For anyone wondering, the final results for 3$ is equal to 6 multiplied by itself: 3,048,038,843,157,366,051,156,451,562,587,960,914,683,843,070,009,432,577,720,371,743,970,260,248,988,253,388,734,087,776,516,378,135,275,070,793,705,483,810,168,992,174,078,819,457,229,916,982,909,272,972,117,802,705,538,546,043,924,392,880,249,338,669,607,593,617,565,773,503,681,777,562,894,648,411,193,991,668,219,111,024,140,760,301,610,404,587,920,245,316,338,496,104,651,309,961,792,720,642,767,736,988,127,034,593,657,473,200,942,901,910,918,843,867,439,542,131,034,903,455,333,688,218,414,524,168,236,538,840,952,909,280,093,318,279,899,396,079,306,730,083,204,209,917,234,906,371,378,011,735,099,583,017,148,436,711,446,474,063,192,400,558,782,588,545,007,076,563,578,617,663,198,436,110,485,321,365,453,724,515,867,573,206,330,463,738,004,176,650,289,141,225,099,730,033,358,933,544,597,672,427,393,530,251,774,965,047,211,695,506,312,946,306,762,294,998,433,468,262,930,173,135,236,327,015,973,465,126,381,014,128,785,885,984,902,934,363,536,428,840,380,015,651,386,113,570,744,138,512,750,613,093,972,490,684,498,513,211,851,441,804,268,854,063,729,284,096,776,385,783,954,252,916,394,679,043,883,583,422,253,317,285,565,365,659,687,584,910,629,328,675,433,251,717,593,584,447,460,445,269,780,478,280,767,340,984,528,217,318,932,238,565,096,396,250,980,632,747,766,225,123,954,084,888,838,021,720,715,193,156,305,805,969,620,729,856 Times.
Wouldn't 3$ be 3! x 2! x 1! = 12 ? Also 6^6^6^6^6^6 should be 10^(10^(10^(10^36305.31580191888))) Or 6 multiplied by itself 10^(10^(10^36305.31580191888)) times, I don't get how multiplying 6 that many times (the small number you gave) will result in such a huge number such as 6^6^6^6^6^6.
I like how maths always keeps surprising you... After so many years of learning, it's a subject that's still interesting. Thank you for another great video
Glass No, you misunderstood his question. What he is asking is if there exists an operator sequence H(n, •) such that, for example, H(1, 3) = 3 + 2 + 1, H(2, 3) = 3·2·1, H(3, 3) = 3^2^1, etc. In other words, the second variable determines the order of the hyperoperator that is applied to all the terms, and the first variable determines the upper bound of the terms, while the lower bound is 1. To my knowledge, no such sequence has been studies.
While it is true that there technically were two definitions of the superfactorial proprosed in 1990, the first definition, denoted by n!s, or denoted by sf(n), which is equal to n!·(n - 1)!···2!·1! is the definition most mathematicians use. Clover's definition is typically called the suprafactorial, or the Clover factorial, or confusingly, the tetrational factorial, which is the name for another operation already. Also, the notation n$ is reserved for Clover's definition, it is never used for the smaller definition. As I said, the smaller definition actually has proposed notations sf(n) and n!s, which is what mathematicians tend to use.
@DejMar I'm completely aware that Clover did not invent the definition of the large superfactorial. That is completely irrelevant to my post, though, since I never made any claims about who invented it, I merely addressed the operator by its commonly used name. In fact, the operation being described can be concisely defined as simply n$ = n!^^n! for all n, since tetration is defined for all natural bases and heights. Simon and Plouffe's definition is not a product of consecutive k^k terms, but rather consecutive k! terms. In other words, sf(n) is defined by sf(0) = 1 & sf(n + 1) = (n + 1)!·sf(n). The product of consecutive k^k terms is not the superfactorial, but the hyper factorial, denoted H(n), with the analytic continuation being the K function.
2:24 but H(0) would give 0^0 which, although there are limits that approach 1, is undefined when trying to solve without limits. Why is it 1 in this case?
