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The idea itself is actually clever. Unfortunately though, when I tried, it did not simplify the proof in any meaningful way. It required just as many steps as in mod 10 method.

I need a bit more clarification as to where you got confused. Binomial distribution is for n independent trials having the same probability p of a certain event, and in our case, the k-th trial is determining whether the success or fail occurs at the k-th panels. Hence, the trials are independent.

You're right. Thanks for the clarification.

I didn't use calculus, so what d'you mean?

This was already a well-known meme on the internet appearing on multiple math-related websites, and I just made a joke video about it in my format and style. I'm not making any claim that I have come up with this idea. It's like I made a video about deriving a quadratic formula and you claiming that I copied someone's content.

True, and perhaps that's why the journey to find them is also interesting - because the answers feel so rewarding.

As a non-native English speaker, I understand your struggle. Thank you for your kind comment.

​@@CornerstonesOfMath Bro, thanks to you I got the best mark. (10/10)

@@centella8 That's great!💯 Glad that I could be of help.

Two comments in a single day? Thanks, you're wonderful too.

Thank you for your kind remarks :)

💙

Well, I don't consider myself some kind of master of thumbnails, but if I may explain: I make thumbnails with Microsoft PowerPoint, using its features of drawing various shapes and color them in various ways. Most of the mathematical equations are also created by the PowerPoint's "insert equation" tool. If I want to be very precise about the graphs in thumbnails, I draw the graphs with Desmos, capture them, and copy+paste them into my PowerPoint slides. In a thumbnail, I try to emphasize the key elements of the problem that is handled in the video. Hope this helped 😀

Well, I'll try my best. I will use the same notations as in the video, where p_k = probability of player k surviving, Nfail = number of fails, p(Nfail=i) = probability where the number of fails is i, X = number of players surviving, and p(X=k) = probability of exactly k players surviving. It is explained in the previous part of the video that p_k = Σ(i=0 to k-1) p(Nfail=i) and p(X=k) = p(Nfail=16-k), which can be also thought as p(Nfail=i) = p(X=16-i). As a first explanation, you can just start from the sum of probabilities p_1 + p_2 + p_3 + ... + p_16 and show that the expression can be rearranged into the expected value of X: p_1 + p_2 + p_3 + ... + p_16 = p(Nfail=0) + [ p(Nfail=0) + p(Nfail=1) ] + [ p(Nfail=0) + p(Nfail=1) + p(Nfail=2) ] + ... + [ p(Nfail=0) + p(Nfail=1) + p(Nfail=2) + ... + p(Nfail = 15) ] = p(Nfail=15) + 2*p(Nfail=14) + 3*p(Nfail=13) + ... + 16*p(Nfail=0) = p(X=1) + 2*p(X=2) + 3*p(X=3) + ... + 16*p(X=16) = E(X) We can write the same thing more elegantly if we know how to use the double sigma notation (if you don't understand, it's just the same thing as above written differently). Since p_k = Σ(i=0 to k-1) p(Nfail=i), p_1 + p_2 + p_3 + ... + p_16 = Σ(k=1 to 16) p_k = Σ(k=1 to 16)Σ(i=0 to k-1) p(Nfail=i) (double sigma notation) = Σ(i=0 to 15)Σ(k=i+1 to 16) p(Nfail=i) (changing the order of sigmas) = Σ(i=0 to 15)Σ(k=i+1 to 16) p(X=16-i) (number of fails being i means number of survivors being 16-i) = Σ(i=0 to 15) (16-i)*p(X=16-i) (inner sigma simply gives (16-i) because p(X=16-i) doesn't depend on k) = E(X) Another way to explain it is by being less precise about the calculation and care more about the concepts. Here, the probability of total k players surviving p(X=k) has the property of probability mass function (PMF) which is discussed in statistics textbooks. But the probability of Player k surviving, p_k, does NOT have the same property (or dimension/unit) as PMF. p_k is expressed as the SUM of such probabilities, more precisely, p_k = Σ(i=0 to k-1) p(Nfail=i), or p_k = Σ(i=0 to k-1) p(X=16-i) hence p_k actually has the equivalent property (or dimension/unit) of the cumulative distribution function (CDF), which is the sum of probability mass function (CDF = Σ PMF). Not exactly the same as the CDF defined from p(X=k) (which is Σ(i=1 to k) p(X=i)), but quite "similar" to that. Therefore, if we add up p_k, it's equivalent to adding up CDFs, which can be thought as Σp_k = Σ CDF = Σ Σ PMF and one of two sigmas simply gives the term having the dimension/unit of people, so Σp_k = Σ PMF*(some variable having the dimension/unit of people, like X) = E(X). This is the best I can do. Hope it helps at least to some degree.

