The Secret Behind -1 Factorial
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- เผยแพร่เมื่อ 20 ต.ค. 2024
- Dive into the mystery of -1 factorial! Explore advanced math topics like integrals and complex numbers as we tackle this perplexing question. Meet the gamma function, the key to extending factorials beyond positive integers. But there's a twist-calculus shows the integral diverges to infinity. Is -1! factorial really infinity, or is it more complex? Unlock the secrets with analytic continuation and decode Wolfram Alpha's intriguing output.
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The video is 23 minutes ago this comment is 2 days ago this channel owner is a time traveler
The infinity with a tilde is "complex infinity"; it's an infinity without a direction (the "north pole" on the Riemann sphere).
You get the same thing by typing 1/0 into WolframAlpha, since 1/0 is defined on the Riemann sphere.
How about a video about the third derivative of the gamma function evaluated at one, an how it relates to the appery's constant, the euler-mascheroni constant and Pi?
I hoped you’d calculate a residue of the pole, or something. 🤔
Res(gamma(z); z = -m
Another reason (and also simpler) why (-1)! is undefined:
We all know:
n! = n(n-1)(n-2)...(3)(2)(1)
But this can be expressed as:
n! = n(n-1)!
If we move (n-1)! to the left, we get:
(n-1)! = n!/n
For example:
n = 3
(3-1)! = 3!/3
2! = 6/3 = 2 ✅
n = 2
(2-1)! = 2!/2
1! = 2/2 = 1 ✅
n = 1
(1-1)! = 0!/1
0! = 1/1 = 1 ✅
If we want to find (-1)! , we substitute n = 0:
(0-1)! = 0!/0 ❗
(-1)! = 1/0 ❗
As you can see, getting (-1)! requires dividing by zero, which is undefined.
Underrated work bro
Sorry but I feel like it wasn’t very clear. What exactly does “complex infinity” mean from wolframalpha? Does it mean like the magnitude of the complex output grows unbounded as the distance between the input and -1 get closer? If someone could explain this I would greatly appreciate it.
Means a number with infinite magnitude and undefined argument.
(-1)! are a Simple pole with residue equal to 1
"complex infinity" is the complex extension of the concept "unsigned infinity" for the reals. "unsigned infinity" is the value at *both* ends of the number line. Imagine tying the infinite real number line into a circle such that both ends meet up again. For example, 1/0 can be defined to be unsigned infinity. It isn't positive infinity since when approach from the left it grows towards negative infinity. And it isn't negative infinity since when approach 1/0 from the right it grows towards positive infinity.
In the complex number, the "complex infinity" is the infinity in all directions at once, as oppose to the infinity in only the 1+i direction (i.e. the infinity with an angle of 45°)
@@megaing1322 beautiful explanation
Look at the function f(x) = e^(x + ix) with real inputs.
It spirals outward around the complex plane, growing in magnitude exponentially but continually cycling through every angle.
The limit of f(x) is “complex infinity”. Infinity without direction, or, alternatively, every direction, depending on how you look at it.
Of course the first pole of the analytic continuation of the gamma function occurs at e^iπ
The main problem about this occurs where lets say you want to try (-n)!
But if its an even amount its result is positive and if its odd its result is also odd... thats one reason why (-n)! is undefined
You don't need the Gamma Function to go negative
n! = (n+1)!/(n+1) -> 0! = 1!/1 = 1 -> (-1)! = 0!/0
It's a bit of a mess but kind off the same
but 1/0 is undefined
@@facts_mathPrecisely, just as Int[0->infinity](e^(-t)*t^(-1)dt is divergent. They are the same thing.
Not a fan of this one. Bri explained factorials and the gamma function a bit (cool) and then said (-1)! Is a special kind of infinity and we can talk a lot about it… then the video ends?
Now he's forced himself into making a video on complex infinity
NICE!
