CHANGE 2/14 



So the double application causes the same change as a single 

 application of the transformation m' = 4ni — 9. 



2/14. Higher powers. Higher powers are found simply by adding 

 symbols for higher transforms, n'", etc., and eliminating the symbols 

 for the intermediate transforms. Thus, find the transformation 

 caused by three applications of «' = 2// — 3. Set up the equations 

 relating step to step: 



n' =2/2 - 3 



n" = 2n' — 3 



;,'" = 2n" - 3 



Take the last equation and substitute for n", getting 



n'" = 2(2// - 3) - 3 

 = 4//' - 9. 



Now substitute for «' : 



n'" = 4(2/7 - 3) - 9 

 = 8/; - 21. 



So the triple application causes the same changes as would be 

 caused by a single application of in' = Sm — 21. If the original 

 was T, this is T^. 



Ex. 1 : Eliminate n' from n" = 3n' and //' = 3//. Form the transformation 

 corresponding to the result and verify that two applications of n' = 3n 

 gives the same result. 



Ex. 2: Eliminate a' from a" = a' + S and a' = a + S. 

 Ex. 3: Eliminate a" and a' from a'" = la", a" = la', and a' = la. 

 Ex. 4: Eliminate k' from k" ^ - W + 2, k' = - 3k + 2. Verify as in 

 Ex. 1. 



Ex. 5: Eliminate tti' from m" = log /;;', m' = log ?n, 

 Ex. 6 : Eliminate p' from p" = (p')'^, p'=p^ 



Ex. 7: Find the transformations that are equivalent to double applications, on 

 all the positive numbers greater than 1, of: 



(i)/;' = 2« + 3; 

 (ii) n' = II- + n; 

 (Hi) n' = I + 2 log n. 



Ex. 8: Find the transformation that is equivalent to a triple application of 

 //' = — 3a/ — 1 to the positive and negative integers and zero. Verify 

 as in Ex. 1 . 



Ex. 9: Find the transformations equivalent to the second, third, and further 

 applications of the transformation //' = 1/(1 + n). (Note: the series 

 discovered by Fibonacci in the 12th century, 1, 1, 2, 3, 5, 8, 13, ... is 

 extended by taking as next term the sum of the previous two; thus, 3 + 5 

 = 8, 5 + 8 = 13, 8 + 13 = . . ., etc.) 



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