$$ 4\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\\ (-\infty,...,-1,0,1,...,\infty)\times(-\infty,...,-1,0,1,...,\infty)\\ \forall\begin{bmatrix}{-1}&{0}\\ {0}&{-1} \end{bmatrix} $$
$\begingroup$
$\endgroup$
6
-
1$\begingroup$ This is beautiful! I think I've half-realized the solution, though there's a piece that's giving me a hard time! $\endgroup$– leoll2Jun 2, 2015 at 11:48
-
$\begingroup$ nice one. i think i've got everything except for the matrix. that's the part giving me a headache $\endgroup$– user12241Jun 2, 2015 at 12:15
-
$\begingroup$ This definitely involves some circular reasoning, but I also haven't figured out the right way to read the matrix. $\endgroup$– Glen OJun 2, 2015 at 12:19
-
2$\begingroup$ Really , nice one! $\endgroup$– Saurabh PrajapatiJun 2, 2015 at 12:48
-
3$\begingroup$ Perfect puzzle! +1 $\endgroup$– BmyGuestJun 2, 2015 at 20:01
|
Show 1 more comment
1 Answer
$\begingroup$
$\endgroup$
5
The first line is
equal to Pi.
The second line is
the integers, Z, multiplied by itself, making Z2.
The third line is
multiplying some matrix ∀ by the negative identity matrix, negating its value. The letter ∀ negated is A.
All together,
Pi + Z2 + A = Pizza.
-
9$\begingroup$ Wow, the last line was very lateral-thinking! I really couldn't realize the meaning of "for every point reflection" $\endgroup$– leoll2Jun 2, 2015 at 12:27
-
2$\begingroup$ When I thought about using the Matrix for making $\forall$ become A, I thought about the 180º rotation Matrix. $\endgroup$– MasclinsJun 2, 2015 at 12:28
-
1$\begingroup$ Ah! Rotation. I figured there was a more specific term than "negating". I should have been thinking more geometrically. $\endgroup$– KevinJun 2, 2015 at 12:43
-
1$\begingroup$ Surely if you're applying a rotation matrix to $\forall$, the matrix ought to be placed before the $\forall$. $\endgroup$ Jun 2, 2015 at 22:07
-
4$\begingroup$ @DavidZhang - $\forall$ must be a row vector, I guess. $\endgroup$– Glen OJun 3, 2015 at 3:06