No need to bring politics into this : - )

marbanak what do you mean? It’s because you’re saying it with excitement, so you use explanation points; making 0! = 1. Screaming 0 loud enough makes 1.

@@i_am_anxious0247 : I suspect, at the time, I was thinking of character-assassination efforts at the federal level. It was quite a drumbeat. Cheers!

marbanak that’s... weird, I guess. Cheers!

@@marbanak are you *actually* implying that zeroes can't be equal to one? that is literally numbers-ist. reported for numerical supremacy and hate speech

Calyo Delphi Yeah, but you could add that nothing to all positive factorials, in which case they should be their current value + 1: 2! should then be 2 + 1 = 3, for example. Yet no one uses that fact to conclude that, but it is not logically sound in that case.

J.J The Great: I don't think so: You cannot NOT arrange any number of items when you have one or more items. As long as the items are present, they are arranged. So the additional +1 is NOT an option for n! as long as n>=1. But when you have NO items, then the option NOT to arrange them is the ONLY arrangement. It's an option that appears ONLY when there are zero items.

You can metaphisically arge that in absence of anything to arrange there is no arrangement possible. And that is a totally valid approach also.

I disagree, having no object to arrange means you have no arrangement of anything whatsoever. So it should be 0 !!

just like null set is a part of a super set.

no programers

Lmfao oh god. That's.. that is so underrated.

Are you looking for null?

and 0!=e

0! is infinite recursion aka undefined just like all negative values

The fact that last no becomes 0\infinity

​@@lolme2646... This guy keeps assuming open 0 in the limit as x ---> 0 for y = x^x being y @(0, delta) = 1 for very small delta. His #4 solution shows this. But L'Hopital's Rule showed f(x) = 1 as x approaches 0 at [0, delta)! y= x^x has a limit of 1 at x=0 not only for series expansion reasons of defining e^x = a series limit but for other mathematics where 0^0 appearance problems.

did you just assume zero is a number!?

@@Blox117 lol

1 - natural (nature) 0 - status (philosophy) -1 - man-made (science)

