For easyness:
D(m+1)=1+2^(m+1)+x^(m+1)
D(m)=1+2^m+x^m
See that for x>1:
D(m+1)<x*D(m)
As: x^(m+1)=x*x^m=(x1)*x^m+x^m, and same for 2^m+1
it's possible to express D(m+1) as a function of D(m)
D(m+1)=D(m)+2^m+(x1)x^m
doing it again:
D(m+1)=2*D(m)+(x2)*(x^m)1
Doing it (x1) times
D(m+1)=(x1)*D(m)+x(x^(m1)2^m1)+3*2^m+2
Doing it x times is unnecessary as D(m+1)<x*D(m)
The expression on the right of D(m) adds (x^m + 3*2^m+2) and detract (2^m*x+x). This is going to be always positive for the presence of x^m in the addition, except if m=1. [This implies that usually (x1)*D(m)<D(m+1)<x*D(m)].
But if m=1, for x{4,5}, the second member is igual or higher and the expression is negative. Concretely for x=4, m=1 the expression is 0, and so verify that D(m+1)=(x1)*D(m)
D(m) 1+2+4=7
D(m+1)=1+4+16=21
Edited on February 9, 2016, 4:15 am

Posted by armando
on 20160208 17:10:31 