# Visually stunning math concepts which are easy to explain

Since I'm not that good at (as I like to call it) 'die-hard-mathematics', I've always liked concepts like the golden ratio or the dragon curve, which are easy to understand and explain, but are mathematically beautiful at the same time.

Do you know of any other concepts like these?

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It looks like mathpop or demand for math entertainment) – rook Apr 2 '14 at 12:59
@ColeJohnson the 'transcendentality" is the beauty of it! – Sabyasachi Apr 4 '14 at 17:07
Why has this got 105k views......................? – bwv869 Apr 8 '14 at 16:49
@TheGuywithTheHat That's the reason for the second spike of visits, on August 27. The comment by LTS is from April; back then the traffic was driven by Ycombinator. As a result, this same question made both April 7 and August 27 the two days with most visits to the site. – Meta Aug 29 '14 at 18:37

Simple,visual proof of the Pythagorean theorem. Originally from Pythagorean Theorem Proof Without Words 6).

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Perhaps I'm trying to oversimplify, but this visual proof would be way easier (perhaps even trivial) if the legs of the big straight angle (S.A.) triangle (hypotenuse = the circle's diameter, third vertex on the top of that leg of length $\;b\;$) were drawn, and then from basic geometry: " In a S.A. triangle, the height to the hypotenuse divides the triangle in two triangles similar to each other and also to the big triangle". Nicely brought. +1 – DonAntonio Mar 31 '14 at 12:31
Another great pictural proof is 4 triangles in a square: mathalino.com/sites/default/files/images/01-pythagora.jpg – Sergey Grinev Mar 31 '14 at 15:10
I don't get the original equality. Are you relying on the similarity of two triangles? How is that obvious from the diagram? – adam.r Mar 31 '14 at 17:10
It's not immediately obvious to me why $\frac{c+a}{b} = \frac{b}{c-a}$ I'm sure it's very simple and I'll kick myself for asking, sorry. – PeteUK Mar 31 '14 at 17:12
Well, I looked at that picture and thought: "Err ... how does this work?" I figured it out, but it was far from obvious for me. – celtschk Apr 3 '14 at 18:33

Here is a classic: the sum of the first $n$ positive odd numbers $= n^2$.

We also see that the sum of the first $n$ positive even numbers $= n(n+1)$ (excluding $0$), by adding a column to the left.

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If you remove the leftmost column, you get a proof that the sum of the first $n$ non-negative even numbers (counting 0) is $n(n-1)$ – becko Apr 3 '14 at 16:29
I read something about how Galileo noted that the distance a falling body covered over a unit time went as the series of odd numbers and was confused until I realized this. – Nick T Apr 3 '14 at 17:37
I remember about 5 years back, it was when we were first learning programming. Just simple kids stuff. Loops. We were given a series to print. $1,4,9,16,\dots$ I was a little overworked so I didn't notice the obvious square pattern. I did notice $1,(1+3),(1+3+5),....$ when we were giving in our work, I saw the guy before me had just done $i*i$. I totally went "hoooly shit" – Sabyasachi Apr 6 '14 at 9:40

My favorite: tell someone that $$\sum_{n=1}^{\infty}\frac{1}{2^n}=1$$ and they probably won't believe you. However, show them the below:

and suddenly what had been obscure is now obvious.

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My first intuitive visualization of this sum was a circle. I wasn't $100\%$ sure that the answer was $1$(long time back, I had never seen an infinite series before, and 1 was just my first immediate thought) :) – Sabyasachi Mar 31 '14 at 17:38
I still don't believe you. – Ojonugwa Ochalifu Apr 2 '14 at 8:51
Another way to think of this is that 1/2 + 1/4 + 1/8 = 0.111...binary = 0.999...decimal = 1. – Justin L. Apr 7 '14 at 3:02
You don't need to divide the square into such complicated fragments, just use vertical lines (assuming the x coordinate ranges from 0 to 1) at x=1/2, x=1/4, x=1/8 etc. Each time you add 1/2^n to the area, and in the limit you obviously get 1. – Maxim Umansky Apr 7 '14 at 4:05
@MaximUmansky: That way you'd just get lines that get closer together and it'll be not as obvious. Here, you see the fractions $\frac{1}{2}$ and $\frac{1}{4}$ in their "standard shape", so what remains must obviously be $\frac{1}{4}$. Then, put the same shapes inside the remaining square which is of the same proportions as the initial one (and it's easily checked that $\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}$); you'll get the next smaller square, hidden deeper in the corner. Repeat, and the square will shrink to a tiny dot (not a whole line, which may intuitively seem larger). – nobody Apr 9 '14 at 20:09

Fractal art. Here's an example: "Mandelbrot Island".

