On Michael Penn's channel, Michael introduced a couple of other number systems such as: The dual numbers R(e) := {a+be : a, b are real} where e is a non-real number such that e^2 = 0, and A real number system extension I can't exactly recall, but it introduced some number w that was the complex cube root of 1, or 1exp(i*2pi/3), or something like that. These are extensions of R, but I think the complex numbers are probably the most used extension of R.
It would be a useful definition if you could calculate the right side numerically. And this mean you have to prove it converges if expressed as some series
@@fullfungo I know, but why can it be written as an equation using Complex numbers? Is it because Euler's Identity requires infinite series that are hidden by the equation?
@@Rudxain “written as an equation” doesn’t make it algebraic. It would have to be a solution to a polynomial with rational coefficients. ln(2), for instance is not a solution to any polynomial, neither is ln(i^(1/i))
Or, you could go the short way with this, remember that i=e^(i*pi/2), represent the problem as finding (e^(i*pi/2))^(1/i), apply your exponent rules, noting that i/i=1, and you're left with e^(pi/2), which is obviously a Real number.
Hold on just a moment my good sir/madam/mathematician. Not all exponent properties hold for all complex numbers, make sure to account for restrictions when constructing your proofs
You could also write the solution as e^(π/2) since it's a negative exponent in the denominator, so you can rewrite it as e^-(-π/2) and then cancel the negatives.
If these rules weren’t applied, all mathematics would fall down literally. For example, sqrt(-1) x sqrt(-1) would have to give -1, right? Well, using laws of indices you can rewrite it as sqrt(-1 x -1), which is the same as sqrt(1), therefore it is 1. But sqrt(-1) x sqrt(-1) doesn’t give 1. That is because it is a negative number in a root that’s not odd. Happens the same with the video
The principal value is e^(pi/2), however roots are generalised, therefore the roots are e^(pi/2 + 2kpi), where k is an integer. So k=0, ans = 4.81whatever k=1, ans = 2575.9something else k=-1, ans = 0.008something small k=2, ans = something massive k=2, ans = something negligible ... and so on But yeah the 4.whatever is quite a real answer.
I love how the title can have 2 meanings, like a pun Either saying that the equation ⁱ√i≈4.81 shouldn't be real Or that ⁱ√i shouldn't be approximately equal to 4.81, a real number
It's Exp(1/i × Ln(i)), where Exp, Ln are complex to complex functions, by definition. So, ...= Exp(-i × (ln 1 + iπ/2 +2πki))= Exp(π/2 + 2πk)= exp(π/2 + 2πk) for any whole k (positive, negative or zero). You can't just apply rules from real exponentiation and logarithms to complex numbers, you should know how they transform. Watch how 2πki emerged when we see complex valued logarithm.
Bringing in the Ln is unnecessary, no? I mean without specifying how many revolutions the initial i is composed of we could just jump straight to i=exp(I pi/2+2i pi k), right?
It is not as simple as you claim. What do you mean by that root? A solution of the equation z^i = i? Well, if so, you will find that it has an infinite number of solutions. e^(pi/2) is only one of them.
This problems makes it more interesting when you realise that it shows you the same solution of i^i is that one,when you multiply i^i with ⁱ√i, one being 1 when you multiply the real solutions of i and the other being also 1
Question: I did, i = exp(i*(pi/2 + 2*pi*N)), therefore i^(1/i) = exp(i*(pi/2 + 2*pi*N)*(1/i) = exp(pi/2 + 2*pi*N). when N = 0, i^(1/i) = exp(pi/2) ~ 4.8105 but I still see infinity number of answers determined by N. What is wrong with my answer ?
