Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

My wife likes to decorate birthday cakes. She told me that she will make a math cake for my birthday and I should provide her a "famous math formula" to be written on the top of the cake.

I realized I can name dozens of physics related famous formulas that one could recognize (Maxwell's equations, Newtons laws, Einstein's $E=mc^2$...) but I couldn't name one that would be more "math related".

Writing some axioms wouldn't work, they take too much space. The famous theorems I know of are not really "a formula" but more like of "statements" that would need some background, or they are not visually appealing (like Fermat's last theorem). (Quests are not math-oriented thus the visual side matters.)

Any ideas what we could put on top of the cake?

share|improve this question

closed as off topic by quid, Henry Cohn, Emil Jeřábek, Felipe Voloch, Todd Trimble Dec 29 '12 at 22:10

Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.

Related (published in the most recent AMM): mpra.ub.uni-muenchen.de/34264/1/Perfect_division1.pdf –  Benjamin Dickman Dec 29 '12 at 10:06
Sorry but this is really off-topic. Voted to close. Not famous but perhaps also fitting given the context: $(x^2 + y^2 -1)^3 - x^2 y^3 =0$ or something like this (see mathworld.wolfram.com/HeartCurve.html) –  quid Dec 29 '12 at 14:05
There used to be a "logic picnic" at the beginning of the fall semester at Berkeley while I was there. John Addison would always bring a "logic cake" full of formulas (the statement of determinacy, or something about P vs NP or ...) that the attendants had to decipher. –  Andres Caicedo Dec 29 '12 at 17:17
I would go for the Pythagorean theorem, with a picture of the right triangle and squares. An evergreen that everybody will appreciate! –  Pietro Majer Dec 29 '12 at 18:27
Someone should email quid's comment (the link) to Frank's wife and delete it before he finds it out so he gets a surprise on what he thought should be a cake with Stokes' theorem. –  Yuichiro Fujiwara Dec 29 '12 at 19:16

19 Answers 19

My all-time favourite formula: Stokes theorem

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

share|improve this answer
Or perhaps $d^2=0$. –  Adam Epstein Dec 29 '12 at 11:20
Or Cartan's structural equation $\Omega=\mathrm{d}\theta+\frac{1}{2}[\theta,\theta]$. –  Qfwfq Dec 29 '12 at 11:33

$e^{i \pi} = -1$

share|improve this answer
$e^{i \pi} + 1 = 0$ might be better. But I expect this question will be closed. –  Benjamin Dickman Dec 29 '12 at 10:02
Should involve pi somewhere in this context –  Geoff Robinson Dec 29 '12 at 15:02

196884 = 196883 + 1

share|improve this answer
I'm sorry, I'm not familiar with this. Could you explain where this equation comes from? –  Joel Reyes Noche Dec 29 '12 at 23:44
@Joel Reyes Noche: en.wikipedia.org/wiki/Monstrous_moonshine –  quid Dec 30 '12 at 0:50
@quid, thanks ! –  Joel Reyes Noche Dec 30 '12 at 7:14

22/7. Because a cake is, approximately, a pi(e).

share|improve this answer
Shouldn't that be $\frac{22}7e$, then? –  Asaf Karagila Dec 29 '12 at 20:10
If you're approximating $\pi$, then why not approximate $e$ also? So $\pi e\approx \frac{22}{7}\frac{19}{7}$. –  Joel Reyes Noche Dec 30 '12 at 7:18

How about the Grothendieck-Hirzebruch-Riemann-Roch formula: ch(f!F) = f*(ch(F)td(Tf))?

share|improve this answer
Yes, if I recall correctly, this one can also be admired on the glass front of the MPIM's reception desk. –  Tobias Fritz Dec 29 '12 at 20:12
@Tobias: Yes, although their formula is not precisely in this particular form. –  Dmitri Pavlov Dec 30 '12 at 9:44

Euler's classical formula for convex polyhedra


where $v$ is the number of vertices, $e$ the number of edges and $f$ the number of faces of a convex triagulated polyhedron in $3$-space.

share|improve this answer
This should be two, not zero, shouldn't it? –  Kofi Dec 29 '12 at 16:16
Oops, I was sloppy: it should be $2$. And, for the formula to always give $2$, the polyhedron must be convex or at least homeomorphic to a sphere. For general polyhedra Euler's equation should read $v-e+f=\chi(P)$ where $P$ is, indeed, the polyhedron ...but this would just be a way to state that the Euler characteristic can be computed by the alternating sum of the number of k-cells of a (cellular) complex, so it would lose its "only combinatorics" appeal –  Qfwfq Dec 29 '12 at 16:36

At Michael Atiyah's 80th birthday conference, the cake had the Atiyah-Singer index formula:

$$\text{Ind}(D) = \int_{T^*M} \text{ch}(\sigma_D) \text{Todd}(TM \otimes \mathbb{C})$$

I can verify that it made the cake even more delicious.

