Back to Riemann, Part 8

Ok, yes, finally, extending the domain for the zeta function to the complex plane.

Up to this point, the conventional form of the zeta function:
z(x) = 1 + 1/2^x + 1/3^x + 1/4^x…

only converged to some constant if x was a real number greater than 1.
By introducing the eta function, we were able to extend the domain to real numbers greater than or equal to 0, but not including 1, which has a pole.

This is the same as saying that the zeta(x) function works for x = a + bi, where b = 0, and x > 1. And the eta(x) function works for x = a + bi, where b = 0, and 0 <= x < 1.

But, say we use the natural logarithm ln(x), instead of the current exponent form.
We know from part 5 that,
2 = e^ln(2)
and, from the rule of exponents that
2^z = e^(z*ln(2))

If z = a + bi,
2^(a + bi) = (2^a) * (2^bi) = (2^a) * e^(bi * ln(2))

e^bi = cos(b) + i*sin(b)

(2^a) * e^(bi * ln(2)) = (2^a) * (cos(b * ln(2)) + i*sin(b * ln(2)))

The reciprocal of 2^(a+bi) = 1/(2^(a + bi)) = 2^(-a -bi)

So, if we look at the above zeta function, the second term is 1/(2^x),
which we now know is:
(1/(2^a)) * (cos(b * ln(2)) – i*sin(b * ln(2)))

The third term is 1/3^x, so that’s
(1/(3^a)) * (cos(b * ln(3)) – i*sin(b * ln(3)))

And we just keep on doing this for every other term in the zeta function.
This is how we extend the domain for the zeta(x) function to the complex plane for all real x > 1.

There are two important points to note here.
First, the complex plane is a mirror around the x-axis. That is, if x1 = 5 + 3i,
then the mirror image point is x2 = 5 – 3i.

1/2^x1 = 1/2^(5 + 3i) = (1/2^5) * (cos(3 * ln(2)) – sin(3 * ln(2))
1/2^x2 = 1/2^(5 – 3i) = (1/2^5) * (cos(3 * ln(2)) + sin(3 * ln(2))

Which is going to hold for all of the other terms in the series. I.e.,
1/3^x1 = 1/3^(5 + 3i) = (1/3^5) * (cos(3 * ln(3)) – sin(3 * ln(3))
1/3^x2 = 1/3^(5 – 3i) = (1/3^5) * (cos(3 * ln(3)) + sin(3 * ln(3))

Because e^i involves rotation around the origin, the rotations generated by each term for x = 5 + 3i are going to be mirrored on the other side of the x-axis for x = 5 – 3i.
That is, zeta(5 + 3i) and zeta(5 – 3i) will also be mirror images of each other.
In short, we only need to calculate zeta(5 + 3i) and we’ll know what z(5 – 3i) is, too.

Second, the eta function.
The eta function is what let us extend the zeta domain to x >= 0.
It also has a complex plane version, and the conversion formula is
zeta(x) = eta(x) / (1 – 2^(1-x))

And, eta(x) was a modified version of the zeta function that converges faster:
eta(x) = 1 – 1/2^x + 1/3^x – 1/4^x + 1/5^x – 1/6^x

To get the eta function we just need to flip the signs of every even cos()/sin() term in the zeta series, and the conversion formula uses 2^(1-x) instead of 2^x, which is the first term in the zeta series. Going from the zeta(x) series to the eta(x) series and back will be easy.

Technically, I could stop here.
The Riemann Hypothesis states that all of the non-trivial zeroes of the zeta function lie on the critical line, x = 1/2.
And, with the eta(x) function, which is valid from x = 0 to x < 1, whose domain has just been extended to include the entire complex plane in that range, I can now write a usable version of the zeta function in Excel to make animated graphs as I navigate around the complex plane (I think; I haven't proved this, yet).

The thing is, I still haven't covered what the trivial and non-trivial zeroes are, I haven't extended the domain of the zeta function to x < 0, and I haven't talked about why the critical strip only extends from x = 0 to x < 1.

So, I'm going to get into some of that next time.

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