What is i?
Рет қаралды 98,761
Күн бұрын0:00 Intro
0:09 i=sqrt(-1)
0:31 solving for i from the Euler's formula
5:42 solving for i involving Lambert W function
Lambert W function explained: • Lambert W Function (do...
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Thank you,
blackpenredpen
@blackpenredpen 3 жыл бұрын
Can u solve for z?
@nuctang 3 жыл бұрын
2nd solution?
@surplusvalue3271 3 жыл бұрын
i can solve but the comment section is too small for it , to fit in .
@hadijaffri9856 3 жыл бұрын
Cant we pur the 2nd equation equal to square root of negative 1 and solve for z? Just a speculation,any math veteran please correct me
@SalimGuntara 3 жыл бұрын
Hospital....where is the hospital???
@sewerynkaminski1116 3 жыл бұрын
what if z = 0?
@surplusvalue3271 3 жыл бұрын
Imagine some kid searching for alphabet 'i' and gets this.
@blackpenredpen 3 жыл бұрын
I don't take any responsibilities for that lol
@JakubS 3 жыл бұрын
mathematics acceleration
@itsiwhatitsi 3 жыл бұрын
The kid will think there are a lot more letter to learn ... reading is hard bruh
@GarGlingT 3 жыл бұрын
We use j my friend.
@rzno3414 3 жыл бұрын
@@mathevengers1131 just email him if you really want him to do something
@nishatiwari9212 3 жыл бұрын
5:15 cos^2 z -1 is always negative and we have sqrt negative which involves z. So "i" is still on both sides, just in disguise.
@antoniomora1621 Жыл бұрын
cos^2(z) can be greater than 1 for certain complex z, so cos^2(z)-1 is not always negative.
@michlop452 Жыл бұрын
@@antoniomora1621 Yeah but there's also a cos(z) term on the outside AND a 1/z term on the outsider-outside, so theres bound to be something complex in there.
@jemandanderes7075 4 ай бұрын
For z = 2nπ cos(z)^2 = 1, so the whole term equals 0, did I do something wrong?
@sulaimanatreshe5559 3 жыл бұрын
Math in the 20th century: Galileo and Newton are wrong. Math in 2021: .._.. prank your friend 5:10
@peggyfranzen6159 3 жыл бұрын
All electrical engineering individuals know this.Thank you.
@ZakEinc 3 жыл бұрын
thats awsome!
@mathevengers1131 3 жыл бұрын
*GOLDEN EQUATION* Sir Steve Chow please read this comment for a golden equation. Your videos are very amazing. I have derived a very beautiful equation in which there are all amazing things like Phi,π,e,i and even Fibonacci series(All five in one equation). I request you to please make a video on it. The beautiful golden equation is difficult to type here but still I will try to type. It is as follow: ϕ(ϕ^(e^(πi)-n)+((-1)^n)(n+1)th term of Fibonacci series) = +((-1)^n)(n+2)th term of Fibonacci series where n is a non-negative integer. You can check it on calculator it will work. If you want proof of it then reply my comment and I will give my number and I can easily explain it's proof on a call as it will be difficult to explain it in comment. Thank you for reading this comment. I hope you will make a video on it.
@utkarshsharma9563 3 жыл бұрын
@@mathevengers1131 dude comment that on the video I don't think he'll be able to notice this in a reply, so I think you should post thus as a comment, not a reply
@sabouedcleek611 3 жыл бұрын
@@mathevengers1131 hey that's pretty neat Using the Fibonacci series' approximation to phi as a connection, and then making one of the -1 into e^i*pi so that they could be all connected together For those interested, a pastebin link of the results up to 100: pastebin [dot] com/Z3WGwf2T Some further reading: www [dot] goldennumber [dot] net/powers-of-phi/ www [dot] maths [dot] surrey [dot] ac [dot] uk/hosted-sites/R.Knott/Fibonacci/propsOfPhi.html#section1.5 (KZfaq is unfriendly to links, oh well)
@blackpenredpen 3 жыл бұрын
5:46 the 3rd way
@user-nv4id1hq2t 3 жыл бұрын
@Ori Bandel dead fish
@hhht7672 3 жыл бұрын
Wouldn't it be possible to generalize the answer like this, i.imgur.com/A7Xc2YZ.jpg ? Apologies for the handwriting and the inconsistencies in how I write the number "2" in advance :,,)
@user-nv4id1hq2t 3 жыл бұрын
@Ori Bandel mine was a reference to LambertW
@ISoldBinLadensViagraOnEbay 7 ай бұрын
I actually have a 4th way e^(i兀/2)=i e^x=lim n →♾[(1+x/n)^n] So i=lim n →♾[(1+i兀／2n)^n]
@ISoldBinLadensViagraOnEbay 5 ай бұрын
Yeah ignore that comment
@reinerwilhelms-tricarico344 3 жыл бұрын
I bet Euler did this kind of tinkering with math for 12 hours every day. 😂
@intellecta2686 3 жыл бұрын
Hahahah intro is too good 👍👍👍
@blackpenredpen 3 жыл бұрын
Thanks!
