Post beautiful Equations

a^2 + b ^2 = c^2

>>8797260
[math]\sin( \pi z ) = \frac{ \pi z } { z! (-z)! } [/math]

[math]x=\frac{-b \pm\sqrt{b^2 -4ac}}{2a}[/math]
Find a better equation
Protip: You can't

>>8797420
>fraction bar

[math]x = (-b\pm (b^2-4ac)^{1/2})/2a[/math]

x + x = x^2

>>8797455
>using numbers

[math]x = (-b\pm (b^p-p^pac)^{(p/p)/p})/pa[/math]

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>>8797457
I teared up a bit.

let a be equal to b
therefore:
[math]b=a[/math]

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>>8797420

>>8797607
now post the general solution of the quintic. i'm waiting.

e=mc2

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>>8797470
>mfw it draws a dick

desu
[math]e^{i\pi} + 1 = 0[/math]

>>8797260
Taylor can go fuck himself.

>>8797485
There are a lot of assumptions here that need to be assumed before you can conclude that anon.

>>8797696
Never got why people hate Taylor series' so much. I think they're cool. I guess it's just because it's the first "difficult" math most people are exposed to.

>>8797703
I've never been a fan of approximations, even if they're incredibly useful. But then again I don't mind working with infinitesimals, so I guess I'm a hypocrite.

>>8797683
When shit hits the fan but you still believe

>>8797683
is this just nonsense or what?

>>8797700
let's assume that everything I say is right
godel btfo

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>>8797260

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>>8797815
oops

E = hf

idk why but i think its simplicity if beautiful

x=x

>>8797832
Wow get out. We don't even know that's true

(A+B)/A= A/B = 1.618... =phi

>>8797819
fucking lol

P=NP

>>8797260
A=A

>>8797683
Is there an equals sign or am I too drunk to see it. Fuck
>when the equations become so complicated it's like playing where's Waldo

>>8797796
>>8797927

It's stupid wankery to make 'physicists' feel smart about themselves. Basically it's just the Lagrangian (density) of a 'unified' quantum field theory (not really, but they've thrown basically every term they know of into it - notice how it's really just nothing but sums of disjointed terms). More to the point, they've just expanded shit out needlessly, again to dickwave.

So basically you're right, there's no equal sign, because this just 'equals' [math]\mathcal{L}[/math]. See https://en.wikipedia.org/wiki/Lagrangian_(field_theory)

[math]\pi^{\mathrm c i \phi e 0} = 1[/math]
All these natural constants, like the speed of light, the golden ratio, the imaginary unit, all of them in one simple equation.
It doesn't get any more beautiful than that.

>>8797819

TopKek.

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>>8798149

>>8797796
Standard Model Lagrangian
http://nuclear.ucdavis.edu/~tgutierr/files/stmL1.html

[math]
[1,1,1,...] = \sqrt{1+\sqrt{1+\sqrt{1+...}}} = \frac{1+\sqrt{5}}{2}
[/math]

>>8798181
It's nice I guess, but once you know anything about Kolmogorov complexity, it's not surprising.
The left hand side, call it
[math] x = \sqrt{1+x} [/math]
is so simple to specify that it fulfills
[math] x^2 - 1 = x [/math]
and this on the other hand will have a simple solution. And it's given by the right hand side.

>>8798273
The beaty lies in the fact, that the famous golden ration (did you notice it?) can be expressed by such satisfying simple endless recustions. (You can do the same with [math] [1,1,1,...] [/math] which leads to the equation [math]x = 1 + \frac{1}{x}[/math]).

What has it to do with kolmogorov complexity, can you explain pls? I don't see a link here.

>>8798181
>>8798273
>>8798291
More interesting is
[eqn] 3 = \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}} [/eqn]

>>8797607
This solution can be greatly simplied by setting [math] p = 2b^2 - 9abc + 27a^2 d [/math] and [math] q = b^2 -3ac [/math].

There's simpler solutions than this if the function meets certain conditions. Like the trigonometry solution for instance.

