The Good Will Hunting Problem

This installment about mathematical problems is going to be on the problem from the movie, “The Good Will Hunting”. The focus will first be on the story behind the actual good will hunting, and secondly on the problem illustrated in the movie. The problem in the movie does not require an extensive amount of work to solve, but it seems to be a more popular problem that anyone could take a crack at. The hope is that this post will be more entertaining, and a little bit easier to read then some of the other posts made previously.

Will Hunting is more of an urban legend then an actual person. There are however several candidates, but there are two main candidates for the actual Will Hunting. One of them is thought to be George Dantzig. One day, while in university, George was late to class. He got there sat down and copied some problems down from the board, these were homework problems related to the lecture. George took them home, and he thought that the last two were rather hard problems, but he worked them out anyways. He turns in the assignment, and nothing happens for six weeks. Then early in the morning he gets a knock on his door, and it is his professor. The professor explains to him that he was only supposed to do the first few problems, the last two were unsolved problems dealing with statistics. George had solved these problems as a part of his homework. Later, when George was completing his PhD, he asked his professor what to write about, and he told him to put those two pieces of paper in a binder, and call it good. The second Candidate for Will hunting is a man by the name of William Sidis. William was a child genius. He was said to have been giving lectures at Harvard at the age of 11. He had a natural ability with math, but he had trouble with the law. He attended communist rallies and assaulted an officer. This lead him to be put in jail. He was released from jail under the condition that he saw a therapist. His therapist being with father. When he was finished seeing his father, he described it as a being a living hell, and swore to be done with academia. William Sidis never worked in a major academic position again. He found office jobs, and in particular he liked running adding machine, because they fascinated him. Though these are the candidates for the real will hunting the movie is not actually based on either, in fact the movie is based on a mathematician by the name of Ramanujan. Ramanujan deserves his own blog post, so in this post he will be covered briefly, but much like Will in the movie he was a math prodigy. Ramanujan was born in India, and his work was impressive considering where he came from. Ramanujan did not have a formal education. When he was in his twenties his potential was seen, and he attended Cambridge. Sadly he died soon after. Much of his work was done concerning partitions. Partitions are the ways you can break down numbers into multiples, and additions.

In the movie, “Good Will Hunting” an MIT professor presents a problem on the board. He claims that it took MIT professors two years to solve the problem. This is probably an exaggeration made for the film, because it is not that difficult to solve. The hardest part about the problem is to understand what the problem is asking. The question is to draw all the homeomorphically, irreducible, trees of size n=10. Breaking that down one piece at a time, trees refers to a network of lines and dots without cycles, meaning no closed polygons. Homeomarphically means you can stretch, and rotate a network tree but it still counts as the same trees. In doing so it puts a finite number on the number of solutions. Next, irreducible means that if there is a tree with a point that could be removed, and not change the tree, then the point should not be there. This is because it could be reduced, so the point was serving no purpose. The last part, n=10, means that the trees must have ten points, that need to be connected with lines using these parameters. There will be an example of one of the solutions in order to fully illustrate the problem at hand. There are ten trees that match the criteria explained, and there will be a link to the ten solutions.

All solutions http://25.media.tumblr.com/tumblr_lm8kawOpTO1qcj2hco1_400.png

Sources https://www.youtube.com/watch?v=iW_LkYiuTKE
              https://www.youtube.com/watch?v=SzjdcPbjaR4
              http://www.eoht.info/page/Good+Will+Hunting+(William+Sidis)

The Riemann Hypothesis Part 2

This blog post is part two of the Riemann Hypothesis, the first part is also on this blog. The first part covered zeta functions, this was in order to give background knowledge allowing for this post to jump right into the problem without a long explanation, and allow for a fuller in depth explanation of the problem. However, I did not mention in the previous post that it is necessary to understand complex numbers, and the complex plane. Complex number were looked over because, that concept is taught to many kids in eighth grade math, so hopefully most of the audience already understands it. The goal of this post is to explain the actual problem, and secondly why the problem is important.

