"What is i^i?" Mathematician: Well, the math involved is actually quite beautiful, so first we employ this powerful technique... Physicist: It's about a fifth.

@Brooke-rw8rc

5 жыл бұрын

... which is basically 3, so pi.

@1224chrisng

5 жыл бұрын

engineer: it's Theta (assuming small angles)

@LudwigvanBeethoven2

5 жыл бұрын

Its a number

@losthor1zon

5 жыл бұрын

@@LudwigvanBeethoven2 - From one who should know his fifths!

@tibi2674

5 жыл бұрын

losthor1zon wish i could give you a cookie right now.

-"Are you ready, kids?" -"0.20787..., captain." - "sqrt(-1) can't hear you." -"0.20787! Captain." -"ooooow"

@musik350

4 жыл бұрын

Not i^i!, that's another number

@smit_1449

4 жыл бұрын

@@musik350 Exclamation mark is used for exclamation here, not factorial. As in AYE, AYE !!!

@ultimategotea

4 жыл бұрын

@@musik350 still works as it is still i^i, just factorial

@jasondeng7677

4 жыл бұрын

"i^i factorial" ????

@IvanToshkov

4 жыл бұрын

@@jasondeng7677 sponge bob pants ^ 2

For the briefest moment, all the Tau fans held their breath that you had woken.

@softlysnowing3959

4 жыл бұрын

yeah i thought that too...

@NStripleseven

4 жыл бұрын

Heh. Yeah.

@michalhoransky1214

4 жыл бұрын

"We should use a different circle constant" Nothing wrong here "and it should be a half of pi" a n g e r

@brianmarco5873

4 жыл бұрын

@Michal Horanský Yep, eta η kzread.info/dash/bejne/Y6WkuMawmd3Kpqw.html

@funkdefied1

4 жыл бұрын

1:56

i^i is a crying face what are you taking about.

@lucasg.5534

4 жыл бұрын

omg

@revimfadli4666

4 жыл бұрын

I guess the face was crying...eye to the eye I know my way out

@pedronunes3063

3 жыл бұрын

i^i

@Daro-Wolfe

3 жыл бұрын

I clicked on the video for this comment

@firkinflamer8604

3 жыл бұрын

Shut up Netflix person >:(

I to the power of I? The result is pretty egotistic, I'd say

@phs125

6 жыл бұрын

Danksa Grabowski i got the reference 😉

@Longuncattr

6 жыл бұрын

Completely despooked.

@GibsonDE

6 жыл бұрын

Danksa Grabowski Gotta keep an I on it

@deeptochatterjee532

6 жыл бұрын

Danksa Grabowski You mean egotistical? I think your comment was a Parker comment

@danksagrabowski2438

6 жыл бұрын

Thanks for pointing out! English is my second language and I didn't know there is a difference between egoistic and egotistic. Good to know :)

"There are some nuances to keep an i on." 3:29

@joshuaquezada9363

3 жыл бұрын

I saw what you did there

@michielschaeverbeke1421

2 жыл бұрын

@@joshuaquezada9363 You mean that the sqrt(-1) saw what he did?

@joshuaquezada9363

2 жыл бұрын

​@@michielschaeverbeke1421 LMAO Didn't mean it that way nice one

The thing about imaginary numbers is that they actually show up in real world physics, making them a real use. They’re in electrical engineering (well, that’s just a secondary method to avoid using complex differential equations, but I’ll count it), particle physics, relativity, and probably a lot more than I know about.

@soup1649

2 жыл бұрын

the schrödinger equation in quantum physics contain an i

@alextaunton3099

2 жыл бұрын

Even in basic electrical work, computing AC phase sucks without complex numbers

@k0pstl939

Жыл бұрын

The fourier transform too

@knutritter461

8 ай бұрын

X-ray analysis of single crystals in chemistry! 😉

@rewazza

4 ай бұрын

@@soup1649 The beauty of i in the schrödinger equation is that it isn't just a shortcut to get to an answer, it is *required* to satisfy its conditions

You know you haven't studied enough for the test when you read a question and on the little helper box it's written: "Approximate i^i to 0.21"

you made a completely imaginary number real. In other words, you made dreams come true

@prdoyle

6 жыл бұрын

Watch me do the same thing! i*i = -1

@vampyricon7026

6 жыл бұрын

+

@GrzegorzusLudi

6 жыл бұрын

But real numbers doesn't exist...

@teachermichaelmaalim6103

5 жыл бұрын

Hahaha. I have plagiarized this. It explains why a dream within a dream looks real! Unfortunately, i^i is an attenuating or decaying factor. It means that as real-time passes by, the real magnitude the dream decreases exponentially!!!

@irrelevant_noob

5 жыл бұрын

Kyazar Shadala Sorry, but i^i is *NOT* a real number. It is a set of real numbers, because powers over imaginary numbers do not generally have a unique result, like we might be used from real numbers... :-B

It's a Parker fifth

@MrQuarris

6 жыл бұрын

When it's close to being a fifth but it's not quite right.

@kindlin

6 жыл бұрын

Yes, that is the joke.

@totaltotalmonkey

6 жыл бұрын

Ephraim Fung I suspect ol Quarris was making an additional joke.

@vampyricon7026

6 жыл бұрын

+

@user-wh5ti3xf4y

6 жыл бұрын

TheGin Uginator14

"We should have a value of pi that is half the normal number!" Me: Ah, yes, quarter tau

@asterixgallier8102

4 жыл бұрын

quau

@TiSapph

4 жыл бұрын

I vote to call it pi-bar. Just like the Planck constant h-bar.

@coastersplus

4 жыл бұрын

if τ = 2π, then clearly -ππ- = π/2 = τ/4

@renedekker9806

4 жыл бұрын

@@TiSapph But h-bar i h divided by 2π. So pi-bar would be 1/2 instead.

