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If we define ln0 as -infinity, it actually makes sense: 0^i=e^(ln0)i=e^(-infi). So it's basically the result of infinitely moving clockwise around the circle |z|=1. It's undefined in a similar way like sine of -infinity

@antoniusnies-komponistpian2172

9 ай бұрын

Or the last digit of a periodic decimal number like 1/7 or 1/11

@TechnoCoderz369

9 ай бұрын

Yeah you're right bro/sis

@lyrimetacurl0

9 ай бұрын

Somewhere along the circle x^2 + y^2 = 1

@TechnoCoderz369

9 ай бұрын

@@lyrimetacurl0 no somewhere along i^x where x is a real number

@gregstunts347

9 ай бұрын

@@lyrimetacurl0 Yes, if the x-axis is the real axis and the y-axis is the imaginary axis.

I love how u combine these beautiful numbers and try to see how they interact. (Note : your edit had become better , but stop please rotating numbers.)

@Qermaq

9 ай бұрын

It's wheel algebra. It makes me dizzy too.

@adityaalghifari3141

9 ай бұрын

Math is my life you know

@arslenedhahri6465

9 ай бұрын

Not ur life only . IT'S ALL THE UNIVERSE !@@adityaalghifari3141

If a real number is an equivalence class of series, you could define a new set of equivalences as two sequences being equivalent if they have the same set of accumulation points. In your case, 0^i would then be the set of numbers on the complex unit circle, because every number on that circle will be an accumulation point as x->0 in x^i. I think it would be interesting if more of these numbers existed that really are sets of numbers

Mathematician: 0^i, what is your phase? 0^i: yes

0^i must be either 0, undefined, or a defined nonzero number. If 0^i = 0, then (0^i)^i = 0^i = 0. But (0^i)^i = 0^(i*i) = 0^-1, which does not equal 0 (and is, in fact, undefined). Thus 0^i does not equal 0. One corollary of Euler's formula is that for every defined, nonzero value of x, x^i = cos(ln(x)) + isin(ln(x)), which is defined and nonzero. If x = 0^i, then x^i = (0^i)^i = 0^-1. And 0^-1 is undefined; thus 0^i cannot be a defined nonzero number. By process of elimination, 0^i is undefined.

@cannot-handle-handles

9 ай бұрын

Very nicely done!

@HiddenPowerIce

9 ай бұрын

You can't use power rules.

@paulchapman8023

9 ай бұрын

@@HiddenPowerIce why not? If (0^i)^i does not equal 0^-1, what does it equal?

@HiddenPowerIce

9 ай бұрын

@@paulchapman8023 Pretty sure in complex numbers z^w = e^wln(z). I agree the answer is undefined thou.

@paulchapman8023

9 ай бұрын

@@HiddenPowerIce when z is a defined nonzero number yes. But is 0 equal to e^ln(0)?

0^0 is "Traditionally undefind" - prove this. In the 1980s everyone knew it was 1. It is only in recent years there is some movement to claim it is undefined.

3:32 - the modulus of re^iθ is only equal to r if θ is a real number. Thus |e^i*ln(x)| is only 1 if ln(x) is real. But ln(0) is not real, as it isn't even defined, so you can't determine the modulus of 0^i, and there's therefore no evidence to suggest that it would lie on the unit circle at all, sadly. You could also consider the limiting case of b^i as b approaches 0, which from the positive direction remains on the unit circle, despite not converging to a any point. However, if you approach 0 from any other direction, b^i leaves the unit circle and diverges in all directions, hence it is impossible to know anything about 0^i.

@therealshavenyak

8 ай бұрын

Thank you. The “its modulus is 1” line of reasoning felt bogus but I couldn’t articulate a rebuttal.

