Відео

Hi @renesperb thanks so much for the alternative perspective and thanks as always for your comments - I love reading your thoughts/insights! 😊 The main reason I avoided that approach is because the formula for the area of a circle is probably non-trivial. On the other hand, the circumference formula is literally the definition of π so the perimeter approach is straight from the definition of π without assuming any background. However, what you are saying is a great observation, of course! 😊 Have an amazing day/evening/night! 😊 (An alternate place where one can use both approaches (perimeter or area) is to prove lim_{x -> 0} (sin x)/x = 1. In the following video, I did it with the perimeter approach: ua-cam.com/video/rHQwhvvSgrI/v-deo.html which seems to be less common than the area approach. However, I also realized you can derive the formula for the area of a circle using lim_{x -> 0} (sin x)/x = 1, so I feel the perimeter approach here is less "circular".)

@@MathMasterywithAmitesh You have done this probably already: if you take the unit circle and its inscribed hexagon you have the inequality π > 3 .

Hi @misbahfirdous4111 thanks so much for your comment and for sharing your thoughts! 😊 Yes, exactly, it's special in some way because it satisfies all the laws! I think that gives a clue that it's secretly something we know! SPOILER below for the secret in case you are interested: The secret is that the operation * in the video is just usual multiplication shifted by 1 because of the formula: a*b + 1 = (a + 1) x (b + 1) In words, if you add one to each of a and b, and multiply them as normal, then you are adding one to a*b. (We can think of it as multiplication in "disguise" where the "disguise" is the function f(x) = x + 1.) An interesting question: can you think of other cool ways to create arithmetic operations for numbers that satisfy all the laws? I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊

Hi @WolfgangFeist thanks for your comment! 😊 I like your approach (the nested interval formulation of the "Bolzano-Weierstrass Theorem") but I think the main drawback is that we still have to prove that a Cauchy sequence converges (let's call this Theorem A). The part of the proof in the video that I'm guessing you find non-intuitive effectively gets pushed to a non-intuitive technical statement needed to prove Theorem A. (Of course, any proof of Theorem A must depend on the completeness axiom of the real numbers.) The argument I know to prove Theorem A is: (1) (a) prove that a Cauchy sequence is bounded (simple to do), (b) prove that a Cauchy sequence has a monotonic subsequence, and then conclude that this bounded monotonic subsequence converges (using the completeness axiom, just like in the video). (2) Prove that if a subsequence of a Cauchy sequence is convergent, then the Cauchy sequence is also convergent (not too hard, but needs to be written out) However, it seems this already involves a lot of work in addition to the work needed to reduce the Bolzano-Weierstrass Theorem to Theorem A. Also, the proof of 1(b) in the above sequence of steps is a technical lemma that is effectively the work done in the video to construct a monotonic subsequence. I feel there isn't really a simpler argument for Theorem A (if you have one, I'd love to hear it! - the main issue is defining what the limit should be which also must use the completeness axiom somewhere). The reason I like the proof in the video is it is completely self-contained based on only the completeness axiom of the real numbers (which manifests as "every bounded monotonic sequence converges"). In terms of possibly helping making the proof in the video more intuitive, the approach of the proof can be split into two steps (1) If the supremum of a sequence is not achieved, then the theorem is easier: just construct a monotonic subsequence of terms approaching the supremum (which must exist, since it is the supremum). (2) If the supremum is achieved, let's say it's achieved by a_{n_1} = max{a_1, a_2, ...}. In this case, take the subsequence a_{n_1 +1}, a_{n_1 + 2}, .... If the supremum of this subsequence is not achieved, then we are reduced to (1). If it's achieved, then say it's achieved by a_{n_2} = max{a_{n_1 + 1}, a_{n_1 + 2}, ...} ≤ a_{n_1}. If we keep going, we have created a bounded monotonically decreasing sequence {a_{n_j}}, that must converge. I would love to hear your thoughts! I hope you have an amazing day! 😊

@@MathMasterywithAmitesh Thanks! Why I came up with the proof using Cauchy: I had in mind to define R as the equivalent classes of Cauchy-series: so Cauchy <=> convergence would be the definition of completeness (as we do in Banach-Spaces e.g). Of course, the equivalences of definitions of completeness will have to be shown at some point. Having R to be the completion of Q just so that all Cauchy sequences converge is {from my point of view} very intuitive. Added after some more thought: The series of the upper boundary beta_k of the nested intervalls is bounded and monotonic (decreasing). So, that one has a limit "a" due to the completeness of R (however originally defined). Now, "a" is less than epsilon/2 {=|beta-alpha|.2^(-k) } away from almost all beta_k , which is less than epsilon/2 away from almost all a_n(k) of the subseries we have choosen; Thus, "a" is also the limit of that subsequence. - Now the monotonic sequence theorem is used instead of Cauchy.