@TiSapph

4 жыл бұрын

@@renedekker9806 I thought that too, but then remembered that we can just call it a Parker Bar and all is good To add to this, we can just use the strike through instead of a proper bar, making it truly a Parker bar: π̶

i^i is a Parker Fifth. It's almost there, but it's cool anyway.

Is there going to be a party about i^i in 2078 ?

@martijnvanweele6204

6 жыл бұрын

Yes there is. At this guy's house, regardless of who lives there by then...

@JochCool

6 жыл бұрын

No, in 2079. Round your numbers properly!

@natehoffmaster6726

6 жыл бұрын

JochCool Obviously no one cared about that since 14 March 2015 was so hyped.

@algorythmh

6 жыл бұрын

Personally, I think 2015 was better: I celebrated at 9:27 anyway so using 16 would have been incorrect

@index7787

6 жыл бұрын

If we last that long

"Eye to pie". My wife thought I was watching pies getting thrown in someone's eye.

@Miju001

6 жыл бұрын

"e to the eye pie" does sound pretty disturbing.

@spencerallbritton9459

6 жыл бұрын

I'm ashamed that I laughed as hard as I did.

@simasjuknelis3043

6 жыл бұрын

Woosh

@theonlygizelli

6 жыл бұрын

nah that would be pie to eye

@landourusnewforme

6 жыл бұрын

Heavyboxes DIY Master i

Come on, StandupMaths. We made this video to go down a rabbit hole. You can’t just say “don’t worry about it.”

@dielaughing73

4 жыл бұрын

Channelling Rick Sanchez there

@KaitlynBurnellMath

3 жыл бұрын

I was a little surprised by this ending to the video, cause like...the real answer is just "all the values are valid" which...really isn't that complicated. But ending the video that way made me second guess myself and run some google searches to see if there was something about 0.208 that made it "more valid" than other answers (not anything I could find, not anymore than arccos(1) = 0 being "more valid" than arccos(1) = 2pi. People might prefer working with 0, but both are valid answers).

@acuerden

3 жыл бұрын

And, of course, by using all the value answers for ln i, (and ln -1) from that method, you extend logarithms to the complex plane. Hell, if you don't mind just keeping track of πi counts, you can start doing natural logs that can handle negative numbers, at the cost of having arbitrarily many natural logs.

@acuerden

3 жыл бұрын

Arbitrarily many for any ln, I mean. Like ln e = 1 or 1 + 2π, or 1 - 2π or 1 + 1074π

@sighthoundman

3 жыл бұрын

@@KaitlynBurnellMath Wellllll........, there is a way that the "standard" solutions are slightly "more valid" than all the others. They are the "principal values" (and in the standard interpretation, the functions are multi-valued, which is why we don't teach this to middle school students: we spend so much time teaching them that functions have to be single-valued, and to then say "not really" would just blow their minds). Of course, when you are using complex numbers for practical problems (fluid dynamics, potential theory, etc.) you should know what branch of the function you are on. The valid solution is the one that represents your situation.

"Although, you could argue I shouldn't just use two pi..." Me: Ayyy "Because it's just the angle 0." Me: AWWWWW YOU DID THAT IN PURPOSE

@alejotassile6441

3 жыл бұрын

Tau team!

"Parker identity"

@commonpepe2270

6 жыл бұрын

because he tried but it's just not as good as eulers?

@TheArezmendi

6 жыл бұрын

Common Pepe precisely.

@klobiforpresident2254

6 жыл бұрын

TheArezmendi But it's a fit, not a near miss.

@TheArezmendi

6 жыл бұрын

Klobi for President I thought it was a miss since there are an infinite amount of answers to i^i .

@Alinoe67

6 жыл бұрын

5:13 i^i = 1/5 , that's a real parker idendity

3:30 Some nuances to keep an _i_ on, you say...

@timpeters7852

6 жыл бұрын

Nick Nirus just about to comment. Well done sir

@BamaFanEdge

6 жыл бұрын

Nick Nirus i see what he did there.

@kcwidman

6 жыл бұрын

Nick Nirus how did you write in italics on a KZread comment?

@kezzyhko

6 жыл бұрын

+Kai Widman __italic__ **bold** --strike through--

@rikwisselink-bijker

6 жыл бұрын

+

Back in the olden days, the most controversial field of mathmatics, more controversial than zero, , more controversial than negative numbers , more controversial than irrational numbers , more controversial than immaginary numbers, was Stand-up Maths' perverse way of saying "on two" when the real way of saying it is "over two."

2:23 Just want to highlight an amazing job pointimg at something that's not there

@irrelevant_noob

5 жыл бұрын

What are you referring to?! He pointed PRECISELY to the "famous one" he was talking about... o.O

@asukalangleysoryu6695

5 жыл бұрын

He means that he can't SEE what he's pointing at, cause the math on the screen was added in post production

@anddero

5 жыл бұрын

That's a Parker pointer.

@user-ur2po3vp2u

4 жыл бұрын

or, you know, you later on edit these things into the position he pointed at

@qwertyTRiG

4 жыл бұрын

And, further on, carefully adjusting his face to be positioned in a clear bit of the screen.

i^i isn't really a good one fifth, but it's a Parker Square of a fifth

@MatanVngsh

6 жыл бұрын

Arnoud van der Lugt honestly, I expected more exciting calculations in this video... But I guess it turned out to be a bit of a Parker square video...

@triruns

5 жыл бұрын

Could have used a fifth of something after this video.

@saichaitanyakudapa9554

5 жыл бұрын

Can we take ln i^i?? Bcoz we don't know whether i^i is positive!!

@alexfenner738

5 жыл бұрын

@@saichaitanyakudapa9554 You absolutely can take the natural log of a negative number. Consider e^i*pi = -1; what we've done is take e and raised it to a (albeit complex) number in order to get a negative number. So, if we rearranged the equation, we get ln(-1) = i * pi. The reason you can't take the natural log of a negative number in the real numbers is because you'll always have that imaginary component. But in the complex numbers it's perfectly valid. Go into google and take the natural log of any negative (real) number. You'll see that you get the natural log of the corresponding positive number as the real component, and 3.141 as the imaginary component.

