Hearted Youtube comments on Mathologer (@Mathologer) channel.

  1. Just to go a step further, the perimeter of the shape described by the wiper arcs is exactly pi times the perimeter of the triangle i.e. (pi * the chord length), whereas the circumference of the circle is (pi * 2 * the radius) and, as noted above, it is easily seen that (2 * the radius) is greater than (the chord length). To check that the perimeter of the SOCW is (pi * the chord length), consider each arc length. If we label the triangle sides A,B,C and their opposite angles a,b,c, then the arc lengths are given by: Aa + Bb + Cc + (B+C)a + (A+C)b + (A+B)c = (A+B+C)(a+b+c) = (chord length)(pi)
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  2. Infinitely mesmerised by the guitar music. Then I consider, is the tablature some kind of code?
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  3. I had worked on this problem in my pre-university days. Though I eventually had to see the solution out (couldn’t solve it), it was my first deep exposure to Ramanujam’s mind and mathematical thinking. (The first time mathematical connect was obviously the introduction to limits concept). That was when I truly understood why he is called “the man who knew infinity”. No wonder, one of the greatest son of Indian soil 🙇
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  4. You can get the non-recursive function from a recursive function using the z-transform. This is used often in Digital Signal Processing problems.
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  5. Don't you get half of your claim, so 50 cents? You only get your 1 dollar if the estate is enough for both creditors so 1,000,001 dollars.
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  6. 7:02 yes it can. It's sufficient to look at the last two digits of a number to check if it's divisible by 4 since 4 divides 100. The last two digits were 81 which is one above a multiple of four.
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  7. I love these visual proofs. The animation of the visual proof at 22:40 is extra fun because two of the triangles are re-scaled. But they are re-scaled instead of having a more complicated animation. Well done!
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  8.  @Noughtgate  the actual amount of elastic force doesn't matter, only the fact that the free portion elongates uniformly while the portion in contact with the wheel remains fixed and does not slip.
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  9. 49:53 the answer is the odd-numbered Fibonacci numbers!
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  10. It was probably only published by someone 10 years ago. People playing around with triangles ABSOLUTELY have discovered this many times over history.
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  11. But the real question is: is that a Math or a Moth shirt?
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  12. My supernatural psychic sense is telling me that your card is... 👸🏻♥️
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  13. One of your finest videos in my opinion, I really enjoyed it!
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  14. I can't believe you mentioned Tree With Deep Roots. Knowing next to nothing about Korean or Korean, I stumbled onto that drama around 2011 and was intrigued by the story of Hangul woven into it. I learned to read it and eventually became somewhat proficient in Korean, all because of a random K-drama! Funny how those things work.
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  15. Lemme check the Nobel prizes... the high point of that guy's career was the photoelectric effect?? Sorry, not impressed.
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  16. It is a riff on "The Claw of Archimedes", a super weapon created by Archimedes, also called the "Ship Shaker". Look it up.
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  17. As a slide rule user, engineering in the 1960s, I found this a fascinating explanation but the graphics are absolutely brilliant, Even knowing what was going to be shown, I was mesmerised by the animation. So beautifully done.
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  18. I want a heart pls 😭😭
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  19. This is why I love your channel, a machine gun of “Aha!” moments makes me feel like Baldrick with a cunning idea. Love it!
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  20. What a joy to see this topic covered with such clarity! I wanted to make my own video about it a few years ago, and I went so far as to create animations in Desmos and tease the topic at the end of my video on Napoleon's Theorem. I came across the idea of using DFT's to explain Petr's theorem in a paper by Gregoire Nicollier, "Some Theorems on Polygons with One-line Spectral Proofs" (Forum Geometricorum Volume 15 (2015) 267-273). It was a lot of fun learning about it and trying to animate it. I just couldn't quite simplify it enough to make it accessible to my audience. So, thanks for the gift of seeing "my video" realized by a real pro!
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  21. 4. The necklace has 12 parts of equal length between the big pearls (? idk), which can be streched into a 3-4-5 triangle to check if an angle is right. Ropes like these were used in ancient Egypt to make right angles, though they weren't so cool-looking, just ropes with knots
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  22. Why not 4k-1 For 3,7,11,19.....