@@WolfgangFeist Thanks so much for sharing that approach! I like that too and I think with a formal approach to math, that's the best definition. 😊 I have to be honest though that even as a mathematician who likes formal/abstract constructions, I prefer to think of real numbers as infinite decimals rather than equivalence classes of Cauchy sequences (with the latter, it just sort of "feels" to me like we're not working with numbers anymore). That's why I prefer the approach in the video for explaining to people learning about sequences/real analysis for the first time because it feels more concrete (we can constructively write down the limit of the subsequence as a supremum/infimum, explicitly connecting to the completeness axiom). On the other hand, many mathematicians would prefer your approach! 😊 I hope you have an amazing day! 😊

@@MathMasterywithAmitesh I agree with the 'infinite decimals' approach as the most intuitive one. Have you thought about formalizing such an approach? That would e.g. mean for Bolzano-Weierstraß to construct a series of finite decimal numbers which converge - to the limit of the subsequence choosen; {If we choose the first intervall [alpha, beta] with alpha, beta from Q, the endpoints will be such a series; but we could do 'better' by using a 10^N-Intervall first and dividing into 10 (instead of 2) parts, always choosing the highest intervall with infinite terms of the original series in it. Now the series of the lower intervall boundaries ist just a series of a cut-off decimal development which converges to an infinite decimal expansion - which is the limit of the choosen subsequence (Nice!). I know, that most mathematicians will not like such a construction at all (because it is dependent on a specific representation of the real numbers). But, to introduce the concepts to beginners, I fully agree: That would be much easier to comprehend.

Hi @RiteshArora-s9x thank you so much for commenting! I will be releasing several more real analysis videos in the coming month. 🥳 I hope you have an amazing day! 😊

Hi @alex_be9698 thank you so much for your comment! I am so happy you liked the video! 😊

Hi @alex_be9698 thank you so much for your kind comment! 😊 It's great to hear from you again! Yes, in hindsight Rolle's theorem definitely feels obvious but it took a long time for people to prove the theorem. I think the idea of recognizing that the derivative is 0 at an extrema is simple but deep! Of course, the fact that there is always a global extrema for a continuous function on a closed interval (i.e., the extreme value theorem) requires real analysis to prove and isn't even taught in most first calculus classes (although it is intuitively obvious). I'll definitely do a video on that proof as well sometime! 😊 I hope you have an amazing day/evening/night, and thanks as always for sharing your comments/thoughts! 😊

@@MathMasterywithAmitesh Thank you very much for your reply. (And of course, as always, for investing your time in sharing your knowledge in public. 👍) It's interesting to the that something, that might look obvious from an intuitive point of view needs advanced skills to be actually proven correctly. Thanks! Have a great evening, too! 🙂

Hi @yashprajapati8857 thanks so much for your comment! 😊 Yes, as you point out, any function of the form f(x) = a*sin(x + b) + c*cos(x + d) (which would also satisfy f''(x) = -f(x)) can be written as A*sin(x) + B*cos(x) for some constants A and B using the trigonometric angle sum formulas that you mention! For example, we can write: a*sin(x + b) + c*cos(x + d) = (a*cos(b))*sin(x) + (a*sin(b))*cos(x) + (c*cos(d))*cos(x) - (c*sin(d))*sin(x) = [a*cos(b) - c*sin(d)]*sin(x) + [a*sin(b) + c*cos(d)]*cos(x) = A*sin(x) + B*cos(x) where we choose A = a*cos(b) - c*sin(d) B = a*sin(b) + c*cos(d) Thanks for sharing your thoughts and I hope you have an amazing day/evening/night! 😊

Hi @poultrymansbiggestfan3133 thank you so much for your comment! 😊 I also love it when complex numbers show up in unexpected places! I feel like a lot of times in math we are introduced to new concepts for the sake of studying them (and they are interesting to study), but it's coolest when we can use the new concepts to answer old problems (that don't reference the new concepts at all in their statements). I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉

@@MathMasterywithAmitesh I definitely agree with you! New concepts in reference to old problems trully make me smile, one could say it is the true essence of mathematics to apply new ideas to other feilds of study. I hope you also have an amazing day/evening/night and Happy New Year!!

