One thing I truly appreciate about these videos (beyond the interesting mathematics) is the legible consistent handwriting and the organization of spacing, coloring, and symbols. The consistent speech patterns, delivery, and general organization of thoughts are clear as well. It's not very often I cannot follow something.
Michael, thank you for making the "click bait" comment up front! We all know how the TH-cam algorithm works, but it's so refreshing to see a presenter call it out so candidly! 👍
The q factorial can be used to define the q binomial [m, n]q, which counts the number of n dimensional subspaces of (Fq)^m (here, q is a power of a prime number and Fq is the field with q elements). :)
One easy way to count the inversions is to go through the numbers in the 1st, 2nd,...,nth place, counting how many numbers before them are larger than the number in the current place. for example, for 4 2 1 3: 4 --> 0 larger 2 --> 1 larger 1 --> 2 larger 3 --> 1 larger --------------------- total: 4 inversions This also proves the formula for the max # of inversions, n*(n-1)/2, as follows: you can at max have 1 inverted (larger) element before the 2nd, 2 before the 3rd, ... (n-1) before the n-th, thus in total: 1+2+...+(n-1) = n*(n-1)/2 inversions.
@@TheEternalVortex42 Yeah, so? It's still a very fast and straightforward way to count the # of inversions. If it ain't broke, don't fix it ;-) (of course, if I were to implement this in code, I would never do it this way; it scales as O(n^2), which means it's a terrible algorithm for large sets. But for small sets, and for manual counting, it's far easier than what Michael showed).
My guess this is some q-analog stuff. There's a "q-analog" to a lot of functions from combinatorics, which are continuous, and the regular ones are the limiting case as q goes to 1. It's also related to "quantum" groups, and possibly quantum mechanics.
@@f5673-t1h If you are referring to the zero ring, arguing that it should be considered a field: the properties that “the field with one element” should have, don’t really fit with the zero ring.
To simplify the calculations at 10:53, we can notice that the quantum factorial is also recursive like the “normal” factorial, where [n]!_q = (1-q^n)/(1-q) * [n-1]!_q instead of multiplying everything from scratch.
The way I proved this is more of a combinatorial proof, i.e. showing there is an explicit bijection between every permutation and every possible term in the q factorial. When you expand the q-factorial = (1+q+...q^(n-1)) * (1+q+...q^(n-2)) * ... (1+q)(1), you are every time choosing an element q^{k_1} from the first bracket, q^{k_2} from the second bracket, etc., such that 0
7:30 : it kinda "extend" my concept of ordered operator, ex: the ordered_plus : a ⨭ b = case 1: |a| + |b| if ab case 3: 0 if a=b (I want that when the order relation (a
Nice! On my combinatorics course I've learned a lot of finite and infinite formal power series and what sequences they generate. This is a very interesting topic!
It is important that to determine inversions by swaps (1) Only swap adjacent numbers (2) use the least possible number of swaps I think (2) is equivalent to (2') only swap adjacent numbers that are in-order
Thank you Michael. Recursive deconstruction and then reconstruction, is a great method to solve problems. The partition function is one good example. In coding, the key for low complexity is not to compute the same entry twice.
You can actually just define the number of permutations of n elements with m inversions using the good ol’ binomial coefficients (or “n choose m”). No limits needed.
Both the notation and the 1-q^n/1-q expression look quite similar to things related to integer partitions, which Michael has covered before. Surely that can’t be a coincidence?
Super cool, Michael! It seems the polynomial records the growth of the number of elements of the symmetric group as word length increases. If you have a group with a canonical set of generators (picture its Cayley graph) what can you say about the generating function of the elements a a given word length?
If it counts the „number of swaps“ then it could also be used to determine the sign in the sum of a determinant, which is exactly the number of swaps counted and positive if it is even and negative if it is odd, right? Or how could we relate this to matrices, I mean a determinant is closely related to permutations and in my first semester linear algebra course we were given the definition with the permutation formular.
