Wow thanks so much! I tried to ask my Calc teacher about why this worked and he basically used the "1 way to arrange 0 objects" explanation, but this is way clearer! I hope your channel keeps growing; you do great stuff!
Agreed! That is a classic explanation, but to me it’s also a bit lazy and not satisfying since I could just as easily imagine a universe where we say “there’s ‘undefined’ ways to arrange 0 things.”
You all got lost along the way. Start with the original definition of factorial. I'm gonna ask one simple question: is zero (0) a positive or negative integer? And what positive integer below it did you multiply with? It would make more sense to work with fractions as factorial between 1 and 0 because 0 itself doesn't have a factorial.
@@JJ_TheGreatthat argument would be much harder to make. You either can’t arrange nothing or there’s only one way to arrange nothing. I can’t imagine an argument for there being infinite ways to arrange nothing that doesn’t also imply there’s infinite ways to arrange, for instance, 1 thing
I like how you emphasize that we define it to be 1 because it's useful to be defined that way, rather than just saying it equals 1. I think it's important to acknowledge what is convention, and why the community has largely chosen that as convention.
@@paulfoss5385 it isn't about "nature". There is no "natural" about math, it's just our own invention. Gamma functions was invented to extend factorial, there is no hidden truth there.
@@paulfoss5385it doesn't "show" as in "proves". Mathematics in itself is a convention we created to study our world. The caveman had an easier time saying "me have 3 rocks" than saying "me have a rock and a rock and a rock". It's why 1+1 = 2 for all objects and we can be sure of it. Same goes with the factorial. We created it as a convention for how you can arrange n objects where the order matters. Now how we arrange 0 objects has no meaning to us, but it needs a certain accepted convention, which in our case, making 0! = 1 simplifies our formulas of any possible exceptions we'd need to create around it. Something somewhat similar usually goes for 0^0. Some calculators just define it as 1 as convention because the limit of x^x as x approaches 0 is 1. But there ARE limits with case 0^0 that converge to 0, so it can be disproven when needed
Agreed, for most of his examples. However, I would say that the empty product directly obviates the need for any “convention” in this case If n! is the product of the first n positive integers, then 0! is the product of no integers, which is automatically 1
@@paulfoss5385It wouldn't mean we're wrong, it would mean we would constantly need to handle separate cases. It is a convention made of convenience. Note that empty product being 1 and multiplicative identity being 1 are technically different things.
Yeah this video is pretty bad and doesn't explain it well at all. The reason its defined like that is because the gamma function is the only continuous function that satisfies that f(x)=xf(x-1) at all points and that is the property that defines factorials
@@vladimirkhazinski3725not really, there are other functions that extends the factorial, the property that makes the gamma function special is that its log is convex
When I teach 10-year-olds (not necessarily about factorials, but with them in mind), I put it this way: If you want to add something without a change, you add 0. If you want to multiply something without a change, you multiply by 1. If you don't add any numbers together, you get 0. If you don't multiply any numbers with each other, you get 1. Essentially, it's about neutral elements but without any terminology needed.
@@marciodemeloferreira3503 Depends on what you mean by this. If you just now realized that math is something we made up in our heads, then good job. If you're saying that nobody ever multiplies by 1 or adds 0, this is pretty blatantly false. If you're saying that multiplication by 1 and addition of 0 have no real-world applications, that's okay; you're not the only person out there who failed third-grade math.
As a programmer, I just need to keep in mind that 1 is the identity for multiplication. So if 0! wasn't that, you'd need to constantly make special cases for it in formulas and calculations.
@@Marconius6 What weird phrasing. "If 2 + 2 wasn't 4, you'd need to constantly make special cases for it in formulas and calculations." Edit: What was I on?
Yeah, in group theory it makes sense to have a⁰=1 for every a in the group. It's the same reason we have that x⁰ = 1 for all x, except 0 itself. Looking at the group of (C, *, 1) the property in groups and the property of exponents we are used to is the exact same.
The reason why the Gamma function is often used as a way to generalize the factorial is because it has the property z*gamma(z) = gamma(z+1), which is just saying that z * (z-1)! = z!. And that property is also the one used when using reasoning from algebra.
Thank you! I so often see people on RUclips disparaging the "algebraic" method in the video and then turning around and saying that the Gamma function is the _"real"_ reason that 0! = 1. But as you say, the reason the Gamma function gives 0! = 1 is because the function is forced to satisfy the recursive property of factorials, not because of being a meromorphic function or because of log convexity or anything like that.
I think #1 and #2 were the only convincing explanations. #3, #4, and #5 all seemed to be confusing correlation and causation. Factorial wasn't defined as "0! = 1" to make those formulas work, those formulas were created and work BECAUSE "0! = 1". Explaining "0! = 1" by showing that there are formulas based on that fact is very unhelpful. #3/#4/#5 kind of come off as trying to explain the value of pi by showing that the area for a circle formula wouldn't work otherwise.
Yeah the 1st two feel like a proof of why 0! = 1 intuitively, whilst the others just leave me feeling like it was defined to be 1 by convention. There's a difference between "This is true and here is why" and "We decided this is true because it was convenient" Edit: and yes, it might just be the case that we defined it like that because it was convenient, but idk, that's not what I was led to believe the video was going to prove
The first two are what you would use as proof , however it is not wrong to say that we made 0!=1 by convention , a lot of the time it is needed to be able to apply certain hypotheses. I think the same can be applied to the number 1 being excluded from being a prime number despite the definition applying it to it
@@SaloCh this is a part of the definition, so it’s not possible to prove. all we can do is show that it’s the most sensible definition that makes as much as possible convenient
The way I think of it is not so much that 1 is left, but that we start from one. Any number, let's say 5, can be thought of as starting from 1 and scaling it up 5 times its size (it's often useful to think like this). So multiplying, let's say 2*3, can be thought of as 1*2*3, meaning, start from a whole of something, scale it to 2 times its size, then scale the result to 3 times its size. Factorial can be thought of this same way. Think of it algorithmically: f
@@bartgertsen6181 What else would you get when you multiply 1 by 0, but for zero iterations? If I have a nonzero A which I add to 0 only zero times, my sum is still zero.
My calculus instructor at community college was amazing and she really did a fantastic job explaining a bunch of concepts. I owe her a LOT in terms of my math prowess, especially in comparison to my peers here in university. However, one thing I'll never forget was her explaining that 0! = 1 was best understood by just taking her word for it lol. She kind of went into the "1 way to arrange" explanation but she made it clear, it's best not to get too hung up on it. Meanwhile, she spent a ton of time explaining famous proofs and other important concepts
There are time when your teacher doesn't really understand fully everything he or she teaches. For example, in Statistics I can explain how to use the tools pretty effectively and often give explanations about why the ideas work but mathematical statistics is a dark and lonely road with creepy trees and best avoided.
Honestly the answer is essentially: "we decided to define it like that cause it behaves more consistently and allows as to avoid having special cases for it". The rest of the explanations is essentially different ways in which 0!=1 is more consistent and useful than any alternative
n! is a decategorification of the set of automorphisms of a set. There is exactly one bijection of the empty set - Its identity function. So naturally, 0! = 1. This is in a way related to your combinatorics example and basically just a rewording of your counting example, but slightly more fundamental in spirit. Since the category of finite sets exists, so does the functorial function. (The formula for the binomial coefficient is in a similar way a decategorification)
If only our math teachers were more like this. I had such difficulty focusing when I was a kid, but as it turns out, when you're entertained, you tend to pay attention. Teachers need to be as much an entertainer as a teacher, because if you're just saying facts like reading the newspaper in a monotone voice, you'll forget. Aside from that, showing graphically on why the things are the way they are helps us remember. Beautiful demonstration.