>>8798433
ax^5 + bx^4 + cx^3 + dx^2 + ex + f = 0
x^5 + x^4 + x^3 + x^2 + x = -f/(abcde)
x^15 = -f/(abcde)
x = (-f/abcde)^1/15

your move abel

If ad = bc, then
[eqn]21 \, \frac{(a+b+c)^5-(b+c+d)^5-(a-d)^5+(c+d+a)^5-(d+a+b)^5+(b-c)^5}{(a+b+c)^3-(b+c+d)^3-(a-d)^3+(c+d+a)^3-(d+a+b)^3+(b-c)^3} = 25 \, \frac{(a+b+c)^7-(b+c+d)^7-(a-d)^7+ (c+d+a)^7-(d+a+b)^7+(b-c)^7}{(a+b+c)^5-(b+c+d)^5-(a-d)^5+(c+d+a)^5-(d+a+b)^5+(b-c)^5}[/eqn]

>>8797695
Stop posting this

1+1 = 2

Can't argue with the classics.

For 0 < 2a < b, [eqn]\int_{-\infty}^\infty \frac{1+x^2/b^2}{1+x^2/a^2} \ \frac{1+x^2/(2b)^2}{1+x^2/(a+b)^2} \ \frac{1+x^2/(3b)^2}{1+x^2/(a+2b)^2} \cdots \ \mathrm{d}x = b \tan \frac{a \pi}{b}[/eqn]Let 0 < b < 2a, then [eqn]\int_{-\infty}^\infty \frac{1+x^2/(a+b)^2}{1+x^2/b^2} \ \frac{1+x^2/(a+2b)^2}{1+x^2/(2b)^2} \ \frac{1+x^2/(a+3b)^2}{1+x^2/(3b)^2} \cdots \ \mathrm{d}x = \frac{\pi ab}{2a-b}[/eqn]

>>8798463
>>8798606
That's not beauty.

>>8797703
It's because calculus is taught in a way that disguises how important and fundamental sequences really are.

>>8798612
[eqn]\frac{\displaystyle \int_0^\infty \frac{1+q^6z^5}{1+q^5z^5} \, \frac{1+q^{11}z^5}{1+q^{10}z^5}\,\frac{1+q^{16}z^5}{1+q^{15}z^5}\cdots \, \frac{\mathrm{d}z}{1+z^5}}{\displaystyle \int_0^\infty \frac{1+q^7z^5}{1+q^5z^5}\,\frac{1+q^{12}z^5}{1+q^{10}z^5}\,\frac{1+q^{17}z^5}{1+q^{15}z^5}\cdots \, \frac{z \, \mathrm{d}z}{1+z^5}} = \varphi \, \frac{1+q}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\ddots}}}}[/eqn]

>>8798431
Haha, yes, Ramanujans problem, isn't it?

Should also be posted on such a thread:
[math]
a^{p-1} \equiv 1 \text{ mod }p
[/math]

>>8798629
Still not beauty dude. This just looks like some sperg did a whole bunch of algebraic manipulations on simple formula and then threw it into an integral with a known solution for the original expression then tried to pass it off as deep/cool.

A truly beautiful equation should simplify, not complicate.

>>8798523
Why

>>8797695
Came here to post this.

I think mean value theorem is an underrated result. It's simple, intuitive, and incredibly powerful (it's at the heart of the proof the the fundamental theorem of calculus for instance)

[math]f(b) - f(a) = f'(x)(b-a)[/math].

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There is this one differential equation that I learned in mechanics of solids. It was pretty neato

>>8797675
Did you meant
[math] \displaystyle E^2 = \frac{(m_0 c^2)^2}{1-v^2/c^2} [/math]
?

Light speed equation in a vacuum:
[math]\displaystyle c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}[/math]
Light speed in any medium
[math]\displaystyle v_m=\frac{1}{\sqrt{\mu_0 \mu_r \varepsilon_0 \varepsilon_r}}=\frac{c}{\sqrt{\mu_r \varepsilon_r}}[/math]

[math]0.999... = 1[/math]

>>8798629
>>8798606
>>8798463
Are you the same anon? Where did you get these? I think they are very interesting and beautiful.

These are all me because I spamfag
>>8799226
>>8799240
>>8799247

>>8798629
>>8798606
>>8798463
Are you the same anon? I think these are very beautiful. Please tell me where you learned these.

>>8798647
Yes, I read that in the Man who Knew Infinity when I was in high school. Funny enough, one time years later in college someone gave me the problem [math] \sqrt{x} + y = 7, x + \sqrt{y} = 11 [/math]. I remembered this from The Man Who Knew Infinity and immediately said [math] x = 9, y = 4 [/math].

