Power series (${\sum }_{n=0}^{\infty }{a}_n(z-c{)}^n$)

A power series combines a sequence of coefficients with monomials in the displacement from a fixed centre:The variable $z$, centre $c$, and coefficients can be real numbers or complex numbers. Its degree-$N$ partial sum $P_N(z)={\sum }_{n=0}^N{a}_n(z-c{)}^n$ is a polynomial, so a convergent power series can turn a complicated function into increasingly accurate finite calculations. Convergence, and the error after stopping, are essential parts of that claim.

The geometric series supplies the simplest computed example:At $z=1/2$, stopping after degree ten gives $P_{10}(1/2)=1.9990234375$, differing from the exact value $2$ by $1/1024=0.0009765625$. At $z=0.99$, the same degree gives only about $10.4662$ instead of $100$. The same formula has very different practical accuracy at different points; the geometric power-series remainder quantifies it.

The structure is useful beyond numerical evaluation. Within its radius of convergence, termwise differentiation of a power series turns a question about a derivative into a question about coefficients. A Taylor series chooses $a_n=f^{(n)}(c)/n!$ when it represents the function locally. A formal power series instead uses the coefficients algebraically, without asserting numerical convergence; the existing generating function examples use that distinction to encode counting problems.

The domain of convergence of a power series is the set of inputs where the series has a finite sum. Inside its radius of convergence the series has absolute convergence, and outside it diverges. At a finite positive radius the real domain may include either, both, or neither endpoint; the radius of convergence alone does not decide endpoint convergence.

If $f(x)={\sum }_{n\ge 0}{a}_n{x}^n$ has $a_0\ne 0$, its reciprocal has a unique formal power series $1/f(x)={\sum }_{n\ge 0}{b}_n{x}^n$. Equating coefficients in the Cauchy product $f\cdot (1/f)=1$ gives the displayed recurrence relation. In particular,For a convergent power series, continuity and $f(0)\ne 0$ give a neighborhood without zeros, where the reciprocal is analytic and this series converges. Its radius need not equal the original radius: zeros of $f$ can introduce closer reciprocal singularities.

For power series $f(z)={\sum }_{m\ge 0}{f}_m{z}^m$ and $g(x)={\sum }_{j\ge 1}{g}_j{x}^j$, their function composition is well defined as a formal power series: the coefficient of $x^n$ receives contributions only from $m\le n$, since $g$ has zero constant term. For $m\ge 1$, expand the coefficient asThe $m=0$ term is one for $n=0$ and zero otherwise. The first four coefficients of the function composition are therefore $f_0$, $f_1{g}_1$, $f_1{g}_2+f_2{g}_1^2$, and $f_1{g}_3+2f_2{g}_1{g}_2+f_3{g}_1^3$. For convergent power series, the same calculation holds wherever $g(x)$ lies inside the radius of convergence of $f$.

Analyticity at a point means that a function agrees in a neighbourhood of that point with a convergent power series. This is stronger than merely having derivatives of every order: the Taylor series of a smooth function can fail to recover its values. For a real variable this property defines a real analytic function. For a complex variable it is equivalent to being a holomorphic function on a neighbourhood of the point.

For the degree-$N$ partial sum $P_N(z)=1+z+\cdots +z^N$ of the geometric series, the finite identity $(1-z)P_N(z)=1-z^{N+1}$ givesFor any fixed $0<r<1$, the triangle inequality then gives the uniform error boundEquality is attained at the positive real point $z=r$, so this is the exact worst-case error on that closed disk. It also proves uniform convergence on each such disk, not just pointwise convergence.

Given a positive tolerance $\epsilon$, choosing $N$ with $r^{N+1}/(1-r)\le \epsilon$ turns a convergence statement into a finite stopping rule. For $\epsilon =10^{-3}$, degree ten suffices throughout $|z|\le 1/2$. Nearer the boundary, $r=0.9$ requires degree $87$ and $r=0.99$ requires degree $1145$. The required degree depends on both the tolerance and the region, not just on the name of the function.

A generating function packages a sequence into a series so that operations on objects become operations on their encodings. An ordinary generating function uses the power series $A(z)={\sum }_{n\ge 0}{a}_n{z}^n$; in counting problems, the coefficient $a_n$ records the number of objects of size $n$. The variable $z$ initially marks size: it need not be assigned a numerical value. Coefficient extraction writes this recovery as $[z^n]A(z)=a_n$. One compact expression can therefore answer infinitely many counting questions, while explaining how their answers fit together.

