmatrix}24&-5\\125&-26\end{bmatrix}, \begin{bmatrix}26&-5\\125&-24\end{bmatrix}, \begin{bmatrix}53&-7\\250&-33\end{bmatrix}, \begin{bmatrix}17&-5\\75&-22\end{bmatrix}, \begin{bmatrix}57&-13\\250&-57\end{bmatrix}, \begin{bmatrix}23&-3\\100&-13\end{bmatrix}, \begin{bmatrix}29&-13\\125&-56\end{bmatrix}, \begin{bmatrix}31&-1\\125&-4\end{bmatrix}, \begin{bmatrix}63&-31\\250&-123\end{bmatrix}, \begin{bmatrix}13&-6\\50&-23\end{bmatrix}, \begin{bmatrix}67&-26\\250&-97\end{bmatrix}, \begin{bmatrix}27&-10\\100&-37\end{bmatrix}, \begin{bmatrix}34&-3\\125&-11\end{bmatrix}, \begin{bmatrix}7&-2\\25&-7\end{bmatrix}, \begin{bmatrix}36&-17\\125&-59\end{bmatrix}, \begin{bmatrix}73&-33\\250&-113\end{bmatrix}, \begin{bmatrix}22&-5\\75&-17\end{bmatrix}, \begin{bmatrix}23&-4\\75&-13\end{bmatrix}, \begin{bmatrix}77&4\\250&13\end{bmatrix}, \begin{bmatrix}39&-5\\125&-16\end{bmatrix}, \begin{bmatrix}8&-1\\25&-3\end{bmatrix}, \begin{bmatrix}41&20\\125&61\end{bmatrix}, \begin{bmatrix}33&-1\\100&-3\end{bmatrix}, \begin{bmatrix}83&-1\\250&-3\end{bmatrix}, \begin{bmatrix}17&1\\50&3\end{bmatrix}, \begin{bmatrix}87&8\\250&23\end{bmatrix}, \begin{bmatrix}44&19\\125&54\end{bmatrix}, \begin{bmatrix}46&-7\\125&-19\end{bmatrix}, \begin{bmatrix}37&-10\\100&-27\end{bmatrix}, \begin{bmatrix}93&-16\\250&-43\end{bmatrix}, \begin{bmatrix}28&-3\\75&-8\end{bmatrix}, \begin{bmatrix}97&-26\\250&-67\end{bmatrix}, \begin{bmatrix}49&-20\\125&-51\end{bmatrix}, \begin{bmatrix}51&-20\\125&-49\end{bmatrix}, \begin{bmatrix}103&7\\250&17\end{bmatrix}, \begin{bmatrix}107&-3\\250&-7\end{bmatrix}, \begin{bmatrix}43&3\\100&7\end{bmatrix}, \begin{bmatrix}54&19\\125&44\end{bmatrix}, \begin{bmatrix}56&-13\\125&-29\end{bmatrix}, \begin{bmatrix}113&-33\\250&-73\end{bmatrix}, \begin{bmatrix}23&-6\\50&-13\end{bmatrix}, \begin{bmatrix}117&-22\\250&-47\end{bmatrix}, \begin{bmatrix}47&-8\\100&-17\end{bmatrix}, \begin{bmatrix}59&-17\\125&-36\end{bmatrix}, \begin{bmatrix}12&-1\\25&-2\end{bmatrix}, \begin{bmatrix}61&20\\125&41\end{bmatrix}, \begin{bmatrix}123&-31\\250&-63\end{bmatrix}, \begin{bmatrix}37&-1\\75&-2\end{bmatrix}, \begin{bmatrix}13&1\\25&2\end{bmatrix}, \begin{bmatrix}27&7\\50&13\end{bmatrix}, \begin{bmatrix}43&4\\75&7\end{bmatrix}, \begin{bmatrix}33&-2\\50&-3\end{bmatrix}, \begin{bmatrix}17&2\\25&3\end{bmatrix}, \begin{bmatrix}18&5\\25&7\end{bmatrix}, \begin{bmatrix}37&17\\50&23\end{bmatrix}, \begin{bmatrix}43&6\\50&7\end{bmatrix}, \begin{bmatrix}22&7\\25&8\end{bmatrix}, \begin{bmatrix}23&11\\25&12\end{bmatrix}, \begin{bmatrix}47&-16\\50&-17\end{bmatrix}\right] \]

\[ 90 \]

