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Hi @kuckkuckrotmg thanks so much for your comment! 😊 I am so happy you found the video interesting! 🥳 I hope you have an amazing day/evening/night! 😊

Hi @genxjack72 thanks so much for commenting, as always, and the interesting question! 😊 I meant that the usual definitions of sin and cos (as well as the standard proofs of trig identities) rely on triangles since the original definitions of sin and cos are based on (right) triangles. You can also prove these angle sum formulas by constructing right triangles and studying them carefully, but that proof can seem a bit complicated. However, in this video, the challenge is to find a proof that avoids the use of triangles altogether (and instead uses the complex numbers). I hope you have an amazing day/evening/night! 😊

Hi @ElectriteRoblox thanks so much for your really kind and supportive comment, and positive feedback! 😊 I am so happy you are loving these videos and learning from them, that means a lot to me! 😊 Yes, I've got lots of videos planned for the future on a wide range of topics and I'm very excited to have you as a viewer! 🥳 I hope you have an amazing day/evening/night! 😊

Hi @RiteshArora-s9x thanks so much for your comment! 😊 I think it's a fundamental statement that likely appears in several books (at least as an exercise) but I don't know of a specific book off the top of my head. I hope you have an amazing day/evening/night! 😊

Hi @WolfgangFeist thanks so much for your comment as always! 😊 Yes, that's true, and an important observation! I could have also emphasized this before the Lemma when stating f(1/2)^2 = a because at that point we already knew this (and then the Lemma is for the most general value of the function). As you say, it is a very important special case of the Lemma. It's interesting since it's not possible to define a^x for all x if a is negative (but still possible for fractional exponents x = p/q in lowest terms, where q is an odd number). Thanks so much for supplementing the video with this observation, and I hope you have an amazing day/evening/night! 😊

Hi @renesperb thanks so much for sharing your beautiful argument! 😊 I love the argument too! 🥳 The one question I would love to hear your thoughts on is: how do we know (a priori) that f is nowhere zero so f'/f is well-defined? Of course, this is true (once we prove f(x) = ce^x for some constant c), but it needs to be justified "a priori" from the equation f'(x) = f(x) since we use it in the argument. The only way I know how to do this directly is already somewhat complicated. If f(x) is 0 at some point, then all its derivatives are 0 at that point because of f'(x) = f(x). If we then show the function is analytic, then that will imply the function is identically 0 everywhere (since the Taylor series is 0). In order to show the function is analytic we can use Taylor's theorem with remainder, since the equation f'(x) = f(x) implies there is a uniform bound on all the derivatives of y on a closed interval (which then implies the remainder -> 0). If you have a simpler method of showing f is nowhere zero (so f'/f makes sense), I would love to hear it! However, in any case, I do love the argument you gave because aside from this point, it does elucidate the statement from the point of view of integration. 😊 I hope you have an amazing day/evening/night! 😊

@@MathMasterywithAmitesh The only way I can see it is to use the uniqueness of the solution of y' = y and y[xo]= yo . If we assume that y1 and y2 are two different solutions ,we may use that (y1 -y2)' = 0 .This means that y2= y1+c , but because of the initial condition we have c =0. Hence if y(xo)=0 then y (x) =0 everywhere.

@@renesperb Hi! Thanks so much for your comment! 😊 Of course, the uniqueness theorem for ODE's also implies the general conclusion as well that every function f(x) with f'(x) = f(x) satisfies f(x) = ce^x for some constant c. In that case, there is no need of any argument, but the point of the proof in the video (and also the proof you gave) is to give an elementary reason in this specific differential equation (and I think your argument is really cool because it also elucidates why e^x shows up in the first place 🥳). In particular, invoking the uniqueness theorem for ODE's seems too high powered for showing that y is everywhere non-zero but so is using Taylor's theorem with remainder (the argument I gave). I think that's the main thing I like about the argument in the video, that it doesn't require knowing anything more about f(x) other than f'(x) = f(x), so it is a complete self-contained proof. Thanks so much for sharing your insights and I hope you have an amazing day/evening/night! 😊

