This is mad! I am a PhD student who is actively studying number walls! Specifically, I have written a program to generate the number wall from a sequence and I have some nice formulas for the number of sequences that have a given window in their number wall. They have some beautiful relationships to diophantine approximation over function fields and linear complexity! It's so excititing to see them be popularised more.
I'm a little hesitant to share my code as it is a valid solution to this videos "Coding Challenge". Sorry for any disappointment! Just to prove I'm not talking total rubbish though, here are some cool number walls created by my code. drive.google.com/drive/folders/15GEhc5FCTcPDSZoEZxBpDcOSfobyoxjX?usp=sharing
Yet another awesome topic that has me saying "how is this the first time I'm hearing about this?"! The mystery of why the number walls will always have integer entries would've been enough to hold my attention, let alone all the other cool properties! Also, fun fact: The characteristic polynomial is the denominator of the generating function of the sequence! The numerator will be a polynomial of degree less than the denominator, determined by the initial terms of the sequence.
May be you know the answer. Supposr we have f(n) = a_1*f(n-1)+...+a_k*f(n-k) where k > 5. We want to express f(n) as function of n. How can we do it in case we can not solve the characteristics equation?
The recursive formula for the squares translates to: x^2 = 3(x-1)^2 - 3(x-2)^2 + (x-3)^2 Expanding the right hand side gives us: (3x^2 - 6x + 3) - (3x^2 - 12x + 12) + (x^2 - 6x + 9) = 3x^2 - 3x^2 + x^2 - 6x +12x - 6x + 3 - 12 + 9 = x^2 The formula works :) The coefficients of the formulas look like the binomial coefficients, they are matching with the rows in pascals triangle. You can generate the formla by expanding the right side of the equation 0 = (x - 1)^n and replacing every x^k by the k-th coefficient c_k. Then solve for the highest power (x^n) and you have the formula for the sums of x^(n-1). Notice, the formula for the squares uses the coefficients of the 3rd row, expanding (x-1)^3 not (x-1)^2.
The polynomial you get for the sequence n^k is just (x-1)^(k+1) (hence the coefficients in the expansion are binomial coefficients). The general solution of such a recurrence is a_0 + a_1 n + a_2 n^2 + ... + a_k n_k (a *very* special case of the general formula, since all the roots are 1) To get the a_i you can just plug in the initial values and solve the linear system. It's a bit tedius for larger k, but say for k = 2 we get a_0 + a_1 n + a_2 n^2 at n=0 -> a_0 = 0 a_1 n + a_2 n^2 at n=1 -> a_1 + a_2 = 1 a_1 n + a_2 n^2 at n=2 -> 2 a_1 + 4 a_2 = 4 So 2(1-a_2) + 4 a_2 = 4 -> a_2 =1 and thus a_1 = 0 so we get n^2 as the solution as we expected
This really reminds me of solving differential equations, especially y'''=3y''-3y'+1 The characteristic equation is r^3-3r^2+3r-1=0 or (r-1)^3=0 We get r=1 with a multiplicity of 3. For d.eq, it meant that the solutions are of form Ax^2 e^x + Bx e^x + C e^x But when it comes to solving functions, the e^(1x) is replaced with 1^n. If it were e^(2x), it would become 2^n. This means that the solutions are now of form A*n^2 * 1^n + B*n * 1^n + C * 1^n = An^2+Bn+C, which is just a quadratic! Note because Sn = 3Sn-1 - 3Sn-2 + Sn-3 is a recurrence relation with three variables on the right, we need three initial values, just like how the fibonacci sequence has F1 = 1 and F2 = 1. Because we are dealing with the square numbers, the three initial values are S0 = 0, S1 = 1, and S2 = 4. This means that C = 0, A+B = 1, and 4A+2B = 4. We get B = 0 and A = 1, giving us Sn = n^2, as it was desired to show. QED.
The equation works - but how well it works ... x ^ 2 = 3 (x-1) ^ 2-3 (x-2) ^ 2 + (x-3) ^ 2 This equation also works: for any k x ^ 2 = 3 (x-k) ^ 2-3 (x-2k) ^ 2 + (x-3k) ^ 2 It is not the end. This also works: f (x) = ax ^ 2 + bx + c and any k f (x) = 3f (x-k) -3f (x-2k) + f (x-3k) Also for higher power exponents
@@marekmatusiak8397 x^m=\sum_{i=0}^{m}=(-1)^i\binom{m+1}{i+1}(x-i-1)^m If you want to shift by k units, here it it shifted by 1 (x+1)^m=\sum_{i=0}^{m}=(-1)^i\binom{m+1}{i+1}(x-i)^m also, = \sum_{i=0}^{m}\binom{m}{i}x^i :) eg. A quintic integer x in terms starting at (x-5): (x+5)^5=\sum_{i=0}^{5}=(-1)^i\binom{6}{i+1}(x-i+4)^5=3125+3125x+1250x^2+250x^3+25x^4+x^5
I love how every time mathologer posts he just shows up with a new video about something I've never heard of and then suddenly I have something new to explore for the next month or so.
Top secret stuff .... been around for 50 plus years 4sure. Like lock combinations that don't have keys. ... its great for art complexities even in cloth manufacturers trades secrets....
I love the feeling of having just about finished watching another great math video, and then I remember this is Matholologer and I've only just finished chapter 2 of 5! You spoil us with knowledge!
Self similarity must be linked to lack of information, in some way that escapes my capacity, but I wish I knew how to evalué and link the information with the self similarity.
Yeah and another thing that immediately came to my mind is the method of finite differences and now I wonder if there is any usable connection between pdes(/(i)bvps?) and those sequences. Meaning e.g. use properties of pdes to find something about the properties of these sequences, or vice versa use properties of these sequences to find out something about pdes(/(i)bvps?) Seems anyway like a not really fitting match up, because you have only one boundary to start with ... but who knows if there isnt any usable stuff on the interior???? Particularly interesting seems to me, if you have such a "stencil" or method of calculation, can you actually show that some pdes are aphysical, meaning end somewhere else than on the boundary? Sure those are only approximations .... and in the end does that mean these approximations break down because you cannot zoom in, or is it that you'd need to also change the starting row if you zoom in...
