i-th root of i
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- Опубліковано 20 вер 2017
- We will solve the most complex math question, i.e. evaluating the principal value of the i-th root of i, namely i^(1/i). In fact, the answer is actually a real number!
Do you enjoy complex numbers and up for a challenge? If so, check out x^x=i • they never teach equat...
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Do you enjoy complex numbers and up for a challenge? If so, check out x^x=i ua-cam.com/video/vCdChDmMYL0/v-deo.html
How is this practically applied to real life????
You can't solve subfactorial of i.
The most surprising thing about this is
I understood more than half of what he explained
I agree
I understood it all!
thats because hes a good teacher
Yeah and I'm terrible at maths.
*Learns about imaginary numbers*
UA-cam:
What's 1/i?
What's e^i?
What's i^i?
What's sqrt(i)?
What's log(i)?
What's i-th root of i?
What's sin i?
What's sin-1(2)?
What's i factorial?
What's integral of x^i?
刘颢云 lolll
Ya got lucky...
I haven't even learned about logarithms and have no clue what e is...
@@Ramu-10 you gotta love when you figure out whats going on
@@Ramu-10 it is euler's number, which approxamatly is 2.71.......
@@Ramu-10 then watch this amazing video about "e" and it's derivation.
ua-cam.com/video/m2MIpDrF7Es/v-deo.html
0:32
"I don't like to be on the bottom, I like to be on the top"
Me: *scrolls down to comments to see if anyone made a joke about that*
A joke would be like what?
Super Psych that is literally exactly what I just did 😂😂😂 you made my day
I assume the top would be the video section, while the bottom is the comment section.
In the video it's already meant to be a joke.
Same here...
0:32 "That's what she said!" xD
Do the e-th root of i and the i-th root of e
The second one is easy:-
e^(i*pi) = -1
ith root of e = e^(1/i)
So (ith root of e)^(pi*i^2) = -1
So (ith root of e)^-pi = -1
Flip over fraction
So (ith root of e)^pi = -1
We know e^(i*pi) = -1
So (ith root of e) = e^i
That's a distance of 1 radian around the unit circle
So (ith root of e) = cos(1)+i*sin(1)
Or about 0.54030 + 0.84147i
just go on wolfram|alpha
Lyri Metacurl
Lol
@@anselmschueler wtf
@@dnpendown3199 Or use Qalculate, the best desktop and command line calculator available!
> root(i,e), root(e,i)
[root(i, e), root(e, i)] = approx. [0.8376315 + 0.54623573i, 0.54030231 - 0.84147098i]
"i" understand this a whole lot better now.
😂😂😂
lmao
😂
@Shri Hari's Animations it's not comedy it's komedi
“i” get it, even though the “roots” of the ha”i”r on my head are gray
“I dont like to be on the top. I like to be on the bottom” - blackpenredpen
lmao
Kinky
Wow, complex roots of imaginary numbers are real......math is so weird!
Sometimes they are, but not usually.
But bro real root of real numbers is also imaginary. For example square root of -1 😂
@Smash Boy Yeah, technically speaking Complex numbers are just an algebraic estructure made from real numbers (just like matrices for example) so it makes sense that if you apply a function to an "imaginary" number (a number on the Y axis of the complex plane) it gives you a "real" number (a number on the X axis of the complex plane).
because “One of the miseries of life is that everybody names things a little bit wrong.” - Richard Feynman 1985.
Purely imaginary in this case, imaginary root of imaginary...
Or you could just
i = e^(iπ/2)
ith root both sides
i√i = e^(π/2)
+2kπ ?
@@arthurmoiret6076 That's the point when he said there are infinitely many solutions to this
I did that too!!!!!
@@arthurmoiret6076 or -2kπ
I'm in love with mathematics now
I was in love with math. Then they said “do specialists”
I just stumbled upon your videos while procrastinating. You're a really warm and enthusiastic talker, keep it up sir! :)
black mic red shirt yay!
As mentioned before, I love this channel. Found it randomly, but so glad I did.
Currently a grad student taking complex analysis (as an elective, took it 2years ago). All of your complex videos are just supplementary/review for me, but they're still very satisfying to watch.
Got any plans for singularities?
