208 Trans. Acad. Sci. of St. Louis. 



The five horizontal strips when thus extended, may be con- 

 sidered as one strip, having an area y* (y 5 — x 5 ), and the 

 area of the vertical strips is x 5 (y 5 — x*). The sum of the 

 two resulting strips is 



y 10 — x 10 = y 6 (y* — x*) + x 5 (y 5 —x s ). 



Equation (3) is readily reduced to this form. It is en- 

 tirely similar to the well-known equation 



y 2 — x 2 = y (y — x) + x (y — x). 



As was suggested in the former paper, the two squares 

 y 2 and x 2 may be changed by applying the multiplying 

 factor to the horizontal sides of the two squares. The 

 areas remaining as in former equations, and adopting the 

 same unit, and the same values for y and x, the area y 10 

 would then be a rectangle, the vertical width of which 

 would be «/ = 5 inches, and the horizontal side would 

 have a length y 9 = 1,593,125 inches or 30.8 miles. The 

 rectangle x 10 would have a vertical dimension x = V* inch, 

 and the horizontal dimension would be x 9 = 0.000,003,81 

 inch. 



In order to adjust Eq. (5) to this change, it must be 

 written, 



y" —x" = if(y — x )+x (y 9 — y*x) 



+ x (y 8 x — y 7 x 2 ) + x (y 7 x 2 — y e x s ) 

 + x (y*x 3 — y 5 x*) + x (y*x 4 — y*x 5 ) 

 + x (y 4 x 5 — 2/V) + x (y 3 x 6 — y 2 x 7 ) 

 + x (y 2 x 7 — yx 8 ) + x (y 8 — x 9 ) 



The five horizontal strip areas of the diagram are now 

 replaced by one, the width of which is y — # — 4.75 

 inches and the length y 9 — 1,593,125 inches or 30.8 miles. 

 The remaining terms are represented by a series of rec- 

 tangles having a common height a? == Va inch, and filling 

 the space x (y 9 — x 9 )=y 4z (1,593,125 — 0.000,003,81) 

 square inches. 



