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ITT we take (to some extent, well known) formulas in math / science

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Thread replies: 31
Thread images: 6

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.
>>
>>8923935
e=mc2
pretty clever
>>
>>8923943
this doesn't even deserve a response
>>
>>8923935
There is no other way to write that equation
>>
>>8923959
what about dA/dt = constant, also known as Kepler's 2nd law
>>
File: sage.jpg (23KB, 700x466px) Image search: [Google]
sage.jpg
23KB, 700x466px
>>8923959
Integrate it a couple times. A = C1 + C2

:-D i'm the greatest
>>
>>8923967
I would personally never write dA/dt = constant, d^2A/dt^2 = 0 is much simpler and that's why it's convention
>>
>>8923977
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
>>
>>8924003
amazing thanks woow
>>
>>8923984
Writing Adot is constant is like writing Adot +1 = constant of whatever, it's exactly the same
>>
>>8923935
A = f * J
>>
File: reading_club.gif (2MB, 600x338px) Image search: [Google]
reading_club.gif
2MB, 600x338px
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.
>>
>>8923977
Leibnz notation is obsolete. Please only use Lagrange.
>>
>>8924036
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]
>>
>>8924053
oops, should be
[math]S = \int d \tau \left( \frac{\dot{X}^2}{\tau^2} - \frac14 \tau^2 m^2 \right)[/math]
>>
>>8923935
$1+1=2-1 \mod 1$
>>
>>8924036
please dont tell me you write things like [math](x^2)'[/math]
>>
What is A??
>>
>>8924059
should be some string theory action
>>
>>8924026
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
>>
>>8923935
I always liked
[math] \mathbf{f} + \mathbf{f}^* =0 [/math] from Kane.
>>
File: initiald.jpg (358KB, 817x818px) Image search: [Google]
initiald.jpg
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>>8924199
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]
>>
>>8925532
replies to
>>8925332
>>
>>
>>8924036
And you are a retard scum of the earth
>>
>>8925265
First letter of the English alphabet you retard
>>
>>8927285
You absolute moron, why did you necro this?
>>
>>8923935
The second derivative of something with respect to time is zero.
>>
>>8927290
What the hell are you talking about some threads on sci move at like one post per week
Thread posts: 31
Thread images: 6


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