The real island of Sark in the (English) Channel Islands looks astonishingly like Mandelbrot island:

Now that I think about it, fractals in general are quite beautiful. Here's a close-up of the Mandelbrot set:

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Are complex numbers easy to explain? – adam.r Mar 31 '14 at 17:15
Probably not. But the idea that an object "looks the same if you zoom in on it" is easy to explain, I think. – user134824 Mar 31 '14 at 17:16
Whereas complex numbers might not be easy to explain, it is easy to explain the idea idea that each pixel is calculated by repeating a simple calculation on two numbers (which are initially the coordinates of the pixel), and the color is a measure of how the resulting pair of numbers "escapes" (grows large). – Kaz Apr 1 '14 at 4:57
@adam.r I don't recall precisely when I learned about complex numbers in school, but it was around the 9th grade (maybe 8th, maybe 10th). To understand the Mandelbrot set, nothing more is needed than basic arithmetic with complex numbers. That should put it within reach for anyone who graduated high school. It's certainly a good candidate for this list. – Szabolcs Apr 1 '14 at 16:12
It is a close up. The Mandelbrot set is not completely self-similar, but there are some self-similarities as you zoom. See this zoom sequence for an example. – user134824 Apr 3 '14 at 3:12

A theorem that I find extraordinarily beautiful and intuitive to understand is Gauss' Theroma Egregium, which basically says that the Gaussian curvature of a surface is an intrinsic property of the surface. Implications of this theorem are immediate, starting from the equivalence of developable surfaces and the 2D euclidean plane, to the impossibility of mapping the globe to an atlas. Wikipedia also provides the common pizza eating strategy of gently bending the slice to stiffen it along its length, as a realization

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A minor correction: strictly speaking, Gaussian curvature is not a topological invariant. – Michael Apr 3 '14 at 22:48
@Michael: Yes, you are right. My mistake. I instead had in mind the surface integral of the gaussian curvature over a closed surface, which is a topological invariant (basically the euler characteristic) – surajshankar Apr 5 '14 at 4:59

I think if you look at this animation and think about it long enough, you'll understand:

• Why circles and right-angle triangles and angles are all related
• Why sine is opposite over hypotenuse and so on
• Why cosine is simply sine but offset by $\pi/2$ radians
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@joeA: it's not as smooth as regenerating at a higher framerate, but gfycat.com allows one to view gifs at different speeds: gfycat.com/TintedWatchfulAxisdeer#?speed=0.25 – Max Apr 2 '14 at 20:51
If high school math just used a fraction of the resources here, we'd have way more mathematicians. – user148298 Apr 3 '14 at 17:00
Is the source of this animation available? (It looks like it's in $\mathrm\TeX$.) – bb010g Apr 3 '14 at 23:31
This is the normal way of introducing sine and cosine at high schools. At least in our country. – Vladimir F Apr 7 '14 at 15:36
@joeA: I've made a programmatic version of this on Khan Academy; feel free to change the speed to your liking: khanacademy.org/cs/circle/4597064320155648 – BHSPitMonkey Apr 16 '14 at 10:48

I do not know if this meets your criteria of "visually stunning", but nonetheless -

I like this proof of Pythagoras' Theorem (image taken from www.wisfaq.nl):

The key to understanding this is to realize that the inner quadrilateral must be a square - the sides are equal in length (obviously) and each of its angles is $90^{\circ}$ because the two angles on either side sum to $90^{\circ}$, and the sum of the three angles is $180^{\circ}$. The area of this square is $c^2$.

The outer square's area is $(a + b)^2$, which is $c^2$ plus $2 a b$ which is the total area of the four triangles, each of area $\frac{1}{2} a b$.

$(a + b)^2 = c^2 + 2 a b$

$a^2 + b^2 + 2 a b = c^2 + 2 a b$

$a^2 + b^2 = c^2$, which is Pythagoras' theorem.

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+1 finally i can prove Pythagoras theorem without looking into internet or books :) – lowtech Jul 31 '14 at 20:58

This is similar to Aky's answer, but includes a second drawing (and no math.)

To me the second drawing is key to understanding why the $\mathrm c^2$ area is equal to the sum of $\mathrm a^2+\mathrm b^2$.

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I've always felt this one worked better in .gif format. – Patrick M Apr 3 '14 at 20:47
This would be amazing if you animated the change from the first to the second by having the triangles on the right flop over onto the ones on the left to create those rectangles. – Nick Retallack Apr 7 '14 at 0:31
I think this one is lacking because it is not explained in the left picture that the blue and red areas are the same size. – Sparr Apr 7 '14 at 15:58
@Sparr they are not. The salmon area is just 4 times the size of the triangle. The only equivalence is bewteen the blue area on the left (c²) and the blue area on the right (a² + b²) which is just Pythagoras' formula. I will upload a better picture to show it. – Tobia Apr 7 '14 at 16:33

This is how I learned Pythagoras' Theorem:

This diagram makes more sense to me than the other ones posted here.

