India: We use colors as variables. Arabia: Well, we don't really want to mess with different pigments while doing math. We're just going to use letters. Europe: The letters are not Christian enough, we will use our own latin and greek letters. Blackpenredpen: 🙂
@2C (02) Chan Kwan Yu This formula will give you principal solution. If you want other solutions you can add 2πn. It will give you infinitely many solutions.
@2C (02) Chan Kwan Yu actually I am sending this same comment from last 20 videos so bprp will read but I think km he's not able to notice this comment so many other comments. I hope he reads this comment.
actually, you can simply extract the +/- sign outside the ln. ln(x - sqrt(x^2 - 1)) = ln[(x - sqrt(x^2 - 1))(x + sqrt(x^2 - 1)) / (x + sqrt(x^2 - 1))] = ln[(x^2 - (x^2 - 1))/(x + sqrt(x^2 - 1))] = ln[1/(x + sqrt(x^2 - 1))] = - ln(x + sqrt(x^2 - 1)). So the +/- can be extracted out of the ln. So in the end it would be +/- arccosh(x).
We don't allow trig functions in the simplification... Otherwise the answer to the whole problem can be trivially reduced at step 2 to "z = arcsin(pi/2+2npi)"
This is completely unrelated, but I was trying to figure out transistors earlier today since one of the bonus problems in my principles of electrical engineering textbook had them in an example of a monostable vibrator (I havent exactly seen a transistor before in problems... or real life... not even sure why it brought them up because the problems were about basics of DC RC circuits) But apparently the way to calculate the voltage across transistors uses the Lambert-W function and I thought back to your "fish" videos you did on the Lambert-W function. Honestly I didn't know it had much use out of "math for fun".
I think it'd be cool if you quickly calculated the solution for n,m=0, to show what value for z that would present. I absolutely love these complicated problems that you keep showing!! Any plans to consider more AIME or even USAMO/IMO videos?
I’m surprised how I’m slowly starting to understand these types of videos as I learn. I still remember how I would not understand any statements in these videos a few years ago.
I love how you use basic examples to train all the little concerns and caveats that must be observed when solving a specific class of problems. Very effective and fun to watch, especially with emoji substitution 😄
@@blackpenredpen *BPRP please please please read this comment.* Your videos are very amazing. I have a request, can you please please please make a video on what I have derived. I have derived a formula for sin inverse of x. The proof is as follow: y=sin^-1(x) sin(y)=x e^(iy)-e^-(iy)=2ix (e^(iy))^2-2ixe^(iy)-1=0 Using quadratic formula: e^(iy)= ix+-√(1-x^2) y= -iln(ix+-√(-(x^2-1)) y= -iln(i(x+-√(x^2-1))) Using ln(ab)=ln(a)+ln(b) y= -i(ln(i))-i(ln(x+-√(x^2-1))) sin^-1(x)= π/2 - iln(x+-√(x^2-1)) To check this formula put x=2 and you will get: sin^-1(2)=π/2-iln(2+-√3) You have proved that sin(π/2-iln(2+-√3)=2 in one of your previous videos. I also request you to put sin^-1(x)=π/180 and put formula of sin^-1(x) which I derived and solve for x so we will get value of sin(1°) or sin(π/180), I had tried to find value of sin(1) this way but I failed. I hope you will make a video on this formula. My name is Kathan Parikh and I am 16 years old. And if you want one more golden equation which includes Phi,π,i,e and even Fibonacci series(All five in one equation) then just reply me so I will give my phone number and you can call me as it is difficult to type the equation, so I will be easily able explain the equation and it's proof to you by sending you a pic or on call.
@2C (02) Chan Kwan Yu This formula will give you principal solution. If you want other solutions you can add 2πn. It will give you infinitely many solutions.
@2C (02) Chan Kwan Yu actually I am sending this same comment from last 20 videos so bprp will read but I think km he's not able to notice this comment so many other comments. I hope he reads this comment.
e^(i·z) = cos(z) + i·sin(z) for all complex z. If you substitute z |-> -z, you derive e^(-i·z) = cos(z) - i·sin(z). Subtract the second equation from the first, and this results in 2·i·sin(z) = e^(i·z) - e^(-i·z). Since 2·i is not equal to 0, you can divide, obtaining sin(z) = (e^(i·z) - e^(-i·z))/(2·i), and this is where that substitution in the video came from. In fact, this formula is often taken to be the definition of sin(z) for all complex z.