You could (and should) write H(n) using product notation: H(n) = product of k=1 to n over k^k. Then by definition H(0) would be an empty product, since starting at 1 you are already above the endpoint (namely 0). It is mathematically useful to define the empty product to be 1 and the empty sum to be 0, because the neutral element for multiplication is 1 (as in a*1 = a) and the neutral element for addition is 0 (a+0=a). You can think of the product sign as meaning: „start at the neutral element and multiply as many elements until you reach the upper bound.“ so if you already are above the upper bound, you do not multiply any elements, so your answer is the neutral element (1). Edit: fixed a typo
Ace O'Spades Most mathematicians define 0^0 = 1, so saying it is undefined is stretching it. Regardless, though, H(0) is not defined as 0^0. It is defined as the empty product. The empty product is 1.
I came up with superfactorial about 5 years ago (I was in 8th grade) and actually called it superfactorial too! Never really found much of a use for it so I scrapped the idea
brushed my teeth, prepared for sleep, and then saw this notification... Read it as subfactorial and that's the reason I clicked on notif, hope I won't be disappointed, lmfao Edit: video was dope... Now show application of those in real life please :)
@@blackpenredpen yeah, I have watched, huge fan of those, thanks to You, that's why I want more :) Also edited my main comment, Liked this video, would like to see application in real life for every of those super and hyper factorials :)
Baba Froga The only applications are analysis. That's about it. Idk why people assume applications exist for everything. 90% of math doesn't have applications besides doing more math
@@angelmendez-rivera351 oh wouldn't agree, you can always see great application for everything... lol even if it is more math, it will lead to somewhere.... anyways "more math" is still real life... lmfao, not imaginary or whatsoever^^
Matteuz That already exists. It's called the "exponential factorial," or "expofactorial" for short. There is no agreed-upon notation, though, but the most common notation I have seen among googologists is n(!1), where n is the input and (!1) denotes expofactorial.
Skallos They do grow very fast. They are just overshadowed by the suprafactorial, which is the second definition that BPRP gave for the superfactorial.
@@angelmendez-rivera351 After watching Numberfile's video on fast growing functions, my expectations for those functions were quite high. They do grow fast, but not as fast as I thought.
Skallos There are no studied functions that grow anywhere nearly as fast as Graham's function or the TREE function. The gap is humongous. It's the gap from 0 to ♾, and then some more.
Try to find the value of this summations: The sum of 1/k! with k=1 to infinity is ? The sum of 1/k$ with k=1 to infinity is ? The sum of 1/H(k) with k=1 to infinity is ?
Yes, there are analytic continuations for the hyperfactorial and for sf(n) = n!·(n - 1)!···2!·1!. There is no known analytic continuation for x$ = x!^^x!, mainly because there is no known, undebatable analytic continuation for x^^x, although this in the research and it seems to exist, so far.
Hey bprp!! I have a question for you. Please answer this _/\_ If n! has 17 zeros then what is the value of n? ( I know the thing about trailing zeros but this is about total number of zeros.)
The second form of 3$ comes out to about 1.03144248E+28, or about 10.3 octillion, or we could just say about 10 thousand trillion trillion, or about 10 thousand trillion in the old British number system.
I would think there should be a factorial like this. [ 6^5^4^2^1 ] I mean that seems to me like the next logical step from multiplying in a decreasing consecutive fashion to doing it exponentially
Ion Grădinaru 1. No, it is not. Virtually every mathematician defines 0^0 = 1. This definition is basically required anyway, or else a lot of fields in mathematics fall apart. 2. H(0) is not identical to 0^0 in expression. H(n) is equal to the product of k^k with k running from 1 to n. So it would be impossible for 0^0 to show up: H(0) is simply the empty product, which is 1.
