@@randomisme4m just as an assumption dude Just like in integration and differentiation equations , we assume some functions like cos x , sin x , etc. as t
take me to the complex world, and take me 3 iterations of automatic differentiation deep down, with 3 recursions of the quotient rule, with complex number division!
I "challenge the teacher all the time" cause they are human too. Hold up, nobody said anything about putting it on the test. I was just playing. You the man. All hail the teacher and his very sufficient answers!!!!
Over 20 years ago, my pre-cal teacher said I was probably the dumbest one in the class on day one because of an unfriendly interaction the year before. After a couple minutes of him smirking while the class laughed, a friend spoke up, “Joe’s probably the smartest one in here. 😂 I never cared what anybody thought of me, so I always played dumb. It was satisfying seeing his smirk melt as he realized they were laughing at him and not me. The best part of the story came later in the semester when a friend overheard him talking to other teachers in the teachers’ lounge. “Joe could probably do every problem in the book.” He never apologized for what he said in the first day, but it meant more to me what he said when I wasn’t around. A true test of a person’s character is what they say and do when nobody is looking or listening. I’ll take a sincere compliment behind my back than an insincere apology to my face any day.
@@josephcoon5809 My teachers never told any body that they are stupid. They were just proud, if there were one child that understood what they were teaching. Most of the time i just talked with them about the current topic in class and prevented that anybody had to do anything even though my math teacher bant me after a while because she could not resume teaching. So she just gave me some tasks and send me out, like I get it you understand everything already but please consider the others. It were a litle bit sad =)
Imagine you’re in high school or college barely learning this and all of the sudden in the exam there’s a problem that says “Prove that Cos(Cos−^1(π)) = π.”
This is awesome. Everyone tends to react to complex numbers with the sentiment that "they make math more confusing" but this problem is an excellent example of how complex numbers can actually make math more sensible.
Complex numbers are only confusing when you look at them in the context of basic math. Once you progress into more involved applications such as electrical engineering, dynamics and vibrations, control systems, etc., complex numbers and Eulers identity SIGNIFICANTLY simplifies the solution process.
Math for me is insurmountable when its imaginary i.e. x, y, z means absolute bolloks to me. However replace it with something relatable like money, BTU, displacement, etc. suddenly I become a savant (to me at least). I could do compounding interest for 30 years inside my head, however I had to re take algebra 101 three times in college because I can't pretend Z matters whatsoever.
I love this video. Because when I saw the cancel with caution video, I had a suspicion that complex numbers might fix the problem. It's great to see it all laid out like that.
Wait, but what if the student gives the complex definition of cosine, but then stops there? The complex definition already shows that pi is in its range, so as long as Euler's contribution is true and the function arccosine is defined as the inverse operation of cosine, then by the definition of arccosine the solution would be correct.
I like this idea. This feels like a trick I would use after I feel comfortable with the math though. Like how after a while you are allowed to just *know* what the derivative of arccos is. I would want to feel comfortable with the complex math behind it all, but once I do, I think your approach is equally as valid Although if I'm being completely honest I would totally use your solution on an exam lol
You would also have to prove that an inverse function exists for the complex cosine (not all functions have inverses), but yes. But perhaps the really important lesson is that real-valued arccosine and the complex-valued arccosine are not, if we're being fully rigorous, the same function.
@@gregconen Sorta. If you're really going the "allow that X is possible" route and that is allowed, even if it doesn't exist otherwise, you're "allowing it to exist [somehow, someway]". The fact it would then makes no sense in any other context than the specific one I just created is already discounted. You should have restricted my powers at the start, but you didn't so whose fault is it really when I use the unlimited power given to me? See, it was you all along. Jokes aside, without such a restriction this is valid to assume it were possible (i.e. if it were to exist it would still preserve all the properties/requirements that would be required and that already exist for the non complex version), as long as you mark clearly that is what you're doing. It is basically like how you can "solve" an infinite sum of all natural integers and get -1/12. It is still not the limit/actual value of the sum, but a "what if" result (what if it were non divergent and we could manipulate it like real non diverging sums). Same idea here is possible without setting a restriction "assuming a well defined complex inverse cosine function exists or were to exist, then by definition taking the cosine of its result ought to return the original value like the non complex version or it would not be a well defined complex inverse cosine function", but the fully sober version basically (I am not sober).
This is not too hard to formalize. If f:ℂ→ℂ, and f⁻¹(x) exists for some x∈ℂ (like 𝜋), and f⁻¹(x) is in the domain of f (which it is for cos, whose domain is all of ℂ), then f(f⁻¹(x)) = x. The last bit is what it means to be an inverse. I don't think the domain bit can be untrue if the other parts are true, but I threw it in there.
Yeah, Mathologer's opinion in his -1/12 video is that when you're using analytic continuation, you must point that out and not just use normal notation like nothing special is happening. So WolframAlpha seems to be wrong here!
I just love Euler's formula. It's so powerful! It lets you save on you memory so you don't have to remember all those trig laws, since they can be so easily derived from it!
I truly do not understand why we putz around with sines and cosines anymore. We invented complex numbers soooooo long ago, and they're a much better way to express the same thing.
@@lossen1984 Well, for example, if "exp(a) = cos a + i sin a", and Re(exp(a)) = Re(cos a + i sin a) = cos a, then... cos (2*a) = Re(exp(2a)) = Re(exp(a)*exp(a)) = Re((cos a + i sin a)*(cos a + i sin a)) = Re(cos^2 a + isin a cos a + i sin a cos a - sin^2 a) = Re(cos^2 a + 2 i sin a cos a - sin^2 a) = cos^2 a - sin^2 a. Or, take sin 2a. Sin 2a = Im(exp(2a)) = Im(cos^2 a + 2 i sin a cos a - sin^2 a) = 2 sin a cos a. See?
The beauty of it is that any of the solutions to z = cos^-1(pi) that you've found will indeed give us cos(cos^-1(pi)) = pi, since things will cancel out on the denominator/numerator and because we have e^(2i*pi*n) = 1 I call this art
Here’s what I’d say. The purpose of precalc is understand how functions work. Cosine’s range by all sense of it’s definition w.r.t to the unit circle (in the real xy plane) does not include pi. So for the purpose of precalc, the teacher is correct and the student was being a prick. However if I was a teacher, put that on a hw/quiz/test, and the student somehow did the complex analysis correctly, I’d give them the credit
The purpose of education is to induce students to learn. Telling them that arcos(π) is undefined is simply a lie. Whatever the purpose of "precalc" might be, it doesn't include telling lies to your students. You might as well tell them that the square root of -1 is undefined as well.
