One of My Favorite Identities Discussed in Class!

追加: 2026-05-10

[00:00:00]goes like this. So, let's look at the partition um version first. So, the number of partitions of n into parts um that are congruent to one or four mod five. Okay, so that's that's one side.

[00:00:37]Equals the number of partitions of n whose consecutive parts consec utive parts differ by at least two.

[00:01:07]So, that's the identity. That's the partition version of the identity.

[00:01:13]Now, let's do an example. So, let's write the partitions of purple type.

[00:01:24]So, what do we got?

[00:01:26]N in So, nine into parts that are congruent to one or four mod nine. One or four mod five.

[00:01:33]So, So, nine.

[00:01:35]And then five one one one. Not five. Oh, wait.

[00:01:41]Sorry, six one.

[00:01:43]Yeah, six That's good. So, I was mixed six plus one plus one plus one.

[00:01:50]What about this one?

[00:01:51]>> And then just the ones, so that's it.

[00:01:53]No, you're missing a bunch.

[00:01:54]>> Oh, right.

[00:01:55]Four plus four plus 1 Right, and then Um, 4 + 4 + 1 and then like that.

[00:02:08]Uh Now, what about this type?

[00:02:15]It's a nine.

[00:02:17]We want to think of this as a like a as a stronger a stronger version of distinction. Remember, like distinct if we just put distinct over here, which would be consecutive parts differ by at least one.

[00:02:30]If we just put distinct over here, we get congruent to 1 mod 3.

[00:02:35]Did we prove that already?

[00:02:37]Yeah, that's Euler's thing.

[00:02:39]Yeah, that's the Yeah, that's a very doable one, right?

[00:02:42]Then, if they I mean, then they're they're all if they differ by at least three, I think you can do something over here as well. There's like, um these generalized to something called the like Andrews Brascoud identities, I think is what they're called, but don't quote me on that. It might be something else. So, we've got nine, 8 + 1 uh 7 + 2 6 + 3 5 + 4 is not allowed. 5 + uh 3 + 1 Now, uh do we know the generating function?

[00:03:13]Uh the purple generating function? For partitions I'm loving this like these aesthetics here.

[00:03:25]So, do you want the product thing?

[00:03:27]Yeah. Okay, so that's the product over uh 1 over 1 minus Q to the 4 5 K mod plus 4 or whatever.

[00:03:39]And then uh 5 plus M to the Yeah, that's that.

[00:03:46]The blue one's a little more slippery.

[00:03:49]Yeah.

[00:03:52]Yeah, I'll write it down, and then we'll come There are a lot of them.

[00:04:01]Okay, so in the spirit of um this Rogers-Ramanujan identity is a is an example of a product sum identity.

[00:04:08]So, in the spirit of that, it should be a sum.

[00:04:11]This should be a product sum identity.

[00:04:13]So, this is the sum is n goes from zero to infinity of um Q to n squared over um 1 - Q 1 - Q squared all the way to 1 - Q to the n. Now, we had like a symbol for that denominator, but I'll just leave it like that, right?

[00:04:34]So, that's that's this. Like, this is not clear to me immediately.

[00:04:41]Like, after seeing these for a long time, it's like get gets kind of clear to me, but it's not clear at all.

[00:04:49]So, it goes like this.

[00:04:53]So, maybe Y and then I don't think I'm going to do all the details here, but it's generally like this.

[00:05:01]So, um blue Q um so, that's going to be the sum over all um this. So, let's see. 1 is less than or equal to lambda 1, which is less than or equal to lambda 2, all the way up to lambda n um where lambda um K - lambda K - 1 is bigger than or equal to 2 of Q to the lambda 1 added up to lambda n. So, this is like a this is a goofy way of doing this is that notice this encodes the partition rule, right? And you're just adding with the partition in the exponent.

[00:05:55]Right? So if you were to add all these together like and reduce the exponent to just a number, you would get the the thing in here.

[00:06:05]And then this goes like the following. So this is going to be the sum over I think it goes like this. 0 less than or equal to mu 1 less than or equal to mu 2 so on and so forth of Q 1 + mu 1 3 + mu 2 all the way up to 2n what is it going to be? Minus 1 or plus 1? Plus 1 plus mu n. Okay, so now why is that?

[00:06:45]Well, if you've got a partition of this blue type, right?

[00:06:49]Well, the smallest part is going to be 1 plus something, right?

[00:06:54]Yeah.

[00:06:55]But then the next smallest part has to be 3 plus something, right? At least.

[00:07:01]And then if you order the plus somethings in this order, you always achieve this difference of two. The third part has to be 5 plus something.

[00:07:09]So on and so forth. Now let's look at what we can do here. This is going to be equal to the sum [snorts] over these mus, right?

[00:07:19]of Q to the 1 + 3 + 5 ending at 2n minus Sorry, 2n + 1.

[00:07:29]times Q to the mu 1 added up to mu n.

[00:07:35]So notice that this is um 1 + 0 plus um 3 + 5, right? So, the mu 2 would be 5.

[00:07:45]Right. Right? This is 1 + 1 plus 3 + 4. This is a 1 + 2 + 3 + 3, right?

[00:07:57]>> now. And now, like check it out, the the one and the threes are just providing us the distance the difference two condition, and then these extra things right here right? That's a That's a partition of well, like one.

[00:08:19]These are the five partitions. Yeah, they're partitions of five Well, not exactly a partition of five, right? Cuz it includes the number zero, right?

[00:08:25]Good question.

[00:08:27]But, I guess it is like I guess it is up all partitions of five.

[00:08:32]You're just allowed to use only a single part as well.

[00:08:53]Oh, no. Now I remember. Why uh Why are we I think this should be minus one. I'm all like dis- discombobulated as to whether this should be plus one or minus one. All right. Well, if it's plus one then if you sum up all those odd numbers, I think you get n + 1 squared.

[00:09:09]What do you think?

[00:09:12]Yeah, I would change it to n squared.

[00:09:16]Right? Is that true?

[00:09:17]I think that's an identity I know. Yeah, if you add up the first half So, 1 + 3 is 4.

[00:09:23]Uh-huh.

[00:09:23]Right? 1 + 3 + 5 is 9, 3 squared.

[00:09:28]Right?

[00:09:29]1 + 3 + 5 + 7 is 16, 4 squared.

[00:09:34]You get a perfect square. The sum of odd numbers is a perfect square.

[00:09:38]Uh-huh. Yeah. You can prove that with induction, right?

[00:09:42]Yeah.

[00:09:43]And then um So, now you've got something like this, right?

[00:09:54]Good.

[00:09:55]And then you can re-index this, right? Now, here's the step I'm going to skip.

[00:10:02]So, you can re-index this and you get Q to the N squared over um 1 minus Q all the way up to 1 minus Q to the N.

[00:10:12]Right? Because this bit right here becomes like the number of partitions of N.

[00:10:20]Sorry.

[00:10:21]Uh the number of partitions of with at most N parts.

[00:10:30]Yeah, the number of partitions with at most N parts.

[00:10:34]So, that's where you kind of like finish.

[00:10:39]Ah.

[00:10:44]Weep.

[00:10:47]So, that's how that goes.

[00:10:48]Um now But, the proof of this is hard like hard, right? Like I said like before.