Hyperfactorial introduction

[@blackpenredpen]

0^0=1 by convention in hyperfactorial as well. : ) Just like in power series, see this video kzbin.info/www/bejne/qHvMnWtrfK14r8U

[@seroujghazarian6343]

I thought 0^0 was undefined

[@angelmendez-rivera351]

Serouj Ghazarian 0^0 is an indeterminate form, which is why limits with this form can equal many different values and exist. This is different from the expression not being defined. If a limit ever is evaluated to equal an undefined arithmetic or algebraic expression, then the limit does not exist. There is a much more rigorous way to state this, but I’m not going to bother with on a short KZbin comment. Needless to say, there is much evidence that suggests that 0^0 is a necessary convention in mathematics, and in most fields of study, 0^0 = 1 is assumed even if it is unprovable.

[@seroujghazarian6343]

@@angelmendez-rivera351 exactly! LIMITS! you know why 0^0 is inderterminate? Because it takes you straight to 0×infinity (lim x->0+ e^(xlnx)) and I'm pretty sure that ln(0) is undefined. Although, if we think like that, 0 itself is undefined

[@seroujghazarian6343]

@@angelmendez-rivera351 also, while you're at it, why don't you take sin(0)/0=1, (cos(0)-1)/0=0, log(b=1)(1)=1 (actually, the last one can be any real number)

[@lucazara9137]

I found this formula for negative hyperfactorial H(-n) = (-1)^(n+1)*(n-1)^(n-1)*H(-n+1)

[@kingbeauregard]

Cool jacket! Did you start a motorcycle gang for mathematicians? You could call yourselves The Unbounded Functions.

[@blackpenredpen]

lol, that will be cool!

[@emethirsch7107]

Speed Without Limit!

[@Uni-Coder]

Painted with black pen

[@blazedinfernape886]

5!= number of seconds in 2 minutes. H(5)=number of milliseconds in a day. Wow that escalated quickly.

[@Gold161803]

I think 10! is the number of seconds in six weeks

[@blazedinfernape886]

@@Gold161803 yup! Numberphile did a video on it.

[@stewartzayat7526]

And what is H(10)?

[@brennanherring9059]

@@stewartzayat7526 Number of Planck times in 11.63 seconds.

[@SeriousApache]

I think with enough brute force it is possible to do H(-1)

[@angelmendez-rivera351]

RUSapache Read my comment above, where I derived a symmetry formula for all integers n, extending H for negative inputs, and I also demonstrated that every half-integers is a root of H(x) given this definition of H.

[@Kitulous]

well, if we go from n to 1 so it looks like this: 1 П k^k k=n Then it would work, actually. H(-1) in this case is (-1)^(-1)*0^0*1^1 We got a 0^0 problem here. But if to fix it knowing that lim x->0 (x^x) =1, then it's (-1)^(-1)*1*1 = -1*1*1 = -1. Seems legit? I think it does.

[@angelmendez-rivera351]

Craftist Except you are not allowed to do that, and that is not the definition of the hyper-factorial. In particular, if you have a product over an ordered sequence, then the indexing is not allowed to happen over a variable boundary if there is no function over the boundary and if the upper boundary is a fixed constant which is lower. By your argument, H(n) = 0 everywhere except for H(1) = 1, which is absurd. No, we simply define H via the recursion formula, and if the value H(x) exists for some x, then it must satisfy the recursion. Then we say that H(0) = 1 by the recursion, and if we use a limit or adopt the convention 0^0 = 1 - and Let me tell you that there is far more reason to adopt than to not adopt it - then we can prove easily that H(-1) by the recursion. However, H(n) for n < 0 could not be calculated using products as you tried it because products are undefined for negative integers. Product and summation notation are a pain in the ass and they only add obstacles unnecessarily to the problem, even if they are more intuitive.

