Strange identity for 1?

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To make the notation more compact, I'll let use the symbol (a ; q)_∞ to represent the infinite product ∏(i = 1, ∞) (1 - aq^(i - 1)). Let

F(x) = ∏(i = 1, ∞) (1 + x^(2i))(1 + x^(3i))(1 - x^(4i - 2))(1 - x^(6i - 3)).

Then

F(x) = (-x^2 ; x^2)_∞ (-x^3 ; x^3)_∞ (x^2 ; x^4)_∞ (x^3 ; x^6)_∞

= [(-x^2 ; x^2)_∞ (x^2 ; x^4)_∞] [(-x^3 ; x^3)_∞ (x^3 ; x^6)_∞]

= [(-x^2 ; x^2)_∞ (x^2 ; x^2)_∞/(x^4 ; x^4)_∞]
× [(-x^3 ; x^3)_∞ (x^3 ; x^3)_∞/(x^6 ; x^6)_∞]

= [(x^4 ; x^4)_∞/(x^4 ; x^4)_∞] [(x^6 ; x^6)_∞/(x^6 ; x^6)_∞]

= 1....Show more

Hey, nice question :)

First of all, we note that the series doesn't converge if |x|>1. It will actually follow from the proof that the radius of convergence is indeed 1.

Now let's try expanding the product. We will get terms of the form (1+x^n) and (1-x^m) for some n,m. To prove that the value of the infinite product is indeed one, it would be helpful if we could somehow cancel some terms. To do this, we will only make use of the identity (1+x^n)(1-x^n) = (1-x^(2n)).

As these products get rather large and messy, we introduce some notation. Instead of writing out all the terms, we only note down the exponent and sign of x. So for example, the product (1+x^4)(1-x^7)(1+x^17)(1-x^321) would become (+4,-7,+17,-321) under our notation. This will make life considerably simpler later.

Note that the least common factor of 2,3,4 and 6 is 12. Hence if we expand the infinite product, the exponents will be periodic with period 12. Meaning, if (1+x^9) is part of the expanded product, then so is (1+x^21) and (1+x^33) and so on.

Let's now find the first few terms of the product! There is just no way around getting our hands dirty a little bit at least... Remember, we only have to find the terms until x^12, it will repeat itself after that!

Under our notation, the product is (-2, +2, -3, +3, +4, -6, +6, +6,+8,-9, +9, -10, +10, +12, +12,...). We get this by explicitly calculating the first few terms. By periodicity, this will continue with (-14, +14, -15,...) etc.

Now it's time to cancel some terms!. Let's start with the smallest exponents. We have a -2 and a +2. As (1+x^2)(1-x^2)=1-x^4, we can cancel the -2 and +2 and get an additional -4 instead. Then we can cancel the -4 and +4 to get a -8. Then we cancel the -8 and +8 and get a -16... This seems to go on forever.

What if we start with the -3 and +3? We cancel them to get a -6. This cancels one of the +6-s, to get a -12. Cancelling this and a +12 we get a -24, and so on. Similarly if we start with a -6, things seem to keep cancelling each other forever. You can play around, cancelling terms, and you will notice that everything cancels out nicely. But why is this true, could it be that we won't be able to cancel, say, the (1+x^(123456)) term?

Let's just write down our product again: P = (-2, +2, -3, +3, +4, -6, +6, +6, +8, -9, +9, -10, +10, +12, +12), plus this shifted by multiples of 12.

Suppose for contradiction that one of our terms didn't cancel in the formal product. Pick the one with the smallest exponent, say (1+x^n) or (1-x^n) didn't cancel. What is n mod 12?

n=1 mod 12 is not possible, as +-1 is not in P.
Can n be 2 mod 12? If this is the case then +n and -n are in P. Moreover, n/2 is not in P, as n/2 is either 1 or 7 mod 12. So we can cancel the +n and -n. So n cannot be 2 mod 12.
Can n be 3 mod 12? Then n is odd, so we can't cancel (1+-x^n) using smaller terms. But +3 and -3 are both in P, so we can cancel. So n is not 3 mod 12.
As 5,7,11 are not in P, n is not 5,7 or 11 mod 12.
n = 9 mod 12 is similar to the n=3 case.
n=10 mod 12 is similar to the n=2 case: 5 and 11 are not in P.

Hence we are left with the n=4,6,8 mod 12 cases.

If n=6 mod 12: the +6 and -6 will always cancel, and we get exactly one contributing term from cancelling +n/2 and -n/2.

What if n=4 mod 12? Note that n/2 is 2 or 8 mod 12, and both 2 and 8 are in P. Now we use the fact that n was the smallest uncanceled term. Since all terms before n have canceled, we conclude that n/2 has canceled as well, to produce a -n term. This we can cancel with our +n term. The argument is similar for n=8 mod 12.

We conclude that such n doesn't exist, and get the required contradiction. Hence in the formal product all terms cancel out, so the product is one (this is true in the formal sense, but we need to be a bit more careful when proving that the value of the product is actually one...)

Going through all these cases might seem a bit tedious (it's not, it's actually really easy), and I think I didn't explain it very well. Probably it's best if you try to go through the cases by yourself - if something isn't clear just shoot me a mail.

So why do all the terms cancel out, what is so special about this product and the set P? Honestly I don't really know. It just works.. Probably you can say more about this, knowing the combinatorial background to your question (which I'd love to hear btw!).

Hope this helps:)...Show more

Well, if he has another identity then yes.. you should look up on its background check to see if he has another ID...Show more