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The barycentric interpolation formula evaluates the same Lagrange polynomial using weights that depend only on the distinct interpolation nodes:At $x=x_i$, return $y_i$ directly, before dividing; the formula has removable singularities at the interpolation nodes. Multiplying every $w_i$ by the same nonzero constant leaves the answer unchanged. Driscoll and Braun derive this formula as a more efficient, numerically stable way to evaluate polynomial interpolation.^[ref]

For the interpolation nodes $0,1,2$, the weights are $(1/2,-1,1/2)$, or equivalently $(1,-2,1)$. With data $(1,2,5)$, at $x=1/2$ the denominator using the latter weights is $2+4-2/3=16/3$ and the numerator is $2+8-10/3=20/3$, giving $p_2(1/2)=5/4$ again. For eleven interpolation nodes in the Runge phenomenon figure, the generator computes the weights by the direct products above and returns stored values exactly at every interpolation node.

The direct products require $O(n^2)$ arithmetic operations once; each subsequent value uses $O(n)$ arithmetic operations. Here $O$ denotes Big O notation. At large polynomial degree or on poorly scaled intervals, forming these products can exceed the range of floating-point arithmetic; normalization or specialized weight formulas may be needed. Stable evaluation also cannot cure sensitivity intrinsic to the chosen interpolation nodes, measured by the Lebesgue constant of interpolation, or the approximation failure in the Runge phenomenon.

A point where a specified complex function fails to be holomorphic, but which is approached by points of its domain. An isolated singularity has a punctured neighbourhood on which the function is holomorphic; a non-isolated singularity has other singularities arbitrarily close to it. The classification as a removable singularity, a pole, or an essential singularity applies to isolated singularities.

If $f$ is a modular form on a finite-index subgroup and $\gamma$ is rational with positive determinant, then $f{|}_k\gamma$ is bounded near infinity. Choose $\sigma \in S{L}_2(\mathbb{Z})$ with $\sigma \infty =\gamma \infty$. The matrix ${\sigma }^{-1}\gamma$ is upper triangular and sends $z$ to $Az+B$ with $A>0$. Thus the existing cusp expansion of $f{|}_k\sigma$ stays bounded under this substitution. Rational conjugation of finite-index modular subgroups gives a positive period for the translate, so boundedness gives a removable singularity in its cusp parameter. Apply this to $\gamma \rho$ for every $\rho \in S{L}_2(\mathbb{Z})$ to obtain holomorphy at every cusp of the new subgroup. Vanishing at the cusps is preserved too.

Let $f,g$ be entire functions, with $g$ not identically zero. If $|f(z)|\le C|g(z)|$ everywhere, then $f/g$ is bounded and holomorphic away from the isolated zeros of $g$. The removable singularity extends it across each zero, giving a bounded entire function. Liouville's theorem makes it constant, so $f=cg$ with $|c|\le C$. If $g$ is identically zero the inequality forces $f=0$. For example, domination by $|\cos z|$ forces a constant multiple of $\cos z$.

Let $f$ be a holomorphic function on $0<|z-a|<R$, and suppose $a$ is an isolated essential singularity. There is a set $E\subseteq \mathbb{C}$ with at most one element such that, for every $w\notin E$ and every $0<r<R$, the equation $f(z)=w$ has infinitely many distinct solutions with $0<|z-a|<r$. Thus each nonexceptional value has preimages accumulating at $a$ in every punctured neighbourhood. The same possible exception works for all these neighbourhoods; it need not be a completely omitted value, since it could be attained only finitely often near $a$. ^[ref]

The basic example is $f(z)=e^{1/z}$ at $a=0$. It is a holomorphic function for $z\ne 0$, and its Laurent serieshas infinitely many nonzero negative-power terms, making zero an essential singularity. Its sole omitted value is zero. For the value one, its preimages are exactlyThey satisfy $e^{1/z_k}=1$ and $|z_k|=1/(2\pi |k|)\to 0$. For any $r>0$, every integer $k$ with $|k|>1/(2\pi r)$ therefore supplies a distinct solution inside $0<|z|<r$.