@allozovsky

2 ай бұрын

Indeed, if 0 = 0·eⁱᵠ, then ln(0) = ln(0) + 𝒊φ "=" −∞ + 𝒊φ, so 0ⁱ = exp(𝒊·ln(0)) "=" exp(𝒊·(−∞ + 𝒊φ)) = exp(−φ − 𝒊·∞) = e⁻ᵠ·(cos (−∞) + 𝒊·sin(−∞)), that is "undetermined times some direction" ¯⁠\⁠_⁠(⁠ツ⁠)⁠_⁠/⁠¯

At this point change your name to BrilliantTheMathGuy

Me who understand nothing: *Cool!*

@-petrichor-7263

9 ай бұрын

Hey I am one of the only ones to get a heart by Bri, nice!

If 0^i exists, let it be x. Now, raise both sides to the i-th power, giving 0^(-1) = 0^(i*i) = x^i. Notice that the left hand side is 1/0 which does not exists, but x^i does exist for every number(including x=0 by assumption), so we get to a contradiction.

4:10 Forget Imaginary numbers! You've invented quantum numbers.

@cheeseburgermonkey7104

9 ай бұрын

Schrodinger's complex number Rolls off the tongue

@artcurious807

9 ай бұрын

0i does have a quantum look to it. maybe by combining different equations using 0i into one equation an 0i wave function will emerge

@AlbertTheGamer-gk7sn

9 ай бұрын

And virtual numbers with negative absolute values.

@40watt53

9 ай бұрын

@@AlbertTheGamer-gk7sn I'm terrified... please tell me these exist.

@AlbertTheGamer-gk7sn

9 ай бұрын

@@40watt53 They exist, and also, fake numbers with negative squares and absolute values exist. Ever heard of quaternions?

That’s impressive, I guess I should erase that question off my math list since it seems to have a whole answer lol 😊

Where you’ve been sir ? We’re missing you❤️

There is an error, for example the absolute value of i^i is not 1, because when You have not real "angles", the value change

@joeyhardin5903

9 ай бұрын

Yes. And if the angle is ln(0) (which isn't real because it's undefined,) then you can't say that 0^i would lie on the unit circle at all. The modulus is undefined, just like the argument.

02:40 - 04:15 You are thinking as 0 aproximated by real numbers.

Isn't there an error here? From what I'm seeing, there is no two-sided limit for y=|x^i| as x->0, because from the left, it's e^(-pi), and from the right it's 1. Unless Wolfram Alpha is giving me some kind of bizarre multiple branches thing?

With numbers, everything is possible 😄

If you plotted every value of 0^z on a four-dimensional Cartesian plot, it would look like a tiny flowerpot (depending on the angle).

What if we used the non-principal value / branches?

0^i=e^(iln(0))=cos(ln0)+isin(ln0) but ln0+ is -♾️ so it would be a point infinitely clockwise on the unit circle.

@divergentmaths

9 ай бұрын

ln 0 = -∞ in classic maths, but in divergent maths ln 0 = -γ where γ is the Euler-Mascheroni constant

@maxvangulik1988

9 ай бұрын

@@divergentmaths how?

@AlbertTheGamer-gk7sn

9 ай бұрын

Well, the cosine and sine of infinity are both indeterminate as sine and cosine are oscillating functions, so we don't know what it would be in infinity.

@divergentmaths

9 ай бұрын

@@AlbertTheGamer-gk7sn The partial sums of Grandi series (the alternate divergent series of 1 - 1 + 1 - 1 + 1 - 1 + ⋯ ) also oscillate between 1 and 0. However, its sum is 1/2 (not indeterminate). Likewise, cos(+∞) = 1 and sin(+∞) = 0.

@AlbertTheGamer-gk7sn

9 ай бұрын

@@divergentmaths Well, take the limit as x goes to infinity of sin(x), cos(x), and (-1)^x. You will see that all 3 limits diverge.

0ⁱ=e⁻ⁱᵞ where γ is the Euler-Mascheroni constant

@divergentmaths

9 ай бұрын

Note that ln 0 = -∞ in classic maths, but in divergent maths ln 0 = -γ where γ is the Euler-Mascheroni constant

@divergentmaths

6 ай бұрын

One could also write: 0ⁱ = cos(γ) - i sin(γ)

@divergentmaths

6 ай бұрын

The proof that ln 0 = -γ is easy. Evaluate the MacLaurin series of ln(1-x) at x = 1. You get ln(0) = - sum from n=1 to +inf of 1/n = -ζ(1) = -γ Therefore γ = -ln(0)

well I think you could make a convincing argument to most people (not mathematicians ofc) that ln(0) = -infinity, by simply showing them a graph of ln(x) on desmos. from that the problem just becomes what sin(-infinity) and cos(-infinity) are, they could be any angle from 0-360 degrees.