If somebody puts cos(arccos(π)) on a test, the answer is π, because in order for arccos(π) to be defined, the (maybe multi-valued) function needs to have π in its domain, and cos will by definition give back the π because cos and arccos are inverses. BUT if somebody asks to solve cos(x)=π it's a different story, because they haven't told you the set of numbers which x is allowed to take. In this case, there's multiple, equally valid things you could write on the test: no solutions in R, infinitely many solutions in C (all results of arccos(π)), but there's more! You see, there are larger number sets than the complex numbers, such as the quaternions. There's simply no ONE valid answer to this question. UNRELATED: As a side note, if somebody asks what is arccos(1), you would answer 0. But if somebody asks what is arccos(π), suddenly there's infinitely many answers? Why can't the arccos function be multi-valued over the reals as well? After all, cos(x)=1 has infinitely many real solutions. Same applies to sine, square roots etc...
@@adb012 You certainly can if you wish. But that leaves you without an answer to the question "What is cos(arcos(π))?" How is that better than having an intuitively correct answer if you _don't_ make that restriction?
Beautiful! I think this explanation illustrates how important it is to know the nature of what you're trying to model. I wonder how many students are taught that imaginary numbers have real-world application. For me, it came quite a bit later in school.
Anyone doing anything remotely connected to electrical-engineering will immediately see the "real value" (huehuehue) of complex numbers. Just take a simple RLC circuit.
Everyone: focusing on the problem Me: "jesus christ how the f does he write a line using 2 different markers and switch between the two every other letter"
@@pauljackson3491 Complex output with real input will always come in conjugate pairs, so the graph will be "even" about the real axis.If you give it non-real input, you might very well end up graphing spirals, which don't have reflection symmetry.
If you're looking for more real world applications that isn't too far outside of calc 1, I suggest learning a bit about hyperbolic trig functions. We use them to get to get to the moon and other planets!
@@Retinetin we've just learned about those! One question I have is where do they originate from? Maybe more specifically, why do coshx and sinhx involve e^x? As opposed to how "regular" trig functions represent the possible ratios of different sides of a right-angle triangle. Sorry if this is not worded well!
If you struggle on Calculus 1... Calculus 2 is universal the worst punishment in any mathematics curriculum. Calculus 1 is derivatives. Calculus 2 is integrals (doing derivatives backwards) Calculus 3 is multivariable (vector) calculus, where you take the concepts of Calc 1 and 2, and treat the non-respective variable as a constant Differential Equations is a toss up Then you go into the upper class versions of math which almost no Engineer goes for because "I don't care about proofs"
@@cpK054L Depends on your curriculum. At my school they cram all the basics into calc 1, spend calc 2 doing parametric equations, vectors, proofs, improper integrals etc. and then they stitch multivar and lin alg together.
@@Robin-en4xs what kind of psycho school are you going to? How are they teaching vectors in a one dimensional course? You need multivariable to explain vectors. And linear algebra for us is taught with diffeq not calculus.
Using the complex world to get around undefined answers is the kind of thing i would do on my calc test because i forgot how to do the problem the normal way.
We were taught a formula for how to calculate roots of complex numbers, but I didn't bother studying for the exam, so in the problem that required you to calculate the root of a complex number, I of course did the only logical thing and said the root is equal to some complex number a + ib and then went on to calculate a and b. Or well I WOULD have, if after substituting in every a for some term dependent on b I wasn't left with a 4th degree polynomial. But at least the prof was impressed and gave me partial credit. And then asked me how the fuck I came up with that on the spot but still only managed to get a 3 (Equivalent to a C)
Actually we don't need to do any of this. We just need to know that pi is in the codomain of cosine (and in the complex world it is), and use the the formula: f (f^-1 (y)) = y
Sir, your whiteboard technique is outstanding! In my teaching assistant years I thought I had a decent one going, but faced with a professional, I now know my folly😃.
@@zachansen8293 Don't beat yourself up over it, some of my peers, who are studying math and physics at uni currently - like me - between the 2nd and 3rd semester, still can't wrap their head around how complex numbers can be a thing. And anything like e^iPi or Eulers Formula might as well be a black box with an input and an output and all the information you get about the process of turning the input into the output is a label on the box saying "Black magic fuckery". Seriously there exists math students who fail to grasp the concept of mapping complex numbers on a 2d-plane, so you have absolutely nothing to be ashamed of
Hello Dear *bprp* . You and your bprp channel, included with your marker, your white board, your mic🎙️, your t-shirt, ..., Give me great feeling. Thank you so much (I love you teacher)
Fricking awesome. Love it. Just imagine being told by a teacher, "I will accept your answer, if you can prove it, but I must warn you that it requires the us of complex number." ... that's biting off more than you can chew.
This video is a bliss, I remember why I used to watch this channel a lot during my high school. Came back from 50k to nearly a million subscribers, u deserve more man. Thanks for all your videos, they are always interesting and enticing you to think more, thanks and keep making more videos as such, and as always, USE THE CHEN LU lol!!!
I love how at the beginning of the video he was like:"you want it to be π?do you wanna have a bad time? because you're really not gonna like what happens next. Don't say I didn't warn you. "
This brings me back to my days of engineering calculus in college. Some of our problems look up an entire page to work out. Sometimes you do all that work and get all the way to the bottom just to discover the answer is 1 or something stupid. Very anticlimactic - but I still loved it.
Yeah if you know arcoshx=ln(x+√(x²-1)) beforehand you can use the fact that cosx=cosh(ix). And i think you meant cos(iz)=cosh(z). Maybe you mixed it up because isin(z)=sinh(iz) and sin(iz)=isinh(z).
@@violintegral Oh sorry yes, that's what I meant. I have indeed lost a point on two on exams because I keep forgetting the i and - in the various cos -> cosh and sin -> sinh extensions.
That was a real journey. Its satisfying when things cancel out to that degree. Even more satisfying when I'm not the one looking for the solution for half an hour.
It's important to have the context of domain and range agreed upon before starting to discuss examples. The complex plane, with its multivalued functions and Riemann sheets, is very beautiful, but students need to learn to walk before they learn to run (or fly). For example, when teaching a young person about square roots, it is natural that the square root of a negative number will be baffling. I don't think it serves a useful purpose to prematurely introduce complex numbers. I think it is more valuable to confront and define the bafflement before one can better appreciate the utility and beauty of complex numbers (as opposed to just seeing them as a "gimmick" to solve previously unsolvable equations, or to solve linear differential equations).
Remember that most of this display of fun with complex numbers was in response to the "But the calculator is always right!" refrain given by so many students. I agree that this argument should never be given to students that haven't first had a firm grounding in complex numbers, and the connections made in Euler's formula. But to not showcase something "gimmicky "? Comlex numbers, at first, were only marginally accepted because they did provide real solutions to otherwise insoluble cubic equations . Pretty good for a "gimmick".
I agree but '*i* might not. Why do they even bother really teaching anything in the Real world. Life is indeed complex and students better get used to the Euler stuff. Ok I jest but please don't disparage the real power of Z.s and sometimes its transformation from complexity to simplicity. Cos(i) is real you know.
