What does i^i = ?

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Spoiler! Value for i^i is below.
Enjoy some more e^iπ with 3blue1brown's first video:
• e to the pi i, a nontr...
Plus the 2017 follow-up video with extra Group Theory:
• Euler's formula with i...
UPDATE: People have pointed out that I've been beaten to the video punch by blackpenredpen! They are also much more thorough about the multiple-value problem.
• i^i
If you don’t believe my value for i^i, I appeal to Wolfram Alpha.
www.wolframalp...
For those of you who just want to see the digits, here they are:
i^i = 0.20787957635076190854695561983497877003387784163176960807513588305541987728548213978860027786542603534052177330723502180819061973037466398699991126317864120573171777952006743376649542246381929737430538703760051890663033049700519005556200475866205294351834431843455027479745344769934714172383230815271481800760921074192047151878353489584821890186029582331295662952070823409567696363742039451439394183861901080820897771751705004348176454751714529894341134142...
CORRECTIONS:
None yet. Let me know if you spot anything!
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MATT PARKER: Stand-up Mathematician
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Опубликовано:

26 сен 2024

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Комментарии : 3,1 тыс.

@TristanBomber 6 лет назад

"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 лет назад

... which is basically 3, so pi.

@1224chrisng 5 лет назад

engineer: it's Theta (assuming small angles)

@LudwigvanBeethoven2 5 лет назад

Its a number

@losthor1zon 5 лет назад

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

@tibi2674 5 лет назад

losthor1zon wish i could give you a cookie right now.

@wouterlahousse9637 5 лет назад

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

@musik350 4 года назад

Not i^i!, that's another number

@smit_1449 4 года назад

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

@ultimategotea 4 года назад

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

@jasondeng7677 4 года назад

"i^i factorial" ????

@IvanToshkov 4 года назад

@@jasondeng7677 sponge bob pants ^ 2

@Ledabot 4 года назад

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

@softlysnowing3959 4 года назад

yeah i thought that too...

@NStripleseven 4 года назад

Heh. Yeah.

@michalhoransky1214 4 года назад

"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 года назад

@Michal Horanský Yep, eta η ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-1qpVdwizdvI.html

@funkdefied1 4 года назад

1:56

@jonw8764 6 лет назад

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

@joshuaquezada9363 3 года назад

I saw what you did there

@michielschaeverbeke1421 2 года назад

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

@joshuaquezada9363 2 года назад

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

@danksagrabowski2438 7 лет назад

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

@phs125 7 лет назад

Danksa Grabowski i got the reference 😉

@Longuncattr 7 лет назад

Completely despooked.

@GibsonDE 7 лет назад

Danksa Grabowski Gotta keep an I on it

@deeptochatterjee532 7 лет назад

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

@danksagrabowski2438 7 лет назад

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 :)

@GogiRegion 6 лет назад

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 года назад

the schrödinger equation in quantum physics contain an i

@alextaunton3099 2 года назад

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

@k0pstl939 Год назад

The fourier transform too

@knutritter461 10 месяцев назад

X-ray analysis of single crystals in chemistry! 😉

@rewazza 6 месяцев назад

@@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

@theginginator1488 7 лет назад

It's a Parker fifth

@MrQuarris 6 лет назад

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

@kindlin 6 лет назад

Yes, that is the joke.

@totaltotalmonkey 6 лет назад

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

@vampyricon7026 6 лет назад

@สนุกพี่เก๋ 6 лет назад

TheGin Uginator14

@TheArezmendi 7 лет назад

"Parker identity"

@commonpepe2270 7 лет назад

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

@TheArezmendi 7 лет назад

Common Pepe precisely.

@klobiforpresident2254 7 лет назад

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

@TheArezmendi 7 лет назад

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

@Alinoe67 7 лет назад

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

@jacksongraham8061 5 лет назад

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

@lucasg.5534 4 года назад

omg

@revimfadli4666 4 года назад

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

@pedronunes3063 4 года назад

i^i

@Daro-Wolfe 3 года назад

I clicked on the video for this comment

@firkinflamer8604 3 года назад

Shut up Netflix person >:(

@kyazarshadala8114 7 лет назад

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

@prdoyle 6 лет назад

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

@vampyricon7026 6 лет назад

@GrzegorzusLudi 6 лет назад

But real numbers doesn't exist...

