Visually stunning math concepts which are easy to explain

up vote 802 down vote favorite
950

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?

soft-question education big-list visualization
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see also, fractals – Sabyasachi Mar 31 '14 at 11:55
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It looks like mathpop or demand for math entertainment) – rook Apr 2 '14 at 12:59
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@ColeJohnson the 'transcendentality" is the beauty of it! – Sabyasachi Apr 4 '14 at 17:07
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Why has this got 105k views......................? – bwv869 Apr 8 '14 at 16:49
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@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. – user147263 Aug 29 '14 at 18:37

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up vote 665 down vote

circle trig animation

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
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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
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Is the source of this animation available? (It looks like it's in $\mathrm\TeX$.) – bb010g Apr 3 '14 at 23:31
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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
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@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

up vote 608 down vote

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:

enter image description here

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
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I still don't believe you. – Ojonugwa Ochalifu Apr 2 '14 at 8:51
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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
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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
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@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

up vote 418 down vote

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

enter image description here

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
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@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
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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
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this is pretty cool too: bl.ocks.org/jinroh/7524988 – ilia choly Apr 7 '14 at 20:54
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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

up vote 414 down vote

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

enter image description here

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.

enter image description here

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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
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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
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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
    
I have noticed an interesting method to calculate a square root based on this fact: disqus.com/home/discussion/geeksforgeeks/… – sergiol Sep 7 '15 at 23:46

up vote 329 down vote

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

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

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Always liked this one – Almo Apr 2 '14 at 14:11
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why area of a^2 is greater tham area of b^2? – jaczjill Apr 3 '14 at 7:34
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@jaczjill Because a and b are not the same number. – Conor Pender Apr 3 '14 at 11:24
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@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
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Have you seen the Montessori binomial cube? – MJD Apr 4 '14 at 14:30

up vote 279 down vote

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

enter image description here

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

enter image description here

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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
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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
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This would be better as an animation (which achieves the zooming effect). – Duncan Apr 7 '14 at 10:31
    
Edited. Do have a look. (It may look horrible, since this is the first time I made an animation.) – Yiyuan Lee Apr 7 '14 at 13:36
    
The animation is worse (I didn't even know what it was supposed to be proving until I looked in the edit history), because it doesn't mark the angles the way your other images did. Maybe if you put both the animation and the other images? – Muhd Apr 11 '14 at 21:41

up vote 207 down vote

While attending an Abstract Algebra course I was given the assignment to write out the multiplication table modulo n. I forgot to do the homework until just before class, but it was so easy to write the program I was able to print the result between classes.

The circular patterns in the tables fascinated me, and compelled me to replace the numbers with colors. The result is a beautiful illustration showing the emergence of primes and symmetry of multiplication.

The colors were chosen to start blue at 1 (cold) and fade to red at n (hot). White is used for zero (frozen), because it communicates the most information about prime factorization.

The interactive version can be found here: https://web.archive.org/web/20140830110358/http://arapaho.nsuok.edu/~deckar01/Zvis.html

Multiplication of the integers modulo 15:

enter image description here

Multiplication of the integers modulo 512:

enter image description here

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This is awesome. Looks very much like a fractal. Is there a name for this fractal? – OmnipotentEntity Apr 7 '14 at 15:52
    
It doesn't have the properties needed to be classified as a fractal, but that doesn't stop us from interacting with it like one. arapaho.nsuok.edu/~deckar01/ZvisFractal.html – Jared Deckard Apr 7 '14 at 17:09
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@OmnipotentEntity it doesn't qualify as having fractional dimension because the underlying space is discrete. If you use a real space instead - the set of x, y such that x * y = 0 mod n, you get a set of hyperboles (linear in dimension). If you take a limit towards infinite squares, then the count of white squares is equal to the length of the edge whenever the length is prime. – Jan Dvorak Apr 7 '14 at 18:48
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Although not colored WolframAlpha also does a nice job of illustrating this: wolframalpha.com/input/?i=Z_200 – Vikram Saraph Apr 9 '14 at 23:05
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The link does not work anymore. – Keba Dec 25 '14 at 1:54

up vote 192 down vote

When I understood Fourier series visually-

Fourier series of square wave

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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
    