a little late but in your final formula you do not treat the case when inside the "ln" the value is negative (ln of a negative value is undefined). This case happens for n
@@jsjsjjsud9556 the equation exp(z)=-x has an infinite solutions : ln(abs(-x))+(2k+1)*i*pi with k integer. So there is no clear extension of the ln function on the real negative. Just writing ln(abs(-x))+i*pi is a shortcut and there are interesting exercises using this abuse to lure the reader
How about: sqrt(happyface^2-1)=sqrt(sin(z)^2-1)=sqrt(-cos(z)^2)=sqrt(i^2cos(z)^2)=i*cos(z) then rewrite in exp form and then try to workout z. Note this was just a quick thought. Probably doesnt take some stuff in to account like multiple branches of solutions. However I though perhaps there was a way to get a bit nicer form. Not sure though, didnt work anything out. The form and considering trig identities just made me think of this.
I am greek and I indeed was not taught complex numbers because when I was at the last grade of high school, complex numbers had stopped being taught already for 2 years... However thanks to uni and youtube videos I think I have a decent understanding of complex numbers!
@@geosalatast5715 yeap,m2! Next semester i have to choose between discrete math or arithmetic analysis(αριθμητική ανάλυση δεν ξέρω αν είναι έτσι στα αγγλικά) or complex analysis.complex analysis looks so interesting but I'm not sure yet
You should have gone a little further and shown that 'n' cannot be between -.152 to -.508 (about)...and since no integers lie within that range, THEN you can deduce that it works for all integers.
I have been subscribing u for 3 years , providing that there are infinite unsolved questions and I am slightly less than being as good as u so I should create a math channel on TH-cam, should not I?
@2C (02) Chan Kwan Yu This formula will give you principal solution. If you want other solutions you can add 2πn. It will give you infinitely many solutions.
@2C (02) Chan Kwan Yu actually I am sending this same comment from last 20 videos so bprp will read but I think km he's not able to notice this comment so many other comments. I hope he reads this comment.
@2C (02) Chan Kwan Yu and this is not the same formula. You can get sin inverse of any number you want with this formula. For example he found sin inverse of π/2 in this video and he took so much time. But with my formula sin inverse of π/2 can be found in some seconds.
You missed infinite answers After all, remembering that we are inside the sin function, at the end you must add +2πk K being an integer Very good video
Do you know what sin(x) means? It's the height of a point at an unit circle (=circle with radius 1 starting at the rightmost point going counterclockwise) with angle x in radians. One way to look at it is to remember what e^xi means. It means rotating a point on the unit circle with am angle of x in radians. For the e^-xi we have the same angle just in reverse direction. So by subtracting this two values, the real parts cancel out as they are the same but the imaginary parts get added together, as they are equal bit with opposite signs. We then divide by 2 because we are twice too high and we divide by I to get rid of the imaginary part.
I was so focussed on the board I didn't see the shirt until I saw a comment 😂Thanks for wearing it!
Tibees gang
😃 😃
😂
😂😂😂
Super 😂💞
India: We use colors as variables.
Arabia: Well, we don't really want to mess with different pigments while doing math. We're just going to use letters.
Europe: The letters are not Christian enough, we will use our own latin and greek letters.
Blackpenredpen: 🙂
Europeans are the pioneers and father of maths . LETS GO EUROPE!!!
@@chronicsnail6675 stfu
@@chronicsnail6675 says the guy who uses indo-arabic numerals
@@chronicsnail6675 A stunning display of ignorance.
@@adrianfrauca8118 and?
You know its terrifying when he giggles time to time.
For happy face^2 it should've been drawn as an actual square face, for the true immersion.
Wow! Didn’t think of it. Nice one.
@@blackpenredpen HOW sin(sin(sin(sin(sin(sin(...))))))=1 ???
what about the cartesian square of a happy face?
happy face^3:
a cube with happy face on every side
Square root of square face is happy face
"This looks... yeah." Sums it up quite nicely.
My guy here rockin' that Tibees merch while solving these unholy equations
You know its worse when he has more than 2 pens
Algebra: letters as variables
Trigonometric algebra: latin letters as variables
Blackpenredpen algebra: emojis
2:52 "I'm just going to put down a happy face....because a fish is too difficult."
🤔 but a fish can be represented with just one letter: 𝛼
Me: using letters as variable
Him:😃
Surprised?
Nope got used to it
yes :D
Well well
@2C (02) Chan Kwan Yu
This formula will give you principal solution. If you want other solutions you can add 2πn. It will give you infinitely many solutions.
@2C (02) Chan Kwan Yu actually I am sending this same comment from last 20 videos so bprp will read but I think km he's not able to notice this comment so many other comments. I hope he reads this comment.
