Aventador versus the cow.

Are cows greener than Lamborghani’s?

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problematic views, futuring maths.

I’ve been going over a couple of ideas for some blogs recently, but haven’t quite had the time or depth to write them all out. I had a look through my ideas, and realised that the reason I haven’t posted them is that I don’t think I can explain all the concepts very simply and easily, and so I choose to not write them up…so I’m going to stop doing that. Very simply, from now on I’m going to give broader explanations of ideas. If you have any questions, feel free to leave a comment and I’ll get back to you.
The first thing I’d like to do is address a question – or perhaps a concept – that most people have in regards to mathematics.
Most people think that they are not logical enough to do mathematics. This puts them off ever even looking at the subject, because they don’t think they’ll understand it.

This view point is note entirely correct. Whilst there is a lot of logic in mathematics and it is certainly helpful to be able to think logically, it is not all there is to it.  Mathematics is – by  necessity – a very creative science.  To stretch mathematics to new areas, you need to be creative in your approach. To attempt a proof of something that no one else has done, you need to be creative.

This being said, there are rules and logical steps that you follow – just like everywhere else. Music is governed by certain laws of physics. Art is governed by visual laws. Poetry and literature are governed by laws of the medium it is written in – even poets like e.e. cummings ascribe to certain laws, albeit they enjoy breaking most of them. Mathematics is the same.

A few people I have told this to have been sceptical of my views. They remember their mathematics at school as being very strict and formal. Very few were privileged enough to be shown the joys of creating new mathematics, or taught to think rather then remember. To people who view mathematics as being a logic and rule based science with no deviations, I suggest you look up some recreational math authors (or contact me for some who I’ve read).

In my mind, creativity is crucial to development in mathematics. You can’t do new maths without creating, you can’t find anything new without creating.
A small example of this: unsolved number theory problems.
Now, most of these problems are very easy to state (for example, the collatz conjecture, which I’ve written about in a previous post). The majority deal with finding out if a certain pattern satisfies every number – this requires us to look at the infinite. How exactly do you approach the infinite without some creativity?

You can’t start counting and eventually get there – infinity is easily viewed as the biggest number you can think of, plus 1. You can’t ever reach it. Your only way of examining it is to simplify it, or to creep up on it and surprise it.

Here’s a devious proof that shows this principle:

assume you want to show that there is a set of consecutive composite numbers of any size.

(so the number n is not prime, nor is n+1, n+2,…for the next s numbers.)

The numbers 8,9,10 show that there exists a set of 3 consecutive numbers. 25,26,27,28 is a set of 4.
But how do you show that there is a set of any integer?
We need to deal with infinity here. We need to find a way to show that there is a set for any number, no matter how big the number is.

here’s the trick: if you take a number n, and you times it by n-1,n-2…2 then the number you get is called the factorial of n, and is shown as n!. so n!=n*(n-1)*(n-2)*…*2

now this is obviously composite. We can also show that n!+2 is composite as n! has a factor of 2, as does 2, so it is composite. We can do this for any factor of n! – so for all numbers from 2 to n-1, we can add it n! and get a composite number. Assume we choose k. then n!=n*(n-1)*…*k*…*2, and n!+k=k( (n*(n-1)*…*(k+1)*(k-1)*…*2+1), so k is a factor.

this means that all numbers from n!+2 up to n!+n are composite. So to find a set of size s, take the s numbers after (s+1)!+1 – they’ll all be composite. (quick question, are there any sets of size 2? Why?)

Thus we have a way of finding any amount of consecutive composite numbers – all the way up to infinity.
It is not at all obvious in the above proof that you need to use a factorial – that required some creativity to come up with. This is why there are still so many unsolved number theory problems – computers can not calculate all the numbers from 1 to infinity, and so we need to use human ingenuity to find shortcuts to infinity.

