on the product of all natural numbers...
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- Опубліковано 21 лис 2024
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I like that you point out that some rules have to be broken to reach valid answers for evaluating functions outside of their domain. You're the first person I've seen who has explicitly stated that fact.
Ikr, too many ppl on youtube don't put that disclaimer 😭
Conjecture: more videos on this sort of thing are presented by British and Australian people than not. They tend to have a much drier wit than the typical American's sense of humor. (Probably due to the extra u in humour.) Perhaps these presenters assume their audience will know that they're being "cheeky" without being overt about it.
The world is the world and people are many and varied and yes I agree there are tensions associated with this topic. But is there any harm in that?
I'm a PhD mathematician. I bet u in 100 years we humans will realize that "those rules" u spoke about in you original comment here should have never been created in the first place. P-series test is false.
I know crazy comment huh. Humans don't understand infinity. I don't understand it. But I do know the sum of the naturals is -1/12. There is no need to put an asterisk on this fact or claim this is only in certain contexts....
No... there is no other context. It's -1/12.
The reflection formula tells us this fact very clearly. If the basil problem is pi^2/6 ... this only is true if the sum of the naturals is -1/12
There is no other context lol.
But don't listen to me... I'm just a cultist.
Just keep ur head in the sand and keep letting the academic elite gaslight you into believing that they understand numbers
Spoil alert: they don't.
I love Micheal Penn. Brilliant man. But he don't understand numbers either. No one does.
I don't understand numbers. But I am a beleiver. Like I said above... I am a cultist. And I am in awe of the magic and mystery of the numbers and I worship the divinely created primes that all numbers come from.
Cultist
In today's video, we're gonna commit MATH CRIMES! Let's begin.
"And that's a good place to stop" out,"And there you have it" in!
Perhaps he felt that it was a good place to stop as soon as he wrote " infinity! "
Im confused, where should i stop?
That's Dr. Penn's call.
No good place to stop when doing infinite products.
sqrt(2 * pi) ~ 2.50662827463, so I guess 2.5 is a good place to stop.
@@Milan_Openfeint hahahaha
Where you had enough fun
I feel like I fell down a rabbit hole and witnessed Riemann and Euler exchanging jokes about transcendental numbers while drinking wine from a Klein bottle.
Genius comment
The final result could also be expressed as "The product of all squares is the circumference of the unit circle", which has the taste of "squaring the circle" ;-)
All these squares make a circle!
@@59de44955ebd Remember, the sum of those squares is 0.
all these squares make a circle……
Deep. Squaring the circle is a fundamental generative process, not any distinct object. Compare with trying to compute the length of the straight line floor from the number theoretically very beautiful arc length of the cycloid (2^3 times the ray of the associated circle).
We could say that the theory of continued fractions - and especially quadratic continued fractions - is generated by squaring the circle.
Ramanujan received his intuitions from a Goddess of continued fractions. Way, way before Gosper gave us the continued fraction arithmetic,
Gosper arithmetic is the holy grail of elementary pure math, "hiding" in plain sight. It's not taught "canonically", the knowledge of the Holy Grail is passed in youtube comments to those who write worthy comments.
"And there you have it."
Me: Oh, well obviously! Very simple and straightforward. *nervous laughter*
The first place where a rule was broken was saying "p=1•2•3•…" because it asserts that there exists some value p (which is assumed to be a real number) that is equal to an infinitely growing product.
Take a plane segment. It has infinite lines. Take a line. It has infinite points. So infinite sums may be finite, just finite in higher dimensions.
@@Miparwo infinite sums (and likewise infinite products) can absolutely have finite results, such as the sum of all powers of 1/2. The problem, though, is that the product of all positive integers is divergent, meaning that the limit of the n'th sub-product doesn't exist as n approaches infinity. Setting it equal to some complex value P is asserting there is some convergence to a limit value, which is false.
