# 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?

• It looks like mathpop or demand for math entertainment)
– rook
Commented Apr 2, 2014 at 12:59
• Something that has always annoyed me is irrational numbers. They are easy to understand (like the proof for $\sqrt{2}$), but not being able to have a rational representation just irks me. And don't even get me started on transcendentals like $\pi$ and $e$. Commented Apr 4, 2014 at 16:12
• @ColeJohnson the 'transcendentality" is the beauty of it!
– Guy
Commented Apr 4, 2014 at 17:07
• There is a considerable overlap with mathoverflow.net/questions/8846/proofs-without-words Commented Jun 20, 2014 at 22:13
• @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
Commented Aug 29, 2014 at 18:37

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

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

• Another nice visualization of a related problem is Szego's problem of the asymptotic of the roots of the truncated polynomials $P_n(z)$ of the Taylor expansion of $e^z$.
– VECH
Commented Aug 5, 2021 at 8:26

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).

• I found this proof delightful. Is there any free resource checking the claim that it was known to the pythagoreans? Commented Jul 1, 2019 at 2:57
• @Santropedro: You might find this paper interesting (p. 16). Commented Jul 1, 2019 at 11:13

The following animation shows how the surface area of a sphere is calculated.

• How did you create that? Commented Nov 16, 2021 at 11:42
• @ArchismanPanigrahi I did not. I saved it from a website. Unfortunately, I don't remember the source. Commented Nov 16, 2021 at 12:08
• I've always liked this animation except I kind of wish they'd kept the original spikes in the sin curve -- it feels a bit off when they get squished together and my brain doesn't quite believe it. Commented Mar 15, 2022 at 19:09

Divergent series can be visual:

From the Wikipedia

Showing that $$(1-1+1-1+\dots)^2=1-2+3-4+\dots$$

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

The exact code for this is extremely simple:

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


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

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

• I'd also note that it's possible to compute ${i \choose k} \mod p$ without computing $i \choose k$. For large $i$ this would matter. Commented Jan 9, 2015 at 14:11
• The basic idea is pretty simple: ${i \choose k} = {i \choose k-1} + {i-1 \choose k-1}$, and this recurrence holds $\mod p$ as well. Commented Jan 9, 2015 at 14:31
• @Coffee_Table: It's literally just the terminal. The code I pasted above write ANSI color codes to the terminal to produce the colored blocks you see above. Commented Feb 10, 2015 at 22:40
• right, thanks. I edited your code to work for Python 3 and I realize now that I made a stupid error when doing so. Commented Feb 10, 2015 at 23:08
• It is a finite version of the Sierpinski triangle. Commented Apr 8, 2017 at 21:07

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:

• 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. Commented Aug 11, 2016 at 5:04
• @enthdegree That's actually not true. It's true for well behaved 2D shapes (and in fact you can use this to define a type of dimension) but in general it's a factor of $(\Delta x)^d$. I say "well behaved shapes" because when you use this with fractals, you get fractions! Commented Apr 6, 2017 at 14:06

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

$$x+\frac{1}{x}\;\geqslant\; 2$$

Two more illustrations from http://www.doubleroot.in

We see $(x+(1/x))^2 \geq 4$:

We know the hypotenuse is always the longest side of a triangle:

• 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
Commented Apr 9, 2016 at 1:12
• @Blue Yes, everyone likes a good diagram (a lot)
– user311151
Commented Apr 9, 2016 at 1:13
• hey, i saw you did some editing to remove my watermark there. but please cite the source atleast? www.doubleroot.in or facebook.com/doubleroot Commented Aug 14, 2018 at 11:05
• @yomayne the original poster is not active anymore, thus there was no reply initially. I added a link to the site you mentioned; since by now you passed 100 points you can edit Community Wiki posts. Please add more precise reference as you see fit. (It seems the site is under a CC licence, so I suppose you it is fine that it is reproduce in principle; if not we can also remove it entirely.)
– quid
Commented Aug 27, 2018 at 15:08

I recently found some stunning visualizations. I preferred to share them all:

$$5)$$ Mean inequalities [from Proof without words]

$$4)$$ Streographic projection [by H.Segerman]

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

$$2)$$ Steiner porism [by Wikipedia]

$$1)$$ Polynomial roots [by J.Baez]

Aren't they incredible?

Ulam Spiral:

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.

The sum of the first $n$ squared numbers:

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

\begin{align}1^2&=1\\2^2&=2+2\\3^2&=3+3+3\\\vdots\ \ &\quad\ \ \vdots\qquad\qquad\quad\ddots\\n^2&=n+n+\dots+n+n\end{align}

Which is the first triangle. The last triangle is given by $\frac12[n(n+1)(2n+1)]$

Thus,

$$3(1^2+2^2+3^2+\dots+n^2)=\frac{n(n+1)(2n+1)}2$$

Please see here for the original post

and here for a more indepth explanation.

