I was not that surprised by the square pattern, nor the linear one. But I was truly amazed when he pointed out that in the linear pattern the natural numbers appear exactly one by one 😇
Cube time! I stopped the video around 5:00 where you said those slices don't fit nicely around the cube. Well I say they fit pretty good if you just grab a couple of little unit helpers. So we get two 1^3 helper cubes to add to 5^3 + 6^3, then a 5 cube plus six 6x6 slabs plus two unit helpers in the corners we get the 7 cube putting everything together! So 2*1^3 + 5^3 + 6^3 = 7^3. So lets go further. Following from our two previous families, the central pattern for cubes looks like its gonna be 6=6*1, 18=6*(1+2), 36=6*(1+2+3), etc. So lets look at the second one. 16^3 + 17^3 + 18^3 ~= 19^3 + 20^3. We're gonna need more helper cubes, this time two sets - one for each big cube on the right. Putting stuff together we get 2*(1^3 + 2^3) + 16^3 + 17^3 + 18^3 = 19^3 + 20^3 makes perfect fit! Now it looks like our helper cubes have a pattern very similar to the central number: 6*1 2*1^3 and 6*(1+2) 2*(1^3+2^3). So now I'm gonna go right into the third cube equation with central number 36=6*(1+2+3) and say that we need 2*(1^3+2^3+3^3) helper cubes. So just jump right to the equation 2*(1^3+2^3+3^3) + 33^3 + 34^3 + 35^3 + 36^3 = 37^3 + 38^3 + 39^3, and it works! If you want to keep the Christmas tree aesthetic, we can call our helper cubes the ornaments :)
For the second puzzle, my answer is 2 I already learned from the video that 10² + 11² + 12² = 13² + 14² so the numerator is just 2(13² + 14²). When (slowly) performing the addition in my head, I was surprised that 13² + 14² actually equals to 365, cancelling out the denominator and leaving 2. Knowing the result, it's fun to think about making an efficient one-page calendar where the front is a 13x13 square and the back is a 14x14 square, with each square containing a date :D
I think I need to build that calendar. Just putting the julian dates into 2 squares. And as Simone Giertz said for her Every Day Calendar, the leap day is the day you should take off, sleep in and have ice cream with a movie.
@@Mathologer I used the exact same reasoning for that challenge! 10^2+11^2+12^2 = 13^2+14^2, (you said that at the beginning of the video) which I already knew was 365.
@@Gunstick Historically, the leap day was when you were getting a new calendar anyway, because it was the end of the year. (Yes, the year originally started with March. It was originally a Roman military calendar that started with the first march of the army in the spring, and stopped after ten months, because winter. January and February were added when the calendar was adapted for civilian use, and then some guy named Caesar added the leap year every fourth year rule so he could be out of the capital on long extended military campaigns, instead of sticking around to declare leap days in person. Starting the new year in January happened *much* later, and was originally a fiscal year. People wanted to do time-consuming year-end things like annual inventory during the winter because business was slow then.)
Mathematicians doesn't believe in "coincidence", these are true mathematical Facts. 😌 3³+4³+5³=6³ but not possible for further case. Pls continue doing such 10-15 min videos also, Sir. 🙏🏼 Long vid are itself goldmine. ❣️❣️❣️
@@arikwolf3777 This is pure arithmetic with everything on the table, not pseudo-numerologic digit sums making the hidden assumption that we work in Base ten, then making all kinds of misleading conclusions about how nine is the most special number blah blah. In REAL numerology we find mystical meaning in rigorous mathematical facts, with all mathematical assumptions as clear as possible, not in half-baked half-truths and falsehoods. Thus e.g. ten is a special number because it is both 10 = 1 + 2 + 3 + 4 and 10 = (1) + (1 + 2) + (1 + 2 + 3), not because our Homo Sapiens human bodies have ten fingers. We need to clearly distinguish cause and effect, and in numerology we (truly or falsely) claim that the causes are the eternal mathematical truths, and the effects are the dense material and dense psychological "coincidences" of our sensorial imperfect world. Thus any arguments for Base ten "coincidences" being relevant for Homo Sapiens humans must be tied to that the number of our fingers/digits equals ten being an EFFECT that is CAUSED by the mathematical and mystical properties of the number ten AS SUCH, and the providentially chosen place of our human evolution in the great non-random Divine scheme of manifestation. Or something similar to that. You cannot just claim "there are only ten digits, how could there be more digits?" or "the human (body, psyche and mind together) is the measure of everything".
There are also mathematically rigorous "exceptional" patterns that are only finite, not infinite (as long as we stay in certain "worlds", like finite dimensional space, locally Euclidean/positive definite metrics, loops being associative (i.e. groups) or Moufang, space being Euclidean or Elliptic not Hyperbolic etc). One such finite, exceptional pattern is the E series of crystallographic root systems, Lie algebras and Lie groups. Note that if we watch the diagrams of the affine ~E series root systems, then ~E_8 has only a 1-fold diagram symmetry (i.e. no symmetry), ~E_7 has only a 2-fold diagram symmetry, ~E_6 has "only" a 3-fold diagram symmetry, ~E_5 = ~D_5 has only a 4-fold diagram symmetry, ~E_4 = ~A_4 has only a 5-fold diagram symmetry, ~E_3 = ~A_1 + ~A_2 has only a 6-fold diagram symmetry. To be rigorous, the symmetry groups of the affine E series root systems are (going from ~E_8 to ~E_3): Triv = Cyc1 = Sym1, Cyc2 = Sym2, Sym3 = Dih3, Dih4, Dih5, Sym2 × Sym3 = Dih6. Of cardinalities 1, 2, 6, 8, 10, 12. So it seems that an order 2 mirror symmetry is "missing" in the diagram automorphism group of the two largest root systems of the affine ~E series. This is to be expected though because the Dih series connection to regular polygons break down for digons/dihedra and monogons/dihedra if we only count allowed permutations of vertices rather than permutations of edges and faces (seeing polygon dihedra as sitting inside spherical space S2 and dihedral groups as sitting inside SO(3)). So it seems that when it comes to diagram symmetries of affine root systems, the ~E series behaves almost inverted to the ~A series who have Dih_(n+1) symmetry for ~A_n (except for ~A_1 which has Cyc2 symmetry, again expected), while the ~D series is more or less constant with the notable exception of ~D_4 having Sym4 spacial 4-fold diagram symmetry, while all the other ~D_n for n >= 2 have Dih4 diagram symmetry.
I thought this vid was going to be more than 30 minutes long, but it's still nice to have an occasional short video from time to time! :) As for the puzzles: 5³ + 6³ ≈ 7³ (just off by two which just so happens to be the cube root of the next number 8) The answer to the second puzzle is 2 since 10² + 11² + 12² = 13² + 14² = 365 (I knew it was 365 earlier since when mentioning this equation some people like to point out that there are about 365 days in a year!) Funnily enough the sequence in the third puzzle goes 5, 6, 6.893..., 7.80557... which seem to round up to 5, 6, 7, and 8. Now I'm curious to know if this rounding pattern continues or if it's just a coincidence . . .
Cannot tell you how much I enjoyed making this short video. The long ones are real killers to make :) Also your 365 remark is spot on. People do find it remarkable that the result is the number of days in a year although that is really just a coincidence :)
the next pattern does in fact round up to 9 and i expect it to continue for several more terms but eventually diverge. it appears to simply be getting further away from the """expected"" value
The rounding pattern continues forever. Indeed, 3^n+...+(n+2)^n < (n+3)^n for all n>3. Moreover, solving 3^n+...+(n+2)^n = x^n, we find that x-(n+2) converges to 1-ln(e-1)=0.458... Fact 1: We have 3^n+...+(n-k)^n < (n-k+1)^n for all integers k >= 0 and n >= k+3. Integrating x^n from 3 to n-k+1, we find that 3^n+...+(n-k)^n < 1/(n+1) * (n-k+1)^(n+1). The result follows, because n-k+1 = 0 and n >= k+3, we have the following. (n-k+1)^n+(n-k+2)^n+...+(n+2)^n < 3^n+...+(n+2)^n < 2(n-k+1)^n+(n-k+2)^n+...+(n+2)^n By Fact 2, dividing everything by (n+2)^n, the left hand side converges to e^(-k-1)+e^(-k)+...+1, and the right hand side converges to 2e^(-k-1)+e^(-k)+...+1. Letting k go to infinity, we find that (3^n+...+(n+2)^n) / (n+2)^n converges to 1+e^(-1)+e^(-2)+... = e/(e-1). Fact 4: For any sequence of positive real numbers a_n converging to a positive real value c, the value n * ((a_n)^(1/n) - 1) converges to ln(c). For any ε>0, we have the following for n large enough. (1 + ln(c - 2ε)/n)^n < c-ε < a_n < c+ε < (1 + ln(c + 2ε)/n)^n ln(c - 2ε) < n * ((a_n)^(1/n) - 1) < ln(c + 2ε) Fact 5: The value (3^n+...+(n+2)^n)^(1/n) - (n+2) converges to 1-ln(e-1). Define a_n = (3^n+...+(n+2)^n) / (n+2)^n, such that (3^n+...+(n+2)^n)^(1/n) - (n+2) = (n+2) * ((a_n)^(1/n) - 1). By Fact 3, the value a_n converges to e/(e-1), so (a_n)^(1/n) - 1 converges to 0. By Fact 4, the value n * ((a_n)^(1/n) - 1) converges to ln(e/(e-1)) = 1-ln(e-1).
If you don't already know that 10² + 11² + 12² = 13² + 14² then you can still calculate it quickly by noticing that it's approximately 5 x 12² = 720, plus an "error term" of (13-12)(13+12) - (12-11)(12+11) + (14-12)(14+12) - (12-10)(12+10) = (13+12-12-11) + 2*(14+12-12-10) = 2+8 = 10. In fact, in general, (a-2)² + (a-1)² + a² + (a+1)² + (a+2)² = 5a² + 10.
I taught woodwork for years, in good selective schools and there were always pupils that thought they could build a hollow cube out of six equal squares of plywood. My boss would actually cut them the squares and watch then trying to assemble them. I regret that it did not occur to me to tell them it was impossible because of Fermat's last theorem, that would have been cool. Martin Gardner showed a nice dissection of a 3, 4 and 5 cubes into a small number of pieces that could be assembled into the bigger cube.
"Martin Gardner showed a nice dissection of a 3, 4 and 5 cubes into a small number of pieces that could be assembled into the bigger cube." Do you still remember where you saw that? Was there any talk about what was being shown being the smallest number of pieces? Also, you may have mentioned that in previous comments, but where are you from ?
@qyrghyz Just noticed that they show a picture of such a dissection on the wiki page dedicated to Euler's conjecture :) en.wikipedia.org/wiki/Euler%27s_sum_of_powers_conjecture
So for that first puzzle... I think my mind just got blown I started with 5 since I was basing on the fact that the first equation of both patterns end and start with 3 (1 + 2 = 3 and 3² + 4² = 5²), so why not do the same for the cube pattern. What I ended up is that 5³ + 6³ = 341, which is almost close to 7³ = 343. I was curious enough, so I tried to expand this by utilizing the same trick from the previous patterns and used 6*(1 + 2) for the next pattern and got 16³ + 17³ + 18³ ≈ 19³ + 20³ (with the difference being 18) Expanding this pattern leaves me with a list of differences and... I don't know about you, but (2, 18, 72, 200, etc.) just screams "I have a pattern"... and it does! Each difference is just twice a triangular number *squared*, and knowing that just blew my goddamn mind... because THAT'S LITERALLY HOW I STARTED working on this pattern. For each sum that uses 6(1 + 2 + ... + n) cubed as a starting point, there is a difference of 2(1 + 2 + ... + n)². How cool is that!? In honor to document this amazing pattern, here's my christmas tree for the cube pattern, with the difference added in. You can think of them like they're ornaments or something xD (the formatting might only work with monitors) 5³ + 6³ = 7³ *- 2* 16³ + 17³ + 18³ = 19³ + 20³ *- 18* 33³ + 34³ + 35³ + 36³ = 37³ + 38³ + 39³ *- 72* 56³ + 57³ + 58³ + 59³ + 60³ = 61³ + 62³ + 63³ + 64³ *- 200* 85³ + 86³ + 87³ + 88³ + 89³ + 90³ = 91³ + 92³ + 93³ + 94³ + 95³ *- 450* ... and here's another one with the sum expanded, just to see the beauty :D 5³ + [6(1)]³ = 7³ - 2(1)² 16³ + 17³ + [6(1 + 2)]³ = 19³ + 20³ - 2(1 + 2)² 33³ + 34³ + 35³ + [6(1 + 2 + 3)]³ = 37³ + 38³ + 39³ - 2(1 + 2 + 3)² 56³ + 57³ + 58³ + 59³ + [6(1 + 2 + 3 + 4)]³ = 61³ + 62³ + 63³ + 64³ - 2(1 + 2 + 3 + 4)² 85³ + 86³ + 87³ + 88³ + 89³ + [6(1 + 2 + 3 + 4 + 5)]³ = 91³ + 92³ + 93³ + 94³ + 95³ - 2(1 + 2 + 3 + 4 + 5)² ...
ALSO just to add, because (1 + 2 + ... + n)² is just 1³ + 2³ + ... + n³, we can replace each squared triangular sum with that so we can get rid of the pesky squares and get a purely cubic pattern :D
@@Exception-mk3xh that cubic pattern does indeed have a geometric interpretation. You just need to write it like this: 5³ + 6³ + 2(1³) = 7³ for it to make sense After wrapping the slices of the cube around the 6 faces (3 around a corner) you have just enough room for the correcting cubes to fit in on the 2 opposite corners
Surprisingly fun for such a short video! Also, as a big fan of empty operations, I was quite happy to see the "0 =" and "0^2 =" at the tops of the trees :)
In the mental arithmatic painting you can use the squre rearangement trick to turn 10²+11²+12² into 13²+14² which turns the expresion into 2×(13²+14²)/365 Which is 2×365/365. There is another square rearangment trick you can use here: X²+x+(x+1)=(x+1)² So the top of the fraction can be written as 10²+11²+12²+13²+14² =10²+11²+12²+13²+13²+13+14. =10²+11²+12²+2×(12²+12+13)+13+14 And so on... Eventually we are going to get 5×10²+4×10 +(3+4)×11+(2+3)×12+(1+2)×13+1×14 Which is equal to 540+77+60+39+14 = 730 = 2×365.
5^3 + 6^3 = 341, which is short of 7^3 by just two. I love that because you'd already mentioned trying to wrap a cube in square slices, I immediately imagined such a wrapping where two opposite corners were missing (though without checking I'm not sure that could really be done without further slicing the slices).
Actually, that almost wrapping that you mention there extends to a generalized cubes pattern. I am saying something about this in the description of this video. Check it out :)
@@Mathologer I just read it - thank you! I also love the hypothetical calendar arrangements. I once designed a calendar based on the Twelve Days of Christmas song, which features tetrahedral numbers - 364 is a tetrahedral number, and is actually the total number of gifts given in the song. I made Dec 25th a special "non calendar" day, and then had a series of months whose lengths were the first 12 triangular numbers, summing 364. A fairly silly calendar, but good fun. I have a spreadsheet that lets me generate it for any given year :) This actually came about because I was writing a novel that featured some magic based on the song... but it never got published (yet).
For the list which begins with 3^2+4^2=5^2. You had a short cut (formula) at the item before the equal sign. I have a short cut (formula) to jump from the last item of the previous line to the first item of the next line. 1st line ends with 5 2nd line begins with 10. Because 5 (last of previous line)/1(number of items on the right side)*2(number of items on the left side) 2nd line ends with 14 3rd line begins with 21. Because 14 (last of previous line)/2(number of items on the right side)*2(number of items on the left side).