@TheRealFlenuan

5 жыл бұрын

a Parker fifth

It is also multivalued, using that i = e^{i pi/2}, e^{i 5 pi / 2}, e^{i 9 pi / 2} etc, we can write that i^i = e^{- (2n+1) pi /2), for n in Z. So really it takes on all sorts of numerical values!

Parker's identity: "i^i is about a fifth, don't worry about the infinite other results"

I fully appreciate your acrobatic algebra you used to solve this. The more I go through college the more I appreciate the things that experienced math users will do to solve things that seem difficult but aren't truly.

Matt: The screen's getting a little crowded here (4:04) Me: Don't worry! I'll give you more space! *changes into full screen mode Doh! didn't work...

@naxxtor

6 жыл бұрын

Dragon Curve Enthusiast if this video was created in an Object based media way, that totally would have worked www.bbc.co.uk/rd/sites/50335ff370b5c262af000004/assets/51b72ca4acfbab4f4d15e967/Objects3.png

@-.._.-_...-_.._-..__..._.-.-.-

6 жыл бұрын

The reason it didn't work is because full screen mode only stretches the image. It does *not* keep the existing image the same size while expanding the image dimension. Hope this helps!

@AdvosArt

6 жыл бұрын

David S. how dense can a person be?

@jansendwan1221

6 жыл бұрын

Ephraim Fung I suspect ol David was making an additional joke.

@bertiewooster4043

6 жыл бұрын

Well... I've met brick walls less dense than him...

"Now we all know we shouldn't use 2pi, we should" "Use Tau!" "Use Zero" Oh. :(

@Victor-tj7gw

5 жыл бұрын

Lmao

@Brooke-rw8rc

5 жыл бұрын

e^(i·τ) = 1 is the most beautiful specific case of Euler's formula. Especially if you leave the unsimplified result, e^(i·τ) = 1+0. It's got the natural base, the TRUE circle constant, the imaginary unit, both arithmetic identities, and the three fundamental operations.

@ijarbis187

5 жыл бұрын

HaleyHalcyon - Gaming Channel no it’s e^0 silly

@thomaskn1012

5 жыл бұрын

@HaleyHalcyon - Gaming Channel No, it's not. e^(i*tau) = 1, but e^1 = 2.71828...

@ishashka

4 жыл бұрын

Zero is a circle constant in a way

after watching that interview, now i cant but imagine how you're fitting everything over the small black background rag. kudos to you, Matt

"i to the i will only leave the world blind." -Gandhi

@Brooke-rw8rc

5 жыл бұрын

Well, around 1/5 of it.

@BradCozine

5 жыл бұрын

@@Brooke-rw8rc I'm going to put that in Wolfram Alpha to see if it says that was funny since I have no clue.

@None_NoneType

5 жыл бұрын

What did it say?

@BradCozine

5 жыл бұрын

@@None_NoneType It told me to ask the Magic 8-Ball.

"There are some nuances to keep an i on." --M.P.

I wasn't expecting that result, and Matt's explanation was fascinating.

Actually, there's an even easier way to come to this conclusion knowing that i = e^(i*pi/2) i^i = (e^(i*pi/2))^i = e^(i*(i*pi/2)) = e^(i^2*pi/2) = e^(-pi/2)

@dqrksun

3 жыл бұрын

Wow geinus

@DirkAlmighty13

3 жыл бұрын

Even more simply, from Euler's identity: e^(iπ) = -1 = i^2 [Euler's identity] e^(iπ)^i = i^2^i [both sides ^i] e^-π = i^(2i) [both sides ^(1/2)] i^i = e^(-π/2) More generally: i^i = e^[(2n+1)π/2] for all integers n

@MitosSuper

2 жыл бұрын

@@DirkAlmighty13 exactly. Not one real number, but infinitely many real number. As many as there are natural numbers. Infinite amount of answers. Mind blowing

@quickplayerhappyerthanmean4508

2 жыл бұрын

@@MitosSuper Do you know what is more amazing? i-th root of i divided by i to the power i is equal to e^π.

@thatwhichislearnt751

2 жыл бұрын

There you used that (a^b)^c = a^(bc), but this is not true, in general, for complex numbers. So, while you got the same result, the argument is incorrect.

I don't call i the imaginary unit, I call it the interesting unit.

@Anankin12

4 жыл бұрын

As opposed to what, 1 the boring unit?

@therealshavenyak

4 жыл бұрын

1 is the loneliest unit.

@neutron417

3 жыл бұрын

True

Pi? What's that? Oh, you mean half Tau.

@godseye8785

5 жыл бұрын

@HaleyHalcyon - Gaming Channel you misspelled tau

@icarokaue7334

4 жыл бұрын

@@godseye8785 you misspelled 0.

So i^i≈1/5 Got it! I'm gonna print that on a T-shirt and walk around maths and physics institutes to trigger some nerds :P

@Xnoob545

5 жыл бұрын

@@Peter_1986 eIHtT dAsH clOsInG bRaCKeT

@omnitroph1501

3 жыл бұрын

But technically it could equal an infinite number of other values.

@DragonWinter36

3 жыл бұрын

@@omnitroph1501 and that’ll just trigger the nerds more

@omnitroph1501

3 жыл бұрын

@@DragonWinter36 you've got a point.

This thing came on my exam yesterday, I couldn't do it and was finding an explanation. I watched the video, and I am satisfied. Thanks mate.

2:15 I already see it, the first equation. Raise both sides to a power of i.

@phmdaemen

3 жыл бұрын

Exactly... why did we not do that?..

@DragonWinter36

3 жыл бұрын

@@phmdaemen because the way Matt did it was way more interesting and told us *why* i^i was about a fifth

@adilmohammed6897

3 жыл бұрын

You deserve a nobel prize

Can't think of a good word for half pi, but how about quartau?

