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  $me^{rry} = x -mas$ !

+ 0 like - 0 dislike

Hi all, I wish ...

\[\begin{eqnarray} y & = & \frac{\ln\left( \frac{x}{m} - sa \right)}{r^2} \\ yr^2 & = & \ln\left( \frac{x}{m} - sa\right) \\ e^{yr^2} & = & \frac{x}{m} - sa \\ me^{yr^2} & = & x - msa \\ me^{rry} & = & x - mas \\ \end{eqnarray}\]

... to the whole PhysicsOverflow community ;-) !

asked Dec 23, 2015 in Discussion by Dilaton (6,040 points) [ revision history ]
recategorized Dec 23, 2015 by dimension10

I'm not sure if creating "fun" and "jokes" as tags for what will probably only be used for Christmas declarations is particularly useful. Also, people will probably come to the "Announcements" section looking for more important stuff, so I think this fits better in "Discussion".

(Also, am I the only one who got confused if the equation had any physical significance? Yeah, I know, taking the \(r^2\)th root within a logarithm is probably not going to be useful.)

@dimension10     I do not understand why Merry is equal to Christmas, to say the least.

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