Beautiful visualization | Sum of first n Hex numbers = n^3 | animation
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- čas přidán 8. 07. 2017
- In this animation I'll show why the sum of first n Hex numbers is equal to nxnxn. Hex number (or centered hexagonal number) is just a number of dots that surround the center dot in a hexagonal lattice.
Hope you like this video.
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Support my animations on:
/ think_twice
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Any further questions or ideas:
Email - thinktwiceask@gmail.com
Twitter - / thinktwice2580
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Overall render time: ~ 15 hours
Programs used:
- Cinema 4D
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Music:
Nocturne op. 9 no. 2
CZcams is like a gold mine. You gotta dig deep to find the treasure..
You goddame right
Your videos deserve at least a million likes.
Thank you:) I don't think that many people on CZcams are interested in this kind of stuff though
Think twice : There are actually more than millions of people liking these kinds of stuff but may be they are not knowing about these channel
I MYSELF CAME AFTER RECOMMEDATION FROM 3BLUE1BROWN
Iftakhar Ahmed yes maybe you’re right. it’s just hard to build an audience.
Amazing work sir please allow us download also
So true!
I love how you synchronized the animation with the music!
Dragon Curve Enthusiast he didn't you did
It would've been perfect had he allowed the Nocturne to finish; it had only about 20 seconds left when he chopped it...
Fred
That is one of the most beautiful proofs I've ever seen.
Excellent! 👏 👏 ☺
(Music: Frederick chopin nocturnes)
Edit: subbed
I think it could've been better this time. Starting from the center of each hexarrengement draw three radial axis evenly spaced. They turn into the outside edges of the shells. From there the remaining three sections form the faces of the shells.
These videos are the most satisfying thing on youtube. The math, the music. The animation is so smooth. Even the color pallet is delightful.
Brett Cooper thank you! I’m glad you enjoyed it:>
wow
It is amazing how changing the perspective allowed to reach this conclusion. You have an interesting way of thinking! And your animations are simply beautiful!
This is beautiful, especially since I rely on hexagons in my cellular automata work. And the graphics bring back fond memories of playing with the "Soma Cube".
This is God's work. Please continue
Amazing as always, unfortunately I had forgotten to turn on the notifications and apparently I lost a lot! Great video, amazing!
This is how you should think, not directly to the solution but think of ways of how can you get better at math with new problems with new solutions!
Nuno Mateus true words
Fantastic animation, and I like your song choice.
Michael Martins thank you
*ITS A FUCKING PIECE OF MUSIC*
Absolutely visually stunning video and I love how quiet and beautiful you made it! Really nice job, definitely need to see some more math CZcamsrs focusing on the visual beauty of mathematics without getting bogged down by long spoken explanations
true
Y'know since it is just a visual perception that a hexagon with a few lines become a cube, but that turns out to be a proof of that. Thats more interesting than i thought it would.
發阿 thanks
I do like the perception trick, however it should be noted that this is an in complete proof. It doesn’t show how an n-sized lattice can be represented as half of the shell of of cube for the general case. Rather it just shows it for a couple cases
You show that the first three "cubes" can be arranged in that pattern where the subcube is missing, but will that hold true for all n? And why?
Each hexagon has a center cube. From the center cube of the nth hexagon, there are n-1 cubes in a row directly to the right of it. There are n-1 more in a row at 120 degrees and another at 240 degrees. The center cube forms the "back" cube, these three rows form the "splines," if you will. The remaining cubes are in three groups, forming three (n-1)x(n-1) squares, which form the sides of the shell.
Think Twice very nice! Though the animation at 1:40 is not reflecting that, which made me suspicious.
ya my animating skills arent that good yet
@@ThinkTwiceLtu Aw come on!
I just discovered your channel, but your demonstrations are so satisfying, keep doing what you’re doing!
You don't understand how useful that video is! At least in my country, there's the Math Olympics, and many other kinds of tests, there's always this kind of logic based questions that make you search for previous tests, there's also tutorials on CZcams, but, you know, this video is really useful if you wanna get past tests like this.
Cool. And you can see that it still goes quadratic in n as it should for an plane: n^3 - (n-1)^3 = 3 n^2 - 3 n + 1.
Brilliant demo! Visual proofs _rock_ !
Such a great video, made me picture Hex numbers in a different way
This is pure beauty.
Please don't stop making these videos , good sir. Will spread the word of your videos! Subscribed!
