# Visually stunning math concepts which are easy to explain [closed]

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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## closed as too broad by Arthur Fischer♦Apr 26 at 14:09

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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......................? –  LTS 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. –  Normal Human Aug 29 '14 at 18:37

Steven Wittens presents quite a few math concepts in his talk Making things with math. His slides can be found from his own website.

For example, Bézier curves visually:

He has also created MathBox.js which powers his amazing visualisations in the slides.

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These are beautiful! –  Cole Johnson May 15 '14 at 23:16
Also, if you read the text out loud you get to say "lerp lerp lerp lerp", which has its own aesthetic appeal. –  msouth Sep 23 '14 at 21:43

Proof that the area of a circle is πr² without words: Proof Without Words: The Circle

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@zinking No, $\pi$ is defined to be the constant that goes in that place in that equation to make it hold. However, I was rather dissatisfied with this "proof". There's lots of distortion involved with deforming the area of the circle into the area of the triangle, that one would have to know calculus to understand why the distortion doesn't matter (at which point, why not just calculate the integral). This is more of a memorization technique to remember the formula. –  Travis Bemrose Apr 13 '14 at 20:02

Take a look at this great example of Fourier series visualisations written in JavaScript.

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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:

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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.

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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

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) –  sebleblanc Apr 3 '14 at 21:55
@sebleblanc It is just amazing that petty much any function can represented with a Taylor series locally –  Josh 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

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

One of my favorites - I've seen it somewhere on the web but can't find it again now, so had to reconstruct myself. It is not as pretty but suffices to convey the idea.

It gives good grasp both for $e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n$ and for $e^{2k\pi i}=1$

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This one (via Proof Without Words) is wonderful but not immediately obvious. Ponder on it and you'll find out how fantastic it is when you get it.

Explanation:
Set the radius to be $1$, then $$HK=2HI=2\cos\frac{\pi}{7}$$ $$AC=2AB=2\cos\frac{3\pi}{7}$$ $$DG=2DF=-2\cos\frac{5\pi}{7}$$ So \begin{align} 2(\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7})&=HK+AC-DG\\ &=HK-(DG-AC)\\ &=HK-(DG-DE)\\ &=HK-EG\\ &=HK-JK\\ &=HJ\\ &=LO\\ &=1 \end{align}

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This is what happens when you take Pascal's Triangle, and color each entry based on the value modulo 2:

The exact code for this is extremely simple:

def drawModuloPascal(n, p):
for i in range(0, n + 1):
print " " * (n - i) ,
for k in range(0, i + 1):
v = choose(i , k) % p
print '\x1b[%sm ' % (';'.join(['0', '30', str(41 + v)]), ) ,
print "\x1b[0m" # reset the color for the next row

Just provide your own choose(n, r) implementation. The image above is a screenshot of drawModuloPascal(80, 2).

You can also do this modulo other primes, to get even more remarkable patterns, but then it becomes much less "easy to explain."

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I'd also note that it's possible to compute ${i \choose k} \mod p$ without computing $i \choose k$. For large $i$ this would matter. –  Michael Lugo Jan 9 at 14:11
The basic idea is pretty simple: ${i \choose k} = {i \choose k-1} + {i-1 \choose k-1}$, and this recurrence holds $\mod p$ as well. –  Michael Lugo Jan 9 at 14:31
@Coffee_Table: It's literally just the terminal. The code I pasted above write ANSI color codes to the terminal to produce the colored blocks you see above. –  Adrian Petrescu Feb 10 at 22:40
right, thanks. I edited your code to work for Python 3 and I realize now that I made a stupid error when doing so. –  Coffee_Table Feb 10 at 23:08

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.

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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

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

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. –  moose 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

Visualisation in ancient times: Sum of squares

Let's go back in time for about 2500 years and let's have a look at visually stunning concepts of Pythagorean arithmetic.