So when doing superfactorial do we use the definition define by sloune and plouffe or pickover?? Or the result will be the same ( which i dont think so ) ?? Thanksp
batu rock The definition by Sloune and Plouffe is the definition 99% of mathematicians who work with this thing use. So that's the definition to go with, and the notation is n!s, where s just means "super." The second definition of the super factorial, the n$ = n!^^n! is sometimes called the suprafactorial by said mathematicians, or just the Clover factorial.
blackpenredpen if I said I referenced your UA-cam channel as one of my maths inspirations on my university application to study maths at Cambridge would that do it?
How to Maths Firstly, I wish you the best luck and thank you, I feel very honored when I know you mentioned me on your college application. Secondly, I just did that on WFA and took me only 5 seconds www.wolframalpha.com/input/?i=integral%20of%20%28tan%28x%29%29%5E%284%2F5%29 Thirdly, after seeing the answer... I would rather do another 100 integrals (not including any radical of tan or cot) than doing that one integral lol.
blackpenredpen Thanks, I really appreciate it I mainly wanted to see if you could even start to attempt it as I would have no clue, I was watching your cube root of tan video and thought “I like this, but is there harder”, and tried some combinations and that monstrosity popped out, if you were willing to put multiple hours into it, do you think you could do it?
How to Maths Well, the killer parts are the factorings and the partial fractions! You can take a look at Chester’s video on the integral of 1/(x^5+1) ua-cam.com/video/nXn2qkY_S3s/v-deo.html I think u will like it.
For anyone wondering, the final results for 3$ is equal to 6 multiplied by itself:
3,048,038,843,157,366,051,156,451,562,587,960,914,683,843,070,009,432,577,720,371,743,970,260,248,988,253,388,734,087,776,516,378,135,275,070,793,705,483,810,168,992,174,078,819,457,229,916,982,909,272,972,117,802,705,538,546,043,924,392,880,249,338,669,607,593,617,565,773,503,681,777,562,894,648,411,193,991,668,219,111,024,140,760,301,610,404,587,920,245,316,338,496,104,651,309,961,792,720,642,767,736,988,127,034,593,657,473,200,942,901,910,918,843,867,439,542,131,034,903,455,333,688,218,414,524,168,236,538,840,952,909,280,093,318,279,899,396,079,306,730,083,204,209,917,234,906,371,378,011,735,099,583,017,148,436,711,446,474,063,192,400,558,782,588,545,007,076,563,578,617,663,198,436,110,485,321,365,453,724,515,867,573,206,330,463,738,004,176,650,289,141,225,099,730,033,358,933,544,597,672,427,393,530,251,774,965,047,211,695,506,312,946,306,762,294,998,433,468,262,930,173,135,236,327,015,973,465,126,381,014,128,785,885,984,902,934,363,536,428,840,380,015,651,386,113,570,744,138,512,750,613,093,972,490,684,498,513,211,851,441,804,268,854,063,729,284,096,776,385,783,954,252,916,394,679,043,883,583,422,253,317,285,565,365,659,687,584,910,629,328,675,433,251,717,593,584,447,460,445,269,780,478,280,767,340,984,528,217,318,932,238,565,096,396,250,980,632,747,766,225,123,954,084,888,838,021,720,715,193,156,305,805,969,620,729,856
Times.
Alex Paradise wow!!!! How did you do it? And this must be pinned!!!!!!
😇
Amazing
Programming
Wouldn't 3$ be 3! x 2! x 1! = 12 ?
Also 6^6^6^6^6^6 should be 10^(10^(10^(10^36305.31580191888)))
Or 6 multiplied by itself 10^(10^(10^36305.31580191888)) times, I don't get how multiplying 6 that many times (the small number you gave) will result in such a huge number such as 6^6^6^6^6^6.
n$ looks like an old code for a string variable to me. I am ancient and lost.
Yeah I thought the same :) Spent too much time coding in BASIC in my youth
You guys are lucky you had access to lowercase letters when you learned BASIC :)
Mário Brito you guys mean Visual Basic? If so, I did that too in high school long time ago
@@blackpenredpen how about basica from Microsoft, or worse TRS-80 BASIC from 1977. I also used a lot of QuickBasic
Just when i already started to learn programming in college lol
I need to ask my employer if I can have a "pay cut" to 3$. Mwahahahaaa!
alkankondo89 Hahahah!!