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:

Magically the Mandelbrot set appears...

• 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. Commented Jun 21, 2016 at 23:46

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.

How to convert a function from Cartesian to polar coordinates:

• This is brilliant! We're reflecting in $y=x$ line as we are changing from second variable (y) to first variable (r). Commented Sep 27, 2020 at 12:17
• This seems relevant - math.stackexchange.com/a/3806702/308392 So we reflect about y = x to find x = g(y) from y = f(x). Now the distance from the origin to the points on the curve is r, if you look at the cartesian plot, as $\theta$ increases we should see r increase. But in your polar plot, we see minimal $r$ appearing periodically. This does not make sense. I don't see how straight grid lines in the cartesian plot become r = constant lines in the polar plot. I think this is wrong. I think if you use the right transformation it will be - desmos.com/calculator/q4gmpdkcw2 Commented Mar 13, 2021 at 9:00

The beauty of watching graphs being constructed has always mesmerized me; I love how such simple figures can be used to make such complicated pictures. And it's especially satisfying with polar graphs.

Even simpler things like conic sections:

This one might not be as easy to explain, but brachistochrones are wonderful things.

The (otherwise also easy to prove) fact that $\sum_\limits{k=1}^n k=\frac{n(n+1)}{2}$ in one picture:

Source of the picture

# 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 ) $$\begin{array}{llr}y&=kx+m& (\text{ line })\\r^2 &= x^2+y^2 & (\text{ circle })\\y &= x^2+ax+b &( \text{ parabola })\end{array}$$

Cool properties consider the rotation $$\left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{rr}\cos(\phi)&\sin(\phi)\\-\sin(\phi)&\cos(\phi)\end{array}\right]\left[\begin{array}{c}x_{new}\\y_{new}\end{array}\right]$$ and then we substitute each $$x^ay^b$$ 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: $$p_a(x,y)= 0, p_b(x,y) = 0$$ then we can multiply them and use the fact that $$b\cdot a = a\cdot b = 0, \forall a \neq 0, \text{iff } b=0$$ 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: $$ax^p + by^p - k^p = 0$$ 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.

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.

• Also, this isn't limited to circles/spheres at all. It holds for squares and cubes and tetrahedra and possibly other shapes, if the formulas are expressed in terms of the radius of the inscribed circle/sphere. Commented Mar 9, 2017 at 2:00
• @user3932000 Pls write an answer expanding your comment Commented Nov 19, 2017 at 15:25

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

Explanation:

Set the radius to be $$1$$, then $$HK=2HI=2\cos\frac{\pi}{7}$$ $$AC=2AB=2\cos\frac{3\pi}{7}$$ $$DG=2DF=-2\cos\frac{5\pi}{7}$$

So

\begin{align} 2(\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7})&=HK+AC-DG\\ &=HK-(DG-AC)\\ &=HK-(DG-DE)\\ &=HK-EG\\ &=HK-JK\\ &=HJ\\ &=LO\\ &=1 \end{align}

• I don't get it. I see a circle divided into 14 segments and 3 oddly placed parallelograms. Explain this please.
– Nick
Commented Feb 22, 2015 at 7:17
– Vim
Commented Feb 22, 2015 at 7:34
• @Nick I've finished. You can take a look at it if you need it.
– Vim
Commented Feb 22, 2015 at 7:45
• @Vim how do we know that, for example, AD and CE are parallel? Commented Mar 14, 2015 at 14:56
• To my mind, however, this proof makes $$\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7}=\frac{1}{2}$$ seem more magical than it really is. One can instead see that the statement is equivalent to $$\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7}+\cos\frac{7\pi}{7}+ \cos\frac{9\pi}{7}+\cos\frac{11\pi}{7}+\cos\frac{13\pi}{7}=0,$$ 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. Commented Dec 6, 2015 at 10:01

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.

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.

• Sounds interesting but I found that link really impenetrable, as a lay person.
– Paul
Commented Apr 3, 2014 at 13:17
• Yeah, great theorem, bad example ;) Commented Apr 3, 2014 at 21:10
• A minor correction: strictly speaking, Gaussian curvature is not a topological invariant. Commented Apr 3, 2014 at 22:48
• @Michael: Yes, you are right. My mistake. I instead had in mind the surface integral of the gaussian curvature over a closed surface, which is a topological invariant (basically the euler characteristic) Commented Apr 5, 2014 at 4:59

Transpose of a matrix column. This gift shows the easiest proof ever made:

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.