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This doesn't explain why the theorem is true for any other triangle though, does it? This is basically just "compute $3^2+4^2$ and $5^2$ to see that $3^2+4^2=5^2$" – Daenerys Naharis Apr 1 '14 at 18:47
The visual is not explaining why (given the three specific squares) the red triangle is a right triangle. – Johannes Apr 1 '14 at 18:50
I do not thing this answer deserves to be downvoted, given the question wasn't asking for visual proofs. It does demonstrate the truth of Pythagoras' Theorem - albeit for a specific example. Someone who hadn't encountered Pythagoras' theorem before but who understood areas might find it interesting and ask whether it was true for other right-angled triangles. +1 from me. – Aky Apr 2 '14 at 6:33
Stop downvoting, people. The question asked for a concept. Not a proof. This is perfectly fine. Mnemonics would be fine too. – smci Apr 3 '14 at 5:51
Clearly this answer requires down-voting. It is a bad answer. The whole raison d'etre of the points are to order the answers by their "goodness" as seen by the community. This is not good. It is not even fair, or mediocre, or ho-hum. It is bad. This is not "easy to explain" because it has a lot of useless information presented (right angle triangle, the a, b, and c, or anything to do with Pythagoras). This doesn't even illustrate that $3^2+4^2=5^2$ except if you count squares, but you can do that with the numbers! And it is certainly not visually stunning... – ex0du5 Apr 3 '14 at 20:17

A well-known visual to explain $(a+b)^2 = a^2+2ab+b^2$:

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why area of a^2 is greater tham area of b^2? – jaczjill Apr 3 '14 at 7:34
@jaczjill Because a and b are not the same number. – Conor Pender Apr 3 '14 at 11:24
@jaczjill the areas of a and b are arbitrary. They could be any size and it would make no difference. It's just an example. – theyetiman Apr 4 '14 at 8:29
Have you seen the Montessori binomial cube? – MJD Apr 4 '14 at 14:30

When I understood Fourier series visually-

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OH. Wow. I just took my signals course and even after I did well on the exam I still didn't actually UNDERSTAND fourier series. Thanks, wow. – VicVu Apr 2 '14 at 22:25
This diagram is a particularly good illustration of the Gibbs Phenomenon, too. Nice! – Steven Stadnicki Apr 4 '14 at 6:09
I don't understand what this is showing: is it showing hte approximation of a square wave by a Fourier series? – ShreevatsaR Apr 5 '14 at 5:18
@ShreevatsaR Yes. It's an illustration of the Gibbs Phenomenon. – Potato Apr 5 '14 at 6:01

A visual explanation of a Taylor series:

$f(0)+\frac {f'(0)}{1!} x+ \frac{f''(0)}{2!} x^2+\frac{f^{(3)}(0)}{3!}x^3+ \cdots$

or

$f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots$

When you think about it, it's quite beautiful that as you add each term it wraps around the curve.

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It is interesting to note that Taylor series are relatively very precise near their center, which makes them especially useful in computer science when you need a precise but still fast approximation. For example, processors do not have a cosine function built-in, but will fly through an equation such as x => 1 - (x^2)/(2!) + (x^4)/(4!) - ... and from this equation, you could build a function that accepts a precision parameter so that the call recurses until higher accuracy is achieved. (note: there are many more efficient ways to do this) – sleblanc Apr 3 '14 at 21:55
@sebleblanc It is just amazing that petty much any function can represented with a Taylor series locally – joshlk Apr 4 '14 at 8:15
@sebleblanc Should have said "[some] processors" instead. x86 processors have had a cosine function for some time now with the FCOS instruction. It doesn't do it with a taylor series as far as I can tell. What I assume it does is use a lookup table followed by some sum/difference formulas. – Cole Johnson Apr 4 '14 at 16:05
@sebleblanc The x86/87 FPU uses CORDIC which lets you compute trigonometric functions in hardware using only bit shifts and addition. – Calmarius Aug 27 '14 at 16:57

This visualisation of the Fourier Transform was very enlightening for me:

The author, LucasVB, has a whole gallery of similar visualisations at his Wikipedia gallery and his tumblr blog.

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Sweet!$\textbf{}$ – Potato Apr 5 '14 at 6:01
@gekkostate You're a bit off. f is the sum of multiple simple waves, all with different frequencies and phase angles. The fourier transform takes a complex wave from a given time period, and gives you the phase angles and frequencies of all of the component waves. f^ is the amplitude of each component wave. – Joel Apr 7 '14 at 2:15
I'm down-voting this because this animation doesn't include phase information. The phase is the difference between a pulse and cw. If you don't want it, you should use another basis (wavelets). – Mikhail Apr 7 '14 at 3:19
this is pretty cool too: bl.ocks.org/jinroh/7524988 – ilia choly Apr 7 '14 at 20:54
I saw this animation a long time ago, and it helped me finally understand what the Fourier transformation does, but I still am lost as to how it works... – Cole Johnson Apr 11 '14 at 1:21

Topology needs to be represented here, specifically knot theory. The following picture is from the Wikipedia page about Seifert Surfaces and was contributed by Accelerometer. Every link (or knot) is the boundary of a smooth orientable surface in 3D-space. This fact is attributed to Herbert Seifert, since he was the first to give an algorithm for constructing them. The surface we are looking at is bounded by Borromean rings.