@Vee Same here :-) – Uwe Keim Apr 4 '14 at 5:43
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This diagram is a particularly good illustration of the Gibbs Phenomenon, too. Nice! – Steven Stadnicki Apr 4 '14 at 6:09
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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
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@ShreevatsaR Yes. It's an illustration of the Gibbs Phenomenon. – Potato Apr 5 '14 at 6:01

up vote 160 down vote

Subdividing circle

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
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@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
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@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
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@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
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@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

up vote 148 down vote

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

enter image description here

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

Jim Wilson University of Georgia

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:

enter image description here

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.

enter image description here

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:
enter image description here

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
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@daviewales It's been started - the mathematics educators SE. – user122283 Apr 5 '14 at 14:45
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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
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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
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@Peter - I'll admit that I'm using the word derive a bit loosely for a mathematics forum (it's a soft answer to a soft question), but it's too bad you see this as "dumbing down" and "a waste of time." Most 7th-grade teachers charged with teaching volume would simply write the formula on the board, and it would be well-forgotten by the end of summer. Her technique might have been weak insofar as mathematical rigor goes, but the pedagogy was very strong. I assure you, this woman was not one to "dumb down" anything; I remember re-learning concepts in 11th-grade that she taught us in Jr High. – J.R. Feb 22 '15 at 9:44

up vote 138 down vote

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

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

Taylor series gif

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

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10  
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
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@sebleblanc It is just amazing that petty much any function can represented with a Taylor series locally – joshlk Apr 4 '14 at 8:15
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@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
    
You are right that X87 co-processors are omni-present, but there are far more embedded CPUs in the world than there are X86/87 CPUs. Just think of how many devices you interact with that are not strictly computers. ///// From my desk, I can see three computer monitors, one TV, two video game consoles, one oscilloscope, one modern refrigerator, two cordless phones, a wireless mouse, a microcontroller dev board, a Raspberry Pi, my cell phone, my credit card. All of these have a processor that does not have the FSIN instruction... If anything, I should have said [most] instead. ;-) – sleblanc Apr 12 '14 at 14:31
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@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

up vote 134 down vote
+50

Simple answer for "what is a radian":

Logarithmic spiral and scale:

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What program produces these images? – rschwieb Aug 11 '14 at 19:49
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@rschwieb They're made by 1ucasvb, he's got an faq here: 1ucasvb.tumblr.com/faq – Alice Ryhl Aug 13 '14 at 14:40
    
@Darksonn Thanks! – rschwieb Aug 13 '14 at 14:49
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I can't believe how long I used radians in college without realizing what a radian really is. – greg7gkb Jan 9 '15 at 0:28
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The exponential curve has curvature-arc length relation like cycloidal? What does the animation show? – Narasimham Apr 9 '16 at 2:31

up vote 131 down vote

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

Source: Wikipedia

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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I do the same thing... except with drawings instead of an actual scale and weights. I need to get this set. – Guillermo Garza Apr 7 '14 at 0:34
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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
1  
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
    
@MeemoGarza Meanwhile you might be interested in this: walter-fendt.de/ph14d/hebel.htm - it's a site created by a physics teacher I had in school. This website contains lots of Java applets that show things in physics. – Martin Thoma Apr 8 '14 at 1:59

up vote 126 down vote

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 $\mathrm a^2+\mathrm b^2$.

enter image description here

Edit: comments requested an animation, so a simple gif is attached... enter image description here

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9  
I've always felt this one worked better in .gif format. – Patrick M Apr 3 '14 at 20:47
1  
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
1  
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
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@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
    
Anyway...THIS was my favourite, so, good that you already did it! – MattAllegro May 12 '14 at 21:49

up vote 122 down vote

enter image description here

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

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56  
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
66  
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
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@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
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@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
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@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

up vote 106 down vote

How about a line integral of a scalar field by http://1ucasvb.tumblr.com:

Line integral of a scalar field

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Awesome — this will be a great introduction to fields & the usage of integrals which I am about to explain to a friend of mine! ;) – Luke Aug 27 '14 at 17:44
    
Nice visual intuition, but why is it true that $|\mathbf{r}'|$ is the right integrator to use? (i.e. the 'flattening/projection to the real line' step) – enthdegree Aug 11 '16 at 5:18

up vote 97 down vote

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

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6  
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. – pasta Apr 5 '14 at 13:55
    