Yeah sure, let's compare someone who had many hours to prepare a 6 minutes video that is only watched by interested people to a teacher who didn't have much time to prepare hours of content and has in front of him 35 immature children that did NOT choose to be here. Seems legit ! And btw : this guy IS saying facts like he's reading the newspaper, almost all the teachers I know are way more entertaining than he is ... What happened there is simple : you got old. And now you realize some stuff was actually more interesting than you thought when you were a teenager ! This "wish our teachers were like this" needs to stop, it's pure nonsense AND insulting.
What makes this confusing is the rule of anything times 0 = 0. Factorial at its base level is multiplication, so most would think 0! = 0. But the consideration that it means the number of ways to arrange something reveals its not just multiplication.
Other things make it nonsensical or argue against it, as well. Multiplication being essentially equivalent to division by the inverse fraction: 1*2=(1/1)/(1/2)=2 etc. Problem: any number divided by 0 is undefined. As such, presumably 0! must surely be undefined if it can in some way be rewritten as some kind of "division by the inverse fraction," putting 0 in the denominator? ----- More fundamentally, factorials are a multiplicative series with a number of positions equal to the number being factorialized. 5! has 5 "positions" in the expression 5x4x3x2x1 4! has 4 "positions" in the expression 4x3x2x1 ... 1! has 1 position: "1" 0!, presumably must therefore have 0 positions. It is basically the "empty/null set" (). There are no numbers being multiplied together. The null set does not have a "value," does it? The null set does not contain 1, or equal "1" does it? Surely its value, per se, is either 0 or more likely "undefined"? Yeah? ----- IMO, 0 is basically an "illegal number" to use in factorials, just as are negative numbers, basically by definition, since a factorial stops once it reaches the integer 1. 0 is **NOT** included in a factorial. It is simply never included in a factorial. The definition should *be* "Factorials are the multiplicative sequential series of [non-zero,] non-negative numbers between the number being factorialized and 1. ([Zero and] negative numbers are not allowed.)" ----- Just my opinions, when thinking rationally about it.
@@KalijahAnderson It is still multiplication, just an empty multiplication involving no things, which evaluates to 1. You're not multiplying by 0, you're multiplying every positive integer less than or equal to 0; there are no positive integers less than or equal to 0, so you have an empty product.
So excellent! All this time I'd only known the combinations reason, but I always assumed 0! was just arbitrarily defined as 1 so the combination formula would always work.
This is… kinda crazy. I’ve always just accepted this without understanding and the way you explained it so simply at first and kept building up is such a nice way to learn the reason behind this! not a math major myself and yet while this is my first time hearing about Gamma and stuff the explanation still made sense.
As someone who has sadly only been able to get up to the calculus level in my math education, or slightly beyond (i'm aiming for math major, but it takes time), i had never really thought about 0! Being 1 due to 1 being the muliplicative identity... but i like that
In a computer program, if you want to compute a product p by computing each factor and incorporating it into the running multiplication leading to the final product p, you will initialize p to 1. I.e. the empty product is 1. Similarly, the empty sum is zero. To calculate 4!, start with the 1empty product 1, and include the factors 4,3,2,1. To calculate 0!, start with 1, and note there are no additional factors.
Very nice video! Here's one more that is more "math major" in construction but "feels like" level 1. The symmetric group on n objects Sn has as its underlying set all bijective self-maps on a set of cardinality n. This is just what a permutation is, so it should come as no surprise that the cardinality of Sn is just n!. Let's consider S0. Well, this is all bijective self-maps on a set of cardinality 0. Only one such set exists: the empty set. There is indeed an empty function from the empty set to itself: it vacuously satisfies univalence and being left-total. It similarly vacuously satisfies bijectivity. Thus at least one object in S0 exists. It is not hard to see that it is the only object, since any other such function would have to vary in at least one output, but there are no outputs to vary. Therefore S0 has cardinality 1. And, as we know, the cardinality of Sn is n!, so we have that 0! = 1
Awesome video, keep it up! I like that the same mathematical fact can be explained in several ways to connect it to all those different settings where this question would arise.
Nice. I was always fighting against this concept ( I wanted 0! to be 0), but this video finally convinced me. As soon as you went into group theory and identity element of multiplication, it was a major eye opener. I don't even remember when was the last time when a single sentence suddenly crushed my world view with "oh, that's why! it's so simple!"
@@jazzabighits4473 There is one way, and here it is: However, I think I may understand the confusion here. Babies develop something called object permanence between the ages of 4 and 7 months. This is the understanding that something can exist even if you don't see it. For instance, if the baby's mother leaves the room, the baby understands that the mother still exists. I assume you have simply failed to develop object permanence as a baby.
@@isavenewspapers8890 So because I logically point out that you can't arrange a group of nothing at all, it's undefined, you think I lack object permanence? A group of 10 things divided into 10 people is 1, a group of 1 into 1 is 1, a group of 1 divided into 0 or a group of zero divided into 1 (or zero or anything) is undefined. But I guess that doesn't matter to you. But babies have it, because they understand zero factorial? Lol alright kid.
I think it's worth mentioning that you could tie Level 2 with Level 5: observe that (-1)! is undefined in the same way that dividing by zero is undefined, diverging to positive infinity approaching from the right, and diverging to negative infinity approaching from the left. This is the same as what we would expect from using the algebraic explanation, 0!/0 = 1/0.
0! should ALSO be "undefined," *is* technically undefined/illegal [since **actual** factorials stop at multiplying by 1 and actually multiplying by zero itself would just always yield zero], if people were actually being honest about it. Zero is **essentially** an illegal integer in factorials. As, presumably, are negative numbers... Yeah? Wolfram|Alpha says this of factorials: "n! is a sequence with integer values for nonnegative n." I would, personally, amend that to "n! is a sequence with [non-zero] integer values for nonnegative n." ----- "Because we say so" and "because it's convenient" are not good enough, and if push comes to shove, there is no actual legitimate proof for the identity 0!=1 because 0!=/=1 they are not, in fact equal. They are only asserted to be so by fiat not because anything actually proves them to be so. ----- 0! is basically the null set, not 1 or any other value. It is undefined. One should think of factorials as being a multiplicative series of sequential integers with a number of positions equal to the integer being factorialized. 1! has 1 position: (_) or (1) 2! has 2 positions: (_x_) or (1x2) 3! has 3 positions: (_x_x_) or (1x2x3) 3! has 4 positions: (_x_x_x_) or (1x2x3x4) 3! has 5 positions: (_x_x_x_x_) or (1x2x3x4x5) How many positions does 0! have? None: () it is the "empty/null set." There are **no numbers** being multiplied together. Its value is not 1, its value is "undefined." It is a non-expression. There is literally **nothing**. ----- Also, multiplication and division are basically interchangeable by using the inverse fraction. So, 1*2=(1/1)/(1/2)=2 1*2*3=((1/1)/(1/2))/(1/3)=6 1*2*3*4=(((1/1)/(1/2))/(1/3))/(1/4)=24 But this is a problem for 0, since: 1) any number divided by zero is ... undefined or basically an illegal operation. 2) Since there are no "positions" for any numbers or fractions in 0! [0 positions], technically one can't even really **do** a division by said "undefined" fraction. Heck, there's not even a "position" **just** for the 0 itself **in order** to invert it let alone enough positions to put it in an expression relating it **with** another number. Again, it's just basically the "null set." No positions, no "value," etc.
It really does feel like an explanation to a third grader...we made it that way...ie because I said so. Unfortunately all of the explanations feel like they are circular arguments. The last one didn't work for negative 1...
