M2C2A: Episode 6⅓: Pi, a three part mathematical journey through the mathematical constant we all know and love, Part 2

[This proof is actually held by vihart, and shared by me, while the extra comes from Spanish Wikipedia, which inspired me to look for more digits of the not periodical sequence, which I found in http://mathworld.wolfram.com/PiContinuedFraction.html.]
Get a square, and draw a circle that touches all sides of the square

Let's say that the square's sides all measure 1.
Now modify the sides like this

so while the perimeter is still 4, now the circle is touched in 8 areas.
Modify it again and again, until you get...
A circle !
But is this REALLY a circle?
Nope!
A real circle has 0 sides, not ∞ like on this infinitely wrinkled circle.
So it is not a real circle, but an infinitely close imitation.
But what is the perimeter (remember this is an infinitely wrinkled circle with no curves, but ∞ sides)?
We started with 4, and by only modifying the sides, we didn't change the perimeter, only to get a final 4 as the perimeter!
So pi isn't four as anyone would have thought with this proof about making an imitation of a circle at first sight.
Extra:
Pi as a continued fraction?
To get pi to an accuracy of 11 digits, try "3+(1/(7+(1/(15+(1/(1+(1/(292+(1/(1+(1/(1+(1/(1+(1/(2+(1/3)))))))))))))))))", and continue the pattern with more digits I found here!
This is a great approximation, but can't you simply say 3.14159265358979.../1?
Not so exiting, but if you aren't worried about all numbers in the equation being integers, more understandable.
-The Roaring Thunder

M2C2A: Episode 6⅔: Pi: A three part mathematic

Me comenta mi Roaring Thunder, que me faltó publicar esta entrega, por lo que arreglo mi error y les comparto el episodio 6 dos tercios
Is pi random?
Nobody knows, but it looks random enough.
First of all: though pi will never change, randomness is still valid.
Second: Let's try to approve or not approve randomness.
One of the properties of randomness is that there are an approximately equal number of appearances of each of the possibilities (here 1, 2, 3, 4, 5, 6, 7, 8, 9 and 0). If you check in the digits of pi, you'll find this property true. Want to check it? Try going to this page and use Ctrl+F. ✓
Another property of randomness, though identical to the first in concept, but referring to the sequences of digits is also something you can try in the page of the last hyperlink, and with some short sequences, you will find it also true. ✓
You may start calling pi random, but we're still not sure. We need to know more digits to see if no sequences appear in the total of all pi digits (after all, you'd never know if a house is a house if you check atom by atom). So final answer?
MAYBE.
-The Roaring Thunder

M2C2A: Episode 14, What is 0^0? Let's unscramble this nonsense once in for all!

What is 0⁰? And I'm not expecting some miracle answer or nothing, I'm expecting the truth. And actually, it's much more complex than you might think (and I'm not talking about i here).

My point is, powers with base 0 (0ⁿ) could be forbidden.

It is known that x^m*xⁿ=x^(m+n).

But also that x^m/xⁿ=x^(m-n).

So if you might call 0² as 0, or 0³, you would be forced to say 0³/0², that is 0¹, would be 0/0.

But there is something wrong about that.

Remember when we talked about 0/0? It could even be 0!

But we know that when we say 0¹, we're talking about a number that when you multiply 1 zero together (that is, leave it as it was), you're talking about 0! So 0/0 is 0 in this case.

But 0⁰ has nothing to forbid it.

The only reason mathematicians called n⁰ as 1, was because n¹ =(n)/n¹, (that is n^(1-1)=n⁰) was always 1... except for 0.

So the only thing that can define without barriers 0^0 is... Our number with infinite answers! 0/0! (or 0j, if you remember one of our last episodes)! But of course because almost everyone else believes 0/0 is undefined, 0⁰ is as undefined as.

-The Roaring Thunder

M2C2A: Episode 13, ∞, what has always been wrong about it

Many say ∞+1, ∞², or ∞-10^(10^(10^(10^10))), are simply ∞, but no.

∞ is like i, you (usually) can't simplify things with it. E.g.
If ∞+∞ is ∞, then we would be obliged to say ∞-∞=∞.

Is it?

We can use something I like to abbreviate as P.I.G (Patterns In Graphs, which can also refer to look for formulas), which is what I'll do.

If we graph x-x=y, we always get 0.

It's a truly linear equation.

We don't expect that when it gets to ∞ it will suddenly rise to ∞ too.
Same story with ∞*∞ and ∞/∞.

What makes sense in both cases, is that neither ∞+∞ or ∞*∞ should be expressed as ∞, but as 2∞, and ∞², respectively.

So ∞+1 should be written as it's written here, and so should 3/∞, 2^∞+1, x^y=y^∞x, or ∞x²+∞x+∞=0, etc.

...or is it?