@Happy_Abe 3 жыл бұрын
Can you explain
@Dionisi0 3 жыл бұрын
2nd answer is wrong, you got a nested i in the radical
@blackpenredpen 3 жыл бұрын
(I know but shhh 🤫)
@jagatiello6900 3 жыл бұрын
i...won't tell
@aizek0827 3 жыл бұрын
What do you mean there's nested i in the radical?? Pls i need to learn
@TACCOFSX 3 жыл бұрын
@@aizek0827 i think the radical is only either a negative number or 0
@vinlebo88 3 жыл бұрын
@@aizek0827 Because √(cos²(z)-1) = √(-sin²(z)) = isin(z)
@SlidellRobotics 3 жыл бұрын
For "second" form, you can simplify cos²Z-1 = -sin²Z, which of course brings out another √-1.
@nemanjalazarevic9249 7 ай бұрын
and then you get e^iz = cos(z) + isin(z)
@tobyzxcd 3 жыл бұрын
3blue1brown’s lockdown series was I think a great introduction to i- one of the ways of thinking of i that stuck with me is a 90 degree rotation in the complex plane
@peggyfranzen6159 3 жыл бұрын
Can't disagree with trigonometric equations!.
@JakubS 3 жыл бұрын
I found 2blue1brown's video really difficult to wrap your head around compared to what I've learned from blackpenredpen
@michalbotor 3 жыл бұрын
the funniest thing about it is that it is a cricular reasoning. ;) in order to explain yourself what i is you need a complex plane, but in order to talk about complex plane you need i...
@tobyzxcd 3 жыл бұрын
@@michalbotor well complex numbers are great for circular reasoning ;-)
@hybmnzz2658 3 жыл бұрын
@@michalbotor he built the complex plane first then explained what multiplication by i does. Besides, one way that you can define complex numbers is as matrices that act on R^2. The field of complex numbers is isomorphic to a certain type of matrices if you want to be fancy.
@muneebahmad5818 3 жыл бұрын
The fact that you could have simply written e^iz=cosz+i(1-cos²z)½ =cosz+(cos²z-1)½ i=1/z ln(cosz+(cos²z-1)½) Without actually solving the quadratic is pretty amazing!
@blackpenredpen 3 жыл бұрын
What is i for u?
@integralboi2900 3 жыл бұрын
The complex number such that |z|=1 and arg(z)=pi/2.
@alexcwagner 3 жыл бұрын
u is blackpenredpen!
@Demki 3 жыл бұрын
For me, i=x+(x^2+1)∈ℝ[x]/(x^2+1), or given another ring C with ring isomorphism φ : ℝ[x]/(x^2+1) → C, i=φ(x+(x^2+1)). Or equivalently, given any ring C with ring epimorphism η : ℝ[x] → C such that ker η = (x^2+1), i=η(x). (here (x^2+1) is the two-sided ideal generated by x^2+1)
@leadnitrate2194 3 жыл бұрын
I am me
@vedants.vispute77 3 жыл бұрын
Height
@romanbykov5922 3 жыл бұрын
This actually doesn't make any sense. Also in the 2nd answer you would almost always have a negative under the sqrt. Or a zero (in the case that z = pi*n). This means you still have an 'i' on both sides. What a nice property of 'i' -- no matter what you do, it won't go :)
@blackpenredpen 3 жыл бұрын
😆 Yea I am aware of that and that’s why i said it’s math for fun and i mean it in the beginning. I actually just wanted to do the 3rd one but it would have been too short for a regular video.
