# Interesting mathematical statements (1 Viewer)

##### -insert title here-
$\bg_white \noindent Leave an intriguing mathematical statement for all of us to be flabbergasted by.$

$\bg_white \noindent I'll start.$

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

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

#### leehuan

##### Well-Known Member
Except why is this in non school lol

#### leehuan

##### Well-Known Member
Collatz conjecture

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

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##### -insert title here-
Except why is this in non school lol
Well it's not exactly a question asking thread, and it's more for entertainment.

$\bg_white Sophomore's Dream$

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

#### KingOfActing

##### lukewarm mess
Collatz conjecture

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

#### leehuan

##### Well-Known Member
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

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

#### leehuan

##### Well-Known Member
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?

##### -insert title here-
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.

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

#### leehuan

##### Well-Known Member
$\bg_white \infty ! = \sqrt{2\pi}$
Ok link me the proof lol

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

#### glittergal96

##### Active Member
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

##### lukewarm mess
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.

1+1=2

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amazing

#### leehuan

##### Well-Known Member
Can we keep our posts restrained to at least MX2 level and not making bad usages of mathematics lmao

#### KingOfActing

##### lukewarm mess
Zeta regularisations are important

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

#### turntaker

-1 x -1 = 2

##### -insert title here-
Ok link me the proof lol

I'm worried about a 1+2+3+4+...=-1/12 here
$\bg_white \noindent Your fear is somewhat correct, as I will now pull out the Riemann Zeta Function, which is used in the proof of \sum_{k=1}^\infty k = -\frac{1}{12}$

It's another zeta regularisation, I'm pretty sure.
Correct, although I don't actually fully understand regularisation yet.

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

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

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

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

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

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

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#### InteGrand

##### Well-Known Member
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.

##### -insert title here-
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

##### Active Member
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.