In a lot of countries (e.g. France), the currency symbol comes after the number, so this might backfire.
@@user-vn7ce5ig1z not in America 🇺🇸🇺🇸🇺🇸🇺🇸🇺🇸🇺🇸🇺🇸🇺🇸
Kiwi yes it does, you just use it wrong
Me : How much does a diamond cost?
Owner: 2019$
Me: Wow what a deal!!😍😍
(A few seconds later)
Me: wait a minute!!!...😱😱
Indian dude counting rice grains on a chessboard: 😱
I think 3$ is enough to make me go 😱😱😱😱
Dollar signs are supposed to go before the value so you only have yourself to blame.
@@TheTdw2000 i think it's how the price is pronounced, the number first then "dollar"
blackpenredpen wait 3$ is equal to 6*2*1 why is that expensive
It's funny that 3$'s last digit is a 6
And the last 3 are 256 which is a power of 2
why
@@oboplays3413 6*6=36, which last digit is 6. Keep multiplying by 6 and you'll always get 6 as the last digit of your number!
@@pablofueros that's actually pretty interesting
You can actually calculate the last digit of any big number as 3^4556789. Using modular arithmetic
Darn I was hoping for h(n$).
@@DudeinatorMC n equals 1. Nah
H(n$) vs H(n)$❓
I like how maths always keeps surprising you... After so many years of learning, it's a subject that's still interesting. Thank you for another great video
At least Sloane and plouffe's actually makes sense.
Pickover seems to have just gone "BIG NUMBERS OHHHHHHHHHHHHHHHHHHHHHHHH"
Hyperfactorials look big but nevertheless grow a lot slower than TREE.
Head Librarian yeah it grows a heck of a lot slower than g. And g to tree is like a snail to the speed of light.
Isn’t the hyper factorial every term of the normal factorial to the tetration of 2?
Yup that's how it is defined
Is there a term that describes a single dimension of hyperoperation applied to factorials? As 3! Is 3x2x1 what would 3+2+1 be defined as?
@@decatessara5029 Well, you can use Gauss's addition formula for this, so you'd just end up with it being the same as a function f(x)=(x/2)(x+1)
Glass No, you misunderstood his question. What he is asking is if there exists an operator sequence H(n, •) such that, for example, H(1, 3) = 3 + 2 + 1, H(2, 3) = 3·2·1, H(3, 3) = 3^2^1, etc. In other words, the second variable determines the order of the hyperoperator that is applied to all the terms, and the first variable determines the upper bound of the terms, while the lower bound is 1. To my knowledge, no such sequence has been studies.
@@angelmendez-rivera351 Ah ok, his question makes more sense now
In here (Turkey) it is 1.30 AM and a new video from blackpenredpen. I clicked to video, and listen...
In here (Germany) it's 23:36. Good night 😊🌃
Also me
@@katarinakraus120 Thanks, same for you :)
@@Ali-mf4sl We are night people 🌕🌕🌕
In Brazil it is 08:06 pm
Let's put euro sign € for Pickover's factorial. Then my salary will seem a lot better:D
Viki V. Actually, the notation 3$ is alright. BPRP was wrong in stating that the other definition uses the notation 3$. In reality, its 3!s.
While it is true that there technically were two definitions of the superfactorial proprosed in 1990, the first definition, denoted by n!s, or denoted by sf(n), which is equal to n!·(n - 1)!···2!·1! is the definition most mathematicians use. Clover's definition is typically called the suprafactorial, or the Clover factorial, or confusingly, the tetrational factorial, which is the name for another operation already. Also, the notation n$ is reserved for Clover's definition, it is never used for the smaller definition. As I said, the smaller definition actually has proposed notations sf(n) and n!s, which is what mathematicians tend to use.
