A CERTAIN SUMMATION ON A SPECIFIED SEQUENCE OF REAL NUMBERS AND APPLICATION TO THE TWIN PRIME CONJECTURE T. AGAMA
Abstract. In this short note we use various geometric figures to prove a certain summation on a specified set of sequences of real numbers. This demonstrate in itself the power and the relevance of geometric figures in producing remarkable identities that may be usefull ouside the realms of geometry. In particular we show that for any function f : N −→ R, then there exist some constant C := C(x, l0 ) > 0 such that X X X 1 f (n)f (n + l0 ) ≥ f (n) f (m), C(x, l ) 0 2≤n≤x m≤n−1 n≤x for some fixed 1 ≤ l0 ≤ x.
1. Introduction and motivation There is no easy way to controll sums of the form X f (n)f (n + h) n≤x
for any f : N −→ C and where h is fixed. Infact proving the twin prime conjecture reduces to establishing a non-trivial lower bound for the sum X Λ(n)Λ(n + 2). n≤x
In this short note we provide a tool for handling these kinds of sums on the average. 2. The area method and {hj }nj=1 be any sequence of real numbers, and let n n P P r and h be any real numbers satisfying rj = r and hj = h, and Theorem 2.1. Let
{rj }nj=1
j=1
j=1
(r2 + h2 )1/2 =
n X
(rj2 + h2j )1/2 ,
j=1
then n X j=2
rj hj =
n X j=2
hj
X j i=1
ri +
j−1 n−1 X X n−j X ri − 2 rj hj+k . i=1
j=1
k=1
Date: August 12, 2018. 2000 Mathematics Subject Classification. Primary 54C40, 14E20; Secondary 46E25, 20C20. Key words and phrases. partition. 1
2
T. AGAMA
Proof. Consider a right angled triangle, say < ABC in a plane, with height h and base r. Next, let us partition the height of the triangle into n parts, not neccessarily equal. Now, we link those partitions along the height to the hypothenus, with the aid of a parallel line. At the point of contact of each line to the hypothenus, we drop down a vertical line to the next line connecting the last point of the previous partition, thereby forming another right-angled triangle, say < A1 B1 C1 with base and height r1 and h2 respectively. We remark that this triangle is covered by the triangle < ABC, with hypothenus constituting a proportion of the hypothenus of triangle < ABC. We continue this process until we obtain n right-angled triangles < Aj Bj Cj , each with base and height rj and hj for j = 1, 2, . . . n. This construction satisfies n n X X h= hj and r = rj j=1
j=1
and 2
2 1/2
(r + h )
n X = (rj2 + h2j )1/2 . j=1
Now, let us deform the original triangle < ABC by removing the smaller triangles < Aj Bj Cj for j = 1, 2, . . . n. Essentially we are left with rectangles and squares piled on each other with each end poking out a bit further than the one just above, and we observe that the total area of this portrait is given by the relation A1 = r1 h2 + (r1 + r2 )h3 + · · · (r1 + r2 + · · · + rn−2 )hn−1 + (r1 + r2 + · · · + rn−1 )hn = r1 (h2 + h3 + · · · hn ) + r2 (h3 + h4 + · · · + hn ) + · · · + rn−2 (hn−1 + hn ) + rn−1 hn =
n−1 X j=1
rj
n−j X
hj+k .
k=1
On the other hand, we observe that the area of this portrait is the same as the difference of the area of triangle < ABC and the sum of the areas of triangles < Aj Bj Cj for j = 1, 2, . . . , n. That is n
A1 =
1X 1 rj hj . rh − 2 2 j=1
This completes the first part of the argument. For the second part, along the hypothenus, let us construct small pieces of triangle, each of base and height (ri , hi ) (i = 1, 2 . . . , n) so that the trapezoid and the one triangle formed by partitioning becomes rectangles and squares. We observe also that this construction satisfies the relation n X (r2 + h2 )1/2 = (ri2 + h2i )1/2 , i=1
Now, we compute the area of the triangle in two different ways. By direct strategy, we have that the area of the triangle, denoted A, is given by X X n n A = 1/2 ri hi . i=1
i=1
A CERTAIN SUMMATION ON A SPECIFIED SEQUENCE OF REAL NUMBERS AND APPLICATION TO THE TWIN PRIME CO
On the other hand, we compute the area of the triangle by computing the area of each trapezium and the one remaining triangle and sum them together. That is, X X n n n n X X A = hn /2 ri + ri + hn−1 /2 ri + ri + · · · + 1/2r1 h1 . i=1
i=2
i=2
i=3
By comparing the area of the second argument, and linking this to the first argument, the result follows immediately.
3. Application to correlated sums Theorem 3.1. Let f : N −→ R, and suppose X f (n)f (n + l0 ) > 0 n≤x
for some fixed 1 ≤ l0 ≤ x, then there exist some constant C := C(x, l0 ) > 0 such that X X X 1 f (n) f (m). f (n)f (n + l0 ) ≥ C(x, l0 ) n 0 such that for 1 ≤ l0 ≤ x, we have XX X f (n)f (n + j) ≤ C(x, l0 ) f (n)f (n + l0 ). j≤x n≤x
n≤x
By inverting these relation, we find that the claimed lower bound is established.
f 2 (n)
4
T. AGAMA
Remark 3.2. Theorem 3.1 is close to proving the twin prime conjecture. The only set-back is the lack of understanding of the nature of the constant C(x, l0 ). It suffices to ensure that the constant grows linearly in x. The next result highlights this notion. Corollary 1. There exist some constant C(2, x) > 0 such that X x2 Λ(n)Λ(n + 2) ≥ (1 + o(1)) . 2C(2, x) n≤x
Proof. The result follows by using variants of the prime number theorem [1] and leveraging the result of Theorem 3.1. 4. Final remarks The lower bound established in Theorem 3.1 proves the twin prime conjecture provided the implicit constant C(2, x) x. Thus the next phase of research is to investigate further this claim. 1. References 1. Montgomery, H.L, and Vaughan, R.C, Multiplicative number theory 1 :Classical theory. vol.97, Cambridge university press, 2006. Department of Mathematics, African Institute for Mathematical science, Ghana E-mail address:
[email protected]/
[email protected]
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