Formal Power Series: Examples and Counterexamples

The series $s={\sum }_{n\ge 0}{x}^n$ has every coefficient equal to $1$. The constant coefficient of $(1-x)s$ is $1$, and for $n\ge 1$ its coefficient is $[x^n]s-[x^{n-1}]s=1-1=0$.

Thus $(1-x)s=1$ by coefficient extensionality. Since $1-x$ has unit constant coefficient $1$, its inverse is unique, so $s=(1-x{)}^{-1}$. In the zero ring both sides are the unique series.

Put $s={\sum }_{j\ge 0}{x}^j$. Its constant coefficient in $(1-x)s$ is $1$, and every positive-degree coefficient is $1-1=0$, so extensionality and inverse uniqueness give $s=(1-x{)}^{-1}$. Hence $(1-x{)}^{-m}$ is the product of $m$ copies of $s$ over every commutative ring. Over a commutative $\mathbb{Q}$-algebra, the formal exponent law gives the same series.

The coefficient of $x^n$ in this product is the number of $m$-tuples $(j_1,\dots ,j_m)$ of nonnegative integers with sum $n$. The stars-and-bars count makes this $\left(\genfrac{}{}{0px}{}{n+m-1}{m-1}\right)=\left(\genfrac{}{}{0px}{}{m+n-1}{n}\right)$. At $n=0$ the unique tuple is all zero, giving coefficient $1$.

Coefficient extensionality now gives the asserted series identity for every $m\ge 1$. When $m=1$ the coefficient is $\left(\genfrac{}{}{0px}{}{n}{n}\right)=1$, recovering the series $s$ from step 1.1; step 2.1 already checks $n=0$.

Let $q=1-2x-2x^2-4x^3-10x^4-28x^5$. Cauchy convolution gives $[x^0]q^2=1$, $[x]q^2=-4$, and coefficients $4-4$, $-8+8$, $-20+16+4$, and $-56+40+16$, all $0$, in degrees $2,3,4,5$ respectively. Hence $q^2\equiv 1-4x(\mathrm{m}\mathrm{o}\mathrm{d}{x}^6)$.

The unique constant-one square root has coefficients determined successively by the equation $v^2=1-4x$ because its unknown degree-$n$ coefficient occurs as $2[x^n]v$. Step 1.1 therefore gives its coefficients through degree $5$.

The equation $x=w-w^2$ is equivalent to $w=x(1-w{)}^{-1}$. Lagrange inversion with $\varphi (u)=(1-u{)}^{-1}$ and $H(u)=u$ gives $[x^n]w=\frac{1}{n}[u^{n-1}](1-u{)}^{-n}$.

Apply the generalized-binomial formula with exponent $-n$ and argument $-u$. The coefficient of $u^{n-1}$ is $(-1{)}^{n-1}(-n)(-n-1)\cdots (-2n+2)\mathrm{/}(n-1)!=\left(\genfrac{}{}{0px}{}{2n-2}{n-1}\right)$. At $n=1,\dots ,6$ this yields $1,1,2,5,14,42$.

Direct substitution of these coefficients gives $w-w^2\equiv x(\mathrm{m}\mathrm{o}\mathrm{d}{x}^7)$, and uniqueness of the compositional inverse confirms the displayed initial segment.

Both $f$ and $g$ have zero constant coefficient and unit linear coefficient. Formal substitution and ring algebra give $f\circ g=g\mathrm{/}(1-g)=x$ because $1-g=(1+x{)}^{-1}$, and $g\circ f=f\mathrm{/}(1+f)=x$ because $1+f=(1-x{)}^{-1}$.

Thus $g$ is a two-sided compositional inverse of $f$, and uniqueness gives the claim. Multiplying ${\sum }_{n\ge 0}{x}^n$ by $1-x$, and ${\sum }_{n\ge 0}(-1{)}^n{x}^n$ by $1+x$, gives constant coefficient $1$ and every later coefficient $0$; extensionality and inverse uniqueness give the two displayed expansions. Equivalently, the inverse equation $(1+x)g=x$ yields $[x]g=1$ and the alternating recursion $[x^n]g=-[x^{n-1}]g$ for $n\ge 2$. These calculations also hold in the zero ring.

The constant coefficient of $x$ is $0$, which is not a unit in the field $K$, so the unit criterion excludes a power-series inverse. Equivalently, every product $xg$ has constant coefficient $0$.

In $K((x))$, negative exponents are permitted and $x{x}^{-1}=1=x^{-1}x$. Thus passing to Laurent series changes the answer.

Steps 1.1 and 1.2 exhibit the claimed contrast.

The outer series $f$ is not a polynomial, and the inner series $1$ does not have zero constant coefficient, so neither admissibility branch applies. More concretely, the proposed constant coefficient would be ${\sum }_{n\ge 0}1$, an infinite sum not defined by the ring operations of $\mathbb{Z}$.

By contrast, $x^2$ has zero constant coefficient, so $f\circ x^2=1+x^2+x^4+\cdots$ is defined; every coefficient has at most one contributor.

Hence $f\circ 1$ is undefined as a formal composition, while the zero-constant substitution in step 1.2 is admissible. This is a local-finiteness obstruction, not a claim about analytic divergence.

Take $N=1$. Every index $i$ contributes the nonzero coefficient $[x^0]f_i=1$, so infinitely many family members have a nonzero coefficient below $N$.

In contrast, the family $(x^n{)}_{n\ge 0}$ is summable: below any fixed degree $N$, only the indices $n<N$ contribute. Its coefficientwise sum is the series with every coefficient $1$.

Step 1.1 violates the defining local-finiteness condition, whereas step 1.2 satisfies it. The nonzero-ring hypothesis is necessary: in the zero ring, the constant series $1=0$ and the original family is summable.

The residue class of $2$ modulo $4$ is nonzero because $2\equiv ̸0(\mathrm{m}\mathrm{o}\mathrm{d}4)$, while its square is $[2{]}_4[2{]}_4=[4{]}_4=[0{]}_4$. The constant-series embedding preserves multiplication, so the two nonzero constant series multiply to the zero series.

Each factor has formal order $0$, whereas the product has order $+\mathrm{\infty }$. This does not contradict the exact product law because $\mathbb{Z}\mathrm{/}4\mathbb{Z}$ is not an integral domain.