Reciprocol Squares and the Value of (pi^2)/6

There is an element of composition involved in math proofs. For the square and circle with the same area, an algebraic approach can be used. For instance, in the formula for area of a circle, 

pi*r^2,

 and the square, 

s^2,

 can be set to be equal to each other: 

pi*r^2=s^2. 

After some manipulation, the value of one side of the square is as follows:

s=r(pi)^(1/2)

If the area of the circle is pi^2, then the area of the square is also pi^2. This is due to the fact that the radius of the circle is pi^(1/2); which makes the area of the circle,

pi*(pi^(1/2))^2,

therefore,

pi*pi,

or, 

pi^2.

Similarly, the square has the following operation,

s=r(pi)^(1/2)

r=pi^(1/2)

therefore,

s^2=((pi^(1/2))*(pi^(1/2)))^2.

 The following is a geometric representation of the the previous algebraic operations:

This construction allows for a base to derive values of different orchestrations related to pi. For instance, the Zeta Function at s=2 is a divergent series with the value at (pi^2)/6. 

1. Draw a square with the area of pi^2, with the center of the square at (0,0).

Named Square A

Derived points,

(pi/2, pi/2), (-pi/2, pi/2), (pi/2, -pi/2), (-pi/2, -pi/2)

 Named points: (a, b, c, d)

2. Draw a circle with the area of pi^2, with an origin at (0,0).

Named Circle A

Derived points,

(pi^(1/2), 0), (-pi^(1/2), 0), (0, pi^(1/2)), (0, -pi^(1/2))

 Named points: (e, f, g, h)

3. Draw 4 circles with their origin as the intersect points of the first circle and the x and y axis.

Named Circles (B, C, D, E); The coordinates for the origin points for each circle can be expressed as the derived points: (e, f, g, h).

The construction of the 4 circles create intersect points between the Circle A and the 4 circles: (B, C, D, E). 

Derived Points of Intersection, named intersect points: intp (a, b, c, d, e, f, g, h, i, j, k, l).

4. Draw 6 triangles that share a corner at the point (0, 0) and the intersect with selected intersect points (i.e. intp a and c and the point (0, 0) create one triangle, then intp c and d and the point (0,0) create the next adjacent triangle). 

Named Triangles: (A, B, C, D, E, F)

Derived Points: the points of the triangle are as follows: 

For all Triangles, the one corner of each triangle is the point (0, 0). 

Triangle A Coordinates: corners (pi^(1/2), 0) and ((pi^(1/2))/2, (pi-pi/4)^(1/2)).

Triangle B Coordinates: corners ((pi^(1/2))/2, (pi-pi/4)^(1/2)) and (-(pi^(1/2))/2, (pi-pi/4)^(1/2))

Triangle C Coordinates: corners (-pi^(1/2), 0) and (-(pi^(1/2))/2, (pi-pi/4)^(1/2))

Triangle D Coordinates: corners (-pi^(1/2), 0) and (-(pi^(1/2))/2, (pi-pi/4)^(1/2)).

Triangle E Coordinates: corners ((pi^(1/2))/2, -(pi-pi/4)^(1/2)) and (-(pi^(1/2))/2, -(pi-pi/4)^(1/2)).

Triangle F Coordinates: corners ((pi^(1/2))/2, -(pi-pi/4)^(1/2)) and  (pi^(1/2), 0) 

5. Draw a Circular Sector with an origin of (0, 0) to the points (pi^(1/2), 0) and ((pi^(1/2))/2, (pi-pi/4)^(1/2)). The yellow shaded region is equivalent to the value (pi^2)/6.

Named Circular Sector A

The interactive proof can be found at the bottom of the page.

How does this proof relate to the Zeta Function?

This proof doesn't prove or disprove the Zeta Function at s=2; this proof mearly shows a geometric representation of the value of the convergent series. However, an algebraic approach can be used to understand the Zeta Function at s=2. 

If the value of each entry in the Zeta Function could be represented as squares, then one could map the entire divergent series on a coordinate plane. For this visual represention, only 5 entries will be used. 

Note: The area of each square is equivalent to the value of each entry for the Zeta Function at s=2. 

This visualization also allows for the mapping of linear equations to the squares like f(x)=x. This causes the opposing corners of the squares ((0, 0) and (1, 1) for the square which has an area of 1) to have two points of intersection for all mapped inputs of the function. 

The interactive proof is found at the bottom of the page. 

This visualization can be done for all values of s when s>1. 

Proof 1

Proof 2