How does one evaluate the following product if the set S happens to be empty?

\begin{aligned} f(n)= n \prod_{x \in S} \left(1-\frac{1}{x}\right) \end{aligned}

Is the value simply n or is it undefined (or zero)??


Edit: It seems rather odd that this question has been rated off-topic for lacking context or other details. I would have thought it rather obvious that it was about how to evaluate the product when there is no x due to an empty set. I would have guessed undefined because one cannot assign a value to $(1-1/x)$. However, as shown by C.Falcon, the convention is $1$. There's no other context or missing details. Feel free to delete if it doesn't meet the relevant standards.

  • $\begingroup$ This resembles the formula for Euler's totient function. If it is so, then $S$ is the divisors of $n$, which cannot be empty. $\endgroup$ – Kenny Lau Jun 27 '16 at 13:37
  • $\begingroup$ Yes, I based my example on that function but my example is not intended to be defined such that S is the divisors of n. Cheers. $\endgroup$ – Molonglo Jun 27 '16 at 13:42
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    $\begingroup$ See also Why is empty product defined to be $1$? and What is the product of the empty set?. $\endgroup$ – Martin Sleziak Jul 3 '16 at 13:56
  • $\begingroup$ @kennyLau In fact, in the formula for $\varphi$ the set $S$ is set of primies dividing $n$, so if you want to use the formula to calculate $\varphi(1)$, you will get empty product. $\endgroup$ – Martin Sleziak Jul 3 '16 at 14:00
  • $\begingroup$ @MartinSleziak Oh, my brain jammed. $\endgroup$ – Kenny Lau Jul 3 '16 at 16:04

An empty product is by convention equal to $1$ (the identity element for the multiplication), therefore whenever $S$ is empty, one has $f:n\mapsto n$.

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    $\begingroup$ It's analogous to a sum indexed by an empty set being zero (the additive identity) by convention. +1 for you. $\endgroup$ – MPW Jun 27 '16 at 13:38

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