Summary of the confusion on problem 2. TL;DR, given the spirit of the question it should have also specified that f(x) > 0 for all x, and I'm clearly prone to deep confusion while trying to juggle the many balls of live broadcasting. The question asks about a function f satisfying f(a + b) = f(a)f(b). The relevant part of the question was whether f(-1) must be 1 / f(1). - It was originally graded in a way to suggest that this _does_ indeed need to be true. - While "explaining" it, I realized that the explanation required that f(0) = 1, and questioned whether this is necessarily true. I even said, "oh, you could just scale it". This thought of scaling is not correct, since (c*f(a))(c*f(b)) = c^2 f(a + b), so the lhs and rhs doesn't scale the same way. - At 27:48, Sam points out that the constant function f(x) = 0 doesn't work, which I misread as asking about whether f(x) = 0 for some x (sorry Sam!). - Crispin points out that f(0) must be 1 because f(x) = f(x + 0) = f(x)*f(0). This is true, as long as f(x) does not equal 0 everywhere, but at that point, I was still weirdly blind to that edge case and enthusiastically accepted this as a reason that the question was originally graded correctly. - At 49:07, more of the discussion that has been happening on twitter is brought to screen, where Eric correctly points out that Crispin's proof doesn't work for the constant function f(x) = 0, which Sam had said all along much earlier and I misread. Thanks for the good discussion, and the tolerance of some live befuddlement.
I have a nice question does all functions that satisfy f(a+b) =f(a)f(b) produces values that are bigger than or equal to zero or can a function with such property have a negative output?
@@oximas Let's assume there is a continuous (you can draw it without lifting your pencil) function f(x) such that f(a+b) = f(a)f(b), and that at some place this function is negative, say f(c) < 0. In the spirit of this post, we can see that f(c+0) = f(c) = f(c)f(0), so f(0) = 1. We know that f(0) = 1 and f(c) < 0, which means that there must be a number d between c and 0 such that f(d) = 0. This means that f(1+d) = f(1)f(d) = 0. But also f(2+d) = 0. And f(3+d). In fact, we would find that no matter what number x is input to the function, f(x) = 0. But this contradicts our assumption that f(c) < 0. The contradiction means that the original assumption must be wrong, so there cannot be a continuous function f(x) such that f(a+b) = f(a)f(b) and f(c) < 0 for any number c. Using this same argument, we can show that the only continuous function such that f(a+b) = f(a)f(b) that can take the value 0 is the function f(x) = 0.
@3Blue1Brown Thank you so much for taking the time to walk through not only the correct answer, but also your own confusion. So often students (and many teachers) see teachers as infallible sages. It's very encouraging to see teachers demonstrating to students the process of admitting their errors but then persisting towards reaching the correct solution. Keep up the great work!
@@oximas For real-valued functions negative values are not possible, because f(x) = (f(x/2))^2 >= 0. If complex values are allowed then you can get negative values, for instance... f(x) = e^(ix) Then f(pi) is negative.
@@tomekczajka ohh i get it so all "real" functions (aka ones that don't contain imaginary values) only give nonnegative values, that's interesting. thanks for the help
Video Timeline 0:00:00 Welcome 0:00:20 Ending Animation Preview 0:01:15 Reminders from previous lecture 0:03:30 Q1: Prompt (Relationship with e^iθ=…) 0:05:40 Q1: Results 0:07:15 WTF, Whats The Function 0:10:00 Exploring exp(x) 0:11:45 Exploring exp(x) in Python 0:14:45 Important exp(x) property 0:15:55 Q2: Prompt (Given f(a+b) = f(a)f(b)…) 0:17:30 Ask: Which is more interesting, special cases or the general case 0:20:00 Q2: Results 0:23:50 Will a zero break Q2? 0:25:40 The e^x convention 0:27:10 Q3: Prompt (i^2 = -1, i^n = -1) 0:27:45 Ask: Zero does not break Q2 0:30:20 Q3: Results 0:31:05 Comparison to Rotation 0:33:00 Visualizing this relationship 0:36:50 The special case of π 0:39:20 Periodic nature of this relationship 0:39:40 Q4: Prompt (e^3i) 0:41:35 Q4: Results 0:43:55 Explaining the celebrity equation 0:45:55 Homework / Things to think about 0:49:15 Ask: Zero does break Q2. 0:50:30 Closing Remarks Water drinks at 0:17:10 & 0:27:45 & 0:40:05 Edits: Moved water drinks to the bottom, spelling errors, these timestamps should be for after the video is trimmed at "Welcome!"
Had a problem where last video's comment became unpinned after updating timestamps for trimmed video. These timestamps are for the trimmed video (assumed it was trimmed at 5:10) if you wish to use it before the video is trimmed, add 5 minutes and 10 seconds to the displayed timestamp.
The memory rule I learned for the digits of e is {Everyone knows how it starts}{Ibsen's birth year}{Again}{The angles of a right isosceles triangle} 2.7 1828 1828 459045
In all my years of highschool and engineering undergrad, the taylor series of e^x being the way to understand exp(x) was never emphasized like in the way here. Euler's formula was just the way to convert between polar and real/Im form of imaginary numbers and do some convenient maths that was noted down and refered to in a formula sheet. Closest was a professor explaining that defining trait of e^x is the function where its differential was the same as its integral, was the same as the value of e^x.
@@ca-ke9493 now that one threw me off in integration. every damn time a question asking integral of e^x, id be relieved, and id put e^x. so many missed marks
I think " Maths for the Curious " is a better title for this series than 'Highschool maths'. As I don't know if this lecture would specifically help in highschool exams etc. But I think it is for anyone who is curious and wants to learn math with a little more intuition and creativity. For example: I work in theoretical physics, and have already learnt lots of Maths. Still going back and understanding math fundamentals in this beautiful way accentuates my understanding. Which is why I believe that " Maths for the Curious " is a better title, age no bar.
@@3blue1brown Hey, Thanx for the reply! Would love it if something similar was implemented in the next lecture onwards. While highschool maths reminds most people of dull problems and unnecessary competition, this on the other hand is truly magical. Love this series and infact for that matter all your material is brilliant, especially the " essence of " series. Keep up the great work!
I am also in High School and i agree. I for example wont have complex numbers in school before uni. Atm tho i think the lessons are in a perfect difficulty. Its quite challenging but if one invests a few hrs most should be able to manage iy
I think it depends where you go to high school. This is in the syllabus of high schoolers in certain parts of Europe and Asia 😅 But regardless of what's in school syllabus, I feel that an appreciation of math like this is all the more important for people in their early to mid teens. So reaching out to high schoolers is a great thing.
I remember a couple years ago, during a particularly boring physics lesson, I messed around with my calculator and typed the sum over the factorials to see what i would get. To my surprise, i found out that the answer was 2.718... so i thought to myself "wait, it can't be", i took the ln of it and sure enough it was 1. I was blown away, and i thought i found a secret way to calculate e. Later I took calc 1 and discovered that i discovered nothing new, but still, that feeling when I accidentally stumbled upon this formula for e was really something else, and throughout the last 2 years of my bachelor's math degree, I only ever felt that way once again, the feeling of pride that i discovered something new and beautiful.
Oh yes, I love epiphanies like this! Aren't they great? I think when you've stumbled upon something in this way you simply can't ever forget it. If only we could learn everything through personal discovery, we'd probably retain it all a lot better.
@Rayan I remember finding the same thing but by counting tiles when i was younger. I realized I could find the next square and made a formula for (x+1)^2 until i realized that there was an (a+b)^2 formula, which was equivalent.
For me, my search for the function equal to the sum of the factorials lead me down the rabbithole of calc. Little did I know that there was no elementary function describing the sum of factorials. It all stemmed from me discovering a neat formula, that the sum of the first n natural numbers is n(n+1)/2. That was really the start of my journey.
His excitement is contagious. And seeing him get flustered because he was live and didn't want to say the wrong thing makes me feel better about how I freeze up or get flustered. We're all human!
Genuine question: Why aren't we taught this concept intuitively at institutions we literally pay thousands for each year? Why is it that we have to come to a free source to learn these things deeply?
Pedagogy is a field that deserves to be studied and improved in it's own right. Unfortunately some institutions just look for teachers with dexterity in the branch of knowledge being taught but not in the knowledge of how to teach.
Being a great teacher and being a great mathematician are different skills and hard to find in the same person. Also, being a great teacher is not particularly valued by society so great communicators who deeply understand a subject frequently find other jobs. Ideally a University should hire mathematicians to do research and not force them to teach and hire teachers to teach.. but that would be too expensive. High schools only hire teachers but there aren't enough talented individuals in the world who deeply understand math, love teaching, and love students. Grant is like the LeBron James of math education and by definition every highschool in the world cant have their own LeBron James.
@@jeffjiang5272 I disagree. Those "good" students usually can't care less about what the teacher is saying as they are probably way ahead of the class. Or atleast that's how it is in high school. The way they teach math is bad because math teachers dont design the curriculum but education specialists who barely know any math.
Because of time.... I can't believe people don't understand this simple thing... This video takes an hour to deeply and beautifully explain the Euler's formula which is one part of complex numbers... While school teachers get about 30 minutes a class to do so
What I admire about you, is how you try to teach from a perspective of someone who doesn't know it. Most often, teachers forget how they felt while learning it and it becomes harder for them to explain than it was to learn.
In the proof of the necessity that f(0) = 1, the hole in the logic was after the step f(0)f(x) = f(x). This results in f(0) = 1 if and only if f(x) is not zero (so that we don't have to divide by 0). This means that, while true that f(x) = 0 is an exception, it's also the only exception. Also at 24:03, where you say that you could scale the function to get a different result for f(0), that wouldn't work because it would no longer satisfy f(x+y)=f(x)f(y). Multiplying the two outputs would result in your scaling factor squared on the right hand side, while the left hand side would only have a single scaling factor. With that said, I really enjoyed this lecture! Can't wait for the next one :)
I’m a middle-aged financial engineer and learn from your lectures- and I was definitely paying attention to math classes in high school. Watching your videos and the beautiful new perspective you cast on sometimes elementary topics is like re-watching a classic movie or re-reading a classic novel and getting whole new appreciation for the material as if you were reading it for the very first time.