share|improve this answer
I wouldn't mind a cake like that for me, even though my name has nothing to do with the Todd class. –  Todd Trimble Dec 29 '12 at 22:08

$\mathrm{P}=\mathrm{NP}$ or $\mathrm{P}\neq \mathrm{NP}$, whichever you prefer.

share|improve this answer
I thought of this but the problem is I just can't decide which one is a theorem. –  user10891 Dec 31 '12 at 13:11

In a different vein from the other answers, how about one of the classic visualizations of the proof of the Pythagorean theorem? It's basically just a bunch of triangles and squares rearranged in a couple ways, and would come out nicely with cake decorator colors. And folks might actually recognize it.

share|improve this answer
Or just a diagram of the Lunes of Alhazen (which I always thought were the lunes of Hippocrates): en.wikipedia.org/wiki/File:Lunules-better.png –  Goldstern Dec 29 '12 at 18:56

A geometric one, where the zero can be made a cake (circle) itself $$ x^2 + y^2 -1 = \Huge \circ $$

share|improve this answer

Gödel's completeness theorem: A (first order) sentence $\varphi$ is provable from the axioms $\Sigma$ iff it holds in every model of $\Sigma$: $$ \Sigma \vdash \varphi \Leftrightarrow \Sigma \vDash \varphi$$

share|improve this answer

Gödels incompleteness theorem in the language of modal logic (where $\Box\varphi$ means that $\varphi$ is provable - say in Peano Arithmetic - and $\bot=\lnot \top$ is any false statement): $$\Box \lnot \Box \bot \Rightarrow \Box \bot.$$

share|improve this answer

How about the snake lemma? It's not a formula, but it could still look great on a cake! Plenty of excellent .tex diagrams here: http://tex.stackexchange.com/questions/3892/how-do-you-draw-the-snake-arrow-for-the-connecting-homomorphism-in-the-snake-l

share|improve this answer

While it's math for a cookie, I'm sure it can be done on a cake as well.

$$|X|\lt|\mathcal P(X)|$$

share|improve this answer

Maybe just make the cake in the shape of a golden rectangle, and use two colors of icing to show the decomposition into a square and a smaller golden rectangle.

share|improve this answer

One which I like much is $$ \exp \left(\begin{bmatrix} . & . & . & . & .\\\ 1 & . & . & . & . \\\ . & 2 & . & . & . \\\ . & . & 3 & . & . \\\ . & . & . & 4 & . \\\ \end{bmatrix} \right)= \begin{bmatrix} 1 & . & . & . & . \\\ 1 & 1 & . & . & . \\\ 1 & 2 & 1 & . & . \\\ 1 & 3 & 3 & 1 & . \\\ 1 & 4 & 6 & 4 & 1 \\\ \end{bmatrix}$$ It is practically easier and a bit more iconic if we reduce it a bit - although for me it is not so pleasing, because the immediate remembering of the Pascal-triangle comes with the 1-4-6-4-1-row: $$ \Large \exp \small \left(\begin{bmatrix} . & . & . & . \\\ 1 & . & . & . \\\ . & 2 & . & . \\\ . & . & 3 & . \\\ \end{bmatrix} \right)= \begin{bmatrix} 1 & . & . & . \\\ 1 & 1 & . & . \\\ 1 & 2 & 1 & . \\\ 1 & 3 & 3 & 1 \\\ \end{bmatrix}$$

With a bit explanation which might be useful for other guests http://go.helms-net.de/math/binomial/index-Dateien/image008.png

share|improve this answer

I think the diagram should be several dotted rays emanating from the same point, arranged so that if you cut along the lines, each piece will have the same volume of cake and of frosting. It is an impressive diagram when the number of pieces is a not too small odd number such as 5, 7, or 9.

(There is also an interactive n player version.)

Gerhard "Save A Piece For Me" Paseman, 2012.12.29

share|improve this answer
I must be missing something. What's wrong with equally spaced rays emanating from the center? –  Steven Landsburg Jan 18 '13 at 2:51
For a round cake, that works Steven. I don't know if it is a sheet cake or has some other shape. If "equally spaced" means dividing the perimeter equally, that works for some polygonal prisms. I like the ones from a 13 by 9 inch pan myself. Gerhard "Has Some To Eat, Too" Paseman, 2013.01.17 –  Gerhard Paseman Jan 18 '13 at 4:14
Gerhard: Thanks. I somehow forgot that cakes are not always round. –  Steven Landsburg Jan 24 '13 at 21:59

Not famous, perhaps, but how about

$$\int_0^a f_A(x)dx = \int_a^1 f_A(x)dx = 1/2$$

from Better Ways to Cut a Cake by Brams, Jones, and Klamler?

share|improve this answer

(comment to D. Pavlov) I once attempted to bake GRR onto cookies (leavened with hartshorn, naturally). It didn't turn out too legible, but probably doable with icing.

share|improve this answer