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The Bernoulli numbers are a sequence of rational numbers that turns integer power sums into exact polynomials. The same numbers describe corrections when a discrete summation is compared with an integral. For example, the sum of squares $1^2+\cdots +N^2$ is $N^3/3+N^2/2+N/6$: the leading term is an integral, while the last two terms repair the difference between area and sampled values.

We use $B_1=-\frac{1}{2}$ throughout. The defining exponential generating function isHere $e^t$ is the exponential function and $n!$ is a factorial; in particular, the coefficient of $t^2$ is $B_2/2!$, rather than $B_2$. At $t=0$ the quotient has limiting value one. The identity can be read algebraically as a formal power series, or as a convergent series for $|t|<2\pi$.^[ref]

To compute the numbers, multiply by $e^t-1$ and compare coefficients. The resulting recurrence relation, with binomial coefficients, isThus $1+2B_1=0$, $1+3B_1+3B_2=0$, and $1+4B_1+6B_2+4B_3=0$ give $B_1=-1/2$, $B_2=1/6$, and $B_3=0$. Continuing givesThe recurrence relation agrees with equation 24.5.3 after replacing its index $n$ by $m+1$.^[ref]

Why do so many values vanish? Put $F(t)=t/(e^t-1)$. Direct substitution gives $F(-t)=F(t)+t$, so $F(t)+t/2$ is an even function. Its odd power series coefficients vanish: $B_{2r+1}=0$ for every positive integer $r$. The exception $B_1$ matters. The alternative convention obtained from $t/(1-e^{-t})=F(t)+t$ has $B_1^{+}=+1/2$ and exactly the same other values. A formula copied between conventions must also retain its summation endpoints.

Historically, integer power sums were studied before the modern Bernoulli number notation. A historical study examines formulas published in 1631 and distinguishes those separate formulas from encoding all exponents using one sequence of constants. This is a useful motivation for the generating function, without assigning all discovery to a single moment. ^[ref]

For a finitely generated torsion Iwasawa module, this is a generator of its characteristic ideal, unique up to a unit of the formal power series ring. It may be chosen as a power of $p$ times a product of distinguished polynomials. The series itself and the ideal it generates are different objects; changing a generator by a unit preserves the ideal. When the finite Euler characteristic of a one-variable Iwasawa module is defined, the inverse p-adic absolute value of its constant term gives that invariant.

For an ideal $H\subseteq R[[X]]$, let $A_n$ consist of coefficients of $X^n$ in members of $H$ with all lower coefficients zero. These are ideals of $R$ and $A_n\subseteq A_{n+1}$, by multiplication by $X$. If $R$ is Noetherian, this chain stabilizes. Choose series whose leading coefficients generate the finitely many distinct coefficient ideals, then cancel coefficients successively. The accumulated multipliers are formal power series, giving an ordinary finite ideal generating set for $H$, rather than merely a dense subideal.

A formal power series with zero constant coefficient has a compositional inverse over $R$ exactly when its linear coefficient is a unit. Starting with $b_1=a_1^{-1}$, the degree-$k$ coefficient of $f(b_{1}T+\cdots +b_k{T}^k)$ is $a_1{b}_k$ plus an expression already determined by lower coefficients, so recursion gives a unique inverse. For a homomorphism of formal group laws, cancellation in the homomorphism identity proves that this inverse is also a formal-group homomorphism.

For $R={\mathbb{Z}}_p[[T]]$ and $\varphi (f)=f((1+T{)}^p-1)$, reduction modulo $p$ is the injective substitution $\overline{f}(T)\mapsto \overline{f}(T^p)$. Thus $\varphi (f)\in pR$ implies $f\in pR$. Dividing by $p$ and repeating proves the displayed equality for every $k\ge 1$. In particular, a ring congruence of Frobenius substitutions on cyclotomic power series reflects back to the original formal power series.