@matthewbertrand4139

6 жыл бұрын

MWSin1 How about one half pi?

@mgb360

6 жыл бұрын

That was an absolutely amazing joke

@L4Vo5

6 жыл бұрын

I'm not a fan of Tau, but I'm willing to use quartau just for the pun.

@lammy3055

6 жыл бұрын

or Hi (Half Pi)

@robertnorth5725

6 жыл бұрын

a word for half pi.....? how bout : BigSlice!?!?!?!?! baaaahahahaaha

YOU CANT JUST TELL US TO NOT WORRY ABOUT IT! WHY ARE THERE INFINITELY MANY VALUES???

@19TonsOfGold

6 жыл бұрын

Because sin(x) and cos(x) are periodic functions

@suave319

6 жыл бұрын

No I mean why are there infinitely many values but only the first one is considered. i.e. the 1/5 one

@e1123581321345589144

6 жыл бұрын

because you can go around a roundabout forever. or at least until you run out of gas...

@ryanmuller9497

6 жыл бұрын

It's the classic inverse function problem - because y=e^(iθ) takes an angle as an argument, it is not just many to one, but infinitely many to one; that is, there are infinitely many values for θ that yield the same value for the function y=e^(iθ). So, when we go to construct the inverse function ln(y)=iθ, we have to account for the fact that θ is actually equivalent to θ+2πk (where k can be any integer, so 2πk represents a whole number of full circle rotations in either the positive or negative direction). Because of this, it would technically be more correct to represent ln(i) as i(π/2+2πk), which would mean that the expression e^(i.ln(i)) is e^(i(i(π/2+2πk)), which simplifies to e^(-π/2+2πk) (because k is any integer, -k is also any integer and thus the negative sign can just be absorbed by k for simplicity). Exponential laws then allow us to express it as e^(-π/2), the principal value found by Matt, multiplied by the factor e^(2πk), which yields the principal value when k=0 and the other possible values when k is non-zero.

@19TonsOfGold

6 жыл бұрын

For the same reason why there's infinitely many angles between negative infinity degrees and positive infinity degrees, but normally you only consider angles between 0° and 360° (or -180° and 180°) - simplicity. Yes arcsin(1) = pi,3pi,5pi... but the arcsin(1) = pi solution is the most useful and most commonly seen.

"Although you could argue I shouldn't just use 2pi for that, should I?" "I mean really because it's also the angle 0" *Top Ten Anime Plot Twists* tau is unloved

Me: expecting a discussion of branch cuts and multivalued functions Video: "Don't worry about it!" ....I feel like I have lost my mathematical innocence.

@andreimaria2137

4 жыл бұрын

Waiting for it and getting that answer... I think I got mathematical blue balls

@marcvanleeuwen5986

3 жыл бұрын

The problem is that i^i is not a function but just an expression, so there is nothing to branch cut out there. The expression does not even invoke ln; that was just a "function" thrown at the expression to try to give it a value. But that attempt is just nonsense: the expression i^i is not defined, and therefore has no value.

@harmonicarchipelgo9351

3 жыл бұрын

@@marcvanleeuwen5986 Hate to break it to you, but complex exponents are defined. You can say that technically it's not a function because it is multi-valued, but the branch cuts are functions. In particular, the principal cut is the standard choice and so it is used as the primary value of the expression, which is thus the one used in the video.

@marcvanleeuwen5986

3 жыл бұрын

@@harmonicarchipelgo9351 The problem is not complex exponents but complex _bases_ (unless the exponent is integer). I didn't say i^i is not a function because of supposed multi-valuedness; it is not a function any more than 3*4^2 is, because there is no argument to apply it to. And branch cuts are not functions, but are used in definitions of functions (to make them well-defined outside the cut). If you want you can say that exp(i ln(i)), which is not the same expression as i^i, has a well defined value if the principal cut is used for defining ln, but nothing justifies saying that this is the value of i^i. See also my comment to the video itself.

@harmonicarchipelgo9351

3 жыл бұрын

@@marcvanleeuwen5986 complex bases aren't defined? What is your basis for that notion? Are you telling me that you don't think i^2.3 is defined? Or i^(-3.5)?

"It's about a fifth..." That's the Parker Square solution

@brokenwave6125

4 жыл бұрын

No...that is just approximation. A mainstay of maths since the dawn of time and something everyone does every day.

actually it can be done in a more simpler way. we have already established e^(i*pi/2) = i now raise both sides to the power i (e^(i*pi/2))^i = i^i e^-pi/2 = i^i it still kind of creeps me out that it has infinite solutions but i take closure in the fact that it has only one principal solution *sigh*

@vampyricon7026

6 жыл бұрын

THIS

@Eurley66

6 жыл бұрын

simple and elegant

@standupmaths

6 жыл бұрын

+shashank Very good point! That is quicker but I wanted to talk about squaring as an example of multiple solutions.

@edmond_ld

6 жыл бұрын

Very elegant solution indeed, it doesn't use the logarithm that is not clearly defined in the imaginary ensemble.

@JNCressey

6 жыл бұрын

==edit, oops wrong== The extra answers come from injecting the complex logarithm into the calculations. Exponentials only have one solution, as they are well defined in being just series of multiplications. ==edit: oops my previous example was wrong, I was being rushed. here's a different example== Think about how doing *A=sqrt(A^2)* introduces a second erroneous answer Say we want to calculate *A= -5+3.* *A^2 = (-5+3)^2 = (-2)^2 = 4* So *A* is the square root of four. Then *A=2* or *A=-2;* there are two square roots of 4. But the original question *A=-5+3* only has one solution.

1:50 I've been playing around with Eulers identity(algebraically not Graphically) and i have come to find that e^(2πi) is indeed equal to 1, but then that means since -e^(πi) also equals 1 then e^(2πi)= -e^(πi), And I didn't know if this was correct because no one around me is so passionate about maths as i am, but now that i have seen this video then my statement has to be true

As a psychologist I must say: I to the power of I gives a whole new personality disorder.