See, cause I'm thinking, "OK, if permutation inversions, from the perspective of ordinal cardinality, are based on POSITION, then NUMBER OF PLACES matters only ONCE." BUT. If permutation inversions are based on NUMBER AND ORIENTATION OF PLACES AWAY FROM NATURAL POSITION, then I can see how we may incline to the inversion-set membership of that permutation... ... EXCEPT that the permutation is applied across the set as a whole via atomic iteration over each member's unique orientation to the rest of the Members in the set, effectively canceling position-away if position-away is still in the same original ordinality. ... right??
q feels like an operator (with 1 being the identity) So you're creating a new permutation by operating q°q°...°q onto a previous permutation Adding the new number and placing it somewhere with up to n-1 inversions Interesting
So dumb question: why do we write 1 + q + q² + … + q^(n-1) as (1 - qⁿ)/(1 - q) only to "simplify" it back to 1 + q + q² + … + q^(n-1) every time we actually use it?
Is there an advantage to writing the q-factorial using the 'sum of finite geometric series' formula for this discussion? In this explanation, Michael is only using the polynomial expansion anyway, so (for this application, at least) why not just use that as the definition?
@10.47 This is what I call "Lost in Defintion" 🤣🤣 What is the point of writing (1-q)^n/(1-q) and then write (1 + ... + q^n) afterwards? This latter one is the real defintion of n, written on the left part. A total waste of time.
Fin badass. But, umm.... am I incorrect in observing that, in the [4]q! calculation, there were errors in the number of inversions in the five four-inversion members? They had 3-inversion characteristics, not 4-inversion. Remember, 312 is a 3-inversion permutation of 123. Likewise, 4213 is still a 3-inversion, NOT a 4-inversion... ... because 2 is still in it's canonical place. The other three permutations in that set were also erroneous: 4132, because 3 is in canonical, ordinal place, 2431, because, again, 3 is in canonical, ordinal place, and so on. Am I correct? Am I missing anything? Even if correct that there is an error in the set, THIS IS A BADASS CONCEPT of permutation predictions and I'M IN LOVE. That is fin sick, dude.
To clarify: "The addition is happening n-1 times to produce a value of n" is correct. Also, instead, you could just change the subject and say "There are n ONES in the equation", which obviously add up to n. But you made teh subject be additions, of which there are only n-1 there by a trivial counting argument.
This reminds me of z-transforms, where you have the lag operator z^{-1} masquerading as a variable in a polynomial. Is there a way to express q as an operator?
@MichaelPennMath Is there any reason for calling q analogue of the factorial as the Quantum factorial ? I mean does it show up in some way in quantum mechanics ?
Sometimes the q in various q-deformations describes how some operators almost commute, where B A = q A B for some operators A and B. That’s kinda quantum-y ?
Every permutation has an unique decomposition in cycles. (I mean by unique here the number, length and elements involved in the cycles). And a cycle of length m has m-1 inversions. Therefore, every permutation has an unique number of inversions. (Butched proof, I could explain further if you ask nicely).
I know it is math channel. But if you click bait a title that has a meaning at least ask a friend why it is name that way. A few sentences of "not exactly" or "generally..." would be fine.
One thing I truly appreciate about these videos (beyond the interesting mathematics) is the legible consistent handwriting and the organization of spacing, coloring, and symbols. The consistent speech patterns, delivery, and general organization of thoughts are clear as well. It's not very often I cannot follow something.
I agree! I wasn't sure what it was about this channel that kept brining me back but this is it!
Couldn't agree more
Michael, thank you for making the "click bait" comment up front! We all know how the TH-cam algorithm works, but it's so refreshing to see a presenter call it out so candidly! 👍
Quantum makes everything Qooler, even Qalculus
I see what you did there.
You mean Quambinatorics?
Check back in Quoctober
Qlever.
Quantum makes Michael stress Patreon subscription
The q factorial can be used to define the q binomial [m, n]q, which counts the number of n dimensional subspaces of (Fq)^m (here, q is a power of a prime number and Fq is the field with q elements). :)
One easy way to count the inversions is to go through the numbers in the 1st, 2nd,...,nth place, counting how many numbers before them are larger than the number in the current place.
for example, for 4 2 1 3:
4 --> 0 larger
2 --> 1 larger
1 --> 2 larger
3 --> 1 larger
---------------------
total: 4 inversions
This also proves the formula for the max # of inversions, n*(n-1)/2, as follows: you can at max have 1 inverted (larger) element before the 2nd, 2 before the 3rd, ... (n-1) before the n-th, thus in total:
1+2+...+(n-1) = n*(n-1)/2 inversions.
This is great!
Isn't this just the definition for # of inversions?
