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On the real roots of ζ(x) = a

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Let \mathbb N be the set of natural numbers and \mathbb Q^+ be the set of positive rational numbers. For every positive integer n \ge 2, there exists a unique real number s_n such that \zeta (s_n) = n. The question is does there exists an n \in \mathbb N such that s_n \in \mathbb Q^+ be rational? In other words, does there exist a positive rational \frac{p}{q} such that \zeta(\frac{p}{q}) \in \mathbb N? At present, we don’t even know if s_n is always rational, let alone irrational or transcendental. Investigation seems to suggest that \zeta(\frac{p}{q}) \notin \mathbb N for any \frac{p}{q} \in \mathbb Q^+. However this has not been completely established. In this post, we shall look at some on the results in this line of research.

Expanding the Riemann zeta function about s=1 gives

\displaystyle{\zeta(s) = \frac{1}{s-1} + \gamma_0 + \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\gamma_n (s-1)^n}

where the constants

\displaystyle{\gamma_n = \lim_{m \rightarrow \infty}\Bigg [\sum_{k=1}^{m}\frac{(\ln k)^n}{k} - \frac{(\ln m)^{n+1}}{n+1}\Bigg]}

are known as Stieltjes constants. Inverting the above series we obtain:

Theorem 1Let x > 1 and \zeta(x) = a; then,

\displaystyle{x = 1 + \frac{1}{a-\gamma _0} - \frac{\gamma _1}{1!(a-\gamma _0)^2} + \frac{\gamma _2}{2!(a-\gamma _0)^3}- \frac{\gamma _3 - 12\gamma _1}{3!(a-\gamma _0)^4} + O\Big(\frac{1}{a^5}\Big)}.

Proof. The proof follows by inverting the Stieltjes series expansion of \zeta (x)\Box

We can rewrite the Stieltjes series expansion of the  Riemann zeta function in the following form which we shall use regularly in our analysis:

\displaystyle{\zeta \Big (1+\frac{1}{s}\Big ) = s + \gamma_0 + \sum_{n=0}^{\infty}\frac{(-1)^n\gamma_n}{n!^n}}.

We want to find an interval where the function \zeta(1+\frac{1}{s}) is strictly decreasing. Clearly \zeta(1+\frac{1}{s}) is continuous and strictly decreasing in the interval (0, \infty). The trivial zeros of \zeta(s) occur at s=-2n where n \in N. In other words, the trivial zeros of \zeta(1+\frac{1}{s}) occur at s = \frac{-1}{2n+1} and all these zeros lie in the interval (-1/3, 0). Hence in this interval, \zeta(1+\frac{1}{s}) will have infinitely many points of inflexion. Excluding the interval (-1/3, 0), we can divide the real line in to two parts (-\infty, -1/3) and (0,\infty); and in each of these two intervals, \zeta(1+\frac{1}{s}) in strictly decreasing. For positive rationals \frac{p}{q}, the scope of our current study is the interval  (0,\infty).

Lemma 2For every finite real a>1, there exists a unique real number c_a, 0< c_a < 1-\gamma _0, such that

\displaystyle{\zeta\Big(1+\frac{1}{a-1 +c_a}\Big) = a}.

Proof. Let lemma follows form the fact that in our interval of interest (0,\infty)\zeta (1+\frac{1}{s}) - s is strictly decreasing. \Box

Theorem 3. Let c_a be as defined in Lemma 2. The asymptotic expansion of c_a is

\displaystyle{c_a = 1-\gamma_0 + \frac{\gamma_1}{a-1} - \frac{\gamma_2 + \gamma_1-\gamma_1\gamma_0}{(a-1)^2}}

\displaystyle{+ \frac{\gamma_3 + 2\gamma_2- 2\gamma_2\gamma_0 + \gamma_1 - 2\gamma_1\gamma_0 + \gamma_1\gamma_0^2 - \gamma_1^2}{(a-1)^3}+ O\Big(\frac{1}{a^4}\Big)}.

Proof. This is obtained by inverting the above Stieltjes series expansion of \zeta (1+\frac{1}{a-1+c_a})\Box

The next result gives us a method to compute c_a numerically. We have

Lemma 4. Let k_r be defined recursively as

\displaystyle{k_{r+1} = a + k_r - \zeta \Big (\frac{a+k_r}{a-1+k_r}\Big )}

where k_0 = 1-\gamma_0. Then \lim_{r \rightarrow \infty}k_r = c_a .

Marek Wolf, has sent me the computed values of c_n for 2 \le c_n \le 1000. Some of these numbers were used in the proof of the following theorem.

Proof. Trivial. \Box

Theorem 5. If 1 \le p-q \le 40 then \zeta (\frac{p}{q}) \notin \mathbb N.

Proof. We have c_2 \approx 0.37240621590. There are exactly twenty three positive fractions \frac{m}{n} with n \le 40 such that

\displaystyle{c_2 \le \frac{m}{n} < 1- \gamma_0}.

For each of these fractions, we find (by actual computation) that there exists a integer r \le 43 such that

\displaystyle{c_r \le \frac{m}{n} < c_{r+1}}.

Since c_n is strictly increasing, it implies that if c_n \in \mathbb Q^+ then the denominator of c_n has to be \ge 41. Hence the theorem follows. \Box

If we continue working with the idea given the proof of Theorem 5, we can prove p-q > n_0 for some large natural number n_o. But this method gives no hope of proving that \zeta (\frac{p}{q}) \notin \mathbb N. Therefore we must look for some other ideas.

Theorem 6. Let \zeta (1+\frac{1}{x_i}) \in \mathbb N, x_i \in \mathbb R^+ , i=1, 2, \ldots , k and a_j \in \mathbb N such that

\sum a_j = \{1, 2, 3, 4, 5, 7, 8, 10, 12 , 15\}.

If y=m+\sum a_k x_k where is any non-negative integer; then \zeta (1+\frac{1}{y}) \notin \mathbb N.

Some applications of Theorem 6 are given below.

Example 7. If x \in \mathbb R^+ and 2 \le n \le 5, n \in \mathbb N then at least one of the two numbers

\zeta (1+\frac{1}{x}) and \zeta (1+\frac{1}{nx})

is not a integer.

Example 8. If x, y \in \mathbb R^+ then at least one of the three numbers

\zeta (1+\frac{1}{x}), \zeta (1+\frac{1}{y}) and \zeta (1+\frac{1}{x+y})

is not a integer.

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