AND ON DEFINITE INTEGRATION. A471 
1] 
facilitate the elimination of the functions F (.), 
First, f(a) =f(2) 
Now, if w takes an increment h, the right hand side of the 
above equation will be increased by some function of 2 and h, 
consequently, we shall have 
f (v--h)=f (#7) LE (a, h)..ccsceeeescienes (1) 
All that we know of the function F (a, h) is, that it must 
vanish when h=o, otherwise, f(#)=f (2) + some quantity. 
Which is absurd. Therefore, we see that F(z, h) must be of 
the form h F\(#,h), where F,(«, h) denotes some other un- 
known function of # and h. 
“. £(@+A)=Ef (x) +h F, (a, h)s..ceseecceeeee (2) 
Now, it is always possible to determine a function of (w-+-h), 
which shall be equal to F, (2, h). 
Therefore, we may put, F, (7+h)=F,(2,h), where F, 
denotes an unknown function of (#--h). 
“. (eh) =f(a) +h Fe(a-th) secceeeseeee (3) 
In asimilar manner we have, F, («-+-h)=F, (x) +h F; (c+h) 
F, («++h)=F;, ()+h F, (x+h) 
F, (@+h)=F, (x) +h F, (x@+h) 
&e. &e. &e. 
Where F, (z), F; (x), F,(x) &c. &e. are unknown functions 
of x. 
If then, we substitute these values, successively, in equa- 
tion (3), we shall obtain 
f(a-+-h) = f(x) + h.P,(x) + 1.F,() + 1.F,(x) + ht. (a) 
+ &e., Ge, Ger EWA E, (MRM) ccvceorccnndscosvssocscecceves (4) 
It, now, remains for us to determine the unknown functions 