\[ \left[\left[3, 11, 10, 11, 5, 3, 3, 11, 5, 11, 5, 3, 10, 3, 11, 10, 10, 11, 3, 10, 3, 5, 11, 5, 11, 3, 3, 5, 11, 10, 11, 3, 3, 11, 10, 11, 5, 3, 3, 11, 5, 11, 5, 3, 10, 3, 11, 10, 10, 11, 3, 10, 3, 5, 11, 5, 11, 3, 3, 5, 11, 10, 11, 3, 3, 11, 11, 5, 3, 3, 11, 5, 11, 5, 3, 10, 3, 11, 10, 10, 5, 10, 5, 10, 10, 5, 5, 10, 10, 5\right], \left[5, 10, 0, 10, 0, 5, 5, 10, 0, 10, 0, 5, 0, 5, 10, 0, 0, 10, 5, 0, 5, 0, 10, 0, 10, 5, 5, 0, 10, 0, 10, 5, 5, 10, 0, 10, 0, 5, 5, 10, 0, 10, 0, 5, 0, 5, 10, 0, 0, 10, 5, 0, 5, 0, 10, 0, 10, 5, 5, 0, 10, 0, 10, 5, 5, 10, 10, 0, 5, 5, 10, 0, 10, 0, 5, 0, 5, 10, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0\right]\right] \]

For the expansions at the remaining cusps, the coefficient ring must be extended. From the data we have, we conclude that an extension by a 10-th root of unity (we call it $\xi$) is enough.

xiord := minRootOfUnity(C,MG)(rrspecs, trfs)
EXTENDEDCOEFFICIENTRING(C, xiord, CX, xi);

\[ 10 \]

ans := [specMAn(trfs,CX,MG)(x) for x in rrspecs];
matrix [qetaGrades x for x in ans]

\[ \begin{bmatrix}3&11&10&11&5&3&3&11&5&11&5&3&10&3&11&10&10&11&3&10&3&5&11&5&11&3&3&5&11&10&11&3&3&11&10&11&5&3&3&11&5&11&5&3&10&3&11&10&10&11&3&10&3&5&11&5&11&3&3&5&11&10&11&3&3&11&11&5&3&3&11&5&11&5&3&10&3&11&10&10&5&10&5&10&10&5&5&10&10&5\\5&10&0&10&0&5&5&10&0&10&0&5&0&5&10&0&0&10&5&0&5&0&10&0&10&5&5&0&10&0&10&5&5&10&0&10&0&5&5&10&0&10&0&5&0&5&10&0&0&10&5&0&5&0&10&0&10&5&5&0&10&0&10&5&5&10&10&0&5&5&10&0&10&0&5&0&5&10&0&0&0&0&0&0&0&0&0&0&0&0\end{bmatrix} \]

The modular polynomial between $x=R(q^5)R(q^{25n})$ and $y=R(q^{25n})$ can easily be computed and agrees with a result computed by Michael Trott in Modular equations of the Rogers-Ramanujan continued fraction, Mathematica Journal, 2004, p.326 (entry "Modular equation of order 10").

modPolynomial ==> modularPolynomial$QEtaModularEquation(CX, An(trfs)(CX))
)time on
mp := modPolynomial(ans,[x,y],[]);
)time off

_{Time: 6.82 (EV) + 0.22 (GC) = 7.04 sec}

We only need the second factor and remember that $x$ actually stands for the product $R(q^5)R(q^{25n})$, i.e., we substitute $u=R(q^5)$ and $v=R(q^{25n})$.

fl := factorList factor eval(mp,[x=u*v,y=v]);
fl.1
modpol := fl.2.factor;

\[ \left[{flag}=\mathtt{\texttt{"}prime\texttt{"}}, {factor}={v}, {exponent}=13\right] \]

f1 := v^4-3*v^3+4*v^2-2*v+1;
f2 := v^4+2*v^3+4*v^2+3*v+1;
f3 := v^10+4*v^9+17*v^8+32*v^7+120*v^6+248*v^5+342*v^4+296*v^3+181*v^2+70*v+10;
f4 := 10*v^10-70*v^9+181*v^8-296*v^7+342*v^6-248*v^5+120*v^4-32*v^3+17*v^2-4*v+1;
assertEquals(modpol,
  u^15 * v^2 * f1^2 * f2_
  + u^10 * f2 * f4_
  + u^5 * v * f1 * f3_
  - v * f1 * f2^2)

\[ \mathtt{true} \]