Hi @ZachWithall-sw4ld thanks so much for sharing this really cool argument! 😊I read it carefully and I love it! 🥳 The one point I was a bit confused by was "(3) multiply both sides by dx/y" because technically it's not possible to separate dx and dy like that in dy/dx. However, what do you think about this approach? If y' = y, then y'/y = 1. If we integrate both sides of this equation with respect to x, then the left-hand side is integral (y'/y dx) = ln |y| + C (u-substitution with u = y) and the right-hand side is integral(1 dx) = x + C'. Therefore, ln |y| = x + C'' for some constant C'', and then we deduce y = e^{x + C''} (which is the desired conclusion on y). I think this is somehow the same as your argument but maybe makes the idea of multiplying by dx/y more rigorous. The one point I think has to be justified right now is why y is non-zero (in the above argument too). You point this out in the last sentence, and it is of course true (once we prove the desired conclusion on y), but it needs to be justified "a priori" from the equation y' = y since we use it in the argument. The only way I know how to do this directly is already somewhat complicated. If y = f(x) is 0 at some point, then all its derivatives are 0 at that point because of y' = y. If we then show the the function is analytic, then that will imply the function is identically 0 everywhere. In order to show the function is analytic we can use Taylor's theorem with remainder, since the equation y' = y implies there is a uniform bound on all the derivatives of y on a closed interval (which then implies the remainder -> 0). I do really like your argument but I'm only not sure how to justify the fact that y is everywhere non-zero in a simple way. If you have any thoughts other than my previous paragraph, please let me know! Thanks so much for sharing your beautiful proof! 😊 I hope you have an amazing day/evening/night! 😊

Thank you so much!!! 😊

Hi @matthewanderson3621 thanks so much for your comment and the amazing video suggestion! 😊 Yes, I think understanding how these differentials work is super cool and there is deep math behind it (e.g., also what is dx in integration?). In the specific example you gave, would a proof of the chain rule be the rigorous explanation you are thinking of for why the dk's cancel (which is the formula you stated) or are you thinking of some other explanation? I'll definitely look into this - I was planning to do a video proving the chain rule soon, but I think explaining what these differentials mean would definitely appeal to a lot of people too as you say! 😊 Thanks again for sharing your video idea, it means a lot! :)

@MathMasterywithAmitesh Yeah the chain rule is what I am talking about but I hope you could share your perspective and explain the intuitive idea behind said rule!

@@matthewanderson3621 Hi! Yes, that sounds like a great idea! I will definitely do a video on how to prove the chain rule! 😊 I think it's a super interesting topic/proof! 😊 Thanks again for the suggestion and have an amazing day! 😊

Hi @RiteshArora-s9x thanks so much for your comment! 😊 Yes, as you say, the fact that for each real number y and for each positive real number x, there is a natural number N such that Nx > y is the Archimedean property of the real numbers! 😊 (In our proof, we use this for x = 1. I would also say it is a property of the real numbers, rather than the natural numbers.) Thanks so much for sharing this insight and I hope you have an amazing day/evening/night! 😊

Hi @pianohouse2725 thanks so much for your comment! 😊 Yes, y = 0 is a solution and that's a great observation. In the video, we claim that every solution is a constant multiple of e^x, i.e., y = ce^x for some constant c. In particular, if we choose c = 0, then this includes the 0 function. However, you are certainly right to point it out as a notable example! Thanks so much for sharing and I hope you have an amazing day/evening/night! 😊

@MathMasterywithAmitesh oh yeah true i didn't check c=0 Btw sir you're welcome 😁

Hi @RiteshArora-s9x thanks so much for sharing! 😊 I didn't know that, I'm not familiar with those exams, but it's a cool puzzle to think about for someone learning calculus and is a special part of the theory of ODE's (so it motivates the tricks there, I guess, and the idea of solving differential equations by solving algebraic equations instead). I hope you have an amazing day/evening/night! 😊

@MathMasterywithAmitesh its enterance exam for admission in B. Tech in IIT. its very hard. You may check math questions

@@MathMasterywithAmitesh putting title IIT advanced exam in title will give boost to views too

@@RiteshArora-s9x Hi! Thanks so much for mentioning this! I will check it out. 😊 I hope you have an amazing day/evening/night! 😊

