I am a mathematician and my significant other is a numerologist.

She warned me that this Valentine's Day is the most important ever because it is 2/14/2014. I can't afford to mess this one up.

I know that if I am to truly show her I care (and not strike out) I must use math and numbers.

Her initials are AB so one thing I have come up with so far is the following strictly algebraic (except for one arccos and the constant $\pi$) equation which plots AB (her initials) in a heart.

$$ \left(\left(\left(x+\frac{7}{4}\right)^2+\left(\frac{2 y}{3}+\frac{1}{4}\right)^2\right)^2+\frac{9}{2} \left(\left(x+\frac{7}{4}\right)^2+\left(\frac{2 y}{3}+\frac{1}{4}\right)^2\right)-4 \left(\left(\frac{2 y}{3}+\frac{1}{4}\right)^3-\left(x+\frac{7}{4}\right)^2 \left(2 y+\frac{3}{2}\right)\right)-\frac{27}{16}\right) \left(\left(\left(x+\frac{7}{4}\right)^2+\left(\frac{2 y}{3}+\frac{3}{4}\right)^2\right)^2+18 \left(\left(x+\frac{7}{4}\right)^2+\left(\frac{2 y}{3}+\frac{3}{4}\right)^2\right)-8 \left(\left(\frac{2 y}{3}+\frac{3}{4}\right)^3-\left(x+\frac{7}{4}\right)^2 \left(2 y+\frac{9}{4}\right)\right)-27\right) \left(\sqrt{\left(-x-\frac{11}{4}\right)^2+\left(\frac{2 y}{3}+\frac{7}{4}\right)^2}+\sqrt{\left(-x-\frac{3}{4}\right)^2+\left(\frac{2 y}{3}+\frac{7}{4}\right)^2}-\frac{5}{2}\right) \left(\sqrt{\left(x-\frac{1}{2}\right)^2+(y-2)^2}+\sqrt{\left(x-\frac{1}{2}\right)^2+(y+2)^2}-\frac{21}{5}\right) \left(\sqrt{1-\left(\left| \frac{x}{5}\right| -1\right)^2}-\frac{y}{5}+\frac{3}{4}\right) \sqrt{\frac{\left| \sqrt{\left(\frac{2 y}{3}+2\right)^2+\left(x+\frac{11}{4}\right)^2}+\sqrt{\left(\frac{2 y}{3}+2\right)^2+\left(x+\frac{3}{4}\right)^2}-\frac{5}{2}\right| }{\sqrt{\left(x+\frac{11}{4}\right)^2+\left(\frac{2 y}{3}+2\right)^2}+\sqrt{\left(x+\frac{3}{4}\right)^2+\left(\frac{2 y}{3}+2\right)^2}-\frac{5}{2}}} \left(\left((\left| y\right| +1)^2+(x-2)^2\right)^2-19 \left((\left| y\right| +1)^2-(x-2)^2\right)\right) \left(\left(\left(\left| y\right| -\frac{29}{20}\right)^2+(x-1)^2\right)^2+18 \left(\left(\left| y\right| -\frac{29}{20}\right)^2+\left(x-\frac{219}{100}\right)^2\right)-8 \left(\left(x-\frac{5}{2}\right)^3-3 \left(x-\frac{39}{20}\right) \left(\left| y\right| -\frac{147}{100}\right)^2\right)-27\right) \sqrt{\frac{\left| \sqrt{(y-2)^2+\left(x-\frac{9}{20}\right)^2}+\sqrt{(y+2)^2+\left(x-\frac{9}{20}\right)^2}-\frac{21}{5}\right| }{\sqrt{\left(x-\frac{9}{20}\right)^2+(y-2)^2}+\sqrt{\left(x-\frac{9}{20}\right)^2+(y+2)^2}-\frac{21}{5}}} \left(\cos ^{-1}\left(1-\left| \frac{x}{5}\right| \right)-\frac{y}{5}+\frac{3}{4}-\pi \right)=0 $$

AB Heart

There is a little noise in the figure because apparently the contourplot function in Mathematica has some issues. I set MaxRecursion to 5 when I plotted it.

I have been experimenting with other ways to express love, beauty and Valentine's Day through math. I have tried circle packing, heart packing, and tessellations and to derive new heart curves. Unfortunately I am not that experienced with Mathematica for some of this stuff.

Wolfram recognizes some common heart curves here. Wolfram Heart Curves

Here is a nice Sierpinski heart from xkcd. xkcd Sierpinski Valentine

The cardioid curve derives its name from the Greek word for heart and the curve is heart like by its cusp. Here someone has modified the envelope of lines method that is sometimes used to produce a cardioid to create an actual heart. How to Create Concentric Circles, Ellipses, Cardioids & More Using Straight Lines and a Circle

Cardioid/Heart from envelope of lines

Does anyone know of other fun math like this for Valentine's Day? Can those interested please post your interpretations of combining math and love for this coming epic Valentine's Day of 2/14/2014?

Happy Valentine's Day 2/14/2014 math.stackexchange!

  • 3
    $\begingroup$ This is impressive. $\endgroup$ – Baby Dragon Feb 8 '14 at 7:43
  • 5
    $\begingroup$ Whatever you do for graphics, be sure to write: $$\text{Happy } \;(\text{valen})\times(\text{day})!$$ $\endgroup$ – Blue Feb 8 '14 at 8:40
  • $\begingroup$ +1, +1, +1 ... this logic is awesome! Since your question got closed, you should post on an answer on this question. math.stackexchange.com/questions/12098/… $\endgroup$ – MacGyver Feb 17 '14 at 16:20

For fellow singles out there, the stable marriage problem gives us hope! Moreover, the Gale-Shapley algorithm guarantees the optimal pairing for those who "propose": mathematical evidence for us to take matters into our hands.

(If only real life is so simple eh)

  • 2
    $\begingroup$ The set of unmatched people is the same in every stable matching. $\endgroup$ – Michael Greinecker Feb 9 '14 at 1:07

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