Just nitpicking, but you got the areas of the triangles wrong. For example, the one of are $\frac{1}{2}$ according to you has actually an area of $\frac{1}{2}\times\frac{\sqrt{2}}{2}\times\frac{\sqrt{2}}{2}=\frac{1}{4}$, so what this figure is proving is that $\sum_{n=2}^{\infty}\frac{1}{2^n}=\frac{1}{2}$. – S4M Apr 7 '14 at 7:12
2  
@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
    
@S4M: Each new triangle bisects the previous one. The full triangle has area 1. The $\frac{1}{4}$ triangle bisects the triangle of area $\frac{1}{2}$ that is left after the $\frac{1}{2}$ triangle and so on. I'm not trying to prove that the right side of the largest bisection line equals the other side (the summation you have), but that the entire collection of triangles approaches 1. – The Code Samurai Apr 8 '14 at 21:17
1  
@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

up vote 93 down vote

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.

Circle perimeter

Source

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4  
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
    
@ColeJohnson I don't know, just wanted to show this awesome animation:D – totymedli Apr 4 '14 at 21:11
    
I just wish someone could make an animation of that. I understand the math behind it, but animations are nice. – Cole Johnson Apr 4 '14 at 22:28
36  
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
    
@ColeJohnson There's a neat way to find the volume of a sphere with Cavalieri's principle and Pythagoras. Just look up Cavalieri's principle on Wikipedia, I think it's on there. – Akiva Weinberger Feb 23 '15 at 1:45

up vote 85 down vote

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

An image of "Mandelbrot island".

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

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

An image of the Mandelbrot set.

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28  
Are complex numbers easy to explain? – adam.r Mar 31 '14 at 17:15
5  
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
6  
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
4  
@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
3  
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

up vote 82 down vote

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

Pythagorean theorem

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10  
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
5  
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
6  
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
50  
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
7  
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

up vote 78 down vote

The magnetic pendulum:

the magnetic pendulum fractal

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.

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How is the shading of the image given? – David Zhang Apr 21 '14 at 14:27
2  
@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
    
Deterministic chaos? – Peter Mortensen Mar 8 '15 at 11:51
2  
@PeterMortensen Absolutely. In fact, you can see how (in certain areas) a minute change in initial conditions can cause a drastic change in outcome. – Peter Mar 9 '15 at 12:36

up vote 78 down vote

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:

Linear interpolation

Enter image description here

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

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2  
These are beautiful! – Cole Johnson May 15 '14 at 23:16
11  
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
    
beautiful, but I don't understand XD – Kokizzu Apr 20 '15 at 11:20
    
See also codegolf.stackexchange.com/questions/21178/… – SztupY Jun 21 '16 at 9:24

up vote 66 down vote

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.

Seifert surface bounding Borromean rings

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38  
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
3  
@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

up vote 61 down vote

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:

enter image description here

enter image description here

enter image description here

enter image description here

enter image description here

enter image description here

enter image description here

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]
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32  
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
    
"Visually stunning math concepts which are easy to explain" – Rahul Apr 8 '14 at 1:06
3  
@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
    
This reminds me of the Radon transform and the Hough transform. – koo Apr 8 '14 at 23:59
1  
Could you post a link to the code here? – Kevin Hwang Apr 9 '14 at 18:46

up vote 60 down vote

Here's a GIF that I made that demonstrates Phi (golden number)

Phi demonstration

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That's Beautiful! – debap Sep 4 '16 at 12:36

up vote 55 down vote

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

Enter image description here

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Really amazing.... – Pacerier Apr 11 '14 at 16:49
    
What is this doing lurking at the bottom of the page? It's brilliant! – P i Feb 25 '15 at 19:39
1  
The link is brilliant! – David Simmons Jun 14 '15 at 9:50
4  
If you had captured an animated gif, this would be the top answer! – Tobia Jul 10 '15 at 0:14

up vote 45 down vote

A very satisfying visualization of the area of a circle.

enter image description here

enter image description here

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2  
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
10  
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
    