That's interesting. Although it is a basic linear equation, I didn't think that this particular type of problems is that popular. Anyway, I'm glad that this video helped :)

Yeah, in these types of problems the difficulty usually comes from finding the functions themselves, since it really puts students' understanding of geometry to the test.😀

Thank you for providing alternative method, which is totally valid. I must have overlooked that method because I had a prior knowledge that this particular problem belonged to the test for students who would NOT major in natural sciences, but rather in humanities, social sciences, arts, etc. Unlike students applying for natural sciences, they did not learn integration by parts during high school years, hence they could only use the method featured in the video. But it is a good thing to discover other methods on RUclips, which is completely out of such context. 😃

I think it could've been better if you elaborated how your end result helps solving this particular problem, because I think it actually does help. That is, for that lengthy (x - y)(x + y)^3 - (x^2 + xy + y^2)^2 part where I have expanded everything, using your result, (x - y)(x + y)^3 - (x^2 + xy + y^2)^2 = (x - y)(x + y)^3 - (x + y)^4 + 2xy(x + y)^2 - (xy)^2 = (x - y - (x + y))(x + y)^3 + 2xy(x + y)^2 - (xy)^2 = - 2y(x + y)^3 + 2xy(x + y)^2 - (xy)^2 = (- 2y(x + y) + 2xy)(x + y)^2 - (xy)^2 = - 2(y^2)(x + y)^2 - (x^2)(y^2) = - (y^2)(2(x + y)^2 + x^2) = - (y^2)(3x^2 + 4xy + 2y^2), so it actually leads to desired simplification without expanding the whole thing.

@@CornerstonesOfMath I didn't go further because I was tired (giving and grading final exams this week) and wasn't entirely sure that continuing would help. I just suspected it. In the past few months, for things I want to film & post to my educational cahnnel, I often ran into things like that.

Spoiler alert: They wouldn't, because I need to make more shorts videos.

You solved the problem while keeping 5! on the left hand side. Good job!

Thank you for the kind comment.

I wonder what method you used to find that one. Anyway, please go check my video for more answers!

@@CornerstonesOfMath it's a bit underwhelming 😅 I just guessed and checked on desmos, although there has to be a more elegant way to find it

Hi. Although my channel focuses on understanding mathematical principles and logic rather than utilizing programming languages, I do understand the importance of programming and I appreciate the examples given for higher number of digits.

Wow indeed. ❤

If you only seek for answers, that would work (although if number of digits becomes large it would be very difficult to identify integer-coordinate points with only eyeballs). I attempted to provide a solution that allows us to apply and review some basic mathematical theories (quadratic equation and properties of integers), which can be applied to other problems of similar type (for example, finding the integer roots of the multivariable quadratic equation).

Thanks! This is one of the most well-known methods to deal with quadratic equation with integer roots.

Thx❤

Well at least I can say that for an arbitrary base n, two-digit number (n - 2)1 base n (n - 2 as n's digit and 1 as 1's digit, base n) ALWAYS satisfies the given property - that is, √(11 base 3) = 1 + 1, √(21 base 4) = 2 + 1, √(31 base 5) = 3 + 1, and so on. This is because (2-digit number (n - 2)1 base n) = n(n - 2) + 1 = (n - 1)^2, so its square root is n - 1 = (n - 2) + 1. I don't know if there are other possible numbers though, and I don't think I will explore that ever (the equation gets way complicated due to the additional variable for the base).