Hahahahah I was JUST writing my comment saying there can be numerous solutions to i^i, when you stopped the fake ending and agreed to it yourself in the video! That is why I love your videos!

@tomarchelone

6 жыл бұрын

Same thing

@standupmaths

6 жыл бұрын

+Abhiram Haritas I know the way you folks think. :]

@balletboy94

6 жыл бұрын

standupmaths how are you able to use a function like log when the domains of eulers formulae and the domains of logs arent the same? Ive seen these types of substitutions done accross my math experience but are there any proofs/axioms that show why such simplifications are allowed in combining functions of different domains.

@matthewbertrand4139

6 жыл бұрын

standupmaths I am only using brackets in smiley faces from now on.

@theMosen

6 жыл бұрын

This vid is totally ripped off of blackpenredpen, who did i^i a few weeks ago, including the joke about ending the video before mentioning the multiple solutions.

5:40 did you just make fun of Tau?

@icedo1013

5 жыл бұрын

Sinom sounded to me like he was doing the opposite and making a case for it.

@sinom

5 жыл бұрын

Ian Vansickle Tau is 2*π and now he says "let's introduce a number that's half of π"

@irrelevant_noob

5 жыл бұрын

Sinom * 5:39 ... Always hated timestamps that are actually AFTER the point being highlighted. -.-

@sinom

5 жыл бұрын

Irrelevant Noob wow I actually thought I edited that (I also hate it) I wanted to change it from 45 to 40 but changed it to 50 by accident.

@JHashcroft

5 жыл бұрын

Having the circle constant be a half turn or a quarter turn in the Argand plane is equally silly. Tau is the one true choice!

I've wondered often about what i^i might be, or even if it is defined. I plugged i^i into my HP 50g calculator and lo and behold it gave the correct answer. Thanks for the video !

I didn't get what happens at 4:24 could someone explain it for me

@isavenewspapers8890

3 ай бұрын

e^(iπ/2) = i Take the natural logarithm of both sides: ln(e^(iπ/2)) = ln(i) The logarithm and exponent cancel out: iπ/2 = ln(i) Now we can use this to substitute for ln(i) in e^(i * ln(i)): e^(i * ln(i)) = e^(i * iπ/2)

Grade A Tau trolling, Matt.

@standupmaths

6 жыл бұрын

+Karl Hite τrolling

@Interfecteris

6 жыл бұрын

This is the best thing on the internets. τauriffic job with that one. The real conundrum is that Tau is twice Pi, but the symbol looks like half of pi. I also like the functions in this video because what is Pi without the e...

@robknightfilms

6 жыл бұрын

The number of legs is the number of that constant you need for a full circle. Pi has two legs, so 1 turn = 2 pi. Tau has one leg, so 1 turn = 1 tau. Debate solved.

you could say that i^i has infinitely many real values....

@jshariff786

4 жыл бұрын

I disagree. Here are the infinitely many ways to represent i (n ranges from 0 to infinity): i = exp(i*pi/2 + 2*n*pi) But when you raise this to the power of i, you get: i^i = exp(i*i*pi/2 + i*2*n*pi) i^i = exp(i^2 *pi/2)*exp(i*2*n*pi) i^i = exp(-pi/2)*1 So yeah...not so much with the infinitely-many real values...

@AntL03

4 жыл бұрын

@@jshariff786 Sorry, you miscalculated. i = exp(i*pi/2 + i*2*n*pi), you forgot the i on the 2*n*pi part. Then i^i = exp(i*ln(i)) = exp(i*i*pi/2+i*i*2*n*pi) = exp(-pi/2-2*n*pi) for any n relative .

@angelmendez-rivera351

4 жыл бұрын

jtron84 If you're not going to do algebra correctly, then you shouldn't be so condescending with how you go about disagreeing with a statement that is, by the way, supposed to be common knowledge. This is something you can literally find in Wikipedia, it's not an obscure mathematical fact.

I’m actually quite proud of myself for figuring this out for myself on the final question on a math test

This is a really helpfull Video. Thank you very much i searched for this information for my studies and i havn't found an better explanation to this topic than this video :)

People think imaginary numbers are useless. However, they are vitally important to electrical engineering. I'm an electrical engineering student. Imaginary numbers make the math for AC circuits much, much easier. Basically, alternating current and alternating voltage are sinusoids, meaning that the graph of the current or voltage is represented by a sine or cosine function. However, a circuit's differential equations are best solved with an e function. They can be solved with a non e function, but it will be unbelievably difficult. Euler's theorem gives us a way to convert a sinusoid to an e function, using complex numbers. When you express a sinusoid as a complex e function, its called a phasor. Side note: electrical engineers use the letter j to be the sqrt(-1) because i = current. e^(jx) = cosx +jsinx This, by the way, is why e^(jpi) = -1 (just plug in pi for x) I'm too tired to go into detail why this theorem works. It's because of a concept called Maclaurin Expansion. Basically you can write any function as an infinite Maclaurin series, for example: e^x = Σ (x^n)/n! , from n = 0 to n = ∞ Basically, the Maclaurin series of e^(jx) = the Maclaurin series of cosx +jsinx

@Safwan.Hossain

5 жыл бұрын

Good to know!

@drewmandan

5 жыл бұрын

And then you start doing Fourier series and suddenly you're right back in the trenches with sin and cosine.

@M4cc4n4

5 жыл бұрын

Yes, this is incredibly important for physics, with applications in circuits, damped harmonic motion, fluid dynamics etc etc

@You_Know_Me

5 жыл бұрын

Yes bro. I m also a electrical eng. student .so i can understand their importance

@ingGS

5 жыл бұрын

They are also useful in Civil Engineering, especially in Structural Dynamics, Vibrations, Soil Dynamics and Earthquakes.

What about i^i^i, or even beyond? Does it follow a pattern or what? Maybe i↑n?

