other zeta functions

The Master Recipe: From an Impulse Train to a Zeta Function

Coming from the study of $\frac{d}{d t} ψ (e^{t})$ in Riemann hypothesis.

Suppose we invent a mathematical universe by creating a "musical score" in time. This precursor signal, $f (t)$ , is an impulse train (Dirac deltas) where we freely choose the time instances at which they occur ( $t_{n}$ ) and the strength or amplitude with which they strike ( $a_{n}$ ).

f (t) = \sum_{n} a_{n} \cdot δ (t - t_{n})

The architecture of number theory tells us that the Laplace Transform of this signal is not the Zeta Function directly, but rather acts as its negative logarithmic derivative. The process for discovering our hidden Zeta, $Z (s)$ , consists of three steps:

1. Laplace Transform:

We apply the transform $L {f (t)} = \int_{0}^{\infty} f (t) e^{- s t} d t$ . Since the Dirac delta "sifts" out everything except the instant $t_{n}$ , the integral evaluates automatically, converting time into an exponential:

L {f (t)} = \sum_{n} a_{n} \cdot e^{- t_{n} s}

2. The Fundamental Equation:

We set this result equal to the logarithmic derivative of our unknown Zeta:

- \frac{Z^{'} (s)}{Z (s)} = \sum_{n} a_{n} \cdot e^{- t_{n} s}

3. Integrate and Exponentiate:

To solve for $Z (s)$ , we integrate both sides from $s$ to infinity. (The integral of $- Z^{'} / Z$ is $- \ln (Z (s))$ ).

Integrating the exponential on the right with respect to $s$ yields a divisor $- t_{n}$ , which cancels the negative signs:

\ln (Z (s)) = \sum_{n} \frac{a_{n}}{t_{n}} e^{- t_{n} s}

We apply the exponential function to eliminate the logarithm and obtain the Universal General Formula for any Zeta Function:

Z_{general} (s) = \exp (\sum_{n} \frac{a_{n}}{t_{n}} e^{- t_{n} s})

Any physical or mathematical system (fractals, quantum billiards, graphs) that produces echoes or bounces in time can be plugged into this formula to reveal its associated Zeta function.

Specialization 1: The Prime Train (The Riemann Zeta Function)

Let us apply the master recipe using the score from the video. The author takes the prime numbers ( $p$ ) and their "harmonics" or powers ( $p^{k}$ ) as the base rhythms.

The timing of each strike is the logarithm of the number: $t_{n} = k \ln (p)$ .
The amplitude of each strike is the logarithm of the base prime: $a_{n} = \ln (p)$ .

Our precursor signal is:

f_{primes} (t) = \sum_{p} \sum_{k = 1}^{\infty} \ln (p) \cdot δ (t - k \ln p)

We plug this into Step 3 of our recipe (the integral). Substituting $a_{n}$ and $t_{n}$ , we see that the multiplying $\ln (p)$ and the dividing $\ln (p)$ cancel out beautifully:

\ln (Z (s)) = \sum_{p} \sum_{k = 1}^{\infty} \frac{\ln (p)}{k \ln (p)} e^{- s k \ln p}

\ln (Z (s)) = \sum_{p} \sum_{k = 1}^{\infty} \frac{1}{k} (e^{- s \ln p})^{k}

Since $e^{- s \ln p}$ is identical to $p^{- s}$ , the equation becomes:

\ln (Z (s)) = \sum_{p} \sum_{k = 1}^{\infty} \frac{(p^{- s})^{k}}{k}

Here comes the algebraic miracle! That inner infinite summation, $\sum \frac{x^{k}}{k}$ , is precisely the Taylor expansion or Maclaurin series of the function $- \ln (1 - x)$ . Substituting this in:

\ln (Z (s)) = \sum_{p} - \ln (1 - p^{- s})

\ln (Z (s)) = \ln (\prod_{p} \frac{1}{1 - p^{- s}})

Taking the exponential of both sides gives us... voilà:

Z (s) = \prod_{p primes} \frac{1}{1 - p^{- s}} = ζ (s)

We have obtained the legendary Euler Product, which is the canonical definition of the Riemann Zeta Function.

Specialization 2: Adding Phases (Dirichlet L-Functions)

Now, how do we construct an L-Function? We subtly modify the prime impulse train. We keep the exact same timings (the rhythm does not change), but multiply the amplitude of each strike by a Dirichlet character $χ (p)$ (which can be $+ 1, - 1, 0$ , or a complex phase).

By the property of multiplicativity, the $k$ -th power of the prime will have its amplitude multiplied by $χ (p)^{k}$ .

Our new "colored" signal is:

f_{L} (t) = \sum_{p} \sum_{k = 1}^{\infty} χ (p)^{k} \ln (p) \cdot δ (t - k \ln p)

We substitute this back into our master integral. Once again, the $\ln (p)$ terms cancel out, and we group the phase $χ (p)$ with the term $p^{- s}$ :

\ln (L (s)) = \sum_{p} \sum_{k = 1}^{\infty} \frac{(χ (p) p^{- s})^{k}}{k}

We apply the exact same Taylor series for the logarithm, $\sum \frac{x^{k}}{k} = - \ln (1 - x)$ , but now with $x = χ (p) p^{- s}$ :

\ln (L (s)) = \sum_{p} - \ln (1 - χ (p) p^{- s})

\ln (L (s)) = \ln (\prod_{p} \frac{1}{1 - χ (p) p^{- s}})

Exponentiating both sides reveals the complete function:

L (s, χ) = \prod_{p primes} \frac{1}{1 - χ (p) p^{- s}}

We have constructed the Dirichlet L-Function.

Dictionary Summary

This derivation demonstrates that the time domain and the complex domain are exact reflections of each other:

Placing pure deltas at the logarithms of the primes yields the Riemann Zeta Function.
Alternating the sign or "coloring" those deltas with a periodic pattern yields L-Functions.
The positions of the "poles" of the initial Laplace Transform correspond precisely to the critical zeros that Riemann hypothesized always have real part $1 / 2$ .