Triple Product
by math_explorer, Jan 8, 2015, 2:04 PM
I don't know why everybody dresses this up with weird constant factors and 
-power substitutions. Math is no fun that way.
Anyway, here are the unbaked versions we want to prove equal.
![\[ \bar\Theta(\eta|q) = \sum_{n=-\infty}^\infty q^{n^2}\eta^n \]](//latex.artofproblemsolving.com/1/e/4/1e4a7f94718b09b566aa87b56a9d01eee20d05d4.png)
![\[ \bar\Pi(\eta|q) = \prod_{n=1}^\infty (1-q^{2n})(1+q^{2n-1}\eta)(1+q^{2n-1}\eta^{-1}) \]](//latex.artofproblemsolving.com/e/e/8/ee8788fc157f6b925356b70d3a7a216341fd08fc.png)
Our goal is to prove them equal for any
. (The zeroth step is to see that they're well-defined for such values, which is not hard because 
and the exponent of 
grows quickly enough to overwhelm any 
terms. By the way, convince yourself that when 
, it makes sense to let 
here.) First, we make some observations:
![\[ \bar\Theta(\eta q^2|q) = \bar\Theta(\eta|q)\eta^{-1}q^{-1} \]](//latex.artofproblemsolving.com/1/f/f/1ffa8da8ea5adc6b7ce6571a2031aa8416eef055.png)
This is true by completing the square:
Not the most rigorous of ways to write a sum, that penultimate line, but you get the point.
Next:
![\[ \bar\Theta(-q|q) = 0 \]](//latex.artofproblemsolving.com/f/3/4/f34e1101733199eac32f7a4fffbed808fbf95835.png)
Because
![\[ \bar\Theta(-q|q) = \sum_{n=-\infty}^\infty (-1)^nq^{n^2+n} \]](//latex.artofproblemsolving.com/d/4/6/d462237b47c7fa0916c7bcec2eab2e1edd16de75.png)
and every
cancels with the term 
, as 
but and 
have different parity.
As an easy consequence of the two statements together,
![\[ \bar\Theta(-q^{2m+1}|q) = 0 \qquad \forall m \in \mathbb{Z} \]](//latex.artofproblemsolving.com/7/1/6/7169aef22c589b46ffc5c7826b86628441e18b1a.png)
What about
?
![\[ \bar\Pi(\eta q^2|q) = \bar\Pi(\eta|q)\eta^{-1}q^{-1} \]](//latex.artofproblemsolving.com/9/c/3/9c3dc7fc8a776c88ae41257fed005a887b04320a.png)
This one ain't hard at all, as it just shifts everything in the product of the second and third factors one term over; you lose a
from the second factor and gain a 
from the third, so you multiply by ![\[ \frac{1 + q^{-1}\eta^{-1}}{1 + q\eta} = q^{-1}\eta^{-1}. \]](//latex.artofproblemsolving.com/0/9/1/091578d8fc7fc85df40e3311029f1022ad32cab2.png)
![\[ \bar\Pi(-q^{2m+1}|q) = 0 \qquad \forall m \in \mathbb{Z} \]](//latex.artofproblemsolving.com/3/5/f/35ffcdc7fa8c23884f3d272d11531214bfc18c0e.png)
To be more precise, these are simple zeroes of as a function of , and it has no other zeroes. The book does not justify this correctly! This is due to Chapter 5, Propotion 3.1, which is actually Book I, Chapter 8, Proposition 1.9: Let 
. Then 
converges, and it converges to 0 iff one of the factors is 0.
Proof: Disregard finitely many terms so
. Then the logarithm can be defined to behave nicely on 
, so we can turn the product into a sum, ![\[ \prod (1+a_n) = e^{\sum \log (1+a_n)}. \]](//latex.artofproblemsolving.com/6/4/c/64ca96ea403974cbfe9064b036132595186dfbd0.png)
converges because 
; if it converges to 
, then the product converges to 
, which is always nonzero. The product can only be zero if we got a zero from one of the terms we disregarded. So we're done.
--- end proof ---
We now have two functions with similar sets of zeroes (we know
's zeroes are a subset of 
's), so we do what every complex analyst loves to do: divide one by the other! (I feel like this is one of the big pieces of complex analysis intuition that deserves to be explicitly stated.)
Fix and consider 
as a function of . It's holomorphic except at 
.
Furthermore, note that![\[ \bar\Theta / \bar\Pi(\eta|q) = \bar\Theta / \bar\Pi(\eta q^{2k}|q) \qquad \forall k \in \mathbb{Z}. \]](//latex.artofproblemsolving.com/7/4/f/74f78656c336480eabeda77eea25bd19feea4136.png)
We can pick 
so that 
(viz. 
). So any value that 
attains for some , it attains for one such in the annulus with inner radius 
and outer radius 
.
In other words, the range of is the same as its range on the annulus. But the annulus is compact and is holomorphic on it, so it's bounded on it, so it's bounded on the entire complex plane... except 
.
But, by Riemann's theorem on removable singularities (Chapter 3, Theorem 3.1), that means is a removable singularity of .
