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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Here is a classic: the sum of the first $n$ positive odd numbers $= n^2$.
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My favorite: tell someone that $$\sum_{n=1}^{\infty}\frac{1}{2^n}=1$$ and they probably won't believe you. However, show them the below:
and suddenly what had been obscure is now obvious.
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Fractal art. Here's an example: "Mandelbrot Island".
The real island of Sark in the (English) Channel Islands looks astonishingly like Mandelbrot island:
Now that I think about it, fractals in general are quite beautiful. Here's a close-up of the Mandelbrot set:
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A theorem that I find extraordinarily beautiful and intuitive to understand is Gauss' Theorema 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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I think if you look at this animation and think about it long enough, you'll understand:
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I do not know if this meets your criteria of "visually stunning", but nonetheless -
I like this proof of Pythagoras' theorem (image taken from www.wisfaq.nl):
The key to understanding this is to realize that the inner quadrilateral must be a square. The sides are equal in length (obviously) and each of its angles is $90^{\circ}$, because the two angles on either side sum to $90^{\circ}$, and the sum of the three angles is $180^{\circ}$. The area of this square is $c^2$.
The outer square's area is $(a + b)^2$, which is $c^2$ plus $2 a b$ which is the total area of the four triangles, each of area $\frac{1}{2} a b$.
$(a + b)^2 = c^2 + 2 a b$
$a^2 + b^2 + 2 a b = c^2 + 2 a b$
$a^2 + b^2 = c^2$, which is Pythagoras' theorem.
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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 $\mathrm a^2+\mathrm b^2$.
An animation:
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A visual explanation of a Taylor series:
$f(0)+\frac {f'(0)}{1!} x+ \frac{f''(0)}{2!} x^2+\frac{f^{(3)}(0)}{3!}x^3+ \cdots$
or
$f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots$
When you think about it, it's quite beautiful that as you add each term it wraps around the curve.
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This visualisation of the Fourier transform was very enlightening for me:
The author, LucasVB, has a whole gallery of similar visualisations at his Wikipedia gallery and his Tumblr blog.
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Topology needs to be represented here, specifically knot theory. The following picture is from the Wikipedia page about Seifert Surfaces and was contributed by Accelerometer. Every link (or knot) is the boundary of a smooth orientable surface in 3D-space. This fact is attributed to Herbert Seifert, since he was the first to give an algorithm for constructing them. The surface we are looking at is bounded by Borromean rings.
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This animation shows that a circle's perimeter is equal 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 is equal to $\pi$ amount of wheels.
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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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A visual display that $0^0=1$. The following is a tetration fractal or exponential map with a pseudo-circle shown in orange. The red area is period $1$ and contains $1$. Example is $1^1=1$. The orange pseudo-circle which contains $0$ is period two. Example is $0^0=1, 0^1=0$.
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Similarly to eykanal's answer, although demonstrating some interesting facts about medians and geometry as well. It demonstrates that $\displaystyle\sum_{n = 1}^{\infty}\frac{1}{2^n} = 1$:
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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.
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, but I can also even tell you the name of the 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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As I was in school, a supply teacher brought a scale to the lesson:
He gave us several weights that were labeled and about four 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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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]
I've built a bunch of interactive explorations over at Khan Academy. A few of my favorites are:
Derivative intuition. Particularly amazing is seeing how $\frac{d}{dx}e^x=e^x$. (Do a few and it should pop up).
Exploring mean and median. Light bulbs are twice as likely to burn out before the average lifetime printed on the package. If that statement surprises you, this exploration points out that mean and median aren't the same thing.
Exploring standard deviation. Standard deviation is a term that gets thrown around a lot. Play around with this to get a more intuitive sense of what it means.
One step equation intuition. Basic introduction to why you can do the same thing to both sides of an equation to solve it.
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Here is a very insightful waterproof demonstration of the Pythagorean theorem. Also there is a video about this.
It can be explained as follows. We seek a definition of distance from any point in $\mathbb{R}^2$ to $\mathbb{R}^2$, a function from $(\mathbb{R}^2)^2$ to $\mathbb{R}$ that satisfies the following properties.
Suppose a function $d$ from $(\mathbb{R}^2)^2$ to $\mathbb{R}$ satisfies those conditions, then for any point $(x, y)$, $d((0, 0), (x, y))^2 = d((0, 0), (x, y))d((0, 0), (x, y)) = d((0, 0), (x, y))d((0, 0), (x, -y)) = d((0, 0), (x^2 + y^2, 0)) = x^2 + y^2$ so $d((0, 0), (x, y)) = \sqrt{x^2 + y^2}$ so for any points $(x, y)$ and $(z, w)$, $d((x, y), (z, w)) = \sqrt{(z - x)^2 + (w - y)^2}$ Now I will show that $d((x, y), (z, w)) = \sqrt{(z - x)^2 + (w - y)^2}$ actually satisfies those properties. It's trivial to show that it satisfies the first 4 conditions. It also satisfies the fifth condition because for any points $(x, y)$ and $(z, w)$, $d((0, 0), (xz - yw, xw + yz)) = \sqrt{(xz - yw)^2 + (xw + yz)^2} = \sqrt{x^2z^2 - 2xyzw + y^2w^2 + x^2w^2 + 2xyzw + y^2z^2} = \sqrt{x^2z^2 + x^2w^2 + y^2z^2 + y^2w^2} = \sqrt{(x^2 + y^2)(z^2 + w^2)} = \sqrt{x^2 + y^2}\sqrt{z^2 + w^2} = d((0, 0), (x, y))d((0, 0), (z, w))$
As a result of this, from now on, I will define the distance from any point $(x, y)$ to any point $(z, w)$ as $\sqrt{(z - x)^2 + (w - y)^2}$ and denote it as $d((x, y), (z, w))$. I will also use $d(x, y)$ as shorthand for $d((0, 0), (x, y))$ Since distance satisfies condition 5, for any right angle triangle at all, not just ones whose legs are parallel to the axes, the square of the length of its hypotenuse is equal to the sum of the squares of the lengths of its legs.
This image shows that using that definition of distance, for any right angle triangle whose legs are parallel to the axes and have lengths $x \in \mathbb{R}^+$ and $y \in \mathbb{R}^+$, the area of a square with the hypotenuse as one of its edges is $(x - y)^2 + 2xy = x^2 + y^2 = (d(x, y))^2$. Combining that result with the fact that distance satisfies condition 5, we can show that for any right angle triangle, even with legs non parallel to the axes, the area of a square with its hypotenuse as its edge has an area equal to the sum of the squares of the lengths of its legs.
Sources:
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I 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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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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A nationwide math contest in Germany recently came up with a task that I found beautiful to explain, because of two points.
You can get an idea, what the proof is, without applying mathematically accurate theory and this intuitive proof is most likely the right way.
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 fulfill 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 fulfill 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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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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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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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:
Multiplication of the integers modulo 512:
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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.
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Steven Wittens presents quite a few math concepts in his talk Making things with math. His slides can be found from his own website.
For example, Bézier curves visually:
He has also created MathBox.js which powers his amazing visualisations in the slides.
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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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The magnetic pendulum:
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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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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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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Math is always fun to learn. Here are some of the images that explain some things beautifully visually
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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 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).
Simple, visual proof of the Pythagorean theorem. Originally from Pythagorean Theorem Proof Without Words 6).