Algebra

Prime Obsession: Bernhard Riemann and the Greatest Unsolved Problem in Mathematics

Prime Obsession: Bernhard Riemann and the Greatest Unsolved Problem in Mathematics

I have just finished reading Prime Obsession by John Derbyshire. A casual math book  which I will pass my thoughts and recommendations about this book in this post.


Prime Obsession is a book about the history surrounding Bernhard Riemann and the Riemann Hypothesis. A hypothesis, which is more than 150 years, and still haven’t been neither proved nor falsified. The Riemann Hypothesis tells us a lot about the Riemann zeta function and how it is linked to the distribution of  primes. This link is the turning point for the whole book.

When I first encountered the book I did not like the writing style of John Derbyshire to be honest. It was in a very conversational tone. But as I progressed through the book I partly forgot about the writing style. More importantly; the content was so great that the writing style for unimportant to me.

The book covers two aspects of the Riemann hypothesis. The even chapters (oh, the author is a mathematician after all – so why not) gives the reader a historical account about European mathematics. They cover many things from a few generations before Riemann all the way to today. I loved these chapters which introduced me to a lot of the persons naming the famous theorems that we are using today such as Gauss, Euler and of course Riemann himself.

The remaining chapters (the odd ones) covers a wide range of mathematical topics. The culmination is the explanation of the Riemann hypothesis and how the non trivial zeros of the zeta function links to the distribution of the primes. Most of the chapters were written in what I would consider being an accessible way which gives a fine introduction to the mathematical topics. It is by no means a textbook that will give you the deeper understanding of the subjects it touches, but it gives you an introduction and tells you what the idea of the subject is.

The only part I didn’t grasp was a chapter on field theory, I thought it lacked something. However, it wasn’t that essential for the main result I think. The main result was a bit complicated to understand as well, but reading the last two odd chapters again helped a lot. I wont say I understand the Riemann hypothesis in the mathematical sense, but I have a conceptual idea of what it does. So this book delivered exactly what it promised to do. Besides that it also inspired me to continue the wonderful journey through mathematics.

If you haven’t read it yet, this is an obvious choice for an item on the wish list.

Posted by Kristian in Other, 2 comments

Project Euler 42: How many triangle words does the list of common English words contain?

Problem 42 of Project Euler presents us with a new string manipulation problem. The problem asks us the following

The nth term of the sequence of triangle numbers is given by, tn = ½n(n+1); so the first ten triangle numbers are:1, 3, 6, 10, 15, 21, 28, 36, 45, 55, …

By converting each letter in a word to a number corresponding to its alphabetical position and adding these values we form a word value. For example, the word value for SKY is 19 + 11 + 25 = 55 = t10. If the word value is a triangle number then we shall call the word a triangle word.

Continue reading →

Posted by Kristian in Project Euler, 8 comments

Project Euler 33: Discover all the fractions with an unorthodox cancelling method

Problem 33 of Project Euler is a really fun little problem.  At least I think so. It reads

The fraction 49/98 is a curious fraction, as an inexperienced mathematician in attempting to simplify it may incorrectly believe that 49/98 = 4/8, which is correct, is obtained by cancelling the 9s.

We shall consider fractions like, 30/50 = 3/5, to be trivial examples.

There are exactly four non-trivial examples of this type of fraction, less than one in value, and containing two digits in the numerator and denominator.

If the product of these four fractions is given in its lowest common terms, find the value of the denominator.

The problem itself is rather easy to solve using brute force since there are less than 90*90 = 8100 possible solutions we would have to check. Each check is fairly easy to perform and thus we would be able to brute force comfortably within the 1 minute range. Continue reading →

Posted by Kristian in Project Euler, 10 comments

Project Euler 28: What is the sum of both diagonals in a 1001 by 1001 spiral?

Problem 28 of Project Euler reads

Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:

21 22 23 24 25
20  7  8  9 10
19  6  1  2 11
18  5  4  3 12
17 16 15 14 13

It can be verified that the sum of the numbers on the diagonals is 101.

What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral formed in the same way?

I am pretty sure it is possible to brute force a solution for this problem, but the thought of having to construct the spiral, so let instead of that let us attack it from a more analytical approach and search for a function of the ring number. This is done in two steps, first we will derive a formula where the calculation for the n’th ring depends on the n-1’th ring. And after that we will derive a formula which is independent on the previous ring. Continue reading →

Posted by Kristian in Project Euler, 35 comments