So we can extend to be entire. And yet, it's still bounded, so by Liouville's theorem (Chapter 3, Corollary 4.5), it's constant!
(Okay, writing out as an exponential term is sort of useful here because it lets you avoid and skip citing the removable singularity theorem. Still, invoking it is pretty clearly motivated. By the way, I must confess that for a long time I had completely forgotten about that theorem and didn't know how to prove this step without composing the exponential term. This is a big sign that there are many things about complex analysis I still Do Not Get™. Dar(j)n.)
But we've only proved is a constant function of ; we need to prove that it's also a constant function of — in fact, equal to 1.
Towards this end, we plug in
and 
. (Motivation? What motivation? This is a functional equation, we don't deal with that sort of stuff.)
![\[\bar\Theta(-1|q) = \sum_n q^{n^2} (-1)^n\]](//latex.artofproblemsolving.com/f/e/3/fe3c952f3383036ad02807fa1b4044f6d45a8e2a.png)
because exponents 
and 
together cover every positive integer;
the last equality by pairing and multiplying
.
Yeah, I really don't know how to motivate this stuff. What's the point of all this? Look carefully — by some crazy coincidence,![\[ (\bar\Theta/\bar\Pi)(-1|q^4) = (\bar\Theta/\bar\Pi)(i|q). \]](//latex.artofproblemsolving.com/3/7/0/370ae90650b837b947378d92ae2307a6e5fe5dab.png)
So let
These points accumulate at. So by analytic continuation, is a constant function of . Yay! Furthermore, the function is equal to 1 at 0, so it's identically 1.
Therefore,
![\[ \bar\Theta(\eta|q) = \bar\Pi(\eta|q). \qquad \blacksquare\underline{\hspace*{0.5cm}}\Box \]](//latex.artofproblemsolving.com/9/7/4/974eb9b0c6b2b0a21dbf5bf808c5c13a54769932.png)
-power substitutions. Math is no fun that way.Anyway, here are the unbaked versions we want to prove equal.
![\[ \bar\Theta(\eta|q) = \sum_{n=-\infty}^\infty q^{n^2}\eta^n \]](http://latex.artofproblemsolving.com/1/e/4/1e4a7f94718b09b566aa87b56a9d01eee20d05d4.png)
![\[ \bar\Pi(\eta|q) = \prod_{n=1}^\infty (1-q^{2n})(1+q^{2n-1}\eta)(1+q^{2n-1}\eta^{-1}) \]](http://latex.artofproblemsolving.com/e/e/8/ee8788fc157f6b925356b70d3a7a216341fd08fc.png)
Our goal is to prove them equal for any
. (The zeroth step is to see that they're well-defined for such values, which is not hard because 
and the exponent of 
grows quickly enough to overwhelm any 
terms. By the way, convince yourself that when 
, it makes sense to let 
here.) First, we make some observations:![\[ \bar\Theta(\eta q^2|q) = \bar\Theta(\eta|q)\eta^{-1}q^{-1} \]](http://latex.artofproblemsolving.com/1/f/f/1ffa8da8ea5adc6b7ce6571a2031aa8416eef055.png)
This is true by completing the square:
Not the most rigorous of ways to write a sum, that penultimate line, but you get the point.
Next:
![\[ \bar\Theta(-q|q) = 0 \]](http://latex.artofproblemsolving.com/f/3/4/f34e1101733199eac32f7a4fffbed808fbf95835.png)
Because
![\[ \bar\Theta(-q|q) = \sum_{n=-\infty}^\infty (-1)^nq^{n^2+n} \]](http://latex.artofproblemsolving.com/d/4/6/d462237b47c7fa0916c7bcec2eab2e1edd16de75.png)
and every
cancels with the term 
, as 
but and 
have different parity.As an easy consequence of the two statements together,
![\[ \bar\Theta(-q^{2m+1}|q) = 0 \qquad \forall m \in \mathbb{Z} \]](http://latex.artofproblemsolving.com/7/1/6/7169aef22c589b46ffc5c7826b86628441e18b1a.png)
What about
?![\[ \bar\Pi(\eta q^2|q) = \bar\Pi(\eta|q)\eta^{-1}q^{-1} \]](http://latex.artofproblemsolving.com/9/c/3/9c3dc7fc8a776c88ae41257fed005a887b04320a.png)
This one ain't hard at all, as it just shifts everything in the product of the second and third factors one term over; you lose a
from the second factor and gain a 
from the third, so you multiply by ![\[ \frac{1 + q^{-1}\eta^{-1}}{1 + q\eta} = q^{-1}\eta^{-1}. \]](http://latex.artofproblemsolving.com/0/9/1/091578d8fc7fc85df40e3311029f1022ad32cab2.png)
![\[ \bar\Pi(-q^{2m+1}|q) = 0 \qquad \forall m \in \mathbb{Z} \]](http://latex.artofproblemsolving.com/3/5/f/35ffcdc7fa8c23884f3d272d11531214bfc18c0e.png)
To be more precise, these are simple zeroes of
as a function of , and it has no other zeroes. The book does not justify this correctly! This is due to Chapter 5, Propotion 3.1, which is actually Book I, Chapter 8, Proposition 1.9: Let 
. Then 
converges, and it converges to 0 iff one of the factors is 0.Proof: Disregard finitely many terms so
. Then the logarithm can be defined to behave nicely on 
, so we can turn the product into a sum, ![\[ \prod (1+a_n) = e^{\sum \log (1+a_n)}. \]](http://latex.artofproblemsolving.com/6/4/c/64ca96ea403974cbfe9064b036132595186dfbd0.png)
converges because 
; if it converges to 
, then the product converges to 
, which is always nonzero. The product can only be zero if we got a zero from one of the terms we disregarded. So we're done.--- end proof ---
We now have two functions with similar sets of zeroes (we know
's zeroes are a subset of 
's), so we do what every complex analyst loves to do: divide one by the other! (I feel like this is one of the big pieces of complex analysis intuition that deserves to be explicitly stated.)Fix
and consider 
as a function of . It's holomorphic except at 
.Furthermore, note that
![\[ \bar\Theta / \bar\Pi(\eta|q) = \bar\Theta / \bar\Pi(\eta q^{2k}|q) \qquad \forall k \in \mathbb{Z}. \]](http://latex.artofproblemsolving.com/7/4/f/74f78656c336480eabeda77eea25bd19feea4136.png)
We can pick 
so that 
(viz. 