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I do not know much about topology so I will take your word for it that this is a beautiful idea/concept. However this picture and your description explain nothing to me. It seems like you missed the "easy to explain" bit in the question. – dfc Apr 7 '14 at 5:25
@dfc I don't know, it seems like you can convey most of the meat here using soap bubbles. – Slade Apr 7 '14 at 20:10

This animation shows that a circle's perimeter equals to $2r*\pi$. As ShreevatsaR pointed out, this is obvious because $\pi$ is by definition the ratio of a circle's circumference to its diameter

In this image we can see how the ratio is calculated. The wheel's diameter is 1. After the perimeter is rolled down we can see that its length equals to $\pi$ amount of wheels.

Source

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But can you show that a sphere with radius, $r$, has a volume of $V_3(r) = \frac{4}{3} \pi r^3$ and a surface area of $SA_3(r) = 4 \pi r^2$ – Cole Johnson Apr 4 '14 at 16:09
That the circle's perimeter is $2\pi r$ is the definition of $\pi$, so I wouldn't say this is an explanation of the fact; rather it's an illustration of what the definition means, and that the value of $\pi$ is about $3.1$. – ShreevatsaR Apr 5 '14 at 5:24

The sum of the exterior angles of any convex polygon will always add up to $360^\circ$.

This can be viewed as a zooming out process, as illustrate by the animation below:

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I always found the easiest way to think about this was that if you move around a convex shape such that you eventually end up at the start, you must have rotated through exactly 360º. If you walk around any shape, and count anticlockwise rotations as negative, but clockwise rotations as positive, I imagine the angles will still add to 360º (or -360º depending on which direction you take). – daviewales Apr 5 '14 at 12:27
I like to view this as a limiting process. Imagine that the picture on the right is the picture on the left zoomed out a great distance. – Steven Gubkin Apr 5 '14 at 14:10
This would be better as an animation (which achieves the zooming effect). – Duncan Apr 7 '14 at 10:31

A visual display that $0^0=1$. The following is a tetration fractal or exponential map with a pseudo-circle shown in orange. The red area is period 1 and contains 1. Example is $1^1=1$. The orange pseudo-circle which contains 0 is period two. Example is $0^0=1, 0^1=0$.

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0^0 = 1 is not even proven, it was just defined that way to make things easier, right? – Michael Apr 4 '14 at 18:59
It depends on the context, but in the grand scheme of mathematics, it is considered undefined. This is because there are completely logical steps that point to it being zero, and equally logical ones that point to it being one. We can't call it both zero and one, so we call it undefined. – recursive recursion Apr 4 '14 at 21:41
There is one problem with this. I have no idea how you make that image mean what you say it means... Visually stunning? Yes. Easy to explain? Maybe. Explained? No. – daviewales Apr 5 '14 at 12:31
So, er, how does this show that $0^0=1$? – David Richerby Apr 5 '14 at 17:55
$x^0 = 1$ because the multiplicative identity is one, $0^x = 0$ because zero is the multiplicative fixed point. That is, if and only if we never, ever, ever tussle with fractional powers; then and only then exponentiation is shorthand for repeated multiplication and therefore $0^0 = 1$. If we at any point use fractional powers, then exponentiation is shorthand for it's natural definition: $a^b = e^{b \ln a}$ and we all know that $\ln 0$ is undefined. There is no dispute, once we make the assumptions explicit the problem goes away. – Karl Damgaard Asmussen Apr 13 '14 at 9:16

Similarly to eykanal's answer, although demonstrating some interesting facts about medians and geometry as well. It demonstrates that $\displaystyle\sum_{n = 1}^{\infty}\frac{1}{2^n} = 1$:

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I think this one better. In the square one you have to make a choice between dividing horizontally or vertically. With the triangle there is only one dividing cut. – nayrb Apr 5 '14 at 13:55
@S4M: The area of the first triangle is defined as $1/2$; how is the calculation wrong when there is no calculation... only definition? – Andrew Coonce Apr 7 '14 at 16:51
@AndrewCoonce I was wrong in my previous comment. I asscumend that the side of the main triangle was $1$. It's all fine if its side is $\frac{\sqrt{2}}{2}$. – S4M Apr 9 '14 at 12:27

When I look up "area of a rhombus" on Google images, I find plenty of disappointing images like this one:

which show the formula, but fail to show why the formula works. That's why I really appreciate this image instead:

which, with a little bit of careful thought, illustrates why the product of the diagonals equals twice the area of the rhombus.