Except the first explanation has the most holes in it. Actually the first explanation would be better suited to proving 0! = 0. “putting no flags on” is not a solution to a permutation. If you have 1 flag there are 1! ways of putting it up (aka 1 way). If not putting up the flag was a solution 1! would be 2 (either you put that 1 flag up, or not)
this came at a perfect time! i was trying to derive some formula i learned in stat today and this one trick I didn't know was the bridge I needed to finally understand it.
I feel like the first example is confusing. For example, with 0!, it’s said there is one way to put zero flags up. But, if that’s true, then 1! would equal 2 because there can be one flag or no flag.
Think of 2 factorial. There is a flag pole with a green flag and a red flag. 2! Is 2 because you can either arrange it with the green flag higher than the red flag, or the red flag higher than the green flag. There are 2 ways you can arrange 2 flags. The options where you remove one or both flags are not included in the way that 2 flags can be arranged. There must be 2 flags. Likewise “no flags” is not an option in 1 factorial; There must be one flag, and there is only one way to arrange that one flag. And for 0! There is only one way to arrange no flags
@Abcde7213 I understand, but I gotta admit it's confusing that "no flags" count as a way to organize 0 flags but doesn't count as a way to organize 1 flag.
I thought about this for a while. And I think the best way to think of it and not be confused is instead of thinking about it from the top down.. so n! = n x (n-1) x (n-2)...x 1. Instead start with 0! and simply define it as 1 by convention. Then 1! REALLY becomes 1 x 1, 2! Becomes 2 x 1 x 1, etc. But there's always that extra times 1 that could be thought of as "and also, multiply it by the empty set identity"
When crossing out products the assumption is that a 1 remains, in how a lot of people write math. No special reason to handle it differently with factorials
I think the reason some people find 0!=1 strange is because they wonder where the 1 has come from. When they see , say, 5! It seems like you are being given 5 cookies, it’s there explicitly stated, and now you’ve got 5 it seems reasonable that you can repeatedly decrement them. Like eating each cookie until they’re all gone. But when the same thought is applied to 0! , It feels like not being given a cookie, taking a bite of it anyway, and then having a cookie. But where did it come from?
That's covered by the first explanation. If you're giving 0 cookies, you have one (1) way to eat it: Not to eat it. That's the only way. And that's one explanation for 0! = 1. Because n! has nothing to do with how many cookies you have, it's about how many way there are to eat them (assuming your cookies are discrete).
I can see what you are saying, and I do agree. But I don’t think that’s how ! Is taught to people at first, it is taught as a little algorithm where you start with N and keep multiplying with the number one lower until you get to 1. If we are saying that ! Is really the number of combinations, then I’d ask what is that other thing, the algorithm that people get taught at school first.
So what i’m really saying is, if the 1st level of encountering N! Is “starting with N keep multiplying by each number lower and stop when you get to 1” ( which, right or wrong, I think is how it is first taught at school) Then 0! Would be 0 * -1 *-2 * -3 etc… never stopping, and simply equal to 0 But of course, that is not the same ! as the one used for’ number of combinations’ , but that’s my point, if it is a different ! Then why wouldn’t the value of 0! also potentially be different?
I would've strengthened the math major argument a bit because it fails to argue why the gamma function is the continuation we want. (1) Yes, there are multiple ways to extrapolate the factorial for nonzero naturals to the reals, but all analytic continuations of the factorial will agree on the integers and indeed predict 0!=1. (2) Though tangential to the original point, if you want to define the gamma function, it's good to point out that it's the only analytic continuation that always obeys the defining equation of the factorial Gamma(x) = (x-1)Gamma(x-1) .
Interesting video! Constructive tip: Either keep your hands in frame all the time or maybe gesture higher so your whole hand enters the frams instead of just your fingers periodically popping up. :)
If we only defined the Gamma function to be Gamma(z) = \int_0^\infty t^z e^-t dt, then we could have had Gamma(n) = n! instead of (n-1)!. Why did that -1 have to creep in there?
Not sure I can get my head around the Algebra part, so perhaps someone can explain a bit... I get that 4!= 4.3.2.1 can then set as 3!=4!/4.... because it seems you're just divding both by four...(same for 3,2 and 1)... but it seems weird that suddenly 0! shows up after 1!.. The 0 was never in the product to begin with so how does it show up now? Would this be the case that we're always just calculating (n-1)! each time instead quotients/products? Im reading this and even realising its a hella complicated thing to explain...
A product series to allow for negative factorials arrives at an imaginary j=0!, as opposed to i. The j operator commonly appears as ±, but it flips signs on every operation due to making selections. Oliver Heaviside used it to describe the field of selections 1892 solution for impedance. Because it initially selects position (+1), it is easy to show 0!=+1. You have to go out of your way to solve for negatives to see it can also be -1. Because it flips signs on each operation, j^2=-1, j+j=0, and j-j=2j. These qualities streamlined Heaviside's solution. The really surprising thing is that you can see it in the math, but no one bothered to document the seemingly paradoxical qualities. I've been working on an imaginary numbers book if you want mind bending reading to put you to sleep at night. ;)
I think we should change the definition of factorial for natural numbers to: 1! = 1 n! = (n+1)!/(n+1) So that we can derive the same formula we have without explicitly stating 0! = 1 as a base case. It would be equivalent, but it would be easier to accept for people.
I think it's helpful to think of factorial as in the same tier as powers, logs, and roots etc., the tier above multiplication and division, on which they are all based.
I think that the cleanest way to define it is, as you alluded to, the empty product: use sets, and, using the property that multiplying all of the elements of a set can be done via sub-sets, extend this to the empty product. x = \Pi_{A} = \PI_{A} \times \Pi_{} = x \times \Pi_{} , so \Pi_{} = 1, and 0! is just the product of the elements of the empty set.
It's also convenient that the same 0! appears when you count how many "words" (real or nonsense) you can spell using exactly 1 A, 2 L's, and 1 B (and 0 C's and 0 D's and...). The alphabet has 26 letters, but the formula can be written without explicitly excluding the unused letters. The "contribute nothing to the multiverse of possibilities" action when multiplying possible outcomes is the multiply or divide by 1 action, as you say.
the last method is very common use for me, it's very common to deduce solutions using correlating functions like that in all of science and especially chemistry and biochemistry I truly am good at procrastinating to be watching this content, but the videos are high quality.
I was thinking the following. The number of bijections between a set of size n and a set of size n is n! So 0! should be the number of bijections from the empty set and itself (this bijection does exist and is equal to the empty set when we consider functions as subsets of the cartesian product). We also know that the number of functions from a set of size n to a set of size m is n^m. So this would also give 0^0 =.1 as being a sensible definition...
Brilliant explanation of a **nonsensical** and **technically wrong** assertion. Convenient, maybe. Correct? No. The null set / empty set does not have a value of 1, it is **at best** "undefined."
i like to think to factorial integer numbers (not negative) as the multiplication of all natural numbers from 1 to that number, so considering there's no natural number smaller than 1, then 0! should be 1
A way that i thought of it is that there are an infinite number of trailing multiply by ones at the end of any number and its still equivalent, so by going from 1! = 1 = 1x1x1... to 0! =1x1x1.../1 = 1. Taking one 1 out of the whole doesnt change the end result
Im really happy that I knew all of this already. I was expecting something new / crazy. Great video btw! 0! being synonymous with a “multiplicative identity” is a cool idea, with potential useful insights in Unital-magmoid Theory: There is probably a lot of research on those very specific Groups anyhow.