@ledouble7337 3 жыл бұрын
@@blackpenredpen Perhaps you've said it was "maths for fun" but fun in maths doesn't mean "calculus for fun with circular reasonnings twice in a row..." So ok if you want to do random maths but I would rather say "calculus for fun" and précise why this is random maths. Because I'm not so sure that every viewer understood that was totally random I would like to precise that writing i=sqrt(-1) can make confusion and you should precise that sqrt(-1) = *+ or -* i So I know it's possible to write i=sqrt(-1) but you should say that is just to make understand i or that if you really want to use this notation rather i it will bring you to error if you don't think about that sqrt(zz') =/= sqrtz * sqrtz'
@ledouble7337 3 жыл бұрын
That is not thé only non sens. How to justify that cosz + isinz is an exponential function without knowing i^2=-1? I don't see so at the beginning It's already circular reasonnig.
@ujueije5762 3 жыл бұрын
He just copying solution problems from Qora or Stack Exchange websites. or Mathematica
@ledouble7337 3 жыл бұрын
@@ujueije5762 It's possible, I'm french I don't know how do you do maths in the us but I think he should do an erratum For "fun" for sure. He can say : OK that was fun but, now, where are the circulars reasonnings ? Or something like that. That's my point of view.
@jonathasdavid9902 3 жыл бұрын
alright, alright the next philosophical question is: Who am √-1 ?
@idolevi612 3 жыл бұрын
The answer is Jonathans David
@antman7673 3 жыл бұрын
Only Eulers Identity can answer that question.
@theblackphilosopher5958 3 жыл бұрын
@@antman7673 brilliant
@trueriver1950 3 жыл бұрын
@@antman7673 Euler's i-dentity
@jonathasdavid9902 3 жыл бұрын
@@idolevi612 😂 c'mon it's just a name.
@stlemur 3 жыл бұрын
For that derivation could you not just start with Euler's identity, take the log of both sides, rearrange to isolate i and then use the Pythagorean identity to rewrite sin z?
@blackpenredpen 3 жыл бұрын
Nice. I didn’t see that. My approach was similar to how I did my previous videos. 😆
@stlemur 3 жыл бұрын
@@blackpenredpen The statement with ln(cos x + i sin x) is Cotes' formula, he came VERY close to Euler's form but didn't make the necessary connection to polar coordinates.
@aidancheung7264 3 жыл бұрын
So in fact, actually we can derive i with any real numbers rather than 3. Tho we have to take care of the positive and negative signs.
@vano__ 3 жыл бұрын
Okay I just noticed that you could just take the euler's formula change isin(z) to √(-1)*√(1-cos²z) which is √(cos²z-1) and its a tremendous shortcut
@Mothuzad 3 жыл бұрын
I came here to say the same. I noticed the difference of 1 - cos squared z is just -sin squared z, and it brings you right back to where you started. These are all basically 1=1 equations, of course, so they can be arbitrarily simplified or complicated.
@danielbenton5817 3 жыл бұрын
Haha I came here to say the same, it rlly annoyed me that he didn't simplify the expression until I realised it brought u right back where u started
@vano__ 3 жыл бұрын
@@danielbenton5817 saaame bro, just can't let it through without even an attempt of simplification alright
@runonwards9290 3 жыл бұрын
The point is he derived it without assuming i= sqrt(-1)
@vano__ 3 жыл бұрын
@@runonwards9290 so how do you know i^2=-1? It's the definition for i, you can't act as it's not here, its just a formula that equals i for any input z
@Nebula_ya 3 жыл бұрын
With the first formula, cos^2 (z) will always be less than or equal to one. When it's less than 1, we have a a square root of a negative number (aka i on both sides again). When it's equal to one, you get 1/z * ln(0), which also doesn't work since ln(0) is undefined 😭
@reidpattis3127 3 жыл бұрын
It’s an old joke, but Here goes: i don’t want to be on the bottom
@Barocalypse 3 жыл бұрын
i like to be on the top
@regularguy9264 3 жыл бұрын
I typically only ever define i squared, but I don't wish to be negative about your video. Thumbs up.
@route66math77 3 жыл бұрын
Haha this is great, I wish there was more room for this sense of playfulness and adventure in the high schools. I feel like we could turn more people onto mathematics.