Something really bothers me. Isn't i (or √(-1)) when squared supposed to be √1 or 1? Because, when you multiply radicals that have the same indices, whether the bases are different or alike, the bases would just multiply with each other while retaining the current index. For instance, √2 • √4 would be equal to √8 √12 • √3 would equate to √36, which is 6 √5 • √5 would be equal to √25, which is 5 Then why isn't √(-1) • √(-1) equal to √1, since (-1)(-1) is 1? But √(-1) • √1 is equal to √(-1) or i, which follows the rule I've previously mentioned
I have a question, like there exists a set of real numbers and another of imaginary numbers so are there more set of numbers like a third dimension of numbers which would extend the complex plane
Kind of. The complex plane allows an extension to the reals that allows us to obtain solutions to problems with negative discriminants. From the little I know about Analytic continuation en.wikipedia.org/wiki/Analytic_continuation it essentially allows us to extend the domain of the complex plane to be able to solve more complex functions. Like the Reimann Zeta Function.
You could also look at this from algebraic eyes and view it as just an extension field. And then you can find other crazy cool extension fields, like making each "number" a polynomial over a field, and then adding, dividing, multiplying polynomials like they're numbers. You then get some very cool and interesting properties out of them. en.m.wikipedia.org/wiki/Field_extension
His result was that the i-th root of i equals 1 / e^( - pi / 2 ). Raising that to the i-th power gives: { 1 / e^( - pi / 2 ) } ^ i = { e^( pi / 2 ) } ^ i = e^( i pi / 2 ) = cos( pi / 2 ) + i sin( pi / 2 ) = 0 + i (1) = i.
There are infinite solutions to this. i = exp(i{pi/2 + 2 pi k}) where k is any integer. Taking this to the power of -i gives us exp(pi/2 + 2 pi k) for all integers k.
Rigorous, maybe not, but the answer is correct. For more rigor, one must understand that imaginary exponents are multi-valued functions and therefore, the answer is not _unique_ --- although all possible answers are real numbers.
But if you multiply this figure by itself (≈ 4.81), surely you would get a real number...?? How can one number be the square root of two distinct others?
Multiplying it by itself would square it. You need to multiply it by itself i times, whatever that might mean. Squaring it would mean that you essentially took the i'th root of -1
@@hypnogri5457 Okay, this is so embarrassing that I'm tempted to delete my comment. But I will let it stand as a testament to the dangers of drinking tea instead of coffee. Thanks for the correction.
Watch This Next!
ruclips.net/video/LzDANtOH6l8/видео.html
blackpenredpen has done this too!
This video really reminds me of the ”i^i”-video by Matt Parker, on his channel: ”standupmaths”:
ruclips.net/video/9tlHQOKMHGA/видео.html
Complex numbers fascinate me, as they have so many cool properties and tie in so well with seemingly unrelated fields of mathematics.
Me too!
Right? Who would have thought i raised to i both of which are imaginary numbers give irrational but real numbers.
Same
Check out quaternions
On Michael Penn's channel, Michael introduced a couple of other number systems such as:
The dual numbers R(e) := {a+be : a, b are real} where e is a non-real number such that e^2 = 0, and
A real number system extension I can't exactly recall, but it introduced some number w that was the complex cube root of 1, or 1exp(i*2pi/3), or something like that.
These are extensions of R, but I think the complex numbers are probably the most used extension of R.
Oh god please don't turn i into another real number OH GOD HE'S DOING IT ANYWAY
🤣
It’s not i though lol
Pretty cool how it loops perfectly
Fun fact, the letter Z is used more in math than the English language itself
but it isn't compare with russian army...
@@whatisblink dude touch grass🥸
lowkey never used it
@@ayoub_mh Complex analysis non-enjoyer moment
@@LittleWhole bro i’m 17
this leads to a wild definition for Pi : π = 2 * ln( ⁱ √ i )
It would be a useful definition if you could calculate the right side numerically. And this mean you have to prove it converges if expressed as some series
Wait, hold up... does this mean that Pi is algebraic but only in the Complex domain?
@@Rudxain pi is not algebraic
@@fullfungo I know, but why can it be written as an equation using Complex numbers? Is it because Euler's Identity requires infinite series that are hidden by the equation?
