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Given that $x^3+x^2=1$ and $x\in\mathbb{R}$, express the infinite product $$(1+x)(1+x^2)(1+x^4)(1+x^8)\ldots$$ in the form $A+Bx+Cx^2$.

In the earlier parts of the question, I have already shown that $$x^4=-1+x+x^2$$ $$x^{-1}=x+x^2$$ $$1-x+x^2-x^3+x^4-x^5+\ldots=x^2$$ $$\frac{1}{1-x}=2+2x+x^2$$

I also know that $$1+x=\frac{1}{1-x+x^2-x^3+x^4-x^5+\ldots}.$$

Can anyone give me a hint? Perhaps there is a way to do it using the previous parts, but I cannot see how.

A. Goodier
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  • The two non-real roots of $x^3+x^2-1$ have absolute value $> 1$, so this infinite product can't converge if $x$ is one of them. You must assume $x$ is the real root. – Robert Israel Jan 03 '18 at 21:41
  • @RobertIsrael Sorry, yes, I should have specified $x\in\mathbb{R}$ – A. Goodier Jan 03 '18 at 21:42

1 Answers1

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Hint. One may recall that $$ (1+x)(1+x^2)(1+x^4)(1+x^8)\cdots=\frac1{1-x}, \qquad |x|<1, $$ then insert $$ x=\frac1{x+x^2} $$ in the preceding identity.

Olivier Oloa
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