*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.
@@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
@@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 (TH-cam is unfriendly to links, oh well)
@@antoniomora4537 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.
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 :,,)
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)
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 :)
😆 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.
@@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'
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 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.
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
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...
@@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.
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!
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 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.
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.
@@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
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 😭
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.
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.
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...
@@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.
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 :)
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?
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.
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.
@@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.
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?
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.
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)
sqrt(cos^2z-1) is just sqrt(-sin^2z), which is just isin(z). So the final answer is just 1/z*ln(cos(z)+/-isin(z)). You could save a lot of trouble and just take the logarithm directly of e^(iz) and dividing by z. Or have I missed something?
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
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
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"
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)
Can u solve for z?
2nd solution?
i can solve but the comment section is too small for it , to fit in .
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
Hospital....where is the hospital???
what if z = 0?
Imagine some kid searching for alphabet 'i' and gets this.
I don't take any responsibilities for that lol
mathematics acceleration
The kid will think there are a lot more letter to learn ... reading is hard bruh
We use j my friend.
@@mathevengers1131 just email him if you really want him to do something
Math in the 20th century: Galileo and Newton are wrong.
Math in 2021: .._.. prank your friend 5:10
All electrical engineering individuals know this.Thank you.
thats awsome!
*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.
@@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
@@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
(TH-cam is unfriendly to links, oh well)
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.
cos^2(z) can be greater than 1 for certain complex z, so cos^2(z)-1 is not always negative.
@@antoniomora4537 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.
For z = 2nπ cos(z)^2 = 1, so the whole term equals 0, did I do something wrong?
5:46 the 3rd way
@Ori Bandel dead fish
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 :,,)
@Ori Bandel
mine was a reference to LambertW
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]
Yeah ignore that comment
I bet Euler did this kind of tinkering with math for 12 hours every day. 😂
Hahahah intro is too good 👍👍👍
Thanks!
Can you explain
What is i for u?
The complex number such that |z|=1 and arg(z)=pi/2.
u is blackpenredpen!
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)
I am me
Height
2nd answer is wrong, you got a nested i in the radical
(I know but shhh 🤫)
i...won't tell
What do you mean there's nested i in the radical?? Pls i need to learn
@@aizek0827 i think the radical is only either a negative number or 0
@@aizek0827 Because √(cos²(z)-1) = √(-sin²(z)) = isin(z)
For "second" form, you can simplify cos²Z-1 = -sin²Z, which of course brings out another √-1.
and then you get e^iz = cos(z) + isin(z)
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 :)
😆 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.
@@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'
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.
He just copying solution problems from Qora or Stack Exchange websites. or Mathematica
@@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.
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
Can't disagree with trigonometric equations!.
I found 2blue1brown's video really difficult to wrap your head around compared to what I've learned from blackpenredpen
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...
@@michalbotor well complex numbers are great for circular reasoning ;-)
@@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.
alright, alright the next philosophical question is: Who am √-1 ?
The answer is Jonathans David
Only Eulers Identity can answer that question.
@@antman7673 brilliant
@@antman7673
Euler's i-dentity
@@idolevi612 😂 c'mon it's just a name.
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!
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?
Nice. I didn’t see that. My approach was similar to how I did my previous videos. 😆
@@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.
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.
It’s an old joke, but Here goes:
i don’t want to be on the bottom
i like to be on the top
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
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.
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
@@danielbenton5817 saaame bro, just can't let it through without even an attempt of simplification alright
The point is he derived it without assuming i= sqrt(-1)
@@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
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 😭
"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 😂
I dont know why I watch your videos, but I love them all.
I typically only ever define i squared, but I don't wish to be negative about your video. Thumbs up.
You explains everything
It was like I am having a Big Mac .
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.
same
Same here
Sqrts aren't defined consistently on the complex world
the lambert W function and logs have the same problems
Not only that, but j^2 = -1 and k^2 = -1 which is how you get quaternions (ok, you also need ijk = -1).
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.
your content is so underrated
I prefer this:
“i” is the second dimension of numbers.
No i is in the second dimension of numbers, however it is not the whole thing.
The element x in the quotient ring R[x]/(x²+1).
hey.