For an elliptic curve over a local field with integral Minimal Weierstrass equation, let $E_0(K)$ consist of points with nonsingular reduction and let $E_1(K)$ be the kernel of $E_0(K)\to {\tilde{E}}_{\mathrm{ns}}(k)$. This definition also applies at a prime of bad reduction of an elliptic curve. The point at infinity is always smooth. With $t=-x/y$ and $w=-1/y$, the Weierstrass equation of an elliptic curve becomesThe unit linear coefficient in $w$ gives a unique integral formal power series $w(t)=t^3+O(t^4)$. Then $x=t/w$ and $y=-1/w$, so the nonidentity points of $E_1(K)$ have $v(x)=-2r$ and $v(y)=-3r$ for $r=v(t)>0$. The local parameter $t$ bijects $E_1(K)$ with ${\mathfrak{m}}_K$ and turns its addition into the formal group of an elliptic curve. At good reduction this agrees with the kernel of reduction of an elliptic curve.

Writing $Y=GZ$ with an invertible matrix-valued function gives $Z^{\prime }=\Lambda Z$ exactly when $\Lambda =G^{-1}AG-G^{-1}{G}^{\prime }$. For a formal power series transformation at an irregular singular point, coefficient comparison recursively removes off-diagonal terms when leading diagonal eigenvalues are distinct. This is a differential-system change of frame, not specifically an electromagnetic gauge transformation.

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.

The ordinary generating function $C(x)={\sum }_{n\ge 0}{C}_n{x}^n$ packages the Catalan recurrence. Its Cauchy product gives the formal power series identityThe constant term is $C_0=1$. Solving the quadratic equation chooseswith the formal power series square root having constant term one. The apparent quotient at $x=0$ has constant term one after cancellation; the other quadratic solution has a pole and cannot be this ordinary generating function. The equation records the empty object or an outer unit with two ordered smaller objects.

For a plane curve defined by a homogeneous polynomial $F$ of degree $d\ge 2$, assume the field characteristic is different from two and $d-1$ is nonzero in the field. A smooth point of a variety is a flex exactly when the determinant of the homogeneous Hessian matrix vanishes there, provided its tangent line is not a component. Put the point at $[0:0:1]$ with tangent $y=0$ and $F_y(p)\ne 0$. The homogeneous identities give $F_{xz}(p)=F_{zz}(p)=0$ and $F_{yz}(p)=(d-1)F_y(p)$, so the determinant is $-(d-1{)}^2{F}_y(p{)}^2{F}_{xx}(p)$. In the completed local ring, write the curve as a formal power series $y=h(x)$. Its linear coefficient is zero and its quadratic coefficient is $-F_{xx}(p)/(2F_y(p))$. Vanishing of that coefficient is exactly intersection multiplicity at least three with the tangent, proving the criterion in these positive characteristics as well as in characteristic zero.

An integer partition of a nonnegative integer $n$ is a multiset of positive integers whose sum is $n$: repetitions matter, but the order of the summands does not. Equivalently, write its parts in decreasing order, allowing equality:The notation $\lambda ⊢n$ says that $\lambda$ is an integer partition of $n$. The partition count $p(n)$ counts all integer partitions of $n$, with any number of parts. Zero has one empty integer partition, so $p(0)=1$. The partition of an integer article develops the same objects in representation theory of the symmetric group.^[ref]

Computed example. The seven integer partitions of five are exactlyThus $p(5)=7$. Sorting each list of positive summands into decreasing order prevents repetitions in this enumeration. Grouping by largest part verifies completeness: largest parts $5,4,3,2,1$ contribute respectively $1,1,2,2,1$ possibilities. For largest part $3$, the remaining $2$ is either $2$ or $1+1$; for largest part $2$, the remaining $3$ uses only $2$ and $1$, giving $2+1$ or $1+1+1$.

An integer partition keeps sizes without named positions. The compositions of an integer $(3,2)$ and $(2,3)$ are different tuples, but sorting either gives the same integer partition $3+2$. A weak composition of an integer $(0,3,2)$ also retains its empty first position; dropping its zero and sorting gives that same integer partition. Conversely, a set partition records which distinct elements belong to each block. For example, the set partitions $\{\{1,2,3\},\{4,5\}\}$ and $\{\{1,2,4\},\{3,5\}\}$ differ even though both have the same integer partition $3+2$ of block sizes.

The Ferrers diagram makes an integer partition into a shape. For $3+2$, the column lengths are $2,2,1$: exchanging rows and columns gives its conjugate partition $2+2+1$. Exchanging again recovers $3+2$, so this operation is a bijection. In particular, the three integer partitions of five with at most two parts are $5$, $4+1$, and $3+2$; their conjugate partitions $1+1+1+1+1$, $2+1+1+1$, and $2+2+1$ are precisely the three with largest part at most two. This visible symmetry explains the equal counts for these two restrictions. The square-cell version is the existing Young diagram.

