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For a sufficiently smooth local map fixing the origin, a local invariant centre manifold is tangent to the generalized eigenspaces with eigenvalues of modulus one. Writing the map as $(u,v)\mapsto (F(u,v),G(u,v))$ and the manifold as $v=h(u)$ gives the invariance equation $h(F(u,h(u)))=G(u,h(u))$. Finite Taylor series coefficients can be found from this equation. If all transverse eigenvalues have modulus less than one, local asymptotic stability reduces to that of the map on the centre manifold.

An even function satisfies $f(-x)=f(x)$; its Laurent series or taylor series contains only even powers.

If $P=z^{d}p$ and $p$ is nonnegative on the unit circle, a root of a polynomial $\zeta =e^{i{\theta }_0}$ there has even multiplicity of a root. Indeed $p(e^{i\theta })$ is a real analytic nonnegative function, whose first nonzero term in its Taylor series at a zero must have even order. The map $\theta \mapsto e^{i\theta }-\zeta$ has a simple zero, and multiplication by $e^{-id\theta }$ is nonvanishing, so the order equals the polynomial multiplicity.

A smooth function is flat at $a$ when every derivative, including order zero, vanishes at $a$. The function $e^{-1/x^2}$ off zero, extended by zero at zero, is flat there but not identically zero near zero. Thus its Taylor series does not determine it. A real analytic function that is flat at a point vanishes on a neighborhood of that point.

If an entire function $f$ is not a polynomial, then there is a point $z_0\in \mathbb{C}$ for which $f^{(n)}(z_0)\ne 0$ for every nonnegative integer $n$. Consequently every coefficient in the Taylor series of $f$ about $z_0$ is nonzero.

For each $n$, the function $f^{(n)}$ is not identically zero, since otherwise $f$ would be a polynomial. Its zero set is closed and has empty interior by the identity theorem, hence is a nowhere dense set. The Baire category theorem says that the complete metric space $\mathbb{C}$ cannot be the union of these countably many zero sets.

Set $S_n=\{x:f^{(n)}(x)=0\}$ and $E_n=\mathrm{int}{S}_n$. The sets $S_n$ are closed, $E_n\subset E_{n+1}$, and Baire's theorem on each interval shows that $\Omega ={\bigcup }_n{E}_n$ is dense. On every connected component of $\Omega$, compactness and the increasing cover by the $E_n$ show that one derivative vanishes locally with a uniform finite order, so $f$ agrees there with one polynomial.

If $F=\mathbb{R}\setminus \Omega$ were nonempty, it would be closed and have no isolated points: polynomial pieces on both sides of an isolated point would have matching derivatives of every order and hence join into one polynomial piece. Applying Baire's theorem to the cover $F={\bigcup }_n(F\cap S_n)$ gives an interval $U$ and an index $k$ for which $F\cap U$ is nonempty and contained in $S_k$. Because $F$ has no isolated points, difference quotients show that every derivative of order at least $k$ vanishes on $F\cap U$. Each polynomial component of $\Omega$ meeting $U$ has an endpoint in $F\cap U$, so its degree is less than $k$. Thus $f^{(k)}=0$ throughout $U$, contradicting $F\cap U\ne \varnothing$. Hence $\Omega =\mathbb{R}$, and its sole connected component carries one polynomial.

For every nonprincipal Dirichlet character, $L(1,\chi )\ne 0$. A nonreal character is covered by Euler product positivity for L-function nonvanishing at height zero, since its squared character is nonprincipal.

For a real character, suppose $L(1,\chi )=0$. The simple zeta pole would cancel, making $F(s)=\zeta (s)L(s,\chi )$ an entire function. Its nonnegative zeta-times-real-L coefficients satisfy $a_n\ge 0$ and $a_{m^2}\ge 1$. Termwise derivatives at two give $(-1{)}^k{F}^{(k)}(2)={\sum }_n{a}_n(\log n{)}^k{n}^{-2}$. The Taylor series of this entire function at two converges at zero. Every term there is nonnegative, and Tonelli theorem identifies its sum asThe last series diverges because all square coefficients are at least one. This contradiction proves the claim. It is the positive-coefficient mechanism behind the more general Landau theorem for a Dirichlet series with nonnegative coefficients.

An odd function satisfies $f(-x)=-f(x)$; its Laurent series or taylor series contains only odd powers.