The Enigma Of The Universe : Mathematical Computation

On the basis of the above definition of the U.S., we can make its exact mathematical computation as follows:

The first thing to be calculated is the number of seeds which are filled in Xo. This number can be calculated if we know-

1. The dimensions of a single mustard (sarṣapa) seed.

2. The number of such mustard seeds contained in the right circular cylinder of diameter 1,00,00 yojanas and height 1000 yojanas.

3. The number of seeds contained in the conical heap.

 Now, we proceed to calculate these three numbers. We find that sarṣapa is the name of a linear measure which is equal to the diameter of one mustard seed (which is also called sarṣapa) and which has a perfectly spherical shape. According to one tradition found in the Trilokasāra,^[1] we get

8 sarṣapas = 1 yava

8 yavas = 1 aṅgula.

1 aṅgula = 64 sarṣapas

Now, as we have already seen the diameters of the continents of oceans are measured by 'paramāṇāngula'; we get, therefore, from the table of space-points (see supra p. 334-336).

1 yojana = (768000 x 500) utṣedha- aṅgula

1 yojana = (768000 x 500 x 64) sarṣapas

= 24576000000 sarṣapas

radius of the circular base of the cylinder

= 50000 yojanas

= 12288 x 1011 sarṣapas

and height of the cylinder = 1000 yojanas

= 24576 x 109 sarṣapas

volume of 1 seed of sarṣapa = 4/3 π (radius)³

radius of the seed of sarṣapa = ½ sarṣapa;

... volume of one seed of sarṣapa

= 4/3 π (½)³ cubic sarṣapas

= π/6 cubic sarṣapas

Now, volume of right circular cylinder

= π x (radius)² x height................. (1)

Now, we should know the height of the right circular conical heap above the cylinder.^[2] According to the Tilokasāra,^[3] if the seeds of tila, sarṣapa, valla, arhada, caṇā (gram), atasi, kulattha, rajāmaṣa etc. are filled in a right circular cylinder in such a way that a conical heap is formed, then

height of the conical heap = circumfernce of the circular base of the cylinder /11

Volume of the right circular cone

= 1/3 π (radius)² x height

^[4] = [1/3 π (radius)²] x

= 1/3 π (radius)² x

= 2/33 π² (radius)³........................................ (2)

Now, putting the values of radius and height in equation no. 1 above, we get the

Volume of right circular cylinder

= π x (12288 x 10¹¹)² x (24576 x 10⁹) cubic-sarṣapas

Volume of one seed of sarṣapa is π/6 cubic sarṣapas. Therefore, the number of sarṣapa seeds in the cylinder

= 6 x (12288 x 10¹¹)² x (24576 x 10⁹)

= 222,651,104,624,64 x 10³¹

= 2.226 x 1044 approximately................... (3)

By substituting the value of radius in equation no. 2, we get the volume of right circular conical heap

= 12/33 x π x (12288 x 10¹¹)³

= 2.119 x 1045 approximately^[5]................... (4)

The number of sarṣapas in the conical heap

= 2.341 x 1045 approximately.^[6]

Thus, the total number of seeds in the cylinderical container of 10000 yojanas diameter and 1000 yojanas height together with  the conical heap is 2.341 x 1045.

We have seen that

(1) Volume of the cylindrical container is 

= πr²h

where r is the radius of the circular base and h is the height of the cylinder

(2) Volume of the conical heap is

= 2/33 π²r²

(radius of the circular base of the cone is also r).

(3) Volume of a single seed is π/6
Thus, the total number of seeds in the cylindrical container with the conical heap above is

Now we know that h, the height of the cylindrical container, is the same in all containers A, B, C and X_o, X₁, X₂...... etc. but the r, the radius is different in them.

Value of h is

= 24576 x 109 sarṣapa

The anavasthita cylinders are made again and again with changed diameters. Let us denote

the original anavasthita cylinder as X_o

firstly made new anavasthita cylinder as X₁

secondly made new anavasthita cylinder as X₂

and so on.