@SHASHANKRUSTAGII 3 жыл бұрын
I dont know why I watch your videos, but I love them all.
@hybmnzz2658 3 жыл бұрын
I thought this was going to be about explaining why we should say i^2=-1 instead of using the square root in the definition. Square roots are not nice because (-i)^2 is also -1.
@pythoncake2708 3 жыл бұрын
same
@alejrandom6592 3 жыл бұрын
Same here
@alejrandom6592 3 жыл бұрын
Sqrts aren't defined consistently on the complex world
@yoavboaz1078 3 жыл бұрын
the lambert W function and logs have the same problems
@Theraot 3 жыл бұрын
Not only that, but j^2 = -1 and k^2 = -1 which is how you get quaternions (ok, you also need ijk = -1).
@sw3aterCS_ 3 жыл бұрын
The element x in the quotient ring R[x]/(x²+1).
@cygntauri2348 3 жыл бұрын
your content is so underrated
@fabianoromanelli9769 3 жыл бұрын
In the 2nd way, you can sostitute square of cos^2 - 1 with square of ( - sinz )^2 that is equal to i*sinz...and if you don't want i on the right side, I think you have to put the condition: z=2kπ with k of the set of integers, so k=-2, -1, 0, 1, 2, 3...
@jordanberman6194 3 жыл бұрын
Wow I really enjoyed watching this video! 😍 Great job ☺️
@peggyfranzen6159 2 жыл бұрын
This guy is pretty cool.
@ozzymandius666 3 жыл бұрын
As a physics man, I hate the W function. Non-analytical functions disturb me at a deep level.
@jamiewalker329 3 жыл бұрын
i is just the co-ordinate (0,1), where co-ordinates are multiplied according to the operation (a, b) * (c, d) = (ac - bd, ad + bc). Thus i*i = (0, 1)*(0, 1) = (-1, 0) = -1 if we decide to write (1,0) = 1.
@tonaxysam 3 жыл бұрын
R^2 is a lot of things, is the complex world, is the vector's world, it's a plane, etc
@jack_papel 3 жыл бұрын
The equation at the bottom of the second method is basically euler's formula. cos²z-1=-sin²z. Rearranging you get e^{iz}=cosz+isinz
@cal9994 3 жыл бұрын
Hey! @blackpenredpen :) Could You Please!! do a video on solving infinitely nested square roots but with increasing powers? And potentially differentiating and integrating them or something? like X*sqrt(X^2*sqrt(X^3*sqrt(X^4..... thank you!! :)) Also, I love what you do on this channel, you're amazing :)
@inchoesce 3 жыл бұрын
"Z can be on the bottom, he likes to be on the bottom anyway... Well I don't know - I have not talked to him for a while." LMAO 😂
@ethanchandler3934 3 жыл бұрын
Sqrt(-1) love your videos
@alkaverma5974 3 жыл бұрын
You know I also thought the same in past but at that time I don't know how to squeeze "i" from euler's equation. Now I know thanks ❤️❤️❤️
@amaymnair4558 3 жыл бұрын
You explains everything It was like I am having a Big Mac .
@miguelalvarez5905 3 жыл бұрын
You could skip the algebra (2nd) by simply ln both sides on any of the +- versions of Euler's formula and then divide by z.
@stancombs4168 5 ай бұрын
In method #2 next to last step: you have sqrt(cos^2(z) - 1) which is equal to sqrt(-sin^2(z)) which is equal i*sin z. So this step is just e^iz=cos(z) +i*sin(z) which is Eulers identity where you started! Beautiful circular calculation.
@sangeetanayak9589 3 жыл бұрын
I'm so glad I saw this!!!, 😄
@amirparsi4165 3 жыл бұрын
I prefer this: “i” is the second dimension of numbers.
@theflaminglionhotlionfox2140 3 жыл бұрын
No i is in the second dimension of numbers, however it is not the whole thing.
@user-nv4id1hq2t 3 жыл бұрын
hey. 5:35 kiddin us? cos2(z) < 1 => cos2(z)-1 so sqrt(cos2(z)-1) already contains i itself.
@blackpenredpen 3 жыл бұрын
Shhh, we have to keep that as a secret.