@daniellambert6207

6 жыл бұрын

+

@fedewar96

6 жыл бұрын

i^(i^i) = i^[e^(-pi/2)] = e^[e^(-pi/2)ln(i)] = e^[e^(-pi/2)*(i*pi/2)] I need someone to check this, it's hard to do calculations in a YT comment.

@Zartymil

6 жыл бұрын

AlFas you can type it on google and it calculates it for you :)

@automatedminer7158

6 жыл бұрын

i ⬆️ i i ⬆️⬆️⬆️ i

@L4Vo5

6 жыл бұрын

More like i^(i^i) = i^(about a fifth) = e^(about a fifth * ln(i)) = e^(about a fifth*i*pi/2) = e^(i*about a tenth of pi) = cos(about a tenth of pi) + i*sin(about a tenth of pi) = about 0.95 + about i*0.31

Great video. Good to see this dealt with , at last. I've been fascinated by Euler's Identity for a long time. Gauss preferred to called imaginary numbers 'Lateral Numbers...quite a good idea. I'd like to see another video on (A+ib)^(A+ib) raising a complex number to the power of a complex number. which can be done using the Euler equation and rules of indices.'

So the fraction of oxygen in the atmosphere is pretty close to i^i. A beautiful way to remember that darned fraction!

i + 1 = 10 If you're using Base-45

@pjgcommunity3557

3 жыл бұрын

I think you mean base-19

@masonhunter2748

3 жыл бұрын

?

@fahrenheit2101

3 жыл бұрын

@@pjgcommunity3557 Not even that. There isn't a base where i + 1 =10 9 is always represented as the digit 9, or a combination of digits such as 1001 in binary I in base 19 is 18 Idk what base 45 would be but if you follow the normal pattern i is still 18.

@dingus42

3 жыл бұрын

​@@fahrenheit2101 In base 45, "i" is the decimal value 44 0 to 9 -> 0 to 9, 10 to 35 -> A to Z, and 36 to 44 -> a to i

@fahrenheit2101

3 жыл бұрын

@@dingus42 Oh. I didn't know it cycled to lowercase.

4:02 how did you know where to stand? Where is the screen visualized immediately??

@stefanlclark

3 жыл бұрын

my guess is similar to this kzread.info/dash/bejne/go2OrJWco8XFoso.html ?

simply amazing! great video

Omg 😮 I was so scared you wouldn't talk about the other values😭

Blackpenredpen did a great video on this.

@samb443

6 жыл бұрын

blackpenredpen always has great videos

@gerstensaft2936

6 жыл бұрын

Maybe he saw his video and want to share this with his community. Who knows. :D

@hdwe1756

6 жыл бұрын

Sen Zen too

@standupmaths

6 жыл бұрын

+PompeyDB Indeed! I've added it to my video description. I only saw it after I'd uploaded my video (wasn't there when I checked KZread for "i^i" a few weeks back).

@mattiasselin4955

6 жыл бұрын

standupmaths i had seen both their videos and was actually a bit disappointed that you didn't bring up the mistake they made. An expression like i^i cannot have multiple values (unlike an equation that can have multiple solutions). The problem lies in the definition of a function. It takes one value in the domain and spits out a result. So ln(z) where z is a complex number should only give you one value. Therefore you need to pick a "branch" (i think that was the word for it) of the natural logarithm.

*nods and pretends to understand*

@hedemegmondom

6 жыл бұрын

At least I'm not alone.

@killslay

6 жыл бұрын

SandyStarchild thank you for making me feel a little less alone in my ignorance. Looking at the comments I thought it was just me

@noredine

6 жыл бұрын

**nods in solidarity**

@Ragnarok540

6 жыл бұрын

This is not hard at all, I think this is way easier than even basic calculus.

@Ragnarok540

6 жыл бұрын

Derivation, I consider Integration is the point where calculus stops being "basic"

When I was studying calculus in year 11, I became fascinated with the function, f(x)= x^x. I asked my Maths teacher how to calculate the first derivative of this function. He was the best teacher I ever had. But this question stumped him. It was 1971. There was no internet. We certainly didn’t have WolframAlpha. All we could conclude was the x < 0 was going to cause real problems. Is this function and it’s derivatives of any interest? Or did I just imagine I’d found something exciting and weird?

@3unruh

4 жыл бұрын

If I remember right, my math teacher gave me that function as a homework in 11. or 12. grade. :) But they forgot to mention that they are only interested in the x > 0 part, so I spent a lot of time trying to figure out the negative part. :D But for x>0, it's actually quite easy to analyze using the right trick: x^x = e^(ln x)^x = e^(x ln x). If we want to differentiate that, we first differentiate x ln x, that gives us ln x + x/x = ln x + 1. And thus e^(x ln x) has the derivative (ln x + 1) * e^(x ln x) = (ln x + 1) x^x. And if you want to find an extremum, for example, we find x where the derivative is 0, i.e., (ln x + 1) * e^{x ln x) = 0. Since the right factor is always >0, this happens only for ln x + 1 = 0, i.e., for ln x = -1, i.e., for x=1/e. (In fact, it's a minimum as can be seen from calculating the second derivative.)

@TheEnderLeader1

4 жыл бұрын

@@3unruh Why did they bother saying that they were only interested in x > 0, when they just could have set you f(x) = |x|^|x|?

@donielf1074

4 жыл бұрын

“Real” problems. Well done. This doesn’t seem that difficult. It’s just (x^x)(ln(x)+1), isn’t it? Rewrite the function x^x as e^(xlnx) and apply chain rule and product rule.

@pieboy2043

4 жыл бұрын

Ross Long that just reflects the x>0 part over the y-axis, what’s more interesting is |x|^x, as for all rational x (I’m not sure about irrational) x^x is always either |x|^x or -(|x|^x), so it’s a way of seeing what it “should be”

@angelmendez-rivera351

4 жыл бұрын

Ross Long Because then you have to deal with differentiating |x|, which is a bigger waste of time than specifying x > 0. Besides, in mathematics, domain restrictions are commonplace and usually necessary.

