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Mathematical Rebus I

Mathematical Rebus III


Rebus' image

$$\det\frac{\partial(x,y)}{\partial(r,\varphi)}\\ \sum_{n=2}^\infty \ddot{\frac{t^n}{n!}} $$

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  • $\begingroup$ Now I've the dilemma of whom to give the Correct Answer. $\endgroup$ – Masclins Jun 17 '15 at 13:07
  • $\begingroup$ If there are multiple answers that both give thorough and accurate answers, then go by timestamp. Point to the word "ago" in "answered x hours ago" and you will see the exact time posted. If in a strange case the answers were posted in the same minute, then I like to give the points to the person who is newer to the site, and probably doesn't have a lot of rep points. $\endgroup$ – JLee Jun 17 '15 at 13:38
  • $\begingroup$ The problem here is that it was half-solved by each. Though I like giving it to the one that's the newest. $\endgroup$ – Masclins Jun 17 '15 at 13:39
  • $\begingroup$ +1 nice puzzle! There's a certain beauty in not needing to have any words or explanation. $\endgroup$ – JLee Jun 17 '15 at 13:41
  • $\begingroup$ I spent way too long uselessly determining that the function in the graph is approximately sec(x-2.2) and I couldn't figure out why it was important. I wish you had done sec(x) instead =\ ... +1 anyway for creativity $\endgroup$ – Sabre Jun 17 '15 at 18:16
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Follow up to rand al'thor answer

The graph shows:

sec(x)

By combining the 2 other answers, we obtain:

sec + r + et = $secret$

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Graph:

work in progress...

$\det\frac{\partial(x,y)}{\partial(r,\varphi)}=r$, because

$x,y$ are Cartesian coordinates and $r,\varphi$ are polar coordinates, in the 2-dimensional plane.

$\sum_{n=3}^\infty \ddot{\frac{t^n}{n!}}=e^t$, because

differentiate $t^n$ twice to get $n(n-1)t^{n-2}$, so the sum is $\sum_{n=3}^\infty \ddot{\frac{t^{n-2}}{(n-2)!}} = \sum_{n=1}^\infty \ddot{\frac{t^n}{n!}}=e^t$.

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