For the blackboard problem, it occured to me that 10²+11²+12²+13²+14²= 5 × 12² + 2 × 1² + 2 × 2² because the double products of (12+x)² and (12-x)² cancel. Knowing 12²=144, it is straightforward that 5 × 12² = 1440/2 = 720. The remaining squares sum up to 10 and we get 730, which is 365 × 2.
Or more simply using the relation in the video, You can calculate quite easily 10squared + 11squared + 12 squared to be 365, AND 13 squared + 14 squared at the same time
Es bewahrheitet sich immer wieder: Die wirkliche Kunst ist es, Schwieriges so einfach zu vermitteln, dass es aussieht als wäre es selbstverständlich. Top gemacht!! Vielen Dank 🙂
I think the answer to the problem on the board is 1 and 364/365. I was adding 100, 121, and 144 when I realized that makes 365, so the problem is just 1 + (169 + 196)/365 Edit: Apparently it's 2. I guess I forgot to completely add a number.
For the second problem: the bogdanov-belski painting. I already knew it was 2. -) But yeah following the video: It's 2x(10²+11²+12²). And 100+121+144=365, goes indeed fast. (memorized) - ) 13²,14² (never memorized those for some reason): I tend to do 10x16+9 and 10x18+16, which is also still fast. (explanation: 13² = (13-3)x(13+3)+3²). So for instance if I have to square 62: 60x64+4=3844, goes decently fast. It's somewhat similar I think to what was described in the video: 7² = 6x8+1 = 5x9+4 = 4x10+9 = 3x11+16 = 2x12+25 = 1x13+36. Note that the added numbers are all squares. Each time you take a column of a squared number and you turn it into a row, you'll lose a square.
1# puzzle: I guess there is a turnaround point at n=3 and the numbers go up from there? Was that the pattern or did I miss something? 2. Well I didn't have a specific tactic to handle this, I just did 100+121+144 in my mind and got 365. Then I've done 169+196 and got 365 again. So, result is 2. 3. Well, you got me there. Thought there was a pattern because 3³+4³+5³=6³, so I thought the next one would be 7⁴ but no it isn't. Big Mathologer always reminding us of the law of small numbers, great! Keep up the wonderful work, and happy holidays!
Great video! Another way I noticed of describing why that (1st power) integer sums pattern works is that the left hand side always begins with a square number n^2, and apart from that term, there are n terms on each side, with each of the terms on the right side each having a term on the left side which it is exactly n larger than. For example, with 9+10+11+12 = 13+14+15, the terms 13, 14, and 15 each have a term on the left hand side they are 3 larger than (10, 11, and 12) making the difference between those be 3 times 3, which equals the square number 9 on the left.
@@jasimmathsandphysics Hey, yeah I've mentioned each of these identities in my videos before, but never noticed this connection between the two of them until seeing this Mathologer video :)
I tried tiling triangles with small triangles. Every row has 2 more small triangles than the row above it, so that the total number of small triangles is 1, 4, 9, 16, 25 etc. (I.e. square numbers.) You can then slice such a tiled triangle so that it envelops another tiled triangle. For example, 9^2 + 12^2 = 15^2. The slices look pretty bad though, and not intuitive at all. 😀
For the cube sequence, presumably the 1st line would be: 5^3 + 6^3 = 7^3 which isn't quite correct, but is very close (left hand side = 341, right hand side = 343)!
Also noticed that in the case of the squared sequence the last number from the previous term and the first number from the current term, are exactly 3, 5, 7, 9, ... apart from each other, which follows the difference of the squares! 0² = 0 3² + 4² =5 ², ----------------------------------->3 - 0 = 3 10² + 11² + 12² = 13² + 14², ---------> 10 - 5 = 5 21² + 22² + 23² + 24² = 25² + 26² + 27², ---------> 21 - 14 = 7 Pretty cool
For people who like to see the algebra version of these patterns, if we say that the largest term to the left of the equals sign is n², then the pattern with squares looks like [ ∑ (n-i)² from k=0 to k ] = [ ∑ (n+i)² from i=1 to k ], where k is the number of terms on the right (or 1 less than the number of terms on the left). In fact, solving that equation gives n = 2k(k+1), or n = 4(1+2+⋯+k), so the only way to make that pattern with is with 4(1+2+⋯+k), as revealed at 1:21. The pattern without squares is [ ∑ (m-i) from k=0 to k ] = [ ∑ (m+i) from i=1 to k ] with m being the last term on the left, and this is exactly equivalent to m = k(k+1) = 2(1+2+⋯+k).
(1) If you consider triangular, square, pentagonal, hexagonal, ... numbers, do you get an infinite pattern of patterns! (2) Note the starting numbers of each row. On the tower with the sums of consecutive integers, the starting numbers are squares. On the tower with the sums of squares, they're alternating triangular numbers: 2n(2n+1)/2
I have a degree in maths and physics and I am working in the field so math is all I do. Yet you manage to keep finding this amazing beautiful things in such simple ideas like Pythagorean triplets and circles which I thought for a long time have nothing to them. Absolutely amazing and beautiful thank you
I love the no comment animation at the beginning. Well paced to be able to capture the sequence. Maybe do some more of this conzent. Not for shorts though as there's not enpugh time.
Some years ago, I noticed the pattern of the first entries: (2n-1)^n + 2n^n = (2n+1)^n. Though; of course, knowing Fermat’s last theorem, I knew the pattern wouldn’t hold exactly, from cubes onward; but still, I was pleased to find that the near-miss -solutions seem to go on forever and ever after. You just need to round off the left sides of the equations to the nearest whole numbers, like so: [(2n-1)^n + 2n^n] = (2n+1)^n. Sadly, my friend and I couldn’t prove this conjecture; maybe, because we didn’t think to consider other entries, and we tried to find an algebraic proof, because of Mathematical rigor.
@@Mathologer Fantastic. This has been bugging me ever since I watched your Moessner's Miracle video. I thought for sure it was going to show up there somehow but it never did. Maybe it's in there somewhere still but I couldn't see it. Looking forward to this video.
The sum of connective odds, starting from 1, is a square (there are nice visual pros for that, btw) 1, 1+3, 1+3+5 etc The sequence you ask about is based on the differences of squares 1,9-1,36-9,100-36 ... numbers are pretty :-)
@@yinq5384 If you mean the sum of integers up to a point (you said of squares) is the square root of the sum of cubes up to the same point, I think you're right.
2. I mentally wrote the numerator as 10^2 + (10+1)^2 + (10+2)^2 + (10+4)^2, and applied the squared sum rule. So I have 5 tens, that's 500 in total. Next I have 1^2 + 2^2 + 3^2 + 4^2, this adds up to 30, and finally 2×10×(1+2+3+4), which is 200. So In total I have 500 + 30 + 200 = 730 = 2×365 It is surprising because 365 just happens to be the number of days in a year.
Full marks :) That's it. Of course, it's even quicker if you apply 10² + 11² + 12² = 13² + 14² Then you just have to calculate one side of the equation.
I first tried using the sum of squares formula but that would have taken more time than just calculating it. I have memorized the squares up to 16^2 very well and the addition wasn't bad, once I got 10²+11²+12² = 365 I knew what the other one would be
Concerning the linear pattern, put the sequence into a table, the perfect squares on the left. Then, you get a beautiful Ulam-like pattern with the primes in the diagonals and in the columns; I especially like the diagonal 5-11-19-29-41-(55)-71-89…. with the differences 6-8-10-12-(14)-16-18… and the second one, 3-7-13-(21)-31-43… So there is something very deep in this about primes in arithmetic progressions as well. Moreover, there is also a hint in the pattern about additive and multiplicative properties of the integers, because just by lining up the additive properties of the integers, you automatically get them ordered by means of the perfect squares. Amazing (how my math teachers missed this revelation)!
The solution to the math problem in the painting is 2. 10²+11²+12² → 100+121+144 = 365 (denominator) Since we know 13²+14² is the same quantity, we're effectively doubling the numerator to get 2/1.
I once noticed the square pattern had triangular numbers somewhat at its heart. Happy to have stumbled upon this video!! The pattern involving the squares seems like another extension of the Pythagoras Theorem just like De Gua’s Theorem.
Puzzle 1: (x+2)^3=(x+1)^3+(x+0)^3, difference is (x^3-3x^2-9x-7) Puzzle 2: This is simple, we know that 10^2+11^2+12^2=13^2+14^2, so we simply evaluate 10^2+11^2+12^2 then multiply it by 2, which gives us 365*2, but we don't have to evaluate that since now we have 365*2/365=2. Puzzle 3: 3^3+4^3+5^3=6^3, this is sorta well known as well. However, the follow-up doesn't have a real non-negative integer solution.
I found a generalization that works for all degrees, not just 1 and 2! It's not as clean, but it's still fun! Theorem: For every d ≥ 1, there is a unique degree-d polynomial P (up to scaling) and a unique constant c such that for every n, P(x - n) + ... + P(x) = P(x + 1) + ... + P(x + n), where x = c·(1 + 2 + ... + n). Furthermore, we must have c = 2d, and P(0) = 0. The patterns shown in the video are the special cases d = 1 and d = 2, for which the polynomial P is quite simple: x and x^2. Here are the first few: - For d = 1, P(x) = x - For d = 2, P(x) = x^2 - For d = 3, P(x) = 6x^3 + x^2 - For d = 4, P(x) = 32x^4 + 16x^3 + x^2 - For d = 5, P(x) = 4500x^5 + 4500x^4 + 930x^3 - 19x^2 - For d = 6, P(x) = 46656x^6 + 77760x^5 + 33696x^4 + 1392x^3 - 511x^2 - ... (Since they're unique up to scaling, I've scaled them up so that the coefficients are integers in lowest terms.) And here's the "Christmas tree" formed for d = 3: (6·0^3+0^2) = (6·5^3+5^2) + (6·6^3+6^2) = (6·7^3+7^2) (6·16^3+16^2) + (6·17^3+17^2) + (6·18^3+18^2) = (6·19^3+19^2) + (6·20^3+20^2) (6·33^3+33^2) + (6·34^3+34^2) + (6·35^3+35^2) + (6·36^3+36^2) = (6·37^3+37^2) + (6·38^3+38^2) + (6·39^3+39^2) .... It's not as pretty, but it works!! I would be very interested to know if there's a geometric interpretation to 6n^3 + n^2, or even the higher-degree ones!
That also looks like a nice idea. Will also ponder a bit more once I got a bit of time. Maybe also check out another kind of generalization that I am writing about in the description of this video :)
By the way, there are various ways to find the polynomials P, but the one I used (and implemented in a program) make use of the following slight extensions: Lemma 1: For every d ≥ 1, every polynomial Q with degree < d, there is a unique monic degree-d polynomial P such that for every n, P(x - n) + ... + P(x) = P(x + 1) + ... + P(x + n) + Q(x), where x = 2d·(1 + 2 + ... + n). Lemma 2: For every d ≥ 1, every polynomial Q with degree ≤ d and every constant c ≠ 2d, there is a unique polynomial P with degree ≤ d such that for every n, P(x - n) + ... + P(x) = P(x + 1) + ... + P(x + n) + Q(x), where x = c·(1 + 2 + ... + n). These immediately imply the theorem, and these in turn can be proven with straightforward recursion/induction, though you'll have to use the fact that the sum of odd powers is a polynomial in 1 + 2 + ... + n = n(n+1)/2, which has a nice proof using central differences and symmetry; it also shows the fact that the sum of even powers is (n+1/2) times a polynomial in n(n+1)/2. (Use central differences instead of forward or backward differences for the cleanest proof :D) So I use these repeatedly to compute the coefficients one by one, starting with the larger terms first. In Lemma 2, the coefficient of x^d in P(x) is p_d := q_d/(c^(d-1) (c - 2d)) where q_d is the coefficient of x^d in Q(x). Then we simply replace P(x) by P'(x) := P(x) - p_d x^d to get a new equation of the form P'(x - n) + ... + P'(x) = P'(x + 1) + ... + P'(x + d) + Q'(x) with a smaller degree, so we can use Lemma 2 recursively. (That's basically also how the proof works.)
Here's the meat of my (Python) code: ``` def get_P(d, c, Q): if Q.deg > d: raise Exception("No answer") # base case if d < 0: return Q # compute the leading coefficient if c == 2*d: if Q[d] != 0: raise Exception("No answer") P_d = 1 # P is monic else: P_d = Q[d] / (c**(d-1) * (c - 2*d)) # compute the new Q by evaluating at several points and then interpolating nQ = Poly.interpolate(*( (s1(n), Q(s1(n)) + P_d * sum((c*s1(n) + i)**d for i in range(1, n+1)) - P_d * sum((c*s1(n) - i)**d for i in range(0, n+1))) for n in range(d+3))) # recursively solve for the remaining coefficients return P_d*x**d + get_P(d - 1, c, nQ) ```
In David Wells's (The Penguin) Dictionary of Curious and Interesting Numbers, in the entry for triangular numbers, there is an analogous pattern: T1 + T2 + T3 = T4; T5 + T6 + T7 + T8 = T9 + T10; T11 + T12 + T13 + T14 + T15 = T16 + T17 + T18, and so on, attributed to M. N. Khatri. I don't understand everything about how this happens, but the fact that the sum of consecutive triangular numbers is a square (whose base is the larger index of the two triangular numbers), plus the fact that the difference of consecutive triangular numbers is the index of the larger triangular number, seems to be at the heart of it.
Ah! Playing around with the triangular-numbers patterns a little more, using the facts I mentioned above, I see that they correspond exactly (bijectively!) to doubled versions of each square pattern. They're just the square patterns in the mirror/double-verse: 6^2 + 8^2 = 10^2; 20^2 + 22^2 + 24^2 = 26^2 + 28^2; and so on. And yet they walk through all the triangular numbers in order without missing a beat. Neat.
I found the pattern. Each of those equations start with an nth triangular number, where n is an even whole number, and 1 is the first triangular number. Each equation has 1 more term on each aide than the previous one. For example: 21 is the 6th triangular number, so if my conjecture is true, this is also true: 24 27 Σ n² = Σ n² n=21 n=25 And since 36 is the 8th triangular number, we see another one of those equations: 40 44 Σ n² = Σ n² n=36 n=41 If you don’t know what the weird symbol Σ means, you should learn it by the time you take algebra 2.
My answer to the quizzes: 1. The theoretical pattern for higher powers would have the highest root of the left hand sum as 2^m * n(1+n)/2 since the sequence for the integers is 2 * n(1+n)/2 while the sequnce for squares is 4 * n(1+n)/2. So you would get for the cubes the left hand sums 7³ + 8³, 22³ + 23³ + 24³, etc., for the tesseracts the sums 15⁴ and 16⁴, 46⁴ + 47⁴ + 48⁴, etc. up to higher dimensions. 2. Although I didn't manage to calculate it in 30 seconds but using the equation you showed, 10² + 11² + 12² + 13³ + 14² can be reduced to either 2(10³ + 11² + 12²) or 2(13³ + 14²), both of which can be calculated in the head with the binomial formula (I used the right sum but left sum might have been easier). The sum you get for the sum itself is 365 which also is the numerator. The fraction thus equals to 2. 3. 3³ + 4³ + 5³ = 6³ which I kind of have guessed. However, 3⁴ + 4⁴ + 5⁴ + 6⁴ doesn't have this kind of pattern since the sum isn't divisible by 7 but rather by the primes 2 and 1129.