EDIT: Some have mentioned in comments that that second diagram is more complicated than it needs to be. Something like this would work as well:

My main objective is to offer students something that encourages them to think about why a formula works, not just what numbers to plug into an equation to get an answer.

As a side note, the following story is not exactly "visually stunning," but it put an indelible imprint on my mind, and affected the way I teach today. A very gifted Jr. High math teacher was teaching us about volume. I suppose just every about school system has a place in the curriculum where students are required learn how to calculate the volume of a pyramid. Sadly, most teachers probably accomplish this by simply writing the formula on the board, and assigning a few plug-and-chug homework problems.

No wonder that, when I ask my college students if they can tell me the formula for the volume of a pyramid, fewer than 5% can.

Instead, building upon lessons from earlier that week, our math teacher began the lesson by saying:

We've learned how to calculate the volume of a prism: we simply multiply the area of the base times the height. That's easy. But what if we don't have a prism? What if we have a pyramid?

At this point, she rummaged through her box of math props, and pulled out a clear plastic cube, and a clear plastic pyramid. She continued by putting the pyramid atop the cube, and then dropping the pyramid, point-side down inside the cube:

She continued:

These have the same base, and they are the same height. How many of these pyramids do you suppose would fit in this cube? Two? Two-and-a-half? Three?

Then she picked one student from the front row, and instructed him to walk them down the hallway:

Go down to the water fountain, and fill this pyramid up with water, and tell us how many it takes to fill up the cube.

The class sat in silence for about a full minute or so, until he walked back in the room. She asked him to give his report.

"Three," he said.

She pressed him, giving him a hard look. "Exactly three?"

"Exactly three," he affirmed.

Then, she looked around the room:

"Who here can tell me the formula I use to get the volume of a pyramid?" she asked.

One girl raised her hand: "One-third the base times the height?"

I've never forgotten that formula, because, instead of having it told to us, we were asked to derive it. Not only have I remembered the formula, I can even tell you the name of boy who went to the water fountain, and the girl who told us all the formula (David and Jill).

Given the upvoted comment, If high school math just used a fraction of the resources here, we'd have way more mathematicians, I hope you don't mind me sharing this story here. Powerful visuals can happen even in the imagination. I never got to see that cube filling up with water, but everything else in the story I vividly remember.

Incidentally, this same teacher introduced us to the concept of pi by asking us to find something circular in our house (“like a plate or a coffee can”), measuring the circumference and the diameter, and dividing the one number by the other. I can still see her studying the data on the chalkboard the next day – all 20 or so numbers just a smidgeon over 3 – marveling how, even though we all probably measured differently-sized circles, the answers were coming out remarkably similar, “as if maybe that ratio is some kind of constant or something...”

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^^1 million upvotes for the stories about maths education. I am writing both of these down for future reference. In fact, I think we should start a website with instructions to help primary teachers teach mathematical concepts in interesting ways. – daviewales Apr 5 '14 at 12:34
@daviewales It's been started - the mathematics educators SE. – Sanath K. Devalapurkar Apr 5 '14 at 14:45
Great stories. I do, however, think your first rhombus is fine. I actually find it easier to understand the result using that image than the second. Either diagonal splits the rhombus into two triangles, each of area $$\frac{1}{2}\frac{d_1}{2}d_2=\frac{1}{2}d_1\frac{d_2}{2}.$$ (Is there some aspect-ratio issue with your image? The white rectangle should be a square, but doesn't look like it on my screen.) – Will Orrick Apr 6 '14 at 2:36
The first rhombus picture is fine. Just draw a rectangle around it, and it's plain to see that the four triangles that form the rhombus cover half of the rectangle. In contrast, the second picture is so confusing that you yourself got it wrong: you write that it "illustrates why the product of the diagonals equals half of the area of the rhombus", while the ratio is the opposite. – LaC Apr 7 '14 at 3:47
I like the pyramid story! There is a more geometrical demonstration that 3 pyramids make a cube:lapasserelle.com/online_courses/maths6/3_pyramids_in_a_cube/… And by slicing pyramids with many cuts parallel to the base plane, it can be shifted, much like a rectangular solid loaf of bread can be shifted to approximate a parallelepiped. The more slices, the better the approximation. This slicing and shifting can also make a parallegram into a rectangle or any triangle into half a rectangle. – HopDavid Apr 7 '14 at 15:22

As I was in school, a supply teacher brought a scale to lesson:

He gave us several weights that were labeled and about 4 weights without labels (let's call them $A, B, C, D$). Then he told us we should find out the weight of the unlabeled weights. $A$ was very easy as there was a weight $E$ with weight($A$) = weight($E$). I think at least two of them had the same weight and we could only get them into balance with a combination of the labeled weights. The last one was harder. We had to put a labeled weight on the side of the last one to get the weight.