• All of these links are broken. Could you provide unbroken links? Commented Jan 2, 2017 at 17:51

I would like to add some explorations of the concept asked by the OP of my own:

1. Visualization of the set of real roots of quadratic equations $ax^2+bx+c=0$, for the specific values of the intervals $a \in [-a_i,a_i]$, $b \in [-b_i,b_i]$, $c \in [-c_i,c_i]$, $a,b,c \in \Bbb N$.

By Cartesian coordinates $(x,y)=(x_1,x_2)$. E.g. $a_i,b_i,c_i=75$:

By Polar coordinates $(\theta, r)=(x_1,x_2)$. E.g. $a_i,b_i,c_i=75$:

Due to the symmetries the opposite patterns $(x,y)=(x_2,x_1)$ and $(\theta,r)=(x_2,x_1)$ are similar.

1. The Chaos Game on the metric space $S^{1} \times [0,\infty)$ with the metric $d((\theta_1,x_1),(\theta_2,x_2)) = d_{S^1}(\theta_1,\theta_2) + |x_1-x_2|$. The distance in $S^1$ is given by the smallest angle measure between $\theta_1$ and $\theta_2$ (this is actually a scaled Euclidean metric on the unit circle itself).

In this version, the points are $(\theta, r)$, (the angle in radians and the radius). And the three attractor points are $A=(0,0)$,$B=(\frac{5\pi}{4},1)$ and $C=(\frac{7\pi}{4},10^4)$.

This is another example locating the attractor points in the same axis: $A=(0,0)$,$B=(\pi-\frac{\pi}{8},1)$ and $C=(\frac{7\pi}{4},10^4)$.

1. Contruction step by step of the Voronoi diagram of the points generated by a classic Chaos Game Sierpinski gasket.

1. And my favorite so far, visualization of the $4$-tuples of the extended Euclidean algorithm in a four dimensional tesseract. The projection of the four dimensional points are shown into a $3D$ visualization adding as a reference a tesseract or hypercube:

• @Leila you are welcome, if you want to know more about them, I have linked the original questions on each point. Commented Apr 6, 2017 at 8:16

The fact that the graph of inverse of a function is nothing more than its image in line $y=x$ but still finding inverse is so difficult is a math concept I really find amazing.

Also inverse of some functions have special name and are really special and useful.

• How is this possible? Mobius strips can exist, but I thought Klein bottles could not? Commented Sep 28, 2022 at 4:44

Consider:

I made this earlier this year in Blender after having spent a few days trying to think of a visual proof of $$a^3-b^3 = (a-b)(a^2+ab+b^2)$$ so that I could make myself a nice paperweight. I think it's quite clear, but I'll explain it anyway.

When it's put together, you see the cube $$a^3$$ with the piece $$b^3$$ cut out of it. This lets you recognize that each block has a dimension of $$(a-b)$$ somewhere, so then I pull the pieces apart and lay them next to each other.

When it's laid down you can see how all of them have the same height $$(a-b)$$ with the red block having a base area of $$a^2$$, the blue block having a base area of $$b^2$$ and the green block having a base area of $$ab$$.

So that shows how $$a^3-b^3 = (a-b)a^2 + (a-b)b^2 + (a-b)ab$$, which is nicer factored as,

$$a^3-b^3 = (a-b)(a^2+ab+b^2)$$

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

The surface obtained by spinning a cube on two diametrically opposite corners:

$$\hskip{4.5cm}$$

All the surfaces are ruled surfaces. The top and bottom are simply conical caps. The curved part in the middle is part of a hyperboloid of one sheet. It can be obtained by taking a cylinder of radius and height $$\sqrt{\frac23}$$ times the edge length of the cube and giving one of the ends a $$60^{\large\circ}$$ twist.

A connection between mathematics and love:

The story goes that a very shy mathematician had fallen in love with a girl, but he did not dare to tell her. Instead, he wrote her a letter with only the following formula: $$y=\pm \sqrt{25-x^2} -\frac{3}{|x|+1}$$

If she was really interested, he counted on her drawing the graph of the formula. How the story ended no one knows ...

So how does one construct such a formula? The first part of the formula (the square root without the fraction) is an ellipse: $$(3x)^2+(5y)^2=15^2$$. Now to get a heart shape the top and bottom of this ellipse must be somewhat lowered and this is accomplished by adding the fraction which is really an adjusted orthogonal hyperbola, $$y=\frac{1}{x}$$, in such a way that it connects to the ellipse. Desmos is a fantastic tool to illustrate this and hence explaining graphs, functions and and a bit of analytical geometry.

Riemann integration has always amazed me. It’s simple, yet extraordinary.