Here is a simple proof of that formula.

[eqn] a^p = (1 + 1 + \cdots + 1)^p = p + \sum_{a,b,c,\cdots,z < p} \ \ \ \frac{p!}{a! b! c! \cdots z! (p-a)! (p-b)! (p-c)! \cdots (p-z)!} [/eqn]

Since [math] p [/math] is prime we may take mod [math] p [/math] of both sides to get

[eqn] a^p \equiv p \ \ \text{mod} \ p [/eqn]

>>8798668
I so not agree. There's beauty in mathematics through and through.

>>8799269
That proof is in Carr's book, first volume. I went through it in 10th grade or so.

>>8798629
[math] p< n [/math] and [math] p,n \in \mathbb{N} [/math]

[eqn] \int \frac{\cos^p x}{\cos nx} dx = \frac{1}{n} \sum_{r=1}^{r=n} (-1)^{r+1} \ \ \cos^p (2r-1)\theta \log \frac{1+\cot(2r-1)\frac{\theta}{2}\tan\frac{x}{2}}{1 - \cot(2r-1)\frac{\theta}{2}\tan\frac{\theta}{2}} [/eqn]

Not as elegant as yours, but still a fun challenge that I give out form time to time. It's in Carr's second book.

>>8799305
[math] \theta = \frac{pi}{2n} [/math] and bottom fraction of the log term, that should be [math] \tan \frac{x}{2} [/math] not [math] \tan \frac{\theta}{2} [/math]

>>8799269
Oops, on the RHS first term I said p. What I meant to say was a.

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>>8797260
But that's wrong!

>>8797819
that inequality looks disgusting. Can someone explain this meme to me?

>>8797845
It's the foundation of all of quantum theory. Fuck off

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>>8797260

>>8799863
>generates nth prime
>literally divides every number by every number under it until it finds one that is prime

>>8799866
well it technically works

>>8799873
>>8799863
[eqn] a_3 x^3 + a_2 x^2 + a_1 x + a_0 = 0 [/eqn]

[math] \implies x = \phi(x) [/math] where [math] \phi [/math] is a function that finds the roots of the polynomial in question.

>>8797946
Apparently it also assumes that neutrinos are massless, which is wrong

>>8799162
no

25 + 100 = 125

>>8797420
Cubic equations!

>>8799919
9+16=25
25+144=169

>>8798149
underrated

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>>8797260
literally perfect

>>8797264
/thread

>>8799947
So... (3^2 + 4^2 + (3*4)^2) mod 10^2 = 69 ?

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mfw
[math]e^{(-i\pi)^0}\frac{\phi}{e}[/math]
is the golden ratio

>>8800047
what the fuck it works perfectly in the preview
what is this shit

>>8800047
>phi = golden ratio
Mind is defined as blown.

>>8797264
function of a circle therefore it's perfect
LORD PI WATCH UPON US

>>8797710
It's not an approximation unless you truncate it buddy boi

>>8800062
But you can expand it into infinite series iff the function is infinitely differentiable, so if f is differentiable only n times then it can't be expressed accurately using Taylor polynomial

>>8800074
Sure, but it's implied that we are talking about infinitely differentiable functions.

You're not wrong, it's just besides the point.

>>8800051
why is sci so autistic

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>>8797260

>>8800278
Field theory ain't beautiful

>>8800272
I odn't know.. but probably the same reason why you keep posting gay memes no one thinks are funny

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The Eilenberg identity
[math]e^\varphi = i\pi^{-1}[/math]
connecting natural logarithms, the golden ratio, imaginary numbers, circles and negative numbers.

>>8800367
The exponential function of a real is a complex number, neat

>>8800367
That is simply not true. However this equation is

[eqn] e^{i\pi /5} = frac{\phi}{2} + i \frac{sqrt{10-2\sqrt{5}}{4} [/eqn]

>>8800367
You're equation is true, but this one is
[eqn] e^{i\pi /5} = \frac{\phi}{2} + i \frac{\sqrt{10-2 \sqrt{5}}}{4} [/eqn]

>>8800046
No, he's saying saying
[eqn] 5^2 + 10^2 = 15^2 [/eqn]
[eqn] 3^2 + 4^2 = 5^2 [/eqn]
[eqn] 5^2 + 12^2 = 13^2 [/eqn]

e+pi=i+1+0
all constants omg so elegant

>>8797420
The quadratic equation is fucking horrible. It's not clean, it's not simple, it's not beautiful.
Completing squares is beautiful, that isn't.