Zero is not a number; it is the empty set and acts as a placeholder nothing more. Numbers that are greater than 0 are numbers. All of those numbers with a negative sign in front are the same exact number with the same exact magnitude just that they span out in the opposite direction as they are pointing 180 degrees or PI radians in the other direction, vector computation proves this. Also the negatives are reflected about the perpendicular bisector that is located at zero on the number line or x axis and the point at zero acts as the point of rotation and the point of symmetry. Zero has no quantitative value therefor it is not a number! Yes operations can be performed on it just as any other number, but the outcomes can vary from one operation to another that doesn't hold to the operations of all other numbers either if they are the set of all real numbers or the set of all imaginary and complex numbers. For example we were all taught that you can not divide by 0 as it is undefined. Let's not look at this as a numerical expression but as a conceptual expression instead. I'll use some regular fractions as examples first to illustrate the point being made. If we take 1 and divide it into 4 such as the fraction (1/4) we are saying that we want to take a slice out of the whole in 4 equal parts and this gives us a ratio proportion of 0.25 or 25% of the whole. These all mean the same thing. If I was to take it's reciprocal and say take 4 and divide it into 1 such as the improper fraction (4/1) we are saying we want 4 equal parts of the original which in turn gives us 4 similar objects of the original. Let's try to do the same thing this time with 0 and 1. I'll start with 0 in the numerator. Let's divide 0 into 4 such as the fraction (0/4) we are saying that we want 0 parts out of the whole which in turn gives us a value of 0. We can conclude that (0/n) = 0 as long as n does not equal 0. Let's now take 4 and divide it into 0 objects this is saying that we want 4 exact objects but there are no objects to take it from. This could yield two valid results. It could either be 4 as we have 4 equal empty sets or it could be 0 as we have no objects; both interpretations are valid assessments. So we could conclude that (n/0) = n or 0 as long as n does not equal 0. We have one case left and that is when we have (0/0) and with this situation we could have a value of either 0 or 1. When we divide anything by 0 (numerator) we end up with 0 which is true. Also when we divide anything by itself we end up with 1 (identity) which is also true. So when we have the indeterminate form (0/0) the answer can be 0 or 1. A little more complex than this but there are other contexts where 0/0 could also possibly equal +/- infinity but that is beyond this discussion. The main reason we consider it to be undefined because we always assume that operations on operands must act like a function with a single input and a single output but we also know that this doesn't always hold in nature, there are many times you can input a single value and get multiple results and when this happens if fails the one to one rule as it then becomes a one to many rule. And since we do most of our computations on electronic or digital devices such as calculators, computers, etc. These devices have physical limitations and we don't know how to represent single operations that yield multiple values due to the fact that the transistors or switches that makes up the logic gates are defined as two state boolean devices either on or off, high or low, open or closed 1 or 0, etc. Yet the evaluation of 0/n, n/0 and 0/0 is well defined and not undefined, but since it has this complexity unlike numbers the empty set or null set or zero vector, zero point or the unit digit place holder it is easier for us to say it is undefined and not have to deal with its complexity. The same situation arises when dealing with the slopes of lines. We know from algebra that a slope with two points p1 and p2 has a slope of m = ( (y2-y2)/(x2-x1)) which is also dy/dx which is also sin(t)/cos(t) with respect to the the angle from the origin above the horizontal or x axis which is also tan(t). We we have 0 slope we have horizontal lines and this holds true because the numerator part the sin(t) or dy = 0. We we have vertical lines we say the slope is undefined because of "division by 0" and in this case the cos(t) or dx = 0 which is exactly where the vertical asymptotes show up in the tangent function. Now intuitively we say that when the ground is level or flat and there is no change in one's height as you move across the xz plane (y axis vertical) that you have 0 or no slope and this makes sense. However when you have no change in the xz plane but your height is constantly moving up or down in discrete intervals I'd don't like the idea the slope is "undefined" I tend to think of the slope as full slope not partial as (m/n) where m and n are not 0, but as in infinite slope. If you are going up you are approaching + infinity and if you are going down you are approaching - infinity. The reason I state this is because if we closely evaluate the slope of a line when m is in the form of the trig functions sin(t)/cos(t) both functions on their own have a domain that accepts all possible values and their range is between [-1,1] as long as they are standard form (meaning there were no transformations applied to them such as scaling, skewing, or rotations). So when we look at the limit as cos(t) approaches 0 we need to look to see what is happening to sin(t) at the same time, it is approaching either +1 or - 1 which implies +/- infinity just as you can see from the vertical asymptote in the tangent function. This is just pure reasoning and logic and basic computations to prove these assessments. Also you can apply the dot product using these points to find the actual value needed to calculate the angle in which they from each other (0,0),(1,0), (0,1),(-1,0),(0,-1) and these are the 5 crucial points that make up the unit circle. You can take any 2 of the points and use the dot product in terms of the cosine function and you would get values of 0, +/- 90, +/-180, +/-270, +360 depending on which direction you chose your points to be in but of course you would have to apply the arccos to get the actual angle after using the dot product. Just something to think about how complex the empty set truly is. Depending on the operation being applied and the context in which it is being used it can yield 0,1, +/- infinity and sometimes it could yield more than one in a single evaluation depending on which direction you are taking the limit from.

Cause Indians are mysterious 😅

Yaroslaw Kaminsky I didn’t because I always put spaces around most operators

Lol i guess this means we should always use spaces in operations

Yaroslaw Kaminsky ₹ 9

I'm in 8th grade and learning C++. Our teacher has recommended us to put spaces between operators to recognize the differences between operators.

I was thinking that the WHOLE time!!!

Well the empty product is the same as 0^0 which is 1.

@@findystonerush9339 0^0 is undefined, but the limit of x^x as x -> 0 is 1. Do not confuse the two.

Although if you use this explanation, (-π)! is also 1, because there are no positive integers less than -π

@@talhochberg5062 "for nonnegative integers n"

@@anon.9303 That's a definition. If you're using that explanation, you can also define the factorial to work on all real numbers (or complex?)