4:00 - 4:14 So basically you're saying 0^i is some point on the unit circle in the complex plane, but there is no specific point, Which I therefore interpret it as "undefined" Not to mention that you have no reason to assume 0^i lies on a unit circle in the first place

@barakeel

3 ай бұрын

2-5 is undefined because negative numbers do not exists

What about i^i?

I thing that quantum mechanics has solution in such crazy mathematical objects

I like your funny words magic man

3:25 Well, if 0 = 0·eⁱᵠ, then ln(0) = ln(0) + 𝒊φ "=" −∞ + 𝒊φ, so 0ⁱ = exp(𝒊·ln(0)) "=" exp(𝒊·(−∞ + 𝒊φ)) = exp(−φ − 𝒊·∞) = e⁻ᵠ·(cos (−∞) + 𝒊·sin(−∞)), that is "undetermined modulus times some direction" ¯⁠\⁠_⁠(⁠ツ⁠)⁠_⁠/⁠¯

Will this affect the economy?

It makes me think... maybe this is new number in new direction? Like i, is "higher" than rest of the numbers 0^i could be "closer to us" (i mean, it could be 3rd direction numberling). What do you think?

@ultrio325

9 ай бұрын

Unfortunately this new numbering system can't have multiplication. this is a consequence of Hurwitz's theorem, and is what Hamilton struggled with until he tried 4D, giving the quaternions.

@antoniusnies-komponistpian2172

9 ай бұрын

No, it's not a new number because it's just a limit of a periodic function. It's rather a set of regular numbers than a new number

So it's more like a waveform?

Sounds like electron clouds could be modelled using such math

0^i just acts like an electron

A bit convoluted but I will take it.

Ambiguity has many modalities. This becomes a transform optive much like modeling by LaPlace Transform ... mathematically. The trick is trying to yranslate it into physical form. You then get bubble gum.

If z approaches 0 along a line at an angle phi, then: 0^i = lim(t->0+) (te^(i phi))^i = lim(t->0+) (e^(ln t) * e^(i phi))^i = lim(t->0+) (e^(ln t + i phi))^i = lim(t->0+) e^i(ln t + i phi) = lim(t->0+) e^(i ln t - phi) = e^(-phi) lim(t->0+) e^(i ln t) = undefined But in some sense 0^i would be all the values in the circle of radius e^(-phi) But phi ranges from 0 to 2pi, so in another sense, 0^i is all the values in annulus e^(-2pi) < |z|

What about 1 or 0 to the i root?

@user-ht9ek6pe4g

2 ай бұрын

0 root i is same as 0^1÷i and 1÷i i is -i

As x approaches 0, the angle of x^i rotates faster and faster, making 0^x meaningless. But for any real x, its modulus is 1.

|e^ix|=1 only works when x is real. ln(0) being undefined means that it'd almost invariably have complex possibilities.

I really like to hear Bri TMG using the word "phase"; I think it is much better than the word "argument". All mathematicians should update their vocabulary.

its parallel to both axis

Theoretically ln(0)= negative infinity

@AlbertTheGamer-gk7sn

9 ай бұрын

Well, the cosine and sine of infinity are both indeterminate as sine and cosine are oscillating functions, so we don't know what it would be in infinity.

0^i cannot be -i, because there’s another number that satisfies this: 0.207879. It also can’t be i, because 4.81 satisfies this. 0^0 is 1, so 0^i can’t be 1. 0^i must be -1

That was great, and 0^0 = 1. :)

Ln(0)=-∞ Sin(-∞)=0 Cos(-∞)=1

0^i=0 Justification: 0^i=(0x)^i=0^i*x^i Let 0^i=z. If z is a nonzero complex number: z=zx^i x^i=1 for all x Clearly false. If z=0: 0=0x^i True. While not rigorous, this should be good justification for 0^i=0.