[@MrRyanroberson1]

Easy. H(1)=1=H(0)×1, so H(0)=0^0= conventionally 1. Then H(0)=0^0 × H(-1)=H(-1), and from there H(-n)=H(n)×H(-1)×-1^(f), and the H(-1) goes away, being equal to 1. f must be worked out to agree with the parity of H(-n), I'm sure there's a closed form. The reason H(n) appears is because on the left side it gets multiplied by values ^ negatives, and so you can see that (-2)^(-2)=1/2^2, with the only difference being a factor of -1 for odd bases, and 1/2^2 on the left becomes 2^2 on the right

[@williamallen9145]

Once upon a time there was a factorial, he drinks a lot of coffee and eats a lot of sugar, he is... HYPER. Haha

[@blackpenredpen]

Pervy Sage lolll

[@kohwenxu]

XD

[@gouravmadhwal5548]

Plz integrate ln(ln(ln(lnx)))

[@angelmendez-rivera351]

Gourav Madhwal Cannot be done in terms of elementary functions, but...

[@DiegoMathemagician]

extreme integration math 1052 xD

[@MichaelProds99]

@@DiegoMathemagician just put the horseshoe dude

[@ZelForShort]

Isn't that horseshoe math?

[@gouravmadhwal5548]

@@ZelForShort yes it is

[@2kreskimatmy]

0^0=1 :thinking:

[@IQuick143cz]

Is there any real continuation of the hyperfactorial? Similar to how we have the Gamma function for the regular factorial.

[@vincentwilliamrodriguez3572]

hyperfactoreos

[@Dreamprism]

Vincent William Rodriguez sounds tasty

[@xusui_2830]

Ever since I started watching you, you've reignited my passion for mathematics. Thank you :)

[@blackpenredpen]

xusui _ : )

[@NonTwinBrothers]

these thumbnails mannnn

[@blackpenredpen]

: ))))))

[@yrcmurthy8323]

Superb right ?

[@blackpenredpen]

@@yrcmurthy8323 thanks!

[@yrcmurthy8323]

*@blackpenredpen No need to say thanks sir,. You deserve an 👏 applause*

[@alex_ramjiawan]

If we reverse the input to n and n+1, etc, and input negative numbers, it equals (H(n))^-1.

[@AndDiracisHisProphet]

so the hyperfactorial of 0 represents the number of milliseconds in one millisecond. also, better notation would be "n?"

[@blackpenredpen]

That's so much better!

[@AndDiracisHisProphet]

Isn't it?

[@blackpenredpen]

AndDiracisHisProphet it is! Hmmm, what should I draw next? I did fish, shark, what's next??

[@stealthpilot2935]

@@blackpenredpen a dolphin

[@AndDiracisHisProphet]

Octopus?

[@yaboylemon9578]

Really digging the new fish/shark themed thumbnails.

[@cfgauss71]

Hyperfactorial is cool...BUT...it is not as cool as your jacket, isn't it?

[@blackpenredpen]

Arthur Clay lol thank you!!

[@Zafarrrrrrrrr]

hyperfactoreos :D

[@hyperupcall]

That's a slick jacket! Love it!

[@blackpenredpen]

Lightning Fast thank you!!!

[@General12th]

I just started the video. I can't wait to learn about hyperfactoreos!

[@General12th]

H(0) = 0^0 * H(-1) 1 = 0^0 * H(-1) 0^0 is undefined, so H(-1) is undefined. The question, then, is what about noninteger values of n? What about complex values of n?

[@ffggddss]

When computing H(5), you can group the factors as follows: H(5) = 5⁵·4⁴·3³·2²·1¹ [2² = 4, so put it in with the 4's] = 5⁵·4⁵·3³ = 20⁵·27 = 3,200,000·27 = 86,400,000 Fred

[@krisbarc4927]

Nice video. What about "tetrafactorial" : T(n) = n^(n-1)^(n-2)^...^1 ?

[@Someniatko]

^...^

[@NoahmassMulti]

That would be 2^10^44 if T(5) was the input

[@NoahmassMulti]

Which is around 10^10^43.6

[@nikodempatrycjuszswiercz4064]

There's another formula for it I've found (using math not google): H(n) = n!/0! * n!/1! * n!/2! * ... * n!/(n-2)! * n!/(n-1)!

[@mr_1wr572]

Cool!

[@nikodempatrycjuszswiercz4064]

This one actually uses regular factorials showing the „hyperness” of the hyperfactorial

[@Cloud88Skywalker]

@@nikodempatrycjuszswiercz4064 This is exactly what I thought he was going to do when he says he's going to rearrange. I did, and you can compact the formula to: *H(n) = (n!)^n / sf(n-1).* sf(n) is the superfactorial, I just looked it up, and it's sf(n) = n! · (n-1)! · (n-2)! · ... · 2! · 1!

