# 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
• I can just see some people say, "Bah, humbug!" ;) Dec 24, 2015 at 13:47
• @TitoPiezasIII - Haha. Well I take all comments constructively and try to address them where possible (hence the edits for context). Dec 24, 2015 at 13:53
• And here's the only thing I could not get of this expression's simplification below :). Dec 24, 2015 at 14:29
• @hypergeometric Did you made it? Very exciting question! Dec 25, 2015 at 11:28

\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!!

• Merry Christmas, my good friend!
– user98186
Dec 25, 2015 at 19:11
• @NimaBavari - Thank you, and to you too! Dec 26, 2015 at 16:11

Notice:

• $$\sum_{n=0}^{\infty}\frac{1}{n!}=\lim_{m\to\infty}\sum_{n=0}^{m}\frac{1}{n!}=\lim_{m\to\infty}\left(1+\frac{1}{m}\right)^m=e$$

• $$\int_{-\infty}^{\infty}e^{-x^2}\space\text{d}x=\lim_{a\to\infty}\int_{-a}^{a}e^{-x^2}\space\text{d}x=\lim_{a\to\infty}\left[\frac{\text{erf}(x)\sqrt{\pi}}{2}\right]_{-a}^{a}=\sqrt{\pi}$$

• $$\text{V}_{sphere}=\frac{4\pi r^3}{3}$$

• $$\text{E}=mc^2$$
• $$\text{EK}=\frac{mv^2}{2}$$