Thanks man...i thought it was very difficult.... you really explained it like a boss

The infinite answer thing is exactly like asin(1). It technically has infinite answers, but most people would give you pi/2.

@MichaelRothwell1

5 жыл бұрын

Yes, but this convention comes at a price. You cannot solve sin(x)=1 by saying x = asin(1) = π/2 as of course there are infinitely many solutions. In the same way, by taking a single choice of value for ln z, you cannot then solve e^w = z by saying w = ln z since again there are infinitely many solutions. In other words, you can't have your cake and eat it.

@alphakrab5022

4 жыл бұрын

@@MichaelRothwell1 For your first example, you can, because asin is defined on [-1;1] and takes value in [-π/2;π/2]. This is actually the only definition of the function asin. But you're right, we can't use ln for complex numbers

@lupsik1

4 жыл бұрын

First part is correct but asin is defined on a restricted domain so that theres only one answer.

@angelmendez-rivera351

4 жыл бұрын

Michael Rothwell But you can have your cake and eat it. Everyone knows how to solve the equation sin(x) = 1 using the arcsin function: you evaluate arcsin(1), and then you append + 2nπ to obtain all the solutions. Similarly, log is multivalued, but you can use log to solve e^w = z by noting that e^(2nπi) = 1, hence w = log(z) + 2nπi

@fedem8229

3 жыл бұрын

@@alphakrab5022 You didn't read it well, he said asin(1)=π/2 is not the only solution, because there are infinitely many solutions for sin(x)=1, but the asin function as it is a function it just spits one value on the interval [-π/2 , π/2]

Could you do a video on Carlyle Circles? I find them very interesting and I would like to see your take on the topic.

This... is beautiful, damn, I almost miss my math class back in college, almost.

The subtle hit against tau is even better when youve seen his argument with steve before xD

1:55 Matt, this is Tau erasure.

@automatedminer7158

6 жыл бұрын

Nothing of value was lost

@Cythil

6 жыл бұрын

But Matt is arguing for us having a Pi/2! Rather then go with 2*Pi = Tau.

@unvergebeneid

6 жыл бұрын

He was trolling tau fans so hard, first with that fake-out and then with calling for a name for pi/2 😆

@Cythil

6 жыл бұрын

Yeah ^_^ Though I quite like Tau.

@rarebeeph1783

6 жыл бұрын

I like both Tau and Pi/2, but not Pi in most cases. Pi is surprisingly difficult to work with in terms of angles.

3:29 There are some nuances to keep an "i" on!

I love this so much. And one inconsequential thing I love is how he says "pi on 2" where Americans would say "pi over 2". It fun encountering new language details like that.

@janmcclure6239

3 жыл бұрын

And the British version is slightly erotic in American

i^i looks like InI. Nice one. I like the way you display equations and graphs against a black background in which the presenter is still visible; it gives "I" power to my eye. I will copy your style on my videos.

Is the logarithm step really needed? e^(pi/2)i=i raised to i should give the same result anyways?

@irrelevant_noob

5 жыл бұрын

Except the "raised to i" part is not well-defined over the set of complex numbers. You do eventually reach the exact same set of numbers that can be considered as possible answers for i^i. :-B

How about i^^i (tetration)? What would an exponent tower with height i look like?

@badjumpcuts6599

6 жыл бұрын

I don't think the hyperoperators are defined for non-integers.

@RubenHogenhout

6 жыл бұрын

What about (a + b*i )^(a + b*i ) ?

@badjumpcuts6599

6 жыл бұрын

RubenHogenhout That would be fine. What I meant was: 5^^3 means that you have a power tower with three 5s in it: 5^(5^5). So a^^b is a power tower with b number of a's What would 10^^(3/4) look like? You would have a power tower where the total number of 10s is 3/4, and I think that's too ambiguous

@RubenHogenhout

6 жыл бұрын

Ok but what is the order of the powers? Because 5^(5^5) is not the same as (5^5)^5 . Thus you start with the last pair?

@jhawk2402

6 жыл бұрын

would 10^^(3/4) be equivalent to 10^(10^10) / 10^(10^(10^10))) OR the 10^(10^(10^10)))th root of 10^(10^10)

I really was disgruntled he wouldn’t bring up the alternate values; that joke got me; -was not disappointed.

well... an i^I will make the whole world... not blind?! love these video's where you explain stuffs like this! I can almost feel the maths waking up inside me again :D

Nice video and channel about popularization of maths! Regarding this video however, I feel that the explaination about the value of i^i is not precise/correct. In fact, with the same argument, I could say that e^{-3i\pi/2} = i, so ln i = -3i\pi/2 and i^i = e^{i ln i} = e^{3\pi/2}, which gives a different value than the one described in the video. The reason is that the logarithm is not uniquely defined in the complex numbers: one must choose a domain (in general a simply connected open subset of the complex plane C not containing the origin 0, for example C minus a half-line starting from 0), and once the domain chosen, the logarithm is defined "up to an additive constant, which is an integral multiple of 2 \pi i. Choosing the value of this additive constant is what is called choosing a "determination" or "branch" of the complex logarithm. In the case of i^i, the choice of different branches of the logarithm give different values for i^i, which is hence not uniquely defined. One could argue that we want to take angles between -\pi and \pi (which works as far as we don't want to compute log(-1)). This is called the "principal branch" of the logarithm (because it coincides with ln on the positive real numbers), but a priori there is no reason for this choice over another, and one could get any value of the form e^{-\pi/2 + 2k\pi}, where k is an integer.

@angelmendez-rivera351

4 жыл бұрын

You are correct, although I should mention that the actual choice of the principal branch of the complex logarithm has it that Arg(z) lies in the semi-open interval (-π, +π], not the open interval.