I just want to say that this series is absolutely awesome. I never thought of myself as a math person much less a math nerd and yet I just sat here for hours and have learned and understood so many things I never grasped before so Thank you! I really wish for this to become a regular thing even after the world returns to "normal". Would I not already be a Patreon this series would have definitely earned it. As a student of I often struggle to understand math and as of now firmly belong in the category of didn't understand it but just went along with it as well as I could but you are really changing my perspective on maths and are showing me that instead of frustrating it can be fun and interesting. EDIT: Okay it seems I accidentally canceled my Patreon subscription last time I cleared up my payments! This grave mistake has been remedied. So in a way this video DID earn my subscription (again).
For those of you who haven't seen it, Mathologer has a wonderful video on Euler's formula: ruclips.net/video/-dhHrg-KbJ0/видео.html A worthy question is to ask what the connection is between the limit he writes and the polynomial here. Perhaps good fodder for the next lecture :)
@@karkaroff1617 If f(n)=0 then we have 0*f(0)=0. And f(0) could 1 or 0 or any other constant. Therefore, if f(n)=0, then equation 3 isn't satisfied. The problem should have stated that f(n) is not 0.
One topic I had always wondered about was a more direct (but challenging) way of defining a^x (a,x∈R, a>0) in terms of Cauchy sequences (a^qₙ) where rational every qₙ is rational and qₙ→x. It would require proving that every such sequence was Cauchy, so by completeness they converge to some real y, and in particular that they converge to the same y. Then (using whatever is your favorite definition of e), exp(x) = e^x, and log(x) = exp⁻¹(x). This makes it a direct extension of exponentiation with rational exponents, which are essentially defined by the requirements that a^x * a^y=a^(x+y) and a^1 = a, along with the convention that a^x > 0. It feels a little more motivated than the backdoor method of, for instance, defining log(x) in terms of an antiderivative, then defining and extending its inverse, and finally proving that the resulting function corresponds with the usual definition for rational arguments. Since it is continuous by construction, these two approaches are equivalent, but it feels kind of . . . slippery.
I'm a mechanical engineering graduate who's in the midst of studying for my math subject GRE in the hopes of pursuing academia and go back to grad school for pure mathematics. Your videos have not only helped me throughout the last year with grasping abstract ideas conceptually, but you've also helped me gain a whole new appreciation for mathematics as a whole. Thank you sir. You are a treasure to this world.
Grant I really hope that there are a good 20-30 more of these lectures in the future, they are so much more insightful that high school math and they have really given me an interest in diving into more complex math, keep up the good work and please don’t stop these lectures once we end lockdown
I don't know why, but this lecture seemed too short to me, even though it was only 10 minutes shorter than the previous ones. Apparently, I've become addicted to your awesome videos, Grant! Well done!
These are really fascinating - I'm a curious high school student and I'm loving seeing these things I just learned in a new way. Thanks for being awesome :) I had actually never seen the Euler Formula, but the connection to the unit circle makes it easier to digest
Fun observation: when you plug in a positive integer N into Maclaurin series form of exp(x) the individual terms are increasing until you get to the Nth term, the N+1th term equals the Nth term (using indexing that calls x^k/k! the k+1th term), and the culmulative sum of the terms passes the halfway point exp(N)/2 of the total sum between the adding on of the Nth and the N+1th terms. You can ask, if you summed up to the Nth term, what fraction of the N+1th term do you need to add on to get to exactly exp(N)/2. The answer appears to be very close to one-third in the limit of large N, but not quite. It's also close to the square of the Euler-Mascheroni constant.
That's actually a really interesting point of view that I haven't seen before, that writing e^i is arguably an abuse of notation for exp(i), and in general the fact that e^x and exp(x) are two different functions that just happen to have the same value everywhere that e^x is defined, e^x meaning "multiply e by itself x times" and exp(x) meaning "do this infinite sum on x." The strict e^x would be undefined on imaginary numbers, but because it's equal to exp(x) whenever it does exist, we just write e^ix to mean exp(ix) without much confusion once you understand the convention. It's pretty similar to the relationship between factorial and the gamma function, which you actually allude to here. x! and Gamma(x+1) are two different functions that just happen to have the same value everywhere that x! is defined (though there is the +1 that makes it annoying). In a strict sense, (1/2)! is undefined; how could you take every integer starting from 1/2 and going down to 1 and multiply them all together? But because Gamma(x), defined by a funny integral, is the same as the factorial where they do exist (offset by 1), we often will write (1/2)! to actually mean Gamma(3/2), and there isn't really any confusion there.
The proof f(x + 0) = f(x) * f(0) => f(0) = 1 only needs the condition f(x) does not equal 0 to be valid. Whether or not f(-1) = 1 / f(1) should count as a valid answer is up for debate.
Yeah, so if for f(a+b)=f(a)*f(b) you put a and b as 0, you get two solutions for f(0), ie, 0 and 1. Whichever you take is up to you. Usually, in math questions like this, the question usually has the assumption that f(0) is not equal to zero. Funny how that was the first thing that popped into my mind, from years of solving objective questions.
There are also undefined values. Technically, the following can work: f(x) = 0 if x >= 0 f(x) = undefined if x < 0 If undefined * 0 = 0 is a valid output (e.g. f(-1)*f(2)=f(1)). Because it also ends up giving undefined * 0 = undefined (e.g. f(-2)*f(1)=f(-1))
Rhitam Dutta In all of the math classes I’ve taken any number to the power of 0 has been 1. Does the possibility of 0 come from that it can be any function that has this property? What’s an example. The last question f(-1)= 1/f(1); I reasoned that a^(-1)=1/a. Assuming f(x) to be a^(x) based on the fact that this function has the given property. Is this correct reasoning?
This video is simply beautiful. I am one those engineers who just accepted euler's formula and used it without ever deeply, intuitively understanding it. This is the best explanation of the formula. Thank you so much. Please keep making more of these videos.
Thanks for doing this. Today I was unable to attend live, but I’d pause and figure out some of the answers at each step. I truly appreciate the simplicity of the expansion series, and the resulting connection to the rotation. As others have mentioned, I do wish this is how the teachers had explained this in math class.
@24:00 F(a+b) = F(a)*F(b) put a=0 and b= -1 thus F(0-1) = F(0)*F(-1), But F(0-1)= F(-1), hence F(0) needs necessarily to be 1. Again F(-1)=F(1-2)=F(1)*F(-2) diviiding both sides by F(-1) F(-1)/F(-1)=F(1)*F(-2)/F(-1) hence 1=F(1)*F(-1-1)/F(-1) once again 1 = F(1)*(F-1)*F(-1)/F(-1) Thus F(1)*(F-1)=1 so, all the three equations are necessarily true for a function with property F(a+b) = F(a)*F(b). Thus option E is correct.
I’ve just finished year 11 in the UK (16 years old) and I’ve never come across these topics before but they are incredibly well explained and very interesting. I’m happy to be enjoying maths more than I usually would and learn some topics which I’m sure I will come across next year in A-Levels.
William McLaughlin hey i am interested what your maths syllabus include cause these topics are actually in my normal math syllabus in India although no Euler formula is shown, it's a plus to know in competitive exams
@@AdityaKumar-ij5ok Another UK student here, just finished with year 11 And I can confirm, the topics are massively disappointing. At the end of the year we did like, 1 lesson on vectors, and that was just how to add them. And in one of the exams we did, the only question about vectors was the very last question. It's mostly just relatively basic geometric proofs, quadratics, and.. that's pretty much it, actually. That is for the higher tier test, by the way - the foundation test is mostly comprised of addition, multiplication, ratios, etc Just like the original comment, I'm also going to take A-Level maths, and _hopefully_ it'll be at least half as interesting as these videos.
My University math professor had a genuine adoration of math. Failing his class is one of the few things I regret in college. His last gift to us on the last day of class was to show us Euler's formula. He talked about how it beautifully encapsulated everything we had studied up to this point and about how beautiful math is.
Videos like these are useful not just for highschool students. For them, sure, they may be helpful in leaming these topics. For people like me, who has studied maths not that long ago and still remembers most of the ideas, watching these videos is very pleasing because I can focus specifically on logical, intuitive details of your explanations without having to struggle memorizing all of this new info from scratch. Yes, we've been taught all these topics, I know how to do all of that mathematically, but it is extremely satisfying to get an intuitive look on things that you already know in the form that is usually taught, re-think them, make the information in my head more organised and open possibilities to use this knowledge in real life.
Thank you so much! It took a while, but at 37:21, the idea of how rotation connects with the definition of i and the complex plane just finally clicked, seeing the exp(i*theta). Your animations are wildly helpful!
OMG you just answered me a question wich confused me for 2 years now and I already gave up to understand why e^i*pi should make sense. It's because it's actually not e^i*pi but exp(i*pi) where exp(x) not necesseraly is equal to the thing we have in mind when we see e^x. THANK YOU!
See, *this* is the confusion caused by the presentation. exp(x) (the infinite one, not the python approximation) *is exactly* e^x ; the Taylor series *is exactly* equal to the function it's derived from, even in the complex plane... It ends up informing our understanding of what exponentiation to a complex value means.
The real significance of the function is that it's the function (well, it and any constant multiply of it, but it's the only of those that also has the algebraic property discussed in the video) whose derivative is itself, and the definition inherently entails calculus.
@@Gold161803 Slight correction to what I said, the constant function 0 ALSO has the algebraic property mentioned in the video AND is it's own derivative, but exp is more interesting for obvious reasons.
As I watched the video, I asked myself the question. Why is e that perfect canonical base for the unit square? It turns out, that any number k bigger than 1 would work since k^x = e^(ln(k)*x) it just slows down (k < e) or speeds up (e < k) the period of the function. e is just the perfect value where e^(i*x) = e^(i*x+2*pi*n) for x in the real domain.
That was the first time I felt really confident to try and go for the question and I am really happy to see I could find the correct answer. (around min 20) That may be silly but I feel like that is my first step to really understand math.
I did A-level maths and and A-level further maths in 1996 at age 18 (high school). We did some complex numbers stuff in further maths.. but you know it’s all learnt to pas an exam then forgotten after the exam. I got an A grade in each. This series is giving me a proper understanding of trig, and the other topics that I never had so firmly ever before. And even little things I never got round to getting comfortable with (- simplifying square roots!) So this level is totally perfect for me ! It’s like I’m taking off from where I was! I did economics at uni and didn’t really learn any maths so I’m loving taking my learning forwards from where I was 24 years ago.
I have an engineering degree and I genuinely thought that e^i.theta was referring to the exponential function, ie multiplying the number e by itself theta (or x) amount of times. Showing that it was the exp function (which is totally different) really opened it up for me.