@@albertwestbrook4813 gimmick as in rather than understanding the solution most of the students will use it as a "gimmick", aka trick to solve the problem. In another word, rather than being a mathematician they become a memorizatician, if that's even a word.
@@chaosplayer9903 In this sense of the word "gimmick", I agree! Short cuts without understanding = sloppy thinking. But the original use of i was correct - it just didn't make sense at the time.
Saw the thumbnail, realized arccos(π) is undefined for real numbers after thinking you can just cancel out cos and arccos at first, clicked on the video and immediately began solving the problem myself when you said we'd need to use complex definitions of cosine. It was a fun problem, really satisfying when the π ± √(π²-1) terms cancelled out. Still crazy to me that we can compute cosine and sine functions using just the quadratic formula and logarithms.
Pro tip for students: surprise lath techers with alternative proofs, not by using a calculator. Way more fun if they have to actually think to verify instead of just "sigh, it's because complex numbers"
The student probably wanted to check the teachers answer because intuitively cos(cos^-1(x))=x because the cosines cancel out, and so it should be true, they then probably checked it with a calculator because for a pre calc student this is definitely a very hard problem and they likely dont know the complex definition of cosine, so they brought it up and learnt new stuff.
This. If you say "But wolframalpha returned this", I roll my eyes and think "Oh hey look they can copy and paste stuff, what a genius", if you do all of that yourself and talk me through the steps you did, I will give you a perfect score that year, and start doing research into scholarship programs you can apply to
it's so beautiful to see the amount of logic thinking involved in proving equations. And to me the more I spend away from these topics, the more impressed I get when I return. Thank you for your continued effort in making great content which is straight to the point using just a couple of pens and a whiteboard.
Nice explaination, Steve! Complex world is full of wonder haha. I think the main point to the precalculus student is to grasp the idea of the real domain vs the complex.
I skipped calc and haven't touched math in 20 years. I just spent 12 minutes watching a video of a man proving a calculator simultaneously right and wrong. I think I might start doing math for fun.
@Epic Game "Complex numbers: ight imma head in" - Complex numbers: great for in-plane rotations; also, a prelude to quaternions (great for rotations in 3D space)...
The thing with this question is if you just give them that arccos is only defined with an input between or equal to -1 and 1, and don’t tell them why, of course they would have questions. But I like to tell students to think of this with the definition of inverse trig functions. Arccos is supposed to give the angle in which we can create a triangle with that angle and get the opposite side to be the (in this case) pi, and have the hypotenuse to be 1. Since the hypotenuse is the biggest side of the triangle, this cannot be. So therefore we cannot define arccos pi
That notation is slightly problematic actually because it is unclear whether it means inverse of cosine or cosine to the -1 or 1/cosine, which is why I personally always write arccos , arctan and arcsin when using the inverses of those functions because that makes it clear that I am inverting the function, not dividing by it
Awesome Vid! My grades have tanked since the beginning of the pandemic and I've been wondering about my future in mathematics, but cool stuff like this reminds me why I pursued math in the first place. Thank you, hope you see this
Had you written 🎂 instead of π and you would have gotten the same result, since you didn't make use of any special properties of π. The only reason for saying that arccos(x) is only defined for x between -1 and 1 is for using real numbers, but cos(acos(x)) is always x if x is not infinity.
@@blackpenredpen Dear sir we know cos^2 pi = (cos pi)^2 in same way Cos^(-1) pi = (cos pi)^-1 =sec pi but why values are different for cos^(-1) pi and sec pi
@@t.minojan7029 Because, cos^(-1) pi = (cos pi)^-1 can't be true. Indeed, cos^(-1) is a inverse function which is different one. The domain is angle and range is ratio.
Another way to prove the undefined is to show the function is not surjective hence it doesn't have Invers and there exist elements in image that doesn't have corresponding elements in the domain.and yes it's not surjective so the Invers doesn't exist so the expression in the thumbnail doesn't exist unless you restrict the function to the defined set.
Either the teacher/professor has to specify which number field you are using or be prepared to accept both answers. Woe to the students who is required to show their work if they put down pi as the answer.
His first answer makes sense though. Cos^-1(x) gives the angle that produces cos(angle)=x. It doesn’t really make sense for anything above 1 or below -1 to work.
Since we never used value of Pi in whole process and assume pi is just another Greek letter then can't we just say that Cos(arccos k) = k , where k is any number on number line ?
yes. also because composition of functions are associative, therefore we can cancel out cos and arccos and be left with k. the issue here was that k was not in the domain of arccos in the real domain, so it was a matter of showing that the function composition is well-defined for k for the complex domain
@@infiinight126 "composition of functions are associative". What does it have to do with the problem? And how can we infer that we can "cancel cos and arccos and be left with k"? You haven't proven your point yet. Because to use and argument of associativity requires three elements, and hier we have only two. So you just compose one more time with arccos? It does not solve the problem neither... Furthermore, you assumed (if I understood what you said) that cos and cos⁻¹ were functions. At any case, for your reasoning to be true, since you assumed cos and cos⁻¹ are "composed", cos⁻¹ and cos have to be functions. But the problem is that cos⁻¹ is NOT defined for |x| > 1, if you do not define it more specifically. Indeed, we saw in the video that there are an infinite number of values of y that verifiy cos⁻¹(x) = y when x = pi for example. And if we suppose that cos⁻¹ is defined for x = pi, y = cos⁻¹(x) can only have ONE value, which is not the case. therefore the foolwing application: R -> R ; x |--> cos⁻¹(x) is totally meaningless. Remember, the formal definition of an application is: An application is the set of three sets E,F, and G of ExF verifying: 1. for all x of E, there exists y of F such as (x,y) is in G and 2. For all ((x,y),(x',y')) of (ExF)², x = x' => y = y'.
@@lesubtil7653 I suppose I wasn't clear. Yes there is 2 functions but u can artificially insert the identity function and then apply the cancellation by proving that they are invertible and inverses of each other. Specifically, we can define the cos by restricting the domain to the principal values([0,pi] -> [-1,1]. As for cos^-1 not being a function, we can simply define the codomain appropriately to make it a function ([-1, 1]->[0,pi]). That is why I said that it is about establishing that the function is well defined.
It is true and in fact is it exactly the point of the arccos, it's the inverse function to the cosine, which means that cos(arccos(x))=x and arccos(cos(x))=x . Of course though, the identity only holds for the values in which the function is defined and is injective, just like any invertible function. In the case of the real cosine, for example, we have to work with the intervals ]0,pi[ and [0,1], while for complex numbers we can relax the restriction and find the identity satisfied on a much larger subset, but the point is still that where the functions are defined, arccos will behave this way almost by definition.