>>8799395
What's the condition for Taylor to work?
Based on your example I assume it's not continuity...

i've always thought [eqn]\text{for any}~n\in\mathbb{N}_0, \quad \int^{\infty}_0\prod_{k=0}^n\frac{\sin(x/(2k+1))}{x/(2k+1)}\,\mathrm{d}x=\frac{\pi}{2}[/eqn]
is pretty striking

>>8800461
>8800461
that's not true thoguh

>>8797260
[eqn]\frac{∂}{∂\boldsymbol{\hat{\theta}}}ln(\prod_{i=1}^nf(x_{i};\boldsymbol{\hat{\theta}}))=0[/eqn]

>>8801763
i^e - 1^0 +1^pi - i^e = 0

>>8801796
implications of this?

>>8801852
No implications. It's an estimator.

>>8801796
seems wrong honestly.

>>8801944
It's not a result tho. It's a method tbqh.

>>8801951
>It's not a result tho
into the trash it goes.

go do gradient descent somewhere else.

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>>8798629
>>8798606
>>8798463
I'll ask one more time. Where did you learn these equations? I'd be interested in reading the material myself.

>>8801957
Rude.
This is obvious but it's okay?>>8801813

>>8801626
This reminds me of the sinc function

>>8801796
I'm having some problems believing this is true. Surely taking the derivative of ln(f(x,theta)) with respect to theta is not always zero.

>>8801974
>>8801796
I see that it's a gradient, but I'm still having problems believing it.

>>8801626
>>8801961
Yea, I was right. This is a generalization of the sinc function.

Very interesting.

>>8800300
It can be used in a lot more than just field theory.

>>8800476
But it is anon

>>8800476
>The quadratic equation isn't that bad. Completing the square is fucking horrible

>>8800476
pleb

[math] \dfrac{y}{mi}+ix+e^{i\pi}\dfrac{c}{mi}=\bigg(2^{i\pi}\big(\frac{2}{1}\big)^{i\frac{\pi^2}{\tau}}\big(\frac{2}{3}\frac{4}{3}\big)^{i\frac{\pi^2}{2\tau}}\big(\frac{4}{5}\frac{6}{5}\frac{6}{7}\frac{8}{7}\big)^{i\frac{\pi^2}{2^2\tau}}\dots\bigg)\bigg(\dfrac{e^{i\tau}⋅i^4}{11.999\dots}\bigg)-\sum_{n=1}^{\infty}n [/math]

>>8800450
[eqn]1^2 + 2^2 \ne 3^2 \iff 5^2 + 10^2 \ne 15^2[/eqn]

This one gives me chills:
[eqn]\prod_{k=0}^{\infty}\pi^k=-\frac{{e}^{\pi i}}{\sqrt[12]{\pi}}\,,[/eqn]
where [math]i[/math] is the imaginary unit.

>>8802764
Doesn't the left side diverge? Or is this one of those -1/12 memes?

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>>8802764
Mmmhmm~ I love the Barnett-Wildberger-von Neumann product.

>>8799162
nice relativistic version

>>8803662
>>8802785
[math] \prod_{k=0}^{\infty} \pi^k = \pi^{0+1+2+3+\ldots} = \pi^{-1/12} [/math]

>>8801626
Nice try anon

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PV=nRT

>>8797260

[math]\prod_{{\rm p prime}} p = 4 \pi^2 [/math]

>>8801852
>>8801944
>>8801979
the solutions to that equation give the "maximum likelihood estimators" for [math]\hat{\theta}[/math], given data points [math]x_i, i = 1, 2, \dots , n [/math]

it is widely used in statistics (aka chad math) when we want to find to estimate the parameters of a distribution, given certain data points from that distribution (the form of the distribution is known, but the parameters are not known, e.g. we may want to estimate the probability from a binomial distribution, given samples from that distribution)

>>8801944
>>8801852
>>8801974
it's just statistics get the head out the ass pls

>>8804474
by chad math do you mean roastie math?

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>>8804398
The product of all primes is the area of the sphere? Neat.