I am not saying here that computers will not be able to prove number theory problems. There is a role for computers in proofs. Sometimes we can reduce a problem down to a set of numbers – so if it satisfies these numbers, then it satisfies all numbers. This is the way the four colour theorem in Graph theory was solved. I wouldn’t be surprised if the Collatz conjecture is similar. but I believe that no computer will ever be able to take a conjecture and prove it from scratch. I may be mistaken, I do think I recall a book that said that Turing proved that computers won’t be able to find solutions to all problems, which leaves the possibility that some problems will be proved by computers, but my current knowledge does not know of any that have been done by a computer.

Now I’m the sort of person who enjoys both logic and creativity.  And I find that maths satisfies both of these loves. I enjoy looking at unsolved problems and attempting to think my way through a new approach. So here are a few that I have been looking at lately/ enjoy thinking about.
Collatz conjecture:

there is a separate blog post about this, but to reiterate: take any number n. if it’s odd, multiply it by 3, and add 1. If it’s even, divide it by 2. Repeat for the new number. Keep doing that, and eventually you’ll reach 1 – or at least, that’s the conjecture (which means it hasn’t been proved).

This problem has been around for around 70 years ( I think), and has some interesting properties that I’ve seen – but there’ll be more on that when I have enough to post!
Goldbach’s conjecture.
Goldbach’s conjecture states that there are an infinite number of “twin primes” where p and p+2 are prime. A few examples are 3,5 ; 101, 103 etc. This one is lots of fun to play with if you enjoy working out primes, and you can start finding links/ideas almost immediately.
The abc conjecture.
Take three numbers (a,b,c) that have no common factors, with a+b=c.  let d be the product of the distinct prime factors of a,b,c.
The conjecture states that d is very rarely much less than c.
This conjecture is quite difficult for a non-mathematician to get their head round at first. It has some interesting and important consequences.
The 196 algorithm
Take any 2 digit number, reverse it, and add the two numbers. Repeat this until the number reached is a palindrome.
196 is the smallest number for which no palindrome has been found. There is no proof yet that this algorithm never gives a palindrome for 196.
To demonstrate:
163+361=524+425=949 which is a palindrome.
These are only a few. There are a lot more, and that’s just in number theory.
Which brings me to another thought: different branches of mathematics.
Something that not many people seem to understand is how diverse mathematics is. They seem to assume that if you have a degree in mathematics, you know all mathematics….this is definitely not true!   
The easiest way to see this is to look at it like this:

Assume that, a long time ago, there was a village in the middle of nowhere with a very detailed language. Now a huge fight erupted, and because of the fight, many people in the village decided to leave and start new villages. These people all formed their own groups which then left the village and headed off in their own direction, never to travel back to the original village.

now over the following thousands of years, this happened many times. Occasionally, splinter groups from one village would join up with another splinter group, forming a new village, but people never travelled to an existing village.

Through this timeline, little changes in the language would become apparent to an outsider. Each village would start forming new words for new ideas or ‘bastardizing’ old words for old ideas or things. Eventually (assuming there had been no outside influence on the language) there would be many different languages, but each language would be similar to each other. An outsider coming in would probably have to learn the basics of the original language, and, from there, have a chance to communicate with the different villagers. But there would be big difficulties. Certain topics and words would be recognisable, but others would be completely alien. The outsider may hear a strange word, and start to investigate it, only to find out it means exactly the same as a completely different word in a different village. Or the outsider may hear a very familiar word, but be shocked to discover it means something completely different.
Mathematics is slightly like this. Certain ideas are the same in all branches, and there are always overlaps. To jump from one branch to another is not always a simple process. Sometimes it requires learning a whole new language, at other times it requires learning a new translation. Sometimes, there are branches that act as an in-between, which can make the jump easier.
At times, mathematicians have found interesting and amazing links when looking at supposedly separate branches. An example of this is complex numbers. When they were first introduced, Riemann began to examine them. He started looking at what happened when he applied certain functions to them. He looked at what happened when he put them into the zeta function…and discovered an amazing link to prime numbers, in the completely separate branch of number theory!
Collaboration amongst mathematicians is always encouraged, but only recently have mathematicians began to collaborate with non-mathematicians. The results have pretty impressive –quantum physicists have discovered a link to the work of Riemann, economists and mathematicians discovered the ideas of game theory which became important to cryptography and computer scientists started to overlap with logicians, set theorists and graph theorists. There have also been interesting and wonderful advances in medicine, psychology and the more ‘physical’ sciences when applied mathematicians have looked at their problems.
We are coming into an age where collaboration is an email away. Mathematicians are finding more and more ways to help their fellow man. Whilst some mathematicians may not enjoy the ‘corrupting of the pure fields of abstract math’ to aid in other subjects, the benefit to all is obvious.