Samuel Pierce yeap, I agree with you and that' s obviously true because the fourth reason doesn't count at all

Yes but the title of the video is why it SHOULD be 1 not any sound reasoning.

This video isn't a proof that 0! = 1, it's reasons why we should and do define it that way.

It wouldn't work. Not with every x, at least. When x=0 every term after the 0^0/0! is equal to 0, so there's no way the sum would converge to 1.

Also, I’d asume that for the definiton of exp(x) as a power series you could not define exp(0), but you should do instead lim(x->0) exp(x)... and in this case you are not using the convention of setting 0^0=1 For me this is an interesting topic, yet I am an engineer so maybe I dont know at all what I am talking about :)

right. he has a video about it.

it's 0!/0 which is +infinity. (-2!) then is (-1)!/(-1) which is +infinity/(-1) = -infinity. (-3!) is (-2)!/(-2) which is (-infinity)/(-2) = +infinity and so forth.

DerToasti not exactly lim(x to 0+) ((x-1)!) Would be lim (x to 0+) (x)!/x = +inf But lim x to 0- gives -inf. If you look at x! For x in (-1,0), (-2,-1), (-3,-2) etc (where (a,b) is the set of numbers between a and b exclusively) You can see that the sign changes every time you move to the next set, so all negative integars factorials are undefined

Recall that the definition of a "factorial" is only defined for a positive integer(including zero). Otherwise, is undefined. -1 is a negative number(integer) thus -1! is undefined.

No, it's (-1)! that's equal to 1/0, -1! would be -1

Yes and that bypasses some peoples' philosophical unease with the question of how many ways there are to arrange "nothing".

B. Xoit "0" way

I didn't understand it so you are probably wromg

@@b43xoit Illogic

@@AlgentAlbrahimi One way, the empty list.

Makes sense, anybody that has extensive knowledge and understanding of mathematics and its mysteries can move almost as fast as the speed if light.

Andrew Criswell well he stated that 0^0=1, and from there algebra dictates 0!=1. Really it's a limit problem that you could work through, but I think you'd have to substitute in the pi or gamma function for 0!, and then you may as well just state 0!=1.

@@semiawesomatic6064 I thought 0^0 is undefined, because you get contradictory outcomes depending on how you approach it.

@@carultch Same with 0! as well.

I agree.

I'm curious about how he'll do that "puzzle".

semi awesomatic The Parker Coin

Did he try using a square? 😃

No, he's using a cylinder.

WarpRulez 👫😂😂😂😂 that's true

Everything that has an exponent of zero will always be one.

If you can cancel something in the integral out before performing the integration, it's a completely valid mathematical step. t^0 is equal to 1, so it can be removed from the expression entirely. Additionally, not removing that t^n from the integral while n=0 would mean requiring that you perform integration by parts.

That's why some people think 0! is undefined as well.

It depends on your definitions, though. If you set up your definitions of multiplication and factorial in a certain way, then 0! = 1 _can_ be proven. It's also the "right" value in the sense that it makes _every_ formula where 0! shows up give the correct value in those situations, and the certain-way-of-setting-up-your-definitions I gave above explains how we know it will _always_ work.

: )

Janox81 Totally agree. If 0! Isn't 1 then neither is 1!

Essentially that's reason 1. Showing n!=n*(n-1)!

No. If 0 wasn't defined, then we would just start at 1 instead of zero which is just as arbitrary of a place to start. If you wanted you could be explicit and specify it's domain as positive integers.

Bravo, nice short correct explanation

THIS! Yes thank you for pointing that out. I think everyone is forgetting the basis of factorials is to multiply all natural numbers before a number no 0 included- there is a limit already to what factors there are for positive numbers, though a pattern is made factorials themselves are not the basis of other factorials. 0 is a neutral number and can have it's own domain for its factorial.

archimidis I see what you mean. To really be rigorous, he should have said the limit as x approaches zero from the positive direction. Since that limit on x^x is equal to 1, I think the proof still stands.