The diagram displays only finitely many preimages; the formula and its limit, rather than the finite picture, establish that infinitely many occur in every punctured neighbourhood. More generally, for any $w\ne 0$, choose an argument $\theta$ of $w$. All its preimages arewith any zero denominator excluded. Such an exclusion occurs only for $w=1$, with exactly one excluded integer $k$ for the chosen argument. The remaining preimages are distinct and tend to zero as $|k|\to \infty$.

The isolated essential singularity hypothesis is crucial. The function $f(z)=z^{-1}$ has a pole at zero; on $0<|z|<r$ its image is only $\{w:|w|>1/r\}$. It omits every value in the closed disc $|w|\le 1/r$, and each attained value has just one preimage. A pole therefore has entirely different local behaviour from an essential singularity.

The Casorati-Weierstrass theorem says that the image near an essential singularity is a dense subset of the complex plane. The Great Picard theorem replaces arbitrary approximation by exact attainment, and then repeats that attainment infinitely often. This is useful when a local restriction on the image can rule out an essential singularity and reduce the possibilities to a pole or a removable singularity. For a meromorphic function viewed as a holomorphic map into the Riemann sphere, the corresponding statement has at most two exceptional sphere values, including infinity in the count; it applies when the map has an isolated essential singularity, even if poles accumulate there. Picard's 1880 memoir proves the local result in section 15, treating the cases with and without accumulating poles separately. ^[ref]

The Great Picard theorem also explains the Little Picard theorem. For a nonconstant polynomial, every value is attained by the fundamental theorem of algebra. For an entire function which is not a polynomial, the function $g(\zeta )=f(1/\zeta )$ has an essential singularity at zero: a removable singularity would force $f$ to be constant by the Liouville theorem, and a pole would force $f$ to be a polynomial. Applying the Great Picard theorem to $g$ leaves at most one finite omitted value for $f$.

A holomorphic function $F$ on a larger open set $V$ whose restriction to $U$ is the original holomorphic function $f$. A removable singularity is a point across which such an extension exists. On a connected larger domain, a holomorphic extension is unique by the identity theorem, provided $U$ is nonempty.

The complex logarithm has infinite-sheeted branching: continuing once around its zero changes its value by $2\pi i$. Multiplication by a function vanishing at that point need not remove the branching. For example $z^m\log z$ with positive integer $m$ tends to zero but changes by $2\pi i{z}^m$ after continuation around zero. This remains a branch point, rather than a removable singularity, because no single-valued holomorphic function exists on a full punctured neighbourhood. The coordinate $w=1/z$ tests logarithmic branching at infinity in the same way.

For $0<b<c<\infty$, the indicator function $f(x)=1_{[b,c]}(x)$ has Mellin transformThe interval stays away from both zero and infinity, so the integral is absolutely convergent for every complex number $s$. Its Mellin fundamental strip is the whole complex plane. Evaluating the antiderivative $x^s/s$ proves the formula for $s\ne 0$; at $s=0$, integrating $1/x$ gives the real natural logarithm $\log (c/b)$. The apparent division by zero is a removable singularity, since $c^s-b^s=s(\log c-\log b)+O(s^2)$ by the Taylor series of the exponential function. The transform is an entire function.

For example, $1_{[1,4]}$ gives $(4^s-1)/s$ with value $\log 4$ at zero. On $s=i\tau$, the Mellin transform sums oscillations over the width $\log 4$ of the interval in natural logarithm coordinates; this is the same cancellation mechanism as in the Fourier transform of an interval indicator.

A removable singularity at $a$ means that there is a holomorphic function on the whole disc which agrees with $f$ away from $a$. Equivalently, the Laurent series has no negative powers. The value at $a$ is filled in by its finite limit.

A removable singularity at $a$ is an isolated singularity for which one can assign a value at $a$ to obtain a holomorphic function on the whole disc. Equivalently its Laurent series has no negative powers. A finite limit $L$ supplies the extension value $f(a)=L$.

A removable singularity is an isolated singularity for which the function on the punctured disk extends to a holomorphic function on the full disk, by assigning or correcting its value at $a$. In Laurent series terms,The remaining power series defines the extension with value $c_0$ at $a$. Conversely, a holomorphic extension has a Taylor series there, so uniqueness of the Laurent series forces all negative coefficients to vanish.