). So any value that 
attains for some , it attains for one such in the annulus with inner radius 
and outer radius 
.In other words, the range of
is the same as its range on the annulus. But the annulus is compact and is holomorphic on it, so it's bounded on it, so it's bounded on the entire complex plane... except 
.But, by Riemann's theorem on removable singularities (Chapter 3, Theorem 3.1), that means
is a removable singularity of .So we can extend
to be entire. And yet, it's still bounded, so by Liouville's theorem (Chapter 3, Corollary 4.5), it's constant!(Okay, writing
out as an exponential term is sort of useful here because it lets you avoid and skip citing the removable singularity theorem. Still, invoking it is pretty clearly motivated. By the way, I must confess that for a long time I had completely forgotten about that theorem and didn't know how to prove this step without composing the exponential term. This is a big sign that there are many things about complex analysis I still Do Not Get™. Dar(j)n.)
is holomorphic in an open set containing the closure of a toy contour 
and 
is inside the set,![\[ f(z) = \frac{1}{2\pi i}\int_C \frac{f(\zeta)}{\zeta - z} d\zeta \]](http://latex.artofproblemsolving.com/e/4/f/e4ff0c7e06aee2bcf17c67d58b06b0487129bfdc.png)
![\[ f'(z) = \frac{1}{2\pi i}\int_C \frac{f(\zeta)}{(\zeta - z)^2} d\zeta \]](http://latex.artofproblemsolving.com/4/9/3/493851578e9156f1d937e6a1b2deb35384bbe1d1.png)
![\[ f^{(n)}(z) = \frac{n!}{2\pi i}\int_C \frac{f(\zeta)}{(\zeta - z)^{n+1}} d\zeta \]](http://latex.artofproblemsolving.com/e/0/9/e09dd5a60228be33f43976faa32b9dad54c3e3c6.png)
equation; by making the contour a big circle, you bound 
within epsilon. Therefore, 
is identically zero; therefore, 
(I made this symbol up) of the original Cauchy's integral formula on a contour around 
not omitting But we've only proved
is a constant function of ; we need to prove that it's also a constant function of — in fact, equal to 1.Towards this end, we plug in
and 
. (Motivation? What motivation? This is a functional equation, we don't deal with that sort of stuff.)![\[\bar\Theta(-1|q) = \sum_n q^{n^2} (-1)^n\]](http://latex.artofproblemsolving.com/f/e/3/fe3c952f3383036ad02807fa1b4044f6d45a8e2a.png)
because exponents 
and 
together cover every positive integer;
the last equality by pairing and multiplying
.Yeah, I really don't know how to motivate this stuff. What's the point of all this? Look carefully — by some crazy coincidence,
![\[ (\bar\Theta/\bar\Pi)(-1|q^4) = (\bar\Theta/\bar\Pi)(i|q). \]](http://latex.artofproblemsolving.com/3/7/0/370ae90650b837b947378d92ae2307a6e5fe5dab.png)
So let
These points accumulate at
. So by analytic continuation, is a constant function of . Yay! Furthermore, the function is equal to 1 at 0, so it's identically 1.Therefore,
![\[ \bar\Theta(\eta|q) = \bar\Pi(\eta|q). \qquad \blacksquare\underline{\hspace*{0.5cm}}\Box \]](http://latex.artofproblemsolving.com/9/7/4/974eb9b0c6b2b0a21dbf5bf808c5c13a54769932.png)
This post has been edited 1 time. Last edited by math_explorer, Mar 2, 2015, 2:14 AM
Reason: fix LaTeX
Reason: fix LaTeX
May 2020
January 2020