Travis Hayes great idea! I will discuss this with peyam bc we are filming together this Friday
in a sense,all of his videos are complex
in two,actually. at least two
I don't believe a word of what you say! You doubtless are as flummoxed as the rest of us mortals. Quit showing off!
This video is amazing. Thank you for the break down!
Drink game:
Take a shot after he says the letter "i"
You'd get drunk after the first 10 seconds.
This is so fantastic and beautiful. Thank you for sharing!!
The reason why i^i = e^(-pi/2) is because i^i = e^(log(i)i) = e^((pi/2)i * i) = e^(-pi / 2). There are infinite answers because the log(z) has infinite answers; as it is defined as ln |z| + arg(z)i, where arg(z) is the angle (which can be, in this case, pi/2, 5pi/2, 7pi/2, ...).
but why log(i)=(pi/2)i ?
J CO Because log(-1) = iπ and -1 = i^2
This guy rekindles our faith in the fact that things that seem to be very tough, are actually very easy, we just need to take one right step, and everything else becomes so easy!
Your voice is a lullaby 😭❤️
You're explanations are simple and crisp, really helping me learn mathematics better
Your*
You are a math god
And you are invisible
Invisible, you scared me
the heck is happenin here????
Hangul Choseong Filler
I FINALLY FOUND YOU (only intellectuals understand)
Who thinks new BPRP is more interesting and better? *raises hand*
Ognjen Kovačević thank you!!!
yay!!
Ognjen Kovačević u know. It's weird that even tho I teach the same way in my classes. But I have been doing different things on YT nowadays and I have been enjoying it way more also.
blackpenredpen wait you're a teacher?! I thought you're just a super smart student!!
e, pi, and i: The Zeus Poseidon and Hades of math
Do Aid correct name of gades?
I don’t even know how to begin to do this
I'm even more amazed that you can formulate these math questions with answers.
You could actually just start with e^(ipi)=-1. Take the square root of both sides: e^(ipi/2)=i. Then take the ith root of both sides. e^(pi/2)= i^(1/i).
such an interesting edge case!
You are the most brilliant math person I have ever come across. I am in awe.
Great work. Thank u
I love it when blackpenredpen features blackpenredpen.
That shirt is crazy!! Also, I love your vids ! 😄
These just make me so cheerful..... I have to limit myself to one a day ( and then I'll start again from the beginning because I'm still forgetting stuff like when I was a kid. )
When I saw the video I took a few minutes to think about it for myself and then come back to see if I was right and I was. Thanks I like keeping my mind sharp with little math riddles like this one.
Euler’s number and pi, name a more iconic duo
Thank you sir🙏🙏🙏
YAY!
This was interesting!
Emmanuel Franklin your name n your profile matches a lot with a malayali. Are you one?
amazing!
I'm going to use this information when I file my taxes.
i blinked and he got pi out of seemingly nowhere
That ball that you always keep in your hand makes you look like an ood from DW xD
Black pen blue pen would make it easier for my eyes to read. Thank you so much for your videos.
0:33 that’s what she said😂😂
That was the question that i had not been able to solve in my exam thanks
same😢
I just wasted 2 minutes being confused by a problem I never even knew existed and still don't understand
Damn, same, I clicked this in my recommended (I've always been horrible at math) I don't get how people in the comments get it so easily
Another way to do this is using De Moivre’s theorem, where i on the complex plane would be pi/2, so theta is pi/2/i, so e^i theta would simplify to e^pi/2
damn I didn't know about imaginary numbers more than how to write them and how to show them on complex plain. yet after watching several other videos of yours I managed to get this one by myself.
I still find it odd that "i" with absolute magnitude of "1" can ever twist itself into something outside of that, as in > 1.
Based
Thanks for your smile too ❤️
Imagine if imaginary roots of other imaginary things could be real. Like trees.
Beautiful!
Amazing!!!
Wait ur a teacher!?
Philip Yao yes i am
blackpenredpen you're so young that we thought you're a 20-years old math youtuber haha
@@blackpenredpen where do you teach I have to go there (つ°ヮ°)つ
Nop, he is a math god.
@@victorvega8061 UC Berkeley
nicely done.... the use of multi color to force focus reminds me of a math professor in the 1960's at the University of Maine.
Thank you Mike!
Really impressive..
