Interesting mathematical statements

**Paradoxica** · 23 Dec 2015, 9:29 PM

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

$\noindent I'll start.$

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

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

**leehuan** · 23 Dec 2015, 9:32 PM

Except why is this in non school lol

**leehuan** · 23 Dec 2015, 9:38 PM

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** · 23 Dec 2015, 9:43 PM

Originally Posted by leehuan

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** · 23 Dec 2015, 9:46 PM

Originally Posted by leehuan

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** · 23 Dec 2015, 9:47 PM

Originally Posted by Paradoxica

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** · 23 Dec 2015, 9:49 PM

Originally Posted by KingOfActing

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** · 23 Dec 2015, 9:59 PM

Originally Posted by leehuan

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** · 23 Dec 2015, 10:02 PM

Originally Posted by Paradoxica

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

Ok link me the proof lol

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

**glittergal96** · 23 Dec 2015, 10:02 PM

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** · 23 Dec 2015, 10:38 PM

Originally Posted by leehuan

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** · 23 Dec 2015, 10:41 PM

1+1=2

**turntaker** · 23 Dec 2015, 10:44 PM

amazing

**leehuan** · 23 Dec 2015, 10:47 PM

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

**KingOfActing** · 23 Dec 2015, 10:50 PM

Zeta regularisations are important

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

**turntaker** · 23 Dec 2015, 10:53 PM

-1 x -1 = 2

**Paradoxica** · 23 Dec 2015, 10:55 PM

Originally Posted by leehuan

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}$

Originally Posted by KingOfActing

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** · 23 Dec 2015, 11:13 PM

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** · 23 Dec 2015, 11:56 PM

Originally Posted by glittergal96

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** · 24 Dec 2015, 12:08 AM

Originally Posted by Paradoxica

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.

**Paradoxica** · 24 Dec 2015, 10:43 AM

Originally Posted by glittergal96

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.

it was a joke -_-
I find that people who reject the axiom of choice are on the same level as those who reject the law of the excluded middle. Half of mathematics is based upon contradiction.

Got the proof for derivative of zeta at zero
$$\noindent $ \eta(s) = \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^s} \Rightarrow \zeta(s) - \eta(s) = 2\sum_{k=1}^\infty (2k)^{-s} = 2^{1-s}\sum_{k=1}^\infty k^{-s} = 2^{1-s}\zeta(s) \Rightarrow \Rightarrow \zeta(s) = \frac{\eta(s)}{1-2^{1-s}} \Rightarrow \zeta(0) = -\eta(0) = -\frac{1}{2}$

$\zeta '(s) = \frac{\eta '(s)(1-2^{1-s}) - \eta(s)2^{1-s}\log{2}}{(1-2^{1-s})^2} \Rightarrow -\zeta '(0) = \eta'(0) + \log{2}$

$\eta '(s) = \sum_{k=1}^\infty (-1)^{k-1}\frac{\log{k}}{k^s} \Rightarrow \eta '(0) = \sum_{k=1}^\infty (-1)^{k-1}\log{k}$

$$\noindent$2\eta '(0) = 2\sum_{k=1}^\infty \log(2k) - 2\sum_{k=1}^\infty \log(2k+1) =\left ( -\sum_{k=1}^\infty \log(2k-1) + \sum_{k=1}^\infty \log(2k) \right ) - \left ( -\sum_{k=1}^\infty \log(2k) + \sum_{k=1}^\infty \log(2k+1) \right ) = \sum_{k=1}^\infty \log \left (\frac{2k}{2k-1} \right ) + \sum_{k=1}^\infty \log \left (\frac{2k}{2k+1} \right ) = \sum_{k=1}^\infty \log \left ( \frac{4k^2}{4k^2 -1} \right ) = \log \left ( \prod_{k=1}^{\infty} \frac{4k^2}{4k^2 -1} \right ) = \log{\frac{\pi}{2}} \Rightarrow \eta '(0) = \frac{1}{2}\log{\frac{\pi}{2}}$

$-\zeta '(0) = \frac{1}{2}\log{\frac{\pi}{2}} + \log{2} = \log{\sqrt{2\pi}}$

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

If you do not know where I got the product identity from, recall the 1995 HSC paper, in which we proved said identity.