There is one very intuitive way to think about it. Going from factorial 3! To 2!, you take away term 3 in 3!. Now you can imagine infinite 1 being multiblied by factorial. So 2! = 2 × 1 x 1 x 1 x .... . If we want 0!, just take away 2 and 1 and we are still left with all the infinite 1 terms . So 0!= 1 x 1 x 1 ... which is obviously 1. Beauty here is that it clearly shows how it fails to see (-1)! With intuitive sense, and perhaps we can only use gamma to extend definition.
So you are suggesting that there are many hints on why, in order to have math as clean and beautiful as possible, we should assume 0!=1, as if we are on one side just choosing it to be so, but under another perspective, discovering some underlying pattern that's already there and choosing convections according to it. That's interesting
Too bad the definition is wrong / nonsensical if one actually examines it correctly. Technically 0! is illegal and/or undefined. Period. Just as is dividing by 0. And "because we say so" is not an actual proof of said identity "0!=1", of which there is, and can be, none. Assertions "by fiat" to the contrary aside...
Level 1 can be described formally with the empty set and vacuous truths. This is what I consider the "morally correct" definition. Any function X->Y can be formally defined as a graph, or subset G of the Cartesian product XxY for which there is no x in X and distinct y,y' in Y such that (x,y),(x,y') are both in G ("vertical line test"). The function is a bijection if (i) there is no distinct x,x' in X and y in Y such that (x,y),(x',y) are both in G (one-to-one), and (ii) there is no y in Y for which there is no x in X such that (x,y) is in G (onto). Notice I rephrased "there exists" statements into "there are no counterexamples" statements. Since {}x{}={}, the only function {}->{} is the empty function, i.e. {} itself, which is in fact a bijection. 0! is the number of bijections on a set of size 0, so 0!=1. This also shows why 0^0=1 too. And justifies the Level 2 and Level 3 descriptions, as well as power series (Level 4) interpreted as generating functions (also cf combinatorial species).
I can't lie, being exposed to calculus is what will make me understand things. Especially how integrating the surface area of a sphere gets its volume.
1:06 okay, but why would you say that there is one way to put 0 flags on a flagpole, and one way to put 1 flag on a flagpole. Wouldn’t it then be more consistent to say that there are 2 ways to put a flag on a flagpole. Positive and negative?
1:00: If 0! is how many ways there are to put zero flags on one pole, then 1! is how many ways there are to put 1 flag on 2 poles (1! = 2), 2! is how many ways there are to put 2 flags on 3 poles (2! = 6), 3! is how many ways there are to put 3 flags on 4 poles (3! = 24), etc., all of which is WRONG. Correct is: 0! is how many ways there are to put zero flags on zero poles, but that is not what he said in the video.
Thanks for this! I get these pretty much up to the gamma funtion. And I even get why Γ(0)=1. But the video explains that there are other ways besides Γ to extend the positive integer factorial function to the reals. So in some sense, isn't the gamma function explanation a bit circular? If we were instead to start from a presupposition that 0! is undefined, we simply wouldn't use Γ to extend ! to reals, right? So I suppose the argument is less "Well, the gamma function makes it one" and more "If we wanted gamma of zero to be undefined, we'd need something way more complicated than gamma to represent it." Am I on the right track there? I'm not trying to challenge the definition that 0!=1, just trying to clearly understand the reasoning.
The Gamma function is the most useful extension to the reals. If you additionally ask that the extension be log convex, then the gamma function is the only way to extend factorial to R. f is log convex just means that log f is a convex function.
For the fourth one, I do not understand how essentially, "it looks nice" is a justification, let alone a proof for why 0! should equal 1. Yes, defining it as such, makes the Taylor series expansion of the exponential function work very well, but that just can not be justification.
It's on ProQuest which requires a university login, but I'll get it added to my website in the next couple days so I can share a link. I don't have a video about it yet, but I think that's a great idea!
Very good video I don't quite get the Combinations at first but it makes sense 😺😺😺 I still do not fully understand the calculus or the math major but I hope to one day
If you continue the diving rule for (-1)! you would get (-1)!=(0!)/0=1/0=inf When you use the binomial formula to get pascal's triangle you divide by inf outside the triangle so it's outside 0 (which makes sense when you apply the "add rule" with 0 and 1 to get 1 on the side of the triangle) And when you use (-1)! in Taylor's formula to get the "more left" terms you would get something divided by inf which is again 0 so you add a bunch of 0's _(Math from Ohio)_ 💀
Wow thanks so much! I tried to ask my Calc teacher about why this worked and he basically used the "1 way to arrange 0 objects" explanation, but this is way clearer! I hope your channel keeps growing; you do great stuff!
Agreed! That is a classic explanation, but to me it’s also a bit lazy and not satisfying since I could just as easily imagine a universe where we say “there’s ‘undefined’ ways to arrange 0 things.”
@@mattholsten7491Agreed! You could argue that there is an infinite number of ways to arrange 0 objects, since they do not exist!
You all got lost along the way. Start with the original definition of factorial. I'm gonna ask one simple question: is zero (0) a positive or negative integer? And what positive integer below it did you multiply with?
It would make more sense to work with fractions as factorial between 1 and 0 because 0 itself doesn't have a factorial.
@@mattholsten7491there’s actually 1 way to do it. Bc you can’t do anything
@@JJ_TheGreatthat argument would be much harder to make. You either can’t arrange nothing or there’s only one way to arrange nothing. I can’t imagine an argument for there being infinite ways to arrange nothing that doesn’t also imply there’s infinite ways to arrange, for instance, 1 thing
I like how you emphasize that we define it to be 1 because it's useful to be defined that way, rather than just saying it equals 1. I think it's important to acknowledge what is convention, and why the community has largely chosen that as convention.
I think the gamma function pretty clearly shows that nature agrees it equals one. If we had a different convention, then we'd simply be wrong.
@@paulfoss5385 it isn't about "nature". There is no "natural" about math, it's just our own invention. Gamma functions was invented to extend factorial, there is no hidden truth there.
@@paulfoss5385it doesn't "show" as in "proves". Mathematics in itself is a convention we created to study our world. The caveman had an easier time saying "me have 3 rocks" than saying "me have a rock and a rock and a rock". It's why 1+1 = 2 for all objects and we can be sure of it. Same goes with the factorial. We created it as a convention for how you can arrange n objects where the order matters. Now how we arrange 0 objects has no meaning to us, but it needs a certain accepted convention, which in our case, making 0! = 1 simplifies our formulas of any possible exceptions we'd need to create around it.
Something somewhat similar usually goes for 0^0. Some calculators just define it as 1 as convention because the limit of x^x as x approaches 0 is 1. But there ARE limits with case 0^0 that converge to 0, so it can be disproven when needed
Agreed, for most of his examples. However, I would say that the empty product directly obviates the need for any “convention” in this case
If n! is the product of the first n positive integers, then 0! is the product of no integers, which is automatically 1
@@paulfoss5385It wouldn't mean we're wrong, it would mean we would constantly need to handle separate cases. It is a convention made of convenience. Note that empty product being 1 and multiplicative identity being 1 are technically different things.
Actual Level 5: "Yeah we just define it like that" - "understandable have a nice day"
Yeah, but you cannot define it other ways because analytic continuation doesn't work then.
@@mateherbay2289You could, but you'd just have to work with the consequences of a function being non-continuous right?
@@yuezienah, for any value you assign to 0! there would be continuos functions that go through n! for all n
Yeah this video is pretty bad and doesn't explain it well at all. The reason its defined like that is because the gamma function is the only continuous function that satisfies that f(x)=xf(x-1) at all points and that is the property that defines factorials
@@vladimirkhazinski3725not really, there are other functions that extends the factorial, the property that makes the gamma function special is that its log is convex
When I teach 10-year-olds (not necessarily about factorials, but with them in mind), I put it this way:
If you want to add something without a change, you add 0.