Man, that's the best. Student "akshully"s the teacher and the teacher's like "Yep, OK, here's the much more difficult world where what you said is true, learn this for the test" as retribution.
Well since u express cos-1(z) as the reciprocal of cos(z), u just need to prove that cos-1 is define for pi. The second part is weird, of course cos(cos-1(pi))=pi if cos-1(pi) exist.
cos^(-1) (z) has nothing to do with the reciprocal of cos(z). That is an unfortunate coincidence that it has a notation that looks like it is implying a reciprocal. The reciprocal of cos(z) is called sec(z) whose full name is secant of z.
I have one tip to make that nasty quadratic formula a little easier to work with. When b is even in ax^2+bx+c, we can use the simplified quadratic formula, which gets rid of that step where you had to factor out a 2 from everywhere. Here it is: when b is even, let b' = b/2, then the quadratic formula is: x = (-b'±√(b'^2-ac)/a. The beauty of this is that when a is 1, there is no denominator, getting rid of any fraction work you would have to do.
Questão muito interessante, legal esse tipo de saída, por mais extensa que seja, é sempre bom verificarmos e relação existente entre campos diferentes. Está de parabéns aí Professor 👏👏👏🤝
It depends on how you define "inverse function". A strict definition might be a function f with a domain X and co-domain Y has the inverse function g with the domain Y and co-domain X, if and only if f(x) = y is equivalent to g(y) = x. Most pedants are going to tell you that there isn't a value for a function whose input isn't in the domain. Personally, I wouldn't find that helpful if the domain and co-domain could be extended to cater naturally for such an input.
beautiful. many high school students shall able to handle this complex calculation skill. sir, you did a great job to lead students going through a tough question using their available skills. you let the students to know this question is not rocket science.
Hello sir, i had a debate about the i-th root of -3 (√.i(-3)) with someone, she told me it was undefined but i found the following answer : (I-th root of -3 will be referred as √) √(-3) = (-3)^(1/i) = (-3)^(-i) = (1)/(-3)^i (-3)^(i) = e^(ln((-3)^i)) = e^(i*ln(-3)) = e^(i*( ln(3) + iπ)) = e^(i*ln(3)-π) So √(-3) = 1 / e^(i*ln(3) - π) Thanks if you take the time to answer my question and sorry if I did any mistake, i'm 14 and i'm not quite sure of myself lol.
I am just happy that i knew it all before i watched the video. And yet i watched the whole video and received the same satisfaction. Math you are such a lovely thing to watch.
Cancel with caution: th-cam.com/video/_8AlSoR-THc/w-d-xo.html
So it does have a solution
Oh now I understand
So it is because Euler's theorem
Thanks brother
How can you set cos(z) =w and use it as a constant in the quadratic formula when z is the variable in the equation?
@@randomisme4m just as an assumption dude
Just like in integration and differentiation equations , we assume some functions like cos x , sin x , etc.
as t
@@amiteshgarg3075 Yeah i get that but doesn't the quadratic formula require a b and c to be constants rather than variables?
I love how threatening this is. “You want to go the complex world? I will take you there.”
But don't regret
take me to the complex world, and take me 3 iterations of automatic differentiation deep down, with 3 recursions of the quotient rule, with complex number division!
And put it on the test
better not
the teacher always laughs in the end
I "challenge the teacher all the time" cause they are human too.
Hold up, nobody said anything about putting it on the test. I was just playing. You the man. All hail the teacher and his very sufficient answers!!!!
😆
Why is it my classmates all hate me?
@@karionwhite2367 they get tired of carrying their own hate around so they give it to you.
Over 20 years ago, my pre-cal teacher said I was probably the dumbest one in the class on day one because of an unfriendly interaction the year before.
After a couple minutes of him smirking while the class laughed, a friend spoke up, “Joe’s probably the smartest one in here. 😂 I never cared what anybody thought of me, so I always played dumb. It was satisfying seeing his smirk melt as he realized they were laughing at him and not me.
The best part of the story came later in the semester when a friend overheard him talking to other teachers in the teachers’ lounge.
“Joe could probably do every problem in the book.”
He never apologized for what he said in the first day, but it meant more to me what he said when I wasn’t around. A true test of a person’s character is what they say and do when nobody is looking or listening. I’ll take a sincere compliment behind my back than an insincere apology to my face any day.
@@josephcoon5809 My teachers never told any body that they are stupid. They were just proud, if there were one child that understood what they were teaching.
Most of the time i just talked with them about the current topic in class and prevented that anybody had to do anything even though my math teacher bant me after a while because she could not resume teaching. So she just gave me some tasks and send me out, like I get it you understand everything already but please consider the others. It were a litle bit sad =)
This dude just pulled an “I will show you the world” song from Aladdin by taking us to the complex world lol.
😂
Couldn't agree more.. Beautiful world
The shinning shimmering splendor of imaginary numbers
like in the song, "It's just for you and me"
"i can show you the world~
rational, real, and the complex~"
I know it’s just his microphone but there’s something so wholesome about him holding a plush pokeball while talking
You never know when a wild Magicarp might appear.
@@whatelseison8970 dunno if I would want to throw a plush pokeball at a magicarp. It might get soggy.
I didnt even know it was a mic. I just saw that he was holding it and accepted it lmao
I thought it was some kind of magic talisman where he gets his math superpowers from.
@@sadiaaa1373 you are so easy
Imagine you’re in high school or college barely learning this and all of the sudden in the exam there’s a problem that says “Prove that Cos(Cos−^1(π)) = π.”
It would be no different than telling me to dunk a basketball. You just can't.
I would just throw my exam in the garbage
Aight bet
Cos(Cos−^1(π)) = Cos((1/cos)π) = Cos(π/cos) = (cos)π/cos = π
@@LKLOCO 🤡
@@LKLOCO what the heck man 😂😂
This is awesome. Everyone tends to react to complex numbers with the sentiment that "they make math more confusing" but this problem is an excellent example of how complex numbers can actually make math more sensible.
You are ysing my pfp stop
Complex numbers are only confusing when you look at them in the context of basic math. Once you progress into more involved applications such as electrical engineering, dynamics and vibrations, control systems, etc., complex numbers and Eulers identity SIGNIFICANTLY simplifies the solution process.
I mean, complex numbers make me think some problems better
They weirdly become your friend in Diff Equ due to spring problems. It just takes some warming up I feel like
Math for me is insurmountable when its imaginary i.e. x, y, z means absolute bolloks to me. However replace it with something relatable like money, BTU, displacement, etc. suddenly I become a savant (to me at least). I could do compounding interest for 30 years inside my head, however I had to re take algebra 101 three times in college because I can't pretend Z matters whatsoever.
Just realized how god tier he is at the color swapping
Thanks
It really helps with understanding how one line of equation translates to the next
I'm frustrated that they don't teach this to all educators. It looks so simple, and unbelievably useful.