**glittergal96** · 24 Dec 2015, 11:00 AM

With all of these zeta function things, people often get confused by what is meant by these divergent sums having values.

The function $\zeta(s)$ is defined by the Dirichlet series $\sum_n n^{-s}$ only where it converges, which is the half-plane where the real part of s exceeds 1.

Elsewhere in the complex plane (apart from at s=1), the zeta function is defined by analytic continuation. In other words, there is a unique "nice" function on the complex plane that extends the series where it converges.

So at points like s=0 it is not like that series is equal to zeta(0), it is just that that sum diverges in the traditional sense and hence is an undefined object. We might as well use that sum notation to instead represent the "nice function" that the convergent sum extends to.

In this sense the statement is more like: If the sum of the natural numbers had to be defined to be something, -1/12 would be a natural value for it to have.

Definitely a lot of the reason that non-mathematicians find this so interesting is that they view the sum as actually converging to -1/12 in a more traditional sense which is of course nonsense. The amount that this fact is thrown around colloquially does not help.

(I can't say I know much about how this fact crops up in physics though.)

**glittergal96** · 24 Dec 2015, 11:09 AM

Some of these statements are pretty fun to prove and not too difficult btw.

People should post them in the undergrad marathon!

**InteGrand** · 24 Dec 2015, 11:12 AM

I think these kinds of divergent sums (and integrals for that matter) come up all in the time in quantum physics. For example, the 1+2+3+4+... one comes up in calculating the Casimir Force in 1D in Quantum Electrodynamics. The 1³ + 2³ + 3³ + ... one comes up in the 3D version of this calculation (its value is ζ(-3) = 1/120 using analytic continuation of the Riemann-Zeta function). As far as I know, experiments have now been done and agree with predictions to a good extent. Here's a derivation on Wiki of the 3D Casimir effect that uses ζ(-3):

https://en.wikipedia.org/wiki/Casimi...regularization .

Is it just a coincidence that using these regularised sums happens to give apparently physically sensible answers??

**Paradoxica** · 24 Dec 2015, 11:16 AM

$Borwein Integral$

$\int_0^\infty \frac{\sin{x}}{x} dx = \frac{\pi}{2}$

$\int_0^\infty \frac{\sin{x}}{x}\frac{\sin{\frac{x}{3}}}{\frac{x} {3}} dx = \frac{\pi}{2}$

$\int_0^\infty \frac{\sin{x}}{x}\frac{\sin{\frac{x}{3}}}{\frac{x} {3}}\frac{\sin{\frac{x}{5}}}{\frac{x}{5}} dx = \frac{\pi}{2}$

$\vdots$

$\int_0^\infty \frac{\sin{x}}{x}\frac{\sin{\frac{x}{3}}}{\frac{x} {3}}\frac{\sin{\frac{x}{5}}}{\frac{x}{5}} \dots \frac{\sin{\frac{x}{11}}}{\frac{x}{11}}\frac{\sin{ \frac{x}{13}}}{\frac{x}{13}} dx = \frac{\pi}{2}$

$$\noindent$\int_0^\infty \frac{\sin{x}}{x}\frac{\sin{\frac{x}{3}}}{\frac{x} {3}}\frac{\sin{\frac{x}{5}}}{\frac{x}{5}} \dots \frac{\sin{\frac{x}{11}}}{\frac{x}{11}}\frac{\sin{ \frac{x}{13}}}{\frac{x}{13}}\frac{\sin{\frac{x}{15 }}}{\frac{x}{15}} dx = \frac{467807924713440738696537864469}{935615849440 640907310521750000}\pi = \frac{\pi}{2}-\frac{6879714958723010531\pi}{93561584944064090731 0521750000} \approx \frac{\pi}{2} - 2.31 \times 10^{-11}$

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