MATH.23:5.1 - From repeated instances to a family of operations

Repeatedly apply f(x)=a*x+b over real numbers. A few calculations give:

f²(x)=a²*x+(a+1)*b,

f³(x)=a³*x+(a²+a+1)*b.

The useful next question is how to represent any number of repetitions without expanding every substitution. The conjectured formula is:

f^n(x)=a^n*x+b*S_n, where S_0=0 and S_n=1+a+...+a^(n-1) for n>0.

Recover the generating rule. Applying f once more changes the coefficient of x from a^n to a^(n+1), and the constant from b*S_n to b*(a*S_n+1). The identity S_(n+1)=a*S_n+1 therefore supplies the induction step. At n=0, f^0 is the identity and S_0=0. The formula is proved for every natural n.

Now vary the problem: combine two potentially different maps f(x)=a*x+b and g(x)=c*x+d. Their composite is:

f(g(x))=(a*c)*x+(a*d+b).

This constructs an operation on coefficient pairs: (a,b) composed with (c,d) gives (a*c,a*d+b). The class is closed under composition. MATH.17 can study this operation; a computational method can use it to combine repetitions.

Order now becomes a useful question. Reversing the maps gives the same linear coefficient but constant c*b+d. Thus they commute exactly when d*(a-1)=b*(c-1). This condition says when reordered operations retain the result.

The progression has produced a general iteration formula, a closed combining operation and a condition for rearrangement. A further question about vector-valued affine maps needs its matrix composition law; scalar commutation cannot be silently carried into that new setting.