@teachermichaelmaalim6103 6 лет назад

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 6 лет назад

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

@andrewchou3277 6 лет назад

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

@martijnvanweele6204 6 лет назад

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

@JochCool 6 лет назад

No, in 2079. Round your numbers properly!

@natehoffmaster6726 6 лет назад

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

@algorythmh 6 лет назад

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

@index7787 6 лет назад

If we last that long

@EleanorDrapeaux 4 года назад

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

@asterixgallier8102 4 года назад

quau

@TiSapph 4 года назад

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

@coastersplus 4 года назад

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

@renedekker9806 4 года назад

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

@TiSapph 4 года назад

@@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

@HeavyboxesDIYMaster 7 лет назад

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

@Miju001 7 лет назад

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

@spencerallbritton9459 7 лет назад

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

@simasjuknelis3043 6 лет назад

Woosh

@theonlygizelli 6 лет назад

nah that would be pie to eye

@landourusnewforme 6 лет назад

Heavyboxes DIY Master i

@SgtAbramovich 6 лет назад

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"

@SomeFreakingCactus 4 года назад

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 года назад

Channelling Rick Sanchez there

@KaitlynBurnellMath 4 года назад

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 4 года назад

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 4 года назад

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

@sighthoundman 3 года назад

@@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.

@AdeonWriter 7 лет назад

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

@Victor-tj7gw 5 лет назад

Lmao

@Brooke-rw8rc 5 лет назад

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 лет назад

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

@thomaskn1012 5 лет назад

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

@ishashka 4 года назад

Zero is a circle constant in a way

@furiondk 6 лет назад

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!

@matthewschad6649 5 лет назад

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

@NickiRusin 7 лет назад

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

@timpeters7852 7 лет назад

Nick Nirus just about to comment. Well done sir

@BamaFanEdge 7 лет назад

Nick Nirus i see what he did there.

@kcwidman 7 лет назад

Nick Nirus how did you write in italics on a RU-vid comment?

@kezzyhko 7 лет назад

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

@rikwisselink-bijker 7 лет назад

@arnoudvanderlugt3230 7 лет назад

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

@MatanVngsh 6 лет назад

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 6 лет назад

Could have used a fifth of something after this video.

@saichaitanyakudapa9554 5 лет назад

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

@alexfenner738 5 лет назад

@@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 лет назад

a Parker fifth

@joeshoesmith 6 лет назад

"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 4 года назад

Tau team!

@Robin_Nixon 7 лет назад

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

@gtziavelis 7 лет назад

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

@sallylauper8222 3 года назад

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."

@dragoncurveenthusiast 7 лет назад

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 7 лет назад

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

@-.._.-_...-_.._-..__..._.-.-.- 7 лет назад

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 7 лет назад

David S. how dense can a person be?

@jansendwan1221 7 лет назад

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

@bertiewooster4043 7 лет назад

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

@bb2fiddler 7 лет назад

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

@brokenwave6125 4 года назад

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

@lancediano8014 3 года назад

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.

@Zeturic 7 лет назад

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

@godseye8785 5 лет назад

@HaleyHalcyon - Gaming Channel you misspelled tau

@icarokaue7334 4 года назад

@@godseye8785 you misspelled 0.

@whitherwhence 6 лет назад

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

@irrelevant_noob 6 лет назад

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

@asukalangleysoryu6695 6 лет назад

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 лет назад

That's a Parker pointer.

@user-ur2po3vp2u 5 лет назад

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

@qwertyTRiG 4 года назад

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

@mussalo 3 года назад

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

@suave319 7 лет назад

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

@Milkyway_Squid 7 лет назад

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

@suave319 7 лет назад

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

@e1123581321345589144 7 лет назад

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

@ryanmuller9497 7 лет назад

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.

