ITT we take (to some extent...

ITT we take (to some extent, well known) formulas in math / science and rewrite them in an unconventional but still mathematically valid way, and other's try and guess what formula/law it is. I'll go first. Pic related.

e=mc2
pretty clever

this doesn't even deserve a response

There is no other way to write that equation

what about dA/dt = constant, also known as Kepler's 2nd law

Integrate it a couple times. A = C1 + C2

:-D i'm the greatest

I would personally never write dA/dt = constant, d^2A/dt^2 = 0 is much simpler and that's why it's convention

A with the double dot is precisely what d^2A/dt^2 means, am I being trolled right now?

d(2K/m)^(.5)/dt = m(dv/dx)v

amazing thanks woow

Writing Adot is constant is like writing Adot +1 = constant of whatever, it's exactly the same

A = f * J

The sequence of polynomials

[math] p_n(x) := \left(1 + \dfrac {x} {n} \right)^n [/math]

have

[math] {\rm e}^x = \lim_{n\to \infty} p_n(x) [/math]

and I recently asked myself how they compare with

[math] q_n(x) := \sum_{k=0}^n \dfrac{1}{k!} x^k [/math]

which have the same limit. And it's not hard: With

[math] \left(x+y\right)^m=\sum_{k=0}^m \dfrac{n!}{k!\,(m-k)!} x^k y^{m-k} [/math]

you find

[math] \sum_{k=0}^n a_k(n)\dfrac {1} {k!} x^k [/math]

with

[math] a_k(n)=\prod_{j=1}^{k-1}\left(1-\dfrac{k-j}{n}\right)\le 1 [/math]

For n to infinity, all [math] a_k [/math] become 1 and you get the classical series expansion.

I found there's also an interesting version of the exponential function that you obtain if you merely require the property [math] \frac{{\rm d}}{{\rm d}x} {\rm e}^{c\, x} = c\, {\rm e}^{c\,x} [/math] to hold at one or a few points, not globally.

Leibnz notation is obsolete. Please only use Lagrange.

You still don't my point: how is it difficult to guess what it is if it's blatantly obvious to every brainlet that it's the same

Here's a more challenging one
[math]S = \int d \tau \left( 2\frac{\dot{X}^2}{\tau^2} - \frac14 \tau^2 m^2 \right) [/math]

oops, should be
[math]S = \int d \tau \left( \frac{\dot{X}^2}{\tau^2} - \frac14 \tau^2 m^2 \right)[/math]

$1+1=2-1 \mod 1$

please dont tell me you write things like [math](x^2)'[/math]

What is A??

should be some string theory action

Pretty cool, what do you mean by an interesting version of the exponential function?
[math]\frac{{\rm d}}{{\rm d}x} {\rm e}^{c\, x} = c\, {\rm e}^{c\,x}[/math] implies your function is [math]{\rm e}^{c\, x}[/math] if its an exponential

I always liked
[math] \mathbf{f} + \mathbf{f}^* =0 [/math] from Kane.

I would give that the D

I mean the polynomial

[math] e_n(x) := \left(1 + \dfrac {1} {1- k\, x_0 \, /\, n }\, k\, \frac{x}{n} \right)^n [/math]

with

[math] e^{k\, x} = \lim_{n \to \infty} e_n(x) [/math]

has, for all n,

[math] \dfrac{d}{dx} e_n(x) = \dfrac{1}{1 + k\, (x-x_0)\, / \, n} \, k\, e_n(x) [/math]

and thus

[math] e_n'(x_0) = k\, e_n(x_0) [/math]

replies to

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And you are a retard scum of the earth

First letter of the English alphabet you retard

You absolute moron, why did you necro this?

The second derivative of something with respect to time is zero.

What the hell are you talking about some threads on sci move at like one post per week