As for the painting, I can't honestly say that I worked it out in my head, nor in any short time. But long-hand, this was my approach. Each square in numerator can be expressed in the form: (10 + a)^2 = 100 + 20a + a^2 for some value of a. Five such squares all together, so we have: [ 5*100 + 20*(0 + 1 + 2 + 3 + 4) + (0^2 + 1^2 + 2^2 + 3^2 + 4^2) ] / 365 = [ 500 + 20*(10) + (30) ] / 365 = [ 500 + 200 + 30 ] / 365 = [ 1530 ] / 365 = 2
@@thecarman3693 I do! I guess my brain & fingers weren't communicating so well last night! Yes, I originally made a mistake - I had the expression 100*(0 + 1 + 2 + 3 + 4) instead of 20*(0 + 1 + 2 + 3 + 4). So my 3rd from last line had been [ 500 + 1000 + 30 ] / 365. I forget to correct my 2nd from last line - whoops!
Easier to grasp, still enjoyable. If you split your usual format into a series of semi-standalone videos of this format.. you could become a youtube GIANT
7:26 since 10^2 + 11^2 + 12^2 = 13^2 + 14^2 we can reduce the equation to 2(13^2 + 14^2) we can then calculate what 13^2 + 14^2 is : ( 169 + 196 ) which gives you... 365 ! Answer ends up being 2(365)/365, cancel out the 365 and you get 2 as the final answer. ( trick to do 169 + 196, just do 169 + 200 and then subtract 4 )
Second puzzle: short answer is no, since I'm not good at mental arithmetic under pressure. But armed with a few factoids and nerves of steel, I could solve the problem in about 30 using the following strategy: first, what's (n-1)^2 plus (n+1)^2 ? We see that the middle terms in the resulting trinomials cancel out, and the end terms add together, giving 2*(n^2 +1^2). Same pattern for n-2. We can write the series in the numerator as (12-2)^2 + (12-1)^2+ ...+ (12+2)^2. Using the above observation we get 5*12^2 plus 10 (=2*2^2 + 2*1^2) in the numerator. In the denominator, we know 365 is also divisible by 5. If you happen to know the other factor is 73(I didn't) = 72+1 then you're home free, otherwise 360=12*30 is pretty common knowledge, so add another 5 to it and you get 5*(6*12) +5 in the denominator. Now, 2*72=144 (gross!) is also pretty common knowledge, so we have twice( 72 + 5) divided by 72+5 =2. I freely admit to having worked this out with an electronic calculator, ahead of time, in about 5 seconds. When you do that, you get 365 for the first three terms in the numerator and another 365 for the two remaining terms, thus proving, by brute force, the original Mathologer proposition.
5³+6³=/=7³ or 125+216=/=343 (it's off by 2). To follow the method: 5³+6³ = 6³+5³ ==> 6³ exists of 6 slabs of 6² or 36; it needs to be wrapped around 5³. Place the 5³ cube on a surface, place four of the 6² squares around it, seen from above you've got 7x7 (note the top layer is just a rim going around, it still needs 25 squares to fill up the middle and get 7x7x6). The height measured from the table is still 6 so place the 5th 6² square on top of what you've got, you still need to fill in half of the edge with 7+6 = 13 squares. To fill up all the remaining gaps you still need 25+13 squares, you only have 6² squares, so there's 2 missing squares. Or put in another way building from 2D: To surround a 5x5 square: you can do that with 6x4 (you only need 4 sides): 25 + 24 =49 =7²: for one slab. Stack 5 slabs above each other: 7²x5. What you used: 6x4x5 (from the 6³cube) + 5³. (What you didn't use up: 6x2x5+6x6 = 96. And you still need 7²x2=98 to get to 7³.). It comes down to the fact that 7x7 isn't equal to 6x8. You miss one square. (twice) 7x7=6x8+1x1. To find that exact line 5³+6³=/=7³. I did... 1D: 1x2 (covers 2 'sides'); 2D: 1x4 (covers 4 sides); 3D: 1x6 (covers 6 sides), and so the number before the equal sign is 6... etc...
I am looking at the super-pattern for the squares... Doesn't look so nice, but 5*(2/1)=10, 14*(3/2)=21, 27*(4/3)=36... Next would be 44*(5/4), so 55²+56²+57²+58²+59²+60²=61²+62²+63²+64²+65². Next one would start at 65*(6/5), or 78.
7:04 2 Specifically, for any natural number n, let t(n) be the nth triangle number (i.e. t(1)=1, t(2)=3, t(3)=6, etc.) So, the linear pattern is sum(i from (2*t(n)-n) to (2*t(n))) of i = sum(j from (2*t(n)+1) to (2*t(n)+n)) of j Squares: sum(i from (4*t(n)-n) to (4*t(n))) of (i^2) = sum(j from (4*t(n)+1) to (4*t(n)+n)) of (j^2) Cubes: [sum(i from (6*t(n)-n) to (6*t(n))) of (i^3)] +2*[sum(k from 1 to n) of k^3] = sum(j from (6*t(n)+1) to (6*t(n)+n)) of (j^3) The two is because a cube has eight vertices and only six faces. Each slice from the largest initial cube covers one face and one vertex, leaving two missing vertices (it also covers two edges, which works out perfectly with the 12 edges of a cube). Each missing vertex needs a smaller cube, with a side length of 1 for the first new cube, 2 for the second new cube, etc.
6:43 Our cases came from a multiple of 2 and the sum of consecutive numbers. In the linear case it was 2(1+2+3+...). In the square case it was 4(1+2+3+...) That leaves us with two choices for higher order cases: follow the multiples of 2 like 6(1+2+3+...) (which corresponds to following sides), or follow the powers of 2 like 8(1+2+3+...) (which corresponds to following the corners). In the multiple case: 5³ + 6³ = 7³ - 2(1)² 16³ + 17³ + 18³ = 19³ + 20³ - 2(1+2)² 33³ + 34³ + 35³ + 36³ = 37³ + 38³ + 39³ - 2(1+2+3)² 56³ + 57³ + 58³ + 59³ + 60³ = 61³ + 62³ + 63³ + 64³ - 2(1+2+3+4)² You can check the power case yourself but it just doesn't work out. Now what happens at the 4th power and higher? 7⁴ + 8⁴ = 9⁴ - (4(1))³ 22⁴ + 23⁴ + 24⁴ = 25⁴ + 26⁴ - (4(1+2))³ 45⁴ + 46⁴ + 47⁴ + 48⁴ = 49⁴ + 50⁴ + 51⁴ - (4(1+2+3))³ 76⁴ + 77⁴ + 78⁴ + 79⁴ + 80⁴ = 81⁴ + 82⁴ + 83⁴ + 84⁴ - (4(1+2+3+4))³ This gives the impression that the square case should actually be written like this: 3² + 4² = 5² - 0(1) 10² + 11² + 12² = 13² + 14² - 0(1+2) ... Unfortunately, this doesn't extend to the linear case, and a close look at the 5th powers makes this extra clear that this is untrue: 9^5 + 10^5 = 11^5 - 2002 28^5 + 29^5 + 30^5 = 31^5 + 32^5 - 162066 57^5 + 58^5 + 59^5 + 60^5 = 61^5 + 62^5 + 63^5 - 2592552 The pattern here is very unclear, and after a bit of checking on wolfram the pattern for the error is very very ugly. Yes, (1+2+3+...+n) is a factor but there's a load of other junk included (500n⁴+1000n³+(752/3)n²+(4/3)n-2/3). 7:06 10² + 11² + 12² = 365. I already know 10² + 11² + 12² = 13² + 14² so that additional 13² + 14² must also be 365. (365 + 365)/365 = 2. 7:29 3³ + 4³ + 5³ = 6³. But 3⁴ + 4⁴ + 5⁴ + 6⁴ ≈ 6.89⁴. And as you increase the power the result still rounds up to roughly the next number. UPDATES: - someone proved the error for puzzle 3 approaches 1-ln(e-1) - oops the error terms are slightly wrong, fixed them
Very, very good :) Glad you had so much fun with these puzzles. Would be interesting to find out how the nice correction term for powers of 3 and 4 works (I suspect there is some nice geometry hiding there) and how it then breaks for 5th and higher powers
@@Mathologer i feel like the best way to incorporate the error term is to move it over to the positive side. So 5³ + 6³ = 7³ - 2(1) becomes 5³ + 6³ + 2(1) = 7³. I don't want to wrap my head around visualizing 4 dimensions so 3 dimensions it is for now. Here's the plan: first rewrite 2(1+2+3+...)² as 2(1³+2³+3³+...) to make things easier to visualize. Take the largest cube on the left and split it into square slices of varying thickness (the same way it's done in the 2d and 1d case). Wrap the square tiles around the cubes, and let the overhangs touch. By now there will be 6 centers covered, 12 edges covered, and 6 corners covered, leaving 2 perfectly cube-shaped slots left. That's where the error terms come in: each one fits perfectly in the cubical slots. Numerical examples: 5³ + 6³ + 2(1³) + 2(1³ + 2³) = 5³ + 6(1)(6²) + 2(1³) = 7³ 16³ + 17³ + 18³ = 16³ + 17³ + (6(1+2))(18²) + 2(1³ + 2³) = 17³ + 6(1)(18²) + 2(1³) + 16³ + 6(2)(18²) + 2(2³) = 19³ + 20³ 33³ + 34³ + 35³ + 36³ + 2(1³ + 2³ + 3³) = 33³ + 34³ + 35³ + 6(1+2+3)(36²) + 2(1³ + 2³ + 3³) = 35³ + 6(1)(36²) + 2(1³) + 34³ + 6(2)(36²) + 2(2³) + 33³ + 6(3)(36²) + 2(3³) = 37³ + 38³ + 39³ And so on. I might email you some visuals so you can better visualise what's happening
@@Mathologer I THINK I CAME UP WITH SOMETHING It's not quite geometric but it's really close Let i be the thickness of the cubical slices, n be the size of the hypercube to be sliced, and k be the number of cubes to be covered. In order for the trick to work: (n-i)⁴ + 8n³i + ERROR = (n+i)⁴ So ERROR = 8ni³, which can be written as 8(n-i)i³ + 8i⁴, for the missing 8 edges and 8 corners. How to convert 64(1+2+3+4+...+k)³ into sum of 8(n-i)i³ + 8i⁴: Note that (1+2+3+...+k)² = (1³+2³+3³+...+k³) Have a series of 3d cubes laid out, of increasing size. Now we have: 64(1+2+3+...+k)(1³+2³+3³+...+k³) Note that 8(1+2+3+...+k) = n. Give each cube depth of n to make it a hypercube. Now we have: 8n(1³+2³+3³+...+k³) Note that 8n can be written as 8(n-i) + 8i. When splitting our series of hypercuboids slice each one into 16 pieces, 8 with depth 8(n-i) for the edges and the other 8 with depth 8i for the corners. I have no idea how to actually place the cubic sheets around the hypercube yet but I at least have a promising start, rearranging the algebra to prepare for a fully geometric proof
3²+4²=5² is also part of another pattern, with a bit of algebra thrown in 4 + n² + (n+1)² = (n-1)² + (n+2)² Set n to 3 and subtract 4 from both sides, and you get 3²+4²=5²
Question: while 10 is numerator, for which both denominators X and Y, that resulting division for each gives other number (denominator) at the zero-symetry? I] 10/X= must be 0.Y0Y0Y0Y0Y0Y..... 10/Y= must be 0.X0X0X0X0X0X..... II] XX and Y-X=10. Y/X=[(a+1)/X + b/X] x (Y-X)= must be 1.Y0Y0Y0Y.... X/Y= is gonna be at the 9 symetry opposite reverse of X (ba) that is 0.ba9ba9ba9ba9ba9.... In Reel and Irreal numbers universe there is only two unique numbers reach this rule, those are 27 and 37. 10/27= 0.37037037037..... 10/27 gives the other one (37) at 0 symethry... 10/37= 0.27027027027..... 10/37 gives the other one (27) at right-left symethry of Zero. 37>27 and 37-27=10 37/27= (27+10)/27=1+0.37037037... =1.37037037... And 27/37= 0.729729729729... 27/37 gives opposite reverse of 27 at symethry of 9! See the symetrycity for 0 and 9... Secret correlation!: Remember for 27 2+7= 9 And for 37 3+7= 10 |27-37|= 10 37{27+10}>27 then 27/27+10/27= 1.37037037... symethry of 0, of 10's 0 (1,0). Numerator 37 comes at 0 symetry. [I] 27 {27+0 or 37+(-10)} < 37 then No 0 symethry but 9 opposite reverse of 27 after 0 decimal = *0.72 9 72 9 72 9 ......* Numerator 27 comes at its opposite reverse at 9-symetry [II] On the other hand we may assume while numerator < denominator after 0 decimal denominator comes with its opposite reverse mutiplied by 10 and 1 substracted That is the say: 37 Opp.Rev. 73 73x(37-27)=73×10=730 730-1=729 {here 1 comes from |27-37|/(37-27)=1 and/or {37} 3+7=10 {27} 2+7= 9 10-9=1 Remember 37 is {2+1}7 And 27 is {3-1}7 {2+0}7 Then 27/37= must be *0.729 729 729....*
For the second puzzle, I did not recognize it as some people did so I did the calculation in my head. The sum is of the form: (x-2)^2+(x-1)^x+x^2+(x+1)^2+(x+2)^2 it can be seen that if you expand the squares, all the terms linear in x cancel out, so you're left with 5x^2+2(1^2)+2(2^2) which simplifies to 5x^2+10, which, for x=12, evaluates to 730 which is twice 365.
For the final puzzle: The solutions can be described as ᵏ√[3ᵏ + 4ᵏ + ... + (k+1)ᵏ + (k+2)ᵏ], which if you calculate for a few values, actually seems to be linear. As k → ∞, the solution approaches k + 2.458675145387.
combining both animations mentally, I've presumed that by doubling the "central" number n×(n+1) in imaginary squares pattern for the linear pattern, in addition to "left" and "right" we could furnish "top" and "bottom" as we did in the squared pattern and... it's true! 1² + 2² × 2 = 3² 4² + 5² + 6² × 2 = 7² + 8² 9² + 10² + 11² + 12² × 2 = 13² + 14² + 15² 16² + 17² + 18² + 19² + 20² × 2 = 21² + 22² + 23² + 24² ... What's next? :)
As a math teacher and amateur programmer I tried to find solutions for n>2: For n=3, the only "real" solution is 3³+4³+5³=6³ but there are other similiar solutions, for example: 4³+...+28³=30³+31³+32³+33³+34³ 8³+...+55³=60³+...+68³ 34³+...+158³=540³ 61³+...+71³=101³+102³+103³ 213³+...+272³=556³+557³+558³+559³+560³ For 3
For cube numbers we get Sum of x^3 = (Sum of x )^2 So : 7:39 1^3+2^3+3^3+4^3+5^3 = (1+2+3+4+5)^2 Wich means 3^3+4^3+5^3 = 15^2 - 1 - 8 3^3+4^3+5^3 = 15^2 - 9 3^3+4^3+5^3 = 15^2 - 3 ^2 There is a way here 3^3+4^3+5^3 = 5^2×3^2 - 3 ^2 3^3+4^3+5^3 = (5^2 - 1)3 ^2 3^3+4^3+5^3 = (24)3 ^2 3^3+4^3+5^3 = (6×4)3 ^2 3^3+4^3+5^3 = 6×2 ^2 × 3 ^2 3^3+4^3+5^3 = 6×(2 × 3) ^2 3^3+4^3+5^3 =6×6^2 = 6^3 3^3+4^3+5^3 = (15-3)(15+3) At this point there is 2 ways 1st 3^3+4^3+5^3 = (5×3-3)(5×3+3)
In fact, there are many unnoticed number theory patterns. For example, when I was memorizing squares, I realized that 12^2 = 144 and 21^2 = 441. The mirrored numbers had mirrored squares! In fact, the same thing happens with 13 and 31! But nothing after that. However, this is just a consequence of addition carrying. In higher bases, as long as there is no carrying, this pattern will continue!