Collatz’ Conjecture

So apart from wondering about whether the English language accepts z’s as a plural or not, this ‘problem’ is intrinsically interesting and very easy to understand:
Start with any natural number (a positive whole number) – I’m going to start with 7 – and  
If your number is odd, multiply it by 3 and add 1. If it is even, divide it by 2.
So as 7 is odd, we multiply it by 3 and add 1, getting 22.
Now we repeat indefinitely, using the new number.
22 is even, so we get 11, odd so 34, 17, 52,26,13,40,20,10,5,16,8,4,2,1,4,2,1,4,2,1….
As you’ll notice, starting with 7, you eventually reach the endless cycle 4,2,1.
Collatz’ conjecture states that for any natural number, by repeating these operations, you will eventually reach the number 1.
It is a conjecture as it has not yet been proven! It is however quite a fun problem to play with. There are a few applications of it, I haven’t read up on it in too much detail, but you can do your own reading on the wiki page (here) or just googling “collatz conjecture” (here).
After playing with it for a bit, you start to notice some patterns:
256,128,64,32,16,8,4,2,1
15,46,23,70,35,106,53,160,80,40,20,10,5,16,8,4,2,1
9,28,14,7,22,11,34,17,52,26,13,40,20,10,5,16,8,4,2,1
The first thing you may notice is that the end sequence is always 16,8,4,2,1 , so long as you don’t start with one of those numbers (if you do, it’s a fairly straightforward sequence). Now, why are these numbers so special?
The generator (the function that produces the next number) either takes the previous number and divides it by 2, or multiples it by 3 and adds 1. thus we can find the previous number by reversing this process, so if we want to find what number leads to 7 we would either multiply it 2 (reverse of dividing it by 2) or minus 1 and divide that by 3 (the reverse of multiplying by 3 and adding 1). From this, we can see that the only numbers that lead to 7 are
2 (3*2+1=7)
And
 14 (14/2=7)
However, we only multiply by 3 and add 1 if the number is odd, thus if we had the number 2 we would not then get 7 but rather 1. From this, we can see that the only number that leads to 7 is 14.
We can use this same process on the number 1 and follow it to see which numbers reach 1 (on the wiki page, I think this is called a recursive build or something).
So: to get to 1 we either divide by 2 or multiply by 3 add 1, so we need to solve
x/2=1
 and
3x+1=1.
We can see that the only numbers that lead to 1 are
 2 (2/2=1)
and
0 (3*0+1=1),
 but 0 is even, hence it will just keep repeating itself. For this reason, when dealing with the Collatz Conjecture, we define 0 to not be a member of the natural numbers. (if it were, then Collatz conjecture would be false! That’s just an aside. As 0 would merely repeat itself, it’s not very interesting, and this is why I’m choosing to ignore it!)
So, the only number that leads to 1 is 2. Using the same method, we can determine that only 4 leads to 2. we can then see that 1 and 8 lead to 4, but as the sequence ends at 1, we find that only 8 leads to 4, and only 16 leads to 8.
Thus, the end sequence will always be 16, 8, 4, 2, 1 (if we don’t start with one of those numbers). What else we can notice is that if we reach the number 16, we will reach 1!
16 is obtained by 5 or by 32. What you may notice already is that any number that is a power of 2 will lead to 1. (a power of 2 is any number n=2k, where k is any natural number, so 4=22, 8=23,256=28). So if a sequence ever reaches a power of 2, it will reach 1.
 Now, as stated, 5 leads to 16. If you multiply 5 by a power of 2, that number will eventually become 5, which will become 16, which will then become 1….
Now, 5 can only be reached by 10, which can be reached by 20 or 3. as 20=5*22, we already knew it would lead to 1, so we ignore it for now. But 3 is a ‘new’ number, and the interesting thing is that 3 multiplied by a power of 2 will also eventually lead to 1!
It seems we’re starting to find some patterns. If you sit down with pencil and paper and, starting at 1, work backwards, using arrows to link numbers, you’ll notice that this pattern seems to happen a lot.  (here’s an image of what you can end up with, and here’s a link of what can happen to you if you do it for too long‼‼)
Now it would be great to introduce some definitions, but it will probably prove easier to look at this through an analogy:
Imagine a tree. Let the number 1 represent the roots of the tree, and let the main trunk represent all the natural numbers that are of the form 2k. so it will look like this:

 Now, as stated above, the number 5 leads into 16, so we can view this as a branch with a root of 5, which links into 16. Our tree will now look like this:



This can then be extended out to show the path of the numbers.
Eventually, you can end up with something that looks like this:
Which you can keep growing!
Now, if we look at the main trunk, we have already noted that it is simply all natural numbers 2 k, which is also 2k× 1. If we look at each branch, you will notice that each one can become a new tree of it’s own, with the main trunk having a root number, and then consisting of all natural numbers of the form 2k× the root.
Therefore, each branch has a root number, and consists of all natural numbers of the form 2k× the root, and has branches extending out of it at some points.
To keep things simple, I will keep the definitions similar to the analogy.
As you may notice, it is easy to find the numbers that form the branch once we know the root number, so the root’s are the most important information in the problem.
We can then number the branches (and hence the roots) by how far away they are from the main trunk. We will number the main trunk 1. All branches that lead into it will be numbered 2, all those that lead into those branches will be numbered 3 and so on. Thus we can refer to the set of branches of number p, and know that they are (p-1) branches away from the main sequence.
So, to clarify: we are talking about roots and branches, where the branches consist of all natural numbers of the form 2k× the root. The roots will then be referred to as the pth root, meaning that this root leads into the p-1 branch, and we will define it as root(p). the branches will then be branch(p), so that branch(p) is the sequence of numbers 2k × root(p).
Thus the number 40 is in branch(2) as it is 23× 5, and 5 is a member of root(2).
Before reading on, make sure you understand the above terms!
Now, via this definition, we can then see that root(p) does not refer to a single number, but rather a sequence of numbers. So therefore, numbers that are members of root(2) are 5,  21, 85, 341 etc (multiplying these numbers by 3 then adding 1 gives 16, 64, 256, 1028 etc respectively).
This therefore causes problems, as we are no longer looking for specific numbers, but rather looking for a pattern of numbers based on the distance away from the main trunk.
Bearing these facts and tools in mind, we can then begin to further explore the Collatz conjecture by looking at the root numbers… however, I have not quite completed this step yet! So, I will end this post here, and update it once I have gotten further in my own personal exploration. In the mean time, I encourage you to do your own searching – have fun!

Right, answers to yesterdays question!

So we started with a single square which had an area of 1. After the first iteration of the pattern, we had four new squares, each having an area of 1/3 X 1/3 = 1/9. On the next iteration we had 20 new boxes, each with an area of 1/9 X 1/9 = 1/81. Write down what you know so far:
Squares total
area sum
area total
1
1
1
5
1+4*(1/9)
1 4/9
25
1+4/9+20*(1/81)
1 56/81