Then he told us how this can be solved on paper without having the weights. So he introduced us to the concept of equations. That was a truly amazing day. Such an important concept explained with a neat way.

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@MeemoGarza although drawings are nice, too, I think that an actual device has some advantages: Pupils rather get excited by a real device than by a drawing; it's easier to accept something that you can touch and play with where you instantly get feedback if what you did was correct; as soon as pupils got familiar with the concept of equations, you can explain the physics behind scales ($a_1 \cdot m_1 = a_2 \cdot m_2$ where $a$ is the length of the part of the scale and $m$ is the mass on that side). That way students can see that math is also about describing the real world in a formal way. – Martin Thoma Apr 7 '14 at 5:58
I agree completely, but I don't have the apparatus. I'll look into purchasing one before the next semester. Thanks for helping me make up my mind! – Guillermo Garza Apr 8 '14 at 0:50

Check out A Mathematical Picture Book at your local library - it has a bit about the Szegő Curve.

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Fourier transform of the light intensity due to a diffraction pattern caused by light going through 8 pinholes and interfering on a wall, for different choices of parameter:

The best thing about them is, they satisfy periodic boundary conditions, and so you can pick one of them and set it as a desktop background by tiling it, resulting in a far more spectacular image than just the single unit cells posted above!

The images seem to be a vast interconnected network of lines once you tile them, but in fact the entire picture is actually just a single circle, which has been aliased into a tiling cell thousands of times.

Here is a video of the first couple thosand patterns: http://www.youtube.com/watch?v=1UVbUWuyNmk

Here is the Mathematica code used to generate and save the images. There are two parameters that are adjustable: mag is the magnification and must be an integer, with 1 generating 600 by 600 images, 2 generating 1200 by 1200 images, etc. i is a parameter which can be any real number between 0 and ~1000, with values between 0 and 500 being typical (most of the preceding images used i values between 200 and 300). By varying i, thousands of unique diagrams can be created. Small values of i create simple patterns (low degree of aliasing), and large values generate complex patterns (high degree of aliasing).