Hi @up1663 thank you so much for your comment! 😊 I am so happy you appreciated the video! I love this argument too because of its accessibility and the way the complex numbers enter the picture (and it's cool we can answer a deep question like this in 10 min). I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉

Hi @GreenMeansGOF thanks so much for sharing! I hope you have an amazing day/evening/night! 😊

Thank you so much!!! I am so happy to read your comment! 😊

Thank you so much!!! 😊

Hi @strikar5552 ! Thank you so much for your comment and feedback! 😊 Yes, you are absolutely right, thank you so much for the correction!! I can't believe I made that mistake 😂 and I am very happy you were paying attention and pointed it out! I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊

Hi @Jack_Callcott_AU thanks so much for your comment! 😊 Yes, I like it since it's not necessarily obvious if someone hasn't seen it before, but it's simple to explain and becomes obvious once they have seen it. I hope you have an amazing day/evening/night! 😊

Hi @meurdesoifphilippe5405 thank you so much for your comment and for sharing your insight! 😊 I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊

Hi @encounteringjack5699 thank you so much for your comment and for sharing your thoughts! 😊 In the video, I don't think we set the limit equal to the result of the derivative. The crux of the argument is (and of course it's a wrong argument somewhere 😝) (1) "If f(x) = arctan(x) + arctan(1/x), then f'(x) = 0, which implies that f(x) is a constant function" - We don't equate the derivative to the limit, we just use f'(x) = 0 for all x to conclude that f(x) is a constant function (2) "Since f(x) is a constant function, lim_{x -> ∞} f(x) = lim_{x -> -∞} f(x), which implies that π/2 = -π/2, or π = 0" - A constant function has the same limits at ∞ and -∞ Of course, there are other parts of the argument than (1) and (2) but that's the crux. Do you see where the error may be? 😊 I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊

@@MathMasterywithAmitesh I see. The problem is that there was an assumption that, because the derivative is zero, then the limit as it goes to positive and negative infinity must be the same. However, that is not true. You could have a piece-wise function happening with the function. As long as the derivative along those areas are zero, the derivative as derived holds true. This can be concluded by the fact the derivatives are opposite in the sign (+,-). Showing that the rates of change are equal, which means any variation in x, apart from x equals 0, the result will be a constant. A graph of the function also shows a piece-wise function, with x = 0 not being defined.

@@encounteringjack5699 Hi! Thanks so much for sharing your detailed thoughts! 😊 Yes, you got it with your second paragraph and your last sentence! 🥳 (I will just add regarding the third paragraph, maybe you meant to say this, but in this case the derivative is 0 everywhere, there isn't any case of the derivatives having opposite sign, but the function does have opposite sign on different pieces 😊)

Hi @thabulos thank you so much for such a kind comment! I am very happy to receive that feedback! 😊 I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊

Hi @joelklein3501 Wow!!! Thank you so much for sharing such an extremely detailed and insightful comment! 😊 I absolutely loved reading your comment, and it beautifully and clearly explains the trigonometric identity from scratch in a way that a precalculus student can understand, as well as explains the identity thoroughly from the perspective of calculus! 🥳 The formatting is very clear and easy to read for anyone as well. 😊 It's the most amazing mathematical comment I have ever received on this channel, thank you SO much for taking the time to share this! 🙏 I hope you have an amazing day/evening/night and Happy New Year!!!!! 🥳🎉🎊

@@MathMasterywithAmitesh Haha, I'm glad to hear that you appreciated it so much. I enjoyed the video, and never actually noticed this arctan identity, so I had fun sketching an explanation. Hope to see more insightful videos from you Good day/evening/night and happy new year!

@@joelklein3501 Thank you so much! 😊 Yes, it's an unusual arctan identity and I loved your explanation right from the basics using right triangles! (Also, although this is not as good as your explanation since it relies on another identity, we can also derive it from the tan angle sum formula tan(a + b) = [tan(a) + tan(b)]/[1 - tan(a)tan(b)].) Thank you so much for your interest - I am looking forward to producing lots more videos this year that I hope you will enjoy watching! 🥳

Hi @mahima_abhishek_ thank you so much for your comment and for sharing! 😊 I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊

@MathMasterywithAmitesh wish you the same 😊