Reflection of a point through a given curve (original)

Hello reader,

This will be my first 'original work' post (NOTE: Chances are, this is not undiscovered knowledge. Somebody, somewhere must have already discovered this. However, I am not aware of any work already done on the problem, and all work has been done independently ie. I worked this out, I didn't copy it from somewhere else). Hope you guys like it :)

I was on a plane back from South Africa (which is an AMAZING place to visit if you ever get the chance) and anxiously awaiting GCSE results. To take my mind of the impending judgement which, on the basis of a poorly designed exams, would give me a grade to decide my aptitude for the given subject for the rest of my life... (-_- ugh...GCSEs)

I started thinking about Maths (as you do) and the problem that had been bugging me for a while. If you draw a curve:

y= f(x)

and a point 

(x₁ , y₁)

you should be able to reflect the point in the curve such that the line connecting the point and the reflected point  is a normal (perpendicular) to the curve (what I like to call a 'proper' reflection) like so:

I got thinking about it because it's easy to estimate the 'proper' reflection point on paper just by drawing (like above), so the maths should be pretty simple too...right?

Well, this is the equation I came up with:

                                           f(x)  + [(x₁ + x)/f '(x)] – y₁ = 0

The solutions (values of x which satisfy the above) to the first order differential equation above can be input into the following formula to give the set of 'proper' reflection points:

                                               (2x – x₁  ,  2 f(x) – y₁)

So, there we are. Sorry if this was a bit confusing, blogger is a bit rubbish for equations and diagrams. Please comment with any difficulties in understanding or general questions and I'll do my utmost to answer. If you have some free time, why not try to see how I proved this solution. The proof of this is unfortunately far too large to contain in this margin (hehehe, sorry, couldn't resist ;) ). Not difficult really, but interesting and with enough pitfalls to dodge to keep you on your toes. 

If you're wondering whether to comment or not... please comment :p

Thanks for reading :)
Anand

Millennium Prize Problems

Just a quick development on the P=NP article for anyone unaware of the Millennium Prize Problems.

In the year 2000, the Clay Mathematics Institute offered a prize of $1,000,000 for a proof for any one of the following 7 open (ie. unsolved) problems:

1.       P versus NP problem – A major unsolved problem in computer science essentially asking whether if a solution to a problem can be verified in polynomial time, can the problem itself be solved in polynomial time i.e. does P=NP?

                                      

2.       Hodge conjecture – A major unsolved problem in algebraic geometry on the subject of the algebraic topology of non singular complex solutions to polynomials.

3.       Poincaré conjecture (solved) – Solved by Grigori Perelman using the principles of Ricci Flow, the problem was on the subject of the topological characterization of spheres.

4.      Riemann hypothesis – A major unsolved problem on the subject of the distribution of the zeros of the Riemann Zeta Function.

                                                                

5.       Yang–Mills existence and mass gap – A major unsolved problem in quantum field theory underlying the Standard Model in physics. It asks for a proof of the existence of a non trivial Yang-Mills.

6.       Navier–Stokes existence and smoothness – A major unsolved problem in engineering and theoretical physics on the subject of the proof of the Navier-Stokes equations which are one of the pillars of fluid dynamics.

7.       Birch and Swinnerton-Dyer conjecture – A major unsolved problem in number theory on the subject of the elliptic curves over given number fields.

Each of these is an extremely difficult (to the extent that many brilliant mathematicians have devoted many years of their lives to solving the problem and still failed to find a complete solution) and very interesting problem and as yet only one has been solved. Furthermore, each problem’s potential solution would have extremely far reaching consequences for their field. 

