INTENSITY OP SUN'S HEAT AND LIGHT. 23 



Subtracting the last of these three equations from the preceding, and that result 

 from the first, and cancelling, 



2m= fudx-{-bu — fu'dx — i u' 



+ fu" d x + i u" 



It is now necessary to determine u' in terms of u, or of x. Recurring to the 

 last segment of the curve above referred to, it is evident that its area above the 

 trapezoid, and denoted by u', is equal to 



v! = CF (X) dx— fF (x _ 1} dx — l (F {x) + jP ( .,_d). 



Developing by Taylor's theorem ; since u = F {x) , 



du d 2 u d 3 u 



■Fe-i) = F^ — j^-\- iJJtf — x.2.3 dx 3 + •••• 



r r du d 2 u 



J F (x _ 1) dx = j F (x) dx — u + YMx~ 1 .2 .3da? + "•■ 



Substituting the two right-hand values in the former equation, the first terms 

 will cancel each other, leaving 



d 2 u d 3 u d i u 



u i2dx* + ^dx 3 'zodx^--'- 



That is, each derived function is equal to — xV^ n °^ the second differential 

 coefficient of the preceding, -\- ^j-th of the third, &c. 



d i u 

 m " = t^JF 4 — •••• 



d' 2 u d 3 u d i u 



du d 2 u d 3 u 



-j-u'dx+fu"dx—.... = J \j^-^ I ^^ J h^----- 

 Substituting these last two values in the equation above, 



du d 3 u 



as was to be demonstrated. Let it now be applied to different examples of series, 

 whose ,a?th term is a function of x. 



I. To find the sum of the arithmetical progression, 



d + 2d + 3d + ...-\-xd = 2,u, 



r o du 



Here u = xd; iudx=hx~d;-j — = d. 



'J dx 



Whence 2 u — i x 2 d + i x d + T \ d + C. 

 If x=l,d — id + id + ^d-\-C. 

 Subtracting, 2 u = i x (x d -\- d) ; which result coincides with the common 

 arithmetical rule. 



II. To find the sum of the geometrical progression, 



ar + ar 2 + ar 3 + .... + af. 