I got to thinking about the basic property of "i". Specifically, is the numeric value of i positive or negative? It seems to me this question cannot be answered. WAAAY back, the question "What is the square root of -1?" couldn't be answered, and this resulted in the complex number system. Maybe, then, an attempt to define the sign of i could result in a new, complementary, number system???

@cannot-handle-handles

9 ай бұрын

One convention for the sign of a non-zero complex number z is z/|z|, so that would mean the sign of i is i.

@jamesharmon4994

9 ай бұрын

@cannot-handle-handles I get what you're saying, and you are correct. It seems I'm not explaining myself very well. By convention, numeric constants are assumed to be positive unless preceeded by "-". But the constant "i" cannot be positive since any positive value when squared is positive. Any negative value when squared is also positive. This leaves the sign of "i" in limbo. Maybe there could be a branch of mathematics about "signless" numbers?? Don't ask me "why?" Who could have imagined the use complex numbers have before "i" was invented?

If you rotate 0 by 90° it still would be 0

It is impossible to find.

Can't we define a domain where the solution can be for every controversial thing?

0^i by desmos is 0

lol he's literally meming now!

exp(-pi^2) how about that

x^0 is not undefined x^0=0/0 which is indeterminate edit: x=0

@I_like_danno_cal_drawings

9 ай бұрын

also if it is undefined it cannot use limits

@AA-100

8 ай бұрын

x^0 is not undefined, x^0 = 1 unless x = 0

@I_like_danno_cal_drawings

8 ай бұрын

x^0=x/x and i meant x=0 for that, i will edit it@@AA-100

Nothing, Exists!

z^w = e^(w(ln|z| + iarg(z))) hence 0^i = e^(i(ln|0| + iarg(0))) lim x→0 ln|x| = -∞ simplified: 0^i = e^(i(-∞)) lim x→-∞ e^(ix) = Ø

79th

Eh. Mathematicians should really stop obsessing over single-valued functions in their ideas of limits. Set-valued functions make perfect sense and are a perfectly valid solution. Lim(sin(1/x);x->0)= [-1,1]. And here: lim(x^i;x->0)=complex unit circle.

@xinpingdonohoe3978

2 ай бұрын

They should stop obsessing over single-valued ideals of multivalued functions in general.

@ClementinesmWTF

2 ай бұрын

@@xinpingdonohoe3978 I absolutely agree. It’s way too limiting (ha?) to try and force that kind of exactness on many things and it opens up a lot of new paths to explore when we stop obsessing over it and accept things can be another way.

0^i is undefined????