For years people have discussed the level of mathematics that can be found within art. Number Walls are proof that a hidden masterpiece can exist if you recognize the pattern of beauty. In the past I watched your videos for the lessons and enthusiasm, given the level of excitement you've displayed concerning Number Walls - I'm afraid you'll forever be the Paint by Number Guy in my book. It's really something to see a scientist morph into an artist in a video. Good job!
The closed formula for the mortal enemy's sequence is: E_n = 0.8469 (1.5747)^n + 0.2636 (1.3802)^n cos(1.7805n + 0.9507). All the decimals are rounded so this won't be very accurate past n = 15 or so, but we can use this to get the asymptotic formula for E_n. Since 1.3802^n
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).
Man if I had a math teacher like you , I would be a rocket scientist by now ! I was in high school in an elite math class but switched to Biology going to university cause of losing interest in math, but i understood 99 % of your video and really got me excited again by math.
My mother was a commercial artist and back in the 1980s she was designing patterns for carpets that were being manufactured using a Wilton loom process. The Wilton loom could accommodate a multi-colored pattern however there were a set of physical rules limiting the transposing of colors in a given row and also a different set of rules determining the transposing of colors from one row to the next. These rules came from the physical limitations of the mechanical loom design. The loom itself was driven from punch cards and this is the earliest use of a mechanical punch cards. I worked on a computer program that could replicate the rules imposed by the loom and could allow for different new patterns to be generated from a few adjustable parameters, such that each of these designs met the loom manufacturing rules.
Thank you the video was fantastic!! I never heard of number walls before, and the graphical presentation was very informative. I look forward to learning more from the links.
So, could the infinite row of zeros at the top be considered to be infinitely high moving upwards too? … thus also being a square window ….. which, for want of a better term, I think should be called the “sky” 😀
I think that actually, the row of zeroes at the top wouldn't extend to an infinite series of rows of zeroes. Since the number wall rule is symmetrical, I believe that above the wall would have to be either symmetric or antisymmetric (all values negated) with the part of the wall below the line. I haven't checked exactly what happens though.
That was my first thought, followed by "I'm sure this would be used to predict primes if there was a recursive formula you could derive from this method...but there isn't."
A nice brief explanation of the determinant is that it is 1 for the identity matrix and additive as function of each is its rows. May sound complicated but if you know these terms you know they can be quickly demonstrated on elementary level. They also easily explain why it's zeroed by linear dependence with a little work
I really enjoy watching your videos Burkard, thanks to you I've rediscovered the thrive and passion to take pen and paper and try to prove things. Thank you so much.
Fascinating. I've been working with linear sequence recurrence for decades, and never heard of this. Once I saw the initial construction, I began thinking about determinants, due to the small cross formula being related to the 2x2 determinant. Computing determinants by the efficient sharing of all previous subdeterminants - we all have seen how to decompose determinants into subdeterminants, but it quickly becomes impractical to compute, even on a computer, since determinants are based on the combination of all possible combinations... This is very nice. Usually to solve for linear filters you need to trot out at least Cholesky decomposition. This is just a simple rule (plus a couple of complicated rules, the big cross and horseshoe).
Came across this from a Fred Lunnon paper about Pagoda Sequence... John Conway, who took an early interest in this topic, christened the zero regions windows, and the table a wall of numbers, [2]. Apparently, on first encountering these results, he transcribed them for safe keeping onto his bathroom wall (the way one does); but having moved house by the time the book came to be written, was obliged to rely on memory, and as a result (to his evident embarrassment) committed two separate typographical errors in restating them.]
This is crazy. I was just exploring recursive sequences and the characteristic polynomials to generate a function that's the sum/product of exponentials and polynomials. I was doing it to generate coefficients to best predict a time series with noise.
Fascinating as always. Thank you Burkard! Lately I've become obsessed with spiral mathematics. The obsession was spurred by a game called Idle Spiral that was released about 4 months ago. I'm no mathematician so figuring out where the game-play mechanics and real math diverge is a challenge for me. But it's fun to try!
We hit on these as recreational math problems one course in university. They were advertised the same as Collatz - dive in and get lost forever in the patterns. I suspect it came up because that was around when computer networks were starting to be used and the horseshoe was published.
So I see a pattern. About 1/2 way through every Mathologer video I comment that this is simply the best and most amazing video yet. And again. No number wall needed to decode the pattern. You are simply the best. And getting better.
How interesting, when I have some spare time I know what I’ll be looking through/trying out. On a slightly off topic note Wonderwall was stuck in my head while watching this
Wow! Pagoda number wall also contains an optical illusion: little squares seem to be rotated to the right or to the left depending on the sorrunding squares!!!
This was a fun coding challenge, it's very satisfying to have the computer do all the work for you. plus you get pretty patterns so I consider that a win-win :D
Math fan slash guitarist here. Starting music is the (in)famous "Asturias", which was inspired IIRC by a legend from Asturia (= Spanish province) about a group of miners that got stuck when then mine collapsed. So, in this "Asturias Leyenda" they kind of hit a wall. Isaac Albeniz was the composer, by the way.
@22:36 Theoretically we don't need other factors to calculate factor at the bottom edge. Since the row has exponential growth and we know first and last term: 992436543 = -243 * x^5 (5 is a number of steps to get from -243 to 992436543) thus: x = (992436543/-243)^(1/5) = -21 this factor from left to right (-243) * (-21) = 5103; 5130 * (-21) = -107163 and so on. But I grant you that the equation with other factors may be easier to use manually, or to get result as a fraction rather than decimal.
When I saw the cross rule it immediately reminded me of matrix multiplication and lo and behold we get into linear algebra and determinants, it would have been nice to have some history of number walls though. The patterns that come from the integers is amazing, I can only think they somehow describe the universe in ways we still do not see.
This and the difference table video have been my favorite videos from you so far. Excellent work. Love it. I swear, all of math is just Pascal's Triangle and the Pythagorean Theorem in disguise.