[@plasmacrab_7473]

You should put the exclamation mark in the exponent so you’d have n^! Now THAT is what I call hyper.

[@norielsylvire4097]

Hyperfactorial of a negative number is equal to zero, as the exponent gets closer to minus infinity, it gets divided by a larger number, which means each index gets smaller and smaller, and tends to zero. And if you multiply a number by zero you get zero.

[@norielsylvire4097]

Sorry, I didn't explain the infinity thing. Take H(-3) for example. H(-3)=-3^-3*(-3-1)^(-3-1)*(-3-2)^(-3-2)•••=-3^-3*-4^-4*-5^-5••• the exponent doesnt get closer to zero but closer to minus infinity, as it gets subtracted one each time. A negative exponent is the same as 1/the number to that exponent. If the exponent is minus infinity it is the same as 1/infinity which tends to zero. If you multiply by zero, you get zero.

[@mayankvats926]

Next is hypersupremeover9000megatron factorial 😂. Love your videos btw!!!

[@karthikrambhatla7465]

Crazy.. heard this for the first time. Thank you❤️. But BlackpenRedPen where this hyperfactorial is used

[@legendgames128]

"He eats a lot of coffee and drinks a lot of sugar." Had me laughing for a while.

[@Hailfire08]

YES DO INFINITE NEGATIVE CONVERGING SUMS - NEGATIVE HYPERFACTORIALS!

[@aditidas9978]

New topic 😊 😊for me.but after i watched this video I did some of the problems on this topic and they seems to be too interesting .thanks,I easily catched the concept right here😊😊😊😊😊😊😊

[@perveilov]

Learn how a factorial can be such a hypebeast

[@shivimish9962]

I was literally thinking about it yesterday, and you make a video today, wow

[@karthikrambhatla7465]

Where did it came across

[@shivimish9962]

@@karthikrambhatla7465 Just popped up when doing maths

[@blackpenredpen]

Wow! What class?

[@shivimish9962]

@@blackpenredpen 12th but it was not in the book😂😀

[@Spartan11199]

Really enjoy these videos man!

[@ZipplyZane]

The natural extension would be for 0^0 = 1, same as 0! = 1. That way you just multiply all the results by one. Therefore, H(0) = 1.

[@ahmedshaikha8938]

Introducing the K-function/Barne’s G-function

[@cellina.starfire]

Isn’t this just every factorial from n! to 1! multiplied together? Pretty cool!

[@nikolamarijanovic6261]

I'm guessing the negatives would work but it depends if they are even or odd.

[@DanDart]

Loving that jacket!

[@blackpenredpen]

Dan Dart thank you!!!

[@Gadget622]

Judging by the title I was thinking it would be something like: f(n) = n ^ f(n - 1) So if n = 4 then f(4) = 4 ^ 3 ^ 2 ^ 1 = 4 ^ 3 ^ 2 = 4 ^ 9 = 262,144 Following the same pattern, f(5) = 5 ^ 262,144, or roughly 6 * 10 ^ 183,230. And f(6) would be far too stupidly big to comprehend.

[@ralfs7762]

Make more videos about factorials!!

[@nietschecrossout550]

Is there an Ultrafactorial? like using tetration: U(5)=high5^low5 * 4^4 * 3^3 * ...

[@caloz.3656]

tetration**

[@nietschecrossout550]

@@caloz.3656 thx, didn't notice it

[@caloz.3656]

@@nietschecrossout550 you're welcome!

[@caloz.3656]

Aden Tate he said tetation.

[@legendgames128]

Hyper factorial has a lot in common with hyper dashing, there is tremendous horizontal ground covered.

[@narmadav4006]

Pl. Integrate If a>1 then dx /( x^(2) + 2ax + 1 ) is equal to

[@harshsrivastava9570]

It is said that bprp is still hearting people and will continue to do so forever...

[@blackpenredpen]

Harsh Srivastava Isn't it?