@42:30 is it because we have defined exponential function in terms of Taylor series rather than power form where we have to define e value as 2.718 in equation..... i mean in Python 42:30 programming function is dependent on x*i variable and not on e value (since taylor series is used) so no matter what value we will defined for x*i program will not depend on e value itself. so 2.718 will not came in picture if Taylor series used, it is like we are defining function with different form.
Hey Grant, I signed up for twitter just to answer that f(a+b)=f(a)f(b) question. But couldn't properly figure out how twitter works :| However, f(0) could either be 0 or 1. Here's the proof. If a=b=0, f(0+0)=f(0)*f(0) => f(0)=f(0)² => f(0) = 0 or 1 if f(0)=0, f(x+0)=f(x)*f(0)=0 So, f(x)=0 is a valid solution. And f(0)=1 gives the exponential solution.
Really nice proof. For those confused on the jump from step f(0)=f(0)^2 to f(0) = 0 or 1, you can substitute f(0) = x, and now x = x^2, and x^2 - x = 0, which is a quadratic with solutions 0 and 1. And just to emphasize that this proof means that either f(x) must be the constant function f(x)=0, or f(0) must =1. EDIT: I just realized that if there is any c such that f(c)=0, then f(x)=0 is true for all x, since there is a number a where x = a + c. f(x) = f(a+c) f(a+c)=f(a)*f(c) f(x)=0 So to sum it all up, if there is any number c such that f(c)=0, then f(0)=0 and f(x)=0 for all x. Otherwise, f(0)=1 and f(x) is never 0.
Fantastic job--I mean the whole experience you created for us. The experience, and your energy made for an engaging live stream in mathematics. What I like most, that I didn't get in high school or college, is the ability to pause, rewind, re-listen and absorb--I always felt rushed. Thank you.
Grant, I just want to say how much joy these bring me every Tuesday and Friday. Is it possible/feasible for you to keep doing this in the long term? Or at least longer than quarantine may last, as it is helping to give a much better understanding of the math concepts that end up getting used in my classes.
The exponential formula only works if x and y multiplicatively commute. So for real and complex numbers it’s ok, but if x and y are matrices, if you try to expand out (x+y)^2 for example, you get x^2+xy+yx+y^2, which is not equal to x^2+2xy+y^2 which is what you need.
Implicitly, the function f is from R to R. I guess he could have stated it explicitly, though. He also missed the special case of f(x) = 0, which satisfies the condition but does not satisfy property (3).
Hi! Please do more of these with homework! I'm currently trying to teach myself math as a hobby and practice sets would be a fantastic resource. I've loved this lockdown live math series and more would be amazing even as covid etc. finally clears up
Another great connection is that we can plug ix into exp and split that into the real and imaginary parts to get the Taylor polynomials for cos and sin.
I was totally enthralled when I realized this is all about straightening out curves or patterns and get a convergence from them to define a coordinate.
Again, for the very last time Let's solve functional equation f(a+b)=f(a)f(b), which is valid for all reals. Let's plug b=0 into it. We get f(a)=f(a)f(0). We can put everything on the left handside and get f(a)-f(a)f(0)=f(a)(1-f(0))=0 If the product is 0, then atleast one of the multiples is 0, so we need to consider 2 cases: 1) f(a)=0 - well, this is the first solution of the equation. Due to it f(1)×f(-1)=0×0=0≠1 2)1-f(0)=0 or f(0)=1. To answer the question, we don't really need to continue and find functions that satisfy this statement, cause we immediately get f(-1)f(1)=f(-1+1)=f(0)=1 So 20:00 f(-1)=1/f(1) is true for all functions f(x)≠0. As there was no remark about that in the question, we need to agree that 3rd statement is NOT true for all such functions. Then answer E is wrong and C is true
Seeing the vectors forming e^πi was mind blowing. Brought me from a "decent understanding" to "complete intuition" You're a God in math, Grant. We love you!!
The beauty of seeing e^x written as exp(x) and the series expansion is that it becomes immediately obvious why the derivative of e^x is simply e^x. Thanks again for making these series! Pointing every high school kid I know this way to learn in maybe a different way to cement ideas and concepts in their heads.
This is not an obvious fact! It's pretty hard to prove that you can differentiate power series in a term by term fashion within their radius of convergence
@@ster2600 you're wrong. It does make it obvious. exp(x) is the sum of x^n/n! terms, plus a constant (1). So the derivative of exp(x) is the sum of the derivative of those terms. The derivative of x^n/n! is nx^(n-1)/n! which is x^(n-1)/(n-1)! Let's call N=n-1 The derivative of x^n/n! is x^N/N! So basically the derivative of each term is the previous term in the series. As it's an infinite series, that makes no difference as n grows. And on the other side, the lower n side, you get the derivative of x which is 1, so you get all your terms back from the original.
@@vincentandrieu5429 I think what Ster Chez was trying to say was that what's true in the finite world isn't always true in the infinite world. Just because the derivative of a finite sum is the sum of each terms' derivative (i.e. the additive property of derivatives), doesn't mean that the derivative of an infinite sum is also the sum of each of it's terms' derivatives. You would have to prove that this is a general property of derivatives (hint: it's not).
I'm watching this, and I think I know it well, but your videos are so good and insightful that it's probably worth watching anyway. I think you've shown me how to love math again, and how much insight can be gained with geometric interpretations.
In poll 2, only the first statement follows from the stated assumption. If you assume real-valuedness you get 2. If you assume non-triviality you get 3. 22:17 f(1/2)^2 = f(1) actually implies $f(1/2)=\pm \sqrt{f(1)}$. If the function must be real valued, you can use f(1/4)^2=f(1/2) to assert using the positive square root. If not, f(x)=exp(i pi x) is a counter example -22:37 the function f(x) =0 satisfies "for all real a, b, f(a+b)=f(a)f(b)". In the tweet as well, f(x+0)=f(x)f(0) is satisfied by f=0
he is just getting deeper and deeper into more complex topics, and I LOVE IT from functions to trig to imaginary numbers and Euler's formula then Taylor series which then touches on calculus and differential equations.
So the main takeaway of this lecture is: "e" does not literally represent the number 2.71828, whereby e^x means 2.71828 being multiplied by itself x amount of times. Rather, *e^x = (1+x/n)^n as n approaches infinity*
I commented that at lecture close the WTF was completely unclear to me. I believe that you have helped me get to grips with this central point. Many, many thanks. David Lixenberg
this episode was a life-saver for me as i was stuck trying to understand the Quantum Fourier Transform (QFT) in my "Intro to quantum computation" course. It involved matrix-vector multiplication where the matrix's entries were n-th (complex) roots of unity & the vector's entries were complex numbers (i.e. qubits' superposition). Now i can finally go ahead & understand period finding & then after that: shor's algorithm!! Thank you Grant!!!!
Great math lessons make me learn something new... but the greatest ones completely recontextualize things I thought I understood beforehand. This video was such an awesome walkthrough and, most importantly, made me understand WHY e^ipi works. Great job!
Man im just a 28 yrs old guy who just started electronics engineery. During my first year at university i realize that I havent got any idea of pure math and they are necesary for everything else. So I start to study by my own as much as i could to cover the blind points in my education. I came from a whole diferent world (i used to study laws and philosophy). A friend of mine luckly send me your "the escense of linear algebra" series. And man I fucking admire your work. You just make me feel pasionate about math and so on. And I supposed to be an engeenier not a mathematician. But here I am. Thank you very much. You not only help me to fill the blinds spaces. Also make me feel pasionate about pure math learning and its beauty and importance unlike my mates whom thogh this is a waste of time. Greatings from Argentina. Sorry about my english. And thanks again. You are incredible.
Solutions to this lesson's homework. 1. To show that the terms of exp(x)*exp(y) have the form x^k*y^m/(k!*m!), one can write the expression as (1+x+x^2/2+...+x^k/k!+...)(1+y+y^2/2+...+y^m/m!+...) By expansion, we can choose one term from the first bracket and another from the second bracket, and the power of x and y would be unique, so the term is just x^k/k!*y^m/m! or x^k*y^m/(k!*m!) 2. To show that exp(x+y) have terms of the form 1/n!*(n choose k) x^k*y^(n-k) Just write exp(x+y)=sum_{n=0}^{infty} (x+y)^n/n! Use (x+y)^n = sum_{k=0}^{n} (n choose k) x^k*y^(n-k) Again, the powers of x and y are always unique. Hence, the coefficient of x^k*y^(n-k) is (n choose k)/n! (remember division by n! from the exp function). Thus, proven. 3. To show that (1) and (2) imply that exp(x)*exp(y)=exp(x+y), see (2). Let n-k=m. We can always pick n=m+k such that we have powers x^k*y^m in the expansion. Additionally, we have the coefficient is (n choose k)/n!=((n!)/(k!*(n-k)!))/(n!) =1/(k!*m!), which is the same as 1. Hence, each term of exp(x+y) corresponds with a term of exp(x)exp(y). Thus, they are equal. 4. For real numbers, this is evidently okay. For complex numbers, we can easily justify that (1) and (2) work because of commutativity, associativity, and distributivity, because then we can do the algebra quite well like with the real numbers. As of matrices, we can define power of matrices and division by scalars, and so exp(A) is defined, given A is a square matrix. However, the property exp(X+Y)=exp(X)exp(Y) holds usually when XY=YX (they commute), such that the binomial theorem can hold (terms in the binomial expansion are stuff like XY and YX, which we could add together if they commuted, but that's not always the case for matrices) Consequently, exp((a+b)X)=exp(aX)exp(bX), where X is a square matrix, and a and b are scalars. Moreover, exp(X)exp(-X)=I, where I is the identity matrix that behaves like 1 in sense of multiplication (behaves like exp(0)). So yes, we can extend the definition for many different objects if we need to, like complex numbers and matrices, and these can benefit us in electric engineering or differential equations. Great homework, great lesson 3b1b! See you in the next lecture!
these lectures are just gems. thanks for what you do. and yeah-- not just high school students--- 39 year old physician with a part-time math interest here.
I had actually been pausing the video to do exactly the same thing. I was pretty pleased to find I had beaten him to showing me the exact same thing. I stumbled a bit more along the way, but (!!!) I realized that you do not have to clumsily type `complex(3,2)` - you can simply say 3+2j
As a calculus teacher, your graphics with the unit circle were outstanding and very insightful. I will definitely use that when I teach this topic this spring!