I just spent twenty or so minutes trying to make a number-wall for the partition function by hand. I may have given myself a mathematical concussion in the process😵Great stuff! Maybe next time I'll start with a simpler sequence though 😂😭
The zero windows seem eerily similar to black holes! Especially with the square rule breaking down at the event horizon (the borders of the windows) and requiring us to use special rules
Thank you for bringing this back. Determinants can be deterring. Ahaha Meanwhile ... Store Associate: "Which gift wrapping paper would you like?" Me: "Please give me a roll of ternary, mod 3 pattern with the maximal zero-free diamond region." Store Associate: "What colour scheme?"
Why do I need a cart? I really like what Master does on UA-cam. The Soviet school of mathematics was very specialized. Now I'm interested to know the context of the work of great mathematicians :). However, it may not matter soon :(. Bureaucracy and ignorance will kill this world, we probably deserve it. The danger of making stupid decisions is very high (sorry, I'm from Russia, Bleak House is waiting for wise words and decisions;).
Amazing thank you! More patterns from numbers! Those recurring patterns with the 0 squares are so much like fractal patterns or that unpronouncable triangle I keep forgetting the name of. Yes they might make good designs for t-shirts! It was just a little confusing how the 'x' you introduced relates to the numbers along the original sequence to convert it to Sn-1 etc.
15:30 I actually felt this mathematicians’ viewpoint overtake my thinking (that is; I thought: ”Aren’t the isolated 0’s just 1*1-squares, and the infinite terminal 0’s just infinite 0-squares?”), *_JUST_* before you started talking about it (the mathematicians’ viewpoint).
I think the long cross rule can be used to compute the two numbers directly below a 2x2 square, but then to compute the two numbers below those 2, you still have to use the horseshoe rule.
I got a mystery sequence in my head which might not even have a formula. And it's base 10 dependant: 6, 2, 5, 5, 4, 5, 6, 3, 7, 6, 8 I should check the Oeis to see if they have a neater formula and no there wasn't. The self similarity might look more beautiful if you zoom out instead of in, since the resolution increases and not decreases.
Haven't watched yet, just wanna say the name reminds me of 6! 3.. no 4! 11? That's Numberwang! Edit: now that I've watched the video, hot diggity damn that was wonderful. The whole process was pretty mysterious, the cross relation vaguely reminded me of the 2×2 determinant but I didn't take this thought far enough. The fact that these algorithms can give you these shifted sequence determinants and thus the recursion formula for any sequence with one is pretty amazing
Your videos are always so entertaining and edjamacational. Every Calculus video I have ever see only mentions what it's used for (For example: to calculate the area of a random blob), but they never use it in the video. It's all talk and no do. Would you please make a Calculus video where an answer is *actually calculated*?
The thing is, calculus is mostly "algebra auto pilot" once you start actually using it. You might like channels like blackpenredpen, which are largely focused on the working out bit.
I'm guessing that 99.99% of professional mathematicians not knowing about number walls is a poor estimate :) - and what a wonderfully taught video. Marco Pizzato is such a Pal you named him twice :)
17:28 The 7 wall contains diagrams of radio telescopes. ;) Look for nearby dots making parabolas with 2 diagonal lines and 1 partial vertical or horizontal. I see one facing up on the left side and one facing to the right somewhat left of the middle. Beautiful video! :)
I'm a bit confused how you generate the numbers to the far left and right. Maybe I missed it in the video. You showed finding one question mark using the cross. But I don't see a simple solution for the locations where you have a question mark both below and to the left/right.
I think since you're starting from an infinitely long sequence, you never really need to generate numbers to the left or right. You only need to go down.
Anyway, here's numberwall. Also, at 22:40, I would think it much easier to remember how to do it if you just walk all four sides clockwise and multiply the ratios.
If you define M x N as the cellwise multiplication of two images, M, N, then after applying the cross rule to seq (assuming no 0s) down = (center**2 - right*left) * 1/up to get an image M, then we'd have shiftUp(M) == (MxM - shiftLeft(M) x shiftRight(M)) x (1/shiftDown(M)) M is a fixed point (i.e. f(M) = M) under the function f that takes an image M to f: M -> shiftDown((MxM - shiftLeft(M) x shiftRight(M)) x (1/shiftDown(M)))
as you explained it i straight away had to implement the number wall formula in a spreadsheet and put in the Thue-Morse sequence. aaand get div/0 errors in my fourth row 😅.
I'm 'just' an engineer (BS, not PhD), so most of these videos eventually break of my desperately clinging fingernails, and I go into 'be entertained' mode. But this video, I actually understood all the way, even though I've never heard of number walls. Nonetheless, TOTALLY BLEW MY FRIKKIN' MIND!!!! Note, you've just ruined every test that has a "what number comes next" question. ;-)
9:20 I’m seeing the coefficients of the binomial cube: 1 3 3 1, in the recursive formula for squares, there. Coincidence? 🤔 9:35 Apparently not, since the coefficients, for the cubes, are the coefficients of the binomial 4th powers: 1 4 6 4 1.
There was mention of number walls for both finite and infinite sequences. On the finite side, I am curious if there’s any pattern that would emerge from any of various starting points in the Collatz sequence - e.g., the 100-ish-digit sequence starting with 27.
I love the music from my country: Asturias by Isaac Albeniz performed by Guitar Classics and Taiyo (Sun) by Yuhi (Evening Sun). I was surprised when it started playing. :)
Can you explain how you calculated the first term in 4th row of your first sequence? Row 1: …. 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, … Row 2: … -1, 0, 1, 0, -1, 1, 1, -1, -1, … Row 3: … 1, 1, 1, 1, 1 , 2, 2, 2 , 1, … Row 4: …. ??, When I use the first cross formula I get ??= -1. You have ??= 1. Thoughts?
I wish you had structured this video to start with the explanation of a number wall as the result of determinant operations instead of using up 30 minutes of my life confusingly going on about unexplained and "apparently arbitrary" cross and horseshoe games. Then you could talk about how the computationally quicker cross and horseshoe rules are derived from the original determinant operations and then how they lead to fibonacci type solutions of sequences. It would have made MUCH more sense.
An extra credit question on my linear algebra final back in the day involved "look-and-say sequences" (A005150 on OEIS is the simplest of them). I wish I could remember how it worked because I feel like it would tie into this number wall idea somehow. The first number in the sequence is 1. Read aloud, that is "one '1'," (11). Read aloud, that is "two '1's" (21). Then "one '2', one '1'" (1211), etc.