It was mentioned in the last post that zeta functions were a big part in understanding the Riemann Hypothesis. Also previously mentioned Euler worked with zeta functions. The difference between Euler’s and Riemann’s work was what they inputted into the zeta function. Euler focused on real numbers, Riemann on the other hand worked with complex numbers. Complex numbers are real number and imaginary number pairs, such as 3+4i, 3 being the real number, and 4i the imaginary.  When Riemann plugged in complex numbers into the zeta function it created a plane rather than a line. but after Riemann, it was understood that there was a whole extra dimension of values that could be put through the zeta function. The zeta function can now be visualized on a plane, the x axis being real numbers, and the y being imaginary numbers. Like all functions the zeta function has a domain, its domain is when x is great than 1. That includes all values both up, and down the imaginary axis, for x values greater than one. Then Riemann proved that the function has a rather nice piece of symmetry. Through analytical continuation Riemann prove you can extend the domain to all values except for x equals one. This leaves a strip between x=0 and x=1, this strip is referred to as the critical strip. Within the strip there is a critical line at x=1/2. That information may be hard to visualize so a graph will be included to help clear up any problems.

Now that the domain of the zeta function has been cleared up, the problem can be looked at. The million-dollar question is where does zeta (s) = 0? S is referring to the complex number as well as the real numbers. Another notation could be seen as s = x + iy. So the search is on for what point on this plane when plugged into the zeta function results in a zero. There are some freebees though, these are called trivial zeros. They are zero that we already know, and really do not care about. Such as every negative even number will result in a zero. The real question is, where are the zeros on the critical strip? Riemann hypothesized that all the zeros in the critical strip would lie on the critical line, at x=1/2. So far no one has found any zero that can disprove Riemann, and nobody has proven him either.

            The question might be asked, who cares? What does this arbitrary function have to with anything important? Something that was not mentioned in the zeta functions paper was that the zeta function could also be written in terms of prime numbers. So the position of these zeros allows for mathematicians to understand the position of prime numbers. For example, if the Riemann hypothesis was proven, it could be known how many prime numbers are there from 1 to a billion, or 1 to a trillion. This function also has ties to physics and quantum mechanics. This tells us that there is a lot of connections in math and physics that have not been made yet. If the function is linked to primes, and quantum mechanics, then how is quantum mechanics related to primes? There are lots of these connections that we do not understand at this moment, so if the Riemann Hypothesis is proven it opens the flood gates of math and physics to new ideas and connections. In doing so, unraveling mysteries and understandings of nature as a whole. That is why it is a million-dollar question, and why people are so interested in it.

        Sources https://www.youtube.com/watch?v=VTveQ1ndH1c
                      https://www.youtube.com/watch?v=sD0NjbwqlYw
                      http://mathworld.wolfram.com/RiemannZetaFunctionZeros.html

The Riemann Hypothesis Part 1

This post’s unsolved math problem is the Riemann Hypothesis. However, The Riemann Hypothesis is a rather complicated, and intricate problem. So the problem will be broken into two posts, one about zeta functions, and the second on the problem itself. This is done to give more background knowledge about the Riemann hypothesis so that from the reader perspective, the information can be consumed more easily. Within this post the goal is to get readers to understand zeta functions, and why they are connected to Riemann’s Hypothesis.

A zeta function, or also known as a Riemann-zeta function, can be first seen in the work of Leonard Euler. Euler used these in his work in the first half of the eighteenth century. A zeta function uses the lower case Greek letter zeta in its notation. It refer to a summation of the reciprocals of all whole numbers raised to the a value. So zeta (x) is equal to 1/(1^x) + 1/(2^x) + 1/(3^x) . . . all the way until infinity. Euler explored these functions and it resulted in some interesting results. For example, Euler evaluated zeta (2), or the same as 1 + 1/4 + 1/9 +1/16 . . . and the result is (pi^2)/6. That might seem a little weird, but at the same time it seems possible. One of the problems people have with zeta functions is that the function produces some strange results. If you evaluate zeta (0) or the equivalent of 1+1+1+1 . . . the answer is -1/2. That seems like an impossible result, because it is a linear growth, that is a non-convergent series, meaning that it does not approach anything but infinity. The hardest part about understanding zeta functions is wrapping your mind around this idea. There are plenty of debates out there that argue about these weird results, but it remains that this is a true result. The main argument is about the equal sign at the end. The argument is over whether the sum actually equals the result, or if it is an equivalent. Some people think that if you sum the zeta function all the way to infinity then you will have the result, but other people think that the result is just an equivalent to the answer, and the latter would be right.

