i-th root of i
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- Опубликовано: 29 авг 2024
- 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 ruclips.net/video/vCdChDmMYL0/видео.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*
RUclips:
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.
ruclips.net/video/m2MIpDrF7Es/видео.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
"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
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 dont like to be on the top. I like to be on the bottom” - blackpenredpen
lmao
Kinky
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 just stumbled upon your videos while procrastinating. You're a really warm and enthusiastic talker, keep it up sir! :)
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
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...
You're explanations are simple and crisp, really helping me learn mathematics better
Your*
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 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!
black mic red shirt yay!
I'm in love with mathematics now
I was in love with math. Then they said “do specialists”
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).
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
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)
This video is amazing. Thank you for the break down!
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 !
I'm even more amazed that you can formulate these math questions with answers.
Drink game:
Take a shot after he says the letter "i"
You'd get drunk after the first 10 seconds.
Your voice is a lullaby 😭❤️
e, pi, and i: The Zeus Poseidon and Hades of math
Do Aid correct name of gades?
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!!
0:33 that’s what she said😂😂
YAY!
This was interesting!
Emmanuel Franklin your name n your profile matches a lot with a malayali. Are you one?
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...☺
That was the question that i had not been able to solve in my exam thanks
same😢
There are other solutions as well: (e^(pi/2+2*n*pi))^i = i for integers n.
0:30 the quote of all time
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
I don’t even know how to begin to do this
This is so fantastic and beautiful. Thank you for sharing!!
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)
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
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. )
Euler’s number and pi, name a more iconic duo
You are the most brilliant math person I have ever come across. I am in awe.
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!
00:32 Lost my concentration there
7
such an interesting edge case!
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.
alternate method:
i^(1/i)=x
i=x^i
now we know e^(i . pi/2) =i
therefore i=e^(i . pi/2) = x^i
so, x=e^(pi/2)
(X^2-5x+5)^(x^2+7x+12)=1
Find all real values of x.
There will be 6 values of x.
Love from India.
0:33 of course you do. Your entire channel is about doing hard work. Lol
I love it when blackpenredpen features blackpenredpen.
That ball that you always keep in your hand makes you look like an ood from DW xD
I'm going to use this information when I file my taxes.
"technically, i^i has infinitely many answers" - that is crazy, considering that i is a number. That's like saying 2^2 or e^e has infinitely many answers! I love your videos about calculations involving i even though I could never solve them on my own. But your explanations are always clear and step by step. Thanks!
Imagine if imaginary roots of other imaginary things could be real. Like trees.
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
i blinked and he got pi out of seemingly nowhere
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 accidentally created a generalized formula for the i-th root of any complex number with a radius of 1 in school which consists of i-th root(z) = e^a, a is equal to the angle of the complex number in polar form.
The first version of my formula is the i-th root(z) = i-th root(r) • i-th root(e^(i•a)), a is still the angle of the number in polar form (only works with the radius of 1).
Also, I removed the i-th root of r from the first version simply because it is useless since it is the i-th root of a number, where the i-th root is not defined. So r must be equal to 1 and the i-th root of 1 must be equal to 1.
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. :)
And 5 years later: Just go straight for Eulers identity: i'th root of i = i^(1/i)=e^(πi/2i)=e^π/2 From 5 steps to 3 steps
Why so complicated though?
Q: i^(1/i)=?
Solution
Rewriting the BASE i on the polar form, by analyzing the complex plane we get that r=1 and θ=π/2+2n, n∈ℤ. Therefore, i=cos(π/2+2n)+isin(π/2+2n). By Euler's formula this can be written as i=e^(i(π/2+2n)).
Keeping the EXPONENT on it's original form, we can conclude that
i=(e^(i(π/2+2n)))^(1/i)=e^(π/2+2n) by simplification
A: i^(1/i)=e^(π/2+2n)
By plugging in n=0 we get the answer i^(1/i)=e^(π/2), as presented in this video.
Black pen blue pen would make it easier for my eyes to read. Thank you so much for your videos.
“I dont like to be on the bottom, i like to be on the top”
- pen guy
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.
0:30 "I don't like to be on the bottom. I like to be on the top."-Blackpenredpen, 2017
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)
By the fundamental theorem of algebra, there are exactly n roots of nth power. So, there are exactly i roots of i :)
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?
" I don't like to be on the bottom, i like to be on the top " (0:31)
- blackpenredpen 2017
The whole point of the question was to explain that e term, and you bring that from within the thin air, solving the sums fast doesnt mean you miss the real part of the solution
Tell them we wrote "i" as "e^i(theta)" and the "i" and "1/i" cancels out
Sir, i have a question,
We know i = √-1
If we have 1/i, can we write it as, 1/√-1, and then
√1/√-1 => √(1/-1) => √(-1/1) => √-1, basically
1/i => i, can we write this? If not then why?
"What we do now? We do our usual business" if math was that easyyy
What would an ith root(x)graph look like?
So much love given to -
i = e^(iπ2(k+¼))
- in all of its forms and so little given to -
i = -e^(iπ2(k-¼))
i^(1/i) = -e^(-π/2 + 2kπ) or e^(π/2 + 2kπ)
You can't just add multiples of 2π to the positive result and declare victory.
You should probably also update this version with e^(pi/2+2*pi*n)
That shirt is crazy!! Also, I love your vids ! 😄
Could also just write e^(pi i/2) = i and take the root of both sides.
Ian But that requires you to know sqrt(i).
Obinna Nwakwue I meant the i-th root.