-------------------------------------------------------------------
--endtest
-------------------------------------------------------------------

Modular Equations for $R(q)$ and $R(q^{5n})$, Gröbner basis method

-------------------------------------------------------------------
--test:RogersRamanujanModularPolynomials-5-GB
-------------------------------------------------------------------

For this demonstration we choose $n=2$. We already know the two relations from the previous sections. \begin{align*} \frac{R(q^2) - R(q)^2}{R(q^2) + R(q)^2} &= R(q) R(q^2)^2 \\ R(q)^5 &= R(q^5)\cdot \frac{1 - 2 R(q^5) + 4 R(q^5)^2 - 3 R(q^5)^3 + R(q^5)^4} {1 + 3 R(q^5) + 4 R(q^5)^2 + 2 R(q^5)^3 + R(q^5)^4} \end{align*}

Let us replace $q$ by $q^5$ in the first relation, combine it with the second relation, use the variable Ri to stand for $R(q^i)$, and then eliminate the variable R5 by computing a Gröbner basis.

rels := [R5*R10^2*(R10+R5^2) - (R10-R5^2),_
  R1^5 * (1+3*R5+4*R5^2+2*R5^3+R5^4) - R5*(1-2*R5+4*R5^2-3*R5^3+R5^4)]

\[ \left[{{R10}}^{2}\, {{R5}}^{3}+{{R5}}^{2}+{{R10}}^{3}\, {R5}-{R10}, -{{R5}}^{5}+\left({{R1}}^{5}+3\right)\, {{R5}}^{4}+\left(2\, {{R1}}^{5}-4\right)\, {{R5}}^{3}+\left(4\, {{R1}}^{5}+2\right)\, {{R5}}^{2}+\left(3\, {{R1}}^{5}-1\right)\, {R5}+{{R1}}^{5}\right] \]

gb := groebnerEliminate(rels,[R5],[R1,R10])$QEtaGroebnerBasisTools
#gb

-- TIME:=0.01

\[ \left[{{R1}}^{5}\, {{R10}}^{15}+\left({{R1}}^{15}+10\, {{R1}}^{10}+{{R1}}^{5}\right)\, {{R10}}^{14}+\left(-4\, {{R1}}^{15}-50\, {{R1}}^{10}+9\, {{R1}}^{5}-1\right)\, {{R10}}^{13}+\left(9\, {{R1}}^{15}+81\, {{R1}}^{10}-5\, {{R1}}^{5}-1\right)\, {{R10}}^{12}+\left(-15\, {{R1}}^{15}-184\, {{R1}}^{10}+85\, {{R1}}^{5}-4\right)\, {{R10}}^{11}+\left(25\, {{R1}}^{15}+274\, {{R1}}^{10}-14\, {{R1}}^{5}\right)\, {{R10}}^{10}+\left(-29\, {{R1}}^{15}-275\, {{R1}}^{10}+31\, {{R1}}^{5}-5\right)\, {{R10}}^{9}+\left(21\, {{R1}}^{15}+285\, {{R1}}^{10}+54\, {{R1}}^{5}-6\right)\, {{R10}}^{8}+\left(-6\, {{R1}}^{15}-54\, {{R1}}^{10}+285\, {{R1}}^{5}-21\right)\, {{R10}}^{7}+\left(5\, {{R1}}^{15}+31\, {{R1}}^{10}+275\, {{R1}}^{5}-29\right)\, {{R10}}^{6}+\left(14\, {{R1}}^{10}+274\, {{R1}}^{5}-25\right)\, {{R10}}^{5}+\left(4\, {{R1}}^{15}+85\, {{R1}}^{10}+184\, {{R1}}^{5}-15\right)\, {{R10}}^{4}+\left(-{{R1}}^{15}+5\, {{R1}}^{10}+81\, {{R1}}^{5}-9\right)\, {{R10}}^{3}+\left({{R1}}^{15}+9\, {{R1}}^{10}+50\, {{R1}}^{5}-4\right)\, {{R10}}^{2}+\left(-{{R1}}^{10}+10\, {{R1}}^{5}-1\right)\, {R10}+{{R1}}^{10}\right] \]

\[ 1 \]

modpol := eval(gb.1, [R1=u,R10=v])