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  23. This kind of single move algorithms often show up in computer science and an often overlooked follow up question is if their were "multithreading" where if two subsequent movements occurred to and from different pegs they could be done simultaneously as 1 move. Obviously this would only apply to larger n but I would be curious if the solutions for that were actually superior given how ordered the solutions seem to be already.
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  24. I was already... not comfortable, but let's say "resigned"... to the fact that there exist very slowly divergent series; but the fact that there are very slowly CONVERGENT series, whose sum is impossible to approximate computationally within a reasonable margin of error, like the no nines series... was a shock!
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  25. What's really ironic is that when I was studying engineering, after 3 years of teaching me how to solve differential equations, they finally admitted that real-world engineering problems rarely have closed-form solutions. So that's when they started teaching numerical methods like Finite Element Analysis and Computational Fluid Dynamics! This video—which drew a parallel between sequences of discrete numbers and continuous functions—reminded me of that! Calculus is soft, silky, continuous, and exact. Numerical methods are rough, gritty, discrete, and approximate. I think that's why math departments continue to teach contrived exercises that resolve to beautiful, closed-form solutions. Whereas engineering departments are forced to tackle real-world problems head-on. Engineers need to arrive at reasonable solutions that are practical to implement; they need not be perfect.
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  26. Fantastic video! A more computationally-efficient method than the determinant method, which also easily and transparently deals with single and multiple zeros: Step 1: After using k delayed copies of the original sequence, instead of repeatedly computing determinants (as in the video), take a rectangular “snapshot” — this is a rectangular matrix of size k-by-n, with n > k. Step 2: Transpose this matrix (so now it’s n-by-k with n > k) and compute its “economy” SVD (singular value decomposition). There are many existing software libraries to do this. This step is the only computationally-intensive step, whose complexity is O(n k^2), which is much less than the determinant method as stated in the video, which is O(n k^3), or even more if care is not taken… Step 3: Now, look at the singular values: If one or more of the smallest singular values is/are zero, then we’re sure this is what Burkard calls “Fibonacci-like sequence”, something usually called “linear recurrence with constant coefficients” (which he mentioned in the video). However, if none of the singular values is zero, then repeat with larger k, i.e., more delayed copies of the original sequence. Step 4: The linear recurrence with constant coefficients is easily determined from the column of the SVD result corresponding to the zero singular value (because this is the vector which can zero-out any set of k consecutive elements of the sequence). Specifically, if the “economy” SVD result of the original n-by-k matrix is U, s, Vt, then the last row of the square k-by-k matrix Vt will be the desired set of linear constant recurrence values. This last row has size 1-by-k, of course. These values will probably need to be scaled by a common multiple if nice integer values are desired! Move the 1st element of this set of k values to the other side of the equation to use as a prediction recurrence equation, i.e., the next value is determined linearly from the previous k-1 values. Remark: The idea of a singular value decomposition (SVD) is mathematically very closely related to a determinant, because it essentially also determines linear combinations of columns (and/or rows). But it is more computationally-efficient in this case, because it can be computed for a rectangular set of values simultaneously, rather than small squares of values one after the other. The second advantage of the SVD is that it can also give us the linear recurrence with constant coefficients “for free” (from the same computational result).
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  27. Don't forget to bring a towel...
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  28. 1:51 Challenge accepted ! (1/11) * 1000^11 + (1/2) * 1000^10 + (5/6) * 1000^9 - 1000^7 + 1000^5-(1/2) * 1000^3 + (5/66) * 1000 = 91409924241424243424241924242500 Pretty faster that way !
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  29. I'm quite happy as long as the longer content still exists. You're one of the few STEM channels that continues to carry videos of 40 minute length or longer. Some of my favorite channels have gone from a 40m to 8/15 minute formats for the sake of the algo and it broke my heart. It's difficult to find a topic you love, and then have to find a new video on that same topic after just 10 minutes. However...it's also nice to sprinkle in those shorter vids. These are the Tough questions that try men's minds. As Felix Unger once said..."Let it be on your head." 😁
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  30. Noticed another "super pattern" for the squares. Looking at the base of the last term of one row, you can derive the base of the first term of the next row. Multiply the base by (n+1)/n, where n = the row number. For example, the base of the last term of the 1st row is 5, and 5 * 2 / 1 = 10, which is the base of the first term of the second row. 14 * 3/2 = 21, 27 * 4/3 = 36, etc...