💗💗
I did some other method : Consider i^(1/i) and get i = (-1)^(1/2) and -1=e^iπ so finally : i=e^(iπ/2) so in the end we have i^(π/2) as the "i"s cancel out !
That was so clean, damn
Tak na dobry sen. U mnie luzik. Dobranoc, kurde jeszcze jasno.
“I dont like to be on the bottom, i like to be on the top”
- pen guy
thank you
0:33 of course you do. Your entire channel is about doing hard work. Lol
What would an ith root(x)graph look like?
wow that is like magic! amazing!
but the real question is... what is i-th root of i based on Wolfram Alpha?
Max Haibara lol!!!!
Just tried it... it says what he said
I just did it before reading the comments, then I did e^(pi/2), then I compared the decimals, and they are the same, so W|A agrees.
@@MattMcIrvin The joke is that you don't need Wolfram Alpha to solve seemingly difficult equations.
Even calculators can compute it, and they give the same answer. (Done on a TI-84)
'THATS IT' PROCEEDS TO VANISH IN THIN AIR
The ad is trying to kill the youtuber 🤣
So you have this chanel too.
Subscribed to this too ❤️❤️🔥🔥😂😂👍👍.
Dude, you rule!
"What we do now? We do our usual business" if math was that easyyy
"Let me do a REAL quick video for you" I see what you did there
0:30 the quote of all time
You can check its true, if you raise both sides to the power i, you get i = e^(i*pi/2) which is the complex polar form of a vector with argument pi/2 and magnitude 1, which is of course just i on the complex plane. The family of solutions are associated with all arguments that equal pi/2 mod 2pi, so any integer number of times around the units circle landing back at i.
There are other solutions as well: (e^(pi/2+2*n*pi))^i = i for integers n.
Nice, thank you
Very good and nice videos with interesting content sir
You re great! :)
That moment when you try to checkmate a grandmaster and he just literally flips the board. Great stuff man, I didn't this was a question that bothered me since 3 minutes ago but thanks for the answer anyways.
Spectacular example solution.
Could one also write i^(1/i) as (e^(pi/2)i)^(1/i), because e^(pi/2) = i according to Euler's formula, allow the i and 1/i to cancel out, and take the remaining e^(pi/2) as a final answer?
You could easily motivate this by analogy with a square root. Like square root of 4 can be found by finding a number a such that:
a^2=4 ==>a=2.
Similarly, the ith root of i means find an a such that:
a^i=i ==> a=e^(pi/2)
hush, hush... i to i.. too shy, shy..
That was pretty neat.
Isn't it
That mic kinda look like a cartoon bomb
You should probably also update this version with e^(pi/2+2*pi*n)
Sir there can also be my method:
Let i^1/i=x
Take ln both sides..
Now 1/i×ln(i)=ln(x)-----(1)
Now as we know e^iπ=-1
And(-1)=i^2
e^iπ=i^2 then take ln both sides then (iπ)=2ln(i)
Which equals (iπ)/2=ln(i)------(2)
Now put (2) in (1)=
{ln (i)=iπ/2}
1/i×iπ/2=ln(x)====
π/2=ln(x) so
e^π/2=x
I hope this can also be a solution...☺
The thumbnail has tortured me after a tough maths test. Stop
Good one!
As far as I know, the transition from sqrt to power imposes restrictions on the argument. And here it is observed?
Professor, I have a question, is it possible to have a real number to the power of a matrix?
My mind is blown and my curiosity satisfied
I bet you are great teacher
I don't even think about this kind of stuff. It's crazy that some dude was like, oh there's an i'th root of I, let's solve what it is...
How much different is it for a question such as i^(a+bi) or i^(rcisArg)
RaspingBubbles6 you can distribut as i^a • i^bi, then write i^bi as i^i^b the write i^i as e to the π/2, there you have it.
00:32 Lost my concentration there
7
I've given this as an interview question (I'm an EE) - it's easy if you just work your way through it.
*SO GOOD*
How or why can you plug in that e to the negative pi halves in the place of i to the i?
Why did you multiply by i over i? You could have substituted e to the i pi/2 as first step. And then cancel out the i and get the result in 2 steps. :)
Always being on top is like having only real numbers. Broaden your horizon!
If I have learned anything from this channel it is that if you something strange to i, e and/or pi are going to pop out of the equation.
The greatest crossover in history.