Suppose |x|<1. Can you give any ideas on how to find the following sum? xx+1+2x2x2+1+4x4x4+1+8x8x8+1+…

Swebaceacichegh

Swebaceacichegh

Answered

2022-01-21

Suppose |x|<1. Can you give any ideas on how to find the following sum?
xx+1+2x2x2+1+4x4x4+1+8x8x8+1+

Answer & Explanation

ma90t66690

ma90t66690

Expert

2022-01-22Added 7 answers

Prove first that for any x(1,1) the identity
11x=k0(1+x2k) (1)
follows from a telescoping product or the fact that every nN+ has a unique representation in base-2. By considering ddxlog(.) of both sides of (1), we get
11x=k02kx2k11+x2k
from which:
k02kx2k1+x2k=x1x
Roman Stevens

Roman Stevens

Expert

2022-01-23Added 10 answers

Evaluating
q02qx2q1+x2q
we obtain
q02qk0(1)kx(k+1)2q=n1xn21n2q(1)n2q1
Now observe that
2qn2q(1)n2q1=p=0v2(n)2p(1)n2q1
where v2(n) is the exponent of the highest power of 2 that divides n. This is
p=0v2(n)12p+v2v2(n)1)+2v2(n)=1
because n2p is even unless p=v2(n)
(This also goes through when n is is odd and we have one value for p, namelly zero.)
Hence the end result is
n1xn=x1x

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