It only seems right to mention Riemann when talking about Riemann zeta functions. Riemann simply expanded on the work of Euler. He expanded the Zeta function’s domain, and graphed it on the complex plane. That will be specifically covered in the next post, but it is good to know how Riemann is involved. The problem itself has to do the graphing of the function itself. Now when Riemann was working with zeta functions, he came across the most controversial/famous result from the zeta function. This being zeta (-1), or equivalent to 1 + 2 + 3 +4 . . . being the sum of all positive numbers. The solution to zeta (-1) is -1/12. It is quite a bizarre answer, but it is actually an important result in several branches of physics. This result was a big part in the explanation of the extra spacial dimensions in string theory. It also allowed for a solution to the Casimir force. It is not important to know about these physics terms, but the point is to show that these results are actually results in the real world.

The last point that needs to be stressed is that these sums are not directly equal to the answers given. As mentioned previously these results are more like equivalent to the sums. If you were to add all the positive numbers together you would get an infinite solution. So mathematicians look at this result, and say that result is useless. Instead of throwing all the math away, mathematicians say what if I could get a meaningful value out of this zeta function. It can be explained with a gold pan analogy. Imagine an infinite amount of dirt, and you use a sifting pan to look through the dirt. The zeta function result is like a nugget of gold that is found in this pile. Even though is it not equivalent to the pile it was the most meaningful thing in the pile. So no some of these zeta function result do not make sense, but they are a legitimate result that gives these summations a meaningful value.

Sources https://www.youtube.com/watch?v=d6c6uIyieoo
              https://www.youtube.com/watch?v=w-I6XTVZXww
              http://mathworld.wolfram.com/RiemannZetaFunction.html

The Collatz Conjecture

In this installment I would like to talk about the Collatz conjecture. As usual I would first like to talk about the contributors, and history of the problem. Then I will talk about the problems itself. Finally, I will talk about what the problem means to math, and why its not that big of a deal. Unlike the problems I have previously mentioned, this problem is incredibly simply, and anyone can understand it in its entirety. In fact, if you can do addition and multiplication, then you can understand the problem.

It seems only right to start with Collatz himself. Collatz has a German mathematician, who was born in 1910. Although Collatz made contributions to math outside his conjecture, such as the Collatz-Wielandt formula, or his contributions to the Perron-Frobenius theorem, his conjecture is the most famous. He died in 1990 at the age of 80. It is not until the 1970’s and 1980’s with the emergence of the personal computer that his conjecture gained popularity. The advancement in computations allowed for checking for counter claims to the conjecture to be much easier. Unlike many of the other problems, the Collatz conjecture is not a millennial problem, so there is not a million-dollar prize for proving or disproving the conjecture. However, Paul Erdos offered 500 dollars for solving it, but he also said it would be pointless to even try to solve it. To quote him he said that, “Mathematics may not be ready for such a problem.”  Now let us look at the problem at which math is not ready for.

That is it, that is the entire problem, only two lines. So to explain the two lines. Start by picking a number. Any whole number like seven. So if it is an even number divide by two, if it is an odd number multiply by 3 and add 1. To carry on with seven, multiply by three to get 21, then add one to get 22. 22 is an even number so I will divied by two. to get 11, then 34,17,52,26,13, and 40. At this point it looks like it just keeps getting bigger and bigger over time. However, 40 is the turning point because the next number is 20, then 10,5,16,8,4,2,1,4,2,14,2,1. Notice once you get to one, then it repeats itself in a loop forever. This is not just for the number 7, in fact go ahead and try any number between 1 and infinity. However it is recommend to choose a small number to save time, but it is possible to try any number. The Collatz conjecture simply says that for any number that these conditions are performed to, it will always end up at one. So far it is known that all whole numbers to 2 raised to the 60^th power have been confirmed to follow this conjecture, but there is no proof that all number follow it. It is one of the hardest conjectures in math to prove, but could be easily explained to a fourth grader.

Now as mentioned that there is no million-dollar prize for proving or disproving the conjecture. This is because millennial problems have deep connections to many parts of math, science, and physics. The Collatz conjecture really does not have those kind of connections. So by solving this you will not cure cancer, or fix the flaws of cold fusion, but it will progress math.  Some of you might ask why multiple by three and add one? Or 3n+1, and the answer to that is if you cannot solve a problem you try to generalize the problem, or solve a similar problem. Mathematicians tried to solve the general form of the equation seen as an+b. The mathematicians found that was an even harder problem to solve. So it remains as 3n+1, if they can solve 3n+1 then they might gain how to solve an+b.