Help me out anyone, I'm probably missing something obvious as usual. In row 3 of the first number wall example, which goes . . . 1 1 1 1 1 2 2 2 1 . . ., I can see how he obtains the second digit. It's the solution to the equation (-1 x 1) + (1 x ?) = 0². But where does the first digit in that row come from? The equation for that would have to have two unknowns, wouldn't it? Another, possibly related, problem is how we continue beyond what's displayed. I understand his example in 2:25, but what about the three dots at the start of row 2: . . . -1 0 1 0 -1 1 -1 -1 . . ., what goes there if we continue to the left? We have the equation (? x 0) + (1 x 1) = -1² where ? can be any number.
Finish the video. He talks on detail about when one of the numbers is 0 which makes it impossible to use the cross-rule to calculate one of the neighbours. He shows another rule to solve that (one for single 0's and one for blocks of 0's).
@@remischmitt9308 Thanks, I should have stayed watching, but I thought I'd stop and consolidate before things got complicated. Anyway, could I now trouble you to check my answers? The dotted gap at the start of row 2 has -1, which I found with the long cross rule.The gap at the start of row 3 has 2.
To the best of the knowledge, it has no easy definition. It can be described as the difference between the (n-1)th and the (n+1)th terms of the paper folding sequence, which is a very well known sequence
@@TheShadowOfMars perhaps I should have said intuitive instead of easy. I am aware of the definition of the paper folding sequence, but I can see why Mathologer didn't think the tangent was necessary
This is mad! I am a PhD student who is actively studying number walls! Specifically, I have written a program to generate the number wall from a sequence and I have some nice formulas for the number of sequences that have a given window in their number wall. They have some beautiful relationships to diophantine approximation over function fields and linear complexity! It's so excititing to see them be popularised more.
hehe is your code open sorced? :D
i will definitely program the number wall to solve sequences for so i can make a "..." function :)
Is you code available anywhere? I could definitely use it to help find some recurrence relations in my mathematical pet projects!
Do the cross and horseshoe formulas hold for any commutative ring?
I'm a little hesitant to share my code as it is a valid solution to this videos "Coding Challenge". Sorry for any disappointment!
Just to prove I'm not talking total rubbish though, here are some cool number walls created by my code.
drive.google.com/drive/folders/15GEhc5FCTcPDSZoEZxBpDcOSfobyoxjX?usp=sharing
@@gamekiller0123 to the best of my understanding they do! Though I work over finite fields and usually take these formulas for granted.
Yet another awesome topic that has me saying "how is this the first time I'm hearing about this?"! The mystery of why the number walls will always have integer entries would've been enough to hold my attention, let alone all the other cool properties!
Also, fun fact: The characteristic polynomial is the denominator of the generating function of the sequence! The numerator will be a polynomial of degree less than the denominator, determined by the initial terms of the sequence.
May be you know the answer. Supposr we have f(n) = a_1*f(n-1)+...+a_k*f(n-k) where k > 5.
We want to express f(n) as function of n. How can we do it in case we can not solve the characteristics equation?
The recursive formula for the squares translates to:
x^2 = 3(x-1)^2 - 3(x-2)^2 + (x-3)^2
Expanding the right hand side gives us:
(3x^2 - 6x + 3) - (3x^2 - 12x + 12) + (x^2 - 6x + 9)
= 3x^2 - 3x^2 + x^2 - 6x +12x - 6x + 3 - 12 + 9
= x^2
The formula works :)
The coefficients of the formulas look like the binomial coefficients, they are matching with the rows in pascals triangle.
You can generate the formla by expanding the right side of the equation 0 = (x - 1)^n and replacing every x^k by the k-th coefficient c_k. Then solve for the highest power (x^n) and you have the formula for the sums of x^(n-1). Notice, the formula for the squares uses the coefficients of the 3rd row, expanding (x-1)^3 not (x-1)^2.
The polynomial you get for the sequence n^k is just (x-1)^(k+1) (hence the coefficients in the expansion are binomial coefficients).
The general solution of such a recurrence is a_0 + a_1 n + a_2 n^2 + ... + a_k n_k (a *very* special case of the general formula, since all the roots are 1)
To get the a_i you can just plug in the initial values and solve the linear system. It's a bit tedius for larger k, but say for k = 2 we get
a_0 + a_1 n + a_2 n^2 at n=0 -> a_0 = 0
a_1 n + a_2 n^2 at n=1 -> a_1 + a_2 = 1
a_1 n + a_2 n^2 at n=2 -> 2 a_1 + 4 a_2 = 4
So 2(1-a_2) + 4 a_2 = 4 -> a_2 =1 and thus a_1 = 0 so we get n^2 as the solution as we expected
This really reminds me of solving differential equations, especially y'''=3y''-3y'+1
The characteristic equation is r^3-3r^2+3r-1=0 or (r-1)^3=0
We get r=1 with a multiplicity of 3. For d.eq, it meant that the solutions are of form Ax^2 e^x + Bx e^x + C e^x
But when it comes to solving functions, the e^(1x) is replaced with 1^n. If it were e^(2x), it would become 2^n.
This means that the solutions are now of form A*n^2 * 1^n + B*n * 1^n + C * 1^n = An^2+Bn+C, which is just a quadratic! Note because Sn = 3Sn-1 - 3Sn-2 + Sn-3 is a recurrence relation with three variables on the right, we need three initial values, just like how the fibonacci sequence has F1 = 1 and F2 = 1.
Because we are dealing with the square numbers, the three initial values are S0 = 0, S1 = 1, and S2 = 4.
This means that C = 0, A+B = 1, and 4A+2B = 4. We get B = 0 and A = 1, giving us Sn = n^2, as it was desired to show. QED.