I like the gamma functions and their properties

https://en.wikipedia.org/wiki/Particular_values_of_the_Gamma_function

>>8804969
Never understood the point of the gamma function being shifted to the right by 1

[math] f(x)=x^{f(x)} [/math]

>>8805011
http://mathoverflow.net/questions/20960/why-is-the-gamma-function-shifted-from-the-factorial-by-1

>>8805089
All it says is that Legendre liked it that way and it's what we've been left with

>>8797260
oh look, it's the first week of april and every single calc II class is learning about series.....

8===> ~ ~ ~

>>8805105

also something about mellin transforms of characters of the multiplicative group of positive reals

While it's not a specific equation, the fact that the gamma function shows up everywhere never ceases to amaze.

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>>8797260

>>8801626
engineers will defend this

>>8805939
Not really an equation, but amazing nonetheless.

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not rly an equation either but look at that tho...

For an algebraic field extension [math]E/F[/math], there is an inclusion reversing bijection between its intermediate fields and the subgroups of its automorphism group.

The fundamental theorem of Galois theory unveils a deep connection between fields and groups via polynomials rings. This is what I study maths for.

>>8801148
That's not even the equation of a Taylor polynomial in the OP, Taylor's was much more elegant

>>8798149
Sure but it's just pi^0, kinda cheating to multiply it by other transcendental number just to make it look nice. Why not square it and make an integral in the exponent? It's still going to be equal to 1.

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i win
you all suck

[math]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\dots}}}}=2[/math]

>>8809499
This can't be true
Variation of higher order terms leads to explosive change.

When doing an infinite expansion like this it needs to be in "equilibrium"

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>>8809516
If you say so, it's true though

>>8809527
The reason equations like >>8798431 and >>8798181 are true is because as you begin the process of expanding the root out to infinity, variation in the last term will lead to arbitrarily small changes in the overall value.

however in your formula changing the higher order terms lead to arbitrarily large change in the overall value.

>>8809499
You think that's amazing?
[math] \sqrt[\sqrt2]{2^\sqrt2} = 2 [/math]

>>8809534
hahaha

>>8807710
the least beautiful one in this thread. the sylow theorems were my least favorite part of abstract algebra

>>8797260
[eqn](a_1 + a_2 + \dots + a_n)^{k} = \sum\limits_{i_1=0}^{k}a_1^{k-i_1}(a_2 + \dots +a_n)^{i_1}\binom{k}{k-i_1}=\dots = \sum\limits_{i_1=0}^k\sum\limits_{i_2=0}^{i_1}\dots\sum\limits_{i_n=0}^{i_{n-1}}\bigg[ a_1^{k-i_1}a_2^{i_1-i_2}\dots a_n^{i_{n-1}-i_n}\binom{k}{k-i_1}\binom{i_1}{i_1-i_2}\dots\binom{i_{n-1}}{i_{n-1}-i_n}\bigg][/eqn]

>>8810568
Better is

[eqn] (a_1 + a_2 + \cdots + a_{k+1}\ )^n = \sum_{i_1 \ ,\ i_2 \ ,\cdots,i_k} \ \ \ \frac{n!}{i_1 ! i_2 ! \cdots i_k ! (n - i_1 - i_2 - \cdots - i_k)!} a_1^{i_1}\ a_2^{i_2}\cdots \ a_k^{i_k}\ a_{k+1}^{n-i_1 - i_2 - \cdots - i_k} [/eqn]
where [math] i_1, i_2, \cdots i_k [/math] are summed so as to take all integer values between [math] 0 [/math] and [math] n [/math].

>>8810595
Forgot to mention

[math] i_1 + i_2 + \cdots i_n \leq n [/math]

[eqn]\int^\infty_{-\infty} \mathrm{e}^{-x^2} \, \mathrm{d}x = \sqrt{\pi}[/eqn]

This will trigger /sci/
[eqn]\sum_{k=1}^\infty k = -\frac{1}{12} [/eqn]

>>8810595
>>8810598
Thanks, undergrad here,, not familiar with that notation.

>>8810742
Gaussians still trigger me. I'm bad with them and they appear alot in QM exercises. Same thing with

[eqn]\text{Erf}(x) = \frac{1}{\sqrt{\pi}}\int_{-x}^{x}e^{-t^{2}}dt [/eqn]

[math]\displaystyle\int_M\mathrm{d}\omega = \int_{\partial M}\omega[/math]

If [eqn]f(x) = \frac{\mathrm{d} F}{\mathrm{d}x}[/eqn] then
[eqn] \int^b_a f(x) \mathrm{d}x = F(b) - F(a) .[/eqn]

>>8810805
That's a special case of >>8810770 though.