@abhipriyeshukla5431 3 жыл бұрын
Wow!!! I love this
@peggyfranzen6159 2 жыл бұрын
It's only mathematics.New math was, well, actually not too bad, it needed an explanation.Ok, moving on then....
@RoMaths 3 жыл бұрын
Nice background music.... Huge appreciation from India.... 🇮🇳🇮🇳🇮🇳
@swift3564 3 жыл бұрын
2:30 if you divide by 2 on both sides, you get something that looks like cosh(x)
@LotusPrincess69 3 жыл бұрын
woah i was just wondering about this and you uploaded about it @-@
@artemqqq7153 3 жыл бұрын
Instead of using natural log, you could just transform your exp(iz) in the left side to cos + i*sin, cos would go away, and you will have I*sin(z) = sqrt(-1)*sin(z), so i=sqrt(-1)
@debtanaysarkar9744 3 жыл бұрын
He is a genius.
@roeemilgrom3720 3 жыл бұрын
i is happy :D
@NonTwinBrothers 3 жыл бұрын
Love those kinda intros
@tuserkampung3522 3 жыл бұрын
i like this video very entertaining me.👍👍
@gtziavelis 3 жыл бұрын
Aye aye, captain!
@flleaf 3 жыл бұрын
can you make video on the fact that you can make 1 to equal -1?
@jlaurenpiano 3 жыл бұрын
For the second answer, if you use the Pythagorean trig identities to simplify √(cos²z-1) you get √(-sin²z) which is just i*sin(z) so you're right back where you started... though the plus or minus is interesting
@forklift1712 3 жыл бұрын
Quite Easily Done!
@michelecastellani1 3 жыл бұрын
I think that in the second one, i is still present in the right part on the equation; infact in the square root there is -sin^2(z) than can be rewrite as sin(z)•i
@screeni4605 3 жыл бұрын
teach me how to be this happy
@blackpenredpen 3 жыл бұрын
Just focus on doing what you love to do.
@mathevengers1131 3 жыл бұрын
@@blackpenredpen *GOLDEN EQUATION* Sir Steve Chow please read this comment for a golden equation. Your videos are very amazing. I have derived a very beautiful equation in which there are all amazing things like Phi,π,e,i and even Fibonacci series(All five in one equation). I request you to please make a video on it. The beautiful golden equation is difficult to type here but still I will try to type. It is as follow: ϕ(ϕ^(e^(πi)-n)+((-1)^n)(n+1)th term of Fibonacci series) = +((-1)^n)(n+2)th term of Fibonacci series where n is a non-negative integer. You can check it on calculator it will work. If you want proof of it then reply my comment and I will give my number and I can easily explain it's proof on a call as it will be difficult to explain it in comment. Thank you for reading this comment. I hope you will make a video on it.
@DeadJDona 3 жыл бұрын
5:00 so 1 under sqrt is cos² + sin², and after substitution you get back to √-sin² => cos + i*sin
@marijul9287 3 жыл бұрын
Could you do examples of Gram-Schmidt process please?
@harshitarora2005 3 жыл бұрын
i=1/n log ( cos n + root cos ^2 n - 1 )
@peggyfranzen6159 2 жыл бұрын
Every electrical engineering student knows.Hey, thank you.Can you do more on complex numbers?
@DarwinSPPD 3 жыл бұрын
Another way would be expressing constant i as a function of F where F is a multivalued solution of equation x^2 = -1 .
@mastermind1258 2 жыл бұрын
Nice :)
@shreyan1362 3 жыл бұрын
Please continue this hahaha series 😭😭😭😭
@bullinmd 3 жыл бұрын
How do you check to ensure the calculations for i are correct?
@AyushKumar-ng6vk 3 жыл бұрын
It's cos^z-1
@ahmjamil0 3 жыл бұрын
Very funny !