No guys you dont understand because 35((818%*+=+*))10^83
The equation works - but how well it works ...
x ^ 2 = 3 (x-1) ^ 2-3 (x-2) ^ 2 + (x-3) ^ 2
This equation also works:
for any k
x ^ 2 = 3 (x-k) ^ 2-3 (x-2k) ^ 2 + (x-3k) ^ 2
It is not the end. This also works:
f (x) = ax ^ 2 + bx + c and any k
f (x) = 3f (x-k) -3f (x-2k) + f (x-3k)
Also for higher power exponents
@@marekmatusiak8397 x^m=\sum_{i=0}^{m}=(-1)^i\binom{m+1}{i+1}(x-i-1)^m
If you want to shift by k units, here it it shifted by 1
(x+1)^m=\sum_{i=0}^{m}=(-1)^i\binom{m+1}{i+1}(x-i)^m also, = \sum_{i=0}^{m}\binom{m}{i}x^i :)
eg. A quintic integer x in terms starting at (x-5):
(x+5)^5=\sum_{i=0}^{5}=(-1)^i\binom{6}{i+1}(x-i+4)^5=3125+3125x+1250x^2+250x^3+25x^4+x^5
I love how every time mathologer posts he just shows up with a new video about something I've never heard of and then suddenly I have something new to explore for the next month or so.
Top secret stuff .... been around for 50 plus years 4sure. Like lock combinations that don't have keys. ... its great for art complexities even in cloth manufacturers trades secrets....
I love the feeling of having just about finished watching another great math video, and then I remember this is Matholologer and I've only just finished chapter 2 of 5!
You spoil us with knowledge!
What was chapter 1?
i can't help but smile whenever self-similarity pops up out of seemingly nowhere!
Self similarity must be linked to lack of information, in some way that escapes my capacity, but I wish I knew how to evalué and link the information with the self similarity.
This seems very similar to second order automata - each cell's value is determined by the three cells over it, and the cell 2 over it.
If you check out some of the references given in the description you'll find that there is indeed a connection ;)
It's a bit like Conway's game of life. Albeit one that has an evaluation direction.
Yeah and another thing that immediately came to my mind is the method of finite differences and now I wonder if there is any usable connection between pdes(/(i)bvps?) and those sequences. Meaning e.g. use properties of pdes to find something about the properties of these sequences, or vice versa use properties of these sequences to find out something about pdes(/(i)bvps?)
Seems anyway like a not really fitting match up, because you have only one boundary to start with ... but who knows if there isnt any usable stuff on the interior????
Particularly interesting seems to me, if you have such a "stencil" or method of calculation, can you actually show that some pdes are aphysical, meaning end somewhere else than on the boundary?
Sure those are only approximations .... and in the end does that mean these approximations break down because you cannot zoom in, or is it that you'd need to also change the starting row if you zoom in...
@@kees-janhermans910Maybe Wolfram rules?
For years people have discussed the level of mathematics that can be found within art. Number Walls are proof that a hidden masterpiece can exist if you recognize the pattern of beauty. In the past I watched your videos for the lessons and enthusiasm, given the level of excitement you've displayed concerning Number Walls - I'm afraid you'll forever be the Paint by Number Guy in my book. It's really something to see a scientist morph into an artist in a video. Good job!
The closed formula for the mortal enemy's sequence is: E_n = 0.8469 (1.5747)^n + 0.2636 (1.3802)^n cos(1.7805n + 0.9507).
All the decimals are rounded so this won't be very accurate past n = 15 or so, but we can use this to get the asymptotic formula for E_n. Since 1.3802^n
Gums drops in a gum disperser..... or atoms ⚛ 🤔 🤣 😏 🙄 electrons protons in the power 🔋 plants??◇operational ◇
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).
Very nice :)
@@Mathologer Thanks :)
Man if I had a math teacher like you , I would be a rocket scientist by now ! I was in high school in an elite math class but switched to Biology going to university cause of losing interest in math, but i understood 99 % of your video and really got me excited again by math.
You are by far the best math channel on UA-cam
Glad you think so and thank you for saying so :)
You can get the non-recursive function from a recursive function using the z-transform. This is used often in Digital Signal Processing problems.
interesting aspect I must say ... 🙂
Maybe indeed useful to try to apply if a problem seems to be of similar kind
Wait so what is the difference between the z-transform and using generating functions to solve the recurrence relation?
@@karolakkolo123 z-transforms are a different name for a sub-type of generating functions
More (fabric designers) should partner with mathematicians. Gorgeous animations!
Yes!! I was thinking of linoleum floor patterns myself.😊👍
Did done been there secrets out
@@richardgratton7557 right on Richard.
My mother was a commercial artist and back in the 1980s she was designing patterns for carpets that were being manufactured using a Wilton loom process. The Wilton loom could accommodate a multi-colored pattern however there were a set of physical rules limiting the transposing of colors in a given row and also a different set of rules determining the transposing of colors from one row to the next. These rules came from the physical limitations of the mechanical loom design.
The loom itself was driven from punch cards and this is the earliest use of a mechanical punch cards.
I worked on a computer program that could replicate the rules imposed by the loom and could allow for different new patterns to be generated from a few adjustable parameters, such that each of these designs met the loom manufacturing rules.
Well, you’ve set me down a rabbit hole again for at least the next week. Amazing video as always!
A like the 🐇 🐰 🐇 🐰 hole.hop hop hop hop hop not lost not lost not lost.♾♾♾♾♾♾♾♾
Thank you the video was fantastic!! I never heard of number walls before, and the graphical presentation was very informative. I look forward to learning more from the links.
Thank you :)
Is differences?
I also see Pascal’s triangle in the Fibonacci, square and cube formulas at 9:47 I.e. 1,1…1,3,3,1…1,4,6,4,1…etc.
Wow... Me too. Seems so obvious.
It's just row n+1 (the one starting 1, n) for x^n (n=0 is trivial!)
Very original video. I watch a ton of math edu content.. and I’ve never heard of math walls. Well done!
Laughing 😆 👽👾😆 🎃😂🤣😅😇🤓😏the spy who loved me....ta da super hero📓
You're a pedagogical genius!
So, could the infinite row of zeros at the top be considered to be infinitely high moving upwards too? … thus also being a square window ….. which, for want of a better term, I think should be called the “sky” 😀
Sooo the sky's the limit?
