Video

Thank you for watching this somewhat lengthy video!

Unfortunately, I don't have anything planned in near future (like in several weeks or a month). But since I have stored many workbooks and handouts that I studied during my school years, I think I can come up with something if I look into it. Also, it would be of great help if you can specify what "coordinate problem" means. Does it include solving complicated geometric problem on a coordinate plane? Does it mean handling graphs of complicated functions using advanced math such as limits and calculus? Does it mean the problem including vectors? Does it mean 3-dimensional coordinate geometry?

@@CornerstonesOfMath Thanks for response !!mostly conic sections and calculus mixed u can do !! U can look at jee advanced problems INDIA it is a good source !! U will catch lot of good subscriber from INDIA !!

@@sigmainclination9483 Thanks. I once took a brief look on JEE Advanced problems and there were indeed pretty challenging and interesting problems in there. Hopefully someday I will cover them.😃

I do use trial and error to guess the factor (like t-2) and synthetic division to do the division, but such processes are simplified in the video due to pacing (many people who learned cubic and quartic equations might already know such methods). In fact, dividing with linear polynomial (like t-2) is so easy that I can just do it in my head by comparing the coefficients, but that seems less universal compared to synthetic division.

Can u tell the source ??

@@sigmainclination9483 Hi❤ It's great to see that this video still gets attention. This is a modification of the problem I saw on the Korean workbook I studied during my high school years (I don't know much about Indian books, but perhaps similar to Black Book). The problem on the book may have originated from other primary sources, such as national exams or college entrance exams, but I couldn't track down any further.

Oh, I just went through the general education curriculum, and since I live in highly-competitive South Korea, I got help from private tuition. I think attending science high school (where students learn science and math more extensively) contributed the most in creating this channel, though. The math I learned then did not diverge too far from regular high school math, but it helped me realize many interesting and intense stuffs can be done even with high school math (which is also the main purpose of this channel).

​​@@CornerstonesOfMaththanks for replying by the way I'm from india 🇮🇳

For more than two numbers, the relation between AM, GM, and HM is not as simple as the two-number case (hence not particularly interesting) and they do not form any special type of sequence.

Thanks for the comment! Such graphical visualization and interactive exercises can indeed be beneficial.

I've come up with a problem on solving | | | |x| - 1 | - 1 | - 1 | = 1 when x is REAL using the basic properties of absolute values (czcams.com/video/i54eOEuYGlQ/video.htmlsi=6Y3qdsQXjOZm9QVc), and someone commented what happens for complex numbers - which I thought was an interesting suggestion, so I made this follow-up video.

Thank you! 💛

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.

@@CornerstonesOfMath But, it would have made use of the first part, that 2 | (x^2 + xy + y^2), so 2 | x and 2 | y. Otherwise, I'm not sure why the first part needed to be included?

​@@Limited_Light Now come to think of it, the first part is only required for deriving a contradiction from Case 4, where both x and y are not divisible by 10. x and y being even limits the rightmost digits to only 2, 4, 6, and 8. Without clarifying first that x and y are even, their rightmost digits can be 1, 2, 3, 4, ..., 9. So maybe I could've organized the proof a bit better, but that first part was indeed being used later on.

@@CornerstonesOfMath I will have to rewatch, particularly the second part.

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 CZcams, 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).