\[ {{u}}^{5}\, {{v}}^{15}+\left({{u}}^{15}+10\, {{u}}^{10}+{{u}}^{5}\right)\, {{v}}^{14}+\left(-4\, {{u}}^{15}-50\, {{u}}^{10}+9\, {{u}}^{5}-1\right)\, {{v}}^{13}+\left(9\, {{u}}^{15}+81\, {{u}}^{10}-5\, {{u}}^{5}-1\right)\, {{v}}^{12}+\left(-15\, {{u}}^{15}-184\, {{u}}^{10}+85\, {{u}}^{5}-4\right)\, {{v}}^{11}+\left(25\, {{u}}^{15}+274\, {{u}}^{10}-14\, {{u}}^{5}\right)\, {{v}}^{10}+\left(-29\, {{u}}^{15}-275\, {{u}}^{10}+31\, {{u}}^{5}-5\right)\, {{v}}^{9}+\left(21\, {{u}}^{15}+285\, {{u}}^{10}+54\, {{u}}^{5}-6\right)\, {{v}}^{8}+\left(-6\, {{u}}^{15}-54\, {{u}}^{10}+285\, {{u}}^{5}-21\right)\, {{v}}^{7}+\left(5\, {{u}}^{15}+31\, {{u}}^{10}+275\, {{u}}^{5}-29\right)\, {{v}}^{6}+\left(14\, {{u}}^{10}+274\, {{u}}^{5}-25\right)\, {{v}}^{5}+\left(4\, {{u}}^{15}+85\, {{u}}^{10}+184\, {{u}}^{5}-15\right)\, {{v}}^{4}+\left(-{{u}}^{15}+5\, {{u}}^{10}+81\, {{u}}^{5}-9\right)\, {{v}}^{3}+\left({{u}}^{15}+9\, {{u}}^{10}+50\, {{u}}^{5}-4\right)\, {{v}}^{2}+\left(-{{u}}^{10}+10\, {{u}}^{5}-1\right)\, {v}+{{u}}^{10} \]

f1 := v^4-3*v^3+4*v^2-2*v+1;
f2 := v^4+2*v^3+4*v^2+3*v+1;
f3 := v^10+4*v^9+17*v^8+32*v^7+120*v^6+248*v^5+342*v^4+296*v^3+181*v^2+70*v+10;
f4 := 10*v^10-70*v^9+181*v^8-296*v^7+342*v^6-248*v^5+120*v^4-32*v^3+17*v^2-4*v+1;
assertEquals(modpol,
  u^15 * v^2 * f1^2 * f2_
  + u^10 * f2 * f4_
  + u^5 * v * f1 * f3_
  - v * f1 * f2^2)

\[ \mathtt{true} \]

With this method, it suffices to find modular equations for $R(q)$ and $R(q^n)$ if $n$ is a prime.

-------------------------------------------------------------------
--endtest
-------------------------------------------------------------------

Modular Equations for $R(q)$ and $R(-q)$

-------------------------------------------------------------------
--test:RogersRamanujanModularPolynomials-minus
-------------------------------------------------------------------

In this section we compute the modular polynomial between $R(q)$ and $R(-q)$.

Again, it is the same as computing the modular polynomial between $R(q^5)$ and $R(-q^5)$.

From the definition of $R(q)$ as a generalized eta-quotient, it is easy to show that \begin{align*} R(-q^5) &= \frac{\eta_{50,15}(\tau) \eta_{50,20}(\tau) \eta_{100,10}(\tau)} {\eta_{50, 5}(\tau) \eta_{50,10}(\tau) \eta_{100,30}(\tau)}. \end{align*} We compute the relation \begin{align*} uv(u-v)^4 - u^2v^2(u-v)^2+2u^3v^3 + (u-v)(1+u^5v^5) &=0 \end{align*} for $u=R(q^5)$ and $v=R(-q^5)$, see Jinhee Yi: "Modular Equations for the Rogers-Ramanujan Continued Fraction and the Dedekind Eta-Function".

expectedpol := u*v*(u-v)^4 - u^2*v^2*(u-v)^2 + 2*u^3*v^3 + (u-v)*(1+u^5*v^5);

We start with the specifications for $R(q)$ and $R(-q)$.

r1spec := eqSPEC([[5,1,1],[5,2,-1]])
r2spec := eqSPEC[[10,1,-1],[10,2,-1],[10,3,1],[10,4,1],[20,2,1],[20,6,-1]]

\[ \frac{{\eta}_{5, 1}\left(\tau\right)}{{\eta}_{5, 2}\left(\tau\right)} \]

\[ \frac{{\eta}_{10, 3}\left(\tau\right)\, {\eta}_{10, 4}\left(\tau\right)\, {\eta}_{20, 2}\left(\tau\right)}{{\eta}_{10, 1}\left(\tau\right)\, {\eta}_{10, 2}\left(\tau\right)\, {\eta}_{20, 6}\left(\tau\right)} \]

Unfortunately, these eta-quotients are not modular functions for $\Gamma_1(20)$.

