Post beautiful Equations

Post beautiful Equations

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en.wikipedia.org/wiki/Lagrangian_(field_theory)
nuclear.ucdavis.edu/~tgutierr/files/stmL1.html
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a^2 + b ^2 = c^2

[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

>fraction bar

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

x + x = x^2

>using numbers

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

I teared up a bit.

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

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

e=mc2

>mfw it draws a dick

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

Taylor can go fuck himself.

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

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.

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.

When shit hits the fan but you still believe

is this just nonsense or what?

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

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oops

E = hf

idk why but i think its simplicity if beautiful

x=x

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

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

fucking lol

P=NP

A=A

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

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 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.

TopKek.

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Standard Model Lagrangian
nuclear.ucdavis.edu/~tgutierr/files/stmL1.html

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

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.

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.

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

No such thing

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.

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
1. [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]2. [eqn]45 \, \frac{(a+b+c)^8+(b+c+d)^8+(a-d)^8-(c+d+a)^8-(d+a+b)^8-(b-c)^8}{(a+b+c)^6+(b+c+d)^6+(a-d)^6-(c+d+a)^6-(d+a+b)^6-(b-c)^6} = 64 \, \frac{(a+b+c)^{10}+(b+c+d)^{10}+(a-d)^{10}-(c+d+a)^{10}-(d+a+b)^{10}-(b-c)^{10}}{(a+b+c)^8+(b+c+d)^8+(a-d)^8-(c+d+a)^8-(d+a+b)^8-(b-c)^8}[/eqn]

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]

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]

That's not beauty.

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

[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]

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]

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.

Why

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].

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

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]

Are you the same user? Regardless, where did you learn these formulas? They are very interesting.

[eqn] a^p = (1 + 1 + \cdots + 1)^p = p+ \sum_{a,b,c,... < p} \frac{p!}{a! b! c! \cdots z! (p-a)! (p-b)! (p-c)! \cdots (p-z)!} [/eqn]
However since p is prime we can conclude [math] a^p \equiv p \text{mod} p [/math] or [math] a^{p-1} \equiv 1 \text{mod} p [/math]

I read that proof in Ramanujan's book several years ago in high school.

Forgive my bad notation of a,b,c, but subscripts on Veeky Forums look really messy

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

Are you the same user? Where did you get these? I think they are very interesting and beautiful.

These are all me because I spamfag

Are you the same user? I think these are very beautiful. Please tell me where you learned these.

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]

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

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

[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.

[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]

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

But that's wrong!

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

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

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

well it technically works

[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.

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

no

25 + 100 = 125

Cubic equations!

9+16=25
25+144=169

underrated

literally perfect

/thread

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

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

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

>phi = golden ratio
Mind is defined as blown.

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

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

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

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

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

why is sci so autistic

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Field theory ain't beautiful

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

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

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

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]

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]