@protocol6 3 жыл бұрын
It's also related to the angle of the elliptical equivalent of the asymptote of a hyperobla. It's i²=cot(-45°), if I remember correctly.This definition only becomes important when you start thinking about values under the radical other than -1 where that angle will vary. In physical terms you can say it's the square root of the negative relation between the units of a and b or i²=-ε(a)/ε(b) for a natural unit system where the ratio is 1. Negating swaps the units and values of a and r so you could also say i²=ε(r)/ε(b) but only if you work around the loss of definite sign when taking the magnitude r=‖a+ib‖. For non-natural unit systems, you'll end up with other values under the radical besides -1 or 1, of course. For instance, you can define special relativistic space-time as d𝜏=‖dt+dx√(1/c)‖ or dt=‖d𝜏+dx√(-1/c)‖ where dt is change in coordinate time, d𝜏 is change in proper time, dx is change in position, and c is the speed of light which relates distances to coordinate time intervals. You can define trig functions for any value under the radical (or parameterized functions) which then makes it computationally efficient to use this form with any given unit system or related pair of units.
@peggyfranzen6159 2 жыл бұрын
Now that's interesting.
@arkay238 3 жыл бұрын
You can get to the second form from Euler’s formula by immediately replacing i*sin(x) with sqrt(-1)*sqrt(1-cos^2(x))
@gary0617 3 жыл бұрын
好棒的推導過程
@blackpenredpen 3 жыл бұрын
李翔老師好！過獎了
@shivamchouhan5077 3 жыл бұрын
I don't understand, please can anyone tell what you are talking about
@gary0617 3 жыл бұрын
@@shivamchouhan5077 Me:Great derivation process!! He:Overrated!! Welcome to Taiwan.^^
@shivamchouhan5077 3 жыл бұрын
@@gary0617 oh thank
@AKABILASETOFICIAL 3 жыл бұрын
If put the 1 that is the only integer, let 1 =n, put in GeoGebra, when n tends to 1, is like the graph is just turning into the proper y axis!
@khanhnguyennam6007 3 жыл бұрын
The first one u had cos^2(z) -1 in the root, which can be interpreted as sin^2(z) so after the square root it’s sin(z)
@icedragonroyal4550 3 жыл бұрын
tell about hyper factorials as a product of super factorials
@RomainPuech 3 жыл бұрын
But if you use z=2pi in the second formula you have i=0... I think it's because the coefficient of the quadratic equation is equal to 0, so maybe to be rigorous we should add "for all e^(iz) with z != pi/2+2kpi k integer"
@raifegeozay687 2 жыл бұрын
this is not the problem actually, here is a simpler one: e^2i*pi=1(eulers formula) take the In both sides 2pi*i=0 which is wrong because neither of factors are equal to 0. the mistake here (and also the same mistake you did) is amussuming the function e^z is one-to-one (true for real numbers but not for complex numbers) the thing i mean is if e^a=e^b, you cant say a=b in complex world (but you can in real world) because if b=a+2n*pi*i(n is integer), e^b =e^(a+2n*i*pi) =e^a*e^(2n*i*pi) =e^a (because e^(2n*i*pi)=1 by eulers formula)
@jimbeasley5319 3 жыл бұрын
It would have been interestng to see the first solution reduced further. I would like to see how you reduce ln(cos z +/- sin z *sqrt(-1))/2 = sqrt(-1)
@agabe_8989 3 жыл бұрын
Friend: What is i? Me: duh. i is you.
@alexcwagner 3 жыл бұрын
what am i?
@blackpenredpen 3 жыл бұрын
U r Alex!
@amitvishwakarma7210 3 жыл бұрын
I thought he is a man...!😄
@shivamchouhan5077 3 жыл бұрын
@@blackpenredpen lol
@shivamchouhan5077 3 жыл бұрын
@@mathevengers1131 wow that's nice, can you give a proof
@mathevengers1131 3 жыл бұрын
@@shivamchouhan5077 I will give proof when bprp will read it. I am trying to contact him by sending the same comment from last 20 videos on both of his channel but he is not reading my comment. I hope he will read it.
@bagochips1208 3 жыл бұрын
i see what you did there
@azerterty1081 3 жыл бұрын
I don't understand why you added the number 1 to the fourth line? thank you !
@marcelolage1395 3 жыл бұрын
The second yiou just do some calculations and arrive at where you began. If you just substituted isinz for sqrt(cosz^2-1) it would be finished
@mikeschmit7125 3 жыл бұрын
wait but the third one isn't yielding the desired answer. W(-pi/2) = ipi/2, thus... (-2/pi)(W(-pi/2)) = (-2/pi)(ipi/2) = -i You can check on wolfram alpha by entering the original equation, it also gives the answer as -i. What's up with that?