WOW I took Dr. Peyam's last class at UCI this past winter on multivariable calculus and "want to find" was the first thing that came to my mind too. I didn't think anybody else knew of him!
This channel is the best channel out here on mathematics. Although I still don't understand some of the harder content, it provides me with a deep intuition of various math topics. I feel like younger students like middle schoolers are also able to understand most of the content, since it is so well explained.
@@Joffrerap I think you are right, (2) does not necessarily follow if f is allowed to have complex values, although the resulting function will not be continuous. More formally, pick a value f1 for f(1), and let a_p be a p'th root of unity for all primes p (for your example we can take a_2 = -1 and a_p = 1 for all other p), and then define f(q) for rational q such that if q = n/d and d = p_1...p_n is the prime factorization of d, then f(q) = f1^(n/d) * a_p_1 * ... * a_p_n. This definition does not depend on the choice of n/d (it doesn't have to be lowest terms) because if you multiply numerator and denominator by any prime p, then the expression gets another factor of a_p^p which is 1 by assumption, and it also satisfies f(q + r) = f(q) * f(r). That covers the rational numbers, but to extend to real numbers you have to use the axiom of choice to find a linearly independent basis for R over Q (a Hamel basis), say b_i in R for I in some uncountable index set. Then any real number x can be uniquely written as a finite sum x = sum_(i in I) q_i * b_i where the q_i are rational numbers, and we let f(x) = prod_(i in I) f_i(q_i), where each f_i is a function on rational numbers satisfying f(q+r) = f(q)*f(r) as above (we can take them all to be the same if we want). If you were to plot this function, it would fill the plane, because it is wildly discontinuous at every point. There are uncountably many such functions, but none of them can be described in a nice way, and the choice of Hamel basis is nonconstructive so you can't calculate values of this function in your computer. The only smooth solutions to the equation are f(x) = 0 and f(x) = a^x for some a, possibly complex.
@@digama0 "This definition does not depend on the choice of n/d (it doesn't have to be lowest terms) because if you multiply numerator and denominator by any prime p, then the expression gets another factor of a_p^p ", why that ? i would think a_p and not a_p^p. also the way you extend f to R seems wrong, you don't use continuity. granted if you assume f continuous, and you know its values on rationals, then for x real, you can define f(x) = lim ( f( qn) ) where qn->x. You "just" have to show it does not depend on the choice of sequence qn.
@@Joffrerap You are right that the definition over Q is incorrect. I will have to think about this some more, but I am reasonably sure that you can arbitrarily choose the values of f(1) and f(1/p^i) for prime powers p^i subject to the restriction f(1/p^(i+1))^p = f(1/p^i), and this uniquely defines a function on Q. Clearly that information is enough to determine f(n/p^i), and to get arbitrary denominators, given a,b coprime, let ma+nb = 1 and then 1/ab = m/b + n/a, so f(1/a) and f(1/b) determine f(1/ab). The extension of the function f I defined on R is not continuous, indeed it's as far from continuous as you can get. The idea behind the Hamel basis is that we can view R as a Q vector space and then all the constraints act independently in each direction so we can make arbitrary choices and it all still works out.
Me doing my own Complex Analysis homework for an actual grade: I have no motivation... Me doing my 3Blue1Brown homework for no reason: This is important to me, so I must do!
Thanks so much for these lectures. I'm 48, and have a patchy engineering / programming / maths history... I wish you'd been my maths* teacher either when I was at school or at college as you've made previously difficult areas seem approachable and actually interesting, and not been afraid to mention areas which you don't find intuitive or agree with the convention. Also seeing you write things out by hand is helpful as I've always had untidy handwriting and got a hard time for it at school, so it's nice to see someone who hasn't got perfect penmanship but is still hyper-intelligent. * Yes, I'm a Brit!
Thank you. Wonderful video, finally understood this important concept. We need more people like you with such a level of understanding and interest teaching in the schools and colleges. Thanks to RUclips, now every person can access such quality education.
12:10 Mathologer video did a video on Euler's formula (e^(pi*i)+1=0) in Homer Simpson speak so the definition of e has to be used to make sense of it. Interesting take on the formula. I feel off you showing the Talor series of e^x and pulling out of nowhere.
That video is truly great. I hear you on the concern of pulling out the series "from nowhere". I suppose for many functions (e.g. sine) we start by seeing how they're defined, and slowly build up intuition from there. That's not a great answer, of course. As soon as you introduce a little calculus, one good motivation for this polynomial is that this is a function that is its own derivative, which you can see as soon as you learn the power rule, which in turn lets you describe lots of phenomena in nature where a rate of change depends on the value that's changing. The important point here is that we don't have to introduce it as a Taylor expansion (as it traditionally is), we could take it as the starting point from e and e^x pop out.
@@3blue1brown I first encounted e in this "function that is its own derivative" manner from my highschool calc1 class! I wasn't aware the tayler expansion that I learned in college calc2 was the traditional introduction of e.
At 14:10 on the video, the exp(3+4) and exp(3) * exp(4) actually diverge at the end and the last 2 digits are flipped. At 14:29, the values diverge even more at the end of the decimal. Not sure if there is something off in the function but that would show they are about the same but not the same, which makes a difference when calculating the effects over extended time or distance.
Grant, this is wonderful. I’m preparing to teach synchronous online classes, and I wanted to see how you conducted a live class. I love this beautiful formula, and the way you shared it with the world in this wonderful, interactive way. Thank you for giving me some ideas, for your lovely personality, your musical taste, and the beautiful math!
I am a critical care physician waging war on the front lines of Covid. Grant, your lectures offer wonder and insight for weary minds, rationally and calm in the midst of tragic irrational chaos. I pray for Covid to end but for these lectures to continue. A global community extends its gratitude for your helping us to "think different".
![Noob To Max With DRAGON REWORK In Blox Fruits [FULL MOVIE]](http://i.ytimg.com/vi/LnBMOinoOvA/mqdefault.jpg)
![The most beautiful equation in math, explained visually [Euler’s Formula]](/img/1.gif)
Summary of the confusion on problem 2. TL;DR, given the spirit of the question it should have also specified that f(x) > 0 for all x, and I'm clearly prone to deep confusion while trying to juggle the many balls of live broadcasting.
The question asks about a function f satisfying f(a + b) = f(a)f(b). The relevant part of the question was whether f(-1) must be 1 / f(1).
- It was originally graded in a way to suggest that this _does_ indeed need to be true.
- While "explaining" it, I realized that the explanation required that f(0) = 1, and questioned whether this is necessarily true. I even said, "oh, you could just scale it". This thought of scaling is not correct, since (c*f(a))(c*f(b)) = c^2 f(a + b), so the lhs and rhs doesn't scale the same way.
- At 27:48, Sam points out that the constant function f(x) = 0 doesn't work, which I misread as asking about whether f(x) = 0 for some x (sorry Sam!).
- Crispin points out that f(0) must be 1 because f(x) = f(x + 0) = f(x)*f(0). This is true, as long as f(x) does not equal 0 everywhere, but at that point, I was still weirdly blind to that edge case and enthusiastically accepted this as a reason that the question was originally graded correctly.
- At 49:07, more of the discussion that has been happening on twitter is brought to screen, where Eric correctly points out that Crispin's proof doesn't work for the constant function f(x) = 0, which Sam had said all along much earlier and I misread.
Thanks for the good discussion, and the tolerance of some live befuddlement.
I have a nice question does all functions that satisfy f(a+b) =f(a)f(b) produces values that are bigger than or equal to zero or can a function with such property have a negative output?
@@oximas Let's assume there is a continuous (you can draw it without lifting your pencil) function f(x) such that f(a+b) = f(a)f(b), and that at some place this function is negative, say f(c) < 0.
In the spirit of this post, we can see that f(c+0) = f(c) = f(c)f(0), so f(0) = 1.
We know that f(0) = 1 and f(c) < 0, which means that there must be a number d between c and 0 such that f(d) = 0.
This means that f(1+d) = f(1)f(d) = 0. But also f(2+d) = 0. And f(3+d). In fact, we would find that no matter what number x is input to the function, f(x) = 0.
But this contradicts our assumption that f(c) < 0. The contradiction means that the original assumption must be wrong, so there cannot be a continuous function f(x) such that f(a+b) = f(a)f(b) and f(c) < 0 for any number c.
Using this same argument, we can show that the only continuous function such that f(a+b) = f(a)f(b) that can take the value 0 is the function f(x) = 0.
@3Blue1Brown Thank you so much for taking the time to walk through not only the correct answer, but also your own confusion. So often students (and many teachers) see teachers as infallible sages. It's very encouraging to see teachers demonstrating to students the process of admitting their errors but then persisting towards reaching the correct solution. Keep up the great work!
@@oximas For real-valued functions negative values are not possible, because f(x) = (f(x/2))^2 >= 0.
If complex values are allowed then you can get negative values, for instance...
f(x) = e^(ix)
Then f(pi) is negative.
@@tomekczajka ohh i get it so all "real" functions (aka ones that don't contain imaginary values) only give nonnegative values, that's interesting.
thanks for the help
Video Timeline
0:00:00 Welcome
0:00:20 Ending Animation Preview
0:01:15 Reminders from previous lecture
0:03:30 Q1: Prompt (Relationship with e^iθ=…)
0:05:40 Q1: Results
0:07:15 WTF, Whats The Function
0:10:00 Exploring exp(x)
0:11:45 Exploring exp(x) in Python
0:14:45 Important exp(x) property
0:15:55 Q2: Prompt (Given f(a+b) = f(a)f(b)…)
0:17:30 Ask: Which is more interesting, special cases or the general case
0:20:00 Q2: Results
0:23:50 Will a zero break Q2?
0:25:40 The e^x convention
0:27:10 Q3: Prompt (i^2 = -1, i^n = -1)
0:27:45 Ask: Zero does not break Q2
0:30:20 Q3: Results
0:31:05 Comparison to Rotation
0:33:00 Visualizing this relationship
0:36:50 The special case of π
0:39:20 Periodic nature of this relationship
0:39:40 Q4: Prompt (e^3i)
0:41:35 Q4: Results
0:43:55 Explaining the celebrity equation
0:45:55 Homework / Things to think about
0:49:15 Ask: Zero does break Q2.
0:50:30 Closing Remarks
Water drinks at 0:17:10 & 0:27:45 & 0:40:05
Edits: Moved water drinks to the bottom, spelling errors, these timestamps should be for after the video is trimmed at "Welcome!"