@@davidlee9870 not enough money in the coloured-dry-erase market i suppose
That's the entire basis for the channel. The math is just a means for him to show it off.
I love this video. Because when I saw the cancel with caution video, I had a suspicion that complex numbers might fix the problem. It's great to see it all laid out like that.
Glad you love it 😃 thanks.
Your marker technique is really impressive!
Thanks.
The satisfaction on his face after cancelling in the last step is just incredible
Wait, but what if the student gives the complex definition of cosine, but then stops there? The complex definition already shows that pi is in its range, so as long as Euler's contribution is true and the function arccosine is defined as the inverse operation of cosine, then by the definition of arccosine the solution would be correct.
I like this idea. This feels like a trick I would use after I feel comfortable with the math though. Like how after a while you are allowed to just *know* what the derivative of arccos is. I would want to feel comfortable with the complex math behind it all, but once I do, I think your approach is equally as valid
Although if I'm being completely honest I would totally use your solution on an exam lol
You would also have to prove that an inverse function exists for the complex cosine (not all functions have inverses), but yes. But perhaps the really important lesson is that real-valued arccosine and the complex-valued arccosine are not, if we're being fully rigorous, the same function.
@@gregconen Sorta. If you're really going the "allow that X is possible" route and that is allowed, even if it doesn't exist otherwise, you're "allowing it to exist [somehow, someway]". The fact it would then makes no sense in any other context than the specific one I just created is already discounted. You should have restricted my powers at the start, but you didn't so whose fault is it really when I use the unlimited power given to me? See, it was you all along.
Jokes aside, without such a restriction this is valid to assume it were possible (i.e. if it were to exist it would still preserve all the properties/requirements that would be required and that already exist for the non complex version), as long as you mark clearly that is what you're doing. It is basically like how you can "solve" an infinite sum of all natural integers and get -1/12. It is still not the limit/actual value of the sum, but a "what if" result (what if it were non divergent and we could manipulate it like real non diverging sums). Same idea here is possible without setting a restriction "assuming a well defined complex inverse cosine function exists or were to exist, then by definition taking the cosine of its result ought to return the original value like the non complex version or it would not be a well defined complex inverse cosine function", but the fully sober version basically (I am not sober).
This is not too hard to formalize. If f:ℂ→ℂ, and f⁻¹(x) exists for some x∈ℂ (like 𝜋), and f⁻¹(x) is in the domain of f (which it is for cos, whose domain is all of ℂ), then f(f⁻¹(x)) = x. The last bit is what it means to be an inverse. I don't think the domain bit can be untrue if the other parts are true, but I threw it in there.
Yeah, Mathologer's opinion in his -1/12 video is that when you're using analytic continuation, you must point that out and not just use normal notation like nothing special is happening. So WolframAlpha seems to be wrong here!
I just love Euler's formula. It's so powerful! It lets you save on you memory so you don't have to remember all those trig laws, since they can be so easily derived from it!
I truly do not understand why we putz around with sines and cosines anymore. We invented complex numbers soooooo long ago, and they're a much better way to express the same thing.
I would love to learn how to derive trig identities from Euler. Can you give me an example of a derivation or point me to a good resource?
@@lossen1984 Well, for example, if "exp(a) = cos a + i sin a", and Re(exp(a)) = Re(cos a + i sin a) = cos a, then...
cos (2*a) = Re(exp(2a)) = Re(exp(a)*exp(a)) = Re((cos a + i sin a)*(cos a + i sin a)) = Re(cos^2 a + isin a cos a + i sin a cos a - sin^2 a) = Re(cos^2 a + 2 i sin a cos a - sin^2 a) = cos^2 a - sin^2 a.
Or, take sin 2a. Sin 2a = Im(exp(2a)) = Im(cos^2 a + 2 i sin a cos a - sin^2 a) = 2 sin a cos a.
See?
The beauty of it is that any of the solutions to z = cos^-1(pi) that you've found will indeed give us cos(cos^-1(pi)) = pi, since things will cancel out on the denominator/numerator and because we have e^(2i*pi*n) = 1
I call this art
youtuuu.tokyo/watch?v=wl6p5AocT5F
@@dakotawoertz5653 This link contains NSFW content.
@@Sir_Isaac_Newton_ Did you click on "report"? I still see the post, 17 hours later. I wonder if reporting it actually does anything.
I call this math
I just reported the link too, I hope it goes away
i finished this video and my reaction was "...wow"
true mathematician's happiness @9:45
feels so good
Holy shit, bprp is really leaning into the Confucius look these days
子曰：「cos(arccos(π))=π 」
He needs to shave it off...
Here’s what I’d say. The purpose of precalc is understand how functions work. Cosine’s range by all sense of it’s definition w.r.t to the unit circle (in the real xy plane) does not include pi. So for the purpose of precalc, the teacher is correct and the student was being a prick. However if I was a teacher, put that on a hw/quiz/test, and the student somehow did the complex analysis correctly, I’d give them the credit
The purpose of education is to induce students to learn. Telling them that arcos(π) is undefined is simply a lie. Whatever the purpose of "precalc" might be, it doesn't include telling lies to your students. You might as well tell them that the square root of -1 is undefined as well.
@@RexxSchneider ... You can say it is not defined in |R though.
If somebody puts cos(arccos(π)) on a test, the answer is π, because in order for arccos(π) to be defined, the (maybe multi-valued) function needs to have π in its domain, and cos will by definition give back the π because cos and arccos are inverses. BUT if somebody asks to solve cos(x)=π it's a different story, because they haven't told you the set of numbers which x is allowed to take. In this case, there's multiple, equally valid things you could write on the test: no solutions in R, infinitely many solutions in C (all results of arccos(π)), but there's more! You see, there are larger number sets than the complex numbers, such as the quaternions. There's simply no ONE valid answer to this question.
UNRELATED:
As a side note, if somebody asks what is arccos(1), you would answer 0. But if somebody asks what is arccos(π), suddenly there's infinitely many answers? Why can't the arccos function be multi-valued over the reals as well? After all, cos(x)=1 has infinitely many real solutions. Same applies to sine, square roots etc...
Depends on the brackets. Do you call cos^-1 on pi or do you combine cos with cos^-1 first?
@@adb012 You certainly can if you wish. But that leaves you without an answer to the question "What is cos(arcos(π))?" How is that better than having an intuitively correct answer if you _don't_ make that restriction?
9:37 and now ladies and gentlemen this is the best part of the night
I love how excited you are at this. It's how I get when I start showing my engineering students structural problems.
That flawless two-marker technique is so satisfying to watch.
Beautiful! I think this explanation illustrates how important it is to know the nature of what you're trying to model. I wonder how many students are taught that imaginary numbers have real-world application. For me, it came quite a bit later in school.