@@ProactiveYellow Take a square, and remove one quadrant. You will be left with 3 smaller squares. Remove the same quadrant from each of the remaining squares, and keep repeating it infinitely many times. If you sum the area removed, the infinite series 1/4 + 3*1/4^2 + ... + 3^(n-1)/4^n + ... converges to 1, so you removed all the area. But you are not left with nothing. The remaining points are a Sierpinski triangle. It has zero 'area', but not zero volume in its dimension.
It shows that infinite sums' convergence should be interpreted in relation to the dimension, and what is meaningless in one dimension. is meaningful in another.
I'm not sure that applies here. We just have scalar products no dimensions involved. Unless you cpunt p factorial as a p dimension object and p plus 1 factorial as a p pmus w dimenssional object. Is that along the lines of what you mean?
@@leif1075 1^(1/4)=i.
thats scalar on scalar function, but gives an imaginary result, which is in another dimension than the real scalars.
It is wrong to assume that operations on scalars only give scalar results, specially when an infinite product requires an infinite variable to express it, which is not a scalar.
Here an interesting pattern: we have a closed formula for the partial sums of 1+2+3+...+n (aka triangular numbers), the famous Sum(n) = n(n+1)/2. And if we interprete this as function on R and integrate it from -1 to 0, we get precisely the -1/12. And (almost) the same also applies to the partial products: let Fac be the factorial function, i.e. Fac(x) = Gamma(x + 1). Then again its integral from -1 to 0 gives us precisely this value sqrt(2 pi) - only that now we have to integrate its logarithm instead. So we have:
1+2+3+... "=" Integral[-1..0](Sum(x) dx) = Integral[-1..0](x(x+1)/2 dx) = -1/12
1*2*3*... "=" Integral[-1..0](log(Fac(x)) dx) = Integral[-1..0](log(Gamma(x + 1)) dx) = sqrt(2 pi)
In case of the sums, the integration from -1 to 0 is definitely not an "coincidence", it works for all Zeta function values of negative integers: for any positive integer k we can always find a closed formula for 1^k + 2^k + ... + n^k, and integrating this "formula" - as function on R - from -1 to 0 then always gives us the value of Zeta(-k).
Isnt everyone else surprised..why is the infinite sum.even involved in a video.about infinite products?
The integral of ln(x!) from -1 to 1 gives us ln(sqrt(2 pi)), we have to then take the exponential to get sqrt(2 pi). So, in effect we have to take the product integral of x! from -1 to 0, which makes sense given it's an infinite product, not a sum.
Neat! I noticed the connection between integral of the generating function of the triangular numbers and the Riemann zeta function a few years back. But I had no idea it could be generalized like that. Well done. =)
The final results reminds me the Stirling approximation: isn't sqrt(2pi) the multiplicative factor of it? It is as if you "remove" the part that varies in such approximation.
thats a very nice observation. it seems reasonable since Stirling's approximation is valid for very large n
Excellent video-you point out the rules we're breaking, you motivate substitutions, and this problem is one I've never seen before
Excellent video! Finally (for me) a clear explanation reconciling 1+2+3+... with -1/12.
A proof by conttadiction that: 1 +0 +1 +0 .... != 1/2
I almost feel guilty to be satisfied watching all those rules being broken
This checks out. Seems like it should be around two and a half.
I like how when you showed the Riemann zeta function, I immediate thought "oh, why dont I just take d/dx ζ(x) at x=0?" I plugged it into my calculator, and sure enough, -ln(2π)/2, indicating that -ln(1)-ln(2)-ln(3)-... Analytically evaluates to -ln(√(2π)), and therefore the product becomes √(2π).
I also appreciate how this aligns with the asymptotic expansion of x! = x^(x+½)*e^-x*√(2π)*(1+1/(12x)+O(x^-3)). And how the constant term there is √(2π).