Start with two contributions: choose $i$ from $\{0,1,2\}$ and $j$ from $\{0,1\}$, with exactly one option at each allowed value. Their individual ordinary generating functions are the polynomials $P(x)=1+x+x^2$ and $Q(x)=1+x$. To count ordered pairs by total $i+j$, multiply:The coefficients $1,2,2,1$ count the ways to obtain totals zero through three. For total two, the possibilities are $(1,1)$ and $(2,0)$; the two corresponding monomials both contribute $x^2$. There are six ordered pairs altogether, agreeing with $P(1)Q(1)=3\cdot 2=6$.

If the first contribution has two distinguishable choices when $i=1$, replace $P(x)$ by $1+2x+x^2$. Its product with $Q(x)$ becomes $1+3x+3x^2+x^3$. The coefficients now carry multiplicities as well as possible totals; the exponent still records cost. This lets the same construction handle several actions with identical costs without losing their separate identities.

In general, the Cauchy product givesFor each split of the total size, the rule of product counts the choices of the two components; summing over those splits counts all possibilities. This requires each combined object to have a unique decomposition into the two components. The same idea works for costs, lengths, or other additive measures, not just the number of pieces. See MIT 18.310, generating functions, sections 1–2.

Now cover a board of length $n$ with tiles of lengths one and two, without gaps or overlaps. Let $t_n$ count these tilings. There is exactly one empty tiling, so $t_0=1$, and $t_1=1$. Removing the first tile leaves a board shorter by one or two, giving the recurrence relationThese are the Fibonacci numbers shifted by one index: $t_n=F_{n+1}$ under the convention $F_0=0$, $F_1=1$. Keeping the empty tiling is essential to both the counting rule and the resulting ordinary generating function.

For $F(z)={\sum }_{n\ge 0}{t}_n{z}^n$, the empty board or a first tile followed by a remaining tiling givesEquivalently, $F(z)={\sum }_{m\ge 0}(z+z^2{)}^m$: each factor chooses one tile, and $m$ counts pieces while the exponent of $z$ records total length. The same ordinary generating function counts ordered histories of two repeatable actions costing one and two units. Histories $(1,2)$ and $(2,1)$ differ despite having the same total cost. MIT's notes use dots and dashes with these costs, including the empty message. ^[ref]

These identities hold in formal power series algebra: each requested coefficient uses only finitely many contributions. The repeated-choice expansion is legitimate because $z+z^2$ has zero constant term; a zero-cost repeatable action would allow infinitely many histories of the same size. Numerical evaluation separately requires a convergent series. Here the radius of convergence is $(\sqrt{5}-1)/2$, the reciprocal of the golden ratio. At $z=1$, the rational function equals $-1$, but the counting series diverges: that value cannot represent its sum.

Other generating functions change the encoding: an exponential generating function divides by factorials, while a Dirichlet series uses $n^{-s}$. A counting generating function also differs from a probability generating function, whose coefficients must be nonnegative and sum to one.

An algebraic generating function is a formal power series $F(x)$ satisfying a nonzero polynomial equation $P(x,F(x))=0$. Over a field of characteristic zero, it is D-finite.

The Euler transform of a sequence $(b_n{)}_{n\ge 1}$ has coefficients $a_n$ defined by ${\sum }_{n\ge 0}{a}_n{x}^n={\prod }_{n\ge 1}(1-x^n{)}^{-b_n}$. In unlabelled counting, $b_n$ may count connected components of size $n$, while $a_n$ counts multisets of components.

A rational generating function is a formal power series equal to a quotient of two polynomials, with denominator nonzero at the origin. Multiplying by the denominator shows that its coefficients eventually satisfy a linear recurrence with constant coefficients.

The binomial transform of a sequence $a_k$ is $b_n={\sum }_{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)a_k$. When $a_k$ counts objects on a specified set of $k$ positions, $b_n$ counts the choices obtained by selecting those positions from $n$ and filling them with such an object. Its inverse is $a_n={\sum }_{k=0}^n(-1{)}^{n-k}(\genfrac{}{}{0ex}{}{n}{k})b_k$, by binomial inversion.

