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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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    $\begingroup$ see also, fractals $\endgroup$ – Guy Mar 31 '14 at 11:55
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    $\begingroup$ It looks like mathpop or demand for math entertainment) $\endgroup$ – rook Apr 2 '14 at 12:59
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    $\begingroup$ @ColeJohnson the 'transcendentality" is the beauty of it! $\endgroup$ – Guy Apr 4 '14 at 17:07
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    $\begingroup$ There is a considerable overlap with mathoverflow.net/questions/8846/proofs-without-words $\endgroup$ – Martin Brandenburg Jun 20 '14 at 22:13
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    $\begingroup$ @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. $\endgroup$ – user147263 Aug 29 '14 at 18:37

66 Answers 66

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This site looks very interesting to learn about algebraic surfaces.

Imagem

http://touch-geometry.karazin.ua/list

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One of my favourites is from Littlewood's Miscellany, where he amicably mentions that "for the professional the only proof needed" for the one-dimensional fixed point theorem is the following figure. The theorem is:

If $f:[0,1]\rightarrow [0,1]$ is continuous and increasing then, under iteration of $f$, every point is either a fixed point or else converges to a fixed point.

enter image description here

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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.enter image description here

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

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In group theory, "visual" explanation of the group $D_4$ (or sometimes called $D_8$, which is dihedral group of order $8$ or degree $4$) was really exciting to me. Even with some elementary knowledge about groups, if someone defines $D_4$ as $$\langle x, a\ |\ a^4 = x^2 = e, axa = x \rangle$$ it might seem meaningless or too abstract to understand. When I first saw this, I simply thought it as something to memorize. But when I saw the visual explanation, it was a "stunning" moment for me because it was really easy to understand and there was nothing to memorize at all:

First let us take a square $S$ with vertices named as $A, B,C, D$.

enter image description here

Then, if we let the element $a$ rotate this square $90^\circ$ clockwise direction and let the element $x$ flip the square through the first diagonal, i.e. $x=y$ line, we can have all $8$ elements of $D_4$ as the following:

enter image description here

$e$: Identity element. It does nothing to $S$.

$a$: Rotates $S$, $90^\circ$ clockwise direction.

$a^2$: Rotates $S$, $90^\circ$ clockwise direction twice, i.e., rotates $S$, $180^\circ$ clockwise direction.

$a^3$: Rotates $S$, $90^\circ$ clockwise direction three times, i.e. rotates $S$, $270^\circ$ clockwise direction.

$x$: Flips $S$ through its first diagonal, i.e., interchanges $A$ and $C$.

$ax$: Flips $S$ through its first diagonal first, then rotates the flipped square $90^\circ$ clockwise direction.

$a^2x$: Flips $S$ through its first diagonal first, then rotates the flipped square $180^\circ$ clockwise direction.

$a^3x$: Flips $S$ through its first diagonal first, then rotates the flipped square $270^\circ$ clockwise direction.


Here, only thing we need to be careful is that rightmost function is applied first (for example, $ax$ means: first flip, then rotate). Now, we can verify the properties that are mentioned in the definition:

First of all, we have $a^4 = e$ because $a^4$ rotates $S$, $90^\circ$ clockwise direction four times, which means $360^\circ$ clockwise direction, which doesn't change the place of any vertex. So it is as same as the identity element $e$.

Secondly, we have $x^2 = e$ because if we flip $S$ through its first diagonal (interchanging $A$ and $C$) and then flip it again, we get $S$ again, which also corresponds to identity element $e$.

Finally, we have $axa = x$, which can be verified by rotating $S$ first, then flipping it and rotating it again. In the end what we get is as same as flipping $S$, which is done by $x$.

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In plane geometry Morley’s theorem is a stunning fact in my opinion: In any triangle, the points of intersections of adjacent trisectors of the angles form an equilateral triangle :enter image description here

In analytical geometry: The generalization of triangles or tetrahedra in n-dimensions is simpleces. And the formula for the simplex volume is a beauty, for example, the volume of four-dimensional simplex which is called pentachoron, pentatope or 5-cell (using the coordinates of its vertices): $$Four\; dimensional\; volume = ±\frac{1}{4!}\;\begin{vmatrix} \;x_1-x_5 && y_1-y_5 && z_1-z_5 && w_1-w_5\;\\ \;x_2-x_5 && y_2-y_5 && z_2-z_5 && w_2-w_5\;\\ \;x_3-x_5 && y_3-y_5 && z_3-z_5 && w_3-w_5\;\\ \;x_4-x_5 && y_4-y_5 && z_4-z_5 && w_4-w_5\;\\ \end{vmatrix}$$ In the case of triangles we get $$A=±\frac{1}{2}\;\begin{vmatrix} \;x_1-x_3 && y_1-y_3\;\\ \;x_2-x_3 && y_2-y_3\;\\ \end{vmatrix}$$

Of course, we can write it even shorter if we use vectors.

And 4-dimensional spheres are truly amazing

https://en.wikipedia.org/wiki/3-sphere?oldformat=true#/media/File:Hypersphere.png

as are tesseracts:

https://en.wikipedia.org/wiki/Tesseract?oldformat=true#/media/File:Net_of_tesseract.gif

And some things are really arcane:

$$\boldsymbol{1+2+3+4+...=-\frac{1}{12}}\\$$

It makes you ponder.

As to calculus, my vote would go to the beauty of Euler’s formula already posted in this thread ($\,\boldsymbol{e^{ix}=\cos{x}+i\sin{x}}\,$)

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    $\begingroup$ **WARNING!** $$-\frac1{12}=\zeta(-1)\ne1+2+3+4+\ldots$$ $\endgroup$ – Simply Beautiful Art Jul 2 '17 at 22:52
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    $\begingroup$ I would remove my downvote if concerns in the previous comment was addressed. $\endgroup$ – Frenzy Li Nov 13 '17 at 18:16
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$$\sum_{i=1}^{\infty}\frac{1}{x^n}=\frac{1}{x-1}$$

In base x this sum equels to 0.1111111.... and if you multiply it by x-1 you get 0.(x-1)(x-1)(x-1).... which equels to 1.

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protected by DeepSea Dec 23 '14 at 4:02

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