Let the

radius of Xo be denoted as r_o

radius of X1 be denoted as r₁

radius of X2 be denoted as r2

and so on.

Radius of A, B and C is also r_o, which is

= 12288 x 10¹¹ sarṣapas.

Let k_o, k₁, k₂...... denote the total number of seeds in X_o, X₁,X₂...... respectively.

Thus,

k_o = 6r_o²h + 4/11 πr_o³ .................. (0)

= 2.341 x 10⁴⁵ approximately.

Now, by dropping one seed each from X_o into the rings of continents and oceans, the last seed will be dropped in the k_o ^th ring with radius

r₁, = 2^(ko-1)r_o = 2^(ko-2) x 2r_o

Now, as 2r_o = 1 lakh yojanas

r₁ = 2^(ko-2) lakh yojanas

When X_o becomes emptied, one seed is dropped in the cylinder A.

Total number of seeds contained in X₁ is

X₁ = 6r₁²h + 4/11 πr₁3.................. (1)

^[7] = 6 [2^2(ko-1)r_o²]h + 4/11 π[2^3(ko-1)r_o³]

Now, by dropping one seed each in the rings ahead the last seed will be dropped in

(k_o+k₁)^th ring.

The radius of this ring will be

r₂ = 2^(ko+k1-2) lakh yojanas

= 2^(ko+k1-1)r_o

In this way, on emptying the X₁, second seed will be dropped in the cylinder A.

Total number of seeds in X2 will be

k² = 6r₂ ² h + 4/11 πr₂³.................. (2)

In the above manner, the last seed of X₂ will be dropped in (k_o+k₁+k₂)^th ring.

Now, third seed will be dropped in the cylinder A and X₃ will be made with radius r₃.

Total number of seeds in X₃ will be

k₃ = 6r₃²h + 4/11 πr₃ ³.................. (3)

This process will continue till A is filled completely with a conical heap above. It means that when ko seeds will be dropped in A, it will be so filled. This will happen when the cylinder

X_ko-1 will be emptied. Its last seed will be dropped in the

Now the first seed will be dropped in the cylinder B, and a new anavasthita cylinder X_ko will be made. Its radius will be

Now, again to fill the cylinder A which is already emptied, the whole process as above will be repeated. It means that when the cylinder X_2ko-1 will be emptied, cylinder A will be filled second time and then the second seed will be dropped in the cylinder B. Thus, by repeating the above process ko times, the cylinder B will be completely filled. This will happen when the cylinder X_2k0-1 will be emptied. Its last seed will be dropped in the

Now the first seed will be dropped in the cylinder C, and a new anavasthita cylinder will be made. Its radius will be and the total number of seeds contained in it will be The values of can be found from the original equations.

Now, by repeating the process again ko times, the cylinder C will be completely filled. This will happen when the cylinder will be emptied. Its last seed will be dropped in the

Now all the three cylinders A, B and C are completely filled. A new anavasthita cylinder will be made. The radius will be given by

The number Utkṛṣṭa Saṃkhyāta (U. S.)

Now, we can see that although the value of kko 3 is definitely defined by the equation (o) to K₀², yet it will not be possible to compute it mathematically in terms of digits. However, by substituting approximate values of k_o etc. in the above equations, we can get a glimpse of the gigentic value of U. S.

By simplifying the equations (1) to (k_{ko 2}) as shown in foot note no. 1 on p. 362, and by neglecting the second term therein, we get

And

Now, by putting the values of k₁, k₂ in these equations, we get,

By putting k′ in place of k_o-1 of and expanding the right side of equation we get

Now, the value of the last term of the right side is the highest and values of other terms are not less than one; hence, by neglecting other terms, we get,

By substituting

Now by putting m in place of 2ⁿ in the above relation we get,

Now, the last term in the above relation will have the highest value. It will be

Thus, by neglecting other terms, we get

here m = 2ⁿ and

We have already found out that k_o>10⁴⁵. By using logarithm,^[8] we get,

Thus the value of is greater than

It means that the value of is greater than the number obtained by raising

to itself 10¹³³ times.^[9]