Interesting mathematical statements (4 Viewers)

Paradoxica · Dec 23, 2015

$\noindent Leave an intriguing mathematical statement for all of us to be flabbergasted by.$

$\noindent I'll start.$

$\lim_{x \rightarrow \infty} e^{e^{e^{\left ( x+e^{-(a + x + e^x + e^{e^x})} \right )}}} - e^{e^{e^x}} \equiv e^{-a}$

$\noindent For all values of "$a$".$

leehuan · Dec 23, 2015

Except why is this in non school lol

leehuan · Dec 23, 2015

Collatz conjecture

$\\ $Let the sequence ${ T }_{ n },{ T }_{ n+1 },\dots $ be defined such that $ \\{ T }_{ k } \in \mathbb {Z}^{+} \, \forall \, k \in \mathbb N \\ $where $ { T }_{ n+1 } = \left\{ \begin{matrix} 3{ T }_{ n }+1 \\ { T }_{ n } / 2 \end{matrix} \right $if ${ T }_{ n }$ is $ \begin{matrix} $odd$ \\ $even$ \end{matrix} \\ $The sequence always converges into 4, 2, 1.$$

Paradoxica · Dec 23, 2015

leehuan said:
Except why is this in non school lol

Well it's not exactly a question asking thread, and it's more for entertainment.

$Sophomore's Dream$

$\int_0^1 \frac{dx}{x^x} = \sum_{k=0}^\infty \frac{1}{k^k}$

KingOfActing · Dec 23, 2015

leehuan said:
Collatz conjecture

$\\ $Let the sequence ${ T }_{ n },{ T }_{ n+1 },\dots $ be defined such that $ \\{ T }_{ k } \in \mathbb {Z}^{+} \, \forall \, k \in \mathbb N \\ $where $ { T }_{ n+1 } = \left\{ \begin{matrix} 3{ T }_{ n }+1 \\ { T }_{ n } / 2 \end{matrix} \right $if ${ T }_{ n }$ is $ \begin{matrix} $odd$ \\ $even$ \end{matrix} \\ $The sequence always converges into 4, 2, 1.$$

The sequence is conjectured to always diverge to 1

leehuan · Dec 23, 2015

Paradoxica said:
Well it's not exactly a question asking thread, and it's more for entertainment.

You could always have posted it under just maths then or extracurricular

Fermat's Last Theorem

$\\ $Consider the equation ${ a }^{ n }+{ b }^{ n }={ c }^{ n } \quad \forall \, \left\{ a,b,c,n \right\} \in \mathbb Z. \\ $If one restricts that $\left| n \right| >2$, then we have no solutions.$

leehuan · Dec 23, 2015

KingOfActing said:
The sequence is conjectured to always diverge to 1

I know lol. This is one of the biggest mind gobbling problems to pure mathematicians apparently; WHY?

Paradoxica · Dec 23, 2015

leehuan said:
I know lol. This is one of the biggest mind gobbling problems to pure mathematicians apparently; WHY?

The thing is, we don't even know how to begin to approach this problem. Paul Erdos has already commented on this problem.
Looks like we'll just have to wait for the next
Euler/Erdos/Tao/Gauss/Noether/Polya/Hilbert/Russell/Lagrange/Riemann/Hardy/Poincare/Fermat/Grothendieck/Newton/Leibniz/Weierstrass/Cauchy/Descartes/Dirichlet/Cantor/Fibonacci/Jacobi/Ramanujan/Hamilton/Godel/Pascal/Apollonius/Laplace/Liouville/Eisenstein/Banach/Peano/Bernoulli/Viete/Fourier/Huygens/Chebyshev/Lebesgue/Turing/Cardano/Minkowski/Littlewood/Legendre/Birkhoff/Lambert/Poisson/Wallis/Tarski/Frege/Hausdorff/Neumann/Galois
to come around and resolve the problem.