I noticed a couple of little patterns not directly mentioned. In the first sequence, the first terms follow the sequence 1**2, 2**2, 3**2, ... In the second sequence, the last term on the left is consecutive multiples of four squared. These observations make it trivial to figure out the Nth row of the patterns in your head.
You are too kind--I went back to check the video to check what I thought I'd seen, only to find you put it right at the head of the comments. It did take me more than the 30 seconds to mentally add 169 and 195. The answer, of course, is 2
Yes, a nice find. Actually both of them: the one I use in the main part of the video and the one I use for the Thank you section at the end of the video :)
I've looked at 7^x + 8^x = 9^x 22^x + 23^x + 24^x = 25^x + 26^x ... and found that interestingly enough, while it isn't 4 exactly, it appears to be approaching 4 as we continue down the line, with its limit being 4.
1: (2 + A^3 + B^3 = C^3) if B has factors 2 and 3, and A,B,C are consecutive integers (need 6 slices to wrap, but they miss 2 corners) 2: Don't know why I didn't reuse the earlier identity, since I know 100, 121, 144 from rote memory but I timed out trying to be clever 3: stumped me for an overarching pattern, 6 and 6.89... are the answers, however... ...a funny thing about the first 2 is if you subtract the exponent (eg 3- exp) from each term instead, the equation still balances, and first term + exponent equals final term without the exponent
While swimming I discovered the difference in consecutive natural number cubes is 6 x triangle number + 1. 2^3 - 1^3 = 8 - 1 = 6*1 + 1 3^3 - 2^3 = 27 - 8 = 6*3 + 1 4^3 - 3^3 = 64 - 27 = 6*6 + 1 Proof is that (x+1)^3 - x^3 = 3x^2 + 3x + 1 Factor 6x out of the first two terms. 6(x)(x+1)/2 + 1. x(x+1)/2 is a triangle number for natural number x. I thought it was cool. Embedded in triangle numbers are square numbers. Embedded in cube numbers are triangle numbers. 6 triangles to cover each face + 1 more to complete the cube.
Yes, nice insight :) There is actually a nice geometrical interpretation of all this. Maybe have a look at the first couple of minutes of my video on Moessner's miracle where I show how to interpret the difference of two cubes of two consecutive numbers as a certain hexagon. That a hexagon can be divided into 6 equilateral triangles then translates into you insight :)
The second puzzle at 7:25, the answer is 2. I did this mentally using the sun of square formula to figure out the sun of the first 14 squares, and the sum of the first 9 squares, and then did a subtraction. It’s (14)(15)(29)/6 - (9)(10)(19)/6, which simplifies to (7)(5)(29) - (3)(5)(19). This becomes (5)(29*7) - (5)(3*19), which is (5)(203 - 57), which is (5)(146). We are dividing the entire thing by 365, which is (5)(73), and 146/73 is 2. So with all of the cancellations, the answer is 2. Not the easiest thing to do in my head, but the answer does reveal itself.
The first numbers of each row in the linear pattern are the square numbers (1, 4, 9, 16...), while the first numbers of each row in the square pattern are every other triangular number (3, 10, 21, 36...) and the last numbers in each row are one less than the other triangular numbers (6, 15, 28, 45, ...).
Strange but true! Didn't know about the infinite pattern of first powers & squares, but I did stumble onto the single equation with cubes many years ago (3³ + 4³ + 5³ = 6³). Tried to find some general rule to extend it, but was never successful. As for the attempt to extend your 1st- and 2nd-power patterns to cubes, the obvious try almost works, for the first line: 5³ + 6³ = 341 = 7³ - 2 Fred
"5³ + 6³ = 341 = 7³ - 2" There is actually something super nice hiding just around the corner. If you are interested have a look at the description of this video.
Me too about the cubes which we studied in school where it was phrased like something like what will be the side/radius length of a cube/sphere made by melting cubes/spheres of side length/radii 3,4 and 5 units.
2nd puzzle is easy. (2*365)/365. And yes, I also noticed those patterns and extended them to 0. Tried to look for other ones with different exponents (after I learned of Fermat last theorem, I looked for non-positive and non-integer exponents), but to no avail. I always liked that first pattern doesn't have a "hole" - so it uses all numbers. I always wondered if the unused numbers in 2nd pattern can be used for something. I guess I will sit down with my pen and paper later this evening to check if there isn't some interesting there. Even if not as beautiful.
For the second puzzle we can use similar graphical proof to get 10²+12² = 2⋅11²+2 (after using 13²+14²=10²+11²+12²), simplifying whole expression to 2⋅(3⋅11²+2) = 2⋅(3⋅121+2)
Alright answer for the puzzles! EDIT: Corrected some typos and semantics First puzzle The n^1 numbers grow like 0 =; 1 + 2 = 3; 4 + 5 + 6 etc The n^2 numbers grow like 0 =; 3^2 + 4^2 = 5^2; 10^2 + 11^2 + 12^2 etc Therefore the n^3 numbers would grow like 0 =; 5^3 + 6^3 = 7^3; 16^3 + 17^3 + 18^3 = 19^3 + 20^3; 33^3 + 34^3 + 35^3 + 36^3 etc... As an extra, the third line would be 16^3 + 17^3 + 18^3 etc The logic for the third line and so on is that, the first term of the third line is equal to the sum of twice the first plus the second term (without raising them) of the second line, for example: 1 + 2 = 3 4 ... 4 = 1.2 + 2 3^2 + 4^2 = 5^2 10^2... 10 = 3.2 + 4 Therefore we can say that the first term of the cubes is 5^3 + 6^3 = 7^3 N = 5.2 + 6 therefore N = 16 For the next line, it's the sum of the first and second term 9 = 4 + 5 - For the first power 21 = 10 + 11 - For the squares So, for the cubes we'd have... N = 16 + 17 = 33 Second Puzzle Since we know from the second line of the squares that 10^2 + 11^2 + 12^2 = 13^2 + 14^2 and that both sides are equal to 365, this means the answer is two (730/365 = 2) Third Puzzle Too lazy to do that by hand!
Problem 1: If the pattern in 1 and 2 dimensions continued, the first 3D example would have used 6 in the bridging term. (6 is the number of faces on a cube). Thus the first identity would have been 5³ + 6³ = 7³. Unfortunately, the RHS is short by 2. Problem 2: Since 10² + 11² + 12² = 13² + 14² = 365, the calculation on the blackboard yields 2. Problem 3: By direct calculation, 3³ + 4³ + 5³ = 6³ (surprise!) and 3⁴ + 4⁴ + 5⁴ + 6⁴= (6.8934...)⁴ (not so pretty). Thanks for sharing another shiny nugget from your mine of mathematical gems.
Question: why does the square of 3.14159...'s quarter fail to satisfy the annulus whose area is π units²? If we plot 4a((√5/2)² - (1/2)²) = 4a = π units², we get four quadrants each of area (π/4)² thus 4(π/4)² area. If we plot a point (0, 1/2) and square 3.14159...'s quarter from it (0, 1/2 + ( 3.14159.../4)²) to represent a point on a circle as it expands according to the square of pi's quarter, it geometrically fails to satisfy the width entirely. Why??? If 3.14159... were correct, should we not expect its own squared quarter to satisfy its own π units² annulus (?)
@@Mathologer 1. Plot x² + y² = 1/4 and x² + y² = 5/4. This produces the annulus 4a((√5/2)² - (1/2)²) = 4a. 2. Find the width w as -1/2 + √5/2. Plot a point (0, 1/2) on the r = 1/2 circle. 3. Square 3.14159...'s quarter off this circumference via. (0, 1/2 + (π/4)²). Arithmetically, the area of each quadrant is (π/4)² thus the annulus is 4(π/4)². What I'm asking is: why is it when we square 3.14159...'s quarter does it fail to reach??? Any one quadrant sweep of width w is equivalent to the square of pi's quarter. In other words: why does (π/4)² = 0.616... instead of the actual geometric width of 0.618... ?
If the exponent is 1 or 2, a^x + b^x = c^x has solutions. But not if x = 3, and not if x = 0 either (as 1+1 != 1). In fact, among the integers, a^x+b^x=c^x has the same solution set as x^2-3x+2=0.
An easy (and general) way for the second exercise is to note that the numerator is of the form: (n-2)² + (n-1)² + n² + (n+1)² + (n+2)² if we were to expand the binomial squares, all the terms in the first power of n would cancel as they have opposite signs in pairs. So what we are left with is: n² + 2² + n² + 1² + n² + n² + 1² + n² + 2² = 5n² + 2*(2² + 1²) for n = 12, this is obviously 5*144 + 2*5 = =1440/2 + 10 = 730, which is twice the numerator of 365. That's what I did in my head in less than 30 seconds. Alternatively, one could do 169 + 196 or 100 + 121 + 144 in less than 30 seconds of mental calculation, and know from the video that it's half the numerator.
7:00 - It would be 5³ + 6³ (341) ~= 7³ (343). The difference is 2 as each 6² slice of 6-cube covers 1 face, 2 sides and 1 corner of the 5-cube. Total 6 faces, 12 sides and 6 corners are covered so it falls short of making a 7-cube by 2 corners (2 * 1³). Next equation would be 16³ + 17³ + 18³ ~= 19³ + 20³. This time the difference is 18 (2 * (1³ + 2³)). 7:25 - It's obvious now. Another way for someone who has not seen this video to solve is to substitute 12 with x. Then the expression becomes (x-2)²+(x-1)²+x²+(x+1)²+(x+2)² / 365. (a+b)²+(a-b)² = 2(a²+b²). So the expression becomes 2(x²+2²)+2(x²+1²)+x² / 365 = 5x² + 10 / 365. Replacing x by 12, it's 5(144)+10 / 365 = 730/365 = 2.
There's a superpattern for the squares too. It's a bit trickier, but 2*5/1 = 10, 3*14/2=21, 4*27/3=36, (n+1)*k_last[n]/n = k_first[n+1]. I haven't proven this, but will leave that as an exercise.
For the pattern of squares first shown at 1:48, in each step n the number of integers skipped to get to the next step n+1 equals 2n + 2. For the equivalent, slightly more complicated pattern of cubes wonderfully sorted out by Exception2001 in a must-read comment, it is 4n + 4. But it isn't n + 1 for the linear pattern. Instead, no integers are skipped. What the hell, cosmos?
Hello Mathologer! I remember seeing one of the videos, explaining the binomial expansion with fractional and negative powers, beautifully explained, but I can't find it any more. Could you point me in the right direction? Also I remember you have a second TH-cam channel, called something like Mathologer2?
6:50 Two other nice patterns in the left-hand terms (ignoring the powers): the top one is the square numbers, and the second one is alternate triangular numbers :)
I was not that surprised by the square pattern, nor the linear one. But I was truly amazed when he pointed out that in the linear pattern the natural numbers appear exactly one by one 😇
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.
Cube time! I stopped the video around 5:00 where you said those slices don't fit nicely around the cube. Well I say they fit pretty good if you just grab a couple of little unit helpers. So we get two 1^3 helper cubes to add to 5^3 + 6^3, then a 5 cube plus six 6x6 slabs plus two unit helpers in the corners we get the 7 cube putting everything together! So 2*1^3 + 5^3 + 6^3 = 7^3.
So lets go further. Following from our two previous families, the central pattern for cubes looks like its gonna be 6=6*1, 18=6*(1+2), 36=6*(1+2+3), etc. So lets look at the second one. 16^3 + 17^3 + 18^3 ~= 19^3 + 20^3. We're gonna need more helper cubes, this time two sets - one for each big cube on the right. Putting stuff together we get 2*(1^3 + 2^3) + 16^3 + 17^3 + 18^3 = 19^3 + 20^3 makes perfect fit!
Now it looks like our helper cubes have a pattern very similar to the central number: 6*1 2*1^3 and 6*(1+2) 2*(1^3+2^3). So now I'm gonna go right into the third cube equation with central number 36=6*(1+2+3) and say that we need 2*(1^3+2^3+3^3) helper cubes. So just jump right to the equation 2*(1^3+2^3+3^3) + 33^3 + 34^3 + 35^3 + 36^3 = 37^3 + 38^3 + 39^3, and it works!
If you want to keep the Christmas tree aesthetic, we can call our helper cubes the ornaments :)
Ornaments, I love it :)
@@Mathologer 💫💫
This also extends to quads or whatever the 4th power is called.
So for cubes we get this
5³ + 6³ = 7² - (2 * (1))²
16³ + 17³ + 18³ = 19³ + 20³ - (2 * (1 + 2))²
33³ + 34³ + 35³ + 36³ = 37³ + 38³ + 39³ - (2 * (1 + 2 + 3))²
56³ + 57³ + 58³ + 59³ + 60³ = 61³ + 62³ + 63³ + 64³ - (2 * (1 + 2 + 3 + 4))²
And for quads, we can do this
7^4 + 8^4 = 9^4 - (4 * (1))³
22^4 + 23^4 + 24^4 = 25^4 + 26^4 - (4 * (1 + 2))³
45^4 + 46^4 + 47^4 + 48^4 = 49^4 + 50^4 + 51^4 - (4 * (1 + 2 + 3))³
76^4 + 77^4 + 78^4 + 79^4 + 80^4 = 81^4 + 82^4 + 83^4 + 84^4 - (4 * (1 + 2 + 3 + 4))³
Can probably generalize that a bit more but its kind of interesting
Edited, improved formatting a bit :)
Great!!!
@@tma8983 Nice, but you can get rid of all those 8/2 and make them a 4, looks nicer, and may help guide us to the solution for the pents.
For the second puzzle, my answer is 2
I already learned from the video that 10² + 11² + 12² = 13² + 14² so the numerator is just 2(13² + 14²). When (slowly) performing the addition in my head, I was surprised that 13² + 14² actually equals to 365, cancelling out the denominator and leaving 2.
Knowing the result, it's fun to think about making an efficient one-page calendar where the front is a 13x13 square and the back is a 14x14 square, with each square containing a date :D
I like your calendar idea :)
I think I need to build that calendar. Just putting the julian dates into 2 squares. And as Simone Giertz said for her Every Day Calendar, the leap day is the day you should take off, sleep in and have ice cream with a movie.
Ypu can also do (12-2)^2+(12-1)+12^2+(12+1)^2+(12+2)^=5×12^2+2×1^2+2×2^2=720+10=730
@@Mathologer I used the exact same reasoning for that challenge! 10^2+11^2+12^2 = 13^2+14^2, (you said that at the beginning of the video) which I already knew was 365.
@@Gunstick Historically, the leap day was when you were getting a new calendar anyway, because it was the end of the year. (Yes, the year originally started with March. It was originally a Roman military calendar that started with the first march of the army in the spring, and stopped after ten months, because winter. January and February were added when the calendar was adapted for civilian use, and then some guy named Caesar added the leap year every fourth year rule so he could be out of the capital on long extended military campaigns, instead of sticking around to declare leap days in person. Starting the new year in January happened *much* later, and was originally a fiscal year. People wanted to do time-consuming year-end things like annual inventory during the winter because business was slow then.)
Mathematicians doesn't believe in "coincidence", these are true mathematical Facts. 😌
3³+4³+5³=6³ but not possible for further case.