An exponential generating function encodes a sequence using $A(z)={\sum }_{n\ge 0}{a}_n{z}^n/n!$. For labelled counting, $a_n$ is the number of structures on a fixed set of $n$ distinct labels. Labels distinguish otherwise similar objects: changing which person belongs to a group can change the structure. Recover the count by coefficient extraction as $a_n=n![z^n]A(z)$. The raw coefficient is $a_n/n!$, not $a_n$.

The factorial makes multiplication account automatically for distributing labels. To combine an $A$-structure on $k$ labels and a $B$-structure on the remaining $n-k$, choose the first label subset in $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ ways, then choose the two structures. The binomial coefficient counts how the two groups of labels interleave within the full label set. ThusThe normalization works because $1/(k!(n-k)!)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)/n!$. This explains the usefulness of exponential generating functions for labelled counting: an ordinary Cauchy product would omit the choices of labels. See Peter Magyar, Math 482, notes for 2/7, labelled product principle.

For a concrete example, there is one undecorated set on any given collection of labels. Its exponential generating function is the exponential function $E(z)=e^z={\sum }_{n\ge 0}{z}^n/n!$. Combine a red set and a blue set, allowing either to be empty:Every label independently receives one of two colours, giving $2^n$ assignments by the rule of product. With three labels, the counts by number of red labels are $1,3,3,1$, totalling eight. Choosing only the two group sizes would give the four possibilities $(0,3),(1,2),(2,1),(3,0)$; it would forget which labels are red.

Indeed, the ordinary generating function for the constant sequence $a_n=1$ is the geometric series $1/(1-z)$. Its square has coefficient $[z^n](1-z{)}^{-2}=n+1$, counting pairs of nonnegative sizes. Both encodings are correct for their respective questions. The decision is whether identity matters when objects are combined.

For unordered sets of labelled components of positive size, an exponential generating function $B(z)$ becomes $e^{B(z)}$: the term $B(z{)}^m/m!$ forgets the ordering of $m$ components. The existing rooted-tree generating function develops the richer example $T=z{e}^T$. A labelled rooted tree consists of its root of a tree and an unordered set of smaller labelled rooted trees, connecting the same label-distribution mechanism to recursive counting.

A power series is lacunary when only a sparse sequence of powers has nonzero coefficients. For ${\sum }_j{c}_j{z}^{m_j}$ with strictly increasing integer exponents, the radius of convergence satisfies $R^{-1}={\mathrm{limsup}}_j|c_j{|}^{1/m_j}$ by the Cauchy-Hadamard theorem. The exponent $m_j$, rather than the term index $j$, controls the coefficient root.

This lacunary power series has radius of convergence one. For $|z|<1$, the inequality $n!\ge n$ for $n\ge 1$ gives absolute convergence by a geometric series comparison. For $|z|>1$ its terms do not tend to zero. At every point of the unit circle the terms have modulus one, so the term test for divergence proves divergence there too. The two initial terms both equal $z$, since $0!=1!=1$; this finite repetition does not affect any convergence conclusion.

For $|x|<1$, the binomial series is

For a complex number $\alpha$, set $\left(\genfrac{}{}{0ex}{}{\alpha }{0}\right)=1$ and $\left(\genfrac{}{}{0ex}{}{\alpha }{n}\right)=\alpha (\alpha -1)\cdots (\alpha -n+1)/n!$ for $n\ge 1$. The displayed power series represents $\exp (\alpha \mathrm{Log}(1+z))$ on the unit disc, using the branch of the complex logarithm with value zero at $1$. It terminates for a nonnegative integer $\alpha$. Otherwise the ratio test gives absolute convergence for $|z|<1$. Termwise differentiation gives $(1+z)F^{\prime }(z)=\alpha F(z)$ and $F(0)=1$, which determine that holomorphic function. With $\alpha =-1/2$ and argument $-z$, its coefficients are the normalized central binomial coefficients.