*WHY ARE THERE LETTERS IN MATH*

@revtheobbyist2222

2 ай бұрын

kid

Reposting and slight editing of recent mathematical ideas into one post: Split-complex numbers relate to the diagonality (like how it's expressed on Anakin's lightsaber) of ring/cylindrical singularities and to why the 6 corner/cusp singularities in dark matter must alternate. The so-called triplex numbers deal with how energy is transferred between particles and bodies and how an increase in energy also increases the apparent mass. Dual numbers relate to Euler's Identity, where the thin mass is cancelling most of the attractive and repulsive forces. The imaginary number is mass in stable particles of any conformation. In Big Bounce physics, dual numbers relate to how the attractive and repulsive forces work together to turn the matter that we normally think of into dark matter. The natural logarithm of the imaginary number is pi divided by 2 radians times i. This means that, at whatever point of stable matter other than at a singularity, the attractive or repulsive force being emitted is perpendicular to the "plane" of mass. In Big Bounce physics, this corresponds to how particles "crystalize" into stacks where a central particle is greatly pressured to break/degenerate by another particle that is in front, another behind, another to the left, another to the right, another on top, and another below. Dark matter is formed quickly afterwards. Mediants are important to understanding the Big Crunch side of a Big Bounce event. Matter has locked up, with particles surrounding and pressuring each other. The matter gets broken up into fractions of what it was and then gets added together to form the dark matter known from our Inflationary Epoch. Sectrices are inversely related, as they deal with all stable conformations of matter being broken up, not added like the implosive "shrapnel" of mediants. Ford circles relate to mediants. Tangential circles, tethered to a line. Sectrices: the families of curves deal with black holes. (The Fibonacci spiral deals with how dark matter is degenerated/broken up and with supernovae. The Golden spiral deals with how the normal matter, that we usually think of, degenerates, forming black holes.) The Archimedean spiral deals with dark matter spinning too fast and breaking into primordial black holes, smaller dark matter, and regular matter. The Dinostratus quadratrix deals with the laminar flow of dark matter being broken up by lingering black holes. Delanges sectrices (family of curves): dark matter has its "bubbles" force a rapid flaking off - the main driving force of the Big Bang. Ceva sectrices (family of curves): spun up dark matter breaks into primordial black holes and smaller, galactic-sized dark matter and other, typically thought of matter. Maclaurin sectrices (family of curves): older, lingering black holes, late to the party, impact and break up dark matter into galaxies. Dark matter, on the stellar scale, are broken up by supernovae. Our solar system was seeded with the heavier elements from a supernova. I'm happily surprised to figure out sectrices. Trisectrices are another thing. More complex and I don't know if I have all the curves available to use in analyzing them. But, I can see Fibonacci and Golden spirals relating to the trisectrices. The Clausen function of order 2: dark matter flakes off, impacting the Big Bang mass directly and shocking the opposite side, somewhat like concussions happen. While a spin on that central mass is exerted, all the spins from all the flaking dark matter largely cancel out. I suspect that primordial black holes are formed by this, as well. Those black holes and older black holes, that came late to the Big Bounce, work together to break up dark matter. Belows method (similar to Sylvester's Link Fan) relates to dark matter flaking off during a Big Bang event. Repetitious bisection relates to dark matter spinning so violently that it breaks, leaving smaller dark matter, primordial black holes, and other matter. Neusis construction relates to how dark matter is broken up near one of its singularities by an older black hole and to how black holes have their singularites sheared off during a Big Crunch. General relativity: 8 shapes, as dictated by the equation? 4 general shapes, but with a variation of membranous or a filament? Dark matter mostly flat, with its 6 alternating corner/cusp edge singularities. Neutrons like if a balloon had two ends, for blowing it up. Protons with aligned singularities, and electrons with just a lone cylindrical singularity? Prime numbers in polar coordinates: note the missing arms and the missing radials. Matter spiraling in, degenerating? Matter radiating out - the laminar flow of dark matter in an Inflationary Epoch? Connection to Big Bounce theory? "Operation -- Annihilate!", from the first season of the original Star Trek: was that all about dark matter and the cosmic microwave background radiation? Anakin Skywalker connection?

@TheSmegPod

9 ай бұрын

what

@johndoyle2347

9 ай бұрын

What what.@@TheSmegPod

Wow!!! I wouldn't be able to come up with those ideas for my math channel!!!😅😅😅 I love this channel!!! I want to make high-quality content as this channel!!!🤩🤩🤩

I guess 0..?

Because 0^i = 0^(0+1i), this is undefined, since 0^0 is undefined. If I would have to choose a defined answer, I would choose = 0, because zero to any power = zero.

@SsvbxxYT

9 ай бұрын

0+1i (which simplifies to i) does not equal 0. If it did, then 0^i would be 1 since 0^0 is commonly defined as 1 (though it too is technically undefined).

@ultimatedude5686

9 ай бұрын

0^0 is not necessarily undefined, and even if you think it is this is a bad argument. is 0^1 undefined because 0^1 = 0^(0 + 1)? You cannot say that zero to any power is zero because that is not true for negative numbers. 0^-1 is 1/0, which is definitely not equal to zero. If this does not work for negative numbers I see no reason to think it would work for imaginary ones.

@129140163

9 ай бұрын

It is my understanding that 0^0 has two possible values: “Undefined” and 1. If 0^0 is undefined, then 0 to any negative power is also undefined. If you make 0^0 equal to 1, then 0 to any negative power becomes ♾️.