Thanks man:) appreciate the support~
There’s just something so beautiful and mesmerizing about complex math, that will most likely never be very useful in most fields, to be displayed visually like this. I love it
Thank you for this video. I've been interested in this sequence for a while and knee its connection to a 2d hexagonal lattice but totally missed the 3d transformation. Really appreciate it
This is sooooo beautiful!
wow that was so beautiful! Thank you so much!
wonderful approach!
most beautiful thing I saw today
this was awesome
Nebula Quiddity trully
@@ThinkTwiceLtu but how on earth do you get hold of this fabulous ideas?
Beautiful!
And people say math isn't art, good job. I feel like I just ascended to another plane of existence.
there is a way to nicely rearange cubes. Look at center cube. Take this cube and cubes on the left, then take cubes on "top" of those (in 2D up-left direction) those cubes will form n*n sized back wall. It can be easily formed just move the cubes to the left to form nice vertical columns(hides upper faces) ,then moved those columns to the front (hide right faces). If you see this, I think that seeing how to make the bottom n*(n-1) sized wall and left (n-1)*(n-1) sized wall will be easy.
You can also rearenge them in such a way: the right cubes will form L shape(in 3D), by moving the upper cubes to the right and the bottom cubes up.
Just beautiful!
This is just beautiful .
So wonderful :) A really nice creative way of seeing this, thanks!
Rob Nicolaides thanks
perfecttt, so satisfying
ahh and chopin....my favorite composer
Nice
Robertas S nice
Think Twice slimcock
The music makes it even better
It's so beautiful, the explanation with the music 10/10
thank you:)
Quite interesting and you've gotten pretty good at the 3D animation.
Memes Read Out loud you taught me well
absolutely insane idea
This is some of the most beautiful math on youtube!
Lovely, quite lovely. Thank you.
Hey this is amazing and so are you! Wow!
so beautiful
just beautiful.
Arjan Kasapi glad you liked it
Beautiful, very very pretty
thank you :)
Bellísimo.
im subbing
after _just one more_ video
'Hexagons are the bestagons'
-Grey
This is such beautiful math!
Dude! So cool
Love it
This is purely genius
Chopin and hexagons, I love it! 😍
beautiful !!!!!! thank you so much so share this beautiful understanding 👏👏👏👏👏👏👏
Hit like just after hearing the music, nice one!
Awesome!
:)
beautiful
Mükemmel!
Insightful.
Very nice.
Amazing
Thanks for the sharing. It is a spectacular visualization.
Jen Kadverson thanks for watching:)
That is truely butiful, as well as the music
thank you:)
A billion likes! Truly amazing
Thank you:)
I know this gets said alot on youtube, but how or why would someone dislike this?? surely must be a missclick. amazing video as always
Very good
I'll never look at 2-D representations of cubes the same way again
me too
Really all your videos are intuitive and very much elegant.
Do upload any fantastic ideas, proof or even beautiful little intuitions you wanna share,please!{ you have platform to show beauty of mathematical ideas unlike all other mathematician}:-)
Nocturne, nice
Great
I'm glad YT recommendations led me here
Visual proofs are the oddly satisfying of mathematics.
Best Music ever!!!
This is genuinely really interesting. Is it the same thing if you use other shapes instead of hexagons?
Morrison Productions yes it's pretty much the same as long as you dont change the number of objects in a hexagonal lattice
+Morrison Productions: What a bizarre question to ask.
If you use other shapes, the numbers all change completely.
A hexagon is surrounded by 6 other hexagons. A triangle will be surrounded by only 3 similar triangles. Not 6
A square will be surrounded by 4 squares (a cross) or 8 squares (a bigger square). But not 6.
1+3 is not the same as 1+4 is not the same as 1+6. Like, what are you even on about, mate. Have you even stopped for a second to think. To just think of a triangle and ask yourself if the first numbers would still be 1, 7, 19, 37, or whether they'd be something else entirely.
This proof is specifically for the sum of the first N Hex numbers. Not the sum of the first N whatever numbers.
Talk about missing the point of the whole video. Talk about not reading the title. Talk about not thinking at all.
So much work went into this animation only for it to completely fly over some people's heads at the most basic level. That saddens me to no end.
You should learn from this channel's host about politeness, not everyone is as wise as you.
I'm also curious about this. I _think_ this has to do with the Schlaffli Symbol for the hexagonal tiling {6,3}, but I'm not perfectly sure, since {4,4} gives n^3/3 - n^2/2 + n/6. I also forget the triangle sum off the top of my head. The hyperbolic tilings allow for most of the other combinations (sans the spherical tilings otherwise known as the platonic solids and some degenerate 2-gon nonsense on spheres as well), so {6,4} should also have a "ring count" number, but with 4 hexagons per vertex, as would {4,6} with 6 squares to the vertex. Finding patterns here would be neat, since the construction of n-sided meta-gons _should_ work even in hyperbolic space.