The Mathematica codes are Product[i, {i, 1, Infinity}, Regularization -> "Dirichlet"] and Sum[i, {i, 1, Infinity}, Regularization -> "Dirichlet"] for inifinite factorial and sum respectively.
this is funny, because yesterday i was playing around with surreal numbers and tried (and failed-ish) to define a factorial on all nonnegative omnific integers to find w! where w is the first transfinite ordinal
That sounds awesome! I've studied the Surreals recently for a college project, but I hadn't heard of "omnific integers". What are those?
@@benjaminhill6171 After quick search it seems these are simply even numbers. Not sure why the fancy name.
@@benjaminhill6171 They're the surreals x such that x = {x-1|x+1}. They can also be defined as the surreals whose Conway normal form has no negative exponents.
(It's chapter 5 in On Numbers and Games.)
@@wyboo2019 I don't know if this would work, but I suggest using the full Stirling asymptotic expansion for the factorial and plugging in omega. Asymptotic expansions converge for infinite values, something the late physicist Martin Kruskal was working on. Just be sure to use the Kruskal-Gonshor definition of the surreal exponential function, not Conway's omega exponential map.
@@tomkerruish2982 Thanks! I've read a bit of Winning Ways for Your Mathematical Plays, but I haven't gotten around to ONaG yet. It took me a whole semester to just understand what Surreals are, along with some very basic surreal analysis and combinatorial game theory, so I feel like I've only scratched the surface of this enormous and amazing new world of math. I'll give it a look!
I love your number theory videos....especially the ones, like this, that I call "funky maths" (sorry, I'm Australian and old). 😂
This would be a lot easier if you just accepted 0 as a natural number.
😂
I do.
"Natural" number is a red herring, hence confusion of many people. 3 is just as imaginary as i.
@inyobill Or... just as real. Some of us are Platonists, and there may even be a few Pythagoreans scattered about.
@tomkerruish2982 feel free to show me a "3", and I'll concede your point.
that's a weird place to find τ
Yes, I found it an interesting coincidence (?) that the final result, sqrt(2 pi) or sqrt(tau), is also the constant factor that occurs in Stirling's approximation to n!. So, one consequence of this result is to say that the "non-constant" factors in Stirling's approximation [ namely sqrt(n)*(n/e)^n ] approach 1 as n approaches infinity. Clearly, another result that needs to be in "strong quotes".
@@richardsandmeyer4431 did you put strong quotes inside strong quotes
Luv this one tu!
And agreed there are tensions about "regularization of that".
But if some others are to be believed stronger tensions arise and, some might say, only exist because physicists can work with these processes to get good, improved or better results.
Disclaimer: I am not a physicist.
That seems to be reason there are hot debates.
A less rigorous approach on math gives more rigorous results in physics hence physicists explore this to see where it may lead rather than sit and debate mathematical robustness of these techniques.
Best summarized by "Heck if it works don't knock it!" in Physics with a "Heck do not rely on that untruth in general" in Mathematics.
And I posit: it is too early to take sides expecially when both sides are relatively perfectly correct.
Human thinkspace tends along lines of: there is only one truth, only one leader, only one champion, ... where in the real world everything that exists is a champion it it own relative right
EDIT: hmmm just wondering...is there a hidden implication that treating this in s and x 2-space has its workings and results from influenced by non-orthogonality in s, x space or some other non-orthogonal base that influences s,x 2-space?
The result is jaw dropping 🤯🤯🤯🤯🤯
"things went wrong a lot of different places"
New life motto :)
definitely not a good place to stop
7:40 "We're breaking all the rules of calculus, but we might as well because none of this is well defined anyway"
(@18:45): This is exactly why I've never liked the math behind the "proof(s)" for (-1/12) being the "solution" to be sum of all natural numbers -- the fact that so many rules need to be at least massaged, if straight-up broken in order to arrive at said alleged solution
In math, the answer to the question "what does it mean?" is "whatever you want it to mean".
