Abraham de Moivre and formula

De Moivre’s formula states that (cos x + isin x)^n = cos nx + isin nx.

Where i is the imaginary number √-1.

Now, for proof and simplicity’s sake, let’s change cosx + isin x to cis x.

This is a proof by induction, as stated by Wikipedia (I use a lot of Wikipedia these days.)

Let the statement (cos x + isin x)^n = cos nx + isin nx be S(n) for n an integer.

For n>0, it is clear that S(1) is true. Now, assume that S(k) is tru for a natural k, or (cis x)^k = cis kx

Consider S(k + 1):

(cis x)^(k + 1) = (cis x)^k * cis x = (cis kx)(cis x)† = (cos kx)(cos x) – (sin kx)(sin x) + i{(cos kx)(sin x) + (sin kx)(cos x)} = cis {(k + 1)x}††

† by the induction hypothesis

†† by the trigonometric identities

Because we have proven that (cis x)^(k + 1) = cis {(k+1)x), we can deduce that S(n) implies S(k + 1).

For S(0), it is clearly true since cos 0 + isin 0 = 1 and any number to the power of 0 (except zero itself) is 1.

For the negatives, let the exponent be -n for a natural n.

(cis x)^-n = (cis x)^n^-1 = (cis nx)^-1 = cos -nx + isin -nx

I don’t know what this is about, but the bold equation is a result of the identity

{\displaystyle z^{-1}={\frac {\bar {z}}{|z|^{2}}},}

where z = cis nx.

Therefore, S(n) holds for any integer n.

QED and for a better understanding, go check out link.

Joseph Liouville and his Numbers (especially his constant)

Number theory. Whoo!

In number theory, let there be a number x where, for any positive integer n, there are infinitely many number pairs (p, q) where q > 1 that:

0 < |x – p/q| < 1/q^n

x is the Liouville number.

Liouville numbers can be described as “almost rational”, and can be approximated by sequences of rational numbers. In other words, Liouville numbers are transcendentals that can be more closely approximated than any other algebraic irrational.

In 1844, Joseph Liouville proved that all Liouville numbers were transcendental. Maybe, if I can find the proof, I might post why, but not now.

One Liouville number is the Liouville Constant, which takes the form of:

0.1100010000000000000000010000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001… I’m not sure if that’s 120 decimal digits.

One thing: Liouville’s Constant (call it L; TapL hahaha) almost satisfies 10x^6 -75x^3 – 190x + 21 = 0, if the solution was rounded, then maybe.

If my mind comes to it, I might go into more detail


Back to another Millennium Prize problem!

There are two types of questions: one, where an algorithm can solve the question in polynomial time (P), and ones, where there is no way to find the answer quickly, but if information is given on showing what the answer is, it can be quickly verified, which is NP, or nondeterministic polynomial time.

The question, stated by Wikipedia, is “if the solution to a problem is easy to check for correctness, must the problem be easy to solve?” or in other words, is P = NP?

If P ≠ NP, then both P questions and NP-complete questions are within the class of NP questions. NP-complete questions are a set of problems where to each of which any other NP-problem can be reduced in polynomial time and the solution can be verified in polynomial time as well. NP problems, therefore, can transform into any NP-complete problems.

Now, if you want, go visit the Wikipedia page on it to understand better about P vs NP.

Poincaré Conjecture… with less detail.

Yes, this was already solved by Grigory Perelman, but I have nothing much, so…

Imagine a spherical item. And, for the easiest solution, let’s use an apple. Wrap the rubber band around the apple. Without tearing it or letting it leave the surface, we can slowly shrink the rubber band until it becomes a single point.

Now, think of a doughnut. Before you get hungry, quickly wrap a rubber band around the doughnut. Now, no matter how hard you try, you can’t shrink the rubber band down to a point without breaking it or making it leave the surface.

The surface of the apple is “simply connected”, while the surface of the doughnut isn’t. Poincaré, about a hundred years ago, knew that a two-dimensional sphere is essentially characterised by this property of simple connectivity, but questioned that of the three-dimensional sphere.

Grigory Perelman already solved it, and the link to the arXiv is here. The upload is by Terence Tao.

Until next time…

Riemann Hypothesis

Yesterday, I said that I would do a post on the Riemann Hypothesis, so here we are now.

There is a function with the name of the Riemann Zeta Function, named after Bernhard Riemann. The Riemann Hypothesis states that the Riemann zeta function has zeroes only at the negative even integers and complex numbers with the real part of 1/2. So, in explanation, ζ (s) is a function where the argument s may be any complex number not equal to one, and its zeroes are at points where s is a negative even integer, like -2, -4, -6, -8 and so on. These are trivial zeroes.

The non-trivial zeroes of the Riemann Zeta Function have a real part of 1/2. So if the hypothesis is true, then all non-trivial zeroes take a form of 1/2 + it, where t is a real number and i is the imaginary number.