So the Collatz conjecture is a conjecture that anyone can understand, but no one can solve. The reason why there is not a large prize for the its solving is it does not need to be solved. Problems like the Riemann hypothesis, and the Navier-Stokes equation have connections to prime numbers and fluid mechanics, unfortunately the Collatz conjecture has no connection. Nevertheless, I encourage you, if you are interested, to go for it. For the million-dollar prize problem that has been solved, the prize was refused. This is because mathematicians do math for the math, not the money.

Sources https://www.youtube.com/watch?v=5mFpVDpKX70
              https://www.youtube.com/watch?v=O2_h3z1YgEU
              http://mathworld.wolfram.com/CollatzProblem.html

The Navier Stokes Equations

This week I would like to talk about the Navier-Stokes existence and smoothness problem. I would first like to talk about the history of the problem, then I would like to talk about the problem itself, then I would like to talk about the physics involved in the problem, and finally I would like to talk about changing the problem to allow for its use.

The problem is named after its contributors Claude-Louis Navier and Georg Gabriel Stokes. Navier was french physicist that lived in the late 1700's and early 1800's. He grew up and took interest in engineering and physics. He made a few contributions to math and science, but his main contribution was the Navier-Stokes equations. Stokes, the other contributor, was born in Ireland. Stokes lived in the 1800's roughly the same time period as Navier. Stokes became the head of mathematics at Cambridge, before dying in 1903. His contributions were similar to that of Navier. They both specifically worked on fluid dynamics. They came up with several equations that to me have too many letters from too many alphabets, but some how someone understands them fully.

The problems itself refers to solving the equations created by Navier, and Stokes. As mentioned before the equations deal with fluid dynamics, but there are a few more conditions dealing with the problems. First the fluids are assumed to be non-compressible. Fluids if you are not aware refer to gases and liquids. These fluids are assumed to be incompressible even though fluids are compressible to a small degree, the change in density, income cases, is so small it is negligible. The second condition is the fluid must be viscous. These equations under those conditions then describe fluid mechanics using newton second law of motion, that being force is equal to mass multiplied by acceleration. As you can imagine this is a big deal. Understanding fluids is what gets your airplanes in the air or your water to your house. These equations sometimes have solutions in certain situation, but mathematicians can not seem to prove that these conditions always exist. There is not a certainty that the smoothness condition can always exist in the three dimensions. On top of that if we assume that the smoothness condition always applies then the question remains, if these smoothness solutions exist do the have bounded energy for their mass? Like many of the problems will be talking about, this problem is a millennial problem. That means if you can solve the problem, then you will receive a million dollars. You will also probably be awarded a Field's Medal, and get to be on the front of a newspaper. Society as a whole will still probably care more about what the Kardashians are wearing instead of the solution to an incredibly important explanation of fluid dynamics, but oh well.

At this point it would be good to mention what the equations actually pertain to in fluid dynamics, but the content of the equations would go over a lot of heads and lose people.  Rather then showing all the equations, it would be more appropriate to give a brief description of the general form of the equations. 

This is the general form of the equation. All of the equations can be derived from this equation.

The first "p" represents the density of the fluid. The "v" represent velocity component. The second "p" refers to pressure. The "t" refers to time. The upside down debt refers to gradients or vector derivatives. Now I wish I could fully explain the equation, but I hope it makes more sense to you as a reader than it does to me.

Now these equations are not useless. In fact they are very important. As mentioned there is not a guarantee on our answers. However, there are still ways of getting numerically practical answers out of the equations. For instance, sometime by assuming factors lie change in density or pressure, as negligible you can get a really close answer. So sometimes you do not need to be able to solve the equation straight up, sometimes you can just get close enough for it to work. Hopefully you as an enthusiast will come up with a solution to get rid of these approximations.

Sources http://www.claymath.org/millennium-problems/navier–stokes-equation
              https://www.grc.nasa.gov/www/k-12/airplane/nseqs.html
              https://math.berkeley.edu/~chorin/chorin68.pdf