@Milkyway_Squid 7 лет назад

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.

@MWSin1 7 лет назад

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

@matthewbertrand4139 7 лет назад

MWSin1 How about one half pi?

@mgb360 7 лет назад

That was an absolutely amazing joke

@L4Vo5 7 лет назад

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

@lammy3055 7 лет назад

or Hi (Half Pi)

@robertnorth5725 7 лет назад

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

@BradCozine 5 лет назад

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

@Brooke-rw8rc 5 лет назад

Well, around 1/5 of it.

@BradCozine 5 лет назад

@@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 лет назад

What did it say?

@BradCozine 5 лет назад

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

@2nafish117 7 лет назад

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 7 лет назад

THIS

@Eurley66 7 лет назад

simple and elegant

@standupmaths 7 лет назад

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

@edmond_ld 7 лет назад

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

@JNCressey 7 лет назад

==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.

@elpain5687 6 лет назад

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 года назад

Wow geinus

@DirkAlmighty13 3 года назад

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 года назад

@@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 года назад

@@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 года назад

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.

@SeeTv. 4 года назад

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

@Anankin12 4 года назад

As opposed to what, 1 the boring unit?

@therealshavenyak 4 года назад

1 is the loneliest unit.

@neutron417 3 года назад

True

@JackFou 7 лет назад

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 лет назад

@@Peter_1986 eIHtT dAsH clOsInG bRaCKeT

@omnitroph1501 4 года назад

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

@DragonWinter36 4 года назад

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

@omnitroph1501 4 года назад

@@DragonWinter36 you've got a point.

@truesoundwave 4 года назад

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

@phmdaemen 4 года назад

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

@DragonWinter36 4 года назад

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

@adilmohammed6897 3 года назад

You deserve a nobel prize

@RSLT 11 месяцев назад

People say that 'i^i' is multi-valued because 'i' can be expressed as 'i = e^(iπ/2 + 2nπi)', where 'n' is an integer.

@julianbell9161 6 лет назад

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 6 лет назад

Good to know!

@drewmandan 5 лет назад

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

@M4cc4n4 5 лет назад

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

@You_Know_Me 5 лет назад

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

@ingGS 5 лет назад

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

@seanl.5181 6 лет назад

"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

@harmonicarchipelgo9351 4 года назад

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 года назад

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

@marcvanleeuwen5986 3 года назад

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 года назад

@@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 года назад

@@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 года назад

@@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)?

@Stjaernljus 7 лет назад

*nods and pretends to understand*

@hedemegmondom 7 лет назад

At least I'm not alone.

@killslay 7 лет назад

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 7 лет назад

**nods in solidarity**

@Ragnarok540 7 лет назад

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

@Ragnarok540 7 лет назад

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

@reynardmeiring9567 Год назад

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

@RandomNullpointer 4 года назад

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

@michaelcrosby7715 6 лет назад

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

@jshariff786 4 года назад

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 года назад

@@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 года назад

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.

@cycklist 7 лет назад

Blackpenredpen did a great video on this.

@samb443 7 лет назад

blackpenredpen always has great videos

@gerstensaft2936 7 лет назад

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

@hdwe1756 7 лет назад

Sen Zen too

@standupmaths 7 лет назад

+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 RU-vid for "i^i" a few weeks back).

@mattiasselin4955 7 лет назад

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.

@kaufinachname9239 4 года назад

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

@ipadair7345 7 лет назад

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

@matthewbertrand4139 7 лет назад

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

@commonpepe2270 7 лет назад

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

@jjtt 7 лет назад

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

@HagenvonEitzen 7 лет назад

or the Parker tau among all taus

@francescosorce5189 7 лет назад

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

@NIMPAK1 7 лет назад

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

@pjgcommunity3557 4 года назад

I think you mean base-19

@masonhunter2748 3 года назад

@fahrenheit2101 3 года назад

@@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 года назад

@@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 года назад

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

@GogiRegion 6 лет назад

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

@MichaelRothwell1 5 лет назад

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 года назад

@@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 года назад

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

@angelmendez-rivera351 4 года назад

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 4 года назад

@@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]

@matingarastudios 4 года назад

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 года назад

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 года назад

@@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 года назад

“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 года назад

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 года назад

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.