Here's a visual proof of

\begin{align*} \left(1^2+2^2+3^2\dots+n^2\right)=\frac{1}{3}(1+2n)(1+2+3\dots+n) \end{align*}

The Pythagoreans used pebbles arranged in a rectangle and linked them with the help of so-called gnomons (sticks) in a clever way. The big rectangle contains $$(1+2n)(1+2+3\dots+n)$$ pebbles. One third of the pebbles is grey, two-thirds are black. The black thirds contain squares with

$$1\cdot1, 2\cdot2, \dots,n\cdot n$$

pebbles. Dismantling the black squares into their gnomons shows that they appear in the grey part. According to Oscar Becker: Grundlagen der Mathematik this proof was already known to the Babylonians (but also originated from hellenic times).

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Math is always fun to learn. Here are some of the images that explain some things beautifully visually

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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. –  user122283 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
Dave and Jill went down the hall / To fill a cube up with water / A pyramid required three trips / And thus they never forgot 'er. –  Blazemonger Jan 8 at 19:49

A very satisfying visualization of the area of a circle.

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I like this, but one tiny criticism is that it's not visually obvious why the unrolled rings should form a triangle. It's obvious if you can see that the unrolled length of the rings is linearly proportional to their radii, but not visually obvious. –  Lqueryvg Sep 1 '14 at 10:52
Is it obvious that you can straighten a ring out and get a rectangle when the ring started out bent? I think a lot of these "bang! pi r squared!" ones leave you with real analysis-induced heebie jeebies :) –  msouth Sep 26 '14 at 0:27

This one is only visually stunning in your imagination, but I like it. The derivative of a circle w.r.t. the radius is the circumference. $$\frac{d}{dr}\pi r^2=2\pi r$$ The derivative of a sphere w.r.t. the radius is the area. $$\frac{d}{dr}\frac{4}{3}\pi r^3=4\pi r^2$$ The derivative of a 4-dimensional sphere w.r.t. the radius is the 3-dimensional area. $$\frac{d}{dr}\frac{1}{2}\pi^2 r^4=2\pi^2 r^3$$ This works because the radius is invariant in n-dimensional spheres. Holding a circle, a sphere or a hypersphere requires your hands to be the same distance apart.

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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

Here is a very insightful waterproof demonstration of the Pythagorean theorem. Also there is a video about this.

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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
@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
@DanielMcLaury Yes indeed I do, tho that demonstration doesn't further anyone's understanding, whereas the above video does. –  B T Aug 28 '14 at 6:29
@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 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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+1 Just brilliant –  Silviu Burcea Apr 1 '14 at 11:45
@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

Allow me to join the party guys...

This is another proof of the Pythagorean theorem by The 20th US President James A. Garfield.

A nice explanation about Garfield's proof of the Pythagorean theorem can be found on Khan Academy.

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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

Logarithmic spiral and scale:

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@rschwieb They're made by 1ucasvb, he's got an faq here: 1ucasvb.tumblr.com/faq –  Kristoffer Ryhl Aug 13 '14 at 14:40
I can't believe how long I used radians in college without realizing what a radian really is. –  greg7gkb Jan 9 at 0:28

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 An iron pendulum is suspended above a flat surface, with three magnets on it. The magnets are colored red, yellow and blue. We hold the pendulum above a random point of the surface and let it go, holding our finger on the starting point. After some swinging this way and that, under the attractions of the magnets and gravity, it will come to rest over one of the magnets. We color the starting point (under our finger) with the color of the magnet. Repeating this for every point on the surface, we get the image shown above. - @DavidZhang, the longer the pendulum swings, the darker the point. I don't know the exact function used for this image, but it you click on the link, you'll find the algorithm. – Peter Apr 21 '14 at 15:27 Check out the "Proofs Without Words" gallery (animated) here: http://usamts.org/Gallery/G_Gallery.php And the related proofs here: http://www.artofproblemsolving.com/Wiki/index.php/Proofs_without_words Many of these are similar to the ones already listed here, but you get a bunch in one place. - Here's a GIF that I made that demonstrates Phi (golden number) - 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. - 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

## protected by GanymedeDec 23 '14 at 4:02

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