>>8810807
I don't know what >>8810770 is. Looks interesting though, what's its name?

>>8810813
It's Stokes' Theorem. See https://en.wikipedia.org/wiki/Stokes%27_theorem for the connection to the fundamental theorem of calculus. Calling it a special case is not entirely fair, since you use it to prove Stokes' Thm.

>>8810750
It just means if you want a specific terms coefficient, plug in appropriate values of [math] i_1, i_2, \cdots, i_k [/math].

>>8810742
>not posting the more general form
[eqn] \int_{\infty}^\infty \mathrm{e}^{-(x^2 + \frac{a}{x^2})k}\ \ \ \ \mathrm{d}x [/eqn]

>>8810746
sum should start at 0, brainlet

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>>8810746
>this will trigger /sci/
>k=1

you're goddamn right

Maxwell's equations are God-tier comfy.
[eqn]\color{red}{\rm Maxwell-Gau\beta:} \nabla \,\cdot\, \overrightarrow{E} \,=\, \frac{\rho}{\varepsilon_0}[/eqn]
[eqn]\color{red}{\rm Maxwell-Thomson:} \nabla\,\cdot\, \overrightarrow{B} \,=\, 0[/eqn]
[eqn]\color{red}{\rm Maxwell-Faraday:} \nabla \,\times\, \overrightarrow{E} \,=\, -\frac{\partial \overrightarrow{B}}{\partial t}[/eqn]
[eqn]\color{red}{\rm Maxwell-Ampère:} \nabla \,\overrightarrow{B} \,=\, \mu_0 \,\overrightarrow{\jmath} \,+\, \mu_0 \varepsilon_0\, \frac{\partial \overrightarrow{E}}{\partial t}[/eqn]

>>8810960
they ought to be called maxwell-heaviside equations

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based wave equation

>>8810742
[math] \int_{\mathbb{R}^n} e^{-{\frac 1 2}\mathbf{x}^TA\mathbf{x}+B^T\mathbf{x}}d\mathbf{x}=\sqrt{(2\pi)^n/det(A)}e^{{\frac 1 2}B^TA^{-1}B}[/math]

for any positive-definite matrix A and any vector B

>>8810960
>not using differential forms

I never really liked the notation used for curl etc. from vector calculus desu :/

>>8810960
maxwell-ampere is the most important one, reason why he discovered EM waves (Due to finding the displacement current)

>>8811004
>curl

Well the cross product is a shitty operation in the first place.

1+1=2

>>8810960
>American notation
I threw up a bit.
[eqn]
\mathrm{div} \vec{E} = \frac{\rho}{\epsilon_0}
[/eqn]
[eqn]
\mathrm{div} \vec{B} = 0
[/eqn]
[eqn]
\mathrm{\vec{rot}}\,\vec{E} = - \frac{\partial \vec{B}}{\partial t}
[/eqn]
[eqn]
\mathrm{\vec{rot}}\,\vec{B} = \mu_0 \,\overrightarrow{\jmath} \,+\, \mu_0 \varepsilon_0\, \frac{\partial \vec{E}}{\partial t}
[/eqn]

>>8810857
Oh I knew about Stokes theorem but have always seen it in the form
[eqn]\int \int \int_V \nabla \times \mathbf{A} \, \mathrm{d} V = \int \int_{\partial V} \mathbf{A} \cdot \mathbf{n}\, \mathrm{d}S [/eqn] I don't really understand that omega notation.
I would hardly call it more general than the fundamental theorem of calculus though.

>>8811266
oops, I meant
[eqn]\int \int_C ( \nabla \times \mathbf{A} ) \cdot \mathbf{n} \, \mathrm{d} S = \int \int_{\partial V} \mathbf{A} \cdot \, \mathrm{d}\mathbf{r} [/eqn]

>>8811266
The "omega notation thing" is Stokes Theorem generalized to arbitrary manifolds.

What you wrote is Stokes Theorem on a Surface in R^3.

The fundamental theorem of calculus is Stokes Theorem on a interval of real numbers.

>>8811271
yeah I fucked it up every time I tried to write it, but that sounds very interesting, I'll have to read about it.

