Beautiful math theorems, equations, proofs

What is your X on the LHS doing here. It's not defined?

Weierstrass's Approximation Theorem

Polynomials, Trig Polynomials, Fourier Series, Lattices, pick your version.
They all express the same idea.

It's the independent variable of the polynomial p(x)

Hey look Heisenberg's uncertainty principle is a kitty

Well I don't understand. On the LHS you have the independent variable X, and not on the RHS. What are the r_k? What are your hypothesis, and your conclusion?

This is a statement of the fundamental theorem of algebra in symbols, which states that a polynomial of degree n will have n complex roots (some of which could be equal, of course). Also keep in mind that the set of real numbers is a subset of the set of complex numbers

Furthermore, it's apparent that r1, r2, ..., rn are the roots of this polynomial which obviously has a degree of n

Ok. Then, look at your equivalence. You're saying : if a polynomial is the null polynomial, then the n numbers r_k are complex numbers, and reciprocically. Is this the fundamental theorem?
And your x_k are still not defined.

>Then, look at your equivalence. You're saying : if a polynomial is the null polynomial, then the n numbers r_k are complex numbers, and reciprocically. Is this the fundamental theorem?
Yes, it is
>And your x_k are still not defined.
Would writing p(xk) as opposed to p(x) remedy this?

>You're saying : if a polynomial is the null polynomial, then the n numbers r_k are complex numbers, and reciprocically
Yes. Any set of complex numbers can be the roots of a polynomial. There are also several proofs proving this theorem true online

Well, I don't think it wouldn't. The problem is that I'm guessing that your polynomial takes n distincts arguments (the x_k), but then it is a different object, it is a multinomial P(X1, X2, ..., Xn). A polynomial has only one independent variable X.

Then, your polynomial has 1 as dominant factor (in front of X^n), which is a limitation.

True true. I could have written a in front of it, but that wouldn't change the roots. Perhaps there is a more rigorous and more formal statement of the fundamental theorem of algebra

Stokes' theorem is pretty fucking dope.

>Any set of complex numbers can be the roots of a polynomial
This is quite trivial. Indeed, if you consider a set of n complex numbers r_k, obviously the polynomial product(X-r_k) has exactly these numbers as root.

It is the other way that is hard to prove, and that I believe is not quite properly stated in this thread.

My understanding of the theorem is : if you consider a polynomial of degree n with complex coefficients, then it is a product of n 1-degree polynomial. And this is the hard bit.

Okay, thanks for clearing this. Nighty night m8.

Generating functions

Picard-Lindelöf

en.wikipedia.org/wiki/Zeta_function_universality

pick any compact subset [math]U [/math] inside the vertical strip [math]1/2 < Real(z) < 1 [/math], whose complement is connected

let [math] F [/math] be any non-vanishing holomorphic function on U [/math]

then there are vertical shifts [math]U +it[/math] such that the Riemann zeta function on [math]U +it[/math] approximates [math] F [/math] on [math]U [/math] arbitrarily well, and there is a positive density of such [math] t [/math]

(the theorem requires [math] F [/math] to be non-vanishing since otherwise the riemann hypothesis would be false)

>something that gave you goosebumps
>preferably algebraic stuff
>mfw

Godel's Incompleteness Theorem

It's amazing to think that a corollary to the theorem is that entropy always increases

underrated

[math]sin^2x + cos^2x = 1^2[/math]
[math]x^2 + y^2 = r^2[/math]
[math]a^2 + b^2 = c^2[/math]

This for me

Minkowski's inequality is beautiful:
For all [math]d \,\in\, \mathbf N^*[/math], for all measurable subset [math]A[/math] of the Lebesgue measured set [math]\left( \mathbf R^d,\, \mathscr B,\, \ell\right)[/math], we have:
[eqn]\forall p \,\geqslant\, 1,\, \forall \left(f,\, g\right) \,\in\, \mathscr L^p \left( A \right),\, \left[ \int_A \left(f \,+\, g\right)^p\, \mathrm d\ell \right]^\frac{1}{p} \,\leqslant\, \left( \int_A f^p\, \mathrm d\ell \right)^\frac{1}{p} \,+\, \left( \int_A g^p\, \mathrm d\ell \right)^\frac{1}{p}[/eqn]