Anyone doing anything remotely connected to electrical-engineering will immediately see the "real value" (huehuehue) of complex numbers. Just take a simple RLC circuit.
Everyone: focusing on the problem
Me: "jesus christ how the f does he write a line using 2 different markers and switch between the two every other letter"
Like the unicorns and teacher, they're both magical !
This channel wasn't named "blackpenredpen" for no reason.
. True Indeed !
Wolfram alpha is weak AF, we need wolfram sigma
Wolfram omega is much better
When we getting Wolfram AlphaZero?
Wolfram Stockfish (Thank you very much for 20 Likes! :))
Lol, legend.
Wolfram Sigma Grindset
It is always insufficient for the student to recite the answer if they do not understand why it is the answer.
cosine is even, odd in the complex world
or maybe its odd, even in the complex world
i odd to know
It's not odd in the complex world
That's odd.
Cosine is even, even in the complex world, oddly.
If a function is always even in R does that mean it is always even in C?
@@pauljackson3491 Complex output with real input will always come in conjugate pairs, so the graph will be "even" about the real axis.If you give it non-real input, you might very well end up graphing spirals, which don't have reflection symmetry.
I'm only on calculus 1, but I love learning outside out my curriculum if only to see how what I'm learning aligns with deeper concepts.
Thank you!!
If you're looking for more real world applications that isn't too far outside of calc 1, I suggest learning a bit about hyperbolic trig functions. We use them to get to get to the moon and other planets!
@@Retinetin we've just learned about those! One question I have is where do they originate from? Maybe more specifically, why do coshx and sinhx involve e^x? As opposed to how "regular" trig functions represent the possible ratios of different sides of a right-angle triangle.
Sorry if this is not worded well!
If you struggle on Calculus 1... Calculus 2 is universal the worst punishment in any mathematics curriculum.
Calculus 1 is derivatives.
Calculus 2 is integrals (doing derivatives backwards)
Calculus 3 is multivariable (vector) calculus, where you take the concepts of Calc 1 and 2, and treat the non-respective variable as a constant
Differential Equations is a toss up
Then you go into the upper class versions of math which almost no Engineer goes for because "I don't care about proofs"
@@cpK054L Depends on your curriculum. At my school they cram all the basics into calc 1, spend calc 2 doing parametric equations, vectors, proofs, improper integrals etc. and then they stitch multivar and lin alg together.
@@Robin-en4xs what kind of psycho school are you going to?
How are they teaching vectors in a one dimensional course? You need multivariable to explain vectors. And linear algebra for us is taught with diffeq not calculus.
It would be nice (both for teachers and casual use) if Wolfram Alpha had a button that set the "master domain" for relevant functions.
At first I was confused I didn’t understand how it could not equal pi. It now makes sense because the inner function is undefined.
You are absolutely remarkable! proud to be attending the same school you once did :)
Thanks.
Using the complex world to get around undefined answers is the kind of thing i would do on my calc test because i forgot how to do the problem the normal way.
We were taught a formula for how to calculate roots of complex numbers, but I didn't bother studying for the exam, so in the problem that required you to calculate the root of a complex number, I of course did the only logical thing and said the root is equal to some complex number a + ib and then went on to calculate a and b. Or well I WOULD have, if after substituting in every a for some term dependent on b I wasn't left with a 4th degree polynomial. But at least the prof was impressed and gave me partial credit. And then asked me how the fuck I came up with that on the spot but still only managed to get a 3 (Equivalent to a C)
Actually we don't need to do any of this. We just need to know that pi is in the codomain of cosine (and in the complex world it is), and use the the formula: f (f^-1 (y)) = y
And how do you know it's in it's codomain one method is to do what he did
@@bloprock5376 Because the function is holomorphic, and trivially so.
Exactly. No need to explicitly define f^-1(y)
I think it's a nice demonstration especially for calculus students, so they can see that you're not just waving your hands
@@saaah707 That's fair
Sir, your whiteboard technique is outstanding! In my teaching assistant years I thought I had a decent one going, but faced with a professional, I now know my folly😃.
I had this exact problem in calculus and I understood nothing, but somehow you made it seem so manageable, thank you for the great explanation.
manageable the same way watching a professional athlete do something looks easy. I still can't do it.
@@zachansen8293 Don't beat yourself up over it, some of my peers, who are studying math and physics at uni currently - like me - between the 2nd and 3rd semester, still can't wrap their head around how complex numbers can be a thing. And anything like e^iPi or Eulers Formula might as well be a black box with an input and an output and all the information you get about the process of turning the input into the output is a label on the box saying "Black magic fuckery". Seriously there exists math students who fail to grasp the concept of mapping complex numbers on a 2d-plane, so you have absolutely nothing to be ashamed of
Hello Dear *bprp* .
You and your bprp channel, included with your marker, your white board, your mic🎙️, your t-shirt, ..., Give me great feeling.
Thank you so much (I love you teacher)
He really was like "call an ambulance, but not for me"
In love with this channel
Fricking awesome. Love it.
Just imagine being told by a teacher, "I will accept your answer, if you can prove it, but I must warn you that it requires the us of complex number." ... that's biting off more than you can chew.
This video is a bliss, I remember why I used to watch this channel a lot during my high school. Came back from 50k to nearly a million subscribers, u deserve more man. Thanks for all your videos, they are always interesting and enticing you to think more, thanks and keep making more videos as such, and as always, USE THE CHEN LU lol!!!
I love how at the beginning of the video he was like:"you want it to be π?do you wanna have a bad time? because you're really not gonna like what happens next. Don't say I didn't warn you. "
This brings me back to my days of engineering calculus in college. Some of our problems look up an entire page to work out. Sometimes you do all that work and get all the way to the bottom just to discover the answer is 1 or something stupid. Very anticlimactic - but I still loved it.
That is a hell of a derivation. I simply applied the cos(iz) = icosh(z) trick and looked up the arccos in Abromovitz and Stegun.
Yeah if you know arcoshx=ln(x+√(x²-1)) beforehand you can use the fact that cosx=cosh(ix). And i think you meant cos(iz)=cosh(z). Maybe you mixed it up because isin(z)=sinh(iz) and sin(iz)=isinh(z).
@@violintegral Oh sorry yes, that's what I meant. I have indeed lost a point on two on exams because I keep forgetting the i and - in the various cos -> cosh and sin -> sinh extensions.
That was a real journey. Its satisfying when things cancel out to that degree. Even more satisfying when I'm not the one looking for the solution for half an hour.
It's important to have the context of domain and range agreed upon before starting to discuss examples. The complex plane, with its multivalued functions and Riemann sheets, is very beautiful, but students need to learn to walk before they learn to run (or fly).
For example, when teaching a young person about square roots, it is natural that the square root of a negative number will be baffling. I don't think it serves a useful purpose to prematurely introduce complex numbers. I think it is more valuable to confront and define the bafflement before one can better appreciate the utility and beauty of complex numbers (as opposed to just seeing them as a "gimmick" to solve previously unsolvable equations, or to solve linear differential equations).