@AlFasGD 7 лет назад

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

@daniellambert6207 7 лет назад

@fedewar96 7 лет назад

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 7 лет назад

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

@automatedminer7158 7 лет назад

i ⬆️ i i ⬆️⬆️⬆️ i

@L4Vo5 7 лет назад

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

@abbe1255 3 года назад

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

@senorkarl 7 лет назад

Grade A Tau trolling, Matt.

@standupmaths 7 лет назад

+Karl Hite τrolling

@Interfecteris 7 лет назад

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 7 лет назад

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.

@discflame 7 лет назад

1:55 Matt, this is Tau erasure.

@automatedminer7158 7 лет назад

Nothing of value was lost

@Cythil 7 лет назад

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

@unvergebeneid 7 лет назад

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

@Cythil 7 лет назад

Yeah ^_^ Though I quite like Tau.

@rarebeeph1783 7 лет назад

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

@doim1676 Год назад

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

@davidroberts6242 3 года назад

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

@joshhickman77 3 года назад

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.

@anonymoususer2756 Год назад

This video made me realise something. Since ln(-1) has infinitely many complex solutions (pii, 3pii, 5pii etc…), you can generalise this and find that ANY logarithm technically has infinitely many solutions. For example: e^(2pii) = 1 e^(2pii) * e^(0.6831) = 2 e^(2pii + 0.6831) = 2 e = 2^1.4427 (2^1.4427)^(2pii + 0.6831) = 2 1.4427 * (2pii + 0.6831) = 2.8854pii + 1 2^(2.8854pii + 1) = 2 So log base 2 of 2 can also be 2.8854pii + 1 as well as 1, and inifinitely many other complex values.

@thierrypauwels 3 года назад

Using his kind of derivation, we can simply prove the following: exp(pi i) = -1, so ln(-1) = pi i 2 pi i = 2 ln(-1) = ln( (-1)^2) = ln 1 = 0 Hence: 2 pi i = 0 which is clearly nonsense. The lesson is that the familiar rules between logarithms and powers are no longer valid when using complex numbers. But he does use them just as if they were valid.

@DarkPortall 3 года назад

You cant actually jump numbera into logarithms if it changes the answer, and jumping a 2 inside a logarithm can. think of 2log(-1). if we don't jump it it's some imaginary number, as there are no real solutions to 10^x=-1, however if we jump it its simply log(1) or 0.

@deepskywalker66 3 года назад

Yeah, math is a Periodical Thingy, SomeWhere it does exist, SomeWhereElse - it DOES NOT ...

@Lincoln_Bio 3 года назад

2 pi i would be a 360 degree rotation around the complex plane, returning to the origin, so yeah it equals zero, checks out tbh, great proof. He literally points out in the video that 2pi i = 0 around the 2 minute mark.

@SamChaneyProductions 5 лет назад

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

@topilinkala1594 2 месяца назад

I personally like the e^ipi-1=0 more because it has five constant mathematicians use all the time 0, 1, i, e and pi.

@kavithathirupathy8580 3 года назад

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

@hurdler 5 лет назад

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 года назад

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 года назад

@@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 Год назад

Excellent!. You get a star.

@jh-ec7si Год назад

Says this at the end of the video

@alamrubilmaruf 6 лет назад

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.

@matteoruggiero5074 5 лет назад

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 года назад

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.

@capjus 5 лет назад

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

@stefanlclark 4 года назад

my guess is similar to this ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-PYZJ3csb_rg.html ?

@javierantoniosilva8477 4 года назад

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

@NortheastGamer 4 года назад

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

@Autogenification 4 года назад

@@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 4 года назад

@@Autogenification Ah yes now I remember thanks!

