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Question: Find the multiplicative inverse of $-1+2x \in \mathbb{Z}[[x]]$.

If I understand correctly, this means $$(-1+2x)(a_0+a_1x)=1$$ where $a_0$ and $a_1$ are integers.

I'm struggling to do this question, all my attempts have led me to believe that it doesn't have a multiplicative inverse. I've also struggled to find any relevant information/context online.

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  • $\begingroup$ Hint : the inverse is NOT a polynomial of degree 1, in fact it is not a polynomial. $\endgroup$
    – Roland
    Feb 18 '17 at 17:27
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From $$(-1+2x)\cdot(a_0+a_1x+a_2x^2+\cdots)=1$$ we can see that it must be true that $a_0=-1$ in order for the constant terms of both sides agree.

But then we must have $(-1)(a_1x)+(-1)(2x)=0$ (hence $a_1=-2$) in order for the degree $1$ terms of both sides to agree.

But then we must have $(-1)(a_2x^2)+(-2x)(2x)=0$ (hence $a_2=-4$) in order for the degree $2$ terms of both sides to agree.

More generally, you can prove by induction that $a_n=-2^n$. Thus, the inverse of $-1+2x$ in $\mathbb{Z}[[x]]$ is $$-1-2x-4x^2-8x^3-\cdots$$ Note that this agrees with the standard answer for a geometric series: $$\frac{1}{-1+2x}=-\left(\frac{1}{1-2x}\right)=-\left(1+2x+4x^2+8x^3+\cdots\right)$$

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Hint:

The inverse is not a polynomial, but a formal power series, that you can deduce from the inverse of $\dfrac1{1-x}$: $$(-1+x)^{-1}=-\sum_{i\ge 0} x^i $$ By substitution you obtain $$(-1+2x)^{-1}=-\sum_{i\ge 0} 2^ix^i.$$

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