15
$\begingroup$

I often hear the phrase "mathematical beauty". That a proof or formula or theorem is beautiful. and I do agree I was awestruck when I first saw Euler's formula, connecting 3 seemingly unrelated branches of mathematics in a single formula $e^{i\pi}=-1$

But beauty is a rather subjective term. When I was taught Linear Algebra the instructor introduced Cayly-Hamilton theorem as beautiful, and I thought it was "nothing special".

I'm interested in theorems that are considered beautiful, and why they are so.

Just as an example to what I think is beautiful, last night a friend told me that the sum of the first $n$ odd numbers is equal to $n^2$. for example if $n=3$ then $1+3+5 =9=3^2$. if $n=5$ then $1+3+5+7+9 = 25 =5^2$ Simplistic. Surprising. Elegant. I liked it a lot.

I would be very much interested in learning more theorems / formulas like that.

$\endgroup$

closed as too broad by William, Hanul Jeon, Hans Lundmark, user98602, Jack M Apr 23 '14 at 21:28

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ For the sum of consecutive odd numbers always being a square, see here. $\endgroup$ – Lucian Apr 23 '14 at 9:31
  • $\begingroup$ Your answer is absolutely beautiful Lucian. $\endgroup$ – Oria Gruber Apr 23 '14 at 12:12
  • $\begingroup$ The reason your instructor finds the Cayley-Hamilton theorem beautiful is perhaps that relatedness of structure between different things (a matrix and its eigenvalues are solutions of the same polynomial in different rings) is a source of beauty. Intuition guides us to hope for it but reason doesn't immediately entitle us to expect it, so it is aesthetically pleasing to find it. $\endgroup$ – Steve Jessop Apr 23 '14 at 12:38
  • 1
    $\begingroup$ Not only was this question not made CW,the OP also accepted an answer as if the question had one perfect reply. $\endgroup$ – rah4927 Apr 23 '14 at 13:42
  • 1
    $\begingroup$ When you write Euler's formula like so $e^{i\pi} + 1 = 0$, it's even more beautiful in the sense that it has all 5 'special' constants in one equation (i.e. $e$, $i$, $\pi$, $1$, $0$). $\endgroup$ – tangrs Apr 23 '14 at 14:06
24
$\begingroup$

$$\int_M d\omega = \int_{\partial M}\omega $$

$\endgroup$
  • $\begingroup$ That's Stokes theorem right? $\endgroup$ – Oria Gruber Apr 23 '14 at 12:14
  • $\begingroup$ ${}{}{}{}{}$Yes. $\endgroup$ – Thomas Apr 23 '14 at 12:35
  • $\begingroup$ This is an excellent answer, but why is it "the" answer? I would like to see what other people have to contribute to this post... and I'm not trying to hate on this post, +1. $\endgroup$ – user142299 Apr 23 '14 at 16:00
12
$\begingroup$

$$\mathcal G(n)=\int_0^\infty e^{-x^n}dx\qquad=>\qquad n!=\mathcal G\bigg(\dfrac1n\bigg)$$ In particular, the Gaussian integral $$\int_{-\infty}^\infty e^{-x^2}dx=\sqrt\pi$$

$\endgroup$
9
$\begingroup$

If $M=M^2$ is a smooth compact $2$-dimensional Riemannian manifold with (smooth) boundary $\partial M$, $K$ denotes it's Gauß-curvature, $k_g$ the geodesic curvature of it's boundary und $\chi(M)$ the Euler-Characteristic, then the theorem of Gauß-Bonnnet states that $$\int_M K dA + \int_{\partial M}k_g ds = 2\pi \chi(M)$$ (There are generalizations of this to higher dimensions. For me the beauty of this particular theorem originates from the fact that is one of the early insights of mathematicians into the deep relationships between topological invariants and analytical quantities)

$\endgroup$
  • $\begingroup$ The analytics are measurable isometrics dependent on first fundamental form. Gauss himself described it as " elegant ", another word for beauty. $\endgroup$ – Narasimham May 13 '15 at 6:04
6
$\begingroup$

Theorem: There exist positive irrational $a,b$ such that $a^b\in\mathbb{Q}$.

$\square$ Consider $\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}}=2$. Then either $\sqrt{2}^{\sqrt{2}}\in\mathbb{Q}$ or $\ldots$ $\blacksquare$

$\endgroup$
  • 3
    $\begingroup$ Eh, I've always made a distinction between things like this, which I would call 'clever' or 'slick', and things I would call beautiful. My beauty is Russellian and Platonic: "stern perfection" that appears less a product of a human mind and more as if it were channelled through a fortunate soul from some greater wellspring of truth and supreme beauty. $\endgroup$ – Eric Stucky Apr 23 '14 at 12:19
6
$\begingroup$

This might sound silly, but the Quadratic Formula was the first formula I ever learned to prove, and I still have a soft spot for it. $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ And do you know what I thought was beautiful? I was so excited when I first learned it that I would solve linear equations as follows: $$ax+b=0$$ $$ax^2+bx+0=0$$ $$x=\frac{-b\pm{\sqrt{b^2}}}{2a}=-{b\over a}$$

$\endgroup$
  • $\begingroup$ Lame's Theorem is pretty theoremoftheday.org/Binomial/Lame/TotDLame.pdf $\endgroup$ – Paul Apr 23 '14 at 8:23
  • 7
    $\begingroup$ You missed out the other solution to your quadratic, namely 0. =) $\endgroup$ – user21820 Apr 23 '14 at 11:59
  • $\begingroup$ @user21820 But at least I got a very early introduction to how extraneous roots happen :) $\endgroup$ – user142299 Apr 23 '14 at 15:59
  • $\begingroup$ @NotNotLogical: Just curious, did you ask why you couldn't use the formula with $0x^2+ax+b=0$? =P $\endgroup$ – user21820 Apr 23 '14 at 23:54
  • $\begingroup$ @user21820 No. :) But it's interesting that, using L'Hopital, the limit as $a\to 0$ is $1/2\cdot (\pm 2c/\sqrt{b^2-4ac})=\pm c/b$ $\endgroup$ – user142299 Apr 24 '14 at 0:20
4
$\begingroup$

Often the beauty of a theorem is measured in terms of the brevity of its formulation. If one has a short easily understandable statement it is often considered a beautiful result particularly if the proof is not obvious or considerably longer than the statement of the theorem. The Cayley-Hamilton theorem is "beautiful" in this sense since the formulation is quite brief whereas the proof is not altogether obvious.

$\endgroup$
3
$\begingroup$

Does the Mandelbrot set being a fractal count as a beautiful theorem?

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.