How to use generating functions with integer partitions -- Number Theory 30

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[MrJepreis]

I finally understand generating functions! Thanks! I wish I had a math lecturer like you 35 years ago.

[iabervon]

I think it would be worth having a lecture on how power series of a formal variable work exactly, and what is necessary for one to be well-behaved. For example, it's not at all obvious that you can multiply by notation that looks like 1 and have things work out correctly just because the notation that looks like a geometric series turns into a rational function. On the other side, it's worth noting that all you really care about with infinite products and sums is that every coefficient of a term in the result comes from a finite portion of the product or sum. We don't care whether the generating function actually converges anywhere, just whether you could write it down and look at each coefficient.

[stewartcopeland]

@ 28:39 Warm-up exercises below:



1) product m = 0 --> infinity (1/(1-q^(5m+1)) * ( 1/(1-q^(5m+4))
2) product m = 0 --> infinity (1/(1-q^(2m+1)) * (1/(1-q^(4m+4))

[MathFromAlphaToOmega]

One nice application (that doesn't use too much fancy math) is that you can find an asymptotic formula for the number of partitions of n into parts of size at most m. We take the product
1/[(1-q)(1-q^2)...(1-q^m)]
and use partial fraction decomposition to write it as
1/m!*1/(1-q)^m+(other terms with lower-order coefficients),
so the main term is
1/m!*(-1)^n(-m choose n),
or roughly n^(m-1)/[m!(m-1)!].

[looney]

Excellent work. My only real point of confusion is using the Pi notation to represent both the number of partitions as well as the generating function. You introduce the Pi(n) notation as representing a number of partitions, and the A(n) notation as representing an arbitrary generating function, but then you sort of assume we understand that Pi(q) is now a generating function. Then at 13:20 I had to do a double take to make sure I understood what was meant when you wrote Pi(n-k-1). I would just stress that the two notations use the same "partition" variable but are fundamentally different.

[zenith]

24:43 can in fact be generalized to claim that # partitions into non-multiples of m equals # partitions where no part is repeated m or more times, which is known as Glaisher's theorem.

[cbraidotti]

In the proof of the last theorem, before inserting an infinite number of ones and then simplifying half of the terms in the denominator with all of the terms in the numerator, should we not check if we can do that? (Something akin to Riemann's rearrangement conditions)

[richardboland]

Euler's partition identity has a red-headed step-child that lives in the shed. A 3rd partition type also equinumerous to partitions into either odd or distinct parts. That is, partitions that may be formed into 'SUDs', symmetric unimodal distributions. e.g., '1, 2, 1' or '2, 6, 6, 6, 2' or '7' or '4, 4, 4'. Note '3, 3, 3, 3' is an SBD, symmetric bimodal distribution. SBDs are equinumerous with partitions into even parts. Questions still in need good answers are: Who proved this result? What papers discuss this theorem?

[unhealthytruthseeker]

The full pgf shows up in conformal field theory as the character of representations of the Virasoro algebra. The "distinct parts" pgf shows up* as a factor in characters whenever you have a fermionic current in you conformal field theory, such as when you have supersymmetry. The "distinct parts" restriction amounts to not being able to use any mode more than once, which is due to the fermionic nature of the modes (can't have more than one fermion in the same state). The absence of the distinct parts restriction happens when you have a bosonic current (bosons can occupy the same state). In this way, integer partitions are closely related to (2D) conformal field theory.

*Technically the distinct parts factor has q^(n/2) rather than q^n, but it's basically the same kind of thing.

[txikitofandango]

Three different uses of π, none of them the circle constant! :)

[CM_France]

Hi,

Very interesting, but it was impossible for me to count how many "and so on and so forth" there were! Many of them!
I just begin to understand the importance of generating functions for counting.

[luisguillermo]

Wow, I was just reading the Wikipedia on partitions exactly when this video came out

[stanleydodds]

I might be nitpicking, but isn't it abuse of notation to say that pi(q) is the limit as m tends to infinity of pi_m(q)? Firstly pi(q) isn't really a function; q is a formal variable, so we are treating it more like an element of the ring Z[[q]], where we haven't even defined a topology or a metric, so I'm not sure limits even make sense in this case without some other definition(s). But even if pi(q) was treated as a function on the set of real or complex numbers, we have to first exclude all of the roots of unity for the functions pi_m(q) to make sense as defined, and then it might converge on this subset pointwise, but I can't see how it converges uniformly, which I would think is an equally valid interpretation of "limit" for a function.

[cicik]

so you are going towards Eulers pentagonal teorem?

[alessioolivieri]

How to solve "partitions of n using even parts an even number of times?"

[nathanisbored]

this and the previous video are missing from the number theory playlist

[konraddapper]

Can any on help me i am struggling to see why
p(q) has a non emty set of q values it converges for