😅
I think that actually, the row of zeroes at the top wouldn't extend to an infinite series of rows of zeroes. Since the number wall rule is symmetrical, I believe that above the wall would have to be either symmetric or antisymmetric (all values negated) with the part of the wall below the line. I haven't checked exactly what happens though.
@@killerbee.13 yeah, like the extension of the Pascal's triangle
Immediate thought: what does the number wall of the prime numbers look like?
Interesting…probably no rows of zeros?
That was my first thought, followed by "I'm sure this would be used to predict primes if there was a recursive formula you could derive from this method...but there isn't."
Depends on the wall created to open() gate() or creating a wall. Does the numbers match????()
I was thinking a number wall of a non-polynomial number... like the digits of e or pi
Here is the number wall of PRIME NUMBERS!
- - - 1 1 1 1 1 1 1 1 1 1 1 1 1 1 - - -
- - - 7 11 13 17 19 23 29 31 37 41 43 47 53 59 - - -
- - - -6 30 -18 42 -30 -22 128 -112 98 90 -78 -70 36 248 - - -
- - - -12 72 -72 72 96 188 480 0 532 384 288 164 352 1184 - - -
- - - 0 144 0 288 144 488 1800 2280 2888 -64 -256 1064 -1952 5312 - - -
- - - 288 576 1152 -1248 -112 4432 -1552 15952 1936 464 3856 -5232 30928 - - -
.
.
.
A nice brief explanation of the determinant is that it is 1 for the identity matrix and additive as function of each is its rows. May sound complicated but if you know these terms you know they can be quickly demonstrated on elementary level. They also easily explain why it's zeroed by linear dependence with a little work
I really enjoy watching your videos Burkard, thanks to you I've rediscovered the thrive and passion to take pen and paper and try to prove things. Thank you so much.
Loved that you used Asturias played by Andres Segovia for this .... Perfect match
Fascinating. I've been working with linear sequence recurrence for decades, and never heard of this. Once I saw the initial construction, I began thinking about determinants, due to the small cross formula being related to the 2x2 determinant. Computing determinants by the efficient sharing of all previous subdeterminants - we all have seen how to decompose determinants into subdeterminants, but it quickly becomes impractical to compute, even on a computer, since determinants are based on the combination of all possible combinations... This is very nice. Usually to solve for linear filters you need to trot out at least Cholesky decomposition. This is just a simple rule (plus a couple of complicated rules, the big cross and horseshoe).
Came across this from a Fred Lunnon paper about Pagoda Sequence...
John Conway, who took an early interest in this topic, christened the zero regions windows, and the table a wall of numbers, [2]. Apparently, on first encountering these results, he transcribed them for safe keeping onto his bathroom wall (the way one does); but having moved house by the time the book came to be written, was obliged to rely on memory, and as a result (to his evident embarrassment) committed two separate typographical errors in restating them.]
Woah! Another CRAZY video!!! These videos i cannot resist to watch:) So beautifully made are they.
This is crazy. I was just exploring recursive sequences and the characteristic polynomials to generate a function that's the sum/product of exponentials and polynomials.
I was doing it to generate coefficients to best predict a time series with noise.
How on earth could you resist to check what happens with the sequence of primes???
Thanks for the video.
I spent the whole video wondering what it would look like with the sequence of primes as a second row... guess I'll have to test it myself some day.
Maybe it will become a mandelbrot fractal?
If the second row is the primes, then third row is oeis A056221
The Prime Wall goes like....
- - - 1 1 1 1 1 1 1 1 1 1 1 1 1 1 - - -
- - - 7 11 13 17 19 23 29 31 37 41 43 47 53 59 - - -
- - - -6 30 -18 42 -30 -22 128 -112 98 90 -78 -70 36 248 - - -
- - - -12 72 -72 72 96 188 480 0 532 384 288 164 352 1184 - - -
- - - 0 144 0 288 144 488 1800 2280 2888 -64 -256 1064 -1952 5312 - - -
- - - 288 576 1152 -1248 -112 4432 -1552 15952 1936 464 3856 -5232 30928 - - -
.
.
.
@@petrospaulos7736 great hint!
And sadly, no zeros in sight?
Fascinating as always. Thank you Burkard! Lately I've become obsessed with spiral mathematics. The obsession was spurred by a game called Idle Spiral that was released about 4 months ago. I'm no mathematician so figuring out where the game-play mechanics and real math diverge is a challenge for me. But it's fun to try!
We hit on these as recreational math problems one course in university. They were advertised the same as Collatz - dive in and get lost forever in the patterns. I suspect it came up because that was around when computer networks were starting to be used and the horseshoe was published.
It's always a good day when mathologer post a video!
Math is just amazing! First time that I've seen or heard of the number wall, very interestting
So I see a pattern. About 1/2 way through every Mathologer video I comment that this is simply the best and most amazing video yet. And again. No number wall needed to decode the pattern. You are simply the best. And getting better.
How interesting, when I have some spare time I know what I’ll be looking through/trying out.
On a slightly off topic note Wonderwall was stuck in my head while watching this
"'Cause after all... that's its number wall..." ("it" being any certain number sequence)
Unexpected symmetries are always delightful.
Wow! Pagoda number wall also contains an optical illusion: little squares seem to be rotated to the right or to the left depending on the sorrunding squares!!!
determinants always seemed rather mysterious to me and it seems like they never fail to amaze me :)
This was a fun coding challenge, it's very satisfying to have the computer do all the work for you. plus you get pretty patterns so I consider that a win-win :D
Math fan slash guitarist here. Starting music is the (in)famous "Asturias", which was inspired IIRC by a legend from Asturia (= Spanish province) about a group of miners that got stuck when then mine collapsed. So, in this "Asturias Leyenda" they kind of hit a wall. Isaac Albeniz was the composer, by the way.
Another amazing video. Great subject to revisit. Math dazzles like a super nova in Mathologer videos.
@22:36 Theoretically we don't need other factors to calculate factor at the bottom edge. Since the row has exponential growth and we know first and last term:
992436543 = -243 * x^5 (5 is a number of steps to get from -243 to 992436543)
thus: x = (992436543/-243)^(1/5) = -21 this factor from left to right
(-243) * (-21) = 5103; 5130 * (-21) = -107163 and so on.
