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This question arose in connection with another problem that I described earlier in Constructing a generating function using a series with all negative and positive powers of a variable. I had certain progress there, in particular I've got certain functional equation which allows to obtain a generating function by substitution of variables into a fixed point of certain operator. It was when trying to simplify the whole thing when the question occurred, and I decided it might be of independent interest so I am posting this.

There is a function of two variables which satisfies certain functional equation. It can be given as follows. Consider the following operator acting on functions of two variables: $$ (\mathbf T F)(z,q):=R_0(z,q)+qR_1(z,q)F(q,q)+q^2R_2(z,q)F(qz,q)+q^3R_3(z,q)F(2q^2,q) $$ with explicitly given rational functions $R_i(z,q)$ (I give them explicitly in the Appendix below). Then my generating function $\Phi(z,t)$ is a fixed point of $\mathbf T$, i. e. $\mathbf T\Phi=\Phi$. Moreover it turns out that the process of iteratively applying $\mathbf T$ starting from $\Phi_0\equiv0$, $\Phi_{n+1}=\mathbf T(\Phi_n)$ converges in the sense that $\Phi_n$ and $\Phi_{n+1}$ coincide up to degree $n$.

What I finally need is to determine somehow $\Phi(q,q)=1+3q+9q^2+23q^3+...$, so I am trying to simplify my functional equation.

Now it happens that $R_2$ is divisible by $z-q$ (see below), so substituting $z=q$ gives $$ R_0(q,q)+qR_1(q,q)\Phi(q,q)+q^3R_3(q,q)\Phi(2q^2,q)=\Phi(q,q). $$ Using this I express $\Phi(2q^2,q)$ through $\Phi(q,q)$ and substitute it into original recursion; but what I obtain gives zero instead of $\Phi(z,q)$ and I don't understand why.

In more detail: from the last equation, $$ \Phi(2q^2,q)=\frac{(1-qR_1(q,q))\Phi(q,q)-R_0(q,q)}{q^3R_3(q,q)}, $$ and this should be satisfied by recursion too. Indeed, I've checked that the series for $$ \Phi_n(2q^2,q)-\frac{(1-qR_1(q,q))\Phi_n(q,q)-R_0(q,q)}{q^3R_3(q,q)}, $$ does not contain any nonzero terms up to $q^n$.

Nevertheless, when I replace the original recursion $$ \Phi_{n+1}(z,q)=R_0(z,q)+qR_1(z,q)\Phi_n(q,q)+q^2R_2(z,q)\Phi_n(qz,q)+q^3R_3(z,q)\Phi_n(2q^2,q) $$ (which gives the correct result) with $$ \tilde\Phi_{n+1}(z,q)=R_0(z,q)+qR_1(z,q)\tilde\Phi_n(q,q)+q^2R_2(z,q)\tilde\Phi_n(qz,q)+q^3R_3(z,q)\frac{(1-qR_1(q,q))\tilde\Phi_n(q,q)-R_0(q,q)}{q^3R_3(q,q)}, $$ I get that $\tilde\Phi_n$ itself contains no nonzero terms up to $q^{n-1}$, so I obtain zero instead of $\Phi(z,q)$.

In other words, I have two recursive equations which both converge in the sense that $n$th and $n+1$st terms coincide up to degree $n$, but when I try to substitute one into another I get the wrong result. Why exactly does this happen and is there a way to correct it?

Added later

I've discovered that if in the above equation relating $\Phi(q,q)$ and $\Phi(2q^2,q)$ I solve out not $\Phi(2q^2,q)$ but $\Phi(q,q)$, then the recursion remains valid. Although this gives me what I need, it is even more puzzling now - why eliminating one term destroys the recursion while eliminating another one from the same equation does not??

Appendix

Explicitly, \begin{align*} R_0(z,q)&=\frac{1+q-q^2-z q-2 zq^2}{\left(1-3 q^2\right) (1-z) (1-2 q z)}\\ R_1(z,q)&=\frac{1-q}{ (1-z)(1-2 q)}\\ R_2(z,q)&=\frac{(1+2 q) (q-z) (1-q z)}{(1-3 q^2) (z-2 q) (1-2 q z)}\\ R_3(z,q)&=\frac{1-2 q^2}{(1-2 q) \left(1-3 q^2\right) (z-2 q)} \end{align*}

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    $\begingroup$ I see one particular source (of all kind) of problems -- that is $(z-2q)$ being not invertible in the ring of power series over indeterminates $z,q$. In particular, $R_2(z,q)$ and $R_3(z,q)$ are not well defined as power series. $\endgroup$ Feb 22 '16 at 23:05
  • $\begingroup$ @MaxAlekseyev There is a transformation which avoids this. One has$$\Phi(q,q)=\bar\Phi(q,\frac12)$$where $\bar\Phi(q,z)$ is the similar fixed point for another operator$$\bar R_0+\bar R_1\bar\Phi(q,\frac12)+\bar R_2\bar\Phi(q,q)+\bar R_3\bar\Phi(q,qz)$$, with$$\bar R_0=\frac{1+q-q^2-2q^2z-4q^3z}{(1-q)(1-3q^2)(1-4q^2z)},$$$$\bar R_1=\frac q{1-2q},$$$$\bar R_2=-\frac12\frac{q^2(1-qz)}{(1-2q) (1-3 q^2) (1-z)}$$and$$\bar R_3=-\frac12\frac{q^2 (1 + 2 q) (1 - 2 z) (1 - 2 q z)}{(1 - 3 q^2) (1 - z) (1 - 4 q^2z)}$$ $\endgroup$ Mar 9 '16 at 10:26
  • $\begingroup$ Then how do you pose your question in terms of this new operator? $\endgroup$ Mar 9 '16 at 10:41
  • $\begingroup$ @MaxAlekseyev Exactly the same happens - if one substitutes $z=\frac12$ in $(\bar{\mathbf T}\bar\Phi)(q,z)=\bar\Phi(q,z)$, the term with $\bar R_3$ drops out; if one now eliminates $\bar\Phi(q,q)$ from the resulting equation, iterations tend to zero; while if one eliminates $\bar\Phi(q,\frac12)$ (and leaves $\bar\Phi(q,q)$), iterations converge to the correct thing. $\endgroup$ Mar 9 '16 at 12:08
  • $\begingroup$ Btw sorry, in the comments somehow $q$ and $z$ exchanged places, $\bar\Phi(q,z)$ in the comments corresponds to $\Phi(z,q)$ in the question. $\endgroup$ Mar 9 '16 at 12:08

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