Pls continue doing such 10-15 min videos also, Sir. 🙏🏼
Long vid are itself goldmine. ❣️❣️❣️
Okay, but do these "facts" work in other base systems?
@@arikwolf3777 yes. Nowhere do any of the equations depend on the digits of the numbers
@@arikwolf3777 This is pure arithmetic with everything on the table, not pseudo-numerologic digit sums making the hidden assumption that we work in Base ten, then making all kinds of misleading conclusions about how nine is the most special number blah blah.
In REAL numerology we find mystical meaning in rigorous mathematical facts, with all mathematical assumptions as clear as possible, not in half-baked half-truths and falsehoods.
Thus e.g. ten is a special number because it is both 10 = 1 + 2 + 3 + 4 and 10 = (1) + (1 + 2) + (1 + 2 + 3), not because our Homo Sapiens human bodies have ten fingers.
We need to clearly distinguish cause and effect, and in numerology we (truly or falsely) claim that the causes are the eternal mathematical truths, and the effects are the dense material and dense psychological "coincidences" of our sensorial imperfect world.
Thus any arguments for Base ten "coincidences" being relevant for Homo Sapiens humans must be tied to that the number of our fingers/digits equals ten being an EFFECT that is CAUSED by the mathematical and mystical properties of the number ten AS SUCH, and the providentially chosen place of our human evolution in the great non-random Divine scheme of manifestation. Or something similar to that. You cannot just claim "there are only ten digits, how could there be more digits?" or "the human (body, psyche and mind together) is the measure of everything".
There are also mathematically rigorous "exceptional" patterns that are only finite, not infinite (as long as we stay in certain "worlds", like finite dimensional space, locally Euclidean/positive definite metrics, loops being associative (i.e. groups) or Moufang, space being Euclidean or Elliptic not Hyperbolic etc).
One such finite, exceptional pattern is the E series of crystallographic root systems, Lie algebras and Lie groups.
Note that if we watch the diagrams of the affine ~E series root systems, then ~E_8 has only a 1-fold diagram symmetry (i.e. no symmetry), ~E_7 has only a 2-fold diagram symmetry, ~E_6 has "only" a 3-fold diagram symmetry, ~E_5 = ~D_5 has only a 4-fold diagram symmetry, ~E_4 = ~A_4 has only a 5-fold diagram symmetry, ~E_3 = ~A_1 + ~A_2 has only a 6-fold diagram symmetry. To be rigorous, the symmetry groups of the affine E series root systems are (going from ~E_8 to ~E_3): Triv = Cyc1 = Sym1, Cyc2 = Sym2, Sym3 = Dih3, Dih4, Dih5, Sym2 × Sym3 = Dih6. Of cardinalities 1, 2, 6, 8, 10, 12. So it seems that an order 2 mirror symmetry is "missing" in the diagram automorphism group of the two largest root systems of the affine ~E series.
This is to be expected though because the Dih series connection to regular polygons break down for digons/dihedra and monogons/dihedra if we only count allowed permutations of vertices rather than permutations of edges and faces (seeing polygon dihedra as sitting inside spherical space S2 and dihedral groups as sitting inside SO(3)).
So it seems that when it comes to diagram symmetries of affine root systems, the ~E series behaves almost inverted to the ~A series who have Dih_(n+1) symmetry for ~A_n (except for ~A_1 which has Cyc2 symmetry, again expected), while the ~D series is more or less constant with the notable exception of ~D_4 having Sym4 spacial 4-fold diagram symmetry, while all the other ~D_n for n >= 2 have Dih4 diagram symmetry.
Oh, and one more thing:
God geometrizes. 😜
I thought this vid was going to be more than 30 minutes long, but it's still nice to have an occasional short video from time to time! :)
As for the puzzles:
5³ + 6³ ≈ 7³ (just off by two which just so happens to be the cube root of the next number 8)
The answer to the second puzzle is 2 since 10² + 11² + 12² = 13² + 14² = 365 (I knew it was 365 earlier since when mentioning this equation some people like to point out that there are about 365 days in a year!)
Funnily enough the sequence in the third puzzle goes 5, 6, 6.893..., 7.80557... which seem to round up to 5, 6, 7, and 8. Now I'm curious to know if this rounding pattern continues or if it's just a coincidence . . .
Cannot tell you how much I enjoyed making this short video. The long ones are real killers to make :) Also your 365 remark is spot on. People do find it remarkable that the result is the number of days in a year although that is really just a coincidence :)
the next pattern does in fact round up to 9 and i expect it to continue for several more terms but eventually diverge. it appears to simply be getting further away from the """expected"" value
The rounding pattern continues forever. Indeed, 3^n+...+(n+2)^n < (n+3)^n for all n>3.
Moreover, solving 3^n+...+(n+2)^n = x^n, we find that x-(n+2) converges to 1-ln(e-1)=0.458...
Fact 1: We have 3^n+...+(n-k)^n < (n-k+1)^n for all integers k >= 0 and n >= k+3.
Integrating x^n from 3 to n-k+1, we find that 3^n+...+(n-k)^n < 1/(n+1) * (n-k+1)^(n+1). The result follows, because n-k+1 = 0 and n >= k+3, we have the following.
(n-k+1)^n+(n-k+2)^n+...+(n+2)^n < 3^n+...+(n+2)^n < 2(n-k+1)^n+(n-k+2)^n+...+(n+2)^n
By Fact 2, dividing everything by (n+2)^n, the left hand side converges to e^(-k-1)+e^(-k)+...+1, and the right hand side converges to 2e^(-k-1)+e^(-k)+...+1.
Letting k go to infinity, we find that (3^n+...+(n+2)^n) / (n+2)^n converges to 1+e^(-1)+e^(-2)+... = e/(e-1).
Fact 4: For any sequence of positive real numbers a_n converging to a positive real value c, the value n * ((a_n)^(1/n) - 1) converges to ln(c).
For any ε>0, we have the following for n large enough.
(1 + ln(c - 2ε)/n)^n < c-ε < a_n < c+ε < (1 + ln(c + 2ε)/n)^n
ln(c - 2ε) < n * ((a_n)^(1/n) - 1) < ln(c + 2ε)
Fact 5: The value (3^n+...+(n+2)^n)^(1/n) - (n+2) converges to 1-ln(e-1).
Define a_n = (3^n+...+(n+2)^n) / (n+2)^n, such that (3^n+...+(n+2)^n)^(1/n) - (n+2) = (n+2) * ((a_n)^(1/n) - 1).
By Fact 3, the value a_n converges to e/(e-1), so (a_n)^(1/n) - 1 converges to 0.
By Fact 4, the value n * ((a_n)^(1/n) - 1) converges to ln(e/(e-1)) = 1-ln(e-1).
If you don't already know that 10² + 11² + 12² = 13² + 14² then you can still calculate it quickly by noticing that it's approximately 5 x 12² = 720, plus an "error term" of (13-12)(13+12) - (12-11)(12+11) + (14-12)(14+12) - (12-10)(12+10) = (13+12-12-11) + 2*(14+12-12-10) = 2+8 = 10. In fact, in general, (a-2)² + (a-1)² + a² + (a+1)² + (a+2)² = 5a² + 10.
@@SmileyMPV n=39 gives sum(k^n,k=3..n+2)^(1/n) = 41.4992... ≈ 41 < 42
I taught woodwork for years, in good selective schools and there were always pupils that thought they could build a hollow cube out of six equal squares of plywood. My boss would actually cut them the squares and watch then trying to assemble them.
I regret that it did not occur to me to tell them it was impossible because of Fermat's last theorem, that would have been cool.
Martin Gardner showed a nice dissection of a 3, 4 and 5 cubes into a small number of pieces that could be assembled into the bigger cube.
"Martin Gardner showed a nice dissection of a 3, 4 and 5 cubes into a small number of pieces that could be assembled into the bigger cube." Do you still remember where you saw that? Was there any talk about what was being shown being the smallest number of pieces? Also, you may have mentioned that in previous comments, but where are you from ?
@@Mathologer Oct 1973 Scientific American, reprinted in "Knotted Doughnuts and Other Mathematical Entertainments".
@@qyrghyz Great, thank you very much for that. Just looked it up in Knotted Doughnuts. Must have seen this before ... :)
@qyrghyz Just noticed that they show a picture of such a dissection on the wiki page dedicated to Euler's conjecture :) en.wikipedia.org/wiki/Euler%27s_sum_of_powers_conjecture
@@Mathologer Ooh Pretty! I missed that.
This channel is mostly above my head but the quality of education here never ceases to amaze me
Glad you enjoy it!
Agreed, I like to take some of the simpler things into 7 to 12 grade math classes just to show them that math can be fun.
So for that first puzzle... I think my mind just got blown
I started with 5 since I was basing on the fact that the first equation of both patterns end and start with 3 (1 + 2 = 3 and 3² + 4² = 5²), so why not do the same for the cube pattern. What I ended up is that 5³ + 6³ = 341, which is almost close to 7³ = 343. I was curious enough, so I tried to expand this by utilizing the same trick from the previous patterns and used 6*(1 + 2) for the next pattern and got 16³ + 17³ + 18³ ≈ 19³ + 20³ (with the difference being 18)
Expanding this pattern leaves me with a list of differences and... I don't know about you, but (2, 18, 72, 200, etc.) just screams "I have a pattern"... and it does!
Each difference is just twice a triangular number *squared*, and knowing that just blew my goddamn mind... because THAT'S LITERALLY HOW I STARTED working on this pattern. For each sum that uses 6(1 + 2 + ... + n) cubed as a starting point, there is a difference of 2(1 + 2 + ... + n)². How cool is that!?
In honor to document this amazing pattern, here's my christmas tree for the cube pattern, with the difference added in. You can think of them like they're ornaments or something xD
(the formatting might only work with monitors)
5³ + 6³ = 7³ *- 2*
16³ + 17³ + 18³ = 19³ + 20³ *- 18*
33³ + 34³ + 35³ + 36³ = 37³ + 38³ + 39³ *- 72*
56³ + 57³ + 58³ + 59³ + 60³ = 61³ + 62³ + 63³ + 64³ *- 200*
85³ + 86³ + 87³ + 88³ + 89³ + 90³ = 91³ + 92³ + 93³ + 94³ + 95³ *- 450*
...
and here's another one with the sum expanded, just to see the beauty :D
5³ + [6(1)]³ = 7³ - 2(1)²
16³ + 17³ + [6(1 + 2)]³ = 19³ + 20³ - 2(1 + 2)²
33³ + 34³ + 35³ + [6(1 + 2 + 3)]³ = 37³ + 38³ + 39³ - 2(1 + 2 + 3)²
56³ + 57³ + 58³ + 59³ + [6(1 + 2 + 3 + 4)]³ = 61³ + 62³ + 63³ + 64³ - 2(1 + 2 + 3 + 4)²
85³ + 86³ + 87³ + 88³ + 89³ + [6(1 + 2 + 3 + 4 + 5)]³ = 91³ + 92³ + 93³ + 94³ + 95³ - 2(1 + 2 + 3 + 4 + 5)²
...
ALSO just to add, because (1 + 2 + ... + n)² is just 1³ + 2³ + ... + n³, we can replace each squared triangular sum with that so we can get rid of the pesky squares and get a purely cubic pattern :D
Very nice :)
@@Exception-mk3xh that cubic pattern does indeed have a geometric interpretation. You just need to write it like this:
5³ + 6³ + 2(1³) = 7³
for it to make sense
After wrapping the slices of the cube around the 6 faces (3 around a corner) you have just enough room for the correcting cubes to fit in on the 2 opposite corners
Well done, was just about to try to figure this out and then realised it might have been done down here
Patterns. Animation. Understanding.
One follows the other. Explanation in its purest form without danger of distraction. This is educational genius.
Surprisingly fun for such a short video!
Also, as a big fan of empty operations, I was quite happy to see the "0 =" and "0^2 =" at the tops of the trees :)
I think you are the first one to agree with me in this respect (and say so :)
That was a wonderfully powerful intro. So beautiful and fully deserving of the absolutely epic music.
Awesome video. From the animation to the soundtrack to the explanation. Just amazing.
In the mental arithmatic painting you can use the squre rearangement trick to turn 10²+11²+12² into 13²+14² which turns the expresion into 2×(13²+14²)/365
Which is 2×365/365.
There is another square rearangment trick you can use here:
X²+x+(x+1)=(x+1)²
So the top of the fraction can be written as 10²+11²+12²+13²+14² =10²+11²+12²+13²+13²+13+14. =10²+11²+12²+2×(12²+12+13)+13+14
And so on...
Eventually we are going to get 5×10²+4×10 +(3+4)×11+(2+3)×12+(1+2)×13+1×14
Which is equal to 540+77+60+39+14 = 730 = 2×365.
Great :)
And thank you so much for getting another video out before Christmas. Especially one with things to ponder over the break!
In principle I've got another short one ready and so who knows, there might even be yet another video before the end of the year :)
@@Mathologer … the gift that keeps on giving :)
5^3 + 6^3 = 341, which is short of 7^3 by just two. I love that because you'd already mentioned trying to wrap a cube in square slices, I immediately imagined such a wrapping where two opposite corners were missing (though without checking I'm not sure that could really be done without further slicing the slices).
Actually, that almost wrapping that you mention there extends to a generalized cubes pattern. I am saying something about this in the description of this video. Check it out :)
@@Mathologer I just read it - thank you! I also love the hypothetical calendar arrangements. I once designed a calendar based on the Twelve Days of Christmas song, which features tetrahedral numbers - 364 is a tetrahedral number, and is actually the total number of gifts given in the song. I made Dec 25th a special "non calendar" day, and then had a series of months whose lengths were the first 12 triangular numbers, summing 364. A fairly silly calendar, but good fun. I have a spreadsheet that lets me generate it for any given year :) This actually came about because I was writing a novel that featured some magic based on the song... but it never got published (yet).
For the list which begins with 3^2+4^2=5^2.
You had a short cut (formula) at the item before the equal sign.
I have a short cut (formula) to jump from the last item of the previous line to the first item of the next line.
1st line ends with 5
2nd line begins with 10.
Because 5 (last of previous line)/1(number of items on the right side)*2(number of items on the left side)
2nd line ends with 14
3rd line begins with 21.
Because 14 (last of previous line)/2(number of items on the right side)*2(number of items on the left side).
For the blackboard problem, it occured to me that 10²+11²+12²+13²+14²= 5 × 12² + 2 × 1² + 2 × 2² because the double products of (12+x)² and (12-x)² cancel. Knowing 12²=144, it is straightforward that 5 × 12² = 1440/2 = 720. The remaining squares sum up to 10 and we get 730, which is 365 × 2.
Or more simply using the relation in the video, You can calculate quite easily 10squared + 11squared + 12 squared to be 365, AND 13 squared + 14 squared at the same time
Es bewahrheitet sich immer wieder: Die wirkliche Kunst ist es, Schwieriges so einfach zu vermitteln, dass es aussieht als wäre es selbstverständlich. Top gemacht!! Vielen Dank 🙂
I think the answer to the problem on the board is 1 and 364/365. I was adding 100, 121, and 144 when I realized that makes 365, so the problem is just 1 + (169 + 196)/365
Edit: Apparently it's 2. I guess I forgot to completely add a number.
For the second problem: the bogdanov-belski painting. I already knew it was 2.
-) But yeah following the video: It's 2x(10²+11²+12²). And 100+121+144=365, goes indeed fast. (memorized)
- ) 13²,14² (never memorized those for some reason): I tend to do 10x16+9 and 10x18+16, which is also still fast. (explanation: 13² = (13-3)x(13+3)+3²).
So for instance if I have to square 62: 60x64+4=3844, goes decently fast.