If you didn't define it before, you can define it however you like. That's what math is actually like. Nothing forces us to choose convergence in the reals as the canonical meaning for infinite sums/products; that's what we *chose* for it to be. Similarly for the analytic continuation of the Rieman Zeta function.
With the important caveat that, once you defined your things, you need to stick with those definitions for the duration of the problem for the answer to mean anything. That's why analytic continuations are generally unique, because at the beginning of the process of defining how to analytically continue a function, definitions were made that are required to be met for the process to work in a way that yields a meaningful result within that particular paradigm.
The cool bit happens when you make 'reasonable' definitions on that front end on how to 'get into' those unavailable regions and give them 'consistent' values, since a set of fairly reasonable extra definitions leads to a very unusual-looking result. :)
@@HeavyMetalMouseYes it's an important distinction to make that even though we can be creative with meanings and interpretations we still have to be consistent as not to be literally wrong.
Whether a series converges or not is a result, not a defonition.
20:00 So therefore -Z'(0) = ln(sqrt(pi/2)) - ln2 = -ln(sqrt(pi)/(2sqrt2)) hence Z'(0) = -ln(sqrt(pi)/(2sqrt2)). = -ln(sqrt(2pi)/4) What am I missing?
I would write that as -ln(sqrt(pi/8)) but yeah I noticed the same mistake.
There was a sign mistake in front of ln2 at 16:19 , so it's -Z'(0) *minus* ln2 = ln(pi/2)/2. The end result is therefore correct
@@AbuMaxime In maths, ever so often, two wrongs do make a right.
On 2nd to last board there is an error in the derivative, it should be plus ln2... Instead of minus, but the final answer seems to be correct.
You are right, I noticed that too.
The reason, the final answer checks out is because at 20:02, when Michael leaves the subsequent manipulation "as a tiny homework exercise", the result would *not* be -ln(√(2π)) but rather -ln(√(π/8)). In order to get 2π instead of π/8 one needs that very sign to be the right way.
In other words, the error in the derivate would have been noticed if Michael had demonstrated this "tiny homework exercise" in the video! 😉
If you want to explain the idea of analytical continuation giving this "1 + 2 + 3 + ... = -1/12"...
Start by talking about 1 + x + x^2 + ... = 1/(1-x). This is true as long as -1 < x < 1. But, if we ignore that and use, say, x = 2, we get 1 + 2 + 4 + 8 + ... = -1. This is obviously absurd, but there is a sense in which it kinda sorta works, if you squint hard enough.
Yeah, we may have it, but we have no good place to stop.
plz make a playlist for analytic number theory
When doing IBP, shouldnt the du be negative?
at about 11:00 i fail to understand why cant s=0 be plugged without IBP
I wonder whose name will be put on the analytically extended "product" like the "Ramanujan sum".
This is weird. For about a week now, this kept cropping into my mind. "Is there an equivalent of the summation of all natural numbers, when considering the product of all natural numbers".
And then this crops up.
Maybe GArak was onto something?
And there you have it?
And this is a good place to stop!
Anyway… Great video, with a lot of incredibile balancing acts and wonderful acrobatics 👍
It's interesting to think about how an infinitely large number might still have properties to it. For example, the sum of all even numbers is an infinitely large even number, but if you add 1 to it, it's an infinitely large odd number. Maybe. I'm not sure if such properties are even rigorous to discuss.
When should we stop?
Try using Euler's Zeta function product and see if the infinite product also equals sum -12 , or not
Where the series converges all of the interchanging is correct, is it not? Is this this formulation equivalent to finding the analytic continuation of your log function and evaluating it?
If the square root of pi is known as the Gaussian Integral, does that make the square root of 2pi the Taussian Integral?
Is there any relation between this result and the Stirling's approximation?
So -1/12 would be the infinite triangular number?
actually it's more satysfying as sqrt(tau)
Analytic continuation is different from zeta (-1): the sum is not defined at s = -1 so 1+2+3+4+ ... not = -1/12!
