You can also do it without thinking about continued fractions. Define H(x) = x + A(x^3); it's easy to show that H(x) = 1 + x / (1 - x^2 / H(x^2) ) (just write both sides in terms of A(x^6) and the rest is algebra), whence H = F. J.P. On Fri, Feb 28, 2014 at 10:50 AM, Joerg Arndt <arndt@jjj.de> wrote:
Define A(x) by A(x) = 1 + x / A(x^2), so A(x) = 1+x/(1+x^2/(1+x^4/(1+x^8/(1+ ...)))).
Define F(x) by F(x) = 1 + x / (1 - x^2 / F(x^2) ), so F(x) = 1+x/ (1-x^2/ (1+x^2/ (1-x^4/ (1+x^4 /(1-x^8/ (1+x^8/ (1-x^16/ ... ))))))).
The following took me a while: Show that F(x) - x = A(x^3)
Spoiler below. I am asking because I'd like to know how you'd rate it in difficulty and time this should/did take.
Regards, jj
... on to the spoiler
... on to the spoiler
... on to the spoiler
******************** SPOILER **********************
Set G(x) = F(x) - x observe that G(x) = 1 + x^3/G(x^2) = 1 + x^3/(1 + x^6 / G(x^2) ) = ... = 1 + x^3/(1 + x^6 / (1 + x^12 / (1 + x^24 / (...) ) ) ) = A(x^3).
Cf. https://oeis.org/draft/A238429 (expansion of F), https://oeis.org/draft/A218031 (expansion of A)
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