Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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)

share|cite|improve this answer
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 $$(\Pi_{i=1}^\infty a_i)^\sim$$

share|cite|improve this answer
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 $$\left(\prod_{n\ge 0}x_n^T\right)^T\;.$$

share|cite|improve this answer
+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.

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.