Integral Madness
Posted on map[Count:April 13, 2014]
| 8 minutes
| 1578 words
| Markus Shepherd
We’ve seen the calculus version
\[ J(x)=\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}\log\zeta(s)x^s\frac{\mathrm{d}s}{s}, \]
of the Euler product, and we know how to express \(\xi(s)\) as a product over its roots
\[ \xi(s)=\xi(0)\prod_\varrho\left(1-\frac{s}{\varrho}\right), \]
where
\[ \xi(s) = \frac{1}{2} \pi^{-s/2} s(s-1) \Pi(s/2-1) \zeta(s) \newline = \pi^{-s/2} (s-1) \Pi(s/2) \zeta(s). \]
High time we put everything together – the reward will be the long expected explicit formula for counting primes!First, let’s bring the two formulae for \(\xi(s)\) together and rearrange them such that we obtain a formula for \(\zeta(s)\):
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From Zeta to J and Back (And Yet Again Back)
Posted on map[Count:April 6, 2014]
| 5 minutes
| 924 words
| Markus Shepherd
We know a lot about the \(\zeta\) and \(\xi\)-functions, we’ve learnt all about the different prime counting functions, most notably \(J(x)\), so it’s high time we found a connection between the two. Probably not too surprisingly, the crucial link is our good friend, the Euler product
\[ \zeta(s)=\prod_{p}(1-p^{-s})^{-1}. \]
What we want to develop now is a version of this product that will suit us to find a formula that magically can count primes.
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Infinity Is Worth No More Than -1/12
Posted on map[Count:February 23, 2014]
| 6 minutes
| 1069 words
| Markus Shepherd
On 16 January 1913, a confused manuscript reached the famous mathematician G. H. Hardy in Cambridge. Other researchers have received similar letters before, and rejected it due to the seemingly incoherent formulae mixed with trivial mathematical results. Professional mathematicians are used to receiving manuscripts by amateurs who believe to have solved famous problems, but this particularly odd scribble caught the eye:
\[ 1+2+3+4+5+6+\ldots+\infty=-\frac{1}{12} \]
Did this amateur mathematician really think that the sum of all natural numbers, a value that will exceed any given boundary at some point, will wind up being a negative fraction?
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Counting Primes Functionally
Posted on map[Count:January 12, 2014]
| 5 minutes
| 1005 words
| Markus Shepherd
After all this playing with the \(\zeta\)-function it is time to return to the overall objective of this whole exercise: counting prime numbers. The idea behind analytic number theory is that primes are unpredictable on the small scale, but actually surprising regular on the large scale. This is why we’ll look at certain functions that behave pretty erratically when we look at every single value, but become smooth and “easy” to calculate once we “zoom out” and consider the global properties, the so-called asymptotic.
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More symmetry and Another Product
Posted on map[Count:December 1, 2013]
| 5 minutes
| 901 words
| Markus Shepherd
We’ve seen that \(\zeta(s)\) satisfies the functional equation
\[ \zeta(1-s)=2^{1-s}\pi^{-s}\cos(\pi s/2)\Pi(s-1)\zeta(s). \]
(Well, it still needs to be proved, but let’s just assume it’s correct for now.) The goal of this post is an even more symmetrical form that will yield the function \(\xi(s)\) which we can develop into an incredibly useful product expression.
On our wish list for \(\xi(s)\) we find three items:
It’s an entire function, i.e., a function that’s holomorphic everywhere in \(\mathbb{C}\) without any poles.
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Are Primes Independent?
Posted on map[Count:November 24, 2013]
| 6 minutes
| 1069 words
| Markus Shepherd
The question may sound silly, but I hope it will become apparent that it’s very reasonable to ask. What we will examine here is the probabilistic interpretation of the prime distribution. So, essentially we ask: “What’s the probability that a randomly chosen number is prime?” Those familiar with some basic probability theory know the notion of independency in this context, so the question I’m basically interested in here is if the probability to find a prime is independent of the preceding or following numbers.
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Perfect Symmetry
Posted on map[Count:October 31, 2013]
| 6 minutes
| 1272 words
| Markus Shepherd
So far, we have seen how the Euler product links the \(\zeta\)-function to the prime numbers. More precisely, it encodes the fundamental theorem of arithmetic. One may also say, it’s the analytic version of it, in a sense that should become clearer shortly.
What we have done so far works perfectly for the real numbers. The sum \(\sum n^{-s}\) that defines \(\zeta(s)\) converges for \(s>1\), that’s how Leonhard Euler found his product, and that’s what Peter Gustav Lejeune Dirichlet used to prove the prime number theorem in arithmetic progressions.
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Does the Euler Product Converge?
Posted on map[Count:October 8, 2013]
| 4 minutes
| 697 words
| Markus Shepherd
Usually, I don’t care too much about convergence as a general overview of the argument is what I aim at here, and otherwise I’ll just trust that things “behave well”. But some words concerning convergence won’t harm.
It’s a well known fact that the harmonic series (which we shortly touched in the previous post) \(\sum1/n\) diverges. I think the best (though not easiest) proof of this to compare it with the corresponding integral:
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Euler Product Revisited
Posted on map[Count:October 6, 2013]
| 4 minutes
| 738 words
| Markus Shepherd
From the previous post we know that the harmless looking series \(\sum n^{-s}\) can be extended to the product \(\prod (1-p^{-s})^{-1}\). At first sight, this does not seem terribly helpful, and it actually makes the rather easy series more complicated. So what’s the big deal?
It’s what has actually been suppressed in the above notation: The sequences we run through. The series runs over all natural numbers (or positive integers, if you prefer), the product runs through all prime numbers.
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In the Beginning, There Was... Euler's Formula!
Posted on map[Count:September 29, 2013]
| 1 minutes
| 206 words
| Markus Shepherd
I will start this blog the way Bernhard Riemann started his paper: with Euler’s product formula that John Derbyshire called the golden key:
\[ \zeta(s)=\sum_{n\ge1}n^{-s}=\prod_{p}(1-p^{-s})^{-1} \]
This holds for any complex number \(s\) with \(\Re s > 1\). If you look up a proof in any modern textbook, you will find a number technical rearrangements that end up in an examination of the absolute convergence on both sides. But actually, the formula is nothing but a fancy way of writing out the Sieve of Eratosthenes.
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