Infinity factorial? Simple. It's the gamma of infinity plus one. :D
I'm confused. when you use the definition if the Riemann function (10:44), it's s (inside the integral) multiplied by something complicated. So the value at 0 should be 0 (before the integration by parts). After the IBP, it's -1/2 probably because the integral diverges if s
The integral representation before doing IBP is only valid for Re(s) > 0. The integral obtained after IBP turns out to be valid for Re(s) > -1. Evaluating the latter at a point where the former is not valid might seem fishy, but it is justified by the identity theorem.
See e.g. these videos: -YSzkfusaWQ&lc and T7aqj1r7CsA (Not the full URLs for obvious reasons, ha)
In the first calculation, if s->0 then Re(s)
The Archimedean Spiral.
Hmm. The product of "all" nats seems to me something very similar to the full hyperoperation tower. Compared to just the sum, with multplication, the second iteration aspect of the hyperoperation tower comes already in play.
So, the question becomes, can we say something sensible and coherent about the top of the hyperoperation tower? How do the (hyper)logarithms and (hyper)roots come in play?
Analytical methods are here clearly just heuristic devises, and "rule breaking" in heuristic sense means just reaching for some perhaps useful hints and insights. A pure mathematician of course ultimately reaches for elementary formalism of solidly coherent intuitive comprehension. "Rule breaking" of reductionistic bottom-up perspective of mathematics is absolutely necessary when reaching for holistic top-down vision, implied by (hyper)logarithms, which travel down the hyperoperation ladders.
A notation sometimes used for the bottom-up hyperoperation ladder is
addition
multiplication
exponentiation
tetration
etc.
For the inverse top down perspective we could similarly write with inverse Dyck pair:
>0<
>1<
>2<
>3<
etc.
In this case I chose to start the numeration from zero for a reason. "Division by zero" is not allowed in the bottom-up perspective of field arithmetics based on addition, but it's vital for the basic nesting algorithm of Stern-Brocot style of deriving number theory in top-down manner. So the zero in top down perspective implies that we are talking about holistic perspective when trying to say something coherent about the top of the hyperoperation tower.
Next hint is also very basic and elementary. The standard order of operations in field arithmetics is the logarithmic top-down order, solve multiplication before addition, exponents before multiplication and before all , solve the the brackets aka Dyck pair(s).
Square root times pi hints first towards some fundamental rotation at the top. A simple bit rotation between Dyck pair and inverse Dyck pair >< goes both L and R directions, and we can view both and >< as different perspectives of the same rotation. Not only that, and >< are also Boolean inverses (aka NOT operations) of each other. The same goes for iterations and >< etc. of any string length with same form.
As it turns out the alphabet of < and > offers sufficient marked characters to generate top down number theory in Stern-Brocot style, < and > interpreted as the numerator element 1/0 and their concatenation interpreted as the denominator element 0/1:
< >
< >
< >
< >
etc.
of which the tally of the countable elements as previously defined gives a two-sided Stern-Brocot type structure of coprime fractions in their order of magnitude. The operator language is chirally symmetric already notationally. We can do something similar with the generator > < for the nesting algorithm of concatenating mediants. Curious readers can check how that behaves by themselves when interpreted numerically in the same manner.
What about the aqrt(2) aspect? If we interprete the rows of the operator language as top down perspective of hyperlogarithms, then the hyper-roots of a two-sided SB-type structure travel along the binary tree nested in the blanks between the mediant words. The zig-zag paths along the binary trees give the theory of continued fractions.
The holistic perspective of number theory does not need to be complex and difficult, on the contrary it needs to be simple and beautiful... and elementary. A word about mereology to conclude. From the holistic perspective, the question is not either holistic or reductionistic perspective (and/or ontology). The holistic perspective is primitive (more than the reductionistic perspective), but it can be truly holistic only by incorporating also the reductionistic perspective of a part. With top down and bottom up perspectives of the hyperoperation structure coherently and defined in elementary computable language, lot of complexity and open questions arise in their middle zone where they meet and mingle, perhaps and probably also solutions to some pesky old conjectures.