$HistoryLength = 0; p = {x, y, L}; nnn = 8; q = 2.0 Table[{Cos[2 \[Pi] j/nnn], Sin[2 \[Pi] j/nnn], 0}, {j, nnn}]; k = ConstantArray[I, nnn]; n[x_] := Sqrt[x.x]; conjugate[expr_] := expr /. Complex[x_, y_] -> x - I y; a = Table[k[[i]]/n[p - q[[i]]], {i, nnn}]; \[Gamma] = Table[Exp[-I \[Omega] n[p - q[[i]]]/c], {i, nnn}]; expr = \[Gamma].a /. {L -> 0.1, c -> 1, \[Omega] -> 100}; ff = Compile[{{x, _Real}, {y, _Real}}, Evaluate[expr], CompilationTarget -> "C", RuntimeAttributes -> {Listable}]; i = 250; mag = 1; d = 6 i mag; \[Delta] = 0.02 i; nn = Floor[Length[Range[-d, d, \[Delta]]]/2]; A = Compile[{{x, _Integer}, {y, _Integer}}, Exp[I (x + y)], CompilationTarget -> "C", RuntimeAttributes -> {Listable}] @@ Transpose[ Outer[List, Range[Length[Range[-d, d, \[Delta]]]], Range[Length[Range[-d, d, \[Delta]]]]], {2, 3, 1}]; SaveImage = Export[CharacterRange["a", "z"][[RandomInteger[{1, 26}, 20]]] <> ".PNG", #] &; {#, SaveImage@#} &@ Image[RotateRight[ Abs[Fourier[ 1 A mag i/ nnn ff @@ Transpose[ Outer[List, Range[-d, d, \[Delta]], Range[-d, d, \[Delta]]], {2, 3, 1}]]], {nn, nn}], Magnification -> 1]  - Those images are incredibly beautiful, but can you explain what exactly they represent and how they were generated? "Fourier transform of the light intensity due to a diffraction pattern caused by light going through 8 pinholes and interfering on a wall" isn't very clear for me. – gregschlom Apr 7 '14 at 9:11 @Rahul: It's aliasing of a circle. Aliasing is easy to explain. Draw a big circle on a clear plastic sheet. Cut the image into little squares. Stack the squares on top of each other, and look at it. That's the image. The different images above were done using little squares with various side-lengths. I can post the code if you'd like, there are literally tens of thousands of visually distinct diagrams which can be formed. – DumpsterDoofus Apr 8 '14 at 1:17 Could you post a link to the code here? – Kevin Hwang Apr 9 '14 at 18:46 I've built a bunch of interactive explorations over at Khan Academy. A few of my favorites are: • Derivative intuition. Particularly amazing is seeing how$\frac{d}{dx}e^x=e^x$. (Do a few and it should pop up). • Exploring mean and median. Light bulbs are twice as likely to burn out before the average lifetime printed on the package. If that statement surprises you, this exploration points out that mean and median aren't the same thing. • Exploring standard deviation. Standard deviation is a term that gets thrown around a lot. Play around with this to get a more intuitive sense of what it means. • One step equation intuition. Basic introduction to why you can do the same thing to both sides of an equation to solve it. - Here is a very insightful waterproof demonstration of the Pythagorean theorem. Also there is a video about this. - This does not actually demonstrate the pythagorean theorem. It is showing that for one single right triangle, the theorem holds. This could be a coincidence. Someone seeing this might think that 3,4,5 is the only pythagorean triple. – Sparr Apr 7 '14 at 16:06 Sparr, you're incorrect. This does indeed demonstrate the pythagorean theorem. What you mean is that it doesn't prove it. All the demonstrations on this page are simply examples, not the infinity of all possibilities. Think before you speak. And all you upvoters, think before you upvote. – B T Aug 27 '14 at 21:12 @BT, do you consider the fact that$1^3 + 2^3 \neq 18^3$a demonstration of Fermat's Last Theorem? – Daniel McLaury Aug 28 '14 at 1:17 @DanielMcLaury Sure it's a demonstration of FLT, but it's not an interesting one. This gif, however, is interesting because it looks cool. – Samuel Yusim Aug 28 '14 at 5:30 @GrumpyParsnip Before you learn why something is true, you might want to do a sanity check to verify that it probably is. This does that, while looking cool. The only thing that I really care about is that it looks cool, though. Mathematics is a very aesthetic thing, so I'm allowed to like something just because it looks nice. – Samuel Yusim Sep 5 '14 at 0:06 I made this a while back to explain sine and cosine. http://math.garbl.es/demos/sine-and-cosine - Great stuff! I'll be using that on my CG classes from now on. – karlphillip Jan 9 '15 at 11:53 How about a line integral of a scalar field by http://1ucasvb.tumblr.com: - Just wanted to point out that The Book of Numbers has a lot of the examples above (as well as many others). - This is from betterexplained.com. It's a really cool website with lots of intuitive explanations of maths concepts. This helped me understand Pythagoras' theorem. Actually my go-to website for intuitive explanations of concepts. These are similar triangles. This diagram also makes something very clear: Area (Big) = Area (Medium) + Area (Small) Makes sense, right? The smaller triangles were cut from the big one, so the areas must add up. And the kicker: because the triangles are similar, they have the same area equation. Let's call the long side c (5), the middle side b (4), and the small side a (3). Our area equation for these triangles is: Area = F * hypotenuse^2 where F is some area factor (6/25 or .24 in this case; the exact number doesn't matter). Now let's play with the equation: Area (Big) = Area (Medium) + Area (Small) F c^2 = F b^2 + F a^2 Divide by F on both sides and you get: c^2 = b^2 + a^2 Which is our famous theorem! You knew it was true, but now you know why. This explains the product rule: - A nationwide math contest in Germany recently came up with a task that I found beautiful to explain, because of two points. 1. You can get an idea, what the proof is, without applying mathematically accurate theory and this intuitive proof is most likely the right way. 2. At any given point of this intuitive proof, you can chime in and ask yourself: But how would I say this in mathematical terms? When you find these terms, eventually you get the proof you were looking for. So here you go: Lea gets the task to write down 2014 numbers. These numbers have to fulfil a specification. For every set of three numbers from that whole set, the arithmetic average of these three must also be within the whole set of 2014 numbers. Your task is to proof, that Lea has to write down the same number 2014 times. Every set of 2014 numbers with any variation in it would not fulfil the specifications. So since we are talking about layman maths here, I'll go with the intuitive way. We have to find a reason, why choosing a set with different numbers would violate the specifications and we have to proof that always taking the same number would not violate