6:15
and that’s why most people don’t have happy faces when they do maths
6:11*
Lol
7:05 Actually, ln(x+sqrt(x^2-1)) = arccosh(x), which can further simply the answer.
This only simplifies one branch of the answer, though.
Probably another branch ln(x-sqrt(x^2-1)) can be written as ln(-1)+arccosh(-x), but the domain might be tricky
actually, you can simply extract the +/- sign outside the ln. ln(x - sqrt(x^2 - 1)) = ln[(x - sqrt(x^2 - 1))(x + sqrt(x^2 - 1)) / (x + sqrt(x^2 - 1))] = ln[(x^2 - (x^2 - 1))/(x + sqrt(x^2 - 1))] = ln[1/(x + sqrt(x^2 - 1))] = - ln(x + sqrt(x^2 - 1)). So the +/- can be extracted out of the ln. So in the end it would be +/- arccosh(x).
We don't allow trig functions in the simplification... Otherwise the answer to the whole problem can be trivially reduced at step 2 to "z = arcsin(pi/2+2npi)"
@@spencergrogin1074except the input is outside the domain of arcsin
your beard looks like a perfect binary tree
"We'll have to go to the complex world"
*[screams in agony]*
The complex realm is called complex for a reason. 😃
This is completely unrelated, but I was trying to figure out transistors earlier today since one of the bonus problems in my principles of electrical engineering textbook had them in an example of a monostable vibrator (I havent exactly seen a transistor before in problems... or real life... not even sure why it brought them up because the problems were about basics of DC RC circuits)
But apparently the way to calculate the voltage across transistors uses the Lambert-W function and I thought back to your "fish" videos you did on the Lambert-W function. Honestly I didn't know it had much use out of "math for fun".
Wow! I didn’t know!
Happy face, meet fish. Fish, meet happy face.
I think it'd be cool if you quickly calculated the solution for n,m=0, to show what value for z that would present. I absolutely love these complicated problems that you keep showing!! Any plans to consider more AIME or even USAMO/IMO videos?
I’m surprised how I’m slowly starting to understand these types of videos as I learn. I still remember how I would not understand any statements in these videos a few years ago.
Though he is doing something very wrong, it seems reasonable
@@karryy01 what he did that was wrog?
Just recently saw that sin(z)=2 video of urs.Amazing. Love ur videos🤗
The video on sin(z)=2 was the first video I saw from your channel! I've been following you since that video was uploaded ;)
Thank you!
I love how you use basic examples to train all the little concerns and caveats that must be observed when solving a specific class of problems. Very effective and fun to watch, especially with emoji substitution 😄
Wow. Gears grinding on this one! Chapeaux.
4:41 “Alright so it looks.... yeah”😭
Loving these Math Content 😊
I remember u used this technique/Other way to write arcsin in ur sin(?)=2 vid. Amazing
Yea. This is a continuation video and also a little celebration (since sinz=2 got over 1M views recently).
@@blackpenredpenI was a follower of your channel since then I think, I really like the content of yours man.
Keep it up!
@@blackpenredpen
*BPRP please please please read this comment.*
Your videos are very amazing. I have a request, can you please please please make a video on what I have derived. I have derived a formula for sin inverse of x. The proof is as follow:
y=sin^-1(x)
sin(y)=x
e^(iy)-e^-(iy)=2ix
(e^(iy))^2-2ixe^(iy)-1=0
Using quadratic formula:
e^(iy)= ix+-√(1-x^2)
y= -iln(ix+-√(-(x^2-1))
y= -iln(i(x+-√(x^2-1)))
Using ln(ab)=ln(a)+ln(b)
y= -i(ln(i))-i(ln(x+-√(x^2-1)))
sin^-1(x)= π/2 - iln(x+-√(x^2-1))
To check this formula put x=2 and you will get:
sin^-1(2)=π/2-iln(2+-√3)
You have proved that sin(π/2-iln(2+-√3)=2 in one of your previous videos.
I also request you to put sin^-1(x)=π/180 and put formula of sin^-1(x) which I derived and solve for x so we will get value of sin(1°) or sin(π/180), I had tried to find value of sin(1) this way but I failed.
I hope you will make a video on this formula.
My name is Kathan Parikh and I am 16 years old.
And if you want one more golden equation which includes Phi,π,i,e and even Fibonacci series(All five in one equation) then just reply me so I will give my phone number and you can call me as it is difficult to type the equation, so I will be easily able explain the equation and it's proof to you by sending you a pic or on call.
@2C (02) Chan Kwan Yu
This formula will give you principal solution. If you want other solutions you can add 2πn. It will give you infinitely many solutions.