@noahali-origamiandmore2050 2 жыл бұрын
This is a very interesting point that you brought up. Nothing done in the video is wrong. There is a way to solve for i and get (2/pi)(W(-pi/2)). However, it is important to keep in mind that the Lambert W function has multiple branches. To write a specific branch, you write the base as a subscript of W (similar to how you write a base of a logarithm). With no base, the principal base is assumed, which is base 0. In WolframAlpha, you type "productlog(base, value)." (2/pi)(productlog(0, -pi/2)) = i, but (2/pi)(productlog(-1, -pi/2)) is also i. Put this into WolframAlpha.
@Grassmpl 3 жыл бұрын
i can be either one of two numbers. No way to tell the difference since they can map to each other via an element in Galois group over the reals.
@mr_lim2009 2 жыл бұрын
My 2 other ways to define i 1) i = ln(-1)÷π 2) i = arcos((e+1/e)÷2) Edit: 3) i = arcos(x)÷ln(sqrt(1+x)-x), for all values of x
@mikeschieffer2644 3 жыл бұрын
So when z = 0 then i is undefined?
@sauravpradhan2349 3 жыл бұрын
In 2nd : The final answer will still get 'i'. As we know, sin^2(Z) = 1 - Cos^2(Z) . Then multiplying both sides by minus sign then.. -sin^2(Z) = Cos^2(Z) -1 . Then, at right hand side of our equation after (+ - ) we will get √-sin^2(Z) which is equal to √-1 or 'i' x sin(z).
@Ironmonk036 3 жыл бұрын
Yo BlackPen/ You know that Formula Board in the background? Where can I order one of those boards?
@blackpenredpen 3 жыл бұрын
Yes. It’s in my Teespring shop. Link in description. Thanks.
@JaydentheMathGuy 3 жыл бұрын
I really thought he was gonna say something like 1 + 1 = 2 and 2 = 1 + 1 at the beginning. i = sqrt -1 and sqrt -1 = i
@blackpenredpen 3 жыл бұрын
Hahaha thanks I know the reference
@AbdulAhad-gc7hc 3 жыл бұрын
For the second approach where he did e^iz =.... MINUS e^i(-z)=..., what gave him the reason to take the difference of the two? Like hw would u no that u start off by doing that?
@peggyfranzen6159 2 жыл бұрын
More about the Taylor series.Thank you.
@rainbowbloom575 3 жыл бұрын
2021: Who am i? 2051: what is i
@zush1 3 жыл бұрын
Isnt there an i on the right anyway as cos^2 is smaller than 1 almost always?
@jamesmclaughlinprimitivele4587 5 ай бұрын
So much clearer now lol
@Maou3 3 жыл бұрын
sqrt(cos²z - 1) is i × sinz So its just like taking ln at the start then moving over the 1/z
@liab-qc5sk 3 жыл бұрын
Real number be like: Nothing beats me but that thing X^2 +1 =0 Is scaring
@KnufWons 3 жыл бұрын
The equation i=(1/z)ln(cos(z) +- sqrt(cos^2(z) -1)) is mathematically identical to Euler’s equation, because cos^2 -1 is equal to -sin^2. A more direct way of deriving that equation would be to take the natural log of cosz + isinz and divide it by z. To get rid of the i on the left, simply square isinz under a square root et voila, sqrt(-sin^2(z)), which in turn is equal to sqrt(cos^2(z) -1).
@tomatrix7525 3 жыл бұрын
Bprp, you should do a draw my life soemtime. Alot of people would be interested
@antoinetarant6025 3 жыл бұрын
simplify with cos²x-1=-sin²x to cancel the square root ?
@aliberro 3 жыл бұрын
The limit as z->0+ of the 2nd form is interesting
@kadirkusmez7724 Жыл бұрын
In the second definition of “i” what is the value of “z” ?
@Zappler3000 3 жыл бұрын
whats the Use of this identy for i?
@davidgillies620 3 жыл бұрын
In the second formulation you can just note that cos^2 z + sin^2 z = 1 so (i sin z)^2 = cos^2 z - 1
@canalMatUem 3 жыл бұрын
O que é a função W?
@dr.rahulgupta7573 3 жыл бұрын
Excellent !! DrRahul Rohtak Haryana India