Before expanding the zeta function to the complex plane, I want to talk about Euler a little more.
Recall the zeta function:
z(x) = 1 + 1/2^x + 1/3^x + 1/4^x…
Euler assumed that x would be small, for -1 < x < 1.
He then defined a function M(x), where there are infinite terms,
M(x) = 1 + x + x^2 + x^3 + x^4…
Then, he multiplied both sides by x,
x*m(x) = x + x^2 + x^3 + x^4 + x^5…
And subtracted both sides, M(x) – x*M(x):
M(x) – x*M(x) = 1 + x – x + x^2 – x^2 + x^3 – x^3 + x^4 – x^4…
M(x) – x*M(x) = 1
M(x)(1 – x) = 1
M(x) = 1/(1 – x)
Ex. M(0.5) = 1/(1 – 0.5) = 2
He followed this up by applying the same approach for x > 1 for the zeta function.
z(x) = 1 + 1/2^x + 1/3^x + 1/4^x…
1/2^x * z(x) = 1/2^x + 1/4^x + 1/6^x + 1/8^x…
1/2^x * z(x) – z(x) = -1 +1/2^x – 1/2^x – 1/^3^x + 1/4^x – 1/4^x – 1/5^x…
z(x)(1 – 1/2^x) = 1 + 1/3^x + 1/5^x + 1/7^x…
This removes all multiples of 2 from the right hand side. Repeating for 1/3^x:
z(x)(1 – 1/2^x)(1 – 1/3^x) = 1 + 1/5^x + 1/7^x + 1/11^x…
Removes all multiples of 2 and 3, etc.
Solving for z(x),
z(x) = (1/(1-1/2^x))*(1/(1-1/3^x))*(1/(1-1/5^x))*(1/(1-1/7^x))…
This is called Euler’s product formula, and it just contains terms with prime numbers in the denominators. And what this does is directly tie the zeta function to the prime numbers. Unfortunately, we’re still faced with the problem of not knowing the distribution of those prime numbers, or what the next one in a very long sequence will be. But, at least now we have a clear connection to the zeta function and the primes.
This probably a good place to talk about zeroes.
For certain polynomials, you can determine their function if you know where they cross the x-axis, and their value when x=0.
Example from the der Veen and de Craats book:
Say you have p(x), with p(2) = p(3) = 0, and p(0) = 1
This is a parabola pointing up, and can be written as:
p(x) = (1 – x/2)*(1 – x/3)
p(x) = 1 – (x*5)/6 + (x^2)/6
Expanding this function for zeroes at all integers 1 through 10,
p(x) = (1 – x/1)(1 – x/2)(1 – x/3)…(1 – x/10)
A simple question now: What’s the coefficient for x?
Answer, this is the component where we just have the a*x term, which is:
x/1 + x/2 + x/3… + x/10
= x*(1 + 1/2 + 1/3 + … + 1/10)
So, Euler assumed that S(x) = sin(pi*x)/pi*x
was a polynomial of infinite degree, with zeroes at …-4, -3, -2, -1, 1, 2, 3, 4…
Basically, all real integers excluding 0. And S(0) = 1.
S(x) = (1 – x/1)(1 – x/(-1))(1 – x/2)(1 – x(-2))(1 – x/3)(1 – x/(-3))…
Because (1 – x/3)(1 – x/(-3)) = (1 – ((x^2)/(3^2)) (1 minus x-squared divided by 3-squared),
S(x) = sin(pi*x)/pi*x = (1 – (x^2)/(1^2))(1 – (x^2)/(2^2))(1 – (x^2)/(3^2))…
If we use the geometric expansion for sin(x),
sin(x) = x – x^3/3! + x^5/5! – x^7/7!…
And substitute pi*x for x,
sin(pi*x) = pi*x – (pi^3)*(x^3)/(3!) + (pi^5)*(x^5)/(5!) – (pi^7)*(x^7)/(7!)…
And divide both sides by pi*x,
sin(pi*x)/(pi*x) = 1 – (pi^2)*(x^2)/(3!) + (pi^4)*(x^4)/(5!) – (pi^6)*(x^6)/7!…
Euler found that the coefficient for x^2 is -(pi^2)/(3!) = -(pi^2)/6.
And, taking the above polynomial for finding the zeroes of S(x), the same coefficient for x^2 is
-x^2(1/1^2 + 1/2^2 + 1/3^2 + 1/4^2)
-1/1^2 – 1/2^2 – 1/3^2 – 1/4^2…
And this is the zeta function for minus zeta(2). That is, the above zeta function, negative, with x=2.
From this Euler concluded that zeta(2) = pi^2/6
der Veen and de Crats say that this isn’t a really rigorous proof, but the end result is correct.
But, by using the same method above, Euler calculated zeta(x) for all even positive integers.
The odd positive integers are much harder, and they don’t work out to be as pretty as for the first few even integers. A little more information on them can be found in the wiki article.
The even numbers can also be found using the Bernoulli formula, where B2n is the 2nth Bernoulli number. (Note that because the odd-numbered Bernoulli values, excluding the first 1, equal zero, the above formula doesn’t work for the odd-numbered Zeta integers.)
Anyway. The above approaches let us get the zeta(x) values for the positive integer values for x, which are all non-zero.