@@Rudxain “written as an equation” doesn’t make it algebraic. It would have to be a solution to a polynomial with rational coefficients.
ln(2), for instance is not a solution to any polynomial, neither is ln(i^(1/i))
It's so much easier though.
i = e^(i*(pi/2))
So i^(1/i) would just remove the i in the above equation, so e^(pi/2)
Or, you could go the short way with this, remember that i=e^(i*pi/2), represent the problem as finding (e^(i*pi/2))^(1/i), apply your exponent rules, noting that i/i=1, and you're left with e^(pi/2), which is obviously a Real number.
Hold on just a moment my good sir/madam/mathematician.
Not all exponent properties hold for all complex numbers, make sure to account for restrictions when constructing your proofs
Great thought
You could also write the solution as e^(π/2) since it's a negative exponent in the denominator, so you can rewrite it as e^-(-π/2) and then cancel the negatives.
Another way is the substitution i=e^(i*pi/2). This immediately cancels out the 1/i exponent, giving e^(pi/2)
immediately after i^i was published i began looking for answers to this
Its crazy how a number like that which has no real numbers can output such an odd specific number
All of this is actually not correct
Because the roots radical rules says that the number has to be real and only negative when the n-root is odd. And i isn’t real
If these rules weren’t applied, all mathematics would fall down literally. For example, sqrt(-1) x sqrt(-1) would have to give -1, right? Well, using laws of indices you can rewrite it as sqrt(-1 x -1), which is the same as sqrt(1), therefore it is 1. But sqrt(-1) x sqrt(-1) doesn’t give 1. That is because it is a negative number in a root that’s not odd. Happens the same with the video
@@nicoburstein3910 wait now i dont know who to believe. Accordingg to you what would be corrext
@@csicee the whole operation can’t be performed since you’re using a function that only works with real numbers, so u can’t solve this using roots
Much simple way write I in euler form and raise it to power 1/i you get e^π/2
i feel so smart by understanding everything that you're doing.
but like, past month, i couldn't..
The principal value is e^(pi/2), however roots are generalised, therefore the roots are e^(pi/2 + 2kpi), where k is an integer.
So
k=0, ans = 4.81whatever
k=1, ans = 2575.9something else
k=-1, ans = 0.008something small
k=2, ans = something massive
k=2, ans = something negligible
...
and so on
But yeah the 4.whatever is quite a real answer.
Moral of the story: even some complex numbers with the right operations turn into real numbers
using eulers formulae
e^(ipi)=-1
raise both sides to 1/2 power,
e^(ipi/2)=i
raise both sides to 1/i power,
e^(pi/2)=i^(1/i)
Still love how i is inverse -i.
I love how the title can have 2 meanings, like a pun
Either saying that the equation ⁱ√i≈4.81 shouldn't be real
Or that ⁱ√i shouldn't be approximately equal to 4.81, a real number
Oh wow you got the joke
Shut up@@antoniusnies-komponistpian2172
However, i does not exclusively equal e^ipi/2 it can equal any varient of e^(i(pi/+2pi*k)) where k is an integer
i can't believe my i's
It's Exp(1/i × Ln(i)), where Exp, Ln are complex to complex functions, by definition.
So, ...= Exp(-i × (ln 1 + iπ/2 +2πki))= Exp(π/2 + 2πk)= exp(π/2 + 2πk) for any whole k (positive, negative or zero).
You can't just apply rules from real exponentiation and logarithms to complex numbers, you should know how they transform. Watch how 2πki emerged when we see complex valued logarithm.
Bringing in the Ln is unnecessary, no? I mean without specifying how many revolutions the initial i is composed of we could just jump straight to i=exp(I pi/2+2i pi k), right?
There are also more values for +2kπ
It is not as simple as you claim. What do you mean by that root? A solution of the equation z^i = i? Well, if so, you will find that it has an infinite number of solutions. e^(pi/2) is only one of them.
Yes, and there are even infinite positive solutions 😅
1/(e^-pi/2) = e^pi/2
the negative exponent and reciprocal cancel.