Other people have already given the derivation of (-1)! by the recurrence relation, so I'm going to ask a different question: Why does Γ(x) have t^(x-1) instead of just t^x? The minus 1 just feels so artificial and all it seems to do is push the gamma function _away_ from the factorial it's used to extend. There is an alternative function Π(x) which is defined for all complex numbers except negative integers, but also has Π(n) = n! for all natural numbers n, (so all numbers the traditional factorial is defined for) rather than (n-1)! for all positive numbers n. The factorials importance in calculus and combinatorics show no sign of a -1 and just use the factorial as is, so Π(x) would appear more natural when trying to extend them compared to Γ(x+1).
Is this question asked in many places? Yes. Have I ever seen a satisfying answer? No.
After careful consideration I have decided to leave -1! Undefined for 2 reasons. First off we know (x-1)! Is x!/x. This is proof for 0! Being 1. But then for -1! We have 0!/0. 0! Is 1 so we have 1/0 and we don’t like that. Secondly, factorials can be considered the amount of possible arrangement of x items. You can arrange 2 items 2 ways (2!) 3 items 6 ways(3!) and 4 items 24 (4!). So how many arrangements can you arrange with -1 items? That doesn’t make a hint of sense. So i’ve Decided to leave it undefined.
Click what video on the screen? Doesn't show up for me. And I can't find a link in the description, either.
At this point, you should change your name to BrilliantTheMathGuy
Well, I think to take the factorial of a negative number, you know how to take the factorial of a number multiply it by any number in its path until you get to 1. Well, to take the factorial of a negative number for example -5, do -5,•-4,•-3,•-2,•-1 and skip zero and then multiply that by one which is just itself. So -5! Is probably 120
Not 120, -120
Very Impressive
(-1)! = 0! / 0
= 1/0
As we don't know what happen when we divide something by zero.
So we can't get answer.
We can divide by zero
Do i!
(-1)!=infinito gorrito
James Anderson said that 1/0=∞. Was he wrong?
-1! = ♾
Hii ssg bro. How are you. I am FrittoBoss do you remember me. I am in the fans and friends video . Thx for uploading more videos 😊.
Integral(tan²x)dx
Well it's not defined so the problem is solved.
I'd argue that (-n)!=-(n!)
Second, but noone honestly gives a shit.
Im gonna watch this video, looks pretty cool
could you explain i! once? a calculater shows me the figure but I wonder how it's possible 😢 sincerely
Calculating (-1)! in casio
Casio calculator: Math ERROR
-1! = -1
i asked my dad the same question, but i never realized that the answer would be this complicated!
This video seems to be “cheating” by telling half or not even half of the story. You bring us to a story with a long ads in between and conclusion the answer is “complex infinity”, and answer you obtained from Wolframalpha?! We already know that and we expected you give us some derivation etc. I think your recent videos fall to similar problem. It give people think all you want to show is the long ads in between a fantastic introduction and sloppy conclusion.
Desmos says -1! Is -1
-3!=-6
(-3)!=undefined
its not a secret anymore you just told everyone smh my h
x!/x=(x-1)!
so 0!=1!/1=1
and (-1)!=1/0 which is undefined
See how you're using real numbers? That's why it's undefined.
1/0 is defined
This video was a big nothing.
7th ig…
Third
1!=1 -1!=-1;
But (-1)! is 0 divided by 0...
x!=x(x-1)(x-2)(x-3)...(3)(2)(1)
(x-1)!=(x-1)(x-2)(x-3)...(3)(2)(1)
x!=x(x-1)!
(x-1)!=x!/x
Plugging in 1...
0!=1!/1
0!=1
Plugging in 0...
(-1)!=0!/0
(-1)!=0/0
I don't think we can easily define 0/0
@@cheeseburgermonkey7104 well it's actually 1/0, since 0! = 1 and not 0 (but 1/0 is undefined too so it doesnt matter too much i guess)
(i)!
What do u think -1! is
-1. ∞
👇. 👇
First
Noone cares 😱😱😱😱😱
You really try to distort all math basics just to get views. Your math and logical mistakes are so obvious that makes me wonder what kind of math you were taught.
this video was so bad literally just made it to get a sponsor
N!=e^2