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  31. 0² + 1² = 1 1² + 2² = 5 0² + 3² = 9 2² + 3² = 13 1² + 4² = 17 ?² + ?² = 21 0² + 5² = 25 Is 21 really a "good" number?
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  32. I’m stunned. What a sublime concept, especially the animations that produce the cool little fractal trees
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  33.  @AlexanderQ689  A short video on change for a change :)
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  34. I am so happy to have access to such great content without any charge. I love mathematics so much and this satiates my curiosity! Looking forward to more of your amazing work ❤️
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  35.  @enderyu  If all the creditors divide their loans into equally-sized chunks, it is equivalent to proportional. Using the example in this video, if creditor A has 10 loans of $10 each, creditor B has 20 loans of $10 each, and creditor C has 30 loans of $10 each (so that's 60 loans of $10 each) and the estate has $200 left, then each $10 loan gets $3.33 paid back. Creditor A gets $33.33 in total, creditor B gets $66.67 in total, and creditor C gets $100 in total.
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  36. 1 cubed plus 2 cubed equals 3 squared. (10sq + 11sq + 12sq + 13sq + 14sq) /365 = 2 because 10 sq + 11sq + 12sq = 13sq + 14sq and 169 + 196 = 365. 3cubed + 4 cubed + 5 cubed = 6 cubed.
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  37. Mathematician looks at 3:10 - "hey, neat, the area proof of pi!" Engineer looks at 3:10 - "hey, look, nuclear reactor tiles!" ;)
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  38. Related, cryptographic hash collisions must exist. Finding them is a bear, so many damn pigeonholes 😆
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  39. Burkard, rather than heap additional praises regarding the dependable brilliance of your channel's content, I thought I might remark on your delivery: Your pacing, pitch, tone and inflections are such that, rather than lulling my mind to sleep in class, you have a special way of always educing and coaxing my brain forward (stubborn mule that it is).
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  40. Best thing about you, you always tell the real mathematician of the theorem. 😀👍🏻 BTW merry Christmas🥳🥳
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  41. This feels like a deliberate deep dive into the slightly shoddy limit taking at the end of the previous video. Brilliant!
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  42. This is mathematical proof that there's the right way and the wrong way of cutting corners!
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  43. Interesting, I noticed years ago that if you free hand draw the best pentagram that you can manage and keep connecting the points of the pentagon in its middle into another pentagram and keep doing this you will quickly end up with horribly crooked pentagrams. I hypothesized (🧐) that what was happening was basically an amplification of whatever small inaccuracies were already in the pentagram that I started with and that if I had drawn a perfect pentagram the inner pentagrams would also be perfect forever.
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  44. Let's collect some interesting comments and remarks as they come in (also check out the description of this video): VincentvanderN I don't know if this is already down here somewhere in the comments, but 2 x 2 = 2 + 2 can be used to show that there are infinitely many prime numbers. We generate a sequence a_1, a_2, a_3 etc by starting with a_1 = 3 and a_{n + 1} = a_n^2 - 2. Now we use our freak equation to show that if q is a prime divisor of some a_m, it cannot be a prime divisor of a_n for any n > m. (And then, as a result neither of an a_n with n < m, because otherwise we could repeat the proof with n in the role of m and get a contradiction.) Here is the argument. Look at the sequence a_m, a_{m+1}, ... modulo q. We have a_m = 0 mod q, a_{m+1} = - 2 mod q, a_{m + 2} = 2 mod q, and then, by the freak equation 2 x 2 = 2 + 2 (in the form 2 x 2 - 2 = 2) we get that a_{m + k} = 2 for all k >= 2. Neat, right? I believe I learned this from Proofs from the Book. charlesstpierre9502 Use 2x2=2+2 to