It's somewhat similar I think to what was described in the video: 7² = 6x8+1 = 5x9+4 = 4x10+9 = 3x11+16 = 2x12+25 = 1x13+36.
Note that the added numbers are all squares.
Each time you take a column of a squared number and you turn it into a row, you'll lose a square.
ditto, but for 10²+11²+12² I had a quick calculation: 10²+11²+12²=(11-1)²+11²+(11+1)²=3x11²+2=3x121+2=365
10^2+11^2+12^2 = 13^2+14^2 was stated at the very beginning!
These visual proofs are so comforting to watch :)
1# puzzle: I guess there is a turnaround point at n=3 and the numbers go up from there? Was that the pattern or did I miss something?
2. Well I didn't have a specific tactic to handle this, I just did 100+121+144 in my mind and got 365. Then I've done 169+196 and got 365 again. So, result is 2.
3. Well, you got me there. Thought there was a pattern because 3³+4³+5³=6³, so I thought the next one would be 7⁴ but no it isn't. Big Mathologer always reminding us of the law of small numbers, great!
Keep up the wonderful work, and happy holidays!
Looks like you really enjoyed this one :)
@@Mathologer yes I did. Well not just this one, the other ones were very good too!
Great video! Another way I noticed of describing why that (1st power) integer sums pattern works is that the left hand side always begins with a square number n^2, and apart from that term, there are n terms on each side, with each of the terms on the right side each having a term on the left side which it is exactly n larger than. For example, with 9+10+11+12 = 13+14+15, the terms 13, 14, and 15 each have a term on the left hand side they are 3 larger than (10, 11, and 12) making the difference between those be 3 times 3, which equals the square number 9 on the left.
Hello 💪I remember you made a short about this
@@jasimmathsandphysics Hey, yeah I've mentioned each of these identities in my videos before, but never noticed this connection between the two of them until seeing this Mathologer video :)
Hi combo class, I enjoy your videos. I know the pattern. Look at the other comment I made on this video.
@@ComboClass Hi, Domotro! In which video did you talk about the linear pattern?
@@wyattstevens8574 It was one of the shorts on my bonus channel @Domotro , it hasn't been in a full episode yet but may appear when it fits
I tried tiling triangles with small triangles. Every row has 2 more small triangles than the row above it, so that the total number of small triangles is 1, 4, 9, 16, 25 etc. (I.e. square numbers.) You can then slice such a tiled triangle so that it envelops another tiled triangle. For example, 9^2 + 12^2 = 15^2. The slices look pretty bad though, and not intuitive at all. 😀
Main thing is that you are having fun :) And definitely a worthwhile thing to try :)
This was fascinating! I enjoyed the shorter form for a change, and I look forward to the next deep dive.
Glad you enjoyed it!
Thanks! This is the kind of videos where, when you want to click LIKE at the end, you realize you already did it earlier because it was so good.
Glad that this video worked so well for you :)
Finally some music and math together! Wonderful work done mathologer!
For the cube sequence, presumably the 1st line would be:
5^3 + 6^3 = 7^3
which isn't quite correct, but is very close (left hand side = 341, right hand side = 343)!
Exactly, the universe is definitely teasing us here with a near miss :)
What a treat! Thank you Mathologer, happy holidays!
you have blown my mind once again.
absolutely loved this video, and the animation was amazing!
amazing work!
Glad you liked it!
Also noticed that in the case of the squared sequence the last number from the previous term and the first number from the current term, are exactly 3, 5, 7, 9, ... apart from each other, which follows the difference of the squares!
0² = 0
3² + 4² =5 ², ----------------------------------->3 - 0 = 3
10² + 11² + 12² = 13² + 14², ---------> 10 - 5 = 5
21² + 22² + 23² + 24² = 25² + 26² + 27², ---------> 21 - 14 = 7
Pretty cool
For people who like to see the algebra version of these patterns, if we say that the largest term to the left of the equals sign is n², then the pattern with squares looks like
[ ∑ (n-i)² from k=0 to k ] = [ ∑ (n+i)² from i=1 to k ],
where k is the number of terms on the right (or 1 less than the number of terms on the left). In fact, solving that equation gives n = 2k(k+1), or n = 4(1+2+⋯+k), so the only way to make that pattern with is with 4(1+2+⋯+k), as revealed at 1:21.
The pattern without squares is
[ ∑ (m-i) from k=0 to k ] = [ ∑ (m+i) from i=1 to k ]
with m being the last term on the left, and this is exactly equivalent to m = k(k+1) = 2(1+2+⋯+k).
(1) If you consider triangular, square, pentagonal, hexagonal, ... numbers, do you get an infinite pattern of patterns!
(2) Note the starting numbers of each row. On the tower with the sums of consecutive integers, the starting numbers are squares. On the tower with the sums of squares, they're alternating triangular numbers: 2n(2n+1)/2
I have a degree in maths and physics and I am working in the field so math is all I do.
Yet you manage to keep finding this amazing beautiful things in such simple ideas like Pythagorean triplets and circles which I thought for a long time have nothing to them.
Absolutely amazing and beautiful thank you
I love the no comment animation at the beginning. Well paced to be able to capture the sequence. Maybe do some more of this conzent. Not for shorts though as there's not enpugh time.
Some years ago, I noticed the pattern of the first entries:
(2n-1)^n + 2n^n = (2n+1)^n.
Though; of course, knowing Fermat’s last theorem, I knew the pattern wouldn’t hold exactly, from cubes onward; but still, I was pleased to find that the near-miss -solutions seem to go on forever and ever after. You just need to round off the left sides of the equations to the nearest whole numbers, like so:
[(2n-1)^n + 2n^n] = (2n+1)^n.
Sadly, my friend and I couldn’t prove this conjecture; maybe, because we didn’t think to consider other entries, and we tried to find an algebraic proof, because of Mathematical rigor.
Can you sometime do a geometric proof like these that shows why:
1^3 = 1
2^3 = 3 + 5 = 8
3^3 = 7 + 9 + 11 = 27
4^3 = 13 + 15 + 17 + 19 = 64
Etc
Yes. In fact, I've got a video for this pretty much ready to go :)
@@Mathologer Fantastic. This has been bugging me ever since I watched your Moessner's Miracle video. I thought for sure it was going to show up there somehow but it never did. Maybe it's in there somewhere still but I couldn't see it. Looking forward to this video.
The sum of connective odds, starting from 1, is a square (there are nice visual pros for that, btw)
1, 1+3, 1+3+5 etc
The sequence you ask about is based on the differences of squares
1,9-1,36-9,100-36
... numbers are pretty :-)
@@Mathologer This is equivalent to ∑(n³) = (∑n²)² right?
@@yinq5384 If you mean the sum of integers up to a point (you said of squares) is the square root of the sum of cubes up to the same point, I think you're right.
2.
I mentally wrote the numerator as 10^2 + (10+1)^2 + (10+2)^2 + (10+4)^2, and applied the squared sum rule. So I have 5 tens, that's 500 in total. Next I have 1^2 + 2^2 + 3^2 + 4^2, this adds up to 30, and finally 2×10×(1+2+3+4), which is 200. So In total I have 500 + 30 + 200 = 730 = 2×365
It is surprising because 365 just happens to be the number of days in a year.
Full marks :) That's it. Of course, it's even quicker if you apply 10² + 11² + 12² = 13² + 14² Then you just have to calculate one side of the equation.
Oh right. Clever solving. 13² + 14² = 169 + 196 = 170 + 195 = 365
This simplifies with the denominator, and you're left with 2 from the other addend.
I first tried using the sum of squares formula but that would have taken more time than just calculating it. I have memorized the squares up to 16^2 very well and the addition wasn't bad, once I got 10²+11²+12² = 365 I knew what the other one would be
Concerning the linear pattern, put the sequence into a table, the perfect squares on the left. Then, you get a beautiful Ulam-like pattern with the primes in the diagonals and in the columns; I especially like the diagonal 5-11-19-29-41-(55)-71-89…. with the differences 6-8-10-12-(14)-16-18… and the second one, 3-7-13-(21)-31-43… So there is something very deep in this about primes in arithmetic progressions as well. Moreover, there is also a hint in the pattern about additive and multiplicative properties of the integers, because just by lining up the additive properties of the integers, you automatically get them ordered by means of the perfect squares. Amazing (how my math teachers missed this revelation)!
This video calmed me down with all its magic glory
The solution to the math problem in the painting is 2.
10²+11²+12² → 100+121+144 = 365 (denominator)
Since we know 13²+14² is the same quantity, we're effectively doubling the numerator to get 2/1.
I once noticed the square pattern had triangular numbers somewhat at its heart.
Happy to have stumbled upon this video!!
The pattern involving the squares seems like another extension of the Pythagoras Theorem just like De Gua’s Theorem.
Woah! I just got done animating a classic visual proof of one of these identities. I’ll tag you when it eventually posts and link to this one. Thanks!
Great :)
@@Mathologer here it is if you’re interested : th-cam.com/video/BBrtHbRcpXs/w-d-xo.html
Puzzle 1: (x+2)^3=(x+1)^3+(x+0)^3, difference is (x^3-3x^2-9x-7)
Puzzle 2: This is simple, we know that 10^2+11^2+12^2=13^2+14^2, so we simply evaluate 10^2+11^2+12^2 then multiply it by 2, which gives us 365*2, but we don't have to evaluate that since now we have 365*2/365=2.
Puzzle 3: 3^3+4^3+5^3=6^3, this is sorta well known as well. However, the follow-up doesn't have a real non-negative integer solution.
I found a generalization that works for all degrees, not just 1 and 2! It's not as clean, but it's still fun!
Theorem: For every d ≥ 1, there is a unique degree-d polynomial P (up to scaling) and a unique constant c such that for every n, P(x - n) + ... + P(x) = P(x + 1) + ... + P(x + n), where x = c·(1 + 2 + ... + n). Furthermore, we must have c = 2d, and P(0) = 0.
The patterns shown in the video are the special cases d = 1 and d = 2, for which the polynomial P is quite simple: x and x^2.
Here are the first few:
- For d = 1, P(x) = x
- For d = 2, P(x) = x^2
- For d = 3, P(x) = 6x^3 + x^2
- For d = 4, P(x) = 32x^4 + 16x^3 + x^2
- For d = 5, P(x) = 4500x^5 + 4500x^4 + 930x^3 - 19x^2
- For d = 6, P(x) = 46656x^6 + 77760x^5 + 33696x^4 + 1392x^3 - 511x^2
- ...
(Since they're unique up to scaling, I've scaled them up so that the coefficients are integers in lowest terms.)
And here's the "Christmas tree" formed for d = 3:
(6·0^3+0^2) =
(6·5^3+5^2) + (6·6^3+6^2) = (6·7^3+7^2)
(6·16^3+16^2) + (6·17^3+17^2) + (6·18^3+18^2) = (6·19^3+19^2) + (6·20^3+20^2)
(6·33^3+33^2) + (6·34^3+34^2) + (6·35^3+35^2) + (6·36^3+36^2) = (6·37^3+37^2) + (6·38^3+38^2) + (6·39^3+39^2)
....
It's not as pretty, but it works!!
I would be very interested to know if there's a geometric interpretation to 6n^3 + n^2, or even the higher-degree ones!
That also looks like a nice idea. Will also ponder a bit more once I got a bit of time. Maybe also check out another kind of generalization that I am writing about in the description of this video :)
@@Mathologer Thanks! And yeah I've seen the various possible generalizations in the description and they're quite interesting!
By the way, there are various ways to find the polynomials P, but the one I used (and implemented in a program) make use of the following slight extensions:
Lemma 1: For every d ≥ 1, every polynomial Q with degree < d, there is a unique monic degree-d polynomial P such that for every n, P(x - n) + ... + P(x) = P(x + 1) + ... + P(x + n) + Q(x), where x = 2d·(1 + 2 + ... + n).
Lemma 2: For every d ≥ 1, every polynomial Q with degree ≤ d and every constant c ≠ 2d, there is a unique polynomial P with degree ≤ d such that for every n, P(x - n) + ... + P(x) = P(x + 1) + ... + P(x + n) + Q(x), where x = c·(1 + 2 + ... + n).
These immediately imply the theorem, and these in turn can be proven with straightforward recursion/induction, though you'll have to use the fact that the sum of odd powers is a polynomial in 1 + 2 + ... + n = n(n+1)/2, which has a nice proof using central differences and symmetry; it also shows the fact that the sum of even powers is (n+1/2) times a polynomial in n(n+1)/2. (Use central differences instead of forward or backward differences for the cleanest proof :D)
So I use these repeatedly to compute the coefficients one by one, starting with the larger terms first. In Lemma 2, the coefficient of x^d in P(x) is p_d := q_d/(c^(d-1) (c - 2d)) where q_d is the coefficient of x^d in Q(x). Then we simply replace P(x) by P'(x) := P(x) - p_d x^d to get a new equation of the form P'(x - n) + ... + P'(x) = P'(x + 1) + ... + P'(x + d) + Q'(x) with a smaller degree, so we can use Lemma 2 recursively. (That's basically also how the proof works.)
Here's the meat of my (Python) code:
```
def get_P(d, c, Q):
if Q.deg > d:
raise Exception("No answer")
# base case
if d < 0:
return Q
# compute the leading coefficient
if c == 2*d:
if Q[d] != 0:
raise Exception("No answer")
P_d = 1 # P is monic
else:
P_d = Q[d] / (c**(d-1) * (c - 2*d))
# compute the new Q by evaluating at several points and then interpolating
nQ = Poly.interpolate(*(
(s1(n), Q(s1(n))
+ P_d * sum((c*s1(n) + i)**d for i in range(1, n+1))
- P_d * sum((c*s1(n) - i)**d for i in range(0, n+1)))
for n in range(d+3)))
# recursively solve for the remaining coefficients
return P_d*x**d + get_P(d - 1, c, nQ)
```
In David Wells's (The Penguin) Dictionary of Curious and Interesting Numbers, in the entry for triangular numbers, there is an analogous pattern: T1 + T2 + T3 = T4; T5 + T6 + T7 + T8 = T9 + T10; T11 + T12 + T13 + T14 + T15 = T16 + T17 + T18, and so on, attributed to M. N. Khatri. I don't understand everything about how this happens, but the fact that the sum of consecutive triangular numbers is a square (whose base is the larger index of the two triangular numbers), plus the fact that the difference of consecutive triangular numbers is the index of the larger triangular number, seems to be at the heart of it.
Yes, I've also seen a visual proof of this in one of the one glance proofs book by Roger Nelsen. I think it was the third book.
Ah! Playing around with the triangular-numbers patterns a little more, using the facts I mentioned above, I see that they correspond exactly (bijectively!) to doubled versions of each square pattern. They're just the square patterns in the mirror/double-verse: 6^2 + 8^2 = 10^2; 20^2 + 22^2 + 24^2 = 26^2 + 28^2; and so on. And yet they walk through all the triangular numbers in order without missing a beat. Neat.
I found the pattern. Each of those equations start with an nth triangular number, where n is an even whole number, and 1 is the first triangular number. Each equation has 1 more term on each aide than the previous one.
For example:
21 is the 6th triangular number, so if my conjecture is true, this is also true:
24 27
Σ n² = Σ n²
n=21 n=25
And since 36 is the 8th triangular number, we see another one of those equations:
40 44
Σ n² = Σ n²
n=36 n=41
If you don’t know what the weird symbol Σ means, you should learn it by the time you take algebra 2.