Remember that most of this display of fun with complex numbers was in response to the "But the calculator is always right!" refrain given by so many students.
I agree that this argument should never be given to students that haven't first had a firm grounding in complex numbers, and the connections made in Euler's formula. But to not showcase something "gimmicky "? Comlex numbers, at first, were only marginally accepted because they did provide real solutions to otherwise insoluble cubic equations . Pretty good for a "gimmick".
I agree but '*i* might not. Why do they even bother really teaching anything in the Real world. Life is indeed complex and students better get used to the Euler stuff.
Ok I jest but please don't disparage the real power of Z.s and sometimes its transformation from complexity to simplicity. Cos(i) is real you know.
@@albertwestbrook4813 gimmick as in rather than understanding the solution most of the students will use it as a "gimmick", aka trick to solve the problem. In another word, rather than being a mathematician they become a memorizatician, if that's even a word.
@@chaosplayer9903 In this sense of the word "gimmick", I agree! Short cuts without understanding = sloppy thinking. But the original use of i was correct - it just didn't make sense at the time.
yeah which is why he just tells them that the answer is undefined
I got goose-pimples when you did the final cancelling. Brilliant.
There a simpler solution: all the trig inverses are defined on length pi intervals, so one can simply move the interval to include pi.
That's their range
Saw the thumbnail, realized arccos(π) is undefined for real numbers after thinking you can just cancel out cos and arccos at first, clicked on the video and immediately began solving the problem myself when you said we'd need to use complex definitions of cosine.
It was a fun problem, really satisfying when the π ± √(π²-1) terms cancelled out. Still crazy to me that we can compute cosine and sine functions using just the quadratic formula and logarithms.
Pro tip for students: surprise lath techers with alternative proofs, not by using a calculator. Way more fun if they have to actually think to verify instead of just "sigh, it's because complex numbers"
The student probably wanted to check the teachers answer because intuitively cos(cos^-1(x))=x because the cosines cancel out, and so it should be true, they then probably checked it with a calculator because for a pre calc student this is definitely a very hard problem and they likely dont know the complex definition of cosine, so they brought it up and learnt new stuff.
This. If you say "But wolframalpha returned this", I roll my eyes and think "Oh hey look they can copy and paste stuff, what a genius", if you do all of that yourself and talk me through the steps you did, I will give you a perfect score that year, and start doing research into scholarship programs you can apply to
it's so beautiful to see the amount of logic thinking involved in proving equations. And to me the more I spend away from these topics, the more impressed I get when I return. Thank you for your continued effort in making great content which is straight to the point using just a couple of pens and a whiteboard.
I think I legitimately did this problem on a test once while getting my physics degree lmao.
I'm glad to see someone explain the answer that every already knew to be intuitively true.
I clapped when I saw the final answer had been written down.
Thanks!
After watching this I can say with confidence: I did not want to go there.
It took me a while to figure out that ball he was holding was his microphone. I thought he was pretending to be an Ood.
Precalculus teacher vs WolframAlpha student? This has to be settled with a dance-off!
Nice explaination, Steve! Complex world is full of wonder haha. I think the main point to the precalculus student is to grasp the idea of the real domain vs the complex.
Thanks.
You make calculus fun, entertaining and understood.
I skipped calc and haven't touched math in 20 years. I just spent 12 minutes watching a video of a man proving a calculator simultaneously right and wrong. I think I might start doing math for fun.
Skipped calc, or stopped taking math classes before calc?
@@error.418 Calc wasn't a requirement in high school, maybe that's changed now. I took physics and math, never bothered with Calc.
@@MrPhyrce Gotcha, thanks for the clarification 👍
and he did it while holding a pokeball
It's all fun and games until matrices attack
"if you would like this on the test, please! let me know!"
bro just won the argument in a single sentence.
Complex numbers: ight imma head in
@Epic Game "Complex numbers: ight imma head in" - Complex numbers: great for in-plane rotations; also, a prelude to quaternions (great for rotations in 3D space)...
Normie
@@nicadi2005 Not just great for rotations; they're better at being rotations than they are at being vectors.
good on that student, we get this cool video about how questionable wolfram is.
also, respect for committing to holding the plushie.
The thing with this question is if you just give them that arccos is only defined with an input between or equal to -1 and 1, and don’t tell them why, of course they would have questions. But I like to tell students to think of this with the definition of inverse trig functions. Arccos is supposed to give the angle in which we can create a triangle with that angle and get the opposite side to be the (in this case) pi, and have the hypotenuse to be 1. Since the hypotenuse is the biggest side of the triangle, this cannot be. So therefore we cannot define arccos pi
The truth? You can't handle the truth! (The truth is in the complex world)
i can handle the truth :P
Math's consistency is just great. Even in the complex world cos^-1 is still the inverse of cos. Beautiful
That notation is slightly problematic actually because it is unclear whether it means inverse of cosine or cosine to the -1 or 1/cosine, which is why I personally always write arccos , arctan and arcsin when using the inverses of those functions because that makes it clear that I am inverting the function, not dividing by it
At first I saw that and Thought Pi, then looked again and remembered the range of arccos(x) haha.
Awesome Vid! My grades have tanked since the beginning of the pandemic and I've been wondering about my future in mathematics, but cool stuff like this reminds me why I pursued math in the first place. Thank you, hope you see this
Had you written 🎂 instead of π and you would have gotten the same result, since you didn't make use of any special properties of π. The only reason for saying that arccos(x) is only defined for x between -1 and 1 is for using real numbers, but cos(acos(x)) is always x if x is not infinity.
I’m just really happy that it is commutative after all if you go into the complex plane.
Here i am, drinking coffe and pretending to understand.
My favorite part of this video is 9:01
“Good news for you guys- negative one, positive one? cancel out” and then his smile at the camera
"Do you want to go there?! I'll take you there!"
Great plot twist. Thank you!
"....But just don't regret it" complex world has never been a regret
i think the most impressive thing is him using the two markers flawlessly in one hand
Does the same thing happen with the sine function?
sin(sin^-1(pi)) = ?
Yes
@@blackpenredpen
Dear sir
we know cos^2 pi = (cos pi)^2
in same way
Cos^(-1) pi = (cos pi)^-1
=sec pi
but why values are different for cos^(-1) pi and sec pi
@@t.minojan7029 Because, cos^(-1) pi = (cos pi)^-1 can't be true.
Indeed, cos^(-1) is a inverse function which is different one.
The domain is angle and range is ratio.