My favorite is 0:07:15
doing gods work
Had a problem where last video's comment became unpinned after updating timestamps for trimmed video. These timestamps are for the trimmed video (assumed it was trimmed at 5:10) if you wish to use it before the video is trimmed, add 5 minutes and 10 seconds to the displayed timestamp.
I lolled at the WTF = what's the function part. I guess he's really prepared, with 69 and WTF.
Great
The memory rule I learned for the digits of e is
{Everyone knows how it starts}{Ibsen's birth year}{Again}{The angles of a right isosceles triangle}
2.7 1828 1828 459045
Leo Tolstoy birth year... etc.
how did I go this long without noticing that there's two 1828s in there
@@xyz39808 you could have gotten four extra digits for free.
@@xyz39808 When I first came across e I assumed the 1828 was recurring.
@@Jehannum2000 762 digits into pi, there are six nines in a row, which is an excellent point to stop memorizing and just say "et cetera".
As a non-native English speaker, I'm really grateful that someone finally explained what WTF means.
i keep asking my friends what "idk" and "idc" means but they just respond that they don't know and don't care :(
@@orang1921this deserves more credit
@@orang1921lol nice joke
😮
@@orang1921 I don't know you and I don't care to know you
When he wrote Wtf = Whats the function
I waited for him to laugh (cause i did) bu he didn’t. This guy is good on camera
That made me giggle too much
@@indianjitsingh8838 my nigga!
Dr Peyam reference (i guess)
I'm glad that he didn't say anything about it. I was watching this
Maybe didn't spend much time in the bar
PLEASE KEEP THESE LECTURES COMING! They're wonderful and are demistifying a lot for me - way past highschool :)
13:15 is where I left off
It's over. At least for now...
In all my years of highschool and engineering undergrad, the taylor series of e^x being the way to understand exp(x) was never emphasized like in the way here. Euler's formula was just the way to convert between polar and real/Im form of imaginary numbers and do some convenient maths that was noted down and refered to in a formula sheet. Closest was a professor explaining that defining trait of e^x is the function where its differential was the same as its integral, was the same as the value of e^x.
@@ca-ke9493 now that one threw me off in integration. every damn time a question asking integral of e^x, id be relieved, and id put e^x. so many missed marks
Ok
I think " Maths for the Curious " is a better title for this series than 'Highschool maths'.
As I don't know if this lecture would specifically help in highschool exams etc.
But I think it is for anyone who is curious and wants to learn math with a little more intuition and creativity.
For example: I work in theoretical physics, and have already learnt lots of Maths. Still going back and understanding math fundamentals in this beautiful way accentuates my understanding.
Which is why I believe that " Maths for the Curious " is a better title, age no bar.
Interesting perspective! And probably true :)
@@3blue1brown Hey, Thanx for the reply! Would love it if something similar was implemented in the next lecture onwards. While highschool maths reminds most people of dull problems and unnecessary competition, this on the other hand is truly magical.
Love this series and infact for that matter all your material is brilliant, especially the " essence of " series. Keep up the great work!
+
I am also in High School and i agree. I for example wont have complex numbers in school before uni. Atm tho i think the lessons are in a perfect difficulty. Its quite challenging but if one invests a few hrs most should be able to manage iy
I think it depends where you go to high school. This is in the syllabus of high schoolers in certain parts of Europe and Asia 😅
But regardless of what's in school syllabus, I feel that an appreciation of math like this is all the more important for people in their early to mid teens. So reaching out to high schoolers is a great thing.
37:22 Grant: π² is same as g
Me: satisfied engineer noises
mathematician: π² = g
engineer: 3² = 10
Heresy!
@@JM-us3fr g = 10
@@JNCressey Cosmologist: e² = 1+- 10^4
@@JNCressey Mathematicians don't believe in g, surely you mean a physicist.
I remember a couple years ago, during a particularly boring physics lesson, I messed around with my calculator and typed the sum over the factorials to see what i would get. To my surprise, i found out that the answer was 2.718... so i thought to myself "wait, it can't be", i took the ln of it and sure enough it was 1. I was blown away, and i thought i found a secret way to calculate e. Later I took calc 1 and discovered that i discovered nothing new, but still, that feeling when I accidentally stumbled upon this formula for e was really something else, and throughout the last 2 years of my bachelor's math degree, I only ever felt that way once again, the feeling of pride that i discovered something new and beautiful.
Oh yes, I love epiphanies like this! Aren't they great? I think when you've stumbled upon something in this way you simply can't ever forget it. If only we could learn everything through personal discovery, we'd probably retain it all a lot better.
@Rayan I remember finding the same thing but by counting tiles when i was younger. I realized I could find the next square and made a formula for (x+1)^2 until i realized that there was an (a+b)^2 formula, which was equivalent.
For me, my search for the function equal to the sum of the factorials lead me down the rabbithole of calc. Little did I know that there was no elementary function describing the sum of factorials. It all stemmed from me discovering a neat formula, that the sum of the first n natural numbers is n(n+1)/2. That was really the start of my journey.
Mine was finding a formula for calculating pi based on the limit of regular polygons.
What was it he second time?
His excitement is contagious. And seeing him get flustered because he was live and didn't want to say the wrong thing makes me feel better about how I freeze up or get flustered. We're all human!
Genuine question: Why aren't we taught this concept intuitively at institutions we literally pay thousands for each year? Why is it that we have to come to a free source to learn these things deeply?
Pedagogy is a field that deserves to be studied and improved in it's own right. Unfortunately some institutions just look for teachers with dexterity in the branch of knowledge being taught but not in the knowledge of how to teach.
Schools have to make it "hard" to distinguish "good" students
Being a great teacher and being a great mathematician are different skills and hard to find in the same person. Also, being a great teacher is not particularly valued by society so great communicators who deeply understand a subject frequently find other jobs. Ideally a University should hire mathematicians to do research and not force them to teach and hire teachers to teach.. but that would be too expensive. High schools only hire teachers but there aren't enough talented individuals in the world who deeply understand math, love teaching, and love students. Grant is like the LeBron James of math education and by definition every highschool in the world cant have their own LeBron James.
@@jeffjiang5272 I disagree. Those "good" students usually can't care less about what the teacher is saying as they are probably way ahead of the class. Or atleast that's how it is in high school. The way they teach math is bad because math teachers dont design the curriculum but education specialists who barely know any math.
Because of time.... I can't believe people don't understand this simple thing... This video takes an hour to deeply and beautifully explain the Euler's formula which is one part of complex numbers...
While school teachers get about 30 minutes a class to do so
What I admire about you, is how you try to teach from a perspective of someone who doesn't know it. Most often, teachers forget how they felt while learning it and it becomes harder for them to explain than it was to learn.
In the proof of the necessity that f(0) = 1, the hole in the logic was after the step f(0)f(x) = f(x). This results in f(0) = 1 if and only if f(x) is not zero (so that we don't have to divide by 0). This means that, while true that f(x) = 0 is an exception, it's also the only exception.
Also at 24:03, where you say that you could scale the function to get a different result for f(0), that wouldn't work because it would no longer satisfy f(x+y)=f(x)f(y). Multiplying the two outputs would result in your scaling factor squared on the right hand side, while the left hand side would only have a single scaling factor.
With that said, I really enjoyed this lecture! Can't wait for the next one :)
This is what I'm looking for. Nicely done
I’m a middle-aged financial engineer and learn from your lectures- and I was definitely paying attention to math classes in high school. Watching your videos and the beautiful new perspective you cast on sometimes elementary topics is like re-watching a classic movie or re-reading a classic novel and getting whole new appreciation for the material as if you were reading it for the very first time.
I just want to say that this series is absolutely awesome. I never thought of myself as a math person much less a math nerd and yet I just sat here for hours and have learned and understood so many things I never grasped before so Thank you! I really wish for this to become a regular thing even after the world returns to "normal". Would I not already be a Patreon this series would have definitely earned it. As a student of I often struggle to understand math and as of now firmly belong in the category of didn't understand it but just went along with it as well as I could but you are really changing my perspective on maths and are showing me that instead of frustrating it can be fun and interesting.
EDIT:
Okay it seems I accidentally canceled my Patreon subscription last time I cleared up my payments! This grave mistake has been remedied. So in a way this video DID earn my subscription (again).
For those of you who haven't seen it, Mathologer has a wonderful video on Euler's formula: ruclips.net/video/-dhHrg-KbJ0/видео.html
A worthy question is to ask what the connection is between the limit he writes and the polynomial here. Perhaps good fodder for the next lecture :)
Please can you tell me which tool is used to make such a great video ?!
Sam was saying that f(x)=0 satisfies f(a)f(b)=f(a+b), but does not satisfy condition 3, and therefore, only answer 1 and 2 are necessarily correct.
23:48 f(0) must be 1, because f(n) = f(n+0) = f(n)f(0) = f(n).1 = f(n)
edit: just when f(n) =/= 0.
@@karkaroff1617 If f(n)=0 then we have 0*f(0)=0. And f(0) could 1 or 0 or any other constant. Therefore, if f(n)=0, then equation 3 isn't satisfied. The problem should have stated that f(n) is not 0.
One topic I had always wondered about was a more direct (but challenging) way of defining a^x (a,x∈R, a>0) in terms of Cauchy sequences (a^qₙ) where rational every qₙ is rational and qₙ→x. It would require proving that every such sequence was Cauchy, so by completeness they converge to some real y, and in particular that they converge to the same y. Then (using whatever is your favorite definition of e), exp(x) = e^x, and log(x) = exp⁻¹(x).
This makes it a direct extension of exponentiation with rational exponents, which are essentially defined by the requirements that a^x * a^y=a^(x+y) and a^1 = a, along with the convention that a^x > 0. It feels a little more motivated than the backdoor method of, for instance, defining log(x) in terms of an antiderivative, then defining and extending its inverse, and finally proving that the resulting function corresponds with the usual definition for rational arguments. Since it is continuous by construction, these two approaches are equivalent, but it feels kind of . . . slippery.
I'm a mechanical engineering graduate who's in the midst of studying for my math subject GRE in the hopes of pursuing academia and go back to grad school for pure mathematics. Your videos have not only helped me throughout the last year with grasping abstract ideas conceptually, but you've also helped me gain a whole new appreciation for mathematics as a whole. Thank you sir. You are a treasure to this world.
