53

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
Hypergeometricx
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2 Answers2

67

$$ \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)

Hypergeometricx
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6

Notice:

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

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

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

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