/sci/ - Science & Math

First for Godel's Theorem

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

>>9265572
5(p) * 500k(y) = 5p00ky

>>9266055
Lol, messed that up. You get the idea

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

>>9265585
Is... is that Wildberger?

https://en.m.wikipedia.org/wiki/Evil_number

>>9265572
Fractional calculus is where all the kool kids are at.

>>9265572
[math]a-b = (\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})[/math]
and
[math](\sqrt[2^n]{a}-\sqrt[2^n]{b})=(\sqrt[2^{n+1}]{a}+\sqrt[2^{n+1}]{b})(\sqrt[2^{n+1}]{a}-\sqrt[2^{n+1}]{b})[/math]

therefore by induction:
[math]a-b = \prod\limits_{i=1}^{\infty}(\sqrt[2^{i}]{a}+\sqrt[2^{i}]{b})\times\lim\limits_{i\rightarrow \infty}(\sqrt[2^{i}]{a}-\sqrt[2^{i}]{b})[/math]

observe that
[math]\lim\limits_{i\rightarrow \infty}(\sqrt[2^{i}]{a}-\sqrt[2^{i}]{b})=0[/math]

hence
[math]a-b=\prod\limits_{i=1}^{\infty}(\sqrt[2^{i}]{a}+\sqrt[2^{i}]{b})\times0 = 0[/math]

meaning that
[math]a=b \quad \forall a,b\in\mathbb{R}[/math]

>>9265572
How about some spooky physics?

>>9267188
>therefore by induction: [omitted]
did you type something wrong here? I don't follow this at all

>>9267188
you can't conclude a=b just because [math]a^{\frac{1}{\infty}} = b^{\frac{1}{\infty}}[/math], which is essentially what you've done

>>9267198
>>9267205

>>9267198
>>9267205
its spooky because its flawed
induction works for all n which are fixed.
but not necessarily for the limit when n to infty.
In this particular case it does not work for the limit.
why not is an exercise to you in real analisys

>>9267188
Can i approach infinity? I feel like infinity is just for real numbers, and only a complex number could approach a complex infinity

>>9267844
Oh wait nvm just skimmed the proof I get it now

https://en.wikipedia.org/wiki/No-ghost_theorem

*ruins your Halloween*

>>9267193
https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19800010907.pdf

Is this spooky math?
[eqn]\forall n\,\in\,\mathbf N,\,\forall A\,\in\, \mathscr M_n\left(\mathbf C\right),\, \left(A^*\,=\,A \,\Rightarrow\, \exists B \,\subseteq\,\mathscr M_{n,\,1}\left(C\right)^n,\, \left(\forall u\,\in\,\mathscr M_{n,\,1}\left(\mathbf C\right),\, \exists! \lambda \,\in\,\mathbf C^B,\, u\,=\,\sum_{v\,\in\,B} \left(\lambda\left(v\right)\,v\right) \,\wedge\, \forall \left(u,\,v\right)\,\in\,B^2,\, \left(u\,\neq\,v\,\Rightarrow\,u\,\bot\,v\right)\,\wedge\, \forall u\,\in\,B,\,\exists\lambda\,\in\,\mathbf R,\,A\,u\,=\,\lambda\,u\right)\right)[/eqn]

>>9267987
>any hermitian matrix has an orthogonal eigenbasis
wow so spooked

>>9268164
>he didn't get it

>>9268167
>with real eigenvalues
that's the easiest part

>>9268173
>that's the easiest part
sure it is, after you need it pointed out

>>9268196
[math]\left \langle u, A u \right \rangle
=\left \langle A^\dagger u, u \right \rangle
=\left \langle A u, u \right \rangle
=\left \langle u, A u \right \rangle^\dagger[/math]
Must hurt to be such a brainlet as to be spooked by this.

>>9268198
I sure am spooked by your retardation.

>>9268184
Euclid BTFO

>>9268480
Actually, no.

>>9265572
Spook math

>>9267188
You have to use coinduction and not induction or else you get an infinite loop in your theorem prover

Observe:

define Nat = Z | S Nat

isFinite :: Nat -> Bool
isFinite Z = True
isFinite (S n) = isFinite n

This doesn't work for:

infinity :: Nat
infinity = S infinity

>>9268542
Wow, that's a lot of precalculus. huh couldn't even have done differential geometry u lil shit?