Well, I wasn't going to dive too deep into this, but since you asked: I haven't tried cases like √(xyz) = x + y + z or √(xyzw) = x + y + z + w, because they already seem too complicated due to the increased number of variables, but I managed to handle √(xyzw) = xy + zw pattern, because this too only requires 2 variables and can be expressed as √(100a + b) = a + b, where a and b are now two-digit numbers maximum. Squaring both sides and doing similar process, we obtain a = - (b - 50) ± √(2500 - 99b) so we have 2500 - 99b = k^2 (k is integer) and 2500 - 99b ≥ 0 (equivalent to b ≤ 25.xxx) This time there are slightly more cases, and it is actually beneficial to use b = (2500 - k^2)/99 = (50 + k)(50 - k)/99 instead, testing integer values k that makes b integers. Then we can find that only possible values are k = 5 and 49, which gives b = 25 and 1, which in turn gives (a, b) = (20, 25), (30, 25), (98, 1) Therefore, there are three cases (if you include single-digit number such as 01): √2025 = 20 + 25 = 45, √3025 = 30 + 25 = 55, √9801 = 98 + 01 = 99. Now I'm wondering if I should make an additional video with these results.

@@CornerstonesOfMath Can't you show that max face sum of a 5 digit number is 54 and even (54)^2<smallest 5 digit number thus it can not have greater that 4 digits. It at least gives us an upper limit. Maybe by using graphs we can show that other number works too and graphs slope of maxfaceSum^2-square is decreasing I think

@@ShivanshPachnanda It can be easily shown that it is impossible from 5 digits because for a 5-digit number abcde, √(abcde) ≤ 9+9+9+9+9 = 45 and thus abcde ≤ 45^2 = 2025. And I think I can show you that even 4-digit numbers (abcd) are impossible since abcd ≤ (9+9+9+9)^2 = 1296, so the value of a is limited to a = 1 only, but then we have 1bcd ≤ (1+9+9+9)^2 = 784, which is impossible. If we use cube root ³√ or 4th root ⁴√ then there can be different stories, but again, I just don't think that this problem is worth expanding.

​@@CornerstonesOfMath edit: I was 1h late with the N>=5 proof there are no solutions for N-digit numbers where n > 2. I checked up to 15 with a python script. 1: [1] 2: [81] 3: [] 4: [] 5: [] 6: [] ... Proof for N >= 5 is easy, because (N * 9)^2 < 10^N if N >= 5 But the cases for 3,4 are less trivial.

I would love to see a video on this conversation thread

Yeah even we don't get to explore the theory of quadratic equations with integer roots, that is perhaps the fastest way to do it.

The answer isn't the point. Learning how to solve the problem is the point.

@@tonyennis1787 I get what you mean but my solution is also a way to solve it. However, the solution shown in the video becomes increasingly better for larger numbers but this "obvious" method is more efficient for 2 digit squares

Thank you, you're noice too

Which one is a, and which one is b?

​@CornerstonesOfMath I have no idea. I guess zombie me at 2:00 PM went crazy. But check this: I will use the notation A = sin(x) B = cos(x) a = sin(y) b = cos(y) So, we have A⁴/a² + B⁴/b² = 1 A⁴b² + B⁴a² = a²b² A⁴b²1 + B⁴a²1 = a²b² A⁴b²(b²+a²) + B⁴a²(a²+b²) - a²b² = 0 A⁴b⁴ + B⁴a⁴ + A⁴a²b² + B⁴a²b² - a²b² = 0 oh I guess that's what zombie me thought! A⁴b⁴ + B⁴a⁴ + (A⁴ + B⁴ - 1)a²b² = 0 A⁴b⁴ + B⁴a⁴ - 2A²B²a²b² = 0 (A²b² - B²a²)² = 0 Then you can go with A²b² = B²a² A²/B² = a²/b² tan²(x) = tan²(y) 1+sec²(x) = 1+sec²(y) sec²(x) = sec²(y) cos²(x) = cos²(y)! Yay!!! I prefer a symmetrical solution like that, instead of writing everything with only cosines or only sines. I feel the other one which is not chosen feels bad. Hahahahahaha. Ok, I am satisfied with my craziness and proud of my algebraic manipulation!!!

@@samueldeandrade8535 Now that is a beautiful solution!

@@CornerstonesOfMath hahaha. Thanks!

That's what I did in the 2nd method.