*That's Surreal!* ❤
Great video BriTheMathGuy!
matt parker's i^i ~= a fifth really saved me here, as i guessed "about 5" and was right
You could say that i-root of i is e^pi/2
Bro just found out what the definition of a root is man.
Well, that's the principle answer. There are infinitely many random answers, all rotations away from each other.
I was going to say "isn't it just 1" but I'm a complete idiot that's the "i-th root" not "divided by i" bro I'm dumb
This problems makes it more interesting when you realise that it shows you the same solution of i^i is that one,when you multiply i^i with ⁱ√i, one being 1 when you multiply the real solutions of i and the other being also 1
Or you could simplify it further to exp(π/2)
Question: I did, i = exp(i*(pi/2 + 2*pi*N)), therefore i^(1/i) = exp(i*(pi/2 + 2*pi*N)*(1/i) = exp(pi/2 + 2*pi*N). when N = 0, i^(1/i) = exp(pi/2) ~ 4.8105 but I still see infinity number of answers determined by N. What is wrong with my answer ?
I totally luv how it loops
for the fraction 1/i, couldn’t you also have used exponent rules to just flip it and raise it to the negative first power?
Yes you can but does that really achieve anything
x^x^-1 is not the same as (x^x)^-1. You have to somehow use the fact that 1/i = -i to progress.
So i√i is e^(π/2)? interesting...
By Euler's identity of i = exp(i*pi/2), i^(1/i) = exp(2*pi)
A complex problem that doesn't seem complex.
I just did ln(i)=0+pi/2
That deserves a "Wow"!
Awesome!
I somehow didn't got the notification of your new video. I dunno why...
Hm. Maybe because it's a short? Who knows! Thanks for watching!
@@BriTheMathGuy Yup maybe
Something really bothers me. Isn't i (or √(-1)) when squared supposed to be √1 or 1? Because, when you multiply radicals that have the same indices, whether the bases are different or alike, the bases would just multiply with each other while retaining the current index. For instance,
√2 • √4 would be equal to √8
√12 • √3 would equate to √36, which is 6
√5 • √5 would be equal to √25, which is 5
Then why isn't √(-1) • √(-1) equal to √1, since (-1)(-1) is 1?
But √(-1) • √1 is equal to √(-1) or i, which follows the rule I've previously mentioned
can you give us an intuitive explanation of i^(1/i).
Thanks
It's wild that a number that doesn't even exist can source coherent math
The imaginary unity exist as much as any real number
But it does exist - the term "imaginary" is however unfortunate and misleading, albeit understandable for historical reasons.
I have a question, like there exists a set of real numbers and another of imaginary numbers so are there more set of numbers like a third dimension of numbers which would extend the complex plane
Kind of.
The complex plane allows an extension to the reals that allows us to obtain solutions to problems with negative discriminants.
From the little I know about Analytic continuation
en.wikipedia.org/wiki/Analytic_continuation
it essentially allows us to extend the domain of the complex plane to be able to solve more complex functions. Like the Reimann Zeta Function.
@@miguelcerna7406 thanks
You could also look at this from algebraic eyes and view it as just an extension field. And then you can find other crazy cool extension fields, like making each "number" a polynomial over a field, and then adding, dividing, multiplying polynomials like they're numbers.
You then get some very cool and interesting properties out of them.
en.m.wikipedia.org/wiki/Field_extension
Write i^(1/i) =e^(i*pi/2i)
How did I end up here, I barely know intervals...
That loop is slick 😎
There is something about these results that not even Richard Feynman could explain intuitively.
It is at this point in math where math makes as much sense as a UFO
simpler to just state that
i=e^(iπ/2)
so i^(1/i) = e^(π/2)=4.81
wait but if you raise that number to the _i_ power, it does not equal _i_
His result was that the i-th root of i equals 1 / e^( - pi / 2 ). Raising that to the i-th power gives:
{ 1 / e^( - pi / 2 ) } ^ i = { e^( pi / 2 ) } ^ i = e^( i pi / 2 ) = cos( pi / 2 ) + i sin( pi / 2 ) = 0 + i (1) = i.