@Septimus_ii 7 лет назад

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

@xshortguy 7 лет назад

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 лет назад

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

@lbranco93 6 лет назад

Polydrome functions

@IronFire116 6 лет назад

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 лет назад

Matthew Miskiewicz That is a good point.

@zaubergarden6900 7 лет назад

are you doing this today because they release the new i-phone?

@eliteextremophile8895 7 лет назад

No. Because they sent him one.

@w2quick 7 лет назад

lets not assosiate math with fanaticism

@ToMeK3001pro 7 лет назад

knowing Matt it may be this

@chemistry4life 7 лет назад

zaubergarden he gotta do i^x then....

@schizophrenicenthusiast 7 лет назад

iPhones are for people who wanna join a cult

@camembertdalembert6323 2 года назад

I don't agree. i^i can have multiple values.

@wdyahnke 7 лет назад

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 7 лет назад

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

@FernieCanto 6 лет назад

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

@bored_person 6 лет назад

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

@bored_person 6 лет назад

Why are you so triggered over mathematics?

@sinom 6 лет назад

5:40 did you just make fun of Tau?

@icedo1013 6 лет назад

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

@sinom 6 лет назад

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

@irrelevant_noob 6 лет назад

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

@sinom 6 лет назад

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 лет назад

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!

@luiscolorado4034 3 года назад

i^i is a set of numbers... the set e^(-pi/2 + 2 k pi) for all integer values of k, all of them reals.

@Eniro20 7 лет назад

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

@badjumpcuts6599 6 лет назад

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

@RubenHogenhout 6 лет назад

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

@badjumpcuts6599 6 лет назад

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 лет назад

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 лет назад

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)

@Grrblt 5 лет назад

Why'd you go the long way around? e^(i*pi/2) = i => e^(i^2 * pi/2) = i^i.

@maksymiliank5135 5 лет назад

because it is easier to understand and explain where did it come from

@angeluomo 5 лет назад

Yes, I was proudly showing this to my son, who is a physics student; and within one second, he pointed out that, if e^(i*pi/2) = i, then you don't need the entire e^ln trick to separate i^i. Just raise e^(i*pi/2) to the ith power and you get e^(-pi/2). Simple.

@maksymiliank5135 5 лет назад

@@angeluomo Yes, but the 'trick' with turning a expression to the form of e^lnx helps you evaluate any complex number raised to a complex power

@angeluomo 5 лет назад

@@maksymiliank5135 Fair point. But it isn't necessary to solve this particular problem. I will remember the trick; however, there is a much easier path to the solution for i^i.

@petormaculan5424 4 года назад

This is just an anecdote, but I dont understand what you are doing to get the answer. However, I understood using this video, so yeahh, this video is necessary for me.

@xxliu95 5 лет назад

I did i^i = (e^(i*pi/2))^i . So in this case we multiply the exponents e^(i*i*pi/2) resulting in the same answer.

@alwinpriven2400 7 лет назад

1:59 he's gonna mention tau! 2:01 did you just call tau 0?! I'm really mad now.

@Werdna12345 7 лет назад

I'm sorry that we take please from your torment. To be honest if Tao was the one that was a homonym to a delicious desert, I would be on Steve Mould's side too

@robknightfilms 7 лет назад

* pleasure

@alwinpriven2400 7 лет назад

Tau* also, Tau>Pi

@alwinpriven2400 7 лет назад

no, 6.28 is definitely bigger than 0.

@Ruminations09 7 лет назад

No, he didn't. He was talking about how that angle is both 0 radians and 2(pi) radians, and thus has 2 different numerical values. His joke was definitely about Tau, but the 0 was not related to that joke.

@pnadk 5 лет назад

Not often do you hear a mathematician say "It's about a fifth."

@red5t653 5 лет назад

Ever heard of the Parker square? It's a magic square, but not quite. This is that Parker. He could've also said that i^i is a Parker fifth, and most people here would get it, I think.

@Djake3tooth 2 года назад

There are multiple values because “the natural logarithm” would actually have to be divided into infinite copies with the imaginary part k*2pi to be a function.