[qModular?(C,MGAMMA1 20) x for x in [r1spec,r2spec]]

-- Chen: Theorem 10.1(3) sspec:=1
-- Chen: Theorem 10.1(3) rspec:=subscript(eta, [5, 1])(tau)/subscript(eta, [5, 2])(tau)
-- Chen: Theorem 10.1(3) alpha:=1/5
-- Chen: Theorem 10.1(3) sspec:=1
-- Chen: Theorem 10.1(3) rspec:=(subscript(eta, [10, 3])(tau)*subscript(eta, [10, 4])(tau)*subscript(eta, [20, 2])(tau))/(subscript(eta, [10, 1])(tau)*subscript(eta, [10, 2])(tau)*subscript(eta, [20, 6])(tau))
-- Chen: Theorem 10.1(3) alpha:=1/5

\[ \left[\mathtt{false}, \mathtt{false}\right] \]

Thus we look for a modular equation of $R(q^5)$ and $R(q^{5n})$.

Those are modular functions for $\Gamma_1(100)$.

rspecs := [dilate(x,5) for x in [r1spec, r2spec]]
nn := lcm [level x for x in rspecs]
MG ==> MGAMMA1 nn
[qModular?(C,MG) x for x in rspecs]
)set stream calc 50
specMA1(C,MG) rspecs.1
specMA1(C,MG) rspecs.2

\[ \left[\frac{{\eta}_{25, 5}\left(\tau\right)}{{\eta}_{25, 10}\left(\tau\right)}, \frac{{\eta}_{50, 15}\left(\tau\right)\, {\eta}_{50, 20}\left(\tau\right)\, {\eta}_{100, 10}\left(\tau\right)}{{\eta}_{50, 5}\left(\tau\right)\, {\eta}_{50, 10}\left(\tau\right)\, {\eta}_{100, 30}\left(\tau\right)}\right] \]

\[ 100 \]

\[ \left[\mathtt{true}, \mathtt{true}\right] \]

\[ {q}-{{q}}^{6}+{{q}}^{11}-{{q}}^{21}+{{q}}^{26}-{{q}}^{31}+{{q}}^{36}-{{q}}^{46}+2\, {{q}}^{51}+O\left({{q}}^{52}\right) \]

\[ {q}+{{q}}^{6}+{{q}}^{11}-{{q}}^{21}-{{q}}^{26}-{{q}}^{31}-{{q}}^{36}+{{q}}^{46}+2\, {{q}}^{51}+O\left({{q}}^{52}\right) \]

We compute the pole orders (qetaGrades) of the respective $q$-expansions at the cusps without computing the actual series. There are 140 cusps for $\Gamma_1(100)$.

gammas := cuspMatrices() $ MG; #gammas
qgrdss0 := [qGrades(C,MG) x for x in rspecs]

\[ 140 \]

\[ \left[\left[0, -2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, -4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 4, 0, 0, 0, -1, 0, 0, 0, 0, -1, 0, 0, 4, 0, 1, 0, 2, 0, 0, 0, -4, 0, 1, 0, 0, -2, 0, -1, 0, 0, -1, -2, 0, 1, 0, -4, 0, 2, 0, 1, 0, 4, -1, 0, 0, -1, 4, 0, 1, 0, 2, 0, -4, 1, 0, -2, -1, 0, -1, -2, 0, 1, 0, -4, 0, 2, 1, 0, 4, -1, 0, 4, 0, 0, -4, 0, 0, 0, -4, 0, 4, 0, 4, 0, -4, 0, -4, 0, 4, 0, 4, 0, -4, -1\right], \left[0, 8, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -8, 0, 0, 0, 1, 0, 0, 0, -1, 0, 0, 0, -1, 0, 0, 0, 0, -1, 0, 0, -1, 0, 1, 0, -8, 0, 0, 0, 1, 0, 1, 0, 0, 8, 0, -1, 0, 0, -1, 8, 0, 1, 0, 1, 0, -8, 0, 1, 0, -1, -1, 0, 0, -1, -1, 0, 1, 0, -8, 0, 1, 1, 0, 8, -1, 0, -1, 8, 0, 1, 0, 1, 0, -8, 1, 0, -1, -1, 0, -1, 0, 0, 1, 0, 0, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, -1\right]\right] \]