make a formula for generating Pythagorean triples. Start with two positive integers m > n. Then (2×2)(mn)² = (2+2)(mn)² (2×2)(mn)² = 2(mn)² + 2(mn)² (2×2)(mn)² = 2(mn)² + 2(mn)² + (m⁴ - m⁴) + (n⁴ - n⁴) (2×2)(mn)² = m⁴ + n⁴ + 2(mn)² - m⁴ - n⁴ + 2(mn)² (2mn)² = (m² + n²)² - (m² - n²)² Set: a = m² - n² b = 2mn c = m² + n² Then: a² + b² = c² This will not work for: aᵏ + bᵏ = cᵏ; k > 2 franknijhoff6009 Hi Burkhard, the 3-variable equation plays a role in integrable systems. In fact, it appeared in Sklyanin's work on quadratic Poisson algebras (around 1982) providing solutions in terms of elliptic functions, and (with an extra constant) as the equation for the monodromy manifold of the Painleve II equation. Sklyanin's paper is: Some algebraic structures associated with the Yang-Baxter equation. Functional.Anal.i Prilozhen 1982 vol 16, issue 4,pp 27-34 ; look at equation (27). Furthermore, in L.O. Chekhov etc al., Painleve Monodromy Manifolds, Decorated Character Varieties and Cluster Algebras , IMRN vol 2017, pp 7639--7691 you can find in Table 1 a close variant of the 3variable equation with extra parameters. I think it would be interesting to explore this idea with exponents. For example 1+1+1+1+2+3=3^2^1^1^1^1 2¹×2¹=2¹+2¹ 3½×3½×3½=3½+3½+3½ 4⅓×4⅓×4⅓×4⅓=4⅓+4⅓+4⅓+4⅓ 5¼×5¼×5¼×5¼×5¼=5¼+5¼+5¼+5¼+5¼ ... tanA + tanB + tanC = tanA x tanB x tanC (that's the identity that features prominently in the Heron's formula video) For completeness sake and for fun let's mention 2⁴ = 4² 2 + 2 = 2 × 2 = 2² log(1+2+3)=log(1)+log(2)+log(3) log(2+2)=log(2)+log(2)=2log(2) https://www.mtai.org.in/wp-content/uploads/2023/09/IOQM_Sep_2023_Question-paper-with-answer-key.pdf ... 29th question in an Indian maths olympiad problem Check out this paper by Maciej Zakarczemny, On the equal sum and product problem (linked to in the description of this video) http://www.iam.fmph.uniba.sk/amuc/ojs/index.php/amuc/article/view/1662 Theorem 5.1. says If there are exactly two sum-equals-product identitites , then one of the following two conditions is true: 1. n - 1 is a prime and 2n - 1 \in { p, p^2, p^3, pq } , 2. 2n - 1 is a prime and n - 1 \in { p, p^2, p^3, pq } , where p, q denote prime numbers. Why don't we start out the sequence of basic sum-equals-product identities with 1=1? Well, N=N for all N :) Very tempting to add the 1=1 at the top. To not do so is very much in line with not calling 1 a prime (saves you a lot of unnecessary heart ache further down the line :) For this video I played around with generating some AI animated photos of Sophie Germain using https://www.myheritage.com/deep-nostalgia as well as with the ChatGPT generated rendering of Sophie Germain in the thumbnail. Here I fed the AI some of the existing pictures of Sophie Germain and asked it to play artist to come up with a picture of her in the same style used in this real picture of Leibniz https://www.alamy.com/gottfried-von-leibniz-image5071937.html). Lots of fun and quite eerie :) Not perfect, but will be interesting to find out what the machines can do in 6 months time (and at what point the machines can make better Mathologer videos than the Mathologer :( In general quite a bit of AI hate in this comment section. Strangely I've never had people complain about me using any of those generic white men with beards images of ancient mathematicians that museums are full of. Sophie Germain primes and safe primes (the 2n+1 primes) are important in cryptography, those involving prime fields. Because if the size field is a prime p, the size of its multiplicative group is p-1. For the group to be secure, the multiplicative group must have a large subgroup of prime size so p-1 must have a prime factor. So, having a prime q such that 2q+1 is also a prime is a good candidate. Even though it is not strictly required as long as p-1 has a big enough prime factor, this is what students are taught as a start. Kaprekar's constant is among the lengths corresponding to exactly two sum-equals-product identities (discussed at the end of this video :) https://en.wikipedia.org/wiki/6174 Cunningham chains are an interesting generalisation of Sophie Germain primes (chains of primes such that the next prime in the chain is always the previous one times 2 plus 1). https://en.wikipedia.org/wiki/Cunningham_chain If you are happy to also play this game with negative integers