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Why does $1/((x-1)(x^2 - 1)) = (x^4-1)/(x-1) \cdot (x^8 - 1)/(x^2 - 1) \cdot (x^{16} - 1)/(x^4 - 1) \cdots$?

I think the equality holds as formal power series. Expanding the LHS, one should get $(1+x+x^2 +\cdots )(1+x^2 + x^4+\cdots).$ But doing cancellation for the RHS gives that it equals $(x^2+1)/(x-1) \cdot (x^4 + 1)(x^8 + 1)\cdots,$ which clearly contradicts the equality. If however, one multiplies both sides by $x^2 - 1$, one gets an obvious equality after cancellation.

I was wondering if someone could explain what I'm doing wrong here (i.e. why I get two different "answers")?

PC1
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Gord452
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  • I assume that $|x|<1$, else this doesn't hold. Just expand the right hand side, there is only $(x-1)^{-1}(x^2-1)^{-1}$ that is left, times a term that tends to $1$ as you add more terms. – PC1 Jul 11 '22 at 01:03
  • If you use stacked fractions using \frac it will be much easier to read the equations. It looks like you want the divide slash to have priority over the multiplication dot, but even so it is hard to read. \frac{x^4-1}{x-1} gives $\frac{x^4-1}{x-1}$ I think it will make seeing the telescope easier. – Ross Millikan Jul 11 '22 at 02:31

2 Answers2

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It is a telescoping product. Everything on the RHS cancels in pairs, except the $1-x$ and $1-x^2$, which appear in the denominator but not the numerator.

RobPratt
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"But doing cancellation for the RHS gives that it equals $(x^2+1)/(x−1)\cdot(x^4+1)(x^8+1)\cdots$, which clearly contradicts the equality."

In fact $(1 + x^2)(1 + x^4)(1 + x^8)(1 + x^{16}) \cdots = 1 + x^2 + x^4 + x^6 + x^8 + x^{10} + \ldots = \frac{1}{1 - x^2}$ -- the first equality is just the generating-functional version of binary expansion, see e.g. this recent question: Demonstrate recursively that $\prod_{k = 0}^\infty (1 + x^{2^k}) = \frac{1}{1-x}$. (Or maybe you mean the fact that you've lost a minus sign? That's because the leftover factor is $x^{2^n} - 1$, which is approaching $-1$ as $n \to \infty$.)

JBL
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