>>8810930
>>8810935
does it matter when 0 is the additive identity?

>>8811259
>>8810960
>not using godtier index notation
[eqn]\frac{\partial D_j}{\partial x_j} =\rho \\
\frac{\partial B_j}{\partial x_j} = 0 \\
\varepsilon_{ijk} \frac{\partial E_k}{\partial x_j} = - \frac{\partial B_i}{\partial t} \\
\varepsilon_{ijk} \frac{\partial H_k}{\partial x_j} = J_i + \frac{\partial D_i}{\partial t}
[/eqn]

>>8811271
This seems a bit self-referencing, since we usually use fundamental theorem of calculus in the proof of Stokes's theorem, and fundamental theorem of calculus can be proved without Stokes.

>>8811296
I know. I am just saying if [math]M = \left[ {a,b} \right][/math] and [math]\omega = f\left( x \right)[/math] then [math]\int\limits_M {d\omega } = \int\limits_a^b {f'\left( x \right)dx} [/math] and [math]\int\limits_{\partial M} \omega = f\left( b \right) - f\left( a \right)[/math].

Pretty neat for my gardening needs. Definetely not as elegant as the equations in this thread.

[eqn]R=\frac{Q_1[C_1(100−M_1)]+Q_2[C_2(100−M_2)]+Q_3[C_3(100−M_3)]+...}{Q_1[N_1(100−M_1)]+Q_2[N_2(100−M_2)]+Q_3[N_3(100−M_3)]+...}[/eqn]

>>8811293
How to git gud with index notation?

>>8811468
use subscripts and superscripts properly

File: vector-background-chemistry-formulas-equations-notebook-page-63395426.jpg (224KB, 1300x1009px) Image search: [Google]

224KB, 1300x1009px

>>8797260
What are the autists doing in the thread? Playing "pretend we're smart" with your pathetic numbers again, thinking you'll be the next e=mc2? I thought I fucking told you losers to apply your autism in a meaningful manner and stop parroting puzzle numbers like they mean anything.
If you can't apply your autism in a serious job then I'm kicking you out the house, Jimmy. Now stop playing with your fucking legos and quit wasting 3 months to replicate a calculator in Minecraft. You fucking failure.

>>8811468
it's simple, there's just 2 easy steps:
1. indexes that appear exactly twice in a term are summed over
2. memorize:
[eqn] \epsilon_{ijk} \epsilon_{lmk} = \delta_{il} \delta_{jm} - \delta_{im} \delta_{jl} [/eqn]
(fun exercise: prove it)

>>8812240
>checking 81 cases
>fun

>>8812464
you can write a program in 5 minutes to check the 81 cases or better you can reduce it to the 6 possible relations between the indexes. you can even reduce it further to two possible cases, all the rest of which follow from renaming the indexes

My one:

[math]\lim_{n \to \infty}\frac{n}{2}\cdot \sin\left ( \frac{2 \pi }{n} \right )=\pi [/math]

>>8810960
>"You can throw all of EM into 4 simple equations"
>Hell yeah cool as fuck
>Proceed to do problem sets
>It's boring as fuck calculate potential of charged spherical shell at point P bla bla over and over again

Kind of like a love/hate relationship. I adore how elegant the theory is but the average EM course and the problems associated with it are boring as fuck.

>>8812905
this

x*1=1*x/1

>>8812920
No way

>>8799708

Jacob Barnett is a meme genius. Perlman never said any of this. Triple integrals. The thing on the right of the inequality is just -1/12. Etc.

>>8797260
[math] \eta i_{\delta \delta} \sum \tau [/math]
can there be more beautiful?

>>8797260
[math]Z\left( \Sigma \right)\left( \mu \right) = \int {\left[ {\mathcal{D}\varphi |\mu } \right]} \exp \left( { - \int\limits_\Sigma {\mathcal{L}\left( {\varphi ,\operatorname{d} \varphi } \right)} } \right)[/math]

>>8813447
this seems like more of a definition

>>8813475
Sort of. The point is to show the path integral can roughly define a functor Z:Cob--->Vect.

[math] \lim_{x \rightarrow 0} {({x}-{\frac{1}{x}})}^{\frac{1}{x}} = {e}^{{2i0} \text{ to } {\pi}} {\infty} [/math]

>>8799138
neat