With the same factors, if $n\le m$, thenhas no negative power and extends holomorphically across $a$. Thus its apparent singularity is a removable singularity, with extension valueFor $m>n$ the extended quotient has a zero of order $m-n$; for $m=n$ the extension is nonzero.

For a bounded entire function, the preceding result with $n=0$ gives Liouville's theorem. If a nonconstant polynomial $p$ had no complex root, $1/p$ would be entire. The leading term gives $|p(z)|\to \infty$ at infinity, so $1/p$ is bounded outside a large disc and is also bounded on the remaining compact disc. Liouville's theorem would make $1/p$, and hence $p$, constant. This proves the fundamental theorem of algebra.

Finally, away from the zeros of $\cos z$, the holomorphic quotient $h=f/\cos z$ has modulus at most $C$. At each zero it therefore has a removable singularity. These isolated points have no finite accumulation, so removing all of them gives a bounded entire function. Liouville's theorem gives a constant $c$ with $|c|\le C$, andConversely every such multiple satisfies the assumed bound.

A removable singularity means that $f$ has a holomorphic extension to the full disk, assigning a finite value at zero. Equivalently its Laurent series has no negative-power terms. One direction follows by the Taylor expansion of the extension; the other by the convergent nonnegative-power series. Thus $\boxed{a_k=0\text{ for all }k<0}$ is the coefficient criterion.

The point is a removable singularity if assigning a value to $f(z_0)$ makes $f$ holomorphic on a full neighbourhood of $z_0$. Equivalently, its Laurent series has no nonzero coefficients of negative powers. The assigned value is its constant coefficient, or equivalently the finite limit ${\lim }_{z\to z_0}f(z)$.

An isolated singularity at $z_0$ means that some radius $r>0$ has $f$ holomorphic on the punctured disk $0<|z-z_0|<r$, while $f$ is not initially defined as a holomorphic function at $z_0$. The adjective isolated refers to the absence of other singularities in this punctured neighbourhood; it permits a removable singularity.

An essential singularity is an isolated singularity that is neither a removable singularity nor a pole. In its Laurent series, infinitely many coefficients of negative powers are nonzero. The alternatives follow because no negative terms give a removable singularity, while finitely many with a lowest nonzero power give a pole.

Suppose, towards a contradiction, that for some $r,w,\epsilon$ the punctured-disk image misses the open disk of radius $\epsilon$ about $w$. Thenis holomorphic and bounded by $1/\epsilon$ on $0<|z|<r$. Part (i) extends it to a holomorphic function $H$ on the full disk. If $H(0)\ne 0$, then $w+1/H$ extends $f$ holomorphically at zero, making zero a removable singularity.

If $H(0)=0$, the function $H$ is not identically zero, since it is nonzero on the punctured disk. The Taylor series and identity theorem give a finite zero order $k\ge 1$, so $H(z)=z^{k}q(z)$ with $q$ holomorphic and $q(0)\ne 0$. Hencehas a pole of order $k$. Both alternatives contradict an essential singularity. ConsequentlyThis is the Casorati-Weierstrass theorem: every complex value can be approximated arbitrarily closely in every punctured neighbourhood of an essential singularity.

Take the Laurent series of $f$ on $0<|z|<r$. For every integer $k\ge 1$, its coefficient of $z^{-k}$ satisfies, on a circle of radius $0<\rho <r$,The coefficient is independent of $\rho$. Letting $\rho \downarrow 0$ therefore gives $c_{-k}=0$ for every $k\ge 1$. The Laurent series is a power series with no negative powers and supplies a holomorphic extension at zero. Thus zero is a removable singularity. This is the Riemann removable singularity theorem, with boundedness used to eliminate the entire principal part.

A removable singularity is an isolated singularity for which the function on the punctured disk extends to a holomorphic function on the full disk, by assigning or correcting its value at $a$. In Laurent series terms,The remaining power series defines the extension with value $c_0$ at $a$. Conversely, a holomorphic extension has a Taylor series there, so uniqueness of the Laurent series forces all negative coefficients to vanish.