@@TheEternalVortex42
Yeah, so? It's still a very fast and straightforward way to count the # of inversions. If it ain't broke, don't fix it ;-)
(of course, if I were to implement this in code, I would never do it this way; it scales as O(n^2), which means it's a terrible algorithm for large sets. But for small sets, and for manual counting, it's far easier than what Michael showed).
Bingo!
You baited me and you threw it right in my face the first 10 seconds. Love it.
My guess this is some q-analog stuff.
There's a "q-analog" to a lot of functions from combinatorics, which are continuous, and the regular ones are the limiting case as q goes to 1.
It's also related to "quantum" groups, and possibly quantum mechanics.
Wait so this might not have an actual application for quantum modeling and quantum cosmology.
The limit as q -> 1 is also related to the mysterious 'field with one element':en.m.wikipedia.org/wiki/Field_with_one_element
@@Sqaarg I maintain that it's not mysterious, and it's just mathematicians being too stubborn to acknowledge a field where 0 = 1.
@@f5673-t1h If you are referring to the zero ring, arguing that it should be considered a field: the properties that “the field with one element” should have, don’t really fit with the zero ring.
To simplify the calculations at 10:53, we can notice that the quantum factorial is also recursive like the “normal” factorial, where [n]!_q = (1-q^n)/(1-q) * [n-1]!_q instead of multiplying everything from scratch.
You don't need any clickbait for me. I watch your videos every day before I go to sleep without any exceptions.
Excellent presentation 👌 Thanks !
The way I proved this is more of a combinatorial proof, i.e. showing there is an explicit bijection between every permutation and every possible term in the q factorial. When you expand the q-factorial = (1+q+...q^(n-1)) * (1+q+...q^(n-2)) * ... (1+q)(1), you are every time choosing an element q^{k_1} from the first bracket, q^{k_2} from the second bracket, etc., such that 0
7:30 : it kinda "extend" my concept of ordered operator, ex: the ordered_plus :
a ⨭ b =
case 1: |a| + |b| if ab
case 3: 0 if a=b
(I want that when the order relation (a
In general, polynomials are quite useful for counting choices.
Nice! On my combinatorics course I've learned a lot of finite and infinite formal power series and what sequences they generate. This is a very interesting topic!
9:41coefficient of q^3 is one,but has three inversions.
There is one permutation in S_3 with 3 inversions.
It is important that to determine inversions by swaps
(1) Only swap adjacent numbers
(2) use the least possible number of swaps
I think (2) is equivalent to
(2') only swap adjacent numbers that are in-order
Thank you Michael. Recursive deconstruction and then reconstruction, is a great method to solve problems. The partition function is one good example. In coding, the key for low complexity is not to compute the same entry twice.
You can actually just define the number of permutations of n elements with m inversions using the good ol’ binomial coefficients (or “n choose m”). No limits needed.
Very fine explanation.
Both the notation and the 1-q^n/1-q expression look quite similar to things related to integer partitions, which Michael has covered before. Surely that can’t be a coincidence?
So an inversion of a permutation σ is a pair (i,j) such that iσ(j) ?
i'm loving the new clickbait michael penn era
pretty cool. Didn't get click-baited tho, I was expecting some q-analog. Please, do q-gamma function next.
When I saw the Bruhat order coming from the q factorial... I was shocked! :D
Super cool, Michael! It seems the polynomial records the growth of the number of elements of the symmetric group as word length increases. If you have a group with a canonical set of generators (picture its Cayley graph) what can you say about the generating function of the elements a a given word length?
If it counts the „number of swaps“ then it could also be used to determine the sign in the sum of a determinant, which is exactly the number of swaps counted and positive if it is even and negative if it is odd, right? Or how could we relate this to matrices, I mean a determinant is closely related to permutations and in my first semester linear algebra course we were given the definition with the permutation formular.
See, cause I'm thinking, "OK, if permutation inversions, from the perspective of ordinal cardinality, are based on POSITION, then NUMBER OF PLACES matters only ONCE."
BUT. If permutation inversions are based on NUMBER AND ORIENTATION OF PLACES AWAY FROM NATURAL POSITION, then I can see how we may incline to the inversion-set membership of that permutation...
... EXCEPT that the permutation is applied across the set as a whole via atomic iteration over each member's unique orientation to the rest of the Members in the set, effectively canceling position-away if position-away is still in the same original ordinality.