Ah!! Thanks.
@DejMar I'm completely aware that Clover did not invent the definition of the large superfactorial. That is completely irrelevant to my post, though, since I never made any claims about who invented it, I merely addressed the operator by its commonly used name. In fact, the operation being described can be concisely defined as simply n$ = n!^^n! for all n, since tetration is defined for all natural bases and heights.
Simon and Plouffe's definition is not a product of consecutive k^k terms, but rather consecutive k! terms. In other words, sf(n) is defined by sf(0) = 1 & sf(n + 1) = (n + 1)!·sf(n). The product of consecutive k^k terms is not the superfactorial, but the hyper factorial, denoted H(n), with the analytic continuation being the K function.
@DejMar Also, I know how repeated exponentiation and tetration works.
I'm trying to enjoy math for fun (I'm not good at it). You sir, are amazing at teaching, enjoying and helping me.
I am not sure if I'll ever need that in my life, but it's very cool.
And the nice thing about the hyperfactorial is the recursive definition works just fine for H(0).
2:24 but H(0) would give 0^0 which, although there are limits that approach 1, is undefined when trying to solve without limits. Why is it 1 in this case?
By definition, just like 0! is defined as 1
You could (and should) write H(n) using product notation: H(n) = product of k=1 to n over k^k. Then by definition H(0) would be an empty product, since starting at 1 you are already above the endpoint (namely 0). It is mathematically useful to define the empty product to be 1 and the empty sum to be 0, because the neutral element for multiplication is 1 (as in a*1 = a) and the neutral element for addition is 0 (a+0=a). You can think of the product sign as meaning: „start at the neutral element and multiply as many elements until you reach the upper bound.“ so if you already are above the upper bound, you do not multiply any elements, so your answer is the neutral element (1).
Edit: fixed a typo
just like √(0), it's 0, of course, but it has no limit from the left in the real numbers
Ace O'Spades Most mathematicians define 0^0 = 1, so saying it is undefined is stretching it. Regardless, though, H(0) is not defined as 0^0. It is defined as the empty product. The empty product is 1.
Angel Mendez-Rivera [citation needed]. I'm not aware of *any* mathematicians who define 0^0 as 1, it leads to such a mess.
I can’t understand English.
However, I can understand this beauty.
I like mathematics very much!
I like this blackpenredpen. I love your videos, all of it. I'm from Philippines and always watching your amazing tutorial. Please notice me.
Thank you! I appreciate your support.
I love this channel for ridiculous maths I had to click
I like TREE(3), it's even bigger and raises even faster :)
TREE(Graham's number)$
@@JorgetePanete yeah, that's scary!
All of these random UA-cam videos with no relevant information for my life are what I live for
I came up with superfactorial about 5 years ago (I was in 8th grade) and actually called it superfactorial too! Never really found much of a use for it so I scrapped the idea
But what is a super-hyper-ultra factorial?
Thank you BrendaWork for sponsoring this video
Also, the dollar sign matters ;)
Pulling that work
The famous multiple question
The port and girl paradox
If you like to keep rolling the track
Brenda work slacks back from your pen
Those 2 definitions for super factorial give different answers, or am I missing something? For example by def.1:
3$ = 3!*2!*1!=12
Эти определения от разных людей и ответы получатся разные
Yeah. So..
Yes, although the notation for the first definition is not 3$, it's 3!s. I'm not sure why BPRP presented it otherwise.
Thanks man! My mind turns from REGULAR to SUPER HYPER after watching this video! Cool level: 2019$ 😎
Mak Vinci Hahahhaha
Btw, I really wonder they came up with all these. I can’t even calculate 3$ with the last definition lol
@@blackpenredpen Yes! Even wolframalhpa cannot handle 6^6^6^6^6^6, so crazy!