Grant I really hope that there are a good 20-30 more of these lectures in the future, they are so much more insightful that high school math and they have really given me an interest in diving into more complex math, keep up the good work and please don’t stop these lectures once we end lockdown
~~~ HOMEWORK ~~~
Time-stamp: 45:56
1. Fully expand exp(x) * exp(y).
2. Expand exp(x + y). Hint: Binomial Formula.
3. Compare the two expansions above.
**4**. Show the properties for complex numbers and matrices.
I don't know why, but this lecture seemed too short to me, even though it was only 10 minutes shorter than the previous ones. Apparently, I've become addicted to your awesome videos, Grant! Well done!
Well, really a lot of the pther lectures were about 70 minutes, but, yeah, this lecture felt very shirt.
Me too. It felt like half of the regular ones.
Lol my watch time this last week has been like 60% 3b1b and that says a lot because I’m predominantly watching yt
It's first 5 or 10 minutes were afk, it is a short one
These are really fascinating - I'm a curious high school student and I'm loving seeing these things I just learned in a new way. Thanks for being awesome :)
I had actually never seen the Euler Formula, but the connection to the unit circle makes it easier to digest
Grant: *drinks water
Me: Write that down!
Will it be on the test?
The animation of the vector interpretation of the terms of the power series... such beauty. Absolutely beautiful.
Fun observation: when you plug in a positive integer N into Maclaurin series form of exp(x) the individual terms are increasing until you get to the Nth term, the N+1th term equals the Nth term (using indexing that calls x^k/k! the k+1th term), and the culmulative sum of the terms passes the halfway point exp(N)/2 of the total sum between the adding on of the Nth and the N+1th terms. You can ask, if you summed up to the Nth term, what fraction of the N+1th term do you need to add on to get to exactly exp(N)/2. The answer appears to be very close to one-third in the limit of large N, but not quite. It's also close to the square of the Euler-Mascheroni constant.
Ho ho ho ho, le Poisson !
Oily Macaroni constant
I love how you just deadpan the WTF acronym
Hi
Grant: "so you see this weird formula, I think the healthy question to ask is WTF..."
Me: :)
Grant: "... - whats the function?"
Me: :O
omg same
@ali PMPAINT yeah lmao... It's even funnier that way to me.
I stopped the video and went down the comments to see If it was only me with dirty head hhhh
Ol
AHHAHAHAHA
That's actually a really interesting point of view that I haven't seen before, that writing e^i is arguably an abuse of notation for exp(i), and in general the fact that e^x and exp(x) are two different functions that just happen to have the same value everywhere that e^x is defined, e^x meaning "multiply e by itself x times" and exp(x) meaning "do this infinite sum on x." The strict e^x would be undefined on imaginary numbers, but because it's equal to exp(x) whenever it does exist, we just write e^ix to mean exp(ix) without much confusion once you understand the convention.
It's pretty similar to the relationship between factorial and the gamma function, which you actually allude to here. x! and Gamma(x+1) are two different functions that just happen to have the same value everywhere that x! is defined (though there is the +1 that makes it annoying). In a strict sense, (1/2)! is undefined; how could you take every integer starting from 1/2 and going down to 1 and multiply them all together? But because Gamma(x), defined by a funny integral, is the same as the factorial where they do exist (offset by 1), we often will write (1/2)! to actually mean Gamma(3/2), and there isn't really any confusion there.
Analytic continuation?
The proof f(x + 0) = f(x) * f(0) => f(0) = 1 only needs the condition f(x) does not equal 0 to be valid. Whether or not f(-1) = 1 / f(1) should count as a valid answer is up for debate.
Yeah, so if for f(a+b)=f(a)*f(b) you put a and b as 0, you get two solutions for f(0), ie, 0 and 1. Whichever you take is up to you. Usually, in math questions like this, the question usually has the assumption that f(0) is not equal to zero. Funny how that was the first thing that popped into my mind, from years of solving objective questions.
There are also undefined values. Technically, the following can work:
f(x) = 0 if x >= 0
f(x) = undefined if x < 0
If undefined * 0 = 0 is a valid output (e.g. f(-1)*f(2)=f(1)). Because it also ends up giving undefined * 0 = undefined (e.g. f(-2)*f(1)=f(-1))
@@chaosredefined3834 he said that the given property holds for all real numbers, which implies that the function is defined for all real numbers
@@rhitamdutta1996 homomorphism moment
Rhitam Dutta In all of the math classes I’ve taken any number to the power of 0 has been 1. Does the possibility of 0 come from that it can be any function that has this property? What’s an example. The last question f(-1)= 1/f(1); I reasoned that a^(-1)=1/a. Assuming f(x) to be a^(x) based on the fact that this function has the given property. Is this correct reasoning?
This video is simply beautiful. I am one those engineers who just accepted euler's formula and used it without ever deeply, intuitively understanding it. This is the best explanation of the formula. Thank you so much. Please keep making more of these videos.
you can easily prove the equality if you derive
Thanks for doing this.
Today I was unable to attend live, but I’d pause and figure out some of the answers at each step. I truly appreciate the simplicity of the expansion series, and the resulting connection to the rotation.
As others have mentioned, I do wish this is how the teachers had explained this in math class.
I really cannot show in words how much I enjoy your lectures. Whenever I see your videos I end up, "WOW, what an amazing way to look at this problem."
@24:00
F(a+b) = F(a)*F(b)
put a=0 and b= -1
thus F(0-1) = F(0)*F(-1), But F(0-1)= F(-1),
hence F(0) needs necessarily to be 1.
Again F(-1)=F(1-2)=F(1)*F(-2)
diviiding both sides by F(-1)
F(-1)/F(-1)=F(1)*F(-2)/F(-1)
hence 1=F(1)*F(-1-1)/F(-1)
once again 1 = F(1)*(F-1)*F(-1)/F(-1)
Thus F(1)*(F-1)=1
so, all the three equations are necessarily true for a function with property F(a+b) = F(a)*F(b).
Thus option E is correct.
I’ve just finished year 11 in the UK (16 years old) and I’ve never come across these topics before but they are incredibly well explained and very interesting. I’m happy to be enjoying maths more than I usually would and learn some topics which I’m sure I will come across next year in A-Levels.
William McLaughlin hey i am interested what your maths syllabus include cause these topics are actually in my normal math syllabus in India although no Euler formula is shown, it's a plus to know in competitive exams
And do further maths. Normal maths has no linear algebra, no complex numbers
Wish I had seen these videos during my education, you have a massive advantage!
@@AdityaKumar-ij5ok Another UK student here, just finished with year 11
And I can confirm, the topics are massively disappointing. At the end of the year we did like, 1 lesson on vectors, and that was just how to add them. And in one of the exams we did, the only question about vectors was the very last question.
It's mostly just relatively basic geometric proofs, quadratics, and.. that's pretty much it, actually.
That is for the higher tier test, by the way - the foundation test is mostly comprised of addition, multiplication, ratios, etc
Just like the original comment, I'm also going to take A-Level maths, and _hopefully_ it'll be at least half as interesting as these videos.
0:45 That animation says it all! And that's why 3Blue1Brown is the best! You taught me what 7 years of Calculus at University didn't. Thank you!
It's like someone built the perfect teacher in a lab. Thank you so much.
I just got my mind blown!! I'm engineer and it's the first time I've seen such explanation about exp(iθ)! Great lesson! Thank you Grant!!
This is some serious educational content, your professionalism and passion to share math knowledge is astonishing. Keep up the good work!
My University math professor had a genuine adoration of math. Failing his class is one of the few things I regret in college. His last gift to us on the last day of class was to show us Euler's formula. He talked about how it beautifully encapsulated everything we had studied up to this point and about how beautiful math is.
You tricked me into listening to a full math lecture by explaining it in an interesting way.
Videos like these are useful not just for highschool students. For them, sure, they may be helpful in leaming these topics.
For people like me, who has studied maths not that long ago and still remembers most of the ideas, watching these videos is very pleasing because I can focus specifically on logical, intuitive details of your explanations without having to struggle memorizing all of this new info from scratch.
Yes, we've been taught all these topics, I know how to do all of that mathematically, but it is extremely satisfying to get an intuitive look on things that you already know in the form that is usually taught, re-think them, make the information in my head more organised and open possibilities to use this knowledge in real life.
Thank you so much! It took a while, but at 37:21, the idea of how rotation connects with the definition of i and the complex plane just finally clicked, seeing the exp(i*theta). Your animations are wildly helpful!
Thank you.
I have a bachelors in math and I think this is the first time I actually fully understood the e^(pi*i) = -1. Thank you!
Man your lessons are incredibly good. It's people like you who change the world. Looking forward for other videos !
I laughed so hard when he said WTF "What's the function ofc" !
lol
Note well the citation source ;)
Well That's Fantastic!
All mathematicians know that WTF stands for "Want To Find." lol
WFT
I saw that!
OMG you just answered me a question wich confused me for 2 years now and I already gave up to understand why e^i*pi should make sense. It's because it's actually not e^i*pi but exp(i*pi) where exp(x) not necesseraly is equal to the thing we have in mind when we see e^x. THANK YOU!
See, *this* is the confusion caused by the presentation. exp(x) (the infinite one, not the python approximation) *is exactly* e^x ; the Taylor series *is exactly* equal to the function it's derived from, even in the complex plane... It ends up informing our understanding of what exponentiation to a complex value means.
My takeaway: exp(x) is very hard to introduce intuitively without the use of calculus. A valiant effort all the same, as always! Well done!
The real significance of the function is that it's the function (well, it and any constant multiply of it, but it's the only of those that also has the algebraic property discussed in the video) whose derivative is itself, and the definition inherently entails calculus.
@@ganondorfchampin my point exactly!
@@Gold161803 Slight correction to what I said, the constant function 0 ALSO has the algebraic property mentioned in the video AND is it's own derivative, but exp is more interesting for obvious reasons.
As I watched the video, I asked myself the question. Why is e that perfect canonical base for the unit square? It turns out, that any number k bigger than 1 would work since k^x = e^(ln(k)*x) it just slows down (k < e) or speeds up (e < k) the period of the function. e is just the perfect value where e^(i*x) = e^(i*x+2*pi*n) for x in the real domain.
That was the first time I felt really confident to try and go for the question and I am really happy to see I could find the correct answer. (around min 20)
That may be silly but I feel like that is my first step to really understand math.
I did A-level maths and and A-level further maths in 1996 at age 18 (high school).
We did some complex numbers stuff in further maths.. but you know it’s all learnt to pas an exam then forgotten after the exam. I got an A grade in each.