Write $a_n=\left(\genfrac{}{}{0ex}{}{2n}{n}\right)/4^n$, $a_0=1$. The binomial series for $\sqrt{1-u}$ has positive numbers $b_n=a_{n-1}-a_n=\left(\genfrac{}{}{0ex}{}{2n}{n}\right)/[4^n(2n-1)]$ as the magnitudes of its nonconstant coefficients. The product $a_n={\prod }_{j=1}^n[1-1/(2j)]$ tends to zero, since $\ln a_n\le -\frac{1}{2}{\sum }_{j=1}^{n}1/j\to -\infty$. Thus $\sum b_n=1$ by telescoping. The Weierstrass M-test gives uniform convergence and absolute convergence on $[-1,1]$. Continuity extends the interior identity to both endpoints. Its radius of convergence is one; outside this interval its terms do not tend to zero.

For a nonnegative integer $L$ and $|z|<1$, this specialization of the binomial series sums the normalizing coefficients of a negative binomial distribution.

A power series $g(x)={\sum }_{\alpha }{b}_{\alpha }{x}^{\alpha }$ with nonnegative coefficients majorizes $f(x)={\sum }_{\alpha }{a}_{\alpha }{x}^{\alpha }$ when $|a_{\alpha }|\le b_{\alpha }$ for every multi-index $\alpha$. Coefficientwise majorization turns estimates for an analytic equation into estimates for a simpler positive-coefficient equation.

The radius of convergence of ${\sum }_{n\ge 0}{a}_n(z-c{)}^n$ is the unique $R\in [0,\infty ]$ separating interior absolute convergence from exterior divergence. It converges absolutely when $|z-c|<R$ and diverges when $|z-c|>R$. At the centre it always has the finite value $a_0$, since all positive powers vanish. The boundary convergence of a power series is a separate question.

For $0<R<\infty$, the interior is a disk in the complex plane and an open interval $(c-R,c+R)$ on the real line. A power series converges uniformly on every smaller closed disk $|z-c|\le r<R$. This gives stable polynomial approximation over a whole compact region, rather than a separate limiting calculation at each point. See Jiří Lebl, Basic Analysis, §11.3.2: convergence of power series.

Four examples reveal what $R$ measures:

The Cauchy-Hadamard theorem computes $R$ directly from the coefficients. The attractive feature is that one growth statistic decides convergence throughout the interior and exterior, including sparse series with many zero coefficients. Finite changes to the coefficients do not alter the radius, though they can change the sum.

Every $R\in [0,1]$ occurs as the radius of convergence of a power series all of whose coefficients are positive integers. For $0<R\le 1$, use the ceiling function to choose $a_n=\lceil R^{-n}\rceil$. ThenFor $|z|<R$, the comparison test for series with a geometric series proves absolute convergence. For $|z|\ge R$, the bound $|a_n{z}^n|\ge (|z|/R{)}^n\ge 1$ violates the fact that terms of a convergent series tend to zero. Hence the radius of convergence is exactly $R$.

For $R=0$, use $a_n=n!$: at every $z\ne 0$ the ratio of successive term absolute values is $(n+1)|z|$, eventually exceeding two. The factorial coefficients therefore give divergence away from zero.

A power series with every coefficient a positive integer cannot have radius of convergence greater than one: $a_n\ge 1$ makes convergence impossible at $z=1$. If zero coefficients are allowed and only finitely many are nonzero, a polynomial instead has infinite radius of convergence.

For a power series centred at $c$ with radius of convergence $R$, its disk of convergence is the open disk $|z-c|<R$, on which the series has absolute convergence. Boundary points may or may not belong to the larger domain of convergence. When $R=0$ this open disk is empty, although the power series still converges at its centre; when $R=\infty$ it is the whole complex plane.

Write the Maclaurin series coefficients as $a_n=f^{(n)}(0)/n!$. When the successive ratios exist eventually and have a limit, the ratio test gives the displayed radius of convergence, including zero and infinity. The unconditional formula is $R^{-1}={\mathrm{limsup}}_{n\to \infty }|a_n{|}^{1/n}$, from the Cauchy-Hadamard theorem. Endpoints must be checked separately.

The existence and convergence of this power series do not by themselves imply that it equals $f$: equality requires analyticity or control of the Taylor remainder. The smooth function $f(x)=e^{-1/x^2}$ for $x\ne 0$, $f(0)=0$, has all derivatives zero at zero, so its Maclaurin series is identically zero despite the nonzero values away from zero.