Hi @alexchan4226 thanks so much for your comment! 😊 Yes, as you say, other than e^x and any constant multiple of e^x (f(x) = ce^x for some constant c), there are none! It's what makes the exponential function so special! 😊 I hope you have an amazing day/evening/night! 😊

Hi @RiteshArora-s9x thanks so much for your comment! 😊 I didn't know that (I'm not aware of the movie) but that's good to know! (I assume you are referring to the quote in the thumbnail "The journey matters, not the destination", which seems to be a quite popular quote, but it's not mine.) Thanks so much for sharing that info and I hope you have an amazing day/evening/night! 😊

@MathMasterywithAmitesh sir i was reading your website. It was written that you completed rudin at age 13 and you proved theorems on your own. But i am not like that. I really struggle in analysis and questions in Rudin are so tough.that without hint for many icouldnot do. How much time do you think per question should be spent for average person ? Thanks

@@RiteshArora-s9x Hi! Thanks so much for your comment! 😊 Firstly, it's ok if it seems tough, it also felt that way to me when I first read the book, it's quite different content and style to pre proof-based math. My suggestion is to use math stackexchange. If you get stuck, you can ask for hints there. Also, I suggest to try proving the propositions, lemmas, theorems etc in Rudin before reading them - those are like exercises with solutions and you can confirm your understanding. It's a barrier at first, but once you overcome it, it definitely gets easier (like with anything). Similarly, in terms of time, it's ok if it takes longer initially, but it will get easier. I remember spending several days (sort of) on one exercise in Rudin (this happened a few times) but once I understood them, my understanding increased so much and things became much faster after that. The general suggestion is math stackexchange is a great resource to support you on your journey. Also, if Rudin is feeling challenging, the book I recommended by Kenneth Ross in the other thread is more leisurely and also great and you could supplement with that (he explains things things in more detail). 😊

@@MathMasterywithAmitesh 🙏

@@MathMasterywithAmitesh sir did you solve every exercise of PMI on your own ? I have never spent more thanhour on question. Maybe that is issue..

Hi @fredrik3685 thanks so much for your comment! 😊 Yes, that is an intuitive way to look at it, that the curve is somehow "forced" once you pick an initial point. (However, you do have to pick an initial value, and then from there you can conclude there is only one curve y = f(x) passing through a specific point such that y' = y always.) Thanks for sharing your intuition and I hope you have an amazing day/evening/night! 😊

Hi @RiteshArora-s9x thanks so much for your comment! 😊 I'm going to be starting a real analysis lecture series on my channel very soon, and I hope that will be helpful (if you subscribe, you will be notified, but also it can be found on the "Playlists" tab on my channel). In terms of books for real analysis (1) "Principles of Mathematical Analysis" by Walter Rudin is an amazing textbook, it's challenging, but if you master it, then it will give you a really strong foundation (2) "Elementary Analysis: The Theory of Calculus" by Kenneth Ross is another wonderful book (I used it a bit myself at some point and I enjoyed reading it - it could be used together with Rudin's book and maybe a bit more leisurely to read) Abstract Algebra: (1) "Abstract Algebra" by Dummit and Foote is quite popular and comprehensive and leisurely to read (I think) - I haven't used it but I know it is very popular I'd recommend looking at those and seeing how you go and also please check out my upcoming real analysis lectures if you are interested in further supplements! I will be beginning with the theory of sequences and convergence and a video will be out in a few days! 🥳 I hope you have an amazing day/evening/night! 😊

@MathMasterywithAmitesh thanks sir.

Hi @owlsmath thank you so much for your really kind and supportive comment! 😊 I am so happy to read it, and it really means so much to me! I am so happy you loved the explanation! Thanks so much for commenting and I hope you have an amazing day/evening/night! 😊

Hi @WolfgangFeist thanks so much for your comment! 😊 Yes, I think the triangle inequality (as well as the name) makes the most sense in R^2, as you say. I wanted this to be a self-contained video on the absolute value in R before moving on to R^2 (since it's already something people have to master and apply when studying inequalities of real numbers before they get to R^2). In some sense, it's also a degenerate triangle in R^2 so may require independent thought even in R^2 (if one of x or y is a multiple of the other, as vectors). However, I'll definitely be doing a sequel at some point on the triangle inequality in R^n for n > 1. I hope you have an amazing day/evening/night! 😊