This series is giving me a proper understanding of trig, and the other topics that I never had so firmly ever before.
And even little things I never got round to getting comfortable with (- simplifying square roots!)
So this level is totally perfect for me !
It’s like I’m taking off from where I was! I did economics at uni and didn’t really learn any maths so I’m loving taking my learning forwards from where I was 24 years ago.
I have an engineering degree and I genuinely thought that e^i.theta was referring to the exponential function, ie multiplying the number e by itself theta (or x) amount of times. Showing that it was the exp function (which is totally different) really opened it up for me.
It ultimately is the same function. Just a different perspective of it.
@42:30 is it because we have defined exponential function in terms of Taylor series rather than power form where we have to define e value as 2.718 in equation..... i mean in Python 42:30 programming function is dependent on x*i variable and not on e value (since taylor series is used) so no matter what value we will defined for x*i program will not depend on e value itself.
so 2.718 will not came in picture if Taylor series used, it is like we are defining function with different form.
Hey Grant,
I signed up for twitter just to answer that f(a+b)=f(a)f(b) question. But couldn't properly figure out how twitter works :|
However, f(0) could either be 0 or 1. Here's the proof.
If a=b=0, f(0+0)=f(0)*f(0)
=> f(0)=f(0)²
=> f(0) = 0 or 1
if f(0)=0, f(x+0)=f(x)*f(0)=0
So, f(x)=0 is a valid solution.
And f(0)=1 gives the exponential solution.
f(x) = 0 is a really boring solution though, "yeah lets have the kernel be the reals LOL what a trolllllllllllllllll"
Perfect, thanks!
Really nice proof. For those confused on the jump from step f(0)=f(0)^2 to f(0) = 0 or 1, you can substitute f(0) = x, and now x = x^2, and x^2 - x = 0, which is a quadratic with solutions 0 and 1.
And just to emphasize that this proof means that either f(x) must be the constant function f(x)=0, or f(0) must =1.
EDIT:
I just realized that if there is any c such that f(c)=0, then f(x)=0 is true for all x, since there is a number a where x = a + c.
f(x) = f(a+c)
f(a+c)=f(a)*f(c)
f(x)=0
So to sum it all up, if there is any number c such that f(c)=0, then f(0)=0 and f(x)=0 for all x. Otherwise, f(0)=1 and f(x) is never 0.
Yes :) when doing this and getting the less boring answer we must assume that f(any real)=0 is not the case (such as for the exp(x) function.
Fantastic job--I mean the whole experience you created for us. The experience, and your energy made for an engaging live stream in mathematics. What I like most, that I didn't get in high school or college, is the ability to pause, rewind, re-listen and absorb--I always felt rushed. Thank you.
Grant, I just want to say how much joy these bring me every Tuesday and Friday. Is it possible/feasible for you to keep doing this in the long term? Or at least longer than quarantine may last, as it is helping to give a much better understanding of the math concepts that end up getting used in my classes.
You sir are a legendary wizard! I wish all of my math teachers taught like you do. Very insightful and satisfying. Keep up the good wizardry!
The exponential formula only works if x and y multiplicatively commute. So for real and complex numbers it’s ok, but if x and y are matrices, if you try to expand out (x+y)^2 for example, you get x^2+xy+yx+y^2, which is not equal to x^2+2xy+y^2 which is what you need.
Implicitly, the function f is from R to R. I guess he could have stated it explicitly, though. He also missed the special case of f(x) = 0, which satisfies the condition but does not satisfy property (3).
For anyone that wonders, if there is a corresponding equation for noncommuting matrices, it's called the Baker-Campbell-Hausdorff formula.
EebstertheGreat oh I guess I should point out that I was answering the homework question number (4*). Sorry!
Hi! Please do more of these with homework! I'm currently trying to teach myself math as a hobby and practice sets would be a fantastic resource. I've loved this lockdown live math series and more would be amazing even as covid etc. finally clears up
Another great connection is that we can plug ix into exp and split that into the real and imaginary parts to get the Taylor polynomials for cos and sin.
I was totally enthralled when I realized this is all about straightening out curves or patterns and get a convergence from them to define a coordinate.
Again, for the very last time
Let's solve functional equation f(a+b)=f(a)f(b), which is valid for all reals.
Let's plug b=0 into it. We get f(a)=f(a)f(0). We can put everything on the left handside and get f(a)-f(a)f(0)=f(a)(1-f(0))=0
If the product is 0, then atleast one of the multiples is 0, so we need to consider 2 cases:
1) f(a)=0 - well, this is the first solution of the equation. Due to it f(1)×f(-1)=0×0=0≠1
2)1-f(0)=0 or f(0)=1. To answer the question, we don't really need to continue and find functions that satisfy this statement, cause we immediately get f(-1)f(1)=f(-1+1)=f(0)=1
So
20:00
f(-1)=1/f(1) is true for all functions f(x)≠0. As there was no remark about that in the question, we need to agree that 3rd statement is NOT true for all such functions. Then answer E is wrong and C is true
Wouldn't b be true? I think the c was cases 1, and 3.
@@hexa3389 No, you missed "none of them".
@@duyaa9526 we have to assume the range of the function will be real as well, or the question is nonsensical
Seeing the vectors forming e^πi was mind blowing. Brought me from a "decent understanding" to "complete intuition"
You're a God in math, Grant. We love you!!
The beauty of seeing e^x written as exp(x) and the series expansion is that it becomes immediately obvious why the derivative of e^x is simply e^x. Thanks again for making these series! Pointing every high school kid I know this way to learn in maybe a different way to cement ideas and concepts in their heads.
This is not an obvious fact! It's pretty hard to prove that you can differentiate power series in a term by term fashion within their radius of convergence
@@ster2600 you're wrong. It does make it obvious. exp(x) is the sum of x^n/n! terms, plus a constant (1).
So the derivative of exp(x) is the sum of the derivative of those terms.
The derivative of x^n/n! is nx^(n-1)/n!
which is x^(n-1)/(n-1)!
Let's call N=n-1
The derivative of
x^n/n!
is
x^N/N!
So basically the derivative of each term is the previous term in the series. As it's an infinite series, that makes no difference as n grows. And on the other side, the lower n side, you get the derivative of x which is 1, so you get all your terms back from the original.
@@vincentandrieu5429 that's only true for finitely many terms, while exp is an infinite series
@@ster2600 What I wrote is not valid for finite series. It's only valid for infinite series, and exp is an infinite series.
@@vincentandrieu5429 I think what Ster Chez was trying to say was that what's true in the finite world isn't always true in the infinite world. Just because the derivative of a finite sum is the sum of each terms' derivative (i.e. the additive property of derivatives), doesn't mean that the derivative of an infinite sum is also the sum of each of it's terms' derivatives.
You would have to prove that this is a general property of derivatives (hint: it's not).
I'm watching this, and I think I know it well, but your videos are so good and insightful that it's probably worth watching anyway. I think you've shown me how to love math again, and how much insight can be gained with geometric interpretations.
0:35 it’s so satisfying that the 3b1b logo perfectly aligns right with the down count of the rhythm.
Alex King this part got cut off :(
In poll 2, only the first statement follows from the stated assumption. If you assume real-valuedness you get 2. If you assume non-triviality you get 3.
22:17 f(1/2)^2 = f(1) actually implies $f(1/2)=\pm \sqrt{f(1)}$. If the function must be real valued, you can use f(1/4)^2=f(1/2) to assert using the positive square root. If not, f(x)=exp(i pi x) is a counter example
-22:37 the function f(x) =0 satisfies "for all real a, b, f(a+b)=f(a)f(b)". In the tweet as well, f(x+0)=f(x)f(0) is satisfied by f=0
43:05
python has a built in round function, no need to import numpy.
great video, thank a lot!!
he is just getting deeper and deeper into more complex topics, and I LOVE IT
from functions to trig to imaginary numbers and Euler's formula then Taylor series which then touches on calculus and differential equations.
So the main takeaway of this lecture is:
"e" does not literally represent the number 2.71828, whereby e^x means 2.71828 being multiplied by itself x amount of times.
Rather, *e^x = (1+x/n)^n as n approaches infinity*
I commented that at lecture close the WTF was completely unclear to me. I believe that you have helped me get to grips with this central point.
Many, many thanks. David Lixenberg
Grant, i just want to thank you for single-handedly making me interested in math again after highschool made it boring.
37:30 Ah, you’ve stumbled across the fundamental theorem of engineering. pi=e=3=sqrt(g)
As an engineer, I see no problems here
As a math nerd, I need to bleach my eyes
Is this a fuzzy math joke?
this episode was a life-saver for me as i was stuck trying to understand the Quantum Fourier Transform (QFT) in my "Intro to quantum computation" course. It involved matrix-vector multiplication where the matrix's entries were n-th (complex) roots of unity & the vector's entries were complex numbers (i.e. qubits' superposition).
Now i can finally go ahead & understand period finding & then after that: shor's algorithm!!
Thank you Grant!!!!
I'm 3 hours late due to a scheduled physics lab but I am very grateful for this 'Lockdown' series !!
Great math lessons make me learn something new... but the greatest ones completely recontextualize things I thought I understood beforehand. This video was such an awesome walkthrough and, most importantly, made me understand WHY e^ipi works. Great job!
The question song at the beginning is nice.
Vincent Rubinetti --- Grant's New Etude
@@swaree you people make the world great.
Man im just a 28 yrs old guy who just started electronics engineery. During my first year at university i realize that I havent got any idea of pure math and they are necesary for everything else. So I start to study by my own as much as i could to cover the blind points in my education. I came from a whole diferent world (i used to study laws and philosophy). A friend of mine luckly send me your "the escense of linear algebra" series. And man I fucking admire your work. You just make me feel pasionate about math and so on. And I supposed to be an engeenier not a mathematician. But here I am. Thank you very much. You not only help me to fill the blinds spaces. Also make me feel pasionate about pure math learning and its beauty and importance unlike my mates whom thogh this is a waste of time. Greatings from Argentina. Sorry about my english. And thanks again. You are incredible.
Solutions to this lesson's homework.
1. To show that the terms of exp(x)*exp(y) have
the form x^k*y^m/(k!*m!), one can write the expression as
(1+x+x^2/2+...+x^k/k!+...)(1+y+y^2/2+...+y^m/m!+...)