This would work perfectly if it wasn't for the definition of ∞, the biggest quantity that can be described, which doesn't have an end.
If ∞+1>∞, then ∞+1 should be infinity!

But if ∞ (∞+1)<∞+1 (∞+2), then ∞+2 should be ∞!
But if this is smaller than ∞^∞(totally infinite!), then ∞^∞ SHOULD BE ∞!!!
So my definition of ∞ is:

nonsense.

M2C2A: Episode 12, The ABACABA, the biggest word in the mathematical dictionary

To write this word is a total mess, but to write the first 63 characters, this you can do.

Write an "a" every 2 spaces or (2¹) spaces. This should look like this:
a_a_a_a_a_a_a_a_a_a_a_a_a_a_a_a_a_a_a_a_a_a_a_a_a_a_a_a_a_a_a_a

Then, from the first unoccupied space, write a "b", every 4 spaces (2²) this time:

aba_aba_aba_aba_aba_aba_aba_aba_aba_aba_aba_aba_aba_aba_aba_aba

Repeat this last step, but with the "c", and now every 8 spaces (2³) .

abacaba_abacaba_abacaba_abacaba_abacaba_abacaba_abacaba_abacaba

If you continue the pattern, this you should get:

abacabadabacaba_abacabadabacaba_abacabadabacaba_abacabadabacaba
abacabadabacabaeabacabadabacaba_abacabadabacabaeabacabadabacaba
abacabadabacabaeabacabadabacabafabacabadabacabaeabacabadabacaba

Of course, if you continue this until the "z", you will get a word (2²⁷)-1 characters long, which is 134,217,727 letters long! 

And if you include all the character in the Unicode (the majority are Chinese characters and some are just representations as " ", but still, no discussing about that), the epic number of 109,449 you get:

5.40785978562894540705755027563290385780481607 × 10³²⁹⁴⁷ letters!!! 

But what does this apply to?

The Sierpiński Gasket!


Every time you see a black square, write "a", if you see a little white triangle, write "b", a bigger white triangle, "c"... etc.



And if all the "a"s in the triangle are one color, and the "b"s other... you get a colorful Sierpiński! (but that's another thing...)

And to really celebrate Sierpiński and the abacaba, I made the "rainbow gasket"!


Cool, huh?
-The Roaring Thunder

M2C2A: Episode 11, The root of a number?!

A square root of a number y is by definition a number x in which x² equals y. And a cube root of a number y is by definition a number x in which x³ equals y.
But we can also do n roots, in which the nth root of y is a number x in which which xⁿ equals y. But there's a special property about roots:

They may have many answers.

For example, the square root of 4.
2² is 4, but also -2². So can 2 and -2 be the roots of 4?
Yes!
Let's now take a different example:
The cube root of 8.
It may be true, but there are 2 more solutions.
First of all, you'll need to know complex numbers.
A complex number has a real part (2, 4, 7982) and an imaginary part, which can be any number times i.
But what is i?
i is the square root of -1, which doesn't exist, but can be used in many ways.
You can add i (23+i), subtract i (9-i), multiply i (23i) and divide i (i/5).
You can get i to the power of a number (i¹=1, i²=-1, i³=-i, and i⁴=1. This pattern repeats, and iⁿ=i^{n mod 4}).
So the second answer of the cube root of 8 is
-1 + 1.7320508075688772i and the third answer is -1-1.7320508075688772i.
And fourth roots?
Let's say we have the fourth root of 81. There are 4 answers for this:
A) 3
B) -3
C) 3i
D) -3i


We know that 3*3*3*3 is 81, and that -3*-3*-3*-3 is 81, but why 3i and -3i?
3i*3i is -9, because if the square root of -1 is i, then i² should be -1. And because 3*3 is 9, 9*-1 is -9, 3i² is -9. But we want 3i⁴, not 3i². So we square -9 to get 81. Same story with -3i.
So really, the root of a number is much more complex than you might think.
 
-The Roaring Thunder

M2C2A: Episode 10, Is 0 a prime???

Let's think of the definition: A number only divisible by 1 and itself. 0/1=0 ✓, but 0/0, itself is equal to...
Yes, calculators will tell you 0/0 is undefined, but if when we say x/y, you're looking for a number z that passes the test zy=x, so if x=0, and y=0, couldn't z be any number? 0×0=0, 0×1=0, 0×π=0, even 0×i^∞=0!
YES!

So 0/0=R∪C, that is, the union of the real and complex numbers.

Any number in or out of the number line can be 0/0, but isn't a number divisible by another when the number is an integer?
So because 0/0 can or can't be an integer, it is only SOMETIMES divisible by itself. So 0/0 is NOT completely divisible by itself, doesn't this seem strange?
We're dealing 0/0 as if we weren't dealing with something comparable with tan(90) or ∞^∞ .
So is it prime?
A little.