If you want to multiply something without a change, you multiply by 1.
If you don't add any numbers together, you get 0.
If you don't multiply any numbers with each other, you get 1.
Essentially, it's about neutral elements but without any terminology needed.
This ist actully a perfect explanation for a 10-year-old. I have 2 questions.
Is he a Genius?
Where do you live?
Horrible
@@AyanSharma-i9fEpic troll
In the real world there is no such thing as multiply by 1 or add zero. That is not a real thing...
@@marciodemeloferreira3503 Depends on what you mean by this. If you just now realized that math is something we made up in our heads, then good job. If you're saying that nobody ever multiplies by 1 or adds 0, this is pretty blatantly false. If you're saying that multiplication by 1 and addition of 0 have no real-world applications, that's okay; you're not the only person out there who failed third-grade math.
As a programmer, I just need to keep in mind that 1 is the identity for multiplication. So if 0! wasn't that, you'd need to constantly make special cases for it in formulas and calculations.
True, so the special case is still there, just implemented in fact(0) 😅
As a programmer, in the preview just true: 0 not equals to 1
Base case
@@Marconius6 What weird phrasing. "If 2 + 2 wasn't 4, you'd need to constantly make special cases for it in formulas and calculations."
Edit: What was I on?
As a software engineer, I can confirm that 0 is indeed not equal to 1.
ya mean not 0 equals 1
Ha, I'ma steal that pun for my students
This is what I thought after seeing the thumbnail
I LOLed so take my upvote.
Maybe not in quantum qubit😅😂
never heard of the "zero product must be 1" before but it makes total sense when you think about it in group algebra.
Yeah, in group theory it makes sense to have a⁰=1 for every a in the group. It's the same reason we have that x⁰ = 1 for all x, except 0 itself. Looking at the group of (C, *, 1) the property in groups and the property of exponents we are used to is the exact same.
The reason why the Gamma function is often used as a way to generalize the factorial is because it has the property z*gamma(z) = gamma(z+1), which is just saying that z * (z-1)! = z!. And that property is also the one used when using reasoning from algebra.
Thank you! I so often see people on RUclips disparaging the "algebraic" method in the video and then turning around and saying that the Gamma function is the _"real"_ reason that 0! = 1. But as you say, the reason the Gamma function gives 0! = 1 is because the function is forced to satisfy the recursive property of factorials, not because of being a meromorphic function or because of log convexity or anything like that.
I think #1 and #2 were the only convincing explanations. #3, #4, and #5 all seemed to be confusing correlation and causation. Factorial wasn't defined as "0! = 1" to make those formulas work, those formulas were created and work BECAUSE "0! = 1". Explaining "0! = 1" by showing that there are formulas based on that fact is very unhelpful.
#3/#4/#5 kind of come off as trying to explain the value of pi by showing that the area for a circle formula wouldn't work otherwise.
There is no correlation here and that is a very unhelpful comparison to make. Second half of your argument sounds awfully platonic too.
Yeah the 1st two feel like a proof of why 0! = 1 intuitively, whilst the others just leave me feeling like it was defined to be 1 by convention.
There's a difference between "This is true and here is why" and "We decided this is true because it was convenient"
Edit: and yes, it might just be the case that we defined it like that because it was convenient, but idk, that's not what I was led to believe the video was going to prove
The first two are what you would use as proof , however it is not wrong to say that we made 0!=1 by convention , a lot of the time it is needed to be able to apply certain hypotheses.
I think the same can be applied to the number 1 being excluded from being a prime number despite the definition applying it to it
Thats cause (no offense) level 3+ requires more then just a basic highschool understanding of math
@@SaloCh this is a part of the definition, so it’s not possible to prove. all we can do is show that it’s the most sensible definition that makes as much as possible convenient
The way I think of it is not so much that 1 is left, but that we start from one. Any number, let's say 5, can be thought of as starting from 1 and scaling it up 5 times its size (it's often useful to think like this). So multiplying, let's say 2*3, can be thought of as 1*2*3, meaning, start from a whole of something, scale it to 2 times its size, then scale the result to 3 times its size. Factorial can be thought of this same way. Think of it algorithmically:
f
I like this explanation. Where can I find more information on it?
same reason 0^0 = 1, because 1 is the multiplicative identity
@@PROtoss987that doesnt make sense to me. How does 0^0 relate to the multiplicative identity. That just means that A*1=A.
@@bartgertsen6181 What else would you get when you multiply 1 by 0, but for zero iterations? If I have a nonzero A which I add to 0 only zero times, my sum is still zero.
Please keep making videos like this. This makes getting back into math easier and more enjoyable.
My calculus instructor at community college was amazing and she really did a fantastic job explaining a bunch of concepts. I owe her a LOT in terms of my math prowess, especially in comparison to my peers here in university. However, one thing I'll never forget was her explaining that 0! = 1 was best understood by just taking her word for it lol. She kind of went into the "1 way to arrange" explanation but she made it clear, it's best not to get too hung up on it. Meanwhile, she spent a ton of time explaining famous proofs and other important concepts
There are time when your teacher doesn't really understand fully everything he or she teaches. For example, in Statistics I can explain how to use the tools pretty effectively and often give explanations about why the ideas work but mathematical statistics is a dark and lonely road with creepy trees and best avoided.
Honestly the answer is essentially: "we decided to define it like that cause it behaves more consistently and allows as to avoid having special cases for it". The rest of the explanations is essentially different ways in which 0!=1 is more consistent and useful than any alternative
"multiplicative identify" really hit the spot for me
Thank you for taking my suggestion and doing a video about it! I see you incorporated the nC0 into the video! It makes more sense now!
Glad you liked it, and thanks for the suggestion!
n! is a decategorification of the set of automorphisms of a set. There is exactly one bijection of the empty set - Its identity function. So naturally, 0! = 1. This is in a way related to your combinatorics example and basically just a rewording of your counting example, but slightly more fundamental in spirit. Since the category of finite sets exists, so does the functorial function. (The formula for the binomial coefficient is in a similar way a decategorification)
If only our math teachers were more like this. I had such difficulty focusing when I was a kid, but as it turns out, when you're entertained, you tend to pay attention. Teachers need to be as much an entertainer as a teacher, because if you're just saying facts like reading the newspaper in a monotone voice, you'll forget.
Aside from that, showing graphically on why the things are the way they are helps us remember.
Beautiful demonstration.
TRUE!!
I could very easily understand how 0! was 1 with this video, but by searching all I learned was that 0! is 1 because it just is..
Yeah sure, let's compare someone who had many hours to prepare a 6 minutes video that is only watched by interested people to a teacher who didn't have much time to prepare hours of content and has in front of him 35 immature children that did NOT choose to be here. Seems legit !
And btw : this guy IS saying facts like he's reading the newspaper, almost all the teachers I know are way more entertaining than he is ...
What happened there is simple : you got old. And now you realize some stuff was actually more interesting than you thought when you were a teenager !
This "wish our teachers were like this" needs to stop, it's pure nonsense AND insulting.
@@Thechessrocker1 has a stroke trying to read the first part
@@Thechessrocker1I agree with this take
What makes this confusing is the rule of anything times 0 = 0.
Factorial at its base level is multiplication, so most would think 0! = 0. But the consideration that it means the number of ways to arrange something reveals its not just multiplication.
Other things make it nonsensical or argue against it, as well.
Multiplication being essentially equivalent to division by the inverse fraction:
1*2=(1/1)/(1/2)=2
etc.
Problem: any number divided by 0 is undefined. As such, presumably 0! must surely be undefined if it can in some way be rewritten as some kind of "division by the inverse fraction," putting 0 in the denominator?