Cusps that correspond to non-poles in the series expansions for both $R(q^5)$ and $R(-q^5)$ can be excluded for the computation of the modular polynomial between the two.

trfs0 := [gammas.i for i in 1..#qgrdss0.1 | qgrdss0.1.i > 0 or qgrdss0.2.i > 0]
#trfs0

\[ \left[\begin{bmatrix}1&0\\50&1\end{bmatrix}, \begin{bmatrix}3&-1\\100&-33\end{bmatrix}, \begin{bmatrix}1&0\\25&1\end{bmatrix}, \begin{bmatrix}3&1\\50&17\end{bmatrix}, \begin{bmatrix}7&3\\100&43\end{bmatrix}, \begin{bmatrix}2&-1\\25&-12\end{bmatrix}, \begin{bmatrix}3&-1\\25&-8\end{bmatrix}, \begin{bmatrix}13&-3\\100&-23\end{bmatrix}, \begin{bmatrix}7&-1\\50&-7\end{bmatrix}, \begin{bmatrix}4&-1\\25&-6\end{bmatrix}, \begin{bmatrix}17&-8\\100&-47\end{bmatrix}, \begin{bmatrix}9&-2\\50&-11\end{bmatrix}, \begin{bmatrix}11&-2\\50&-9\end{bmatrix}, \begin{bmatrix}23&-3\\100&-13\end{bmatrix}, \begin{bmatrix}6&-1\\25&-4\end{bmatrix}, \begin{bmatrix}13&-6\\50&-23\end{bmatrix}, \begin{bmatrix}27&-10\\100&-37\end{bmatrix}, \begin{bmatrix}7&-2\\25&-7\end{bmatrix}, \begin{bmatrix}8&-1\\25&-3\end{bmatrix}, \begin{bmatrix}33&-1\\100&-3\end{bmatrix}, \begin{bmatrix}17&1\\50&3\end{bmatrix}, \begin{bmatrix}9&-4\\25&-11\end{bmatrix}, \begin{bmatrix}37&-10\\100&-27\end{bmatrix}, \begin{bmatrix}19&-8\\50&-21\end{bmatrix}, \begin{bmatrix}21&-8\\50&-19\end{bmatrix}, \begin{bmatrix}43&3\\100&7\end{bmatrix}, \begin{bmatrix}11&-4\\25&-9\end{bmatrix}, \begin{bmatrix}23&-6\\50&-13\end{bmatrix}, \begin{bmatrix}47&-8\\100&-17\end{bmatrix}, \begin{bmatrix}12&-1\\25&-2\end{bmatrix}, \begin{bmatrix}13&1\\25&2\end{bmatrix}, \begin{bmatrix}14&5\\25&9\end{bmatrix}, \begin{bmatrix}16&7\\25&11\end{bmatrix}, \begin{bmatrix}17&2\\25&3\end{bmatrix}, \begin{bmatrix}18&5\\25&7\end{bmatrix}, \begin{bmatrix}19&3\\25&4\end{bmatrix}, \begin{bmatrix}21&5\\25&6\end{bmatrix}, \begin{bmatrix}22&7\\25&8\end{bmatrix}, \begin{bmatrix}23&11\\25&12\end{bmatrix}, \begin{bmatrix}24&-1\\25&-1\end{bmatrix}\right] \]

\[ 40 \]

Unfortunately, if we look at the pole orders of the remaining cusps, neither $R(q^5)$ nor $R(-q^5)$ has positive pole order at all these cusps.

matrix [qGrades(C,MG)(x,trfs0) for x in rspecs]

\[ \begin{bmatrix}-2&1&-4&2&1&4&4&1&2&-4&1&-2&-2&1&-4&2&1&4&4&1&2&-4&1&-2&-2&1&-4&2&1&4&4&-4&-4&4&4&-4&-4&4&4&-4\\8&1&1&-8&1&-1&-1&1&-8&1&1&8&8&1&1&-8&1&-1&-1&1&-8&1&1&8&8&1&1&-8&1&-1&-1&1&1&-1&-1&1&1&-1&-1&1\end{bmatrix} \]