It exists in the [math]\ell^p[/math] spaces too:
[eqn]\forall p \,\geqslant\, 1,\, \forall \left( u,\, v\right) \,\in\, \left( \ell^p \right)^2,\, \left[ \sum_{n \,=\, 0}^\infty \left(u_n \,+\, v_n \right)^p \right]^\frac{1}{p} \,\leqslant\, \left( \sum_{n \,=\, 0}^\infty {u_n}^p \right)^\frac{1}{p} \,+\, \left( \sum_{n \,=\, 0}^\infty {v_n}^p \right)^\frac{1}{p}[/eqn]
This guarantees the existence of a normed vector space that has the same norm as [math]\left\_| \cdot \right\_|_2[/math] but in infinite dimension. It would be the most natural space structure extending the physical world to an infinite dimension (it would also be a reflexive Banach space which suggests physics could be done in it).

I always thought the better Euler's identity to meme would have been the product form for the zeta function

It's has a similar "what the fuck" appeal when you first see it and it's probably the most elementary way to extract stuff about primes out of the zeta function so it's actually useful

Poncelet's closure theorem.

According to Marcel Berger, the most beautiful result about conics.

geogebra.org/m/WYd5CExw

sin(πx)=πx/[(x!)(-x!)]

bump

scribd.com/doc/233602815/Barnetts-Identity-Pdf1

Is there anything more beautiful than the Barnett identity?

The hierarchy of the Cantorian cardinalities -- the fact that there isn't just one infinity, there's actually an infinite number of infinities, each provably smaller than the next. They are all easily constructed by taking the set of integers and using it as the starting point for iterating the powerset function. Even nicer is the theorem (still unproven) that there are no levels of infinity that cannot be generated using that powerset iteration technique (a.k.a. the "generalized continuum hypothesis"), and the fact that nobody has yet found a counterexample.

>The irrationality of square root of 2.

I like the fact that an irrational number to the power an irrational number can be rational. The proof is extremely elegant:

If sqrt(2)^sqrt(2) is rational, then you're done.

If not, then take that irrational number to the power of sqrt(2):

(sqrt(2)^sqrt(2))^sqrt(2)

= sqrt(2)^(sqrt(2)*sqrt(2))

= sqrt(2)^2

= 2

which is rational.

Either way, you get a rational result, without needing to know which branch is actually true.

Here's my favorite proof in mathematics.

It turns out that all positive integers are interesting.

Proof:

Assume the contrary. Then there is a lowest non-interesting positive
integer. But, hey, that's pretty interesting! A contradiction.
QED

>en.wikipedia.org/wiki/Zeta_function_universality

It's...beautiful.

>I just finished my first set theory class!

>sqrt2^sqrt2^sqrt2

You sure that order of operations works?
wolframalpha.com/input/?i=sqrt2^sqrt2^sqrt2

of course it does
(x^a)^b=x^(ab),

(x^a)^b is not x^(a^b)

Easily the spectral theorem for self adjoint operators:
For every self adjoint operator [math] A [/math] on a hilbert space [math] H [/math], we can write
[eqn]
A = \int_{\sigma (A)} \lambda~ d E_\lambda
[/eqn]
where [math] \sigma(A) [/math] is the spectrum of [math] A [/math] and [math] E_\lambda [/math] is the spectral measure of [math] A [/math]

You can easily get the spectral theorem for self adjoint matrices from that by definining [math] \sigma(A) = \{\lambda_1,...,\lambda_n \} [/math] and [math] E_\lambda [/math]
as the counting measure that maps an eigenvalue [math] \lambda_i [/math] to [math] v_iv_i^T [/math], where [math] v_i [/math] is the corresponding eigenvector. That gives us
[eqn] A = \int_{\sigma (A)} \lambda~ d E_\lambda = \sum_{i=1}^n \lambda_i v_iv_i^T [/eqn]

For a prime [math] p [/math] in [math] \mathbb{N} [/math], [math] p = a^2 + b^2 [/math] for some [math]a,b \in \mathbb{N} [/math] iff [math] p \sim 1 [/math] in [math] \mathbb{Z} / 4 \mathbb{Z} [/math].

Of course, for [math] p \neq 2 [/math]...

>It's amazing to think that a corollary to the theorem is that entropy always increases

Citation? Sounds like you've been reading crackpot stuff.