@@timh.6872 Man! I think something pretty serious is going on in your head... I wish if you could lay it flat on layman's language or at least have suggested a clue ( maybe a channel) to do some study and understand what is hiding behind your notational talking. I would be grateful if you do it now.
It's a kind of Stendhal Syndrome I sort of cry watching it. I Did it twice!
Transformation between the flat cubes and the cube shell shouldn't be taken at random, we can easily generalize this step and the visualisation should show it.
2:18 which is in reality a parabola: 3n^2-3n+1.
the derivation of this stuff in the video is n*6. and indeed the number increases with n*6 ( 1 ... 7 ... 19 ... 37 ... )
however, your derivation is 6n -3
and still your absolute figure is correct. why?
Which is in reality a telescopic sum..
@@MrTiti please, clarify your query, I cannot get it
Beautiful!
But you didn't let the Nocturne finish! What's wrong with you? It was almost done! And the ending is utterly beautiful!! That's one of my lifelong favorite Chopin pieces! Op. 9, No. 2, Nocturne in E♭ major!
OK, the math was beautiful anyway - you're forgiven! . . .
BTW, I noticed this relation some years ago, just from the algebra:
Hex(n+1) = 6Tri(n) + 1 = 6·½n(n+1) + 1 = 3n² + 3n + 1 = (n+1)³ - n³
So thanks for showing it visually!! But I'm gonna have to go back over it several times; I don't yet see how you've shown that the 3-faced cubic shell will always result from rearranging the hexagon.
At the same time, I *can* visualize a hexagon of dots distorting into a 3-face cubic shell...
Fred
Really interesting, but it also helped me fall asleep
magic
i gasped when i realised they could all fit together before you said asjsjsjs
MosT bEauTifuL tHINGs evEr
x^2 + 2x + 1 is the previous series..
This one is x^3 + 3x^2 + 3x + 1... or use binomial expansion with (x+1)^n... however...a beautiful animation that helped create more vivid images...
The hexagon function can be said to be 6x - 5...however, to add the sum, sigma should be used. That sigma can be simplified to.. -5(randomInt) + 6 × sigma(n=1 to randomInt) n
very clever
Александр Куприянов спасибо:)
😍😍😍 I wish you were my Maths teacher
thank you for the animations. since it is hex numbers it should have gone up to 6
I love to read all the positive comments here !!
Another way to get the next number is to take the previous answer plus the interation number times 6.
So the 3rd iteration is the second iteration (7) plus (2x6)= 19
Then the 4th iteration is the 3rd interation (19) plus 3x6 = 37.
I'm not that mathy, so idk how to get any random n value without having to already know the previous answer though.
0:50
After I noticed the typo my OCD immediately started firing up. The rest of the video is absolutely amazing tho
I love you
From this video, we know that a hex number is the difference between two consecutive cubes. The nth hex number can be found with this way too:
(2n-1)n+(n-1)
..the expression 3x^2 + 3x + 1 exactly describes a way that a cube grows, adding 1 to its edge length each time: E.g., start with the unit cube, then add on this many extra unit cubes for each successively larger cube: 7 -> 19 -> 37 -> 61 -> 91 - > 127 etc. (these are the differences of consecutive cubes.) Geometrically, one can imagine 'pasting' 3 'slabs', each of face area x^2, plus 3 'columns' of x unit cubes, plus a single unit cube to fill the remaining void, completing the slightly larger cube.
Great Animation! Similar to 3 blue 1 brown, your video gave more importance to the visualization than the formula. If you haven't already seen 3b1b videos, i strongly suggest you check them out.
interesting!
So what sum of numbers will add up to n^4? He showed n^2 for odd numbers and here n^3 for hex numbers.
SwordQuake2 T E S S E R A C T S
SwordQuake2 Rhombic dodecahedral numbers
Pedro Nunes Sums that give sixth powers require tesseracts, but not those that give fifth powers.
I was thinking of the center piece plus 6 triangle numbers of degree n-1:
6 × (n-1) × n/2 + 1
This is the same formula in disguise:
= 3 × (n-1) × n + 1
= 3 × (n² - n) + 1
= 3n² - 3n + 1
= _(n³ - n³)_ + 3n² - 3n + 1
= n³ - ( n³ - 3n² + 3n - 1 )
= n³ - (n-1)³
Is it correct to say that you can modelize this through this recursive function ;
f(1)= 1
f(n) = f(n-1) + 6(n-1) ?
or
f(0)=1
f(n) = f(n-1) + 6n ?
mathematical jawbreaker
This is how the mind of a 2D person differs from a 3D person .