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In physics, do we have \begin{align} \frac{q^{-1}y-1}{y-1}-\frac{qy^{-1}-1}{q(y^{-1}-1)} =(1-q^{-1}) \delta(y)? \end{align} If $y \not\to 1$, then \begin{align} \frac{q^{-1}y-1}{y-1}-\frac{qy^{-1}-1}{q(y^{-1}-1)} = 0. \end{align} If $y \to 1$, then \begin{align} & \frac{q^{-1}y-1}{y-1}-\frac{qy^{-1}-1}{q(y^{-1}-1)} \\ & =(1-q^{-1})\left(\frac{1}{1-y} + \frac{1}{1-y^{-1}}\right) \\ & = (1-q^{-1})\left(\sum_{n \geq 0} y^n + \sum_{m \leq 0} y^m\right) \\ & =\left (1-q^{-1}\right)\left(\sum_{n \in \mathbb{Z}} y^n + 1\right) \\ & = \left(1 - q^{-1}\right)\left(\delta(y)+1\right). \end{align} Where is the mistake in my computations? Thank you very much.

Edit: q is a complex number and y is a variable. This type of computations happen in the paper.

I think that maybe the following computations are correct. \begin{align} & \frac{q^{-1}y-1}{y-1}-\frac{qy^{-1}-1}{q(y^{-1}-1)} \\ & =\left(1-q^{-1}y\right)\left(\frac{1}{1-y} + \frac{y^{-1}}{1-y^{-1}}\right) \\ & = \left(1-q^{-1}y\right)\left(\sum_{n \in \mathbb{Z}} y^n\right) \\ & = \left(1-q^{-1}y\right)\delta(y) \\ & = \left(1-q^{-1} \right)\delta(y). \end{align}

LJR
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  • Can you explain what $q$ and $y$ represent? – Kwin van der Veen Aug 21 '16 at 09:16
  • @fibonatic, thank you very much. I edited the post. q is a complex number and y is a variable. This type of computations happen in the paper. – LJR Aug 21 '16 at 09:20
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    Something is not very healthy here. A Dirac delta is only really well defined under an integral. It is difficult to see how the left-hand side in your first expression, which seems like a straightforward algebraic expression, can lead to a expression that contain a Dirac delta function. –  Aug 21 '16 at 10:51
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    @flippiefanus I think OP means the kronecker delta, not the Dirac delta – Courage Aug 21 '16 at 11:49
  • There's nothing in this question that makes it specific to physics. – David Z Aug 21 '16 at 12:41
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    gauge theory couplings is an application. Back to physics.S.E ? –  Aug 21 '16 at 12:49
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    @LJR: At a first view I can already spot a problem: if $y \to 1$ AND $y \ge 1$, then the series $\sum y^n$ does NOT converge, therefore it is not a replacement for $\frac 1 {1-y}$. – Alex M. Aug 21 '16 at 12:50
  • "In physics", I could not say, but in mathematics, $$\frac{q^{-1}y-1}{y-1}-\frac{qy^{-1}-1}{q(y^{-1}-1)}$$ is undefined when $y=1$ and, for every $y\ne1$, this is $$ \frac{q^{-1}y-1}{y-1}-\frac{qy^{-1}-1}{q(y^{-1}-1)} =\frac{y-q}{q(y-1)}-\frac{q-y}{q(1-y)}=0.$$ – Did Aug 21 '16 at 12:58
  • My respsonse: it is false in mathematics, so to find out whether it is OK in physics, ask in http://physics.stackexchange.com/ ... but it seems they have already rejected it by migrating it here. So you are out of luck. – GEdgar Aug 21 '16 at 13:09
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    @AlexM.: It's even worse: Since the sum goes over $n\in\mathbb Z$, it never converges: If $|y| \ge 1$, the positive exponent side diverges, if $0<|y|\le1$,the negative exponent side diverges, and if $y=0$ the negative exponent terms are not even well defined. – celtschk Aug 21 '16 at 13:10
  • *Typo in my comment: Replace "when $y=1$" by "when $y=0$ or $y=1$"' and "for every $y\ne1$" by "for every $y\ne0,1$", since for $y=0$ the expression $$\frac{q^{-1}y-1}{y-1}-\frac{qy^{-1}-1}{q(y^{-1}-1)}$$ is undefined as well. – Did Aug 21 '16 at 13:53

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