Visually stunning math concepts which are easy to explain

Question

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?

Answer 1

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 $e^{2k\pi i}=1$

Answer 2

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).

Answer 3

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

Answer 4

Divergent series can be visual:

From the Wikipedia

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

Answer 5

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."

Answer 6

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:

Answer 7

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:

Answer 8

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

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

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

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

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$2)$ Steiner porism [by Wikipedia]

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

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Aren't they incredible?

Answer 9

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.

Answer 10

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,

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

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Please see here for the original post

and here for a more indepth explanation.

Answer 11

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...

Answer 12

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.

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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.

Answer 13

How to convert a function from Cartesian to polar coordinates:

Answer 14

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

Answer 15

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.

Answer 16

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.

Answer 17

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.

Answer 18

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.

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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

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

Answer 19

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.

Answer 20

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

Answer 21

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.

Answer 22

Gluing two Mobius strips along their edges is a Klein bottle.

Source

Answer 23

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.

Answer 24

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.

Answer 25

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,

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

Answer 26

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.

Answer 27

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.

2. 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 $C=(\frac{7\pi}{4},10^4)$.

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

4. 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:

Answer 28

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.

Answer 29

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

Answer 30

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