There’s actually a formula for the Riemann Zeta Function:

{\displaystyle \zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}={\frac {1}{1^{s}}}+{\frac {1}{2^{s}}}+{\frac {1}{3^{s}}}+\cdots }

For more info, visit this website, Wikipedia.

Maybe, I’ll go more in depth about the Riemann Hypothesis, but not now.

RSA Factoring Challenge

The RSA Factoring Challenge is a set of factorisation challenges that involve numbers with hundreds of digits. The highest one cracked was RSA-250, which was 250 digits long, 829 bits, and was factored in February 2020 by Fabrice Boudot, Pierrick Gaudry, Aurore Guillevic, Nadia Heninger, Emmanuel Thomé, and Paul Zimmermann (what’s with the cool names?). The number was:


The factorisation was:

6413528947707158027879019017057738908482501474294344720811685963202453234463 0238623598752668347708737661925585694639798853367 × 3337202759497815655622601060535511422794076034476755466678452098702384172921 0037080257448673296881877565718986258036932062711

According to Wikipedia, the factorisation process utilised approximately 2700 CPU core-years, using a 2.1Ghz Intel Xeon Gold 6130 CPU as a reference. The program used to factorise RSA-250 was the Number Field Sieve algorithm, using the open source CADO-NFS software.

The dedication of the factorisation was towards Peter Montgomery, an American mathematician known for his contributions toward computational number theory and cryptography. He passed away on February 18, 2020.

The highest unsolved RSA Factoring Challenge is RSA-2048, with a whopping 617-digit number (2048 bits), with the number being:


And if the factorisation was found, the prize would have been US$200,000 but the challenge ended before anyone was awarded the prize. RSA-2048 may not be factorable, unless in the future, considerable advances in integer factorisation or computational power are made.

There are many other challenges, like RSA-617 with 617 digits, which still has not been factored, and the smallest RSA challenge number is RSA-260, which is:


So, if RSA-260 is cracked, I will be posting about it.

I think tomorrow, I’ll do the Riemann Hypothesis if I have enough time.

Behind the scenes of the proof that level 5 is impossible.

Level five of what? Geometry Dash, of course.

JK, Back on Track is harder.

I’m talking about Conway’s Soldiers.

And today, just for comfort, instead of calling the soldiers soldiers, I will be calling them pegs.

Conway’s Soldiers is simple. There are an infinite number of pegs behind a line that divides the land between the pegs and the enemy. The pegs may jump another peg vertically or horizontally, but never diagonally.

A number, Φ, exists so that:

{\displaystyle \varphi ={\frac {{\sqrt {5}}-1}{2}}\approx 0.61803\,39887\ldots }
If you couldn’t tell, Φ is the reciprocal of the golden ratio.

Select the target square, which, in our case is any square on level 5, and mark it with Φ^0, or just 1 for simplicity. All the other squares on the enemy territory will be marked with Φ^n, where n is the Manhattan distance to Φ^0. Also, for simplicity, Φ^0’s square will be called T, for target.

We can define a “score” for the configuration of the pegs. If two pegs are lined up to reach T after one jump, the score is Φ^1 + Φ^2. The score can change in three ways:

  • If a jump moves a peg towards T, the change in score is Φ^(n – 2) – Φ^(n – 1) – Φ^n. That is because The original pegs are at Φ^(n – 2) and Φ^(n – 1), and when the peg jumps, the jumped peg, Φ^(n – 1) is removed, hence subtracting the score, and Φ^(n – 2) moves to Φ^n, hence subtracting Φ^(n – 2) and adding Φ^n, which leads to 0.
  • If a jump doesn’t impact the value of the jumping peg, the score is changed by -Φ^(n – 1).
  • If a jump moves a peg away from T, the score is changed by -2Φ^(n + 1).

Therefore, the score can never increase.

Now, let’s think about it at a different approach. If there was a row of pegs right behind T, the score would be:

{\displaystyle \varphi +2\varphi ^{2}+2\varphi ^{3}+\ldots }
and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on…

The score of a line of pegs n spaces away would be {Φ^(n – 1)} * (Φ + 2Φ^2 + 2Φ^3 + 2Φ^4 + …). Two things:

\sum _{{n=2}}^{{\infty }}\varphi ^{n}=1
and also,
{\displaystyle \varphi ^{2}=1-\varphi }

That means that if S(1) was the starting score for a board with T at level one, S(1) = 5 + 3Φ. Since the difference of two squares directly next to each other is logarithmically Φ, S(n + 1) = S(n) * Φ.


  • S(2) = 3 + 2Φ
  • S(3) = 2 + Φ
  • S(4) = 1 + Φ
  • S(5) = 1

Now, let there be an end configuration score E, where E = 1 + ϵ, where 1 is the contribution of the peg on T, and ϵ the contribution of the peg behind the line. ϵ is positive, but can be as small as possible.