The rings get smaller as you move in, narrowing to a point. It doesn't matter what the proportion is. It's a simple fact Demonstated by two things. 1. The circles smoothly decrease in length. 2. They narrow to a point. A better way to do this image would be to stack the rings. Then it makes more sense.... Although one can always just rotate by 90 degrees. XD – TheGreatDuck Apr 12 '16 at 18:20

up vote 43 down vote

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

enter image description here enter image description here enter image description here enter image description here enter image description here

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In E-M wave propagation, I thought the E and M components were supposed to be $90$ degrees out of phase. The figure appears to have them in phase or $180$ degrees out of phase, depending on how you interpret the directions of the axes. – David K Aug 28 '14 at 21:25
    
@DavidK That's only for circularly polarized photons. This is illustrating a plane-polarized wave, whose components are in phase but offset $90^\circ$. – Mario Carneiro Jan 9 '15 at 1:42
    
@MarioCarneiro Quite right, a quick review of wave equations verify the diagram as shown above. I have this mental image of one of my high school teachers walking across the room waving one arm up and down and one arm side to side to simulate the $E$ and $B$ components, but apparently I don't actually remember it very well after all. – David K Jan 9 '15 at 2:56
6  
When I read the number 142857 it was like a dejavu for me and I was like "OH MY GOED!! I'VE KNOWN THIS NUMBER!" – Qwerty Jan 11 '15 at 4:20
1  
@NateGlenn It's really just a visualisation of long multiplication. (Try writing out the long multiplication of $13 \cdot 12$ and see how it corresponds to the diagram.) – Martin Ender Mar 16 '16 at 15:57

up vote 38 down vote

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):

Pythagoras' Theorem

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  
+1 finally i can prove Pythagoras theorem without looking into internet or books :) – lowtech Jul 31 '14 at 20:58
    
yes it is the most common proof, the easiest to remember – user1952009 Apr 22 '16 at 21:13

 

up vote 28 down vote

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*}

                                       enter image description here

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 red, two-thirds are blue. The blue thirds contain squares with

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

pebbles. Dismantling the blue squares into their gnomons shows that they appear in the red 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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27 down vote

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.

                                                enter image description here

It gives good grasp both for ex=limn→∞(1+xn)n and for e2kπi=1

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up vote 26 down vote

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

Pascal's Triangle 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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4  
I'd also note that it's possible to compute (ik)modp without computing (ik). For large i this would matter. – Michael Lugo Jan 9 '15 at 14:11
    
@MichaelLugo I did not know that! Thank you for giving me something interesting to read up on :) – Adrian Petrescu Jan 9 '15 at 14:17
5  
The basic idea is pretty simple: (ik)=(ik−1)+(i−1k−1), and this recurrence holds modp as well. – Michael Lugo Jan 9 '15 at 14:31
1  
@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 '15 at 22:40
1  
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 '15 at 23:08

up vote 23 down vote

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.

Pythagoras' theorem

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:

betterexplained

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This depends on the idea that the area of all shapes increases on the order of the square of a scale parameter. It's true but it takes a lot of machinery to prove if you don't already believe it to start with. – enthdegree Aug 11 '16 at 5:04

up vote 19 down vote

Francis Galton's Bean machine is interesting as it demonstrates Central Limit Theorem :

enter image description here

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1  
It's one of the best answers. – Seifolahi Mar 16 '16 at 3:42
3  
it probably demonstrates it... – enthdegree Aug 11 '16 at 5:41

up vote 18 down vote

It's not exactly stunning, but it is interesting and visual and simple enough for an elementary school child:

There are only 5 platonic solids.

Numberphile has a great video explaining it: https://www.youtube.com/watch?v=gVzu1_12FUc

In short, the reason is that there are only enough space for 3, 4, or 5 equilateral triangles at a corner; only enough space for 3 squares at a corner; and only enough space for 3 pentagons at a corner; and not even enough space for 3 hexagons at a corner, so there are only 5.

Although I guess it was stunning enough for the ancient Greeks to decide that they were the geometric basis of the five elements of the universe: earth, fire, wind, water, aether.

