Hence the sum of the series =‡n3-}n, and the sum of the series = n3-n=the single obeliscal area.

Fig. 3. The sectional areas along the sectional axes, 1, 3, 5, 7, 9, 11, &c., and between the ordinates 0 and 1, 1 and 2, 2 and 3, &c., are

circumscribing parallelogram = axis x ordinate

Parabolic area=216=144.

144-143=1

=36 × 6=216.

Obeliscal area =}n3-n=363-16=143.

Though the actual difference between every two corresponding obeliscal and parabolic sectional areas equals unity; yet the relative difference between two such areas will be greater nearer the apex, and less as the ordinates recede from the apex.

Generally the corresponding areas of the nth section will be as 1. 2n-1° : 1 2n−1 ̄ + 1 n

2

2

2

611.

2661.

When n=6, the areas will be as 601 When n=12, the areas will be as 264 The sum of the two ordinates: = the axis of an obeliscal sectional area. As the successive sectional axes, or distance between the two ordinates, are continually increasing by 2, while the difference between the two ordinates, unity, - remains the same, it follows that the opposite sides of the single obelisk (fig.6.), will continually approach to parallelism, but which they can never attain; for how great soever the sectional axes, or the sum of the two ordinates may be, still their difference will equal unity, so the sides of a sectional obeliscal area can never become parallel to the axis.

Fig. 6.

The two sides of an obeliscal sectional area are always equal, and the two ordinates are always parallel. If the two ordinates were also equal, then the four sides would form a rectangular parallelogram, the opposite sides of which would be parallel to each other, as are the ordinates.

An ordinate equal the mean ordinate of any obeliscal sectional area will always correspond to an axis equal to the distance from the apex to the point of bisection of that sectional axis, less unity, a constant quantity.

For the sectional axis intercepted by the n-1 and th ordinates 2n-1, the half of which = n- = the mean of the two ordinates n-1 and n.

So the whole axis from the apex to the point of bisection of the sectional axis will

1

2

But the axis corresponding to the ordinate n- will = n— 2 =n2-n+, which is less than n2-n+ by. Hence the mean ordinate of the 1st sectional area, which, will be at the distance from the apex-1=1 unity; so that an ordinate drawn at from the apex, and made =

an ordinate to the parabola.

unity, will be

The parabolic area of the 1st section will be to the corresponding obeliscal area::::: 4:3.

parabola exceed

unity.

The ordinates of the parabolic and obeliscal area are equal at the beginning and end of each section, but the intermediate ordinates of the parabola are greater than the corresponding intermediate ordinates of the obeliscal area. This difference of the ordinates makes a sectional area of the the corresponding sectional obeliscal area by If the double ordinates, like the velocity ordinates, were made ordinates of an obeliscal area; then the successive sectional areas would equal 12, 32, 52, 72 (Figs. 3, 4, 5), or equal twice the single obeliscal series of sectional areas of Figs. 1. or 6. Then each parabolic sectional area will exceed the corresponding obeliscal sectional by 3 of 1.

The Construction and Summation of Obeliscal Series.

The sum of the series 1+2+3+4, &c.= } n + 1. n. Fig. 7-2. The number of squares of unity=1+2+3+ 4+5+6

= the area of the triangle + 6

=16×6+16

= {/ n x n + 1 n

= } n + 1. n.

=

For axis=sum of the series 1+2+3+4, &c. = Į n + 1. n; but here the ordinate of unity more than the number of terms, or side of the last square. Or ordinate=n+ {}}

The circumscribing parallelogram will = 1 n+1 × nxn+1/

nate.

= axis

x ordi

Also by construction the sum of the areas limited by the ordinates will equal the sum of the corresponding squares } n + 1. n. n + 1/.

=

For the straight line joining the two ordinates 8 and 71 cuts off a triangle from the square of 8 = the triangle added to the same square; consequently the area contained by this

straight line, the sectional axis 8, and the two ordinates will the square of 8.

These series of areas would form an obeliscal area = the sum of the corresponding squares.

Fig. 7. The ordinate of the series of squares = n + 1/2, the square of which = n+1.n+= twice the axis of the squares +1.

In order to construct a parabolic area, the axis should vary as the square of the ordinate. If n+, the ordinate of the series of squares, be made the ordinate of a parabolic area, the corresponding axis should = 1 (n + 1. n + 1) 1 n + 1. n + 1/ or = ordinate2 of the parabolic area = the axis of the squares +

=

Hence the parabolic area will have an axis greater than the series of squares by unity; or equal

ordinate2 =

The apex of the parabola will be in the produced axis of the squares at the distance of above the first square. The 144 nth ordinate of the squares, which n+, will be common to both areas; but the parabolic area being curvilinear, the ordinate will continually vary as axis from the apex to the nth ordinate, which parabolic area so generated will be to the corresponding series of n squares,

or 1 (n3 + 11 n2 + 1⁄2 n + 1) : } (n3 + 11 n2 + 1 n)

Difference =

3

1 ( 1 n + 1 ) =2n+1

Fig. 7. The difference between the two areas at the 8th ordinate, which are as 204-708 204,