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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?

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    $\begingroup$ Related (published in the most recent AMM): mpra.ub.uni-muenchen.de/34264/1/Perfect_division1.pdf $\endgroup$ Commented Dec 29, 2012 at 10:06
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    $\begingroup$ 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) $\endgroup$
    – user9072
    Commented Dec 29, 2012 at 14:05
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    $\begingroup$ 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. $\endgroup$ Commented Dec 29, 2012 at 17:17
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    $\begingroup$ I would go for the Pythagorean theorem, with a picture of the right triangle and squares. An evergreen that everybody will appreciate! $\endgroup$ Commented Dec 29, 2012 at 18:27
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    $\begingroup$ 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. $\endgroup$ Commented Dec 29, 2012 at 19:16

18 Answers 18

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My all-time favourite formula: Stokes theorem

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

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    $\begingroup$ Or perhaps $d^2=0$. $\endgroup$ Commented Dec 29, 2012 at 11:20
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    $\begingroup$ Or Cartan's structural equation $\Omega=\mathrm{d}\theta+\frac{1}{2}[\theta,\theta]$. $\endgroup$
    – Qfwfq
    Commented Dec 29, 2012 at 11:33
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$e^{i \pi} = -1$

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    $\begingroup$ $e^{i \pi} + 1 = 0$ might be better. But I expect this question will be closed. $\endgroup$ Commented Dec 29, 2012 at 10:02
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    $\begingroup$ Should involve pi somewhere in this context $\endgroup$ Commented Dec 29, 2012 at 15:02
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196884 = 196883 + 1

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  • $\begingroup$ I'm sorry, I'm not familiar with this. Could you explain where this equation comes from? $\endgroup$
    – JRN
    Commented Dec 29, 2012 at 23:44
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    $\begingroup$ @Joel Reyes Noche: en.wikipedia.org/wiki/Monstrous_moonshine $\endgroup$
    – user9072
    Commented Dec 30, 2012 at 0:50
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22/7. Because a cake is, approximately, a pi(e).

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    $\begingroup$ Shouldn't that be $\frac{22}7e$, then? $\endgroup$
    – Asaf Karagila
    Commented Dec 29, 2012 at 20:10
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    $\begingroup$ If you're approximating $\pi$, then why not approximate $e$ also? So $\pi e\approx \frac{22}{7}\frac{19}{7}$. $\endgroup$
    – JRN
    Commented Dec 30, 2012 at 7:18
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How about the Grothendieck-Hirzebruch-Riemann-Roch formula: ch(f!F) = f*(ch(F)td(Tf))?

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  • $\begingroup$ Yes, if I recall correctly, this one can also be admired on the glass front of the MPIM's reception desk. $\endgroup$ Commented Dec 29, 2012 at 20:12
  • $\begingroup$ @Tobias: Yes, although their formula is not precisely in this particular form. $\endgroup$ Commented Dec 30, 2012 at 9:44
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Euler's classical formula for convex polyhedra

$$v-e+f=2$$

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.

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  • $\begingroup$ This should be two, not zero, shouldn't it? $\endgroup$ Commented Dec 29, 2012 at 16:16
  • $\begingroup$ 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 $\endgroup$
    – Qfwfq
    Commented Dec 29, 2012 at 16:36
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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.

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    $\begingroup$ I wouldn't mind a cake like that for me, even though my name has nothing to do with the Todd class. $\endgroup$
    – Todd Trimble
    Commented Dec 29, 2012 at 22:08
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$\mathrm{P}=\mathrm{NP}$ or $\mathrm{P}\neq \mathrm{NP}$, whichever you prefer.

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  • $\begingroup$ I thought of this but the problem is I just can't decide which one is a theorem. $\endgroup$
    – user10891
    Commented Dec 31, 2012 at 13:11
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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.

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

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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): $$\Box \lnot \Box \bot \Rightarrow \Box \bot.$$

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

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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.

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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: https://tex.stackexchange.com/questions/3892/how-do-you-draw-the-snake-arrow-for-the-connecting-homomorphism-in-the-snake-l

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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

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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?

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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

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  • $\begingroup$ I must be missing something. What's wrong with equally spaced rays emanating from the center? $\endgroup$ Commented Jan 18, 2013 at 2:51
  • $\begingroup$ 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 $\endgroup$ Commented Jan 18, 2013 at 4:14
  • $\begingroup$ Gerhard: Thanks. I somehow forgot that cakes are not always round. $\endgroup$ Commented Jan 24, 2013 at 21:59
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(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.

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