Those numbers have more digits that there are elementary particles in the entire observable universe.
brushed my teeth, prepared for sleep, and then saw this notification... Read it as subfactorial and that's the reason I clicked on notif, hope I won't be disappointed, lmfao
Edit: video was dope... Now show application of those in real life please :)
aha, yes ..wanted to say, pls more video on combinatorics and subfactorial, yolo
Been there, done that ua-cam.com/video/dH7kt-xAlRA/v-deo.html
Thank you!!
@@blackpenredpen yeah, I have watched, huge fan of those, thanks to You, that's why I want more :) Also edited my main comment, Liked this video, would like to see application in real life for every of those super and hyper factorials :)
Baba Froga The only applications are analysis. That's about it. Idk why people assume applications exist for everything. 90% of math doesn't have applications besides doing more math
@@angelmendez-rivera351 oh wouldn't agree, you can always see great application for everything... lol even if it is more math, it will lead to somewhere.... anyways "more math" is still real life... lmfao, not imaginary or whatsoever^^
Wish I had a teacher like you! 💖
0.5 super-factorial?
I was expecting something like n^(n-1)^(n-2)^...^2^1 lmao
Matteuz lolll that will be so cool
Matteuz That already exists. It's called the "exponential factorial," or "expofactorial" for short. There is no agreed-upon notation, though, but the most common notation I have seen among googologists is n(!1), where n is the input and (!1) denotes expofactorial.
Angel Mendez-Rivera wow, that’s cool!
maybe notate it "nᵎ" or "n‽"
That’s why we put dollar symbol before the number to specify the cost
what are the uses for these operations?
I have seen the first kind of superfactorial befre but not the hyperfactorial or the double up arrow version.
As math progresses, coding will be more integrated into it
Amazing. This reminds me of numberphile's video on Graham's Number. they used like 4 arrows i believe.
I was quite surprised at that simple relation, I thought that surely the superfactorial and hyperfactorial grow much faster than that.
Skallos They do grow very fast. They are just overshadowed by the suprafactorial, which is the second definition that BPRP gave for the superfactorial.
@@angelmendez-rivera351 After watching Numberfile's video on fast growing functions, my expectations for those functions were quite high. They do grow fast, but not as fast as I thought.
Skallos There are no studied functions that grow anywhere nearly as fast as Graham's function or the TREE function. The gap is humongous. It's the gap from 0 to ♾, and then some more.
Would you use it in superpoker
This is very cool BPRP! Will you be continuing with more big functions/operations? I would like to see them 100% Maybe fast growing hierarchry? :)
There is also primorial #n - product of first n prime numbers.
Krutoy Gadget
Wow!!! I didn’t know about it. Thanks for letting me know and I will look into it
Actually, I'm fairly certain it is n#, not #n.
@@angelmendez-rivera351 Yes, in Wikipedia it is n#, but in one book it was written #n.
Why is hyperfactorial of 0 equal to 1? At 2:22
Amazing 😍. I know the superfactorial and hyperfactorial from the Google but I didn't see the relationship between them. This is wonderful ❤️
How would i write down a normal factorial but with only even numbers? Like n*(n-2)*(n-4)*...*2
There is a !! for odd numbers. For even numbers you can just factor out the 2 so 6*4*2 =2^3(3*2*1)
So factorial for evens is just 2^n*n!
@@Green_Eclipse I haven't thought about that! Thanks :D
I’ve only seen this term in Yugioh lol.
Is there a practical need for super or hyperfactorials? Normal factorials are used in combinations, permutations, taylor seriers, etc.
holy moly , that second super factorial grow very quick, 6^6^6 already have 36306 digits
that should be called ultrafactorial instead
Lol!! superultrafactorial
David Franco,6^6^6=6^((7777-1)×6)
any app for it ???
Is there a name for n^(n-1)^(n-2)^...^3^2^1 ?
Henrix98 Yes. It is called the expofactorial, and it is sometimes denoted by n(!1), although there is no actual universally agreed upon notation.
Try to find the value of this summations:
The sum of 1/k! with k=1 to infinity is ?
The sum of 1/k$ with k=1 to infinity is ?