"whereas 0^0 is undefined." I mean, that's why he said it's "conventionally accepted," for convenience's sake. He was very careful during part 4 if you'd notice to avoid saying "equals," even when he was saying 0! should be 1, not 0! equals 1.

archimidis heap I also argree with you

A^x/A^x = A^x/A^x When dividing variables with powers, subtracting the exponents gives the correct answer. So, subtract the exponents on the left. x-x=0 Now we have, A^0=A^x/A^x Anything over itself is 1 Apply this to A^x/A^x A^0=1 Therefore, any number to the 0th power is 1. This also applies to 0^0, making it 1.

I was thinking the same (almost) about the fourth suggestion. I thought 0^0 was indeterminate number, not necessarily undefined , nice video though!

reason1: ok reason2: ya I know this reason3: WTF

陈明天 not funny the original comment was likes 10 times more funnier!?!?.

@@user-ot4rp8yn8r thanks for saying the exact same joke

@@theflaminglionhotlionfox2140 erm I m not talking joke...? I think that is my condition of understanding for 4 proofs.

So that we can put all the terms into one sigma notation.

kzfaq.info/get/bejne/qLCZn5tmrK6pqZc.html

Good point. Although a slight tweak to the last reason would have also worked. If we calculate e instead (e^1), then we have as the first term 1^0/0!, so we can avoid the 0^0 ambiguity. So e=1/0! + 1/1! + 1/2! + 1/3! + …. Rearranging, we can get 1/0! = 1/1! + 1/2! + 1/3! + … - e. Taking the reciprocal of both sides, we get 0! = 1/(1/1! + 1/2! + 1/3! + … - e). Just doing the first several terms, we will see that the answer converges to 0! = 1.

: )

Really?

0.

You can't put in 0 for x because (-1)! is undefined. You *can* substitute 1 for x, though, which is what ZipplyZane did.

Yes, but your "contradiction" relies on -1! having a defined and finite value. [0! = 0*(0-1)! = 0*(-1!) which is not zero, but undefined since -1! is undefined in the reals]

+Denis Fluttershy There's a reason why the domain of the factorial is the natural numbers... and that of the Gamma function is the complex numbers.

No we are saying that -1! is undefined because it is... differently from 0! which is DEFINED to be 1 (and this can be justified since no matter what definition of "!" one uses there is no pole or division by zero involved in the value of 0! or Gamma(1))

Sorry - we will disagree deeply on the statement "We know some fundamental things from life and we do math with these thing." Mathematical objects/concepts have no obligation to have a connection to "life" (or the physical world), and very often they do not. What's the physical meaning of the pi-th derivative of e^i? Yet it has a mathematical meaning and a value. We can write n! = Gamma (n+1). It's the absolute truth for n in N, and if you try to find the value of Gamma for integer negative values of n you will find it's undefined. So Gamma (0) is undefined (in R), as is Gamma (-1), Gamma (-2) and so on. Which is why above I was saying that -1! is undefined because it is: no matter what (commonly used) definition of factorial you try to use, you simply cannot define a value for it that makes sense given other simple mathematical concepts (other "non log-convex" Gamma functions excepted... which aren't necessarily all that simple)

Wolfgang Wilhelm yes. And I have another video just on negative factoreo

Could there also be a factorial of complex numbers?

PS: thx for all the vids - they are very interesting :)

Wolfgang Wilhelm yes if you extend the factorial again. I may work that out in the future

Yes there is a complex factorial (extended) because GAMMA function is meromorphic in the whole complex plane. Beware the negative real part though, you will need the Euler's mirror to compute it (the functional equation) The integral shown only converges for Re(z) >0 where z = a+b*i Also, (-1)! = GAMMA(0) = undefined (a simple pole with residue 1)

Rahul Majumder You are right. I actually have another video on negative factorials

blackpenredpen yeah I watched it, by Pi function...

Yes.

(-1)! = infinity.

