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All content in this area was uploaded by Georgi Guninski on Jan 29, 2023

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Content uploaded by Georgi Guninski

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All content in this area was uploaded by Georgi Guninski on Jan 29, 2023

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Note: simple real function with zeros greater than

one the primes

Georgi Guninski∗Stefcho Guninski

February 6, 2023

Initial revision: Jan 28 2023 Current version 1.3

Abstract

We found simple real function with zeros greater than one the primes:

j1(x) = (sin(πx))2+ (sin(πΓ(x)+1

x))2. For x > 0, j1(x) is continuous. The

proof is based on Wilson’s theorem.

1 Functions with zeros at primes

The main idea of this note was posted by the ﬁrst author (AKA joro) on math-

overﬂow on Oct 18, 2022 [1].

For natural n > 1 by Wilson’s theorem [2] nis prime iﬀ n|((n−1)! + 1).

For natural nwe have Γ(n)=(n−1)!.

Deﬁne j0(x) = Γ(x)+1

x. Natural x > 1 is prime iﬀ j0(x) is integer. The

solutions to sin(y) = 0 are y=πy0for integer y0.

Deﬁne j1(x) = (sin(πx))2+ (sin(πj0(x)))2= (sin(πx))2+ (sin(πΓ(x)+1

x))2

Theorem 1 (Lemma1).The zeros of j1(x)greater than one are the primes.

Proof. j1(x) is the sum of two real squares and it vanishes when both of them

vanish. (sin(πx))2vanishes at integers. The second square (sin(πΓ(x)+1

x))2

vanishes at primes when xis natural by Wilson’s theorem.

In other words this is Wilson’s primality test over the reals.

2 Properties of j1(x)

For x > 0, j1(x) is continuous.

For integer n > 1, j1(n) = 0 if nis prime and j1(n) = sin(π

n)2if nis

composite. At integers, sagemath computes exact closed form.

Computations with sagemath [3] computes the derivative:

j0

1(x)=2πcos (πx) sin (πx) +

2πΓ(x)ψ(x)

x−πΓ(x)

x2−π

x2cos πΓ(x)

x+π

xsin πΓ(x)

x+π

x

∗email :gguninski@gmail.com,email :guninski@guninski.com

1

Figure 1: j1(x)

3 Generalizations

Sums of real squares have great expressive powers, since the sum is zero iﬀ all

the squares are zero.

To get a function with zeros at the twin primes, deﬁne

jtwin = (sin(πx))2+ (sin(πj0(x)))2+ (sin(πj0(x+ 2)))2

Let g(n) be polynomial with integer coeﬃcients. To get a function with zeros

primes of the form g(n) (it is an open problem if there are inﬁnitely many such

primes) deﬁne jg(x) = (sin(πx))2+ (sin(πj0(g(x))))2

4j1(z)as a complex function

j1(z) may be extended as a complex function.

In this case it acquires additional complex zeros like near 1.2825900... +

6.414635304...i

The complex zeroes ρof j1(x) satisfy sin(πρ) = ±i·sin(π(Γ(ρ)+1)

ρ)

Warning about numerical instability.

When working symbolically at the integers, sagemath correctly compute the

zero at the prime 41.

If we work numerically in mpmath with precision 40 decimal digits, we get

the large error 0.537256282697..., Increasing the mpmath precision to 80 decimal

digits give small error at 41.

Another sort of numerical instability is many oscillations in short interval.

2

Figure 2: Here is X-Ray, <(j1(z)) = 0 is plotted red and =(j1(z)) = 0 is plotted

blue. The intersections of the curves are zeros.

Figure 3: Oscilations near 8

3