then you get more solutions. In particular, since (-1)+(-1)+1+1=0 and (-1)x(-1)x1x1=1, you can splice these two blocks into a/any sum-equals-product identity of length n to arrive at a sum-equals-product identity of length n+4. And for complex integers we've also got things like this: (1 - i) + (1 + i) = 1 - i + 1 + i = 1 + 1 -i + i = 2 + 0 = 2 = 1 + 1 = 1 - (-1) = 1 - i^2 = (1 - i)(1 + i) 2xy - (2+x+y) +1 = N-2 <=> 2xy - 2 - x - y +1 = N-2 <=> 2xy - x - y +1 = N <=> 4xy - 2x - 2y +2 = 2N <=> 2x(2y - 1) - 2y +2 = 2N <=> 2x(2y - 1) - (2y - 1) = 2N-1 <=> (2x - 1)(2y - 1) = 2N-1 Probably the easiest way to demonstrating this equivalence is to go backwards and start by expanding (2x-1)(2y-1)... Relevant Project Euler problem: 88 https://projecteuler.net/problem=88 Relevant Online Encyclopedia of integer sequences: A033178 @Frafour Tangentially related to 2x2 = 2+2: there is this video of Gromov (Gromov: 4 = 2 + 2 as "proof of Donaldson's theorem") https://youtu.be/bgePMb8wxv0?si=DmTf2wzbIr1F21f8 observing that 4 = 2+2 in 3 ways. That is, there are 3 ways to partition a 4-element set in two 2-element subsets. The fact that 3 is smaller than 4 is unique to 4, and produces a surjection from A_4 to A_3 - the alternating groups on 4 and 3 elements. This shows that A_4 is not simple, 4 is the only number where this happens, and this is responsible for many weird things in dimension 4. In another direction, I first encoutered Sophie Germain primes accidentally when learning group theory in my undergrad. There is an elementary proof of simplicity of the Mathieu groups M_11 and M_23, which actually gives a general simplicity criterion for subgroups of S_p, p prime https://www.jstor.org/stable/2974771?origin=crossref. When I read this I wondered if this ever works for proving simplicity of the alternating group A_p. If I remember correctly, it works precisely when p is a Sophie Germain prime! @YSCU261 The roots of a quadratic equation of the form x^2-bx+c=0 satisfy the following equations : r1+r2=b r1*r2=c So we have that the solutions of xy = x+y can be expressed as the solutions to the quadratic equation x^2-bx+b for all b this can be extended to x+y+z=xyz and so on with vieta's formulas
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  45.  @cheater00  Every positive integer is a some of 4 squares. https://en.wikipedia.org/wiki/Lagrange%27s_four-square_theorem A positive integer is a some 3 squares if and only if it not of the form 4^a(8b+7). The "only if" direction is easy. https://en.wikipedia.org/wiki/Legendre%27s_three-square_theorem Finally, a positive integer is a some of 2 squares if and only if the exponent of each of its prime factors of the form 4k+3 in its prime factorization is even. (This could have been mentioned in the video.) https://en.wikipedia.org/wiki/Sum_of_two_squares_theorem
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  46.  @Josh-dj1bl  You change the structure by permuting infinite terms, that doesn't guarantee the sum stays the same. And there are already uncountably many real numbers that must be covered by those permutations. My intuition is that it is countably infinite
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  47. I understand every word you say, but your sentences are mostly beyond my understanding! I am amazed at how you can explain these issues (I last studied Math over 50 years ago, but remain fascinated).
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  48. Ja bitte!
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  49. I remember going through a full and rigorous proof of the euler characteristic formula in graph theory, and all I recall was it being quite a doozy! Enjoyed your mathologerized version very much.
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  50. Fascinating. With the crushed cans and cylinders at 22:00, it's like the can or cylinder is trying to retain its original surface area while being crushed smaller, so it tends towards a Schwarz lantern shape. For what you said at the end, I didn't feel that anything necessary to understanding was left out. Perhaps because the "staircase diagonal = 2" idea is familiar to me, I actually felt there was more explanation than necessary, but perspectives may differ so don't take that as a criticism.
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