PS: note also that the continued fraction of e has simple dyadic periodic structure. With close resemblance with the Wallis product. Looking at the both sides of the zig zag paths of e in the tree nested in two-sided SB-structure gives a better view than the one-sided "canonical" limitation of the perspective.
fun idea, next time you do a video like this. Start the video with the image of a giant bold quotation mark, then at the end do the same. So that way the whole video is in heavy quotation marks lol.
I thought this video was going to be about infinite sets and cardinal numbers
So if I reshape a unit circle into a square, the side length of this square is infinity factorial. That's wild!
When would anyone doing work for which someone would be willing pay, use the information in this video?
I'd like to understand this in the context of the hyperreals of non-standard analysis.
In the sense that str(1+2+3+…)=-1/12 can we also say that str(1×2×3×…)=sqrt(2pi)?
16:19 minus sign error?
Well spotted. It also explains the apparent mistake when deriving the -sqrt(2pi) result.
I know this entire construction is very hand-waivy (at least in this video), but why couldn't we use the formula at 10:46 and immediately plug in s = 0? Then we would get that Zeta(0) = 0.
A hypothesis, if you were to pull the s out of the integral, plugging in s = 0 would result in 0*infinity?
It's 1/12^2
😊
Separate but related question, what's the cardinality of the infinity from multiply all the (positive) natural numbers? It's a countably infinite product of finite objects so I'd be satisfied to hear it's still countable but some part of me can't help but wonder if the infinite product is somehow analagous to a power-set (not really sure why that's what my intuition is suggesting to me, it's probably complete nonsense) which would then yield the power-set of of the natural numbers which is the Reals.
A number doesn't have cardinality -- a set does. So you're really asking about the cardinality of a set with inf! elements. Creating such a set is easy: {1, 2!, 3!, 4!, ... }. The bijection is with N is clear: 1->1!, 2->2!, ... so the cardinality of the set would be aleph_0.
A quick way to prove it:
Let F_M={a: M -> M | all k: a_k N | all k: a_k
@@ArtemKreimer Thanks for clarifying my language re cardinality and sets, I couldn't think of a non-stupid way to say "size of infinity". I don't completely follow how the set {1!, 2!, 3!,...} has inf! elements though, surely it's just the set {1, 2, 6, 24, 120,...} which clearly bijects with with the natural numbers as you described but I don't see how it corresponds with inf!.
@@ArtemKreimer The set {f(k) | k in S} doesn't have f(|S|) elements, it has |S| elements. (assuming f is an injective function). By your logic, a set with 2^|N| elements would be {2^k | k in N}, so 2^|N| = |N|. But that's a boring answer, and goes against the 2^|N|=|P(N)|=|R| that most people would rather go with.
From Stirling's approximation, n! ~ (n/e)^n > 2^n so it's really something like aleph 1 (or rather 2^omega with some corrections).
If the sum of all naturals exists then it is equal to -1/12
If the product of all naturals exists then it is equal to sqrt(2.pi)
Everything is in the "if"
Proposal: rename "heavy quotes" to "bold quotes" for extra confusion!
And what's the factorial of aleph one?
I imagine that any definition you define to give it would depend on whether you take the continuum hypothesis as an axiom but I might be wrong. It also might not have a good regularisation. I'm also curious though.
Mu
Aleph 2. Aleph 2 is the cardinality of all well ordered sets.
@@kazedcat Eh?
Is not the axiom of choice equivalent to the statement that all sets can be well ordered?
@@drdca8263 The axiom of choice says that there is always a selection formula for any element of an infinite set. Well ordered just means that the set has order from least to greatest and that the least element exists.