You have i = e^(pi i/2)
rt_i (i) = rt_i ((e^(pi/2)^i) = e^(pi/2)
Ian That's exactly what I thought of once I saw the last step.
This way is much simpler.
Ian but that would require you know that e^(pi i/2) is equal to -minus one- i, and how do you know that? tho I guess here you have to know what i^i is so not really that much of a difference
ApplepieFTW actually e^(pi i/2) = i, but I catch your point. The problem is that roots and noninteger powers are defined in terms of the exponential function on C, so...
'THATS IT' PROCEEDS TO VANISH IN THIN AIR
I *really* think you need to explain the multivalued nature of complex exponentiation. It is really beautiful math but it seems like people resort notational tricks which work sometimes, but not always. And it doesn’t foster understanding
Simplest example:
-1=i*i=sqrt(-1)*sqrt(-1)=
=sqrt((-1)*(-1))=sqrt(1)=1
Well, _i_ root of _i_ is pretty easy to calculate.
The _a_ root of _b_ can be rewritten as _a_ ^(1/ _b_ )
So, _i_ root of _i_ = _i_ ^( 1/ _i_ )
Now, 1/ _i_ = -i. You can figure that out geometrically / visually with ease.
_a_ ^ ( - _b_ ) = 1 / ( _a_ ^ _b_ )
So, _i_ root of _i_ = _i_ ^( 1/ _i_ ) = _i_ ^ (- _i_ ) = 1/ ( _i_ ^ _i_ )
To calculate _i_ ^ _i_ , we can use one of euler's formulas. For starters, _a_ ^ _b_ = _e_ ^( ln( _a_ ) _b_ ), thus _i_ ^ _i_ = _e_ ^( ln( _i_ ) _i_ )
ln( _a_ + _bi_ ) = ln( length( _a_ + _bi_ ) ) + angle( _a_ + _bi_ ) _i_ , so _e_ ^( ln( _i_ ) _i_ )
= _e_ ^( ( ln( length( _i_ ) ) + angle( _i_ ) _i_ ) _i_ )
= _e_ ^( ( ln( 1 ) + angle( _i_ ) _i_ ) _i_ )
= _e_ ^( ( 0 + angle( _i_ ) _i_ ) _i_ )
= _e_ ^( ( angle( _i_ ) _i_ ) _i_ )
= _e_ ^( ( 1/2 _pi i_ ) _i_ )
= _e_ ^( 1/2 _pi i i_ )
= _e_ ^( -1/2 _pi_ ) becuase _i i_ = -1
Now, _e_ ^( -1/2 _pi_ )
All of that says _i_ ^ _i_ = _e_ ^( -1/2 _pi_ )
We still have to find 1/that.
1/( _a_ ) = _a_ ^(-1), so 1/( _e_ ^( -1/2 _pi_ ) ) = _e_ ^( -1/2 _pi_ ) ^(-1). Stacked exponents multiply, so _e_ ^( -1/2 _pi_ * -1 ). The negative cancel, resulting in:
_i_ root of _i_ = _e_ ^( 1/2 _pi_ ). The last part here can be calculated.
In conclusion, the _i_ root of _i_ = _e_ ^( 1/2 _pi_ ) ~~ 4.8105
You say "1/ i = -i. You can figure that out geometrically / visually with ease." But I fail to see that visually... I only get that algebraically
How or why can you plug in that e to the negative pi halves in the place of i to the i?
Tak na dobry sen. U mnie luzik. Dobranoc, kurde jeszcze jasno.
Another way is to get this result is to convert i into it’s polar form, you get there a lot quicker
Thanks for your smile too ❤️
Of cooourse, (e^(pi÷2))^i is just equal to e^(i(pi÷2)) which is i in polar form.
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.
Great work. Thank u
Omg, u did it so easily. The way that I've done was making:
Y=i^(1/i)
ln(Y)=ln(i)/i
then, i = e^(iπ/2)
ln(Y)=ln(e)×iπ/2i
ln(Y)=π/2
Y=e^(π/2)
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...
is this really the right answer?
look at this:
i = exp(pi * i / 2) = exp(5* pi * i /2)
but now the i-th root of it can be
exp(pi / 2) or exp(5* pi /2) or any other power of the form (4n+1) * pi / 2
Considering that on a Cartesian plane moving by pi is 180° so since it is pi/2 which is 90° therefore you can have the statement pi*(4n-3)/2 where n is any real number ;)
"Let me do a REAL quick video for you" I see what you did there
This guy : make a complex math things
Me : 1/1-
Isnt it actually e^(pi/2+(2)(pi)(k)) where k is any integer simce there are infinite angles for theta when conveting to polar form
hush, hush... i to i.. too shy, shy..
Doesn't this give a problem? i=e^(pi/2+2 k pi)i for any integer k, thus we obtain e^(pi/2)*e^(2 k pi) for any integer k, thus we obtain a lot of different values?
Thank you sir🙏🙏🙏
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.
So you have this chanel too.
Subscribed to this too ❤️❤️🔥🔥😂😂👍👍.
Let's see, this is math, where we always try to answer a new problem by casting it into terms of one we've already solved . . .
i-root of i = i^(1/i) = i^(-i) = 1/(i^i) and we've already visited i^i; it's = e^(-½π).
So the (principal) i-root of i = e^(½π) ≈ 4.810477...
I like your video. But I was wondering why and how multiplying i/i is a valid mathematical operation, when 0/0 in general is undefined. Any thoughts?
i/i=1 my dude
Any non-zero number divided by itself is 1, and i is a non-zero number.
Bprp: "Now what? Do the usual business."
Me (being a 2 year old): Shit myself.