The sum of 1/H(k) with k=1 to infinity is ?
The first one is easy: e, but there is a way to find the exact value of the other two ?
In the second I got approximately 1,58683
Simply awesome
Why is the first definition different than the second?
It’s a very rough calculation but 3$ has a magnitude on the order of 10^10^10^36305. So needless to say, the super factorial grows very quickly
DragonFire17 yup, it’s some serious stuff!
2 is very special. 2$ (second form) is the same as H(2) or 2² or 2×2 or 2+2. Daaaaamn
What do you compute with super and hyper factorial?
I've seen all before! Are there any uses?
Are there any extensions to the rationals/reals/negative integers? How about derivatives?
Drew Michael Look up Barnes G function and K function
what is d(n$)/dn and dH(n)/dn?
Very Interesting!
I have to ask the question: are you planning on naming your next bunny factoreo?
I don’t own a bunny. That’s peyam’s bunny. But I sure will if I get one
@@blackpenredpen Oh, I saw a few videos with you and a punny. You were just hanging out with Dr. Peyam. My bad.
@@blackpenredpen Either way, one of you should name their bunny factoreo. It seems very fitting. 😁
Dacota Sprague yea definitely! I do want a bunny myself in the future. It’s so cute!
Nice. My mind is blown yet again :-)
What's the point of the last super factorial, if it's even impossible to calculate 3$?
What's the point of Tree?
@@4snekwolfire813 just to promote MrBeast of course!
Imagine Graham's number after seeing the 3$ example...
For the hyper factorial, isn't 0^0 indeterminate? You mentioned H(0) = 1, but from the formula 0^0 can be many things, like e, pi, etc...?
Keep letting us know this type of new concept! Please....
Is there any real application to these hyper-super-large number?
What is the use of super and hyperfactorials?
"To make You tube videos", said this channel's owner somewhere in the comments
Wait, just to be sure, the two forms of the superfactorial do not have the same meaning right? Like they can't be equated right
AAK Music They do not.
every single time I see these super powers and what not I feel like its made up lol
So which definition of n$ is bigger? And are there analytic continuations of n$ like the Gamma function for n!?
Logan Kageorge
Definitely the one by Pickover
Yes, there are analytic continuations for the hyperfactorial and for sf(n) = n!·(n - 1)!···2!·1!. There is no known analytic continuation for x$ = x!^^x!, mainly because there is no known, undebatable analytic continuation for x^^x, although this in the research and it seems to exist, so far.
Where is this applied? Or we don't do that here? Cuz factorials atleast have a use...
Hey bprp!! I have a question for you. Please answer this _/\_
If n! has 17 zeros then what is the value of n?
( I know the thing about trailing zeros but this is about total number of zeros.)
It's beautiful!
Wait. Isn’t hyper factioral smaller than super factorial for all values greater and equalt to 4?
Hey, Can you do a video about the Inverse function of the Factorial (for all numbers which the original function is defined at?)
Like the Inverse of the Gamma/Pi functions
This looks like a Numberphile content topic ;d
Intuitively, superfactorial makes me think of something like:
n$ = n^(n-1)^(n-2)^(n-3}^...^3^2^1
The second form of 3$ comes out to about 1.03144248E+28, or about 10.3 octillion, or we could just say about 10 thousand trillion trillion, or about 10 thousand trillion in the old British number system.
Wait a second is it supposed to evaulate left to right or right to left, because I could be wrong, if I was doing to operations in the wrong order.
Hi, I am a math teacher as well. can we collaborate.
Uses of super and hyper factorial
To make UA-cam videos : )
@@blackpenredpenyeah
Are you going to change your name to blackpenredpengreenpen?
I would think there should be a factorial like this. [ 6^5^4^2^1 ] I mean that seems to me like the next logical step from multiplying in a decreasing consecutive fashion to doing it exponentially
6^5^4^3^2^1 is equal to 6!1
When would you ever need a super or hyper factorial lol
When you want to make a Numberphile video but have no topic.