So for example the set of all integers is not well ordered because even though it has order there is no least element. But you can always pick a number in the set of all integers.
6:11
Please would you check how to correctly write zeta ?
Your version resembles more like "f".
Capital Zeta is Z in Greek. So the llwer-case resembles a Z.
Your upper scroll should be the right-edge of a (~)horizontal stroke ; your vertical descent should head to the left, and then only turn back horizontally to the right, before a downtail.
⁻/_,
more evidence that tau is more fundamental to math than pi
Easy to prove that every digit of the answer written in any base is zero, so shouldn't the answer be zero?
I like this, but I feel like this isn't a useful definition because it has no real meaning? Not sure if I'm missing it, but I dont think there would be much calculations that could use this. You could show that any product of finite consecutive numbers is (other than like n=2) is greater than this value so it as a definition doesn't seem to convey any actual usefulness. I'm just finishing my undergrad so don't mind me, I am curious to be told way this is not correct.
There are tons of applications in quantum physics where applying this kind of formulas yields meaningful results. It's also math for fun.
I think there is some math that needs to be discovered/formalized. Some ceturies from now we'll say 1+2+3+...=infinity-1/12 and 1x2x3x...=exp(infinity)+sqrt(2pi) or something similar. For now, we can't work with infinity very well.
16:15 The product Rule at the end is aplied wrongly, there should be a plus instead of a minus, that is not corrected
The product of all natural numbers? Well, zero is a natural number, so of course the product is zero.
Natural numbers start at 1.
@@aarongoldsmith9967 “strictly positive integers” is a silly datatype. Why would you want a semigroup instead of a proper monoid?
@@drdca8263we want ∞! Factorial starts from 1.
@@drdca8263 It doesn't matter what you think or want. When people say "natural numbers" they mean the numbers 1,2,3,4,... so this is all silly babble in the first place!
@@aarongoldsmith9967 That’s not true in general? Plenty of people mean the set {0,1,2,…} .
20:10 I'm not entirely sure if Zeta'(0) is truly equal to -ln(sqrt(2pi)) bc ln2 - ln(pi/2)/2 should be ln( 2 * sqrt2 / sqrt(pi) ) instead.
Obviously, 0 is a natural number
No, it is *not* universally accepted that way.
Then the product of all natural numbers is "zero times infinity"...
@@PawelS_77 in a fight between an exact 0 vs a limit to infinity, the exact 0 wins.
And in a fight (product) between an exact zero, 0/1, and an exact infinity, 1/0, it is a draw because: 0/1 * 1/0 = 0/0 = nullity.
So after 21 minutes of breaking rules we have calculated something that has no meaning? In any case I liked the video
And I wish you would do more videos on Zeta Riemann Function
some meaning: inf! is a carry from factorial number system.
But this number system has no carry, it's closed, "frozen"
Indeed, (n+1)digit(dimension) has (n+1)-arity.
Nothing happens when moving from one to another digit.
(IMG/KER=dimension/curvature=quality/quantity=1=constant).
Carry doesn't go out of that number system and circulated inside.
And it is reciprocal to radius of that circle.
inf!=sgrt(2*pi)=2*pi*1/sgrt(2*pi).
Bad math
If you take the definition of the natural numbers that includes 0, the answer is easy: every product with a factor 0 is 0.
First
And so I grant you 10 internet points
@@sewerynkaminski1116 Can you loan them to me?
It should be considered a new type of number, with that coefficient. Like i*π is not π.
Imagine if the coefficients of imaginary numbers were confused with real numbers. That's what happening here.
The sum is not −⅟₁₂.
Is −⅟₁₂K, where K is a new kind of unit.
It's a lot weirder it's CN²-1/12+Kε. So-1/12 is actually the real number component and the other components are just ∞ and 1/∞ . If you want to make a new number it will be like -1/12 +C∞ +Kε
This is math murder happening in front of our eyes