@2C (02) Chan Kwan Yu actually I am sending this same comment from last 20 videos so bprp will read but I think km he's not able to notice this comment so many other comments. I hope he reads this comment.
Yhank you for that video. It was quite interesting! Even the solution was a little bit too "complex" :-)
Finally, emojis in math
x/0 should be the poop emoji :-)
This is what I call fun! From now on I'll use pi*(4n+1)/2 instead of =)
Great! [ π(4n+1)/2, m є Z, n є Z ] is funnier (idk why I used [] )
i love your t-shirt!! tibees
ah yes me at late in the evening pretending to understand these kind of maths
Actually it was basic
I really love complex formula for sin(z), really fun to do
Two..?
Two...
And then it got intense when another pen approached the board
The simplest case n=m=0 gives rather nice and tidy solution Z = π/2 - i*ln(π/2 ± √(π²/4-1)) which is quite close to just π/2-i
Love the Tibees shirt!
Ah ! With this t-shirt, I finally understand : you want a beard as long as Tibee's hair !
😆
Fun fact: my beard is growing at a logarithmic rate.
@@blackpenredpen legend says if you live to infinity years old your beard size will approach a constant called pen’s constant
The videos on the shorts channel (bprp fast) are so fast that seeing you teach this at normal pace feels very strange .
z is Arg, so we can z+ 2πl is the answer (l is integer)
Love the Tibees shirt!!
Thanks, Carter.
8:16 He's so done with it, lmao
"Technically, should have written pi m." ...
pi m...
as in... Dr Peyam?
XD
That’s deep
LMAO that's not anything new...!
Can you do IITJEE math questions? Those are terrifying when you read them but are fun to solve and give you tons of views
I'm so happy the only part I didn't get was the intrusion of (e^iz - e^-iz)/2i and of course the ln log, my weakness.
e^(i·z) = cos(z) + i·sin(z) for all complex z. If you substitute z |-> -z, you derive e^(-i·z) = cos(z) - i·sin(z). Subtract the second equation from the first, and this results in 2·i·sin(z) = e^(i·z) - e^(-i·z). Since 2·i is not equal to 0, you can divide, obtaining sin(z) = (e^(i·z) - e^(-i·z))/(2·i), and this is where that substitution in the video came from. In fact, this formula is often taken to be the definition of sin(z) for all complex z.
@@angelmendez-rivera351 Oh, so that's where it comes from, thanks for breaking it down for me
Do you have any provision for Maths question doubt solving? I seriously need it for my JEE preperation.
Now this video has 100k views. So do another video like this
That stress at the end as ya were running out of space 😂
Well, that was my guess to begin with. It's quite intuitive.
a little late but in your final formula you do not treat the case when inside the "ln" the value is negative (ln of a negative value is undefined). This case happens for n
ln of a negative value is just ln(abs(x))+i*pi it isn't undefined
@@jsjsjjsud9556 the equation exp(z)=-x has an infinite solutions : ln(abs(-x))+(2k+1)*i*pi with k integer. So there is no clear extension of the ln function on the real negative. Just writing ln(abs(-x))+i*pi is a shortcut and there are interesting exercises using this abuse to lure the reader
How about:
sqrt(happyface^2-1)=sqrt(sin(z)^2-1)=sqrt(-cos(z)^2)=sqrt(i^2cos(z)^2)=i*cos(z) then rewrite in exp form and then try to workout z. Note this was just a quick thought. Probably doesnt take some stuff in to account like multiple branches of solutions. However I though perhaps there was a way to get a bit nicer form. Not sure though, didnt work anything out. The form and considering trig identities just made me think of this.
Holy moly! I thought this was impossible.
thats awesome
The infamous C and R axes
😂
I understood all of that and was able to follow along, but, if I were given that problem to solve from scratch I'd not have a chance.
Why don't you use the simplified quadratic formula for even b coefficient?
Whaaaaah... long time no see..
U got me at the first minute not gonna lie
The sin(?)=2 vid is such a good video, lol
It was a bit messy ... but i understood... thanks again !
inb4 this question comes out for my finals in 3 weeks
So it's like a warped 2d lattice of points in the complex plane!
Could use arcosh for the second logarithm?
6:12
Regret at its peak
8:15 lol
Yeah. Bravo
I think it would have been a lot better if you cancelled out the 1/2 before substituting :)
10:39: He: Very Nice
Me: 😫
Fun fact: complex nunbers are not taught anymore in greek high schools
@@tzonic8655
Unfortunately 😪😪😪
I am greek and I indeed was not taught complex numbers because when I was at the last grade of high school, complex numbers had stopped being taught already for 2 years... However thanks to uni and youtube videos I think I have a decent understanding of complex numbers!