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up vote 15 down vote

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.

ddrπr2=2πr
The derivative of a sphere w.r.t. the radius is the area.
ddr43πr3=4πr2
The derivative of a 4-dimensional sphere w.r.t. the radius is the 3-dimensional area.
ddr12π2r4=2π2r3
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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up vote 12 down vote

Divergent Series can be visual:

enter image description here

from the Wikipedia

showing that (1−1+1−1+…)2=1−2+3−4+…

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up vote 11 down vote

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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Sounds interesting but I found that link really impenetrable, as a lay person. – Paul Apr 3 '14 at 13:17
    
Yeah, great theorem, bad example ;) – Carsten S Apr 3 '14 at 21:10
3  
A minor correction: strictly speaking, Gaussian curvature is not a topological invariant. – Michael Apr 3 '14 at 22:48
1  
@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

up vote 11 down vote

There's also some really cool art in Polynomiography. Dr. Bahman Kalantari seems to have made really interesting visualizations of polynomials, and considering these functions are everywhere, it might be cool to check them out.

Polynomiography

A Polynomial

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up vote 11 down vote

Allow me to join the party guys...

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

enter image description here

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

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up vote 11 down vote

The Julia set of a complex number c is a fractal (for each c you have one) that has a weird property: they visually look like the Mandelbrot set around that point c. This becomes clear in this illustration I made for a school project, which consists of tiny images of Julia sets:

mosaic

Magically the Mandelbrot set appears...

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1  
Mandelbrot defined his Set as the set of values for which the Julia set is connected; its formula is intimately related to that of a Julia set. – Anton Sherwood Jun 21 '16 at 23:46

up vote 10 down vote

enter image description here

This is @Blue's very nice visual proof from trigonography.com that

x+1x⩾2
shareciteimprove this answer
 
1  
+1. Thanks for the compliment! :) – Blue Apr 9 '16 at 1:00
    
BTW: It's cool (and surprising) that you know me from trigonography.com and not here. Maybe I should update the site more often, now that I know I have an audience. :) – Blue Apr 9 '16 at 1:12
1  
@Blue Yes, everyone likes a good diagram (a lot) – user311151 Apr 9 '16 at 1:13

up vote 9 down vote

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 ddxex=ex. (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.

shareciteimprove this answer
 
    
All of these links are broken. Could you provide unbroken links? – Drew Christensen Jan 2 at 17:51

up vote 8 down vote

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

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up vote 8 down vote

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.

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up vote 6 down vote

Just wanted to point out that The Book of Numbers has a lot of the examples above (as well as many others).

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up vote 6 down vote

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. enter image description here

Explanation: enter image description here
Set the radius to be 1, then
HK=2HI=2cosπ7
AC=2AB=2cos3π7
DG=2DF=−2cos5π7
So
2(cosπ7+cos3π7+cos5π7)=HK+AC−DG=HK−(DG−AC)=HK−(DG−DE)=HK−EG=HK−JK=HJ=LO=1
shareciteimprove this answer
 
    
I don't get it. I see a circle divided into 14 segments and 3 oddly placed parallelograms. Explain this please. – Nick Feb 22 '15 at 7:17
    
@Nick I'll add it in my answer. – Vim Feb 22 '15 at 7:34
    
@Nick I've finished. You can take a look at it if you need it. – Vim Feb 22 '15 at 7:45
    
@Vim how do we know that, for example, AD and CE are parallel? – Joffan Mar 14 '15 at 14:56
1  
To my mind, however, this proof makes
cosπ7+cos3π7+cos5π7=12
seem more magical than it really is. One can instead see that the statement is equivalent to
cosπ7+cos3π7+cos5π7+cos7π7+cos9π7+cos11π7+cos13π7=0,
which expresses that the mean horizontal coordinate of seven equally spaced points on the unit circle is 0, a fact that follows from rotational symmetry. – Will Orrick Dec 6 '15 at 10:01

up vote 6 down vote

The sum of the first n squared numbers:

enter image description here

The first 3 triangles are the same, just rotated. Also, notice that

122232⋮  n2=1=2+2=3+3+3  ⋮⋱=n+n+⋯+n+n

Which is the first triangle. The last triangle is given by 12[n(n+1)(2n+1)]

Thus,

3(12+22+32+⋯+n2)=n(n+1)(2n+1)2

Please see here for the original post

and here for a more indepth explanation.

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up vote 5 down vote

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.

