Originally Posted by
kathrynmath Let 0 < x < 2 be a non-dyadic value expressed as x =summation(d_n/2^n)
; dn = 0 or 1 for all n.
Show that for gm defined as gm=summation(1/2^nh(2^nx) from 0 to m one has
g'm(x) = (-1)^d0 + (-1)^d1 + .... + (-1)^dm
where h=|x|.
I started looking at some example. First, I had dn=0 if n was odd and 1 if n was even.
Then x=1/4^n=4/3
Alternatively dn=0 if n is even and 1 if n is odd.
Then x=1/2^(2n+1)=2/3
So g_0'(4/3)=h'(4/3)=1
g_0'(2/3)=h'(2/3)=1
g_1'(4/3)=h'(4/3)+h'(2*4/3)=-1+h'(2+2/3)=-1+h'(2/3)=-1+1=0
g_1'(2/3)=0
Then g_1'(x)=h'(summ(dn/2^n)+h'(2*sum(dn/2^n))