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More fundamentally, factorials are a multiplicative series with a number of positions equal to the number being factorialized.
5! has 5 "positions" in the expression 5x4x3x2x1
4! has 4 "positions" in the expression 4x3x2x1
...
1! has 1 position: "1"
0!, presumably must therefore have 0 positions. It is basically the "empty/null set" (). There are no numbers being multiplied together. The null set does not have a "value," does it? The null set does not contain 1, or equal "1" does it? Surely its value, per se, is either 0 or more likely "undefined"? Yeah?
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IMO, 0 is basically an "illegal number" to use in factorials, just as are negative numbers, basically by definition, since a factorial stops once it reaches the integer 1. 0 is **NOT** included in a factorial. It is simply never included in a factorial. The definition should *be* "Factorials are the multiplicative sequential series of [non-zero,] non-negative numbers between the number being factorialized and 1. ([Zero and] negative numbers are not allowed.)"
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Just my opinions, when thinking rationally about it.
@@KalijahAnderson It is still multiplication, just an empty multiplication involving no things, which evaluates to 1. You're not multiplying by 0, you're multiplying every positive integer less than or equal to 0; there are no positive integers less than or equal to 0, so you have an empty product.
You look so young yet you've achieved so much. Very impressive sir.
Saying that 1 is the neutral element to a product is just the most simple way to explain it, and the most logical way too, in my opinion.
Fantastic delivery and clear explanation. I wish I had you as my calc professor in college!
Wow. So crisp and clear. Thanks!
So excellent! All this time I'd only known the combinations reason, but I always assumed 0! was just arbitrarily defined as 1 so the combination formula would always work.
@@FlatEarthMath That's not what "arbitrarily" means.
This is… kinda crazy. I’ve always just accepted this without understanding and the way you explained it so simply at first and kept building up is such a nice way to learn the reason behind this! not a math major myself and yet while this is my first time hearing about Gamma and stuff the explanation still made sense.
As someone who has sadly only been able to get up to the calculus level in my math education, or slightly beyond (i'm aiming for math major, but it takes time), i had never really thought about 0! Being 1 due to 1 being the muliplicative identity... but i like that
In a computer program, if you want to compute a product p by computing each factor and incorporating it into the running multiplication leading to the final product p, you will initialize p to 1. I.e. the empty product is 1. Similarly, the empty sum is zero. To calculate 4!, start with the 1empty product 1, and include the factors 4,3,2,1. To calculate 0!, start with 1, and note there are no additional factors.
Very nice video!
Here's one more that is more "math major" in construction but "feels like" level 1.
The symmetric group on n objects Sn has as its underlying set all bijective self-maps on a set of cardinality n. This is just what a permutation is, so it should come as no surprise that the cardinality of Sn is just n!. Let's consider S0. Well, this is all bijective self-maps on a set of cardinality 0. Only one such set exists: the empty set. There is indeed an empty function from the empty set to itself: it vacuously satisfies univalence and being left-total. It similarly vacuously satisfies bijectivity. Thus at least one object in S0 exists. It is not hard to see that it is the only object, since any other such function would have to vary in at least one output, but there are no outputs to vary. Therefore S0 has cardinality 1. And, as we know, the cardinality of Sn is n!, so we have that 0! = 1
Awesome video, keep it up! I like that the same mathematical fact can be explained in several ways to connect it to all those different settings where this question would arise.
The new math channel in the market ❤
"Level 2" is my favourite. Explains the reasoning very logically.
Currently working on my chemistry PhD thesis and focusing on anything but that. Interesting video!
Nice. I was always fighting against this concept ( I wanted 0! to be 0), but this video finally convinced me. As soon as you went into group theory and identity element of multiplication, it was a major eye opener. I don't even remember when was the last time when a single sentence suddenly crushed my world view with "oh, that's why! it's so simple!"
It is 0, don't fall for this. There are 0 ways to arrange 0 things.
@@artursruseckis4242 What did you even expect to gain by fighting against this?
@@jazzabighits4473 There is one way, and here it is:
However, I think I may understand the confusion here. Babies develop something called object permanence between the ages of 4 and 7 months. This is the understanding that something can exist even if you don't see it. For instance, if the baby's mother leaves the room, the baby understands that the mother still exists. I assume you have simply failed to develop object permanence as a baby.
@@isavenewspapers8890 So because I logically point out that you can't arrange a group of nothing at all, it's undefined, you think I lack object permanence? A group of 10 things divided into 10 people is 1, a group of 1 into 1 is 1, a group of 1 divided into 0 or a group of zero divided into 1 (or zero or anything) is undefined. But I guess that doesn't matter to you.
But babies have it, because they understand zero factorial?
Lol alright kid.
@@isavenewspapers8890 Maybe a simple series will help you understand:
3! = 3x2x1 = 6
2! = 2x1 = 2
1!= 1x1 = 1
0! = 0x1 or 0x0 = 0
really great math video. it definitely deserves more.
I think it's worth mentioning that you could tie Level 2 with Level 5: observe that (-1)! is undefined in the same way that dividing by zero is undefined, diverging to positive infinity approaching from the right, and diverging to negative infinity approaching from the left. This is the same as what we would expect from using the algebraic explanation, 0!/0 = 1/0.
0! should ALSO be "undefined," *is* technically undefined/illegal [since **actual** factorials stop at multiplying by 1 and actually multiplying by zero itself would just always yield zero], if people were actually being honest about it. Zero is **essentially** an illegal integer in factorials. As, presumably, are negative numbers... Yeah?
Wolfram|Alpha says this of factorials:
"n! is a sequence with integer values for nonnegative n."
I would, personally, amend that to "n! is a sequence with [non-zero] integer values for nonnegative n."
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"Because we say so" and "because it's convenient" are not good enough, and if push comes to shove, there is no actual legitimate proof for the identity 0!=1 because 0!=/=1 they are not, in fact equal. They are only asserted to be so by fiat not because anything actually proves them to be so.
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0! is basically the null set, not 1 or any other value. It is undefined.
One should think of factorials as being a multiplicative series of sequential integers with a number of positions equal to the integer being factorialized.
1! has 1 position: (_) or (1)
2! has 2 positions: (_x_) or (1x2)
3! has 3 positions: (_x_x_) or (1x2x3)
3! has 4 positions: (_x_x_x_) or (1x2x3x4)
3! has 5 positions: (_x_x_x_x_) or (1x2x3x4x5)
How many positions does 0! have? None: () it is the "empty/null set." There are **no numbers** being multiplied together. Its value is not 1, its value is "undefined."
It is a non-expression. There is literally **nothing**.
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Also, multiplication and division are basically interchangeable by using the inverse fraction.
So, 1*2=(1/1)/(1/2)=2
1*2*3=((1/1)/(1/2))/(1/3)=6
1*2*3*4=(((1/1)/(1/2))/(1/3))/(1/4)=24
But this is a problem for 0, since:
1) any number divided by zero is ... undefined or basically an illegal operation.
2) Since there are no "positions" for any numbers or fractions in 0! [0 positions], technically one can't even really **do** a division by said "undefined" fraction. Heck, there's not even a "position" **just** for the 0 itself **in order** to invert it let alone enough positions to put it in an expression relating it **with** another number. Again, it's just basically the "null set." No positions, no "value," etc.
Lol. Blew me away on the first simple explanation. I thought I understood math
If you can explain the concept to a 3rd grader, you're the expert.
It really does feel like an explanation to a third grader...we made it that way...ie because I said so. Unfortunately all of the explanations feel like they are circular arguments. The last one didn't work for negative 1...