[@lilyyy411]

"Eats a lot of coffee and DRINKS a lot of sugar" That's how hyper it is

[@swenji9113]

Well if you believe in your formula you don't have to write H(n) in terms of H(n-1) to get H(0). By convention, an empty product (product of zero terms) is equal to the neutral element for the multiplication law, so it is 1. In the same way, an empty sum is by convention equal to 0, which is the neutral element for addition. These conventions ensure that any reccurence formula you will get for your sum or product remains true at 0

[@JBaker452]

Think a bit about the Q function; if it does not blow your mind then think harder ;-) I have more to say about it later

[@angelmendez-rivera351]

Anyway great video! Now do superfactorials!

[@Gamma_Digamma]

Woah, I'm drooling

[@MrRyanroberson1]

Well for negatives actually, due to the recursive, H(n)=H(n+1)/(n+1)^(n+1), and for negatives the n+1 is negative or zero, and you can flip the denominator and get a multiplying chain. H(-5)=H(5)×H(-1) (from odd H to odd H the ± is the same), and now we just need to define H(-1), and since H(0)=H(1)/1, we can understand 0^0 to be 1. This leaves 0^0 H(-1) = H(0), and therefore H(-1) is also 1, since 0^0 is 1

[@MrRyanroberson1]

The whole "from odd H" is wrong, there's a rule but it doesn't resolve itself to me at the moment

[@jzanimates2352]

Hey blackpenredpen. Recently I have been watching dr peyam’s show, and mainly his videos on the half derivative of x and imaginary derivative of x. I would really like to see you try things like half intergrals and imaginary intergrals. Please like if you agree.

[@FGj-xj7rd]

More like hypersharktorial.

[@yrcmurthy8323]

The thumbnail says it

[@alextaunton3099]

Is there an integral-based formula for a hyperfactorial like there is for a normal factorial (i.e. the pi function for a normal factorial)?

[@mike4ty4]

How about integrating (x^3)/cos(2x) by expressing it with De Jonquiere's Polylogarithm (the Li_s(z) function)? I wanna see you labor and sweat with tears, see you have to use the full rainbow of pens, and/or see to it that they run out of ink in the middle of a derivation, and see you to have to recruit extra blackboards! I want drama! :g:

[@Rundas69420]

Even here mah boi the gamma function is present. Because H(z) can be defined as follows: H(z)=K(z+1) where K is the, oh wonder, K-Function. And K(z) is defined as K(z)=(2pi)^((1-z)/2)*exp[(z choose 2) + integral from 0 to (z-1) of ln(Γ(t+1))dt]. Boi Edit: So you just need to plug in z=z+1 to get a definition for H(z) in terms of the Gamma-Function. I'm using z here because you can plug everything, even complex numbers, into the Gamma-Function.

[@martinianofaure4270]

You can't have H(-1). Following that argument, H(0)=0^0*H(-1), and since 0^0 is undetermined, H(-1) is undetermined too.

[@rafaeldubois8192]

Should try ¡n! Instead of H(n)

[@blackpenredpen]

Rafael Dubois Good idea

[@Tcrrxzz]

H(-1) approaches infinity

[@JLConawayII]

We need a hypergamma function!

[@alan2here]

Not the next hyper-operation up? Lets try a few ranks. F(0, 4) = 5 F(1, 4) = 10 F(2, 4) = 24 F(3, 3) = 9 or maybe = ^{3, 2, 1} = elements of {9, 8, 1}

[@lilyyy411]

my favorite note is H flat

[@maxwibert]

Is there such a thing as a hyper-gamma function? By that I mean an analytic extension of the hyperfactorial to some well behaved subset of R or C?

[@shivammalluri6403]

Why don't you try the integral of the tetration of sin(x)

[@blackpenredpen]

Shivam Malluri ok

[@shivammalluri6403]

@@blackpenredpen great👍

[@Bodyknock]

I’m trying to think of where hyperfactorial would actually come up in a real life combinatorics problem but can’t think of one and didn’t find anything off-hand on Google. (Comparatively factorial n! is extremely useful and subfactorial !n is the number of derangements of n objects). I’m wondering if this came up in the course of a real world calculation or combinatorial proof or if it was just a function someone thought up that just seemed like it might have some interesting numerical properties.