@nachoImagine 7 лет назад

7:05 "Don't worry about it.." Well, I do worry about it, ¿i^i could have many real numbers as a result? Like... WTF? I need to know this!

@limecyanizer4394 5 лет назад

e^(-pi/2)=0.20787957635 e^(-5pi/2)=0.0003882032 e^(-9pi/2)=0.0000007249 e^(-13pi/2)=0.000000000000135

@endogeneticgenetics 5 лет назад

Seriously, wtf was that...?!

@chibill_mc 5 лет назад

LimeCyanizer but which is the real one :p we need to do the version with the square root to find out xD

@jcespinoza 4 года назад

I also DO worry about it! :(

@agsystems8220 4 года назад

Does it worry you that x^2=1 also has multiple solutions (+/-1)? Exponentiation already doesn't always have an inverse, meaning we can't just find x in the previous equation by taking the square root. Taking things to the power of i has similar dangers to taking to fractional powers, such as 1/2. In order to make it a function at all we need to arbitrarily select one of the possible solutions as the answer. We do the same here. It just means we need to be careful when we actually use the function, because much of the time it will no do all the work, and we will need to narrow down the solution to our actual problem further using more specific tools. The problem with imaginary powers is that i=(1/i^3), so even integer powers of i are sort of fractional powers, and you run into the same lack of true inverse issues you do with square roots.

@earthbjornnahkaimurrao9542 7 лет назад

its strange that something with magnitude of 1 raised to a power with magnitude of 1 would be so much less than 1

@thoughtpolease7183 3 года назад

I don't know the term "magnitude" but I was thinking this same thing. It is really unintuitive

@ND62511 6 месяцев назад

The argument for why it makes most sense for i^i = e^-π/2 operates under similar logic as to why e^iπ = -1. Both are the initial value of a series of valid answers i^i = e^-(2πn + π/2) -1 = e^(2πin + iπ) Where n = integer So I feel like Matt’s initial answer makes the most sense.

@unnamed7225 3 года назад

On the first day of school My mom: How was school? Me: I made some friends My friends: *5:57*

@ateium2409 3 года назад

@anshumanagrawal346 3 года назад

😂😂😂😂

@steliostoulis1875 7 лет назад

*insert parker square joke here*

@newmanhiding2314 2 года назад

Engineers trying to find the value of 1/5: “Write that down! Write that down!”

@ElagabalusRex 7 лет назад

I don't understand why people hate the word "imaginary". Whenever they suggest geometrically-inspired alternatives like "orthogonal" or "planar", they end up describing ℝ x ℝ, which is not the same as ℂ.

@Pytha974 7 лет назад

Well ℝ x ℝ is isomorphic to ℂ so you can argue that it's the same.

@volbla 7 лет назад

Because it implies, especially to laymen and students, that they are not as "valid" as Real numbers, which isn't true. Just like the name Real implies that thosr number make sense, which also isn't true. It was a simpler time when these names were made up.

@matthewbertrand4139 7 лет назад

ElagabalusRex ...how did you type the characters for the sets of all real and complex numbers I have to know

@MultiQuanzaAVQ 7 лет назад

Ok, in this chat thread of 6 people lets all vouch to call them complex numbers.

@ZipplyZane 7 лет назад

Simple. Because the name "imaginary" implies those numbers can't describe anything that actually exists. It demotes them to purely theoretical values that can't describe anything in the real world. Fortunately, there's not must use for imaginary numbers by themselves. So, if i is involved, just refer to the number as "complex."

@blisles7626 4 года назад

It intriguing how different people say fractions etc. I would say pi over two, while my friend says pi on two and my lecturer says pi upon two yet it is definitely the same thing.

@Anankin12 4 года назад

I think pi on two could be wrong, since I believe there is something with factorials which is said like that. Other ones fine with me

@killianobrien2007 2 года назад

For small natural denominators like that I would say "pi halves" but I would just use over everywhere else

@noahosborne6454 2 года назад

"Are you ready kids?" ""About a fifth captain!"