But I grant you that the equation with other factors may be easier to use manually, or to get result as a fraction rather than decimal.
When I saw the cross rule it immediately reminded me of matrix multiplication and lo and behold we get into linear algebra and determinants, it would have been nice to have some history of number walls though. The patterns that come from the integers is amazing, I can only think they somehow describe the universe in ways we still do not see.
Almost as excited to see the t-shirt as I am to hear the math!
Your ironic t-shirts are the second best part of your videos
This and the difference table video have been my favorite videos from you so far.
Excellent work. Love it.
I swear, all of math is just Pascal's Triangle and the Pythagorean Theorem in disguise.
I just spent twenty or so minutes trying to make a number-wall for the partition function by hand. I may have given myself a mathematical concussion in the process😵Great stuff! Maybe next time I'll start with a simpler sequence though 😂😭
Actually I could use some math triage here. I got a fraction in my counting and I'm not sure where I went wrong
Here is my number-wall so far
* 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 *
* 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 *
* 0 0 1 1 2 3 5 7 11 15 22 30 42 56 77 *
* 0 0 1 -1 1 -1 4 -6 16 -17 34 -24 84 -98 * *
* 0 0 1 0 0 -1 2 -4 14 -17 34 -76 112 * * *
* 0 0 1 0 0 -1 0 2 8 11 -4 82 * * * *
* 0 0 1 -1 1 -1 1 -1 3 -9 27 * * * * *
* 0 0 1 0 1 0 2 -1 0 0 * * * * * *
* 0 0 1 -1 1 -2 4 -1 0 * * * * * * *
* 0 0 1 0 1 -3 7 -1 * * * * * * * *
* 0 0 1 1 1 1 (11.5)
11.5? Did I miscalculate somewhere?
The computer is your friend.
The partition function was the first function I thought of as well!
what's partition function?
@@NoNameAtAll2 The "hardest what-comes-next" video can show you. It's the number of ways to break the index up into equal or smaller integer parts.
The zero windows seem eerily similar to black holes! Especially with the square rule breaking down at the event horizon (the borders of the windows) and requiring us to use special rules
Hi,
Wonderful!
The guitar piece is called "Asturias" and is from Isaac Albéniz.
Thank you for bringing this back. Determinants can be deterring. Ahaha
Meanwhile ...
Store Associate: "Which gift wrapping paper would you like?"
Me: "Please give me a roll of ternary, mod 3 pattern with the maximal zero-free diamond region."
Store Associate: "What colour scheme?"
Thank you so much for "I definitely don't expect you to have understood everything"! (24:38) And thanks for the awesome video!
You can do this with symmetric polynomials. Hn(eigenvalues) are xn and En(eigenvalues) are the char poly coefficients
There are still many more new things to discover. Great Video.
What a wonderful discovery these walls! Great music choice
Thanks for such a masterpiece.
love your shirts as well.
love this kind of intricate pattern appearing out of logic
In diagonal recording, this is even more amazing.
Why do I need a cart? I really like what Master does on UA-cam. The Soviet school of mathematics was very specialized. Now I'm interested to know the context of the work of great mathematicians :). However, it may not matter soon :(. Bureaucracy and ignorance will kill this world, we probably deserve it. The danger of making stupid decisions is very high (sorry, I'm from Russia, Bleak House is waiting for wise words and decisions;).
Amazing thank you! More patterns from numbers!
Those recurring patterns with the 0 squares are so much like fractal patterns or that unpronouncable triangle I keep forgetting the name of. Yes they might make good designs for t-shirts!
It was just a little confusing how the 'x' you introduced relates to the numbers along the original sequence to convert it to Sn-1 etc.
Mathologer, your shirt-gamecis criminally underrated!
I just want to share some love and say I always enjoy your musical choices but this time Asturias was a very good fit for a less-known topic.
This is absolutely amazing. Numbers are just way cool. I really need to go study this stuff more.
15:30 I actually felt this mathematicians’ viewpoint overtake my thinking (that is; I thought: ”Aren’t the isolated 0’s just 1*1-squares, and the infinite terminal 0’s just infinite 0-squares?”), *_JUST_* before you started talking about it (the mathematicians’ viewpoint).
I think the long cross rule can be used to compute the two numbers directly below a 2x2 square, but then to compute the two numbers below those 2, you still have to use the horseshoe rule.
I got a mystery sequence in my head which might not even have a formula. And it's base 10 dependant: 6, 2, 5, 5, 4, 5, 6, 3, 7, 6, 8
I should check the Oeis to see if they have a neater formula and no there wasn't.
The self similarity might look more beautiful if you zoom out instead of in, since the resolution increases and not decreases.
This is awesome. Thank you!
I always love your videos, a great way to relax on a Sunday afternoon
This is very useful for a problem I'm working on right now. Thanks you so much for making this!
Haven't watched yet, just wanna say the name reminds me of
6! 3.. no 4! 11? That's Numberwang!
Edit: now that I've watched the video, hot diggity damn that was wonderful.
The whole process was pretty mysterious, the cross relation vaguely reminded me of the 2×2 determinant but I didn't take this thought far enough.
The fact that these algorithms can give you these shifted sequence determinants and thus the recursion formula for any sequence with one is pretty amazing
Your videos are always so entertaining and edjamacational.
Every Calculus video I have ever see only mentions what it's used for (For example: to calculate the area of a random blob), but they never use it in the video. It's all talk and no do.
Would you please make a Calculus video where an answer is *actually calculated*?
The thing is, calculus is mostly "algebra auto pilot" once you start actually using it. You might like channels like blackpenredpen, which are largely focused on the working out bit.
I'm guessing that 99.99% of professional mathematicians not knowing about number walls is a poor estimate :) - and what a wonderfully taught video. Marco Pizzato is such a Pal you named him twice :)
The number wall for the Catalan series is crazy.
17:28 The 7 wall contains diagrams of radio telescopes. ;) Look for nearby dots making parabolas with 2 diagonal lines and 1 partial vertical or horizontal. I see one facing up on the left side and one facing to the right somewhat left of the middle.
Beautiful video! :)
I'm a bit confused how you generate the numbers to the far left and right. Maybe I missed it in the video. You showed finding one question mark using the cross. But I don't see a simple solution for the locations where you have a question mark both below and to the left/right.