Except the first explanation has the most holes in it. Actually the first explanation would be better suited to proving 0! = 0. “putting no flags on” is not a solution to a permutation. If you have 1 flag there are 1! ways of putting it up (aka 1 way). If not putting up the flag was a solution 1! would be 2 (either you put that 1 flag up, or not)
@@chrisparkin6894how is the second one circular? It's basically the defining property of the factorial: (x+1)*x!=(x+1)!
I know what I’m going to reteach in BC calc tomorrow!
this came at a perfect time! i was trying to derive some formula i learned in stat today and this one trick I didn't know was the bridge I needed to finally understand it.
I feel like the first example is confusing. For example, with 0!, it’s said there is one way to put zero flags up. But, if that’s true, then 1! would equal 2 because there can be one flag or no flag.
Think of 2 factorial. There is a flag pole with a green flag and a red flag. 2! Is 2 because you can either arrange it with the green flag higher than the red flag, or the red flag higher than the green flag. There are 2 ways you can arrange 2 flags. The options where you remove one or both flags are not included in the way that 2 flags can be arranged. There must be 2 flags. Likewise “no flags” is not an option in 1 factorial; There must be one flag, and there is only one way to arrange that one flag. And for 0! There is only one way to arrange no flags
@Abcde7213 I understand, but I gotta admit it's confusing that "no flags" count as a way to organize 0 flags but doesn't count as a way to organize 1 flag.
1:55 - 2:44 That is a whole level of explanation by itself, and I also think it is the most convincing one as well
I thought about this for a while. And I think the best way to think of it and not be confused is instead of thinking about it from the top down.. so n! = n x (n-1) x (n-2)...x 1. Instead start with 0! and simply define it as 1 by convention. Then 1! REALLY becomes 1 x 1, 2! Becomes 2 x 1 x 1, etc. But there's always that extra times 1 that could be thought of as "and also, multiply it by the empty set identity"
I never even learned about factorials until taylor polynomials which really makes no sense, I wish I learned about them earlier they are really cool
Bruh , i read it as 0 is not equal to 1 as in coding ,lol 😂
Lol me too!
Same. Python pops up everywhere I go.
I read it as not 0 is equal to 1 because the spacing was between ! and = not 0 and !
That’s True
C# babyyyy
4:04 when cancelling out 5! on the numerator and the denominator, it is important to show the 1 that is still remaining in both.
When crossing out products the assumption is that a 1 remains, in how a lot of people write math. No special reason to handle it differently with factorials
I think the reason some people find 0!=1 strange is because they wonder where the 1 has come from.
When they see , say, 5! It seems like you are being given 5 cookies, it’s there explicitly stated, and now you’ve got 5 it seems reasonable that you can repeatedly decrement them.
Like eating each cookie until they’re all gone.
But when the same thought is applied to 0! , It feels like not being given a cookie, taking a bite of it anyway, and then having a cookie.
But where did it come from?
That's covered by the first explanation. If you're giving 0 cookies, you have one (1) way to eat it: Not to eat it. That's the only way. And that's one explanation for 0! = 1. Because n! has nothing to do with how many cookies you have, it's about how many way there are to eat them (assuming your cookies are discrete).
I can see what you are saying, and I do agree.
But I don’t think that’s how ! Is taught to people at first, it is taught as a little algorithm where you start with N and keep multiplying with the number one lower until you get to 1.
If we are saying that ! Is really the number of combinations, then I’d ask what is that other thing, the algorithm that people get taught at school first.
So what i’m really saying is, if the 1st level of encountering N! Is “starting with N keep multiplying by each number lower and stop when you get to 1”
( which, right or wrong, I think is how it is first taught at school)
Then 0! Would be 0 * -1 *-2 * -3 etc… never stopping, and simply equal to 0
But of course, that is not the same ! as the one used for’ number of combinations’ , but that’s my point, if it is a different ! Then why wouldn’t the value of 0! also potentially be different?
This is great - thank you! WONDERFUL presentation; subscribed!
Glad you liked it!
Dr Sean you’re the best.
This guy's a doctor?! He looks 20! Amazing video
Gotta love how wet start with "it makes sense" and gradually go to "if it doesn't, it fucks up other theories"
But it doesn't mess up other theories.
I would've strengthened the math major argument a bit because it fails to argue why the gamma function is the continuation we want. (1) Yes, there are multiple ways to extrapolate the factorial for nonzero naturals to the reals, but all analytic continuations of the factorial will agree on the integers and indeed predict 0!=1. (2) Though tangential to the original point, if you want to define the gamma function, it's good to point out that it's the only analytic continuation that always obeys the defining equation of the factorial Gamma(x) = (x-1)Gamma(x-1) .
Interesting video!
Constructive tip: Either keep your hands in frame all the time or maybe gesture higher so your whole hand enters the frams instead of just your fingers periodically popping up. :)
If we only defined the Gamma function to be Gamma(z) = \int_0^\infty t^z e^-t dt, then we could have had Gamma(n) = n! instead of (n-1)!. Why did that -1 have to creep in there?
Thank you, Dr Sean!
Not sure I can get my head around the Algebra part, so perhaps someone can explain a bit... I get that 4!= 4.3.2.1 can then set as 3!=4!/4.... because it seems you're just divding both by four...(same for 3,2 and 1)... but it seems weird that suddenly 0! shows up after 1!.. The 0 was never in the product to begin with so how does it show up now?
Would this be the case that we're always just calculating (n-1)! each time instead quotients/products?
Im reading this and even realising its a hella complicated thing to explain...
A product series to allow for negative factorials arrives at an imaginary j=0!, as opposed to i. The j operator commonly appears as ±, but it flips signs on every operation due to making selections. Oliver Heaviside used it to describe the field of selections 1892 solution for impedance. Because it initially selects position (+1), it is easy to show 0!=+1. You have to go out of your way to solve for negatives to see it can also be -1. Because it flips signs on each operation, j^2=-1, j+j=0, and j-j=2j. These qualities streamlined Heaviside's solution. The really surprising thing is that you can see it in the math, but no one bothered to document the seemingly paradoxical qualities. I've been working on an imaginary numbers book if you want mind bending reading to put you to sleep at night. ;)
I think we should change the definition of factorial for natural numbers to:
1! = 1
n! = (n+1)!/(n+1)
So that we can derive the same formula we have without explicitly stating 0! = 1 as a base case. It would be equivalent, but it would be easier to accept for people.
Looking into this Ty
I think it's helpful to think of factorial as in the same tier as powers, logs, and roots etc., the tier above multiplication and division, on which they are all based.
I think that the cleanest way to define it is, as you alluded to, the empty product: use sets, and, using the property that multiplying all of the elements of a set can be done via sub-sets, extend this to the empty product. x = \Pi_{A} = \PI_{A} \times \Pi_{} = x \times \Pi_{} , so \Pi_{} = 1, and 0! is just the product of the elements of the empty set.
i always imagined it as a 1 going into a machine that would get multiplied by other numbers and for 0 it wouldn't get multiplied by anything
Very nice video, congrats! I find it concise and entertaining.
It's also convenient that the same 0! appears when you count how many "words" (real or nonsense) you can spell using exactly 1 A, 2 L's, and 1 B (and 0 C's and 0 D's and...). The alphabet has 26 letters, but the formula can be written without explicitly excluding the unused letters.
The "contribute nothing to the multiverse of possibilities" action when multiplying possible outcomes is the multiply or divide by 1 action, as you say.
the last method is very common use for me, it's very common to deduce solutions using correlating functions like that in all of science and especially chemistry and biochemistry
I truly am good at procrastinating to be watching this content, but the videos are high quality.
honestly, the counting one immediately made sense to me.
