170 Mr. P. Nicholson on the 



Examples. 

 Ex. 1. Transform tlie quadratic function ax'^-\-hx-\-c into 

 another which sliall have x — e instead of x. 

 Here A=a, B = i, C = r; therefore Ae = ae 

 B, = B+cA 



B,= B,+<'A C,= C-i-t'B,. 

 Now by substituting the real quantities we shall have 

 B.= 6 +ea 



B.= B.+m C,=c-\-eB„ 

 whence B^ = b + ^2ae, C, = c-{-be + ae- ; therefore 



«.r - + bx + cz=a[x — ey + {b + 2ae)[x — e) + c-{-be-\- ae^ : 



or since v=x—e 

 ax^ + bx-\-c = av- + {b + 2ae)v + {c-\-be-\-ae-). 

 Ex.2. Transform the cubic function x^ — bx^ -{- ex — d inio 

 another in x—e = iK 



Here A=l, B=-b, C = c,D=-d, 



B.= -i+ e 



B,= -b + 2e C.,=c— be+ <?- 



B.= —b + 3e C,=c—2be-irSc"- Y)=—d-\-ec — he'--\-e\ 

 Whence x' —bx^-\-cx —d=v^ -Jr^ — h ■{- 3e)v^-\-{c — 2be-{.5e^)v-\- 

 {—d-\'CC—be'^+e^) 



In the derivative table the values of B , Bj are found by 

 adding the quantity e successi\'ely. Again, by multiplying the 

 value of Bi by e, and adding the product to c, the third co- 

 efficient of the original ecjuation, gives the value of C.^; and 

 multiplying the value of B^ by e, and adding the product to 

 C , gives the value of C^ ; and by multiplying the value of C.^ 

 by c, and adding the pi'oduct to —d, the absolute number of 

 the given function, gives the value of D„ the absolute number 

 of the transformed function, which is of the same form as the 

 original; instead of demonstrating by successive steps, as has 

 been done, the method may be derived from two consecutive 

 equations, as is in the following 



Gc7icral Theorem. 

 If x — e=^v or x = v-^c 

 and A.»''-'-f...4-K... =1V-'+Qi)"-2+Rt;"-='+...+«, (A) 

 and Ax" +...4-K.r + L= PV' + QV + llV--+.. + a'7;4-i3',(B) 

 then will P'- P, Q' = Q + t^P, R'=:R+t'Q &c. and /3' = L+^«. 



'Demonstration. 

 For multiply the first side of equation (A) by .r, and the 

 second side by its equal v-\-e, uul the product is llie ecjuation 



Aj:-" 