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2  
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
6  
Sort the list in non-decreasing order as a1,a2,…,a2014. Since the average of a1, a2, and a3 is among these, it must equal a2, so a2=a1+d and a3=a1+2d for some d. Since the average of a2, a3, and a4 is among the list, is must equal a3. Proceeding in this manner we see that ak=a1+(k−1)d for some d. Now note that the average of a1, a2, and a4, which is a1+43d, is also in the list. It follows that d=0. – Daniel McLaury Aug 28 '14 at 1:36

up vote 5 down vote

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

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but he has to prove C=2πr first – zinking Apr 8 '14 at 6:23
2  
@zinking No, π 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

up vote 5 down vote

Ulam Spiral:

enter image description here

Discovered by Stanislaw Ulam, the Ulam Spiral or the Prime Spiral depicts the certain quadratic polynomial's tendency to generate large number of primes.Ulam constructed the spiral by arranging the numbers in a rectangular grid . When he marked the prime numbers along this grid, he observed that the prime numbers thus circled show a tendency to occur along diagonal lines. A 150x150 Ulam Spiral is shown below where the dots represent the occurance of prime numbers. The high density along the diagonal lines can be seen as represented by the darker shade of blue. enter image description here

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up vote 4 down vote

A visual display that 00=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 11=1. The orange pseudo-circle which contains 0 is period two. Example is 00=1,01=0.

pseudo-circle

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10  
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
4  
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
30  
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
4  
So, er, how does this show that 00=1? – David Richerby Apr 5 '14 at 17:55
7  
x0=1 because the multiplicative identity is one, 0x=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 00=1. If we at any point use fractional powers, then exponentiation is shorthand for it's natural definition: ab=eblna and we all know that ln0 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

up vote 3 down vote

Polynomials can describe geometric objects

In high school we learn that some low order polynomials can describe geometric shapes:

Basic shapes we all recognize ( as intro )

y=kx+mx2+y2=r2y=x2+ax+b(line)(circle)(parabola)

Cool properties consider the rotation

[xy]=[cos(ϕ)−sin(ϕ)sin(ϕ)cos(ϕ)][xnewynew]
and then we substitute each xayb and carry out the multiplications and we will still have a polynomial. By similar reasoning we can do scaling and translation and still remain a polynomial. If we rewrite the polynomials to be expressions equal to 0:
pa(x,y)=0,pb(x,y)=0
then we can multiply them and use the fact that
b⋅a=a⋅b=0,∀a≠0,iff b=0
This gives us ability to combine shapes into one and the same representation. We can also do something of the opposite: equation systems which can get the intersection. Example is intersection of two lines is an equation system of two lines. The interesection of a sphere and a plane is a point or a circle. This is also where the expression conic section comes from: an intersection between a cone and something!

And still after all this which is so visually accessible and easy to explain in one sense, still involves challenges in modern math of algebraic geometry has had lots of development even in the last 50 years.

below: xp+xp−kp=0 for p=6. When p grows it will get closer and closer to a rectangle. To the right is the "fifth heart curve" (source: Wolfram Alpha) is an 8th degree polynomial.

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up vote 1 down vote

I recently find some stunning visualizations... I prefer to share them all:

5) Mean inequalities [from Proof without words]

enter image description here

4) Streographic projection [by H.Segerman]

enter image description here

3) Farey-Ford Tessellation in non-euclidean geometry [by F.Bonahon]

enter image description here

2) Steiner Porism [by Wikipedia]

enter image description here

1) Polynomial Roots [by J.Baez]

enter image description here

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up vote 0 down vote

In the figure below : Can Mickey Mouse Divide by 7

To find the remainder on dividing a number by 7, start at node 0, for each digit D of the number, move along D black arrows (for digit 0 do not move at all), and as you pass from one digit to the next, move along a single white arrow.

For example, let n = 325. Start at node 0, move along 3 black arrows (to node 3), then 1 white arrow (to node 2), then 2 black arrows (to node 4), then 1 white arrow (to node 5), and finally 5 black arrows (to node 3). Finishing at node 3 shows that the remainder on dividing 325 by 7 is 3.

If you try this for a number that is divisible by 7, say 63, you will always end up in node 0. Therefore it can also be used to test divisibility by 7. Incase while traversing the digits of number 'n', you end up in the node 0, 'n' is divisble by 7 else not.

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posted @ 2017-02-11 18:07  Mr.Rico  阅读(799)  评论(0)    收藏  举报