I think since you're starting from an infinitely long sequence, you never really need to generate numbers to the left or right. You only need to go down.
Thanks for using Albeniz's beautiful "Asturias (Leyenda)" at 1:00, and of course for the video as well. :)
Great stuff as usual!
9:45 for power n, the formula F_n is the sum from k = 0 to n of (n + 1 choose k)(-1)^(n + k) F_{n + 1 - k}
Anyway, here's numberwall.
Also, at 22:40, I would think it much easier to remember how to do it if you just walk all four sides clockwise and multiply the ratios.
this video is just amazing , thank you !!! (and I think you can be prod of your job when Notch is supporting you for doing this amazing stuff !)
Great content to have breakfast with :). Would be great to see the prime numbers number wall :0
If you define M x N as the cellwise multiplication of two images, M, N, then after applying the cross rule to seq (assuming no 0s)
down = (center**2 - right*left) * 1/up
to get an image M, then we'd have
shiftUp(M) == (MxM - shiftLeft(M) x shiftRight(M)) x (1/shiftDown(M))
M is a fixed point (i.e. f(M) = M) under the function f that takes an image M to
f: M -> shiftDown((MxM - shiftLeft(M) x shiftRight(M)) x (1/shiftDown(M)))
as you explained it i straight away had to implement the number wall formula in a spreadsheet and put in the Thue-Morse sequence. aaand get div/0 errors in my fourth row 😅.
I always give u a LIKE before I look your video.
I'm 'just' an engineer (BS, not PhD), so most of these videos eventually break of my desperately clinging fingernails, and I go into 'be entertained' mode. But this video, I actually understood all the way, even though I've never heard of number walls. Nonetheless, TOTALLY BLEW MY FRIKKIN' MIND!!!!
Note, you've just ruined every test that has a "what number comes next" question. ;-)
10:06 Yes, they do, Burkard. Yes, they do. 😅
9:20 I’m seeing the coefficients of the binomial cube: 1 3 3 1, in the recursive formula for squares, there. Coincidence? 🤔
9:35 Apparently not, since the coefficients, for the cubes, are the coefficients of the binomial 4th powers: 1 4 6 4 1.
I started making walls in Google Sheets before you got to the zero issue section and encountered this myself!
Very good :)
There was mention of number walls for both finite and infinite sequences. On the finite side, I am curious if there’s any pattern that would emerge from any of various starting points in the Collatz sequence - e.g., the 100-ish-digit sequence starting with 27.
Amazing video as always!
15:52 Those "divisible-by-prime squares" can be viewed as "p-adic 0-squares" right?
The end Animation made it worth it
I like your teachings Sir.
Really amazing I must play with this algorithm
I love the music from my country: Asturias by Isaac Albeniz performed by Guitar Classics and Taiyo (Sun) by Yuhi (Evening Sun).
I was surprised when it started playing. :)
I don't understand a word of it but it's still fascinating!
this is great! wow!
Can you explain how you calculated the first term in 4th row of your first sequence?
Row 1: …. 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, …
Row 2: … -1, 0, 1, 0, -1, 1, 1, -1, -1, …
Row 3: … 1, 1, 1, 1, 1 , 2, 2, 2 , 1, …
Row 4: …. ??,
When I use the first cross formula I get ??= -1. You have ??= 1.
Thoughts?
I wish you had structured this video to start with the explanation of a number wall as the result of determinant operations instead of using up 30 minutes of my life confusingly going on about unexplained and "apparently arbitrary" cross and horseshoe games. Then you could talk about how the computationally quicker cross and horseshoe rules are derived from the original determinant operations and then how they lead to fibonacci type solutions of sequences. It would have made MUCH more sense.
An extra credit question on my linear algebra final back in the day involved "look-and-say sequences" (A005150 on OEIS is the simplest of them). I wish I could remember how it worked because I feel like it would tie into this number wall idea somehow.
The first number in the sequence is 1. Read aloud, that is "one '1'," (11). Read aloud, that is "two '1's" (21). Then "one '2', one '1'" (1211), etc.
Help me out anyone, I'm probably missing something obvious as usual. In row 3 of the first number wall example, which goes . . . 1 1 1 1 1 2 2 2 1 . . ., I can see how he obtains the second digit. It's the solution to the equation (-1 x 1) + (1 x ?) = 0². But where does the first digit in that row come from? The equation for that would have to have two unknowns, wouldn't it?
Another, possibly related, problem is how we continue beyond what's displayed. I understand his example in 2:25, but what about the three dots at the start of row 2: . . . -1 0 1 0 -1 1 -1 -1 . . ., what goes there if we continue to the left? We have the equation (? x 0) + (1 x 1) = -1² where ? can be any number.
I was about to ask the same question
Finish the video. He talks on detail about when one of the numbers is 0 which makes it impossible to use the cross-rule to calculate one of the neighbours. He shows another rule to solve that (one for single 0's and one for blocks of 0's).
@@remischmitt9308 Thanks, I should have stayed watching, but I thought I'd stop and consolidate before things got complicated.
Anyway, could I now trouble you to check my answers? The dotted gap at the start of row 2 has -1, which I found with the long cross rule.The gap at the start of row 3 has 2.
Did I miss the part where he explained what the “pagoda sequence” is?
To the best of the knowledge, it has no easy definition. It can be described as the difference between the (n-1)th and the (n+1)th terms of the paper folding sequence, which is a very well known sequence
@@Stekey21 The paper-folding formula is: f_0 = 0; f_(2n) = f_n; f_(4n+1) = 0; f_(4n+3) = 1. And the Pagoda formula is simply: p_n = f_(n+1) - f_(n-1).
@@TheShadowOfMars perhaps I should have said intuitive instead of easy. I am aware of the definition of the paper folding sequence, but I can see why Mathologer didn't think the tangent was necessary
Maybe have a look at the links in the description of this video :)
22:49 opportunity missed, multiply all factors going clockwise(or anti clockwise) multiplied equals 1, that's a nice one
The great mathematicians were quite adept in recognizing patterns and formulating them into algorithms/equations.