@dinamosflams 3 года назад

ah yeas, as the old proverb says: "an eye to an eye *is about a fitht* "

@c31979839 4 года назад

We need to start a petition to change the name of imaginary numbers.

@Brindlebrother 4 года назад

the legendary numbers

@marcelosantos5683 4 года назад

I think someone actually had the idea to call them "lateral numbers" but that didn't catch, i don't know

@muriel5935 4 года назад

Numbers²

@kamo7293 4 года назад

@@marcelosantos5683 im pretty sure taht was euler or gauss who called them lateral. edit: gauss, it was gauss

@Gaspar.Albertengo 4 года назад

They are called "complex numbers", and live onto the "complex plane"... there is no need to change anything, just use the things already there

@GDinferno Год назад

Wow, coolest thing is in my opinion the inverse! If you take the superroot of this answer (superroot is the inverse of tetration), you get i, from a positive, entirely real number!

@skepticmoderate5790 7 лет назад

I actually tried to solve it. I didn't get very far.

@standupmaths 7 лет назад

+skeptic moderate The important thing is: you gave it a go.

@parkermowery6784 7 лет назад

standupmaths Sounds like a bit of a Parker Square of a method. And yes I can make that joke because people make it with me more than you'd expect. You reach pretty far!

@Thisath100 7 лет назад

Most of the time, I genuinely don't understand 3/4th of the video but I just love watching it because, hey, it's Matt. And still, I learnt i^i= about a fifth. #feelclever

@standupmaths 7 лет назад

+Thisath Ranawaka That's all you need to remember!

@JugglingBlog 7 лет назад

@standupmaths what would be the best way to get my maths knowledge up to a level where I can study IT in university? I already tried it and, since I've been doing nothing but programming all my life, everything was easy except the maths portion. I've been doing fine in school, but at university they kept throwing complex numbers, sets and other stuff at the students, with way too little time to learn it. I don't really need a degree, but I'd just like to have it in my pocket to prove it to myself that I can do it, but I need something to help me start. I was googling all the stuff we needed to prove, but after many hours I still wasn't able to find a single good explanation. Are there any books you would recommend for university grade maths that explain the topics in an understandable way?

@KarenSDR 5 лет назад

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 года назад

And the British version is slightly erotic in American

@gmozzi5827 7 лет назад

Nice video! but it's a somewhat convoluted way to get to i^i value if you've already shown Euler identity. You could've simply substituted like this: (i)^i= (e^(i*pi/2))^i= e^(i*i*pi/2) = e^(-pi/2)~ 1/5, basically the logarithm step wasn't really crucial.

@SirRichard94 7 лет назад

Gabriele Mozzicato but logs are FUN do you hate fun?

@gmozzi5827 7 лет назад

Unnecessary logs aren't ;)

@thetrickster42 7 лет назад

Gabriele Mozzicato he did it to draw out people screamong at their screens that there was more than one solution ;) It also helps lead people deeper into complex numbers, and they might look up logs of a complex number, branch cuts and other things people are talking about in the comments

@gmozzi5827 7 лет назад

He already showed that i can be expressed in multiple ways trough Euler formula, the log isn't needed at that point. I agree that showing it might get people interested in complex logarithms, but to me doing it while proving something feels clunky, proofs shouldn't have unnecessary details

@lammy3055 7 лет назад

shush

@leonardromano1491 7 лет назад

Well you could argue that log(i)=i(pi/2 +2pi*k) for all whole numbers k. Therefore i^i =exp(-(pi/2)+2kpi) has infinitely many solutions on R⁺.

@Friek555 6 лет назад

THIS! I'm very disappointed that this video didn't talk about this. pi/2 is just one of infinitely many "logarithms" of i

@michaelnguyend.624 6 лет назад

But i^i is a constant. So i^i cannot have many definition. So in this case it must be undefined...

@visceralconfidence2987 2 года назад

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

@PaveDearce 7 лет назад

I always thought that the beauty of Euler's identity is that it relates the five most important numbers that we've discovered in one equation: e, i, pi, 1, and 0.