5:35 kiddin us?
cos2(z) < 1 => cos2(z)-1 so sqrt(cos2(z)-1) already contains i itself.
Shhh, we have to keep that as a secret.
Sqrt(-1) love your videos
As a physics man, I hate the W function. Non-analytical functions disturb me at a deep level.
I'm so glad I saw this!!!, 😄
How do you check to ensure the calculations for i are correct?
Wow I really enjoyed watching this video! 😍 Great job ☺️
This guy is pretty cool.
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...
Yo BlackPen/ You know that Formula Board in the background? Where can I order one of those boards?
Yes. It’s in my Teespring shop. Link in description. Thanks.
Nice background music....
Huge appreciation from India.... 🇮🇳🇮🇳🇮🇳
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.
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 ❤️❤️❤️
So when z = 0 then i is undefined?
The equation at the bottom of the second method is basically euler's formula. cos²z-1=-sin²z. Rearranging you get e^{iz}=cosz+isinz
can you make video on the fact that you can make 1 to equal -1?
He is a genius.
Could you do examples of Gram-Schmidt process please?
2:30 if you divide by 2 on both sides, you get something that looks like cosh(x)
whats the Use of this identy for i?
teach me how to be this happy
Just focus on doing what you love to do.
@@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.
i is happy :D
woah i was just wondering about this and you uploaded about it @-@
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 :)
Wow!!! I love this
It's only mathematics.New math was, well, actually not too bad, it needed an explanation.Ok, moving on then....
O que é a função W?
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?
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.
R^2 is a lot of things, is the complex world, is the vector's world, it's a plane, etc
Isnt there an i on the right anyway as cos^2 is smaller than 1 almost always?
Love those kinda intros
simplify with cos²x-1=-sin²x to cancel the square root ?
In the second definition of “i” what is the value of “z” ?
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.
For the second way, what if z=0?
Quite Easily Done!
what am i?
U r Alex!
I thought he is a man...!😄
@@blackpenredpen lol
@@mathevengers1131 wow that's nice, can you give a proof
@@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.
i like this video very entertaining me.👍👍
hyperbolically, if j^2=+1 and e^(jz)=cosh(z)+j⋅sinh(z), then:
what is j?
I think it's Jay
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?
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.
5:20 choosing z=2pi makes it evaluate to 0. Why?
Wolframalpha evaluates the 3rd solution as -i
What could be the issue?
i see what you did there
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)
I don't understand why you added the number 1 to the fourth line? thank you !
Aye aye, captain!
Hey there is a mistake in final result domain of cos theta is not satisfied
Friend: What is i?
Me: duh. i is you.
Wait if go with 2pi in the 2nd i will get i=0, isn't that wrong?
I don’t think the 2nd equation is right. Unless there’s a certain z value that makes it true
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))
Every electrical engineering student knows.Hey, thank you.Can you do more on complex numbers?
sqrt(cos^2z-1) is just sqrt(-sin^2z), which is just isin(z). So the final answer is just 1/z*ln(cos(z)+/-isin(z)). You could save a lot of trouble and just take the logarithm directly of e^(iz) and dividing by z. Or have I missed something?
tell about hyper factorials as a product of super factorials
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
5:00 so 1 under sqrt is cos² + sin², and after substitution you get back to √-sin² => cos + i*sin
Another way would be expressing constant i as a function of F where F is a multivalued solution of equation x^2 = -1 .
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
How about i=ln(-1)/π ?
i=sin/sn
Right?
I expect a Pokemon to come out of that pokeball and teach us math.
Nice :)
You:
What is *I* ?
My English teacher:
New grammar level here
Please continue this hahaha series 😭😭😭😭
It's cos^z-1
Very funny !
e^iz×e^-iz=1 how is this can anyone explain please?
What Euclidean geometry say about his fifth postulate?
Is it true if i say, i is not sqrt(-1), but i²=-1 ??
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
Hahaha thanks I know the reference
Can you prove that i is constant?
2021: Who am i?
2051: what is i
@blackpenredpen but "what is square root of two?"
So i can be any z?
"i am equal to" was a good joke.
Have you ever Read " Rene Descartes?" Now that's also brillance!
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"
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)
But i=e^i(π/2+2πn) 😅 so we could add infinitely many solutions to the 3rd way isn't it ?