... right??
q feels like an operator (with 1 being the identity)
So you're creating a new permutation by operating q°q°...°q onto a previous permutation
Adding the new number and placing it somewhere with up to n-1 inversions
Interesting
So dumb question: why do we write 1 + q + q² + … + q^(n-1) as (1 - qⁿ)/(1 - q) only to "simplify" it back to 1 + q + q² + … + q^(n-1) every time we actually use it?
Fascinating
9^x(ln2,25*4^x-ln1,5*6^x)/(4^x-6^x+9^x)² =0 x=log(2/3)0,5+*->x max=1/(0,5²-0,5+1) =4/3
Is there an advantage to writing the q-factorial using the 'sum of finite geometric series' formula for this discussion? In this explanation, Michael is only using the polynomial expansion anyway, so (for this application, at least) why not just use that as the definition?
If you keep it as a fraction, you can write the numerator as a q-Pochhammer symbol, which is presumably useful in some contexts I guess
@10.47 This is what I call "Lost in Defintion" 🤣🤣
What is the point of writing (1-q)^n/(1-q) and then write (1 + ... + q^n) afterwards? This latter one is the real defintion of n, written on the left part. A total waste of time.
Oh boy
Fin badass.
But, umm.... am I incorrect in observing that, in the [4]q! calculation, there were errors in the number of inversions in the five four-inversion members?
They had 3-inversion characteristics, not 4-inversion.
Remember, 312 is a 3-inversion permutation of 123.
Likewise, 4213 is still a 3-inversion, NOT a 4-inversion...
... because 2 is still in it's canonical place. The other three permutations in that set were also erroneous:
4132, because 3 is in canonical, ordinal place, 2431, because, again, 3 is in canonical, ordinal place, and so on.
Am I correct? Am I missing anything?
Even if correct that there is an error in the set, THIS IS A BADASS CONCEPT of permutation predictions and I'M IN LOVE. That is fin sick, dude.
2:00 You meant to say n-1 times right? There are n-1 additions there: 1+1 is one addition, 1+1+1 is two, etc.
To clarify: "The addition is happening n-1 times to produce a value of n" is correct. Also, instead, you could just change the subject and say "There are n ONES in the equation", which obviously add up to n. But you made teh subject be additions, of which there are only n-1 there by a trivial counting argument.
Ohhh, it's the NUMBER OF SWAPS in the permutation validating an inversion, not the number of sent-codomain positions.
OK, got it!
Hmm.
What does this mean?
Excellent presentations. It would be nice if he could better motivate problems.
Any practical significance?
This reminds me of z-transforms, where you have the lag operator z^{-1} masquerading as a variable in a polynomial. Is there a way to express q as an operator?
I wonder where to use this knowledge.
@MichaelPennMath
Is there any reason for calling q analogue of the factorial as the Quantum factorial ? I mean does it show up in some way in quantum mechanics ?
Sometimes the q in various q-deformations describes how some operators almost commute, where B A = q A B for some operators A and B.
That’s kinda quantum-y ?
And the q-gamma function?
Where the hell did all this math come from?
Why do the coefficients seem to make a parabola?
It turns out that these polynomial are what is called "unimodal", which means that they go up and then go down, so have a single mode. Look up this.
4:10 You forgot the subscript q from that [1]. Video ruined.
I love q-analogs.
How do we know the # of inversions is unique?
Every permutation has an unique decomposition in cycles.
(I mean by unique here the number, length and elements involved in the cycles).
And a cycle of length m has m-1 inversions.
Therefore, every permutation has an unique number of inversions.
(Butched proof, I could explain further if you ask nicely).
@@Redentor92 Just took a class on differential geometry. This makes more sense now.
He's so mathematical that he apathetically describes himself doing clickbait. hahahaah
This is weird... All I can see is L'Hoptal.
wow, another YT video that explains factorial :)
Whats the use of it? Some examples please
that title and thumbnail is worthy of an NFT project :D
I know it is math channel.
But if you click bait a title that has a meaning at least ask a friend why it is name that way. A few sentences of "not exactly" or "generally..." would be fine.
how did this channel get so clickbaity i used to love the channel
You must be one of the Q from the Star Trek Q Continum planet😂
I wike math
Measuring a Quantum-equation (x²-2x-3=0) falls into one of its solutions: |Ψ> = 0.707..| -1> + 0.707..|3>
🥸
i have watched till the "if you've been sticking that long till now" and i have subsrcribed botgOFmyTH-camacountz :)