Xmas Maths 2015

Simplify the expression below into a seasonal greeting using commonly-used symbols in commonly-used formulas in maths and physics. Colours are purely ornamental!


\begin{align}
\frac{
\color{green}{(x+iy)}
\color{red}{(y^3-x^3)}
\color{orange}{(v^2-u^2)}
\color{red}{(3V_{\text{sphere}})^{\frac 13}}
\color{orange}{E\cdot}
\color{green}{\text{KE}}
}
{
\color{orange}{2^{\frac 23}}
\color{green}{c^2}
\color{red}{e^{i\theta}}
\color{orange}{v^2}
\color{green}{(x^2+xy+y^2)}}
\color{red}{\sum_{n=0}^{\infty}\frac 1{n!}}
\color{orange}{\bigg/}
\color{orange}{\left(\int_{-\infty}^\infty e^{-x^2} dx\right)^{\frac 23}}
\end{align}

NB: Knowledge of the following would be helpful:

Basic Maths:

  • Taylor series expansion
  • Normalizing factor for the integral of a normal distribution
  • Rectangular and polar forms for complex variables
  • Volume of a sphere

Basic Physics:

  • Kinematics formulae for motion under constant acceleration
  • Einstein’s equation
  • One of the energy equations

Answer


\begin{align}
&\frac{
\color{green}{(x+iy)}
\color{red}{(y^3-x^3)}
\color{orange}{(v^2-u^2)}
\color{red}{(3V_{\text{sphere}})^{\frac 13}}
\color{orange}{E\cdot}
\color{green}{\text{KE}}
}
{
\color{orange}{2^{\frac 23}}
\color{green}{c^2}
\color{red}{e^{i\theta}}
\color{orange}{v^2}
\color{green}{(x^2+xy+y^2)}}
\color{red}{\sum_{n=0}^{\infty}\frac 1{n!}}
\color{orange}{\bigg/}
\color{orange}{\left(\int_{-\infty}^\infty e^{-x^2} dx\right)^{\frac 23}}\\
&=
\frac{
\color{green}{(x+iy)}
\color{red}{(y^3-x^3)}
\color{orange}{(v^2-u^2)}
\color{red}{(3V_{\text{sphere}})^{\frac 13}}
\color{orange}{E\cdot}
\color{green}{\text{KE}}
}
{
\color{red}{e^{i\theta}}
\color{green}{(x^2+xy+y^2)}
\color{orange}{\cdot2^{\frac 23}}
\color{green}{c^2}
\color{orange}{v^2}
}
\color{red}{\sum_{n=0}^{\infty}\frac 1{n!}}
\color{orange}{\bigg/}
\color{orange}{\left(\sqrt{\pi}\right)^{\frac 23}}\\
&=
\color{green}{\left(\frac{x+iy}{e^{i\theta}}\right)}
\color{red}{\left(\frac{y^3-x^3}{x^2+xy+y^2}\right)}
\color{orange}{(v^2-u^2)}
\color{red}{\left(\frac {(3V_\text{sphere})^\frac 13}{\left(2\sqrt{\pi}\right)^{\frac 23}}\right)}
\color{orange}{\left(\frac{E}{c^2}\right)}
\color{green}{\left(\frac{\text{KE}}{v^2}\right)}
\color{red}{\sum_{n=0}^{\infty}\frac 1{n!}}
\\
&=
\color{green}{\left(\frac{re^{i\theta}}{e^{i\theta}}\right)}
\color{red}{\left(\frac{(y-x)(y^2+xy+x^2)}{x^2+xy+y^2}\right)}
\color{orange}{(v^2-u^2)}
\color{red}{\left(\frac {3\cdot \frac 43 \pi r^3}{4\pi}\right)^\frac 13}
\color{orange}{\left(\frac{mc^2}{c^2}\right)}
\color{green}{\left(\frac{\frac 12 mv^2}{v^2}\right)}
\color{red}{(e)}
\\
&=
\color{green}{\left(r\right)}
\color{red}{\left(y-x\right)}
\color{orange}{(2as)}
\color{red}{\left(r^3\right)^\frac 13}
\color{orange}{\left(m\right)}
\color{green}{\left(\frac 12m\right)}
\color{red}{(e)}
\\
&=
\color{green}{\left(r\right)}
\color{red}{\left(y-x\right)}
\color{orange}{(as)}
\color{red}{\left(r\right)}
\color{orange}{\left(m\right)}
\color{green}{\left( m\right)}
\color{red}{(e)}
\\
&=
\color{orange}{\left(m\right)}
\color{red}{(e)}
\color{green}{\left(r\right)}
\color{red}{\left(r\right)}
\color{red}{\left(y-x\right)}
\color{green}{\left(m\right)}
\color{orange}{(as)}
\end{align}

Merry Christmas, everyone!!


The following links might be helpful.
Complex numbers and polar coordinates
Difference of two cubes
Kinematics formulae for constant acceleration in a straight line
Volume of a sphere
Einstein’s mass-energy equivalence
Kinetic energy
Taylor/Maclaurin series expansion of e
Gaussian integral (normalizing factor for the normal distribution)

Attribution
Source : Link , Question Author : Hypergeometricx , Answer Author : Hypergeometricx

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