Saturday, August 26, 2006

Euler Product Formula

One of the major breakthroughs in Analytic Number Theory came with Leonhard Euler's Product Formula (see Theorem 4 below). Associated with this result is the Riemann zeta function which is the foundation of perhaps the most important unsolved mathematical problem in number theory today, the Riemann Hypothesis.

One might ask if Leonhard Euler came up with the zeta function why is it commonly called the Riemann Zeta Function. The answer has to do with the development of the function. While it is true that Euler discovered it, he used it with integer values. Johann Dirichlet would later extend it to real numbers, and Bernhard Riemann extended it to complex numbers. The most interesting properties stem from the complex numbers so it makes sense that it is currently named the Riemann zeta function (see Ed Sandifer's article for details).

So, how did Euler come up with the Euler Product Formula? The Harmonic Series was shown to be divergent (have an infinite sum) by Nicolas Oresme in the 1300s (this is considered one of the high points of mathematics in the middle ages, see Lemma 3 below). Euler wondered if the prime harmonic series (1 + 1/2 + 1/3 + 1/5 + ...) was also divergent. He then began studying the zeta function ζ(s) where:

ζ(s) = 1 + (1/2s) + (1/3s) + ...

For Euler, s is an integer. It turns out that when s=1, ζ(s) = the harmonic series and is therefore divergent. When s ≠ 1, it turns out the the zeta function is convergent, that is, it has a finite limit. In this way, Euler came up with his Product Formula. He then used this result to show that there are an infinitude of primes and the prime harmonic series is infinite. Unfortunately, while his conclusion about the prime harmonic series is correct, his proof today is not considered valid (see Ed Sandier's article for more information). See here for details on the proof for the divergence of the prime harmonic series.

Lemma 1: Infinite Geometric Series

if x ≠ 1, then:

1/(1-x) = 1 + x + x
2 + x3 + ... + xn

Proof:

(1) Let s = 1 + x + x2 + x3 + ... + xn

(2) xs = x + x2 + x3 + ... + xn

(3) s - xs = 1

(4) s(1 - x) = 1

(5) s = 1/(1-x)

QED

Lemma 2: [1/(2n+1)] + ... + [1/(2n + 1 + 2n - 1)] ≥ (1/2)

Proof:

(1) At each n, we have a set of 2n elements.

For example, when n=0, we have 1 element (1/2)

When n=1, we have 2 elements (1/3, 1/4)

When n=2, we have 4 elements (1/5, 1/6, 1/7, 1/8) and so on.

(2) If we pair up two lists of elements in reverse order:

[1/(2n + 1)], [1/(2n+2)], ..., [1/(2n+1+2n-1)]
[1/(2n+1+2n-1)], [1/(2n+1+2n-2), ..., [1/(2n + 1)]

We can see that we have 2n columns where each pair adds up to [1/(2n + 1) + 1/(2n+1) ]

(3) This means that sum of the series of [1/(2n + 1)] thru [1/(2n+1)] is:

(1/2)(2n)[1/(2n + 1) + 1/(2n+1)] =

= 2n-1(2n+1 + 2n + 1)/[(2n+1)(2n + 1)] =

= (2n+1 + 2n + 1)/[(22)(2n + 1)]

(4) Now (1/2) = [(2)(2n + 1)]/[(22)(2n + 1)]

(5) So, this lemma is proven since (2n + 1) ≥ 2 → (2n+1 + 2n + 1) ≥ (2n+1 + 2).

QED

Lemma 3: Harmonic Series Diverges

Let H = 1 + 1/2 + 1/3 + 1/4 + 1/5 + ... + 1/n

Then H has no limit and proceeds toward infinity

Proof:

(1) It is clear that the following sum S diverges and proceeds toward infinity:

S = 1 + 1/2 + 1/2 + 1/2 + ...

(2) Let H1, H2, etc. describe the elements that make up H. (So, that H1 = 1, H2 = 1/2, etc.)

(3) At each each point 2n+1, it is possible to group the elements into:

1/(2n + 1), 1/(2n+2), ..., 1/(2n+1+2n-1)

So that we have:

At [1/(20 + 1)], we have: 1/2

At [1/(21 + 1)], we have: 1/3, 1/4

At [1/(22 + 1)], we have: 1/5, 1/6, 1/7, 1/8

and so on.

(4) Then using Lemma 2 above, we can be sure that the sum of each of these partitions is greater or equal to (1/2)

(5) This gives us that the sum of H ≥ sum of S.

QED

Theorem 4: Euler Product Formula





Proof:

(1) 1/(1-x) = 1 + x2 + x3 + ... xn [See Lemma 1 above]

(2) We can see that (1-x)-1 = 1 + x2 + x3 + ... xn

(3) Let x = (1/ps) where p is a prime.

(4) Then we have: (1 - 1/ps)-1 = 1 + 1/ps + 1/p2s + ... + 1/pns

(5) And, ∏(p) [(1 - 1/ps)-1] = ∏(p)[1 + 1/ps + 1/p2s + ... + 1/pns]

where ∏(p) is multiplication of the primes p.

(6) ∏(p)[1 + 1/ps + 1/p2s + ... + 1/pns] =

(1 + 1/2s + 1/22s + ... + 1/2ns)(1 + 1/3s + 1/32s + ... + 1/3ns)(1 + 1/5s + 1/52s + ... + 1/5ns)*...*(1 + 1/ps + 1/p2s + ... + 1/pns) =

= (1*1*1*...*1) + ([1/2s]*1*1*...*1) + (1*[1/3s]*1*...*1) + ([1/22s]*1*1*...*1) + (1*1*[1/5s]*...*1) + ... + ([1/2ns]*[1/3ns]*[1/5ns]*...*[1/pns]) =

= 1 + 1/2s + 1/3s + 1/4s + 1/5s + ... + 1/(2ns*3ns*5ns*...*pns) [By the Fundamental Theorem of Arithmetic]

= ∑(1,∞)ns

QED

Corollary 4.1: There are an infinitude of primes

Proof:

(1) Assume that there are a finite number of primes.

(2) Then there exists an integer p such that p = the maximum prime.

(3) Then, ∏(1,p) [1/(1 - 1/p)] is finite number.

(4) But if s=1, then ∏(1,p) [1/(1 - 1/p)] = ∑(1,∞) 1/n. [See Theorem 4 above]

(5) Now, ∑(1,∞) 1/n is the Harmonic Series which diverges (that is, goes to infinity). [See Lemma 3 above]

(6) Therefore, we have a contradiction and we reject our assumption in step #1.

QED

References

4 comments:

Bruno said...

Who is "Reimann"? Is he a relation of Tolkein?

Larry Freeman said...

Here is a link to Reimann's biography.

Ben said...

I think you should add in the infinite geometric series section that it only works if |x|<1 and as n tends towards infinity even though it would be obvious to those who are familiar with summation since I like rigorous proofs.

Sotirakis said...

Looking at the proofs, I believe that the old saying about the GP and the Specialist runs true. The GP knows less and less about more and more until he knows nothing about everything, whereas the specialist knows more and more about less and less until he knows everything about nothing. The proofs are a contradiction to basic mathematical ideology. Should anybody be willing to accept them join the blind leading the blind. Look carefully and wonder!!