@@ghostboi71 He is just criticizing that we write "sin^-1" instead of "arcsin". He is correct on this though.
ok i understand it
Thank you
That’s fantastic to watch this instead of using hard formulas to solve it❤
Another way to prove the undefined is to show the function is not surjective hence it doesn't have Invers and there exist elements in image that doesn't have corresponding elements in the domain.and yes it's not surjective so the Invers doesn't exist so the expression in the thumbnail doesn't exist unless you restrict the function to the defined set.
that was amazing!! i'm still in high school and not taking these stuff yet but it's so fun to watch them on youtube
That was fun! 💕
I hope you put this on the test.
Rip students who will have this question in their test
Either the teacher/professor has to specify which number field you are using or be prepared to accept both answers. Woe to the students who is required to show their work if they put down pi as the answer.
His first answer makes sense though. Cos^-1(x) gives the angle that produces cos(angle)=x. It doesn’t really make sense for anything above 1 or below -1 to work.
Very cool. It makes me want to visualize the complex COS transform.
Wait till I challenge master yoda to a math duel
Dead people can't fight eachother
@@midnightsnacker3101 o yeah?
shortcut: they've inverses and that's why you get pi since they cancel each other out by definition
Since we never used value of Pi in whole process and assume pi is just another Greek letter then
can't we just say that
Cos(arccos k) = k , where k is any number on number line ?
yes. also because composition of functions are associative, therefore we can cancel out cos and arccos and be left with k. the issue here was that k was not in the domain of arccos in the real domain, so it was a matter of showing that the function composition is well-defined for k for the complex domain
@@infiinight126 "composition of functions are associative". What does it have to do with the problem? And how can we infer that we can "cancel cos and arccos and be left with k"? You haven't proven your point yet. Because to use and argument of associativity requires three elements, and hier we have only two. So you just compose one more time with arccos? It does not solve the problem neither...
Furthermore, you assumed (if I understood what you said) that cos and cos⁻¹ were functions. At any case, for your reasoning to be true, since you assumed cos and cos⁻¹ are "composed", cos⁻¹ and cos have to be functions.
But the problem is that cos⁻¹ is NOT defined for |x| > 1, if you do not define it more specifically. Indeed, we saw in the video that there are an infinite number of values of y that verifiy cos⁻¹(x) = y when x = pi for example. And if we suppose that cos⁻¹ is defined for x = pi, y = cos⁻¹(x) can only have ONE value, which is not the case. therefore the foolwing application: R -> R ; x |--> cos⁻¹(x) is totally meaningless.
Remember, the formal definition of an application is: An application is the set of three sets E,F, and G of ExF verifying: 1. for all x of E, there exists y of F such as (x,y) is in G and 2. For all ((x,y),(x',y')) of (ExF)², x = x' => y = y'.
@@lesubtil7653 I suppose I wasn't clear. Yes there is 2 functions but u can artificially insert the identity function and then apply the cancellation by proving that they are invertible and inverses of each other. Specifically, we can define the cos by restricting the domain to the principal values([0,pi] -> [-1,1]. As for cos^-1 not being a function, we can simply define the codomain appropriately to make it a function ([-1, 1]->[0,pi]). That is why I said that it is about establishing that the function is well defined.
It is true and in fact is it exactly the point of the arccos, it's the inverse function to the cosine, which means that cos(arccos(x))=x and arccos(cos(x))=x . Of course though, the identity only holds for the values in which the function is defined and is injective, just like any invertible function. In the case of the real cosine, for example, we have to work with the intervals ]0,pi[ and [0,1], while for complex numbers we can relax the restriction and find the identity satisfied on a much larger subset, but the point is still that where the functions are defined, arccos will behave this way almost by definition.
Man, that's the best. Student "akshully"s the teacher and the teacher's like "Yep, OK, here's the much more difficult world where what you said is true, learn this for the test" as retribution.
If you’re going through complex world, keep going
(After Winston Churchill)
Or else you will become imaginary forever!
You're a very engaging instructor. This was mind bending.
But if we set pi as a variable, then doesn't it prove that for all real and unreal numbers , cos(cos^-1(x))=x
Yup
For complex
The joy of being dragged into the complex world. This guy is a gem!
Well since u express cos-1(z) as the reciprocal of cos(z), u just need to prove that cos-1 is define for pi.
The second part is weird, of course cos(cos-1(pi))=pi if cos-1(pi) exist.
cos^(-1) (z) has nothing to do with the reciprocal of cos(z). That is an unfortunate coincidence that it has a notation that looks like it is implying a reciprocal. The reciprocal of cos(z) is called sec(z) whose full name is secant of z.
I have one tip to make that nasty quadratic formula a little easier to work with. When b is even in ax^2+bx+c, we can use the simplified quadratic formula, which gets rid of that step where you had to factor out a 2 from everywhere. Here it is:
when b is even, let b' = b/2, then the quadratic formula is: x = (-b'±√(b'^2-ac)/a. The beauty of this is that when a is 1, there is no denominator, getting rid of any fraction work you would have to do.
"I'ld be glad to put this on the test for you. " -- Evil, Evil! Evil! What a difference a domain makes! Brains beat calculators every time.
Your videos are awesome and have inspired me a lot. I just wish that you were my teacher, especially for Calculus and advanced mathematics.
Questão muito interessante, legal esse tipo de saída, por mais extensa que seja, é sempre bom verificarmos e relação existente entre campos diferentes. Está de parabéns aí Professor 👏👏👏🤝
That was beautiful. Simple and a nice way to so how different it is to work with complex numbers
does this work for other inverse functions whos input isn’t in the domain?
Unlike their real versions, complex sine and cosine are not bounded.
It depends on how you define "inverse function". A strict definition might be a function f with a domain X and co-domain Y has the inverse function g with the domain Y and co-domain X, if and only if f(x) = y is equivalent to g(y) = x.
Most pedants are going to tell you that there isn't a value for a function whose input isn't in the domain. Personally, I wouldn't find that helpful if the domain and co-domain could be extended to cater naturally for such an input.
beautiful. many high school students shall able to handle this complex calculation skill. sir, you did a great job to lead students going through a tough question using their available skills. you let the students to know this question is not rocket science.
Thank you!
Hello sir, i had a debate about the i-th root of -3 (√.i(-3)) with someone, she told me it was undefined but i found the following answer :
(I-th root of -3 will be referred as √)
√(-3) = (-3)^(1/i)
= (-3)^(-i)
= (1)/(-3)^i
(-3)^(i) = e^(ln((-3)^i))
= e^(i*ln(-3))
= e^(i*( ln(3) + iπ))
= e^(i*ln(3)-π)
So √(-3) = 1 / e^(i*ln(3) - π)
Thanks if you take the time to answer my question and sorry if I did any mistake, i'm 14 and i'm not quite sure of myself lol.
(-3)^(1/i) is not (-3)^(-i)
I am just happy that i knew it all before i watched the video. And yet i watched the whole video and received the same satisfaction.
Math you are such a lovely thing to watch.