@AlbertTheGamer-gk7sn

9 ай бұрын

@@129140163 Well, 0 to any negative power really is infinity. Therefore, 0^0 is equal to 1.

@AlbertTheGamer-gk7sn

9 ай бұрын

Well, it is undefined as the sine and cosine diverge as they go to infinity, but since it is an oscillating function, it doesn't diverge to any point, so it is indeterminate.

Isn't it undefined ? Like 0^i = 0^(i*i/i ) = 0^(-1/i) = 1/(0)^i Therefore undefined..

@singh.ayushman

9 ай бұрын

0¹=0^(2-1)=0²/0¹ therefore undefined.

@AlbertTheGamer-gk7sn

9 ай бұрын

Undefined as sine and cosine diverge as they approach infinity, but since they oscillate, they don't diverge to a specific point.

But can you prove that: 0^i ≠ 1^i

1+(-1)=0 -1=e^((2k+1)iп); 1=e^(2kпi) 0^i=(e^((2k+1)iп)+e^(2kпi))^i 😁

Hoiiiii :)

Bro lost me about 30 seconds in 💀

☝️🤓 Um.. Actually... According to Worfamalpha,0^i is undefined...

0^i = e^(i*ln0) = cos(ln0)+isin(ln0) = [cos(1)+isin(1)]^ln0 => [e^(1*i)]^ln0 = [cos(1)+isin(1)]^ln0 => e^i = cos(1)+isin(1) ?? ln0=-infinity ?Yeah raise that to minus infinity makes sense to try and solve that??xD? Anyway, (e^i)^(-infinity) = (cos(1)+isin(1))^(-infinity) => { 0^i =[-1,1] - i*[-1,1] }!!!...its undefined or you can say what i say its an infinite horizontal strip from -infinity to +infinity with answers of amplitude from -1 to 1 xD thats why its undefined ...pick and choose xD your poison xD thats why all type of equations get annihilated when raised to the power of infinity and all answers can be produced from infinity xD hence undefined!!! INFINITE POWER!!!!!!MUAHAHAHA

@silouy4259

9 ай бұрын

My nose is bleeding what did i just red

@TruthOfZ0

9 ай бұрын

@@silouy4259 I said it was poisonous xD

@cheeseburgermonkey7104

9 ай бұрын

Thanks for the major stroke

@TruthOfZ0

9 ай бұрын

@@cheeseburgermonkey7104 Im a man who can slap but can also stroke...i said it was poisonous xD na na na nah

@cheeseburgermonkey7104

9 ай бұрын

@@TruthOfZ0Do you know anything about math above middle school, or are you just copying what the video said?

1k liker

What if it is not a point but the whole circle?

First and pin me

@lockmancoal8508

9 ай бұрын

No

@someone-wv3ds

9 ай бұрын

No

@anncherian

9 ай бұрын

Ok

@Ros2fi0

9 ай бұрын

Dumb

I liked my own comment👇

Complex numbers are fake invented math because (1) the definition of a complex number contradicts to the laws of formal logic, because this definition is the union of two contradictory concepts: the concept of a real number and the concept of a non-real (imaginary) number-an image. The concepts of a real number and a non-real (imaginary) number are in logical relation of contradiction: the essential feature of one concept completely negates the essential feature of another concept. These concepts have no common feature (i.e. these concepts have nothing in common with each other), therefore one cannot compare these concepts with each other. Consequently, the concepts of a real number and a non-real (imaginary) number cannot be united and contained in the definition of a complex number. The concept of a complex number is a gross formal-logical error; (2) the real part of a complex number is the result of a measurement. But the non-real (imaginary) part of a complex number is not the result of a measurement. The non-real (imaginary) part is a meaningless symbol, because the mathematical (quantitative) operation of multiplication of a real number by a meaningless symbol is a meaningless operation. This means that the theory of complex number is not a correct method of calculation. Consequently, mathematical (quantitative) operations on meaningless symbols are a gross formal-logical error; (3) a complex number cannot be represented (interpreted) in the Cartesian geometric coordinate system, because the Cartesian coordinate system is a system of two identical scales (rulers). The standard geometric representation (interpretation) of a complex number leads to the logical contradictions if the scales (rulers) are not identical. This means that the scale of non-real (imaginary) numbers cannot exist in the Cartesian geometric coordinate system.