My answer to the quizzes:
1. The theoretical pattern for higher powers would have the highest root of the left hand sum as 2^m * n(1+n)/2 since the sequence for the integers is 2 * n(1+n)/2 while the sequnce for squares is 4 * n(1+n)/2. So you would get for the cubes the left hand sums 7³ + 8³, 22³ + 23³ + 24³, etc., for the tesseracts the sums 15⁴ and 16⁴, 46⁴ + 47⁴ + 48⁴, etc. up to higher dimensions.
2. Although I didn't manage to calculate it in 30 seconds but using the equation you showed, 10² + 11² + 12² + 13³ + 14² can be reduced to either 2(10³ + 11² + 12²) or 2(13³ + 14²), both of which can be calculated in the head with the binomial formula (I used the right sum but left sum might have been easier). The sum you get for the sum itself is 365 which also is the numerator. The fraction thus equals to 2.
3. 3³ + 4³ + 5³ = 6³ which I kind of have guessed. However, 3⁴ + 4⁴ + 5⁴ + 6⁴ doesn't have this kind of pattern since the sum isn't divisible by 7 but rather by the primes 2 and 1129.
As for the painting, I can't honestly say that I worked it out in my head, nor in any short time. But long-hand, this was my approach.
Each square in numerator can be expressed in the form:
(10 + a)^2 = 100 + 20a + a^2
for some value of a. Five such squares all together, so we have:
[ 5*100 + 20*(0 + 1 + 2 + 3 + 4) + (0^2 + 1^2 + 2^2 + 3^2 + 4^2) ] / 365
= [ 500 + 20*(10) + (30) ] / 365
= [ 500 + 200 + 30 ] / 365
= [ 1530 ] / 365
= 2
Don't you mean [730] / 365 = 2 ??
@@thecarman3693 I do! I guess my brain & fingers weren't communicating so well last night!
Yes, I originally made a mistake - I had the expression 100*(0 + 1 + 2 + 3 + 4) instead of 20*(0 + 1 + 2 + 3 + 4). So my 3rd from last line had been [ 500 + 1000 + 30 ] / 365. I forget to correct my 2nd from last line - whoops!
Easier to grasp, still enjoyable. If you split your usual format into a series of semi-standalone videos of this format.. you could become a youtube GIANT
7:26 since 10^2 + 11^2 + 12^2 = 13^2 + 14^2 we can reduce the equation to 2(13^2 + 14^2) we can then calculate what 13^2 + 14^2 is : ( 169 + 196 ) which gives you... 365 !
Answer ends up being 2(365)/365, cancel out the 365 and you get 2 as the final answer. ( trick to do 169 + 196, just do 169 + 200 and then subtract 4 )
Second puzzle: short answer is no, since I'm not good at mental arithmetic under pressure. But armed with a few factoids and nerves of steel, I could solve the problem in about 30 using the following strategy: first, what's (n-1)^2 plus (n+1)^2 ? We see that the middle terms in the resulting trinomials cancel out, and the end terms add together, giving 2*(n^2 +1^2). Same pattern for n-2. We can write the series in the numerator as (12-2)^2 + (12-1)^2+ ...+ (12+2)^2. Using the above observation we get 5*12^2 plus 10 (=2*2^2 + 2*1^2) in the numerator. In the denominator, we know 365 is also divisible by 5. If you happen to know the other factor is 73(I didn't) = 72+1 then you're home free, otherwise 360=12*30 is pretty common knowledge, so add another 5 to it and you get 5*(6*12) +5 in the denominator. Now, 2*72=144 (gross!) is also pretty common knowledge, so we have twice( 72 + 5) divided by 72+5 =2.
I freely admit to having worked this out with an electronic calculator, ahead of time, in about 5 seconds. When you do that, you get 365 for the first three terms in the numerator and another 365 for the two remaining terms, thus proving, by brute force, the original Mathologer proposition.
Video still on before start ads and i already love it. That image next to 'skip ad' is 👌
Great animation, very good explanation
5³+6³=/=7³ or 125+216=/=343 (it's off by 2).
To follow the method: 5³+6³ = 6³+5³ ==> 6³ exists of 6 slabs of 6² or 36; it needs to be wrapped around 5³.
Place the 5³ cube on a surface, place four of the 6² squares around it, seen from above you've got 7x7 (note the top layer is just a rim going around, it still needs 25 squares to fill up the middle and get 7x7x6).
The height measured from the table is still 6 so place the 5th 6² square on top of what you've got, you still need to fill in half of the edge with 7+6 = 13 squares. To fill up all the remaining gaps you still need 25+13 squares, you only have 6² squares, so there's 2 missing squares. Or put in another way building from 2D: To surround a 5x5 square: you can do that with 6x4 (you only need 4 sides): 25 + 24 =49 =7²: for one slab. Stack 5 slabs above each other: 7²x5. What you used: 6x4x5 (from the 6³cube) + 5³. (What you didn't use up: 6x2x5+6x6 = 96. And you still need 7²x2=98 to get to 7³.). It comes down to the fact that 7x7 isn't equal to 6x8. You miss one square. (twice) 7x7=6x8+1x1.
To find that exact line 5³+6³=/=7³. I did... 1D: 1x2 (covers 2 'sides'); 2D: 1x4 (covers 4 sides); 3D: 1x6 (covers 6 sides), and so the number before the equal sign is 6... etc...
Now we only need somebody to animated your dissection :)
actually after a few experiments I found some examples satisfying a generalized form for cubes (just adding one term to Fermat equation) :
3^3 + 4^3 + 5^3 = 6^3
3^3 + 10^3 + 18^3 = 19^3
4^3 + 17^3 + 22^3 = 25^3
5^3 + 30^3 + 40^3 = 45^3
6^3 + 8^3 + 10^3 = 12^3
7^3 + 14^3 + 17^3 = 20^3
8^3 + 48^3 + 64^3 = 72^3
....
I am looking at the super-pattern for the squares... Doesn't look so nice, but 5*(2/1)=10, 14*(3/2)=21, 27*(4/3)=36... Next would be 44*(5/4), so 55²+56²+57²+58²+59²+60²=61²+62²+63²+64²+65². Next one would start at 65*(6/5), or 78.
Good idea to look for the super pattern there too :)
7:04 2
Specifically, for any natural number n, let t(n) be the nth triangle number (i.e. t(1)=1, t(2)=3, t(3)=6, etc.)
So, the linear pattern is sum(i from (2*t(n)-n) to (2*t(n))) of i = sum(j from (2*t(n)+1) to (2*t(n)+n)) of j
Squares: sum(i from (4*t(n)-n) to (4*t(n))) of (i^2) = sum(j from (4*t(n)+1) to (4*t(n)+n)) of (j^2)
Cubes: [sum(i from (6*t(n)-n) to (6*t(n))) of (i^3)] +2*[sum(k from 1 to n) of k^3] = sum(j from (6*t(n)+1) to (6*t(n)+n)) of (j^3)
The two is because a cube has eight vertices and only six faces. Each slice from the largest initial cube covers one face and one vertex, leaving two missing vertices (it also covers two edges, which works out perfectly with the 12 edges of a cube). Each missing vertex needs a smaller cube, with a side length of 1 for the first new cube, 2 for the second new cube, etc.
6:43 Our cases came from a multiple of 2 and the sum of consecutive numbers. In the linear case it was 2(1+2+3+...). In the square case it was 4(1+2+3+...) That leaves us with two choices for higher order cases: follow the multiples of 2 like 6(1+2+3+...) (which corresponds to following sides), or follow the powers of 2 like 8(1+2+3+...) (which corresponds to following the corners).
In the multiple case:
5³ + 6³ = 7³ - 2(1)²
16³ + 17³ + 18³ = 19³ + 20³ - 2(1+2)²
33³ + 34³ + 35³ + 36³ = 37³ + 38³ + 39³ - 2(1+2+3)²
56³ + 57³ + 58³ + 59³ + 60³ = 61³ + 62³ + 63³ + 64³ - 2(1+2+3+4)²
You can check the power case yourself but it just doesn't work out.
Now what happens at the 4th power and higher?
7⁴ + 8⁴ = 9⁴ - (4(1))³
22⁴ + 23⁴ + 24⁴ = 25⁴ + 26⁴ - (4(1+2))³
45⁴ + 46⁴ + 47⁴ + 48⁴ = 49⁴ + 50⁴ + 51⁴ - (4(1+2+3))³
76⁴ + 77⁴ + 78⁴ + 79⁴ + 80⁴ = 81⁴ + 82⁴ + 83⁴ + 84⁴ - (4(1+2+3+4))³
This gives the impression that the square case should actually be written like this:
3² + 4² = 5² - 0(1)
10² + 11² + 12² = 13² + 14² - 0(1+2)
...
Unfortunately, this doesn't extend to the linear case, and a close look at the 5th powers makes this extra clear that this is untrue:
9^5 + 10^5 = 11^5 - 2002
28^5 + 29^5 + 30^5 = 31^5 + 32^5 - 162066
57^5 + 58^5 + 59^5 + 60^5 = 61^5 + 62^5 + 63^5 - 2592552
The pattern here is very unclear, and after a bit of checking on wolfram the pattern for the error is very very ugly. Yes, (1+2+3+...+n) is a factor but there's a load of other junk included (500n⁴+1000n³+(752/3)n²+(4/3)n-2/3).
7:06 10² + 11² + 12² = 365. I already know 10² + 11² + 12² = 13² + 14² so that additional 13² + 14² must also be 365. (365 + 365)/365 = 2.
7:29 3³ + 4³ + 5³ = 6³. But 3⁴ + 4⁴ + 5⁴ + 6⁴ ≈ 6.89⁴. And as you increase the power the result still rounds up to roughly the next number.
UPDATES:
- someone proved the error for puzzle 3 approaches 1-ln(e-1)
- oops the error terms are slightly wrong, fixed them
Very, very good :) Glad you had so much fun with these puzzles. Would be interesting to find out how the nice correction term for powers of 3 and 4 works (I suspect there is some nice geometry hiding there) and how it then breaks for 5th and higher powers
@@Mathologer i feel like the best way to incorporate the error term is to move it over to the positive side. So 5³ + 6³ = 7³ - 2(1) becomes 5³ + 6³ + 2(1) = 7³.
I don't want to wrap my head around visualizing 4 dimensions so 3 dimensions it is for now.
Here's the plan:
first rewrite 2(1+2+3+...)² as 2(1³+2³+3³+...) to make things easier to visualize.
Take the largest cube on the left and split it into square slices of varying thickness (the same way it's done in the 2d and 1d case).
Wrap the square tiles around the cubes, and let the overhangs touch. By now there will be 6 centers covered, 12 edges covered, and 6 corners covered, leaving 2 perfectly cube-shaped slots left. That's where the error terms come in: each one fits perfectly in the cubical slots.
Numerical examples:
5³ + 6³ + 2(1³) + 2(1³ + 2³) = 5³ + 6(1)(6²) + 2(1³) = 7³
16³ + 17³ + 18³ = 16³ + 17³ + (6(1+2))(18²) + 2(1³ + 2³) = 17³ + 6(1)(18²) + 2(1³) + 16³ + 6(2)(18²) + 2(2³) = 19³ + 20³
33³ + 34³ + 35³ + 36³ + 2(1³ + 2³ + 3³) = 33³ + 34³ + 35³ + 6(1+2+3)(36²) + 2(1³ + 2³ + 3³) = 35³ + 6(1)(36²) + 2(1³) + 34³ + 6(2)(36²) + 2(2³) + 33³ + 6(3)(36²) + 2(3³) = 37³ + 38³ + 39³
And so on.
I might email you some visuals so you can better visualise what's happening
@@Mathologer I THINK I CAME UP WITH SOMETHING
It's not quite geometric but it's really close
Let i be the thickness of the cubical slices, n be the size of the hypercube to be sliced, and k be the number of cubes to be covered.
In order for the trick to work:
(n-i)⁴ + 8n³i + ERROR = (n+i)⁴
So ERROR = 8ni³, which can be written as 8(n-i)i³ + 8i⁴, for the missing 8 edges and 8 corners.
How to convert 64(1+2+3+4+...+k)³ into sum of 8(n-i)i³ + 8i⁴:
Note that (1+2+3+...+k)² = (1³+2³+3³+...+k³)
Have a series of 3d cubes laid out, of increasing size.
Now we have: 64(1+2+3+...+k)(1³+2³+3³+...+k³)
Note that 8(1+2+3+...+k) = n. Give each cube depth of n to make it a hypercube.
Now we have: 8n(1³+2³+3³+...+k³)
Note that 8n can be written as 8(n-i) + 8i. When splitting our series of hypercuboids slice each one into 16 pieces, 8 with depth 8(n-i) for the edges and the other 8 with depth 8i for the corners.
I have no idea how to actually place the cubic sheets around the hypercube yet but I at least have a promising start, rearranging the algebra to prepare for a fully geometric proof
3²+4²=5² is also part of another pattern, with a bit of algebra thrown in
4 + n² + (n+1)² = (n-1)² + (n+2)²
Set n to 3 and subtract 4 from both sides, and you get 3²+4²=5²
Last problem:
Modulo 10, x⁴ can only be 0,1,6 or 5. The four monomes yield 1+6+5+6 =8(mod10), hence there is no integer solution.
Vielen Dank für das tolle Video, es ist in höchstem Maße faszinierend. Es löst Glücksgefühle aus. Unglaublich...
Question: while 10 is numerator, for which both denominators X and Y, that resulting division for each gives other number (denominator) at the zero-symetry?
I] 10/X= must be 0.Y0Y0Y0Y0Y0Y.....
10/Y= must be 0.X0X0X0X0X0X.....
II] XX and Y-X=10.
Y/X=[(a+1)/X + b/X] x (Y-X)= must be 1.Y0Y0Y0Y....
X/Y= is gonna be at the 9 symetry opposite reverse of X (ba) that is 0.ba9ba9ba9ba9ba9....
In Reel and Irreal numbers universe there is only two unique numbers reach this rule, those are 27 and 37.
10/27= 0.37037037037.....
10/27 gives the other one (37) at 0 symethry...
10/37= 0.27027027027.....
10/37 gives the other one (27) at right-left symethry of Zero.
37>27 and 37-27=10
37/27=
(27+10)/27=1+0.37037037...
=1.37037037...
And
27/37=
0.729729729729...
27/37 gives opposite reverse of 27 at symethry of 9!
See the symetrycity for 0 and 9...
Secret correlation!:
Remember for 27
2+7= 9
And for 37
3+7= 10
|27-37|= 10
37{27+10}>27 then 27/27+10/27=
1.37037037... symethry of 0, of 10's 0 (1,0). Numerator 37 comes at 0 symetry.
[I]
27 {27+0 or 37+(-10)} < 37 then
No 0 symethry but 9 opposite reverse of 27 after 0 decimal =
*0.72 9 72 9 72 9 ......*
Numerator 27 comes at its opposite reverse at 9-symetry
[II]
On the other hand we may assume while numerator < denominator after 0 decimal denominator comes with its opposite reverse mutiplied by 10 and 1 substracted
That is the say: 37 Opp.Rev. 73
73x(37-27)=73×10=730
730-1=729 {here 1 comes from
|27-37|/(37-27)=1 and/or
{37} 3+7=10
{27} 2+7= 9
10-9=1
Remember 37 is {2+1}7
And 27 is {3-1}7 {2+0}7
Then
27/37= must be *0.729 729 729....*
Wouldn't mind these shorter videos coming more often between masterclasses. Merry Christmas professor.
For the second puzzle, I did not recognize it as some people did so I did the calculation in my head. The sum is of the form:
(x-2)^2+(x-1)^x+x^2+(x+1)^2+(x+2)^2
it can be seen that if you expand the squares, all the terms linear in x cancel out, so you're left with
5x^2+2(1^2)+2(2^2) which simplifies to 5x^2+10, which, for x=12, evaluates to 730 which is twice 365.