Yeah they doesn't seem useful
Gabriel Antas They are useful in combinatorics and physics.
They're used in defining certain mathematical constants iirc
@@angelmendez-rivera351 can you show me some examples? I've still haven't found one
Those definitions for n$ aren't equal, unless somehow 6^^6 = 12.
What about TREE factorial
Hmmmm maybe let’s not go there lol
As far as I am concerned, there is no function with this name. What would you define this as?
I propose naming the product n$ * H(n) as the "Ultra Factorial" of n.
H(0) shouldn't be undefined?
Ion Grădinaru Why should it be?
0^0 is undefined
Ion Grădinaru 1. No, it is not. Virtually every mathematician defines 0^0 = 1. This definition is basically required anyway, or else a lot of fields in mathematics fall apart. 2. H(0) is not identical to 0^0 in expression. H(n) is equal to the product of k^k with k running from 1 to n. So it would be impossible for 0^0 to show up: H(0) is simply the empty product, which is 1.
What about "n?" ?
It is "easy" to expand the regular and super ! to the complex plane. The hyper! should be something like (Gamma(n+1))^(n+1)/n$ rigth?
gaier19 Yes
No
So when doing superfactorial do we use the definition define by sloune and plouffe or pickover?? Or the result will be the same ( which i dont think so ) ?? Thanksp
batu rock The definition by Sloune and Plouffe is the definition 99% of mathematicians who work with this thing use. So that's the definition to go with, and the notation is n!s, where s just means "super." The second definition of the super factorial, the n$ = n!^^n! is sometimes called the suprafactorial by said mathematicians, or just the Clover factorial.
Lucky that it uses only the tetration and not the Nth operation
Anyone know what the ? Function is. I've never been able to find it.
You mean the question mark function? I have seen it before it’s a very weird thing. Lol
@@blackpenredpen yes, all I know is it grows very quickly, aside from that I have no clue as to what it does.
Isn't the first súper factorial the same as n!^1*(n-1)^2*(n-2)^3...*3^(n-2)*2^(n-1)*1?
Now we need H(tree(3))
Or maybe not.
Try (H(H(tree(3))))^H(tree(3))
The number of digits of the number of digits of the number of digits of the number of digits (4 times) of 3$ is 36,306. A very Big number
I have seen the hyperfactotial but not the superfactorial.
Solve integration of e^x×logx and sin^1/3 x
I have here your greatest challenge, one which I don’t believe you can complete
Integrate (tanX)^(4/5)
Good luck
That’s very easy. Go watch all my videos first as a pre req then I will do that for you. : )
blackpenredpen if I said I referenced your UA-cam channel as one of my maths inspirations on my university application to study maths at Cambridge would that do it?
How to Maths
Firstly, I wish you the best luck and thank you, I feel very honored when I know you mentioned me on your college application.
Secondly, I just did that on WFA and took me only 5 seconds www.wolframalpha.com/input/?i=integral%20of%20%28tan%28x%29%29%5E%284%2F5%29
Thirdly, after seeing the answer... I would rather do another 100 integrals (not including any radical of tan or cot) than doing that one integral lol.
blackpenredpen
Thanks, I really appreciate it
I mainly wanted to see if you could even start to attempt it as I would have no clue, I was watching your cube root of tan video and thought “I like this, but is there harder”, and tried some combinations and that monstrosity popped out, if you were willing to put multiple hours into it, do you think you could do it?
How to Maths
Well, the killer parts are the factorings and the partial fractions! You can take a look at Chester’s video on the integral of 1/(x^5+1) ua-cam.com/video/nXn2qkY_S3s/v-deo.html I think u will like it.
But is there some convinient way to say: "/sum_{k=1}^{n} k"
"n*(n+1)/2",
"1+2+3+...+n",
"n+1 choose 2",
or most conveniently just "the nth triangular number"
@@philipphoehn3883 thank you :-)
Ok cool but what do I do with this now?
Overflow occurred in computation.
Amazing