Suppose $A\ne 0$ and $R>0$, and $H$ is an analytic function on a disk of radius strictly larger than $R$. The Taylor series at zero of $f(z)=A\log (1-z/R)+H(z)$ has radius of convergence $R$. Its coefficients are the sum of $-A/(n{R}^n)$ and the more rapidly decreasing coefficients of $H$. At $x=-R$, the logarithmic part is an alternating harmonic series and converges conditionally; at $x=R$ it diverges. The power series for $H$ converges absolutely at both endpoints and therefore cannot change either conclusion. The real convergence interval is $[-R,R)$.

For $f(z)=\log (2-e^z)$, take $R=\log 2$ and $H(z)=\log [(2-e^z)/(1-z/R)]$, with its branch normalized by $H(0)=0$. The quotient has nonzero removable value $2R$ at $z=R$. Its next zeros are at $R\pm 2\pi i$, so a holomorphic logarithm exists on a larger disk. This shows why the Taylor series converges at $x=-\log 2$ even though its singular positive endpoint is excluded.

Knowing the radius of convergence does not settle convergence on its boundary. The three power seriesall have radius one, but behave differently on $|z|=1$.

For the geometric series, boundary terms all have complex modulus one, violating the term test for divergence. For the second series, absolute values give the divergent harmonic series at every boundary point. At $z=1$ the series itself is harmonic and diverges. At any other point on the unit circle, its geometric partial sums obeyThe Dirichlet test with the decreasing sequence $1/n\to 0$ therefore gives conditional convergence. In particular, $z=-1$ gives the negative of the alternating harmonic series, with sum $-\ln 2$. For the third series, the convergent p-series ${\sum }_{n\ge 1}1/n^2$ bounds the absolute values at every boundary point, giving absolute convergence by the comparison test for series.

At $z=-1$, the rational function $1/(1-z)$ is well defined and equals $1/2$, yet its geometric series diverges there. A formula for a function and a particular series representation can have different domains. Likewise, applying termwise differentiation of a power series preserves the radius but need not preserve the boundary convergence.

If the terms of a power series at a nonzero point $z$ are bounded by $M$, then $|a_n{w}^n|\le M(|w|/|z|{)}^n$. A convergent geometric series and the comparison test for series give absolute convergence at every point strictly inside that circle. Consequently terms at every point outside the radius of convergence must be unbounded. This conclusion is stronger than merely saying that those terms fail to tend to zero.

Inside both convergence discs, the triangle inequality gives absolute convergence of the sum. Suppose $R<S$ and the sum had radius $T>R$. At any modulus between $R$ and $\min (T,S)$, subtracting the absolutely convergent second series from the absolutely convergent sum would give absolute convergence of the first series outside its radius of convergence, a contradiction. Interchanging the two series covers $S<R$.

Let a power series ${\sum }_{n\ge 0}{a}_n{z}^n$ have finite radius of convergence $R$. If its terms were bounded in complex modulus at a point $z$ with $|z|>R$, say $|a_n{z}^n|\le M$, choose $w$ with $R<|w|<|z|$. ThenThe comparison test for series with a convergent geometric series would give absolute convergence at $w$, contradicting its position outside the convergence disc. Hence the terms are unbounded at every such $z$. The argument also works when $R=0$ and gives a stronger obstruction than the usual term test for divergence.

For a series ${\sum }_n{a}_n{x}^{m_n}$ with distinct increasing integer exponents, define the ordinary coefficients to be $a_n$ at $m_n$ and zero elsewhere. The Cauchy-Hadamard theorem then givesThe exponent in the root is the actual degree $m_n$, not the index $n$. For $a_n=(n!{)}^2$ and $m_n=n^2$, $0\le 2\log (n!)/n^2\le 2\log n/n\to 0$, so $R=1$. Boundary behavior is a separate question.

For a nonzero polynomial $p$, the power series ${\sum }_{n\ge 0}p(n)z^n$ has radius of convergence $1$. If $p$ has degree $d$ and leading coefficient $c\ne 0$, then $p(n)\sim c{n}^d$ and $|p(n+1)/p(n)|\to 1$, so the ratio test applies inside the unit disk. At every point of its boundary, the terms fail to tend to zero. The zero polynomial is the exceptional case with radius of convergence $\infty$.