$\infty ! = \sqrt{2\pi}$

leehuan · Dec 23, 2015

Paradoxica said:
$\infty ! = \sqrt{2\pi}$

Ok link me the proof lol

I'm worried about a 1+2+3+4+...=-1/12 here

glittergal96 · Dec 23, 2015

If you have a small ball in 3 dimensional space, it is possible to decompose it as a union of a finite number of sets, which can be moved by rotations and translations such that the pieces never overlap and such that the final object constructed is an arbitrarily large ball.

Colloquially, one can cut a pea into a finite number of pieces and reassemble it into something the size of the sun.

KingOfActing · Dec 23, 2015

leehuan said:
Ok link me the proof lol

I'm worried about a 1+2+3+4+...=-1/12 here

It's another zeta regularisation, I'm pretty sure.

turntaker · Dec 23, 2015

1+1=2

turntaker · Dec 23, 2015

amazing

leehuan · Dec 23, 2015

Can we keep our posts restrained to at least MX2 level and not making bad usages of mathematics lmao

KingOfActing · Dec 23, 2015

Zeta regularisations are important

1 + 2 + 3 + ... =/= -1/12, but is rather 'assigned' that value

turntaker · Dec 23, 2015

-1 x -1 = 2

Paradoxica · Dec 23, 2015

leehuan said:
Ok link me the proof lol

I'm worried about a 1+2+3+4+...=-1/12 here

$$\noindent Your fear is somewhat correct, as I will now pull out the Riemann Zeta Function, which is used in the proof of $ \sum_{k=1}^\infty k = -\frac{1}{12}$

KingOfActing said:
It's another zeta regularisation, I'm pretty sure.

Correct, although I don't actually fully understand regularisation yet.

$\zeta(s) = \sum_{k=1}^\infty k^{-s}$

$\noindent The derivative of the zeta function evaluated at $s = 0$ is $-\frac{1}{2}\log(2\pi). $ I am still scouring the internet for a proof of this fact, as I cannot seem to find a proof of it anywhere.$

$$\noindent Differentiating with respect to $s$, we have: $\zeta'(s) = \sum_{k=1}^\infty (-\log{k}) k^{-s} \Rightarrow \Rightarrow \Rightarrow \Rightarrow \zeta'(0) = -\sum_{k=1}^\infty \log{k}$

$$\noindent Combining the above, we have: $ -\sum_{k=1}^\infty \log{k} = -\log{\sqrt{2\pi}} \Rightarrow \Rightarrow \Rightarrow \Rightarrow \Rightarrow \Rightarrow \Rightarrow \Rightarrow \sum_{k=1}^\infty \log{k} = \log{\sqrt{2\pi}}$

$\log{1} + \log{2} + \log{3} + \dots + \log{\infty} = \log{\infty !} = \log{\sqrt{2\pi}}$

$\therefore \infty ! = \sqrt{2\pi}}$

InteGrand · Dec 23, 2015

I like the 1 + 2 + 3 + 4 + … = -1/12 result, and find it also quite amazing that this is used in physics and gives some experimentally verifiable results. There's a lot of 'weird' stuff like this in this series of lectures on Mathematical Physics by Carl Bender that can be found on YouTube.

Also, I think this thread should be in the maths Extracurricular Topics forum.

Paradoxica · Dec 23, 2015

glittergal96 said:
If you have a small ball in 3 dimensional space, it is possible to decompose it as a union of a finite number of sets, which can be moved by rotations and translations such that the pieces never overlap and such that the final object constructed is an arbitrarily large ball.

Colloquially, one can cut a pea into a finite number of pieces and reassemble it into something the size of the sun.

Only if I accept the axiom of choice. : PPPPPPP

glittergal96 · Dec 24, 2015

Paradoxica said:
Only if I accept the axiom of choice. : PPPPPPP

Even if you don't accept the axiom of choice (which is a bit limiting, but some minority of mathematicians don't), you would not be able to prove that such a reassembling of the pea into the sun is impossible. (Because the axiom of choice is consistent with the other axioms of set theory.)

This is still pretty unintuitive.

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