@@geosalatast5715
Είναι κρίμα Ένας τόσο ωραίος τομέας των μαθηματικών να διδάσκεται μόνο στο πανεπιστήμιο...
@@geosalatast5715 yeap,m2! Next semester i have to choose between discrete math or arithmetic analysis(αριθμητική ανάλυση δεν ξέρω αν είναι έτσι στα αγγλικά) or complex analysis.complex analysis looks so interesting but I'm not sure yet
Does the solution plot to a point or to some interesting pattern? Too lazy to do this myself.
6:10, there are like 26 letters in english alphabet, around a infinite many symbols and other stuff, and he choose a happy face 🤣🤣🤣🤣🤣
That's impressive
I'm 100th like of your video 😜
I don't even know the first step but it's seems like very cool solutions so here's a like
BPRP: 2:58
also BPRP: 3:19
«I don't know if you use happy face in math, is very hard»😂
Something very serious is going on when blue pen is involved.
i just checked it with grapher and it seems like the function doesn’t intersect y = 1 at all. what’s the deal? i plugged y = sin(sin(x)) in desmos
There is no real solution, only complex solutions.
Does this have more solutions than sin(z)=2 or its the same infinite?
Good question! I believe they are both “countable” infinity.
God I am such a nerd for enjoying this, but non-nerds will never see the beauty in this accomplishment.
You should have gone a little further and shown that 'n' cannot be between -.152 to -.508 (about)...and since no integers lie within that range, THEN you can deduce that it works for all integers.
You actually became much more funnier
Mr. bprp can u help me : integral of cos (cot x - tan x) dx from 0 to pi
A tibees T-shirt! cool!
Apparently it’s also asin(pi/2) and -asin(pi/2)+pi
One can only imagine
Nice
Gahhh!!! I was wishing he would say go pokemon go... Nd the end XD
I have been subscribing u for 3 years , providing that there are infinite unsolved questions and I am slightly less than being as good as u so I should create a math channel on TH-cam, should not I?
You should surely I will support it
You should
If you have nothing better to do and are confident in your teaching skills, then go for it! :D
sin sin sin sin sin x = 1
Lol. I pass.
@@blackpenredpen this is an imo compendium question ... I am not joking
@2C (02) Chan Kwan Yu
This formula will give you principal solution. If you want other solutions you can add 2πn. It will give you infinitely many solutions.
@2C (02) Chan Kwan Yu actually I am sending this same comment from last 20 videos so bprp will read but I think km he's not able to notice this comment so many other comments. I hope he reads this comment.
@2C (02) Chan Kwan Yu and this is not the same formula. You can get sin inverse of any number you want with this formula. For example he found sin inverse of π/2 in this video and he took so much time. But with my formula sin inverse of π/2 can be found in some seconds.
thumbnail excited me to here
You missed infinite answers
After all, remembering that we are inside the sin function, at the end you must add +2πk
K being an integer
Very good video
Anyone know some good videos to understand the use of complex numbers here? I have no clue where he got eiz - e-iz / 2i from either.
Thankyou
Do you know of Euler's formula? Euler's formula states exp(i·z) = cos(z) + i·sin(z). You can use this to derive the formula he used there.
Do you know what sin(x) means? It's the height of a point at an unit circle (=circle with radius 1 starting at the rightmost point going counterclockwise) with angle x in radians.
One way to look at it is to remember what e^xi means. It means rotating a point on the unit circle with am angle of x in radians. For the e^-xi we have the same angle just in reverse direction.
So by subtracting this two values, the real parts cancel out as they are the same but the imaginary parts get added together, as they are equal bit with opposite signs. We then divide by 2 because we are twice too high and we divide by I to get rid of the imaginary part.
@@lily_littleangel thank you for the help!
@@angelmendez-rivera351 hi, it’s the formula I haven’t come across, thanks
8:50 at this point the happy face 😀is now becoming 😭
8:51 2 Pie Em? emmm... Payem! Payem! calling Payem!
easy. inverse sine (pi/2)
Beard game going stronggggg!!!
6:11
Because it's hard to be happy when doing maths
jkjkjk i luv maths
I love when the answer looks more like a question then the question did... ;p
z = arcsin(π/2)
Sin after sin
I have endured
Yet the wounds I bear
Are the wounds of COMPLEX ANALYSIS
does this prove than if I commit two sins in a row, then that still counts as one?
10:38 for a given level of "nice" :D
Hello i am programmer and wish to learn this. how can I learn this starting from scratch