By expansion, we can choose one term from the first bracket
and another from the second bracket, and the power of x and y
would be unique, so the term is just x^k/k!*y^m/m! or x^k*y^m/(k!*m!)
2. To show that exp(x+y) have terms of the form 1/n!*(n choose k) x^k*y^(n-k)
Just write exp(x+y)=sum_{n=0}^{infty} (x+y)^n/n!
Use (x+y)^n = sum_{k=0}^{n} (n choose k) x^k*y^(n-k)
Again, the powers of x and y are always unique. Hence, the coefficient of x^k*y^(n-k) is
(n choose k)/n! (remember division by n! from the exp function). Thus, proven.
3. To show that (1) and (2) imply that exp(x)*exp(y)=exp(x+y), see (2).
Let n-k=m. We can always pick n=m+k such that we have powers x^k*y^m in the expansion.
Additionally, we have the coefficient is (n choose k)/n!=((n!)/(k!*(n-k)!))/(n!)
=1/(k!*m!), which is the same as 1. Hence, each term of exp(x+y) corresponds with a term
of exp(x)exp(y). Thus, they are equal.
4. For real numbers, this is evidently okay.
For complex numbers, we can easily justify that (1) and (2) work because of commutativity, associativity, and distributivity,
because then we can do the algebra quite well like with the real numbers.
As of matrices, we can define power of matrices and division by scalars, and so exp(A) is defined, given A is a square matrix.
However, the property exp(X+Y)=exp(X)exp(Y) holds usually when XY=YX (they commute), such that the binomial theorem can hold
(terms in the binomial expansion are stuff like XY and YX, which we could add together if they commuted, but that's not always the case for matrices)
Consequently, exp((a+b)X)=exp(aX)exp(bX), where X is a square matrix, and a and b are scalars.
Moreover, exp(X)exp(-X)=I, where I is the identity matrix that behaves like 1 in sense of multiplication (behaves like exp(0)).
So yes, we can extend the definition for many different objects if we need to, like complex numbers and matrices, and these can benefit us in
electric engineering or differential equations.
Great homework, great lesson 3b1b! See you in the next lecture!
thanks a lot !
Please do more of these live vids. They are awesome! No need for lockdown, just do it!
Hey Grant, please make WTF T-shirts !
Yes please. I won't buy one cause I'd be bullied to death at school but it'll be cool.
I'd buy it.
Buyer beware!
yeah "WTF?" on the front and "What's the Function?" on the back!
Should be something like this:
WTF:
What's the Function?
this is incredible. One of the most worthwhile math videos I've ever seen.
these lectures are just gems. thanks for what you do. and yeah-- not just high school students--- 39 year old physician with a part-time math interest here.
I wish that internet in the home was a thing when I was a kid. This channel would have been amazing when I was in high school / college.
WTF LOL
also I loved that you used python and showed it to us
I had actually been pausing the video to do exactly the same thing. I was pretty pleased to find I had beaten him to showing me the exact same thing. I stumbled a bit more along the way, but (!!!) I realized that you do not have to clumsily type `complex(3,2)` - you can simply say 3+2j
WHAT'S THE FUNCTION LOL.
As a calculus teacher, your graphics with the unit circle were outstanding and very insightful. I will definitely use that when I teach this topic this spring!
Grant: WTF= what's the function?
Dr.Peyam: WTF= want to find
Us: WTF..
= We're thy fans!
WOW I took Dr. Peyam's last class at UCI this past winter on multivariable calculus and "want to find" was the first thing that came to my mind too. I didn't think anybody else knew of him!
This channel is the best channel out here on mathematics. Although I still don't understand some of the harder content, it provides me with a deep intuition of various math topics. I feel like younger students like middle schoolers are also able to understand most of the content, since it is so well explained.
27:16 what if f(1/2)= -sqrt( f(1) )?
i guess then f(1/4) would have to be imaginary which is ok.
You know that, for all x, f(x) = f(x/2 + x/2) = f(x/2)^2 >= 0, therefore f is positive.
But you're right he was a bit quick on that.
@@MrTimy06 exept for imaginary numbers. i^2 < 0
@@Joffrerap I think you are right, (2) does not necessarily follow if f is allowed to have complex values, although the resulting function will not be continuous. More formally, pick a value f1 for f(1), and let a_p be a p'th root of unity for all primes p (for your example we can take a_2 = -1 and a_p = 1 for all other p), and then define f(q) for rational q such that if q = n/d and d = p_1...p_n is the prime factorization of d, then f(q) = f1^(n/d) * a_p_1 * ... * a_p_n. This definition does not depend on the choice of n/d (it doesn't have to be lowest terms) because if you multiply numerator and denominator by any prime p, then the expression gets another factor of a_p^p which is 1 by assumption, and it also satisfies f(q + r) = f(q) * f(r).
That covers the rational numbers, but to extend to real numbers you have to use the axiom of choice to find a linearly independent basis for R over Q (a Hamel basis), say b_i in R for I in some uncountable index set. Then any real number x can be uniquely written as a finite sum x = sum_(i in I) q_i * b_i where the q_i are rational numbers, and we let f(x) = prod_(i in I) f_i(q_i), where each f_i is a function on rational numbers satisfying f(q+r) = f(q)*f(r) as above (we can take them all to be the same if we want).
If you were to plot this function, it would fill the plane, because it is wildly discontinuous at every point. There are uncountably many such functions, but none of them can be described in a nice way, and the choice of Hamel basis is nonconstructive so you can't calculate values of this function in your computer. The only smooth solutions to the equation are f(x) = 0 and f(x) = a^x for some a, possibly complex.
@@digama0 "This definition does not depend on the choice of n/d (it doesn't have to be lowest terms) because if you multiply numerator and denominator by any prime p, then the expression gets another factor of a_p^p ", why that ? i would think a_p and not a_p^p.
also the way you extend f to R seems wrong, you don't use continuity.
granted if you assume f continuous, and you know its values on rationals, then for x real, you can define f(x) = lim ( f( qn) ) where qn->x. You "just" have to show it does not depend on the choice of sequence qn.
@@Joffrerap You are right that the definition over Q is incorrect. I will have to think about this some more, but I am reasonably sure that you can arbitrarily choose the values of f(1) and f(1/p^i) for prime powers p^i subject to the restriction f(1/p^(i+1))^p = f(1/p^i), and this uniquely defines a function on Q. Clearly that information is enough to determine f(n/p^i), and to get arbitrary denominators, given a,b coprime, let ma+nb = 1 and then 1/ab = m/b + n/a, so f(1/a) and f(1/b) determine f(1/ab).
The extension of the function f I defined on R is not continuous, indeed it's as far from continuous as you can get. The idea behind the Hamel basis is that we can view R as a Q vector space and then all the constraints act independently in each direction so we can make arbitrary choices and it all still works out.
For me this is mindblowing and eye opening, I studied math long time ago and have never understood it so deeply. Thank you Grant
Me doing my own Complex Analysis homework for an actual grade: I have no motivation...
Me doing my 3Blue1Brown homework for no reason: This is important to me, so I must do!
Thanks so much for these lectures. I'm 48, and have a patchy engineering / programming / maths history... I wish you'd been my maths* teacher either when I was at school or at college as you've made previously difficult areas seem approachable and actually interesting, and not been afraid to mention areas which you don't find intuitive or agree with the convention. Also seeing you write things out by hand is helpful as I've always had untidy handwriting and got a hard time for it at school, so it's nice to see someone who hasn't got perfect penmanship but is still hyper-intelligent.
* Yes, I'm a Brit!
2:51 - Your 8s look like 9s. How many points did you lose on high school math tests because of this?
Haha, I had to go back and look. Yeah you are right!
Thank you. Wonderful video, finally understood this important concept. We need more people like you with such a level of understanding and interest teaching in the schools and colleges. Thanks to RUclips, now every person can access such quality education.
Imagine if this guy taught all your math classes since elementary school, jeez. These kids would be solving PDEs by 8th grade lol.
I have been wondering this forever, and asked many math teachers to explain, to no avail. Thank you!
12:10 Mathologer video did a video on Euler's formula (e^(pi*i)+1=0) in Homer Simpson speak so the definition of e has to be used to make sense of it. Interesting take on the formula.
I feel off you showing the Talor series of e^x and pulling out of nowhere.
That video is truly great.
I hear you on the concern of pulling out the series "from nowhere". I suppose for many functions (e.g. sine) we start by seeing how they're defined, and slowly build up intuition from there. That's not a great answer, of course. As soon as you introduce a little calculus, one good motivation for this polynomial is that this is a function that is its own derivative, which you can see as soon as you learn the power rule, which in turn lets you describe lots of phenomena in nature where a rate of change depends on the value that's changing. The important point here is that we don't have to introduce it as a Taylor expansion (as it traditionally is), we could take it as the starting point from e and e^x pop out.
@@3blue1brown I first encounted e in this "function that is its own derivative" manner from my highschool calc1 class! I wasn't aware the tayler expansion that I learned in college calc2 was the traditional introduction of e.
At 14:10 on the video, the exp(3+4) and exp(3) * exp(4) actually diverge at the end and the last 2 digits are flipped. At 14:29, the values diverge even more at the end of the decimal. Not sure if there is something off in the function but that would show they are about the same but not the same, which makes a difference when calculating the effects over extended time or distance.
They are exactly the same. Computers just don't have access to infinite accuracy, so take everything they compute with a grain of salt.
I hope that lockdown never ends. Learning so much these days!
i enjoy that too but bruh, i don't want to stay at home forever. :D
I understand your sentiment, but you do realise there are people starving to death because they have no jobs?
@@SoumilSahu I think you get the point of the comment, he didn't mean that ofc
You realize that you can watch educational YT videos even when there is no lockdown right?
@@Trucmuch but there is no time to think through things
Grant, this is wonderful. I’m preparing to teach synchronous online classes, and I wanted to see how you conducted a live class. I love this beautiful formula, and the way you shared it with the world in this wonderful, interactive way. Thank you for giving me some ideas, for your lovely personality, your musical taste, and the beautiful math!
The problem is that you need the derivative of exp(x) in order to obtain the polynomial. So your definition of exp(x) depends on exp(x) !
Or you could ask yourself, the derivative of what polynomial equals itself? And that will be exp(x).
I am a critical care physician waging war on the front lines of Covid. Grant, your lectures offer wonder and insight for weary minds, rationally and calm in the midst of tragic irrational chaos. I pray for Covid to end but for these lectures to continue. A global community extends its gratitude for your helping us to "think different".