As this shape is created using a pattern, there will be a pattern for the total area (that is a colloquial logic proof, but you use those principles if you want a ‘solid’ proof). All that we have to do is find the pattern that is forming from these areas, so I’m going to rewrite the area sums in a slightly different way for the 3rd iteration: 1 + (51-50)/91 + (52-51)/92. (I presume most people know 52=5X5 = 25).
The amount of boxes is also in a pattern, they are going up in powers of 5.
So, for the first iteration, the area = 1.
Second iteration = 1 + (51-50)/91.
Third iteration = second iteration + (52-51)/92
Fourth iteration = third iteration + (53-52)/93
So what will the sum be for any iteration?
Let’s try and find the sum for the nth iteration. As you will notice, the area of any iteration is equal to the area of all the previous ones + the area of the new one, meaning that the sum can be written as a pattern (which we’ve already shown) but this means that we can now use the summation symbol…
(∞,i=1) ∑ i
This symbol is an easier way of writing out the sum 1+2+3+4+….all the way to infinity (and beyond!). very simply, the lower limit (i=1) is the number you start with, ∞ is the upper limit (So you continue adding up to this number) and the pattern between each number is (in this case) 1.
If we had
(∞,i=1) ∑ (i ^2)
It would mean add up the numbers 1 + 4 +9 +16….all the way to infinity, whereas
(5,i=1) ∑( i^2)
Is equal to 1+22+32+42+52
=1+4+9+16+25
=55.
So I hope you now understand summations.
Getting back to our problem: we are trying to find the sum of all the areas, we know that the area of the nth iteration is equal to the sum of all the previous iterations plus the new one, and we’re noticing that the 4th iteration is equal to the previous 3 plus (53-52)/93. Notice now that there are three powers, two of which are 3, the other is a 2. 3 is 4-1, 2 is 4-2. It is therefore possible to write this iteration as (54-1-54-2)94-1.
If you now replace the 4 with an n, you get the general form: (5n-1-5n-2)/9n-1.
Realising this, we can make the pattern for the area, the nth iteration =
(5n-1-5n-2)/9n-1+(5n-2-5n-3)/9n-2+…+1
So if we write it with a summation symbol:
1+(n,i=2) ∑((5^(n-1)-5^(n-2))/(9^(n-1))
Will give you the total area for whatever value of n you use.
But we can simplify it down. Looking at just the part in the summation symbol : (5n-1-5n-2)/9n-1 we can simplify it to 5n-1(1-5-1)/9n-1. 1-51 = 4/5, so =(4/5)*(5n-1)/(9n-1), so the summation can be written as
1+(4/5) (n,i=2)∑(5^(n-1)/9(n-1))
Note that this series is for the first iteration of the pattern (the first time the squares are added).
We could simplify it to
1+(4/5) (n,i=1) ∑(5^n/9^n)
(I appologise for the hideous format of the summations, i struggle with HTML….)
But the original question was what is the total area that this shape will make? To do this, we need to look at what the sum looks like at infinity, ie, we want to look at the area at the millionth iteration of the pattern (just an example, we actually want to use a number that is even bigger) so we want to find the answer to the equation
1+(4/5)(∞,i=1) ∑ (5^n/9^n)
So let’s have a closer look at the summation.
The first term is going to be 5/9, the second will be 25/81, 125/243…etc which is a geometric sequence.
A geometric sequence is a sequence where each term is a multiple of the term before, so 1,2,4,8,16,32,64….would be a geometric sequence. Adding up terms like these can be rather difficult without using a formula, so it’s rather…lucky?… that someone has found them ( I may at some stage do a blog on the history of the arithmetic and geometric sequences) and the formula for finding the sum is as follows: given that the ratio between the terms of a geometric series lies between -1 and 1, the sum of the terms to infinity can be expressed as a/(1-r), where a is the first term in the sequence and r is the ratio. If we had the series 1; 1/2 ; 1/4 ; 1/8 etc, a would be 1, and r would be 1/2 so the equation would be 1/(1/2)=2.
There is an interesting story about this particular series as well, but that will be another time….
Now, using the above formula for the equation given, we can solve just the summation as follows:
(5/9)/(1-(5/9)) = (5/9)/(4/9) =5/4
Substituting it in, we get 1 +(4/5)X(5/4)=2 …and this is the total area of the shape!