😀😀😀😀😀😀😀😀😀😀😀😀 😕

How to arrange your future

Taylor series

It is undefined

@@aleksapetrovic7088 Infinity. Stop disrespecting our creator by calling it "undefined". Infinities gave us our dimensions; we must respect infinities. If we were living in Minecraft, we would call circles "undefined". Since we are living in a world with polar coordinates, the premium package with the spherical bundle, we are accustomed of seeing circles, so they are not "undefined". Also, infinities are everywhere, we cannot move without them, and the Big Bang couldn't happen without them, without them, we would continue to be lumped together in singularity. There are an infinite number of points in a unit line segment alone, and given the fact that infinities are required to extrude something to the next dimension or travel through time, we should divide by 0, spread our wings, learn how to fly, and do the impossible.

Infinity. Stop disrespecting our creator by calling it "undefined". Infinities gave us our dimensions; we must respect infinities. If we were living in Minecraft, we would call circles "undefined". Since we are living in a world with polar coordinates, the premium package with the spherical bundle, we are accustomed of seeing circles, so they are not "undefined". Also, infinities are everywhere, we cannot move without them, and the Big Bang couldn't happen without them, without them, we would continue to be lumped together in singularity. There are an infinite number of points in a unit line segment alone, and given the fact that infinities are required to extrude something to the next dimension or travel through time, we should divide by 0, spread our wings, learn how to fly, and do the impossible.

Well (i)! = Π(i) = int (x^i e^(-x) ) = int (e^(i*log(x) -x) = int (cos(log(x)) e^-x) + i* int (sin(log(x)) e^-x) ≈ 0.50 - 0.15i Pretty horrible. Although (i)! !!!! Is pretty bad too... But repeated factorials slowly brings the answer towards 1+0i. Similarly how repeated factorials of a number between 0 and 1 approaches the fixed point 1.

The Gamma function, and the Pi function, can be used as analytic continuations to the factorial for ALL complex numbers (excluding negative integers). Pi(t) = int(x from 0 to inf) (x^t e^-x) dx which exists for complex numbers. So we can say that i! Exists as does i! ! ! ! ! And is approximately 0.9923-0.0003i Assuming the !!!! Can be assumed to be 4 factorials instead of 2 double factorials. (n!!= n(n-2)(n-4)... (3.5±0.5)(1.5±0.5) depending if n is even or odd... It can be analytically continuated to the complex numbers using x!!=(Pi(x/2)*2^((x+1)/2)) / pi^0.5 (((i!)!!)!!) ≈ 0.99+0.01i )

Neo Matrix defined*

A Wild Triple Factorial has appeared

You ought to try the Digamma function.

AndDiracisHisProphet I'm not sure why stage 4 needed x=0. From the Taylor expansion we get e^x = 1 + x/1 + x²/(1*2) + ... + x^k/k! +... So e^x = 1+ sum(n from 1 to inf) (x^n/n!) But this isn't very neat, we would want 1 in the sum as well. So would want 1 = x^0/0! For all x. Hence we want 1 = 1/0! (for x not 0 for now), so 0! Should be 1. Can then deal with 0^0 afterwards, lim (x->0+) x^0 = 1 and lim (x->0-) x^0 = 1, so in this case our 0^0 = 1.

Thank you, Anti-joke-chicken

AndDiracisHisProphet Sorry :(

I was thinking the same thing

@@wolffang21burgers Thanks. You cleared that up for me. I was wondering why 0^0 = 1 by convention?

vlatko the limit as x-> 0 of x^x is equal to one

: )

Ethanol 314 you can take the derivative or integral of the gamma function, and the Pi function. n! Is not a function to take a derivative or integral of

Yes you can if you use a function, like GAMMA(n), the log derivative is called the digamma function and you can find ALL the minimum and maximum of the GAMMA function.Really powerful function, because you can compute partial harmonic series, Here is an example 1+1/2+1/3+...+1/65536 = ?? well easy with digamma(65537)+ gamma (Euler's number = 0.577215664901)

Thanks!

You dont know about the concept of limits if you say for example 1/infinity its like saying 0,000....(many zeros)001 then you can say its technically 0 because even in real life if you cut 1 apple for example this many times you wouldnt be able to see anything remaining.

Same with 0^0. Why bother to leave things undefined? It's just saying, "Screw this challenge, I will not face my fears", but that is against the mission of humanity of breaking away.