For the final puzzle:
The solutions can be described as ᵏ√[3ᵏ + 4ᵏ + ... + (k+1)ᵏ + (k+2)ᵏ], which if you calculate for a few values, actually seems to be linear.
As k → ∞, the solution approaches k + 2.458675145387.
combining both animations mentally, I've presumed that by doubling the "central" number n×(n+1) in imaginary squares pattern for the linear pattern, in addition to "left" and "right" we could furnish "top" and "bottom" as we did in the squared pattern and... it's true!
1² + 2² × 2 = 3²
4² + 5² + 6² × 2 = 7² + 8²
9² + 10² + 11² + 12² × 2 = 13² + 14² + 15²
16² + 17² + 18² + 19² + 20² × 2 = 21² + 22² + 23² + 24²
...
What's next? :)
As a math teacher and amateur programmer I tried to find solutions for n>2:
For n=3, the only "real" solution is 3³+4³+5³=6³ but there are other similiar solutions, for example:
4³+...+28³=30³+31³+32³+33³+34³
8³+...+55³=60³+...+68³
34³+...+158³=540³
61³+...+71³=101³+102³+103³
213³+...+272³=556³+557³+558³+559³+560³
For 3
Looks like you are having fun there. That first example is very pretty :)
For cube numbers we get
Sum of x^3 = (Sum of x )^2
So :
7:39
1^3+2^3+3^3+4^3+5^3 = (1+2+3+4+5)^2
Wich means
3^3+4^3+5^3 = 15^2 - 1 - 8
3^3+4^3+5^3 = 15^2 - 9
3^3+4^3+5^3 = 15^2 - 3 ^2
There is a way here
3^3+4^3+5^3 = 5^2×3^2 - 3 ^2
3^3+4^3+5^3 = (5^2 - 1)3 ^2
3^3+4^3+5^3 = (24)3 ^2
3^3+4^3+5^3 = (6×4)3 ^2
3^3+4^3+5^3 = 6×2 ^2 × 3 ^2
3^3+4^3+5^3 = 6×(2 × 3) ^2
3^3+4^3+5^3 =6×6^2 = 6^3
3^3+4^3+5^3 = (15-3)(15+3)
At this point there is 2 ways
1st
3^3+4^3+5^3 = (5×3-3)(5×3+3)
The missing numbers in the tree of sums of squares have a nice pattern to: 1*1 , 2*1 ; 2*3 ... 3*3 ; 5*3 ... 5*4 ; 7*4 ... 7*5 ; 9*5 ... 9*6 etc.
In fact, there are many unnoticed number theory patterns. For example, when I was memorizing squares, I realized that 12^2 = 144 and 21^2 = 441. The mirrored numbers had mirrored squares! In fact, the same thing happens with 13 and 31! But nothing after that. However, this is just a consequence of addition carrying. In higher bases, as long as there is no carrying, this pattern will continue!
I noticed a couple of little patterns not directly mentioned.
In the first sequence, the first terms follow the sequence 1**2, 2**2, 3**2, ...
In the second sequence, the last term on the left is consecutive multiples of four squared.
These observations make it trivial to figure out the Nth row of the patterns in your head.
You are too kind--I went back to check the video to check what I thought I'd seen, only to find you put it right at the head of the comments. It did take me more than the 30 seconds to mentally add 169 and 195. The answer, of course, is 2
:)
I'm a bit of a math nerd, but the best part of the video is still the music! It's so good!
Isn't it :)
Yes, a nice find. Actually both of them: the one I use in the main part of the video and the one I use for the Thank you section at the end of the video :)
I've looked at
7^x + 8^x = 9^x
22^x + 23^x + 24^x = 25^x + 26^x
...
and found that interestingly enough, while it isn't 4 exactly, it appears to be approaching 4 as we continue down the line, with its limit being 4.
1: (2 + A^3 + B^3 = C^3) if B has factors 2 and 3, and A,B,C are consecutive integers (need 6 slices to wrap, but they miss 2 corners)
2: Don't know why I didn't reuse the earlier identity, since I know 100, 121, 144 from rote memory but I timed out trying to be clever
3: stumped me for an overarching pattern, 6 and 6.89... are the answers, however...
...a funny thing about the first 2 is if you subtract the exponent (eg 3- exp) from each term instead, the equation still balances, and first term + exponent equals final term without the exponent
7^3 + 8^3 = 9^3
Because the pattern is 2^1, 2^2, 2^3 for each pattern
Actually, I'd say the first equation should be 5^3 + 6^3 = 7^3 :)
@@Mathologer well, you're still two short 😀
While swimming I discovered the difference in consecutive natural number cubes is 6 x triangle number + 1.
2^3 - 1^3 = 8 - 1 = 6*1 + 1
3^3 - 2^3 = 27 - 8 = 6*3 + 1
4^3 - 3^3 = 64 - 27 = 6*6 + 1
Proof is that (x+1)^3 - x^3 = 3x^2 + 3x + 1
Factor 6x out of the first two terms.
6(x)(x+1)/2 + 1.
x(x+1)/2 is a triangle number for natural number x.
I thought it was cool.
Embedded in triangle numbers are square numbers. Embedded in cube numbers are triangle numbers. 6 triangles to cover each face + 1 more to complete the cube.
Yes, nice insight :) There is actually a nice geometrical interpretation of all this. Maybe have a look at the first couple of minutes of my video on Moessner's miracle where I show how to interpret the difference of two cubes of two consecutive numbers as a certain hexagon. That a hexagon can be divided into 6 equilateral triangles then translates into you insight :)
The second puzzle at 7:25, the answer is 2.
I did this mentally using the sun of square formula to figure out the sun of the first 14 squares, and the sum of the first 9 squares, and then did a subtraction. It’s (14)(15)(29)/6 - (9)(10)(19)/6, which simplifies to (7)(5)(29) - (3)(5)(19). This becomes (5)(29*7) - (5)(3*19), which is (5)(203 - 57), which is (5)(146). We are dividing the entire thing by 365, which is (5)(73), and 146/73 is 2. So with all of the cancellations, the answer is 2. Not the easiest thing to do in my head, but the answer does reveal itself.
Very good. Even quicker if you apply 10² + 11² + 12² = 13² + 14² Then you just have to calculate one side of the equation.
The first numbers of each row in the linear pattern are the square numbers (1, 4, 9, 16...), while the first numbers of each row in the square pattern are every other triangular number (3, 10, 21, 36...) and the last numbers in each row are one less than the other triangular numbers (6, 15, 28, 45, ...).
Well spotted :)
Strange but true!
Didn't know about the infinite pattern of first powers & squares, but I did stumble onto the single equation with cubes many years ago (3³ + 4³ + 5³ = 6³).
Tried to find some general rule to extend it, but was never successful.
As for the attempt to extend your 1st- and 2nd-power patterns to cubes, the obvious try almost works, for the first line:
5³ + 6³ = 341 = 7³ - 2
Fred
"5³ + 6³ = 341 = 7³ - 2" There is actually something super nice hiding just around the corner. If you are interested have a look at the description of this video.
Me too about the cubes which we studied in school where it was phrased like something like what will be the side/radius length of a cube/sphere made by melting cubes/spheres of side length/radii 3,4 and 5 units.
2nd puzzle is easy. (2*365)/365.
And yes, I also noticed those patterns and extended them to 0. Tried to look for other ones with different exponents (after I learned of Fermat last theorem, I looked for non-positive and non-integer exponents), but to no avail. I always liked that first pattern doesn't have a "hole" - so it uses all numbers. I always wondered if the unused numbers in 2nd pattern can be used for something. I guess I will sit down with my pen and paper later this evening to check if there isn't some interesting there. Even if not as beautiful.
For the second puzzle we can use similar graphical proof to get 10²+12² = 2⋅11²+2 (after using 13²+14²=10²+11²+12²), simplifying whole expression to 2⋅(3⋅11²+2) = 2⋅(3⋅121+2)
Alright answer for the puzzles!
EDIT: Corrected some typos and semantics
First puzzle
The n^1 numbers grow like
0 =; 1 + 2 = 3; 4 + 5 + 6 etc
The n^2 numbers grow like
0 =; 3^2 + 4^2 = 5^2; 10^2 + 11^2 + 12^2 etc
Therefore the n^3 numbers would grow like
0 =; 5^3 + 6^3 = 7^3; 16^3 + 17^3 + 18^3 = 19^3 + 20^3; 33^3 + 34^3 + 35^3 + 36^3 etc...
As an extra, the third line would be 16^3 + 17^3 + 18^3 etc
The logic for the third line and so on is that, the first term of the third line is equal to the sum of twice the first plus the second term (without raising them) of the second line, for example:
1 + 2 = 3
4 ...
4 = 1.2 + 2
3^2 + 4^2 = 5^2
10^2...
10 = 3.2 + 4
Therefore we can say that the first term of the cubes is
5^3 + 6^3 = 7^3
N = 5.2 + 6 therefore N = 16
For the next line, it's the sum of the first and second term
9 = 4 + 5 - For the first power
21 = 10 + 11 - For the squares
So, for the cubes we'd have...
N = 16 + 17 = 33
Second Puzzle
Since we know from the second line of the squares that 10^2 + 11^2 + 12^2 = 13^2 + 14^2 and that both sides are equal to 365, this means the answer is two (730/365 = 2)
Third Puzzle
Too lazy to do that by hand!
Problem 1: If the pattern in 1 and 2 dimensions continued, the first 3D example would have used 6 in the bridging term. (6 is the number of faces on a cube). Thus the first identity would have been 5³ + 6³ = 7³. Unfortunately, the RHS is short by 2.
Problem 2: Since 10² + 11² + 12² = 13² + 14² = 365, the calculation on the blackboard yields 2.
Problem 3: By direct calculation, 3³ + 4³ + 5³ = 6³ (surprise!) and 3⁴ + 4⁴ + 5⁴ + 6⁴= (6.8934...)⁴ (not so pretty).
Thanks for sharing another shiny nugget from your mine of mathematical gems.
Also check out the description of this video for some extra fun :)
Question: why does the square of 3.14159...'s quarter fail to satisfy the annulus whose area is π units²?
If we plot 4a((√5/2)² - (1/2)²) = 4a = π units², we get four quadrants each of area (π/4)² thus 4(π/4)² area.
If we plot a point (0, 1/2) and square 3.14159...'s quarter from it (0, 1/2 + ( 3.14159.../4)²) to represent a point on a circle as it expands according to the square of pi's quarter, it geometrically fails to satisfy the width entirely.
Why??? If 3.14159... were correct, should we not expect its own squared quarter to satisfy its own π units² annulus (?)
I am afraid that that after just glancing over what you've written here I do not know what you are trying to say here :)
@@Mathologer
1. Plot x² + y² = 1/4 and x² + y² = 5/4. This produces the annulus 4a((√5/2)² - (1/2)²) = 4a.
2. Find the width w as -1/2 + √5/2. Plot a point (0, 1/2) on the r = 1/2 circle.
3. Square 3.14159...'s quarter off this circumference via. (0, 1/2 + (π/4)²).
Arithmetically, the area of each quadrant is (π/4)² thus the annulus is 4(π/4)².
What I'm asking is: why is it when we square 3.14159...'s quarter does it fail to reach???
Any one quadrant sweep of width w is equivalent to the square of pi's quarter.
In other words: why does (π/4)² = 0.616... instead of the actual geometric width of 0.618... ?
If the exponent is 1 or 2, a^x + b^x = c^x has solutions. But not if x = 3, and not if x = 0 either (as 1+1 != 1). In fact, among the integers, a^x+b^x=c^x has the same solution set as x^2-3x+2=0.
An easy (and general) way for the second exercise is to note that the numerator is of the form:
(n-2)² + (n-1)² + n² + (n+1)² + (n+2)²
if we were to expand the binomial squares, all the terms in the first power of n would cancel as they have opposite signs in pairs. So what we are left with is:
n² + 2² + n² + 1² + n² + n² + 1² + n² + 2² = 5n² + 2*(2² + 1²)
for n = 12, this is obviously 5*144 + 2*5 = =1440/2 + 10 = 730, which is twice the numerator of 365.
That's what I did in my head in less than 30 seconds. Alternatively, one could do 169 + 196 or 100 + 121 + 144 in less than 30 seconds of mental calculation, and know from the video that it's half the numerator.
Very good :)
@@Mathologer Merry Christmas and Happy New Year, Burkhard - and thank YOU for all the excellent content!
I prefered doing 10²+11²+12² = 3 * 11² + 1 + 1 = 363+2 = 365 ( the relationship between 121 and 363 led me to this).
Formula: sum(2n^2+k)^2 LHS k=n=>2n. RHS: k=2n+1=>3n
7:00 - It would be 5³ + 6³ (341) ~= 7³ (343). The difference is 2 as each 6² slice of 6-cube covers 1 face, 2 sides and 1 corner of the 5-cube. Total 6 faces, 12 sides and 6 corners are covered so it falls short of making a 7-cube by 2 corners (2 * 1³). Next equation would be 16³ + 17³ + 18³ ~= 19³ + 20³. This time the difference is 18 (2 * (1³ + 2³)).
7:25 - It's obvious now. Another way for someone who has not seen this video to solve is to substitute 12 with x. Then the expression becomes (x-2)²+(x-1)²+x²+(x+1)²+(x+2)² / 365. (a+b)²+(a-b)² = 2(a²+b²). So the expression becomes 2(x²+2²)+2(x²+1²)+x² / 365 = 5x² + 10 / 365. Replacing x by 12, it's 5(144)+10 / 365 = 730/365 = 2.
Very good. Maybe also read through the description of this video for some more nice insights :)
There's a superpattern for the squares too. It's a bit trickier, but 2*5/1 = 10, 3*14/2=21, 4*27/3=36, (n+1)*k_last[n]/n = k_first[n+1]. I haven't proven this, but will leave that as an exercise.
For the pattern of squares first shown at 1:48, in each step n the number of integers skipped to get to the next step n+1 equals 2n + 2.
For the equivalent, slightly more complicated pattern of cubes wonderfully sorted out by Exception2001 in a must-read comment, it is 4n + 4.
But it isn't n + 1 for the linear pattern. Instead, no integers are skipped. What the hell, cosmos?
dang ! you're a mighty force of bringing Enlightenment in this Age of Confusion !
Outstanding animation, doctor! 👍👍👍👍
, yeah, same Timelord math trick indeed 🤓🖖🤓🖖
I bought a few packs of Rogers sugar cubes to have fun with watching this video. Happy New Year!
1:44
The last number of one series and the first number of the next are in type:
ab, a(b+1)
Hello Mathologer! I remember seeing one of the videos, explaining the binomial expansion with fractional and negative powers, beautifully explained, but I can't find it any more. Could you point me in the right direction? Also I remember you have a second TH-cam channel, called something like Mathologer2?
6:50 Two other nice patterns in the left-hand terms (ignoring the powers): the top one is the square numbers, and the second one is alternate triangular numbers :)
Animation of rearranging of squares to show that 1+2=3 just blows my mind
Why can't you start the pattern a^3+b^3=c^3+d^3 or a^3+b^3+c^3=d^3+e^3 or similar?
I am not going to stop you. However, the natural choices for these numbers suggested by everything we did in this video do not work :)
5^3 + 6^3 = 7^3 - 6 (covered corner points) + 8 (actual corner points)
10^2 + 11^2 + 12^2 + 13^2 + 14^2 = 10^2 * 5 + 21 * 4 + 23 * 3 + 25 * 2 + 27 = 730
730 / 365 = 2