@endercool8139

Ай бұрын

"Fake invented math" like math in general isnt invented by us humans for a better understanding of the world ?wdym fake ,by your logic all math is fake ,why should the complex numbers only be fake?complex numbers are nor real nor fake they re just another helpful tool we have to better understand the world .but its fine you can have your wrong opinion

If you really think 0^0 is always 1 no matter what, then why don't you publish a research paper? Send it to peers and it will be proved once for all. And if you can't do it then please don't claim that 0^0=1.

@Bodyknock

9 ай бұрын

He didn’t say he thinks 0^0 “is always 1 no matter what”. He knows it’s undefined and why its undefined. Instead he’s saying that IF we wanted to literally give 0^0 a value by definition then he thinks that the value which would make the most sense to define it as would be 0. He didn’t give the reasons in this video why he thinks that 0 would be the best value to pick if you hypothetically wanted to define a value for 0^0 but I recall he had another video where he does. This isn’t much different than extended the definitions of other functions which originally weren’t defined beyond a certain domain. The square root of -1 originally wasn’t defined, for instance, but i was invented specifically so that it would have a value by definition and then from there Complex numbers followed. Factorials originally were only defined for the positive integers, but by definition were extended through the gamma function to non-integers. Divergent sums like 1+2+3+4+… have no defined value normally, but they can be given defined values, like -1/12 for that particular sum, which then make sense in an extended number system that follows. So nobody is claiming 0^0 is normally defined. What he’s saying is that if you do choose to give it a value of 1 by definition then the resulting extended system you get is worth pursuing.

@Ninja20704

9 ай бұрын

He never claimed that it was true. And for the record you cannot prove a definition. You can only justify why you choose a definition. He linked his original video on this topic on why some formulaes and results in algebra and calculus need 0^0=1 and would not work if we insisted that it was undefined. And in many areas of math, saying that 0^0 should be 1 is the only value that can keep things consistent. I highly suggest you actually watch that video before making such bold statements.

@happygood18

9 ай бұрын

@@Ninja20704 Didn't he do in his 0^0 video? And 0^0=1 is does not look like any definition. And his justification is rather personal than universal. And because of him, I saw many of the followers are telling that 0^0 is always 1 no matter what, and they are trying to prove it like there is no tomorrow. And also there are many areas of math where 0^0 would lead to a straight up disaster.

@happygood18

9 ай бұрын

@@Bodyknock Nowhere he said in that video that it's undefined. He actually defined it 1 and argued for that. Well, 1 is correct in many areas, for example binomial distribution but not all. In many other areas, 0^0 would lead to an error. Complex numbers exists just like how negative numbers exists. It never contradicts with something unlike 0^0=1 which works in some field but is an error in others. Hence, mathematicians let it go undefined for the time being. All real numbers added, equals to -1/12 is only possible we break certain rules of a series. For example, 1-1+1-1+1-1+....=? If n is odd, then the sum of this infinite series is 1, If n is even then it is 0. Now if I say it's actually 1/2. Then we could make something like 1+2+3+4+...=-1/12. But this is not correct in usual sense. But breaking some laws will make it -1/12. Anyway, whatever he did, because of him, most of his viewers are now fighting to prove that 0^0=1 no matter what, like there is no tomorrow.

@Bodyknock

9 ай бұрын

@@happygood18 0:37 “ If x = 0 then 0^x is traditionally undefined.” He literally says it in the first minute of the video. And yes, when you use a system where divergent sums like 1+2+3+4+… have a defined value then that system loses certain properties the usual Real Numbers under standard arithmetic possess. But that’s fine, it doesn’t make those alternate extended systems not worth pursuing because they do reveal interesting properties about these sums that simply leaving them undefined doesn’t. Finally if someone thinks the usual definition of 0^0 is 1 that’s on them. That’s not what is said in this video though.

0=0*e^(i2nπ) therefore 0^i=0