them. The later one is rather easy. Take any arbitrary number three times. The arithmetic average will be the same number, which is in your set already. That wasn't too bad, right? But what about sets with not all the same numbers? We are mathematicians, so we'll just do what we always do: Chop the problem into pieces we can solve. The first piece is where we have two equal numbers and one other number in our set. Let's assume, the single number is bigger than the two equal numbers. What will that do to our arithmetic average? Right, it will be below the middle between the bigger and the smaller number. We can write that arithmetic average down and specifications are ok. But now we have created another set of three numbers. The single, big number (I'll call it a), one of the two equal numbers (that would be b) and the arithmetic average of a, and b (I'll call that one c). So now we would have to also add the arithmetic average of a, b and c. A quick sketch will show you, that this new number is also slightly below the middle between a and b. And like that we will always have to add a new number. The arithmetic average of a, b and the new number will never reach the middle. Something, that you can also verify with a few sketches. So we would have to add infinitely more numbers, but we wanted only 2014. Apparently, no two numbers can be equal. So what if all numbers are different? There is one special case. Let's call our numbers a, b and c again. If b is equally far away from a and c (so b could be 3, a could be 1, then c would be 5). In that case, b is the arithmetic average. But we have to have 2014 different numbers. As soon as we add a fourth number d, it's spoiled. d could be 7, to be still in a distance of 2 to c, but then the set a, b and d would not contain its own arithmetic average. So we know, that within a set of 2014 numbers, we would have sets of three, where these three numbers don't include their own arithmetic average, no matter what we do. And now we look back at our idea about the set with two equal numbers. We see: As soon as we have a bigger and a smaller number and the number in between those is not exactly in the middle, we can once again start with our endless construction of arithmetic averages. We always replace the number between the bigger and the smaller one by the arithmetic average of the three and we can never reach the middle, but it will always get closer to the middle (thus be another number). And as I said, making this proof mathematical will not alter it. It will be all the same, but with more equations and sequences. Since we excluded the option of making anything infinite, it is correct as it stands here. This one made me realize: Proofs are not the miracles or the magic they seemed to be for me during high school. Of course, there are hard proofs (and things you can't proof, there is a proof for that), but often you only have to think clearly and to chop the problem into the right pieces. - Hmm, I don't think "induction by contradiction" ("contradiction by induction"?) is valid even though our intuition would like it. In other words, you've shown that you can't inductively build a set with the desired property, having smaller sets along the way that also have the property. But this doesn't exclude the possibility that the property might not hold for all smaller sets, yet 2014 is the first time you have a set large enough for the property to hold. – Travis Bemrose Apr 13 '14 at 20:36 Sort the list in non-decreasing order as$a_1, a_2, \ldots, a_{2014}$. Since the average of$a_1$,$a_2$, and$a_3$is among these, it must equal$a_2$, so$a_2 = a_1 + d$and$a_3 = a_1 + 2d$for some$d$. Since the average of$a_2$,$a_3$, and$a_4$is among the list, is must equal$a_3$. Proceeding in this manner we see that$a_k = a_1 + (k-1)d$for some$d$. Now note that the average of$a_1$,$a_2$, and$a_4$, which is$a_1 + \frac{4}{3} d$, is also in the list. It follows that$d = 0$. – Daniel McLaury Aug 28 '14 at 1:36 This is a neat little proof that the area of a circle is$\pi r^2$, which I was first taught aged about 12 and it has stuck with me ever since. The circle is subdivided into equal pieces, then rearranged. As the number of pieces gets larger, the resulting shape gets closer and closer to a rectangle. It is obvious that the short side of this rectangle has length$r$, and a little thought will show that the two long sides each have a length half the circumference, or$\pi r$, giving an area for the rectangle of$\pi r^2$. This can also be done physically by taking a paper circle and actually cutting it up and rearranging the pieces. This exercise also offers some introduction to (infinite) sequences. - This doesn't work for me :( Sadly, I find it non-obvious that the length of the bottom really will converge helpfully to$\pi r$. – Nicholas Wilson Apr 8 '14 at 16:50 @NicholasWilson By definition, the circumference of a circle is π times the diameter of the circle. Here, we've cut the circle into slices, half flipped one way, half flipped the other. Therefore half the circumference (πd) is on the bottom. Half of the diameter is the radius, r. Does that help? – ghoppe Apr 8 '14 at 19:29 @ghoppe Nice try! But it's all wiggly. We're lucky that$\sin{\theta} \sim \theta$as$\theta \rightarrow 0$, so the actual perimeter of the bottom (which is known to be$\pi$) does converge to the width of the rectangle. But - that nice property of$\sin\$ is exactly what we're trying to prove! – Nicholas Wilson Apr 9 '14 at 8:20
@ghoppe What would you visually infer about the perimeter of a Koch snowflake? I've had people assure me "It must be bounded!" Visual inferences are susceptible to error :( You need to invoke some limit arguments to convince me those arcs on the segments do converge. Similarly, inscribing polygons to determine the circumference sounds very dangerous (think what would happen if you tried that with a Koch snowflake) -- but bounding the area above and below by inscribed/exscribed polygons is definitely a sound proof. – Nicholas Wilson Apr 9 '14 at 17:56
@NicholasWilson coming back to this discussion rather late, but perhaps I can convince you. The length of the bottom is always πr, it doesn't change with iteration so all we have to show is that the shape ends up as a rectangle. This is equivalent to stating that the line between two points on a circle tends to the tangent as the two points get closer together, which I believe is at least one definition of the tangent. – Ben Rowland Dec 16 '14 at 22:16

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