It does.
This is amazing actually except in terms of functions √x ≠x½
So by logic, 4.81 ^ i should (roughly) equal i right?
I feel like we're missing an infinite family of solutions here
Lol
wow, this is the most perfectly loop i ever seen on youtube!!! literraly.
- Can I get i to 1 over i bananas?
- how many???
- little more than 4 but not 5
A rule says that the x rote of x equals x to the power of 1/x. That should be true with i too.
i over i
i to the i
An eye for an eye
Too many i's
.
.
Aye aye, captain!
Imagine that
Well, we do know e^iπ/2 is equal to i, so if you accept that i-th root makes sense, you shouldn't have trouble saying that (e^π/2)^i = i
huh, did you just record the last sentence and then take the second half and move it to the beginning of the video? in any case, nice editing :D
i=-1^(1/2)
e^(ipi)=-1
i=e^(ipi/2)
i^(1/i)=e^(ipi/2i)=e^(pi/2)
i^(1/i) = (e^(i*pi/2))^(1/i) = e^((i*pi/2)*(1/i)) = e^(pi/2)
How do you know what formula to use? I don’t get it
An I to the I leaves everybody blind.
In another math channel, this video would have taken 10+ minutes.
Probably because that other channel wouldn't have said, in the middle of the derivation, "refer to my other video where I do most of the work".
Is there a more visual proof? Like i*i rotatesa point by 90° so i^1/i also does something?
My head can't take this.
Aaaaa, won't you please not leave a number with a negative exponent in the denominator??? But converted it into decimal instead 😳
we can also get this by manipulating this equation: e^(I*pi) = -1
Hm, but what about the principal branch, that the square root symbol is defined as.
There are infinite solutions to this. i = exp(i{pi/2 + 2 pi k}) where k is any integer. Taking this to the power of -i gives us exp(pi/2 + 2 pi k) for all integers k.
i am confused
feels like you just punctured another universe
*i* have no idea what is going on
First Please dont transform e^-pi/2 to a decimal number Please just transform it to e^pi/2. And the cool kids don’t use i but j.
Eye to eye 👀
wow! Its so beautiful like ruler's equation of e^(i×pi)+1=0
I still have no idea how that works ngl
Damn, that’s real
If you asked me if I liked this videos, I'd just have to say "i i captain!"
aye aye*
@@apenasmeucanal5984 seems like you just missed the joke
I trust you that this works, but it still feels like a notation trick. I don't understand complex numbers enough to know if this is rigorous.
Rigorous, maybe not, but the answer is correct. For more rigor, one must understand that imaginary exponents are multi-valued functions and therefore, the answer is not _unique_ --- although all possible answers are real numbers.
Too many eyes. Too many eyes
This video was music to my i's.
Holy shoot it took me like 5 seconds to see that it looped damn that was seamless
Excellent video
Ok so i to the i of the i is i in the i of the i of i inside the i of the i?
But if you multiply this figure by itself (≈ 4.81), surely you would get a real number...?? How can one number be the square root of two distinct others?
Multiplying it by itself would square it. You need to multiply it by itself i times, whatever that might mean.
Squaring it would mean that you essentially took the i'th root of -1
@@hypnogri5457 Okay, this is so embarrassing that I'm tempted to delete my comment. But I will let it stand as a testament to the dangers of drinking tea instead of coffee. Thanks for the correction.
@@GlorifiedTruth Would you like a cup of coffee? ☕ :)
Eye to eye gives you a pie sometimes...with a girl
Very interesting that a calculation involving only the square root of -1 can give you over 4!
Omg it’s looped perfectly
Doesn't 2 perfectly match the equation tho?
e^(π/2)
i^i can be many things, not just e^{-π/2)
Points for that looping...
This problem reminds me of my selfish friend Susan. Who is always talking about herself. "i this i that i i i"
Short and Sweet!!!
is it rational or irrational?
Interesting question, but probably irrational since it has e and π and even a squareroot in it.
Yeah, this is my RUclips channel