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This question is related to notation of infinite product.

We know that, $$ \prod_{i=1}^{\infty}x_{i}=x_{1}x_{2}x_{3}\cdots $$

How do I denote $$ \cdots x_{3}x_{2}x_{1} ? $$

One approach could be $$ \prod_{i=\infty}^{1}x_{i}=\cdots x_{3}x_{2}x_{1} $$

I need to use this expression in a bigger expression so I need a good notation for this. Thank you in advance for your help.

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What about simply $\,\,...x_3x_2x_1\,\,$? – DonAntonio May 24 '12 at 21:43
What is the difference between the two products? – Michael Greinecker May 24 '12 at 21:46
Presumably, the OP is regarding non-communicative multiplication. – Phira May 24 '12 at 21:47
I think the important question is "What do you mean by $\cdots x_3 x_2 x_1$?" – user17762 May 24 '12 at 21:48
I would interpret the "backwards product" as the limit of $x_n x_{n-1} \cdots x_1$ as $n \to \infty$, so for commutative multiplication this is the same as the usual "forward product". – mrf May 24 '12 at 22:31

Is there any reason to avoid the obvious $\;\; \displaystyle\prod_{i=-\infty}^{-1} x_{-i} \;\;$ ?

(as opposed to dropping the negative signs, like in the approach you suggested)

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Yes, this is possible! Thanks. – Uday May 24 '12 at 22:06

Sometimes in Clifford algebra when they do products backwards they talk of the "reverse" of the product. I've seen this denoted various ways with tidles: $\widetilde{abc}=cba$ or $(abc)^{\sim}=cba$. If you like them you could consider $$\widetilde{\Pi_{i=1}^\infty a_i}$$ or $$(\Pi_{i=1}^\infty a_i)^\sim$$

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I will check if this standard notation and use this. Thanks. – Uday May 24 '12 at 22:04

If they’re matrices, you can of course simply use $$\left(\prod_{n\ge 0}x_n^T\right)^T\;.$$

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+1. Nice trick... – user17762 May 25 '12 at 5:41
Similar to the above, we can also define $\left(\displaystyle \prod_{n \geq 0} x_n^{-1}\right)^{-1}$ assuming it makes sense to talk of $x_n^{-1}$ – user17762 May 25 '12 at 19:24

In the theory of non-autonomous abstract evolution equations, it is quite costumary to use the followiong notation:

For a family of operators $U_0,U_1,\ldots,U_{n-1}\in\mathcal{L}(X)$, we denote the "time-ordered" product of these operators by \begin{equation*} \prod_{p=0}^{n-1}U_p:=U_{n-1} U_{n-2} \cdots U_1 U_0\quad\mbox{and}\quad\prod_{p=n-1}^{0}U_p:=U_0U_1\cdots U_{n-2} U_{n-1} . \end{equation*}

See Pazy, Page 130.

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(With tongue in cheek:) what about this? $$\left(x_n\prod_{i=1}^\infty \right)\;$$

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