Suppose real coefficients satisfy $c\le a_n\le C(1+n{)}^m$ for all sufficiently large $n$, where $c,C>0$ and $m\ge 0$. Then the power series $\sum a_n{z}^n$ has radius of convergence one. For $|z|<1$, the comparison test for series and the ratio test dominate it by a polynomial times a geometric sequence. For $|z|\ge 1$, the terms have modulus at least $c$ eventually and violate the term test for divergence. Finite changes to the coefficients do not affect the conclusion.

For the power series ${\sum }_{n\ge 0}{a}_n(z-c{)}^n$, the Cauchy-Hadamard theorem giveswith $R=\infty$ when the limit superior is zero, and $R=0$ when it is infinite. Away from the centre, the root test explains the formula: the limit superior of the term roots at displacement $z-c$ is $|z-c|/R$, interpreted as zero for $R=\infty$ and infinite for $R=0$. Values below one give absolute convergence; values above one imply that the terms fail to tend to zero. A value of one leaves the boundary question open. At the centre the series converges regardless of $R$.

For $a_{2n}=1$ and $a_{2n+1}=0$, the coefficient roots alternate between one and zero, so their ordinary limit does not exist. Their limit superior is nevertheless one, giving $R=1$ for ${\sum }_{n\ge 0}{z}^{2n}=1/(1-z^2)$ inside the unit disk. A ratio of consecutive coefficients would be undefined at the zero coefficients; the root-limsup formula handles the gap directly.

In the first part of his 1892 study of functions given by Taylor series, Jacques Hadamard introduced an upper-limit formulation for sequences and used coefficient roots to determine the radius of convergence. Sections 4–6 separate finite positive, zero, and infinite root-growth cases; the later parts study singularities and continuation. See Hadamard, Essai sur l'étude des fonctions données par leur développement de Taylor, Journal de mathématiques pures et appliquées, fourth series 8 (1892), pp. 107–109. This original text explains the coefficient-growth question without implying that the study of power series began there.

If $L=\mathrm{limsup}{a}_n^{1/n}$ for positive coefficients, then the coefficients $a_n^2$ have root limsup $L^2$. Coefficients $a_n^{a_n}$ instead have root termsso their radius depends on the growth of $a_n\log a_n$ relative to $n$.

If the ordinary power series ${\sum }_{n\ge 0}{b}_n{w}^n$ has radius $R\in (0,\infty )$, thenconverges for $\mathrm{Re}z<\log R$ and diverges for $\mathrm{Re}z>\log R$. Behavior on the boundary depends on the coefficients.

The Cauchy product has coefficients $c_n={\sum }_{k=0}^n{a}_k{b}_{n-k}$. Absolute convergence permits regrouping and makes its sum the product of the two sums.

Inside its radius of convergence, a power series $f(z)={\sum }_{n\ge 0}{a}_n(z-c{)}^n$ can be differentiated term by term:The differentiated series has the same radius of convergence. The conclusion concerns the interior, including uniform convergence on smaller closed disks; it does not assert that differentiated boundary series converge wherever the original did. For example, ${\sum }_{n\ge 1}{z}^n/n$ converges conditionally at $z=-1$, but its differentiated series ${\sum }_{n\ge 1}{z}^{n-1}$ does not.

Applied to the geometric series, it gives the useful identityDifferentiation turns the plain coefficients into integer weights. This is one way generating functions translate a weighted counting question into an operation on a known series. A proof controls the original and differentiated tails on smaller disks; the Weierstrass M-test provides the needed uniform bounds. See Lebl, Basic Analysis, §11.3: power-series differentiation and analyticity.

Suppose an analytic function near $c$ is not identically zero there. Its Taylor series has a first nonzero coefficient $a_m$, andThe term $a_m(z-c{)}^m$ is its leading Taylor term. Dividing by $(z-c{)}^m$ gives a limit of $a_m\ne 0$ as $z\to c$, so this term determines the leading local magnitude and angular behaviour. The remainder uses big O notation; it is smaller by one power near the centre.

For a real-valued analytic function of a real variable, an even integer $m$ gives the same nearby sign on both sides of $c$, while an odd integer $m$ gives opposite signs. For example, $f(x)=x^2(1+x)$ has leading Taylor term $x^2$ and is positive on both sides of zero sufficiently nearby; $g(x)=x^3(1+x)$ has leading Taylor term $x^3$ and changes sign there. The finite factor $1+x$ stays positive for $|x|<1$. The ordinary constant term is the leading one when $f(c)\ne 0$.