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Let $N$ be an integer. Let us consider a Laurent polynomial in $q$ given by $\sum_{s=N}^{M} \gamma_s q^s$. Then consider the expression $(\sum_{s=N}^{M} \gamma_s q^s) \times \sum_{i=0}^{\infty}q^i$.

Then I'm unable to see why for $d \geq M-N$, the coefficient of $q^d$ is $\sum_{s=N}^{M} \gamma_s$ (which is asserted in a literature)?

I'm not sure if I'm allowed to distribute individual terms.

Any suggestion or remark is highly valued

Mittens
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Sherlock
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2 Answers2

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It seems the condition on $d$ needs to be revised somewhat. In the following we use the coefficient of operator $[q^k]$ to denote the coefficient of $q^k$ of a series. Given the Laurent polynomial \begin{align*} \sum_{s=N}^{M} \gamma_s q^s \end{align*} we have $M\geq N$. We write the condition $d\geq M-N$ as \begin{align*} \color{blue}{d=M-N+e}\tag{1} \end{align*} with $e$ a non-negative integer.

We obtain \begin{align*} \color{blue}{[q^d]}&\color{blue}{\sum_{s=N}^M\gamma_sq^s\sum_{i=0}^{\infty}q^i}\\ &=\sum_{s=N}^M\gamma_s[q^{d-s}]\sum_{i=0}^{\infty}q^i\tag{2.1}\\ &=\sum_{s=0}^{M-N}\gamma_{s+N}[q^{d-s-N}]\sum_{i=0}^{\infty}q^i\tag{2.2}\\ &=\sum_{s=0}^{M-N}\gamma_{M-s}[q^{d-M+s}]\sum_{i=0}^{\infty}q^i\tag{2.3}\\ &\,\,\color{blue}{=\sum_{s=0}^{M-N}\gamma_{M-s}[q^{e-N+s}]\sum_{i=0}^{\infty}q^i}\tag{2.4}\\ &\stackrel{?}=\sum_{s=0}^{M-N}\gamma_{M-s}\tag{2.5}\\ \end{align*}

Comment:

  • In (2.1) we use the linearity of the coefficient of operator and apply the rule $[q^{k-l}]A(q)=[q^k]q^lA(q)$.

  • In (2.2) we shift the index to start with $s=0$.

  • In (2.3) we change the order of summation $s\to M-N-s$.

  • In (2.4) we apply (1).

Let's have a closer look at $[q^{e-N+s}]$ in (2.4). Since for each $s$ we want to select the summand $\gamma_{M-s}$ in order to obtain (2.5), we need a non-negative exponent of $q$. We therefore have the condition \begin{align*} \color{blue}{e-N+s\geq 0\qquad\qquad 0\leq s\leq M-N}\tag{3} \end{align*}

  • $N\leq 0$: Since $e\geq 0$, (3) is fulfilled whenever $N\leq 0$.

  • $N>0$: The situation is different when $N$ is positive. Setting >$s=0$ we get from (1) and (3) \begin{align*} e&\geq N\\ d=M-N+e&\geq (M-N)+N\\ \color{blue}{d}&\color{blue}{\geq M} \tag{4} \end{align*}

Conclusion: In case $N>0$ we need $d\geq M$ instead of $d\geq M-N$. The condition on $d$ can therefore be stated as \begin{align*} \color{blue}{d\geq \max\{M,M-N\}} \end{align*}

Plausibility check $N=2, M=3, d=1$:

Here we have $N>0$ and violate (4) since $d=M-N=3-2=1$ is less than $M=3$. We obtain \begin{align*} [q^1]&\left(\gamma_2q^2+\gamma_3q^3\right)\sum_{i=0}^{\infty}q^i\\ &=\left(\gamma_2[q^{-1}]+\gamma_3[q^{-2}]\right)\sum_{i=0}^{\infty}q^i\\ &=0 \end{align*} We see although we have $1=d\geq M-N=1$, we get $0$ instead of $\gamma_2+\gamma_3$.

Markus Scheuer
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  • I have a very elementary query, In $(2.1)$, the infinite sum $\sum_{i=0}^{\infty}q^i$ is distributed over finitely many sum $\sum_{s=N}^M\gamma_sq^s$. Could you please hint at why this can be done? – Sherlock Jun 23 '23 at 17:17
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    @Sherlock: Assuming $|q|<1$ the geometric series $\sum_{i=0}^{\infty}q^i=\frac{1}{1-q}$ converges. Multiplication of this series with $\gamma_{s_0}q^{s_0}$ is admissible and summing up a finite number of terms also. – Markus Scheuer Jun 23 '23 at 18:17
  • @epi163sqrt Would you be interested in this MSE link? The first case can be done by sum of residues and the second by basic formal power series. – Marko Riedel Jun 23 '23 at 19:31
  • @MarkoRiedel: Many thanks, Marko. I've answered it this moment. :-) – Markus Scheuer Jun 23 '23 at 19:38
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Hint: $$\left(\sum_{s=N}^{M} \gamma_s q^s\right) \times \sum_{i=0}^{\infty}q^i = \sum_{\ell=0}^{M-N} \sum_{i=0}^\infty \gamma_{N + \ell} q^{N+i+\ell} = \sum_{\ell=0}^{\infty}\sum_{i=0}^{\infty} \gamma_{N+\ell}q^{N+\ell+i} \mathbf 1_{0\le \ell\le M-N} = \sum_{d=0}^{\infty} \left(\sum_{\ell=0}^{d} \gamma_{N+\ell}\mathbf 1_{0\le \ell\le M-N}\right) q^{N+d}$$

Can you finish the proof?

Kroki
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  • could you please elaborate on the equality of last two expressions (How $i$ is absorbed in the expression)? – Sherlock Jun 21 '23 at 18:50
  • What don't you understand? – Kroki Jun 21 '23 at 18:54
  • I just mentioned in my previous comment. How are we exactly relating $l+i$ and $d$? – Sherlock Jun 21 '23 at 19:00
  • $\ell + i = d$ this is the relation. – Kroki Jun 21 '23 at 19:01
  • I am still confused about the last equality. If we expand the L.H.S we get $q^N(\sum_{i=0}^{\infty} \gamma_N q^i+....+\sum_{i=0}^{\infty}\gamma_M.q^{M-N+i})$. On the other hand expanding R.H.S we get $q^N(\sum_{d=0}^{\infty}(\sum_{s=N}^{M}\gamma_s) q^d)$. I don't see how they are equal. – Sherlock Jun 22 '23 at 08:14
  • Your RHS is not correct. – Kroki Jun 22 '23 at 14:23