Now, for S(5), which is one, since ϵ > 0, E > S(5). But the score never increases, so therefore that must mean that S(5) ≥ E. We have reached a contradiction, since if a > b, then b ≥ a cannot be true.


Credits to Wikipedia for the information. For detail, go check out the page.


Dante was elected as mayor in Hypixel Skyblock. The admins warned us, telling us to not vote for him. We all thought, “shut up admins! New update! NEW MAYOR! VOTE FOR THE NEW MAYOR DANTE!!!!!” But the admins were right. We should not have voted for him.

How strong are Dante’s debuffs?

First debuff: More tax!

The auction house and bazaar taxes have been doubled. The NPC buy prices have been raised by 10%. Five coins/minute for talking, two coins for travel. WTF? That means bazaar flipping is 9% less effective. 9% might not seem like a lot, but when it comes to bazaar flipping, it’s lots of money. Kinda.

Second debuff: Official buildings!

“Official buildings”, like AH, Bazaar Alley, THE BANK and more cannot be accessed without either a Happy Mask or a Dante Talisman, which the Happy Mask is the cheaper option as it is only 100 coins, but takes up the helmet slot. Dante Talisman is not only common, it is one million coins. ONE MILLION! Plus, how are the new players supposed to progress now?

Third debuff: Cancelled events!

Oringo, Winter Island and Spooky Festivals have been cancelled. WTF WHY ARE THEY CANCELLED THEY’RE LITERALLY SOME OF THE BEST MONEY-MAKING METHODS!!!!! What’s dangerous about Oringo? Literally all he does is give out pets that buff you, it’s not like they end up killing you.

Extra debuffs: Mayor elections no more!

Dante elections have replaced the normal mayor election. That means that Dante will be around for a long time, unless the Resistance does something about it. I miss doing slayers while Aatrox was mayor, or waiting for that moment when Diana was mayor so that I could finally start doing mythological events. When am I going to be able to do that?

Literally, Dante is so useless, Barry is better than him, and Barry’s never been elected once. Neither has Diaz, and Diaz’s perks are actually 100x better than Dante’s perks. I swear, I want to join the Resistance and be the one to behead him. Period.

Second update: “You’re clicking too fast!”

This one I personally hate. I was doing an experiment, when suddenly, bam! I was kicked out of the experiment menu and in the chat, there was this message: “You’re clicking too fast!” Many people have been annoyed about this, complained in the forums, and Donpireso has said that they are working on a fix, but it’s unclear when the fix is. It’s made people lose so much money, it’s gone into the millions. One guy had very OP loot from experiments then lost it all because of “You’re clicking too fast!” This is all because of autoclickers that use them for some reason, but please disable this during experiments, starring dungeon items, and other clicking-related things.

Why am I writing about this? I dunno… content drought.

Also, the nerfs that Hypixel admins made doesn’t impact me much, but other people must be mad about it. Bye!

Sigma proof… explained

Unrelated, but a GD level called Sigma: https://geometry-dash-fan.fandom.com/wiki/Sigma

Yesterday, I posted a proof on why the sigma sum for phi was true, but I kind of rushed it, so I will be going into details.

If you can do a little bit of “hard maths”, then it is easy to prove that:

{\displaystyle \varphi ^{2}=1-\varphi }
is true.

is true.

\sum _{{n=2}}^{{\infty }}\varphi ^{n}=1

That means that the sum is Φ^2 + Φ^3 + Φ^4 + Φ^5 + Φ^6 + …, but since Φ^2 = 1 – Φ, that means that Φ^3 = Φ^2 * Φ, or Φ-Φ^2. If we keep multiplying, we begin to see a pattern here, where the exponent keeps going up by one, and in the full sum, there is, for any positive integer k, a Φ^k and a -Φ^k, which implies that they can be cancelled out. But there is no term to cancel the one in Φ^2, so one is left, and Φ^∞, since Φ is smaller than one, the limit as x approaches infinity of Φ^x converges to 0. That means the final sum is 1-0, or just 1.


Conway’s Soldiers… part 2.

{\displaystyle \varphi ={\frac {{\sqrt {5}}-1}{2}}\approx 0.61803\,39887\ldots }
\sum _{{n=2}}^{{\infty }}\varphi ^{n}=1
The sigma sum

I will prove that the sigma sum is true given the value of phi, which is also the reciprocal of the golden ratio.

Remember how Φ^2 = 1 – Φ? The sigma sum states that 1 – Φ + Φ – Φ^2 + Φ^2 – Φ^3 + Φ^3 – Φ^4 + …, that means the powers of Φ keep cancelling themselves out, and if a number x is smaller than one, the limit as n goes to infinity of x^n approaches zero. And since Φ is less than one, Φ^∞ must converge to zero.

That means the final sum is 1 + Φ^∞ after all the cancellation, and since Φ^∞ converges to zero, the final sum is 1 + 0, or just one.