DID THIS VIDEO JUST CONVINCE ME THAT MATHEMATICS IS EQUALLY FUN AND INTERESTING AS SCIENEC;)

Didn't even get this idea.. ! Omg! That's great I'm so exited....!

holy moly that went way over my head - but was entertaining

it makes me incredibly upset that the result doesn’t have a magnitude of one. it just feels wrong in my gut.

@joshhickman77

3 жыл бұрын

Because the magnitude of i is 1, which is less than e, when raising it to a complex power it spirals in instead of spiraling out. e balances the spiral so it goes in a circle instead. That's the intuition here, I think.

i^i: *exists* Keanu Reeves in Matrix: JESUS CHRIST THAT THING'S REAL?!?!

@NortheastGamer

3 жыл бұрын

Can you remind me what part of the movie this is from?

@Autogenification

3 жыл бұрын

@@NortheastGamer it's the scene in the first film, after Neo's interrogated at his office job by a Smith agent; he's in the car with Morpheus and Trinity and they get that machine insect thing out of his body that the Smith agents planted but Neo thought that the interrogation was a bad dream, hence shouts that comment

@NortheastGamer

3 жыл бұрын

@@Autogenification Ah yes now I remember thanks!

I started crying because it was so beautiful

That is...really amazing and surprisingly simple!!

at 1:54 I got really excited when he said he shouldn't use 2Pi, I thought he was going to break out Tau, but he didn't! OPPORTUNITY MISSED THERE MATT!

@matthewbertrand4139

6 жыл бұрын

wdyahnke Not an opportunity missed at all. Matt hates tau, and so do I. It's stupid.

@FernieCanto

6 жыл бұрын

"Matt hates tau, and so do I. It's stupid." People feel triggered by Tau? That's odd.

@bored_person

6 жыл бұрын

The purpose of tau is pedagogical, I thought that was common knowledge.

@bored_person

6 жыл бұрын

Why are you so triggered over mathematics?

Just have a question about unicity there: Let k be an integeer, e^(i*(2k+1/2)*pi) = 1 Why would ln(i) = e^(i*pi/2) more than any other value? Is it a convention or something?

@ivanerofeev1269

3 жыл бұрын

Mathematicians can mean different things by a^b. In many applications (like Wolfram Alpha/Mathematica) a^b = e^(b log a) where log a is a principal value of logarithm a (=ln|z| + iθ where −π

“Don’t worry about it.” Kind of reminds me of a Parker Square!

can someone explain me what is 1/4th power of -ve1 because 1/2nd power of - ve 1 is called or defined as i........ so the question is what's the 1/4th power of - ve 1 or 1/8th power of the same. please reply someone

0:34 "But *i* believe" got u there ^^

But where did the music go?

Is it not very problematic using log when you have complex input when it comes to whats okey to "plug in"?

My favorite use of i is in calculating the Mandelbrot Set (hope I spelled that correctly.) You get some very beautiful graphics from that. Writing a program to do that, even having the algorithm was something of a challenge for me as I don't have much experience using i in calculations.

Actually ln(i) could equal i*5pi/2 or i *9pi/2 or i(4k+1)pi/2 in general. And most complex analysis books note that the natural logarithm function from C to C are multivalued. So there are infinitely many possible values of i^i

@sebasbot01

2 жыл бұрын

I was wondering whether we can even call that a function, as functions have exactly one value per input. Wouldn't it be better to just define the angle you can put into the logarithm as from 0 to 2 pi?

@xinkeguo-xue

2 жыл бұрын

@@sebasbot01 You're right. It's not a function. A multi-valued function in math is actually not a function. They just both have the word "function" in their name.

@stevemenegaz9824

Жыл бұрын

Excellent!. You get a star.

@jh-ec7si

10 ай бұрын

Says this at the end of the video

Let's start a movement to call Pi/2 *Anti-Tau*

@matthewbertrand4139

6 жыл бұрын

Ipad Air Let's start a movement to get tau erased from the universe.

@commonpepe2270

6 жыл бұрын

tbh it would be more sensible if Tau equalled Pi/2. the letter even looks like half a Pi...

@jjtt

6 жыл бұрын

Sorry, it already has a name, "eta". Search "eta circle constant" in Google

@HagenvonEitzen

6 жыл бұрын

or the Parker tau among all taus

@francescosorce5189

6 жыл бұрын

Why eta... psi would have been better (tau has 1 "leg", pi has two and psi has three, it would make more sense...)

That was the greatest! Thanks man!

presumably if you can take a number to the power of i then it follows that you can take the i-th root of any number? what's that operation like what's the i-th root of 5?

So i^i has infinitely many results, all of which are real? That's just weird

@xshortguy

6 жыл бұрын

If you want it to have any meaning, you have to make sure the operation is well-defined. Kind of the same reason why the positive square root is always chosen when doing the square root operation.

@jwmmath

6 жыл бұрын

...yep...all because of the cyclic nature of unit circle, (integer) powers of i, ... "trivializing" a great constant!

@lbranco93

6 жыл бұрын

Polydrome functions

@IronFire116

6 жыл бұрын

David Cox Only infinitely many answers as any other equation. 1=1 can be written as 1=cos(0), but it's the same result. Same with i^i.

@dejanatanackovic909

6 жыл бұрын

Matthew Miskiewicz That is a good point.

“On the left is completely imaginary ... on the right is completely real” Matt Parker gets political

@aaronleperspicace1704

5 жыл бұрын

😂😂😂 and he's right

@aaronleperspicace1704

4 жыл бұрын

@Eric Lee because his statement was so right, it didnt need anyone commenting that it's right, right?? I agree!

@aaronleperspicace1704

4 жыл бұрын

@Eric Lee Hahaha! You must be soooo much fun at parties

@aaronleperspicace1704

4 жыл бұрын

@Eric Lee because of your incredible sense of humour.

What if I take j as a basis vector of length 1 in a 2d vector space then what’s j^j ? j^2 is for example j

Why are properties such as ln(a^b)=bln(a) preserved when we extend the domain of the ln function to the complex numbers?