"You can only arrange 0 objects one way...there are No Objects arranged. That's it. One way."
I was thinking the following. The number of bijections between a set of size n and a set of size n is n! So 0! should be the number of bijections from the empty set and itself (this bijection does exist and is equal to the empty set when we consider functions as subsets of the cartesian product). We also know that the number of functions from a set of size n to a set of size m is n^m. So this would also give 0^0 =.1 as being a sensible definition...
Seeing the gamma function gave me flashbacks to statistics class! 😅
Absolutely brilliant! Subbed
Brilliant explanation of a **nonsensical** and **technically wrong** assertion.
Convenient, maybe.
Correct? No.
The null set / empty set does not have a value of 1, it is **at best** "undefined."
0! = 0!
"But why" don't question math, that just gives you more work. First ten of the video watched, got all the information I needed, now by
i like to think to factorial integer numbers (not negative) as the multiplication of all natural numbers from 1 to that number, so considering there's no natural number smaller than 1, then 0! should be 1
This goes back to the argument of whether zero exists😂
The Bausffs is doing side quests being a math professional outside of inting top lane
Why does he look like he just got out of his emo phase (no hate im in mine 💀)
Very interesting, thanks!
Thanks! Please make a video about why 5th order polynomials are not solvable by radicals. Best regards!
Thanks for the video idea! I added it to my list
A way that i thought of it is that there are an infinite number of trailing multiply by ones at the end of any number and its still equivalent, so by going from 1! = 1 = 1x1x1... to 0! =1x1x1.../1 = 1. Taking one 1 out of the whole doesnt change the end result
log(n!)=log(1)+log(2)+...+log(n)
*Mathematicians and programmers agree that 0!=1*
Clever one 😂
Great video and explanation
Love these videos!
This guy’s voice is in my head all these years when I study maths and I am Greek 😂🤷🏼♂️
Im really happy that I knew all of this already. I was expecting something new / crazy. Great video btw!
0! being synonymous with a “multiplicative identity” is a cool idea, with potential useful insights in Unital-magmoid Theory:
There is probably a lot of research on those very specific Groups anyhow.
There is one very intuitive way to think about it. Going from factorial 3! To 2!, you take away term 3 in 3!. Now you can imagine infinite 1 being multiblied by factorial. So 2! = 2 × 1 x 1 x 1 x .... . If we want 0!, just take away 2 and 1 and we are still left with all the infinite 1 terms . So 0!= 1 x 1 x 1 ... which is obviously 1. Beauty here is that it clearly shows how it fails to see (-1)! With intuitive sense, and perhaps we can only use gamma to extend definition.
So you are suggesting that there are many hints on why, in order to have math as clean and beautiful as possible, we should assume 0!=1, as if we are on one side just choosing it to be so, but under another perspective, discovering some underlying pattern that's already there and choosing convections according to it. That's interesting
As my teacher in highschool said : "A definition, you don't have to understand it, you have to learn it"
Too bad the definition is wrong / nonsensical if one actually examines it correctly. Technically 0! is illegal and/or undefined. Period. Just as is dividing by 0.
And "because we say so" is not an actual proof of said identity "0!=1", of which there is, and can be, none. Assertions "by fiat" to the contrary aside...
Level 1 can be described formally with the empty set and vacuous truths. This is what I consider the "morally correct" definition.
Any function X->Y can be formally defined as a graph, or subset G of the Cartesian product XxY for which there is no x in X and distinct y,y' in Y such that (x,y),(x,y') are both in G ("vertical line test"). The function is a bijection if (i) there is no distinct x,x' in X and y in Y such that (x,y),(x',y) are both in G (one-to-one), and (ii) there is no y in Y for which there is no x in X such that (x,y) is in G (onto). Notice I rephrased "there exists" statements into "there are no counterexamples" statements. Since {}x{}={}, the only function {}->{} is the empty function, i.e. {} itself, which is in fact a bijection. 0! is the number of bijections on a set of size 0, so 0!=1.
This also shows why 0^0=1 too. And justifies the Level 2 and Level 3 descriptions, as well as power series (Level 4) interpreted as generating functions (also cf combinatorial species).
Am I the only one who thinks that the math major explanation is more understandable than the calculus one?
I can't lie, being exposed to calculus is what will make me understand things. Especially how integrating the surface area of a sphere gets its volume.
“0! =1”
“source?”
“I made it up”
1:06 okay, but why would you say that there is one way to put 0 flags on a flagpole, and one way to put 1 flag on a flagpole. Wouldn’t it then be more consistent to say that there are 2 ways to put a flag on a flagpole. Positive and negative?
1:47 could also be -1. -1!, 0! and 1! Equals 1 hmmm interesting
1:00: If 0! is how many ways there are to put zero flags on one pole, then 1! is how many ways there are to put 1 flag on 2 poles (1! = 2), 2! is how many ways there are to put 2 flags on 3 poles (2! = 6), 3! is how many ways there are to put 3 flags on 4 poles (3! = 24), etc., all of which is WRONG.
Correct is: 0! is how many ways there are to put zero flags on zero poles, but that is not what he said in the video.
Thanks for this! I get these pretty much up to the gamma funtion. And I even get why Γ(0)=1. But the video explains that there are other ways besides Γ to extend the positive integer factorial function to the reals. So in some sense, isn't the gamma function explanation a bit circular? If we were instead to start from a presupposition that 0! is undefined, we simply wouldn't use Γ to extend ! to reals, right? So I suppose the argument is less "Well, the gamma function makes it one" and more "If we wanted gamma of zero to be undefined, we'd need something way more complicated than gamma to represent it." Am I on the right track there? I'm not trying to challenge the definition that 0!=1, just trying to clearly understand the reasoning.
The Gamma function is the most useful extension to the reals. If you additionally ask that the extension be log convex, then the gamma function is the only way to extend factorial to R. f is log convex just means that log f is a convex function.
What a great video!
1:04 that'd be like saying _"there's an answer to any number divided by 0. Just don't divide it and get the same result"_
Gosper's approx for 0 gives sqrt(pi/3) ~ 1.02, very close to 1.
I found the counting definition to be the most logical and the algebraic definition to be a convenient 'it just is'.
If you think about it, they all start with 1 times sometimes, why would 0! Start vir 0 times something?
For the fourth one, I do not understand how essentially, "it looks nice" is a justification, let alone a proof for why 0! should equal 1. Yes, defining it as such, makes the Taylor series expansion of the exponential function work very well, but that just can not be justification.
Is your thesis available? I’d like to read it. Or do you have a video on it? Or could you make one?
It's on ProQuest which requires a university login, but I'll get it added to my website in the next couple days so I can share a link. I don't have a video about it yet, but I think that's a great idea!
cool video, I'd like to see 0^0=1 next
Another justification I've seen:
x! = x(x-1)!
If x=1 then
1!=1(1-1)!
1=1*0!
1=0!
Very good video
I don't quite get the Combinations at first but it makes sense 😺😺😺
I still do not fully understand the calculus or the math major but I hope to one day
Not sure if it's relevant but the definition of derivatives makes h=0 which is in the numerator.
tldr:
x in N, x! = 1 if x is 0, else x! = product(n) from i = 1 to n
If you continue the diving rule for (-1)! you would get (-1)!=(0!)/0=1/0=inf
When you use the binomial formula to get pascal's triangle you divide by inf outside the triangle so it's outside 0 (which makes sense when you apply the "add rule" with 0 and 1 to get 1 on the side of the triangle)
And when you use (-1)! in Taylor's formula to get the "more left" terms you would get something divided by inf which is again 0 so you add a bunch of 0's
_(Math from Ohio)_ 💀