P = NP (And Some Basic Computer Science)

The computer science and maths problem P = NP is one that many people have heard of, mostly because a $1,000,000 prize has been offered for a solution (that's right, though this is probably the hardest way imaginable to become a millionaire!). The Clay Mathematics institute named it as one of their seven 'Millennium Prize Problems', and as such it has gained a certain notoriety. Another, less provable but I think equally true reason it is famous is because its name ( ie. P=NP) is not difficult to remember, at least compared with other such problems (eg. Navier - Stokes existence and smoothness). But anyway, onto the maths:

P=NP is a problem to do with the complexity of certain questions that can be solved by computers. Say you wanted to write a program that would give you all the even numbers from a set of numbers that you give it. For example, say you give it x numbers. For each number, it would divide by 2 and then if there is no remainder, the program tells you the number is even. Therefore, for each number, the program takes one step to 'solve' it, and then moves onto the next number. The running time for this program is therefore x steps. However, if you wanted all the numbers which are not divisible by 2 or 3, then the same program would take 2 steps per number, giving a running time of 2x.

Now this is a simplistic introduction to the concept of 'running time', and by extension 'complexity'. However, running time is not the same as complexity. If you gave the 'even number program' described above 10¹⁰⁰ numbers, it would have a very long running time (10¹⁰⁰ steps). However, complexity is about the equation which gives us the running time. In the case of our even number program, the running time = X steps, with X being the number of input numbers. This is considered an 'easy' problem.

Now some problems have a complexity equation that is a polynomial (ie. the sum of powers of x). These problems are considered 'easy' as the running time increases relatively slowly as x increases. Other problems do not, and rather have a complexity equation that is an exponential function (eg. 2^x). These problems are considered 'hard' as the running time increases very quickly as x increases (see how big 2¹⁰ is, and then 2¹⁰⁰ if you don't believe me).

So where does P=NP fit into all this computational complexity theory? Well P-problems (or Polynomial problems / easy problems) and NP-problems (Not Polynomial problems / exponential problems / hard problems) both exist, and P problems are a subset of NP problems (ie. all 'easy' problems have ways of solving them which are 'hard', but not all 'hard' problems have ways of solving them which are 'easy'). 

Once the problem is solved however, another problem to verify the solution arises, and this second problem is also either 'easy' or 'hard'.

P=NP states that all problems that can be verified in polynomial time (ie. the second problem is 'easy') can also be solved in polynomial time (ie. the first problem has an 'easy' way of solving it, even if people don't know how yet). For example, if you make a program that, given a set of integers, finds if some nonempty subset of them sum to 0? For instance, does a subset of the set {−2, −3, 15, 14, 7, −10} add up to 0? The answer "yes, because {−2, −3, −10, 15} add up to zero" can be quickly verified with three additions (three steps). However, there is no known algorithm to find such a subset in polynomial time (there is one, however, in exponential time, which consists of 2ⁿ-1 steps) (Source of Example: Wikipedia)

P=NP would state that there must exist an algorithm to find the subset in polynomial time, as all problems that can be verified 'easily' can also be solved 'easily'.

There is great debate as to whether P=NP is true, with many mathematicians and computer scientists working on the project. The current majority lies in the P does not equal NP camp, who believe there exist problems that can be difficult to solve but easy to check (like the one above), but more than enough believe that P does indeed equal NP. A proof either way would have huge practical and theoretical implications in many fields including, of course, mathematics which is why this is hailed as one of the most important problems of the modern age. Any takers?

Hello World

Hello World,

Welcome to My Maths Blog (not the most inventive name I'm aware but it's late and it's the best I've got). My name is Anand (recently found out this is Sanskrit for 'joyful' :) ) and I am 16 years old. A quick bit about myself: I live in Hertfordshire, UK and attend a local school called Haberdashers' Aske's Boys' School. My favourite subject is mathematics (no, really), and has been for the last 2 years. A few non-academic interests of mine include sports (namely hockey, cricket and football), music, theatre and film, novels, economics and philosophy. I support Saracens RC, Spurs FC and Southgate Hockey Club (who I also play for in the junior section)....right, that's enough about me for now.

I've created this blog because I want to share both some of the 'original' (someone else has no doubt come up with them already, but by original I mean I haven't read/learnt it somewhere else) ideas I think of from time to time but seldom write down on paper, and the interesting mathematical tidbits I come across as I try to learn as much as I can about this awesome subject. Hopefully, this should be fun :)

How I foresee this will work is every so often I will post an article on some point of interesting maths. Comments either with questions (which I will do my utmost to answer), points of interest, developments or even just thanks would be greatly appreciated. So here goes...