a , z k ( vT x)3 , z k + i (x : \ )3 + (y k + i" y k )x 



f (x) = 6d + od + oT 



k k k 



( Vi-\ )xd k , VWJktiJk' (z kVi-Vi x k )d k ... 



6 + T k 6 •<*> 



In order to solve equation (6), the second derivatives (\r\ + -\) 

 must be detemnined at the end points of the interval . This is accomplished 

 by requiring that the first derivative, equation (2), be continuous at each 

 given data point, i.e., 



lim f'(x) = f'fy). 



This means that when equation (2) is written for the intervals 

 (x, ,/X, ) and (>c ,x ), the two values obtained for f (x, ) must agree. 

 Thus, from equation (2), we have for the interval (x _,/X, ), 



' , \ (x k - \-i >2 t ^-"k-i 1 (z k'Vi )d k-i m 



f ^>=^^ ^ 1 < 7 > 



and for the interval 



^VW' 



■ , -\ { \ + i~\ )2 , (y k + i- y k } ( vr z k )d k m 



f(x k )= - 2d + d " 6 {8) 



k k 



By equating (7) and (8), the formula for determining the values of the unknown 

 second derivatives becomes 



Z k-1 d k-1 . Z k (d k-1 +d k } . Vl d k K+}' Y k ) (y k" y k-1 } m 



6 3 6 d. d. . * [y) 



k k-l 



