Mondayman
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Ah Indeed it was! When loading latex my page tends to jump and skip posts.ProfuselyQuarky said:Hey, that was mine . . .![]()
Ah Indeed it was! When loading latex my page tends to jump and skip posts.ProfuselyQuarky said:Hey, that was mine . . .![]()
You can take it, if you desire :)Mondayman said:Ah Indeed it was! When loading latex my page tends to jump and skip posts.
collinsmark said:For the last several years, I've been partial to the Gaussian integral.
\sqrt{\pi} = \int_{- \infty}^\infty e^{-x^2}dx
- Visually, while one side is edgy with the square root sign, and other the side is very curvy.
- It is a relationship between \pi and e, two of the most important mathematical constants.
- Everybody loves \pi, and e is not far behind.
- One side involves a square root, while the other side has a square.
- Taking an exponent to another exponent is always cool.
- The function e^{-x^2} defines the general shape of the "bell curve" and is extremely important within probability theory and statistics, including the Central Limit Theorem (a profound theorem within probability theory).
- While the equation is strictly mathematical, it does have many applications in physics (the probability/statistics go without saying). The e^{-x^2} bell curve shape is the general "shape" (complex envelope, if you prefer) of a wavefunction that minimizes the Heisenberg uncertainty.
- Edit: and the analytical proof of the Guassian integral is beautiful in its own right, but I'll leave the proof out of this post.
It's a post modern interpretation of the OP's requirements. It's not actually designed to win, but to encourage thinking outside the box.DaTario said:But we are seeing equations involving different quantities of physics (forces, angular momentum, photographic blurring, etc) the candidate should be asked to present the equation.
Besides the fact that the photo also presents inequalities.![]()
Hepth said:I like a function that I found on my own, and saved me a ton of headaches in a project. It allows one to reduce expressions with derivatives of delta functions (assuming an integral over u):
$$
F(u)\delta^{(n)}(u) = \sum_{i=0}^n (-1)^{n-i} \left(\frac{n!}{i!(n-i)!}\right) F^{(n-i)}(0) \delta^{(i)}(u) $$
where
$$f^{(i)}(u) \equiv \frac{\partial^i}{\partial u^i} f(u)$$
Honestly I'm not sure if it has a reference anywhere, possibly a hepth original?
haha i actually deleted it as i realized it wasn't so aesthetically pleasing...micromass said:Why do you hate binomial coefficients?
You're too late. @Samy_A already entered that one. (You've got to check the whole thread before posting an entry...debajyoti said:Cauchy's Integral Formula:
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micromass said:The only thing prettier than this is the proof of this equality.
atyy said:Can you prove it without going to 2 dimensions?
micromass said:Yes, that's definitely possible.
atyy said:Give me a clue?
These may be of interest here, too,kith said:Regarding aesthetics in quantum mechanical expressions, we had a thread from a tattoo artist in the Quantum Physics forum not too long ago which may be of interest to the readers here.