We realize, however, that if we start with $R(q^5)$ and $R(q^5)^4 R(-q^5)$, we can remove more cusps and end with an element that has proper poles at all the remaining cusps.

rrspecs := [dilate(x,5) for x in [r1spec, r1spec^4*r2spec]]

\[ \left[\frac{{\eta}_{25, 5}\left(\tau\right)}{{\eta}_{25, 10}\left(\tau\right)}, \frac{{{\eta}_{25, 5}\left(\tau\right)}^{4}\, {\eta}_{50, 15}\left(\tau\right)\, {\eta}_{50, 20}\left(\tau\right)\, {\eta}_{100, 10}\left(\tau\right)}{{{\eta}_{25, 10}\left(\tau\right)}^{4}\, {\eta}_{50, 5}\left(\tau\right)\, {\eta}_{50, 10}\left(\tau\right)\, {\eta}_{100, 30}\left(\tau\right)}\right] \]

qgrdss := [qGrades(C,MG) x for x in rrspecs];
trfs := [gammas.i for i in 1..#qgrdss.1 | qgrdss.1.i > 0 or qgrdss.2.i > 0];
#trfs
qgrdss := [qGrades(C,MG)(x,trfs) for x in rrspecs]

\[ 25 \]

\[ \left[\left[1, 2, 1, 4, 4, 1, 2, 1, 1, 2, 1, 4, 4, 1, 2, 1, 1, 2, 1, 4, 4, 4, 4, 4, 4\right], \left[5, 0, 5, 15, 15, 5, 0, 5, 5, 0, 5, 15, 15, 5, 0, 5, 5, 0, 5, 15, 15, 15, 15, 15, 15\right]\right] \]

For the expansions at the remaining cusps, the coefficient ring must be extended. From the data we have, we conclude that an extension by a primitive 10-th root of unity (we call it $\xi$) is enough.

xiord := minRootOfUnity(C,MG)(rrspecs,trfs)
EXTENDEDCOEFFICIENTRING(C, xiord, CX, xi);
ans := [specMAn(trfs,CX,MG)(x) for x in rrspecs];
matrix [qetaGrades x for x in ans]

\[ 20 \]

\[ \begin{bmatrix}1&2&1&4&4&1&2&1&1&2&1&4&4&1&2&1&1&2&1&4&4&4&4&4&4\\5&0&5&15&15&5&0&5&5&0&5&15&15&5&0&5&5&0&5&15&15&15&15&15&15\end{bmatrix} \]

The modular polynomial between $x=R(q^5)$ and and $y=R(q^5)^4 R(-q^5)$ can easily be computed.

modPolynomial ==> modularPolynomial$QEtaModularEquation(CX, An(trfs)(CX))
mp := modPolynomial(ans,[x,y],[])

\[ {{y}}^{6}+\left(-{{x}}^{5}-1\right)\, {{y}}^{5}+5\, {{x}}^{5}\, {{y}}^{4}-10\, {{x}}^{10}\, {{y}}^{3}+5\, {{x}}^{15}\, {{y}}^{2}+\left(-{{x}}^{20}+{{x}}^{15}\right)\, {y}-{{x}}^{20} \]

Let us express that polynomial in the original functions $u=R(q^5)$ and $v=R(-q^5)$. The first factor is a power of $y=v$ and can be ignored.

modpol := (eval(mp,[x=u,y=u^4*v])/u^19) :: Pol(ZZ)
expectedpol

\[ {{u}}^{5}\, {{v}}^{6}+\left(-{{u}}^{6}-{u}\right)\, {{v}}^{5}+5\, {{u}}^{2}\, {{v}}^{4}-10\, {{u}}^{3}\, {{v}}^{3}+5\, {{u}}^{4}\, {{v}}^{2}+\left(-{{u}}^{5}+1\right)\, {v}-{u} \]

\[ -{{u}}^{5}\, {{v}}^{6}+\left({{u}}^{6}+{u}\right)\, {{v}}^{5}-5\, {{u}}^{2}\, {{v}}^{4}+10\, {{u}}^{3}\, {{v}}^{3}-5\, {{u}}^{4}\, {{v}}^{2}+\left({{u}}^{5}-1\right)\, {v}+{u} \]

assertEquals(modpol, -expectedpol)

\[ \mathtt{true} \]

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--endtest
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