[@tientranquang0807]

Wait, isn't H(n)=[(n!)^n]/[(n-k)!] ? (k

[@dwaraganathanrengasamy6169]

lim n -> -1 H(n)=1 where we approach from right. Is it right...?

[@angelmendez-rivera351]

Dwaraganathan Rengasamy Yes

[@michalnemecek3575]

I have a different formula for the hyperfactorial. assume n = 5 for this example Step 1: H(5) = 5^5*4^4*3^3*2^2*1^1 Step 2 (still equal to H(5), but I won't put it there because of the width of a space): 5*5*5*5*5* 4*4*4*4* 3*3*3* 2*2* 1 Step 3: first column is 5!/0!, second one is 5!/1!, third is 5!/2! and so on... Step 4: H(5) = (5!/0!)*(5!/1!)*(5!/2!)*(5!/3!)*(5!/4!) Step 5 (again, not writing H(5)=): 5 Π 5!/(k-1)! k=1 Step 6: replace 5 with n n H(n) = Π n!/(k-1)! k=1 EDIT: in step 6 I absolutely had to write H(n)=

[@toddbiesel4288]

H(5) = 86,400,000. Also, how about using ¡ as a factorial symbol?

[@manuelepedicillo864]

1:54 what happened?

[@rchishray7814]

Write it as !!n instead of H(n)

[@jorgelenny47]

So 0 to the 0th power is 1... Intresting as bronzes

[@bernhard5295]

Found a way to express hyperfac through superfac H(n)= n!^n/S(n-1). S(n)= 1!*2!*...*n!

[@vp_arth]

hyperfactorial is about hype

[@Ready4Music]

We have factorial, hyperfactorial, and… Hold up… Superfactorial??? 🤔😮 Me: Calling Superman for explaining this type of factorial. 😁

[@davidugono3727]

There's also subfactorial.

[@rafciopranks3570]

Check out Foias Ewing constant.

[@Theraot]

Now I wonder if there is a factorial for each version of street fighter.

[@Theraot]

hyperfactorial superfactorial turbofactorial alphafactorial arcadeeditionfactorial ...

[@flyingpigeon2812]

Im lovin the smilie's 😂😂😂

[@yassinenacif418]

Explostion in my mind : 5:26

Can you make a video with hyperfactorial arrow notation please

[@welltypedwitch]

So... What exactly would you use this for? Apart from H(5) of course

[@alexhenson]

H(n)=(n!)^n *drops pen on the ground*

[@JoshuaHillerup]

More evidence that in natural numbers 0^0 is 1. It's in say real numbers where it becomes undefined.

[@STEVE_K_J]

So we have factorial of 1/2, subfactorials of 1/2 but what about the hyperfactorial and superfactorial of 1/2 ?!

[@paulthompson9668]

Analytic continuation of hyperfactorial?

[@Larry640]

I'm curious as to what the use of this might be, as basically any value n greater than 10 is already incomprehensibly large. As in near universe-breaking

[@shre6619]

We need hyperfactorial of i

[@jainpriyanka14978]

Sir i think we can do its write .....

[@shubham1999]

Can you tell what is method of Multipliers in linear PDE?

[@benedettopagano3536]

Can't we all just agree that 0^0 should be 1?

[@Brianlotito]

I think that we can do hyperfactorial of a negative number. How? Idk, I just think that it is posibble

[@JTX8000]

HYPER GAMMA FUNCTION

[@aLumpOfParticles]

How about H(1/2)? 😁😁

[@michaelgandil1897]

Isn't the conclusion in the end mean that 0^0 equals 1?

[@cotasamnemano366]

H(-1)=undefined

[@Math-bz8bw]

Hi. Can you write H (n) such as integral

[@AKA-f7p]

H(-1)=1

[@fCauneau]

Seems that H(n)=n!^n/P(j!) for j=1to(n-1), no ? sorry : I wrote P for "product"

[@maxwibert]

6 weeks = 10! seconds, as a matter of fact

[@caloz.3656]

drinks a lot of coffee, eat a lot of sugar...

[@blacker4404]

What about H(1/2)?

[@anilsharma-ev2my]

Are you able to make an app over it ??

[@saxbend]

So another definition would be n!^n ÷ ( (n-1)! (n-2)! (n-3)! ...2! 1!)