# Visually stunning math concepts which are easy to explain

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

Do you know of any other concepts like these?

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It looks like mathpop or demand for math entertainment) –  rook Apr 2 at 12:59
@ColeJohnson the 'transcendentality" is the beauty of it! –  Sabyasachi Apr 4 at 17:07
Why has this got 105k views......................? –  LTS Apr 8 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. –  Thursday Aug 29 at 18:37

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 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 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 at 12:34
@daviewales It's been started - the mathematics educators SE. –  SDevalapurkar Apr 5 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 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 at 3:47
I like the pyramid story! There is a more geometrical demonstration that 3 pyramids make a cube:lapasserelle.com/online_courses/maths6/3_pyramids_in_a_cube/… And by slicing pyramids with many cuts parallel to the base plane, it can be shifted, much like a rectangular solid loaf of bread can be shifted to approximate a parallelepiped. The more slices, the better the approximation. This slicing and shifting can also make a parallegram into a rectangle or any triangle into half a rectangle. –  HopDavid Apr 7 at 15:22

A very satisfying visualization of the area of a circle.

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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 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 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 at 16:06
@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 at 5:30

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 at 11:45
@congusbongus I'm not sure how this was generated, but I wonder is there is a way to slow down and or stop the animation? +1 Fabulous. –  joeA Apr 1 at 21:47
@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 at 20:51
If high school math just used a fraction of the resources here, we'd have way more mathematicians. –  user148298 Apr 3 at 17:00
Is the source of this animation available? (It looks like it's in $\mathrm\TeX$.) –  bb010g Apr 3 at 23:31

A visual explanation of 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, its 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 at 21:55
@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 at 16:05

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

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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 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 at 14:10
This would be better as an animation (which achieves the zooming effect). –  Duncan Apr 7 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 –  Darksonn Aug 13 at 14:40

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 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 at 1:17 Could you post a link to the code here? – Kevin Hwang Apr 9 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. - 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. - 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 bezier curves visually: He has also created MathBox.js which powers his amazing visualisations in the slides. - These are beautiful! – Cole Johnson May 15 at 23:16 Here's a GIF that I made that demonstrates Phi (golden number) - proof that the area of a circle is πr² without words: https://www.youtube.com/watch?v=whYqhpc6S6g - 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 at 17:38 I still don't believe you. – Ojonugwa Ochalifu Apr 2 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 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 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 at 20:09 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 - 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: http://arapaho.nsuok.edu/~deckar01/Zvis.html Multiplication of the integers modulo 15: Multiplication of the integers modulo 512: - This is awesome. Looks very much like a fractal. Is there a name for this fractal? – OmnipotentEntity Apr 7 at 15:52 @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 at 18:48 Although not colored WolframAlpha also does a nice job of illustrating this: wolframalpha.com/input/?i=Z_200 – Vikram Saraph Apr 9 at 23:05 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. - 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 smaller, the resulting shape gets closer and closer to a rectangle. It is obvious that the short side of this rectangle has length$r$, and a little thought will show that the two long sides each have a length half the circumference, or$\pi r$, giving an area for the rectangle of$\pi r^2$. This can also be done physically by taking a paper circle and actually cutting it up and rearranging the pieces. This exercise also offers some introduction to (infinite) sequences. - This doesn't work for me :( Sadly, I find it non-obvious that the length of the bottom really will converge helpfully to$\pi r$. – Nicholas Wilson Apr 8 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 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 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 at 17:56 Archimedes was a clever man, and his inscribed polygons were for determining the area. For the same reason, his argument about the area of the sphere has to be very ingenious - not as straightforward as a simple polar integral! – Nicholas Wilson Apr 9 at 18:00 This visualisation of the Fourier Transform was very enlightening for me: The author, LucasVB, has a whole gallery of similar visualisations at his Wikipedia gallery and his tumblr blog. - Sweet!$\textbf{}$– Potato Apr 5 at 6:01 @gekkostate You're a bit off. f is the sum of multiple simple waves, all with different frequencies and phase angles. The fourier transform takes a complex wave from a given time period, and gives you the phase angles and frequencies of all of the component waves. f^ is the amplitude of each component wave. – Joel Apr 7 at 2:15 I'm down-voting this because this animation doesn't include phase information. The phase is the difference between a pulse and cw. If you don't want it, you should use another basis (wavelets). – Mikhail Apr 7 at 3:19 this is pretty cool too: bl.ocks.org/jinroh/7524988 – ilia choly Apr 7 at 20:54 I saw this animation a long time ago, and it helped me finally understand what the Fourier transformation does, but I still am lost as to how it works... – Cole Johnson Apr 11 at 1:21 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. - Sort the list in non-decreasing order as$a_1, a_2, \ldots, a_{2014}$. Since the average of$a_1$,$a_2$, and$a_3$is among these, it must equal$a_2$, so$a_2 = a_1 + d$and$a_3 = a_1 + 2d$for some$d$. Since the average of$a_2$,$a_3$, and$a_4$is among the list, is must equal$a_3$. Proceeding in this manner we see that$a_k = a_1 + (k-1)d$for some$d$. Now note that the average of$a_1$,$a_2$, and$a_4$, which is$a_1 + \frac{4}{3} d$, is also in the list. It follows that$d = 0\$. –  Daniel McLaury Aug 28 at 1:36

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

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## protected by Community♦Apr 8 at 16:42

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