Counting by a generating function. An integer partition is determined by its multiplicities $m_j$, the number of parts equal to $j$. Its size condition is ${\sum }_{j\ge 1}j{m}_j=n$. Consequently its ordinary generating function isProof outline. Choose the term $q^{j{m}_j}$ from the $j$th geometric series. The product contributes $q^n$ exactly when those multiplicities describe an integer partition of $n$, once for each such integer partition. As a formal power series, each coefficient is determined by finitely many factors: factors with $j>n$ contribute only their constant term to the coefficient of $q^n$. For $n=5$, multiplication through $j=5$ gives $1+q+2q^2+3q^3+5q^4+7q^5$ up to degree five, independently recovering $p(5)=7$. Restricting the factors gives the generating function for partitions with bounded parts. This is useful because restrictions on summands become choices of factors, while conjugate partitions turn restrictions on numbers of rows into restrictions on row lengths.^[ref]

The Laurent series field $k((t))$ consists of formal series ${\sum }_{n\ge N}{a}_n{t}^n$ with finitely many negative-degree terms. Its $t$-adic valuation is the least exponent with nonzero coefficient.

Newton's identities let two different summaries of the same polynomial roots talk to each other. Elementary symmetric polynomials record products of distinct root positions; power sums record sums of individual powers, $p_r={\sum }_{i=1}^n{x}_i^r$. Either can be easier to obtain in a particular problem. The identities turn one into the other, often avoiding any need to find the individual polynomial roots.

Write $e_0=1$ and $e_m=0$ for $m>n$. For every positive integer $m$, the identity isIt is a polynomial identity over the integers, so it remains true on substitution in any commutative ring. To solve it for $e_m$ by division, however, $m$ must have a multiplicative inverse. Over a field of characteristic zero, the first $n$ power sums recover every elementary symmetric polynomial. The same recovery works when the positive characteristic of a field is greater than $n$; small characteristic needs care.^[ref]

The first three relations are $e_1=p_1$, $2e_2=e_1{p}_1-p_2$, and $3e_3=e_2{p}_1-e_1{p}_2+p_3$. When division by two and three is permitted, this becomesFor the three roots $1,2,3$, their power sums are $p_1=6$, $p_2=14$, and $p_3=36$. The formulas give $e_2=(36-14)/2=11$ and $e_3=(216-252+72)/6=6$, recovering the polynomial $z^3-6z^2+11z-6$ from aggregate data.

The beauty is that the summaries do not depend on the order of the polynomial roots. In symmetric polynomials, one can work with quantities that survive every relabeling; in linear algebra, traces of matrix powers give the same kind of aggregate information about eigenvalues. The identities convert those summaries into coefficients of a characteristic polynomial.

A short derivation packages the identities into a formal power series. Let $E(t)={\prod }_i(1+x_{i}t)={\sum }_{m\ge 0}{e}_m{t}^m$. Since its constant coefficient is one, it has an inverse as a formal power series, and its formal derivative satisfiesMultiply by $E(t)$ and compare the coefficient of $t^{m-1}$. This gives Newton's identities without a convergence assumption or division by $m$. The division issue arises only when using them to recover $e_m$. The logarithmic-derivative approach is also developed in Newton's Identities (January 2019 notes) by Milan Mossé (§4, pp. 6–8).

The alternate name Newton-Girard identities recalls Albert Girard's earlier work. A modern historical account credits formulas through sixth powers to Albert Girard in 1629 and discusses their later appearance in Isaac Newton's Arithmetica universalis.^[ref] This citation refers to that account; the original historical editions have not been inspected here.

This is the power-series specialization of a generating function.

The ordinary generating function of a sequence $(a_n{)}_{n\ge 0}$ is the formal power series $A(z)={\sum }_{n\ge 0}{a}_n{z}^n$. Algebraic identities for $A$ encode recurrences and counting constructions for the coefficients.

At $2$ there is good reduction, and the surjectivity of good reduction over a local field gives an exact sequenceThe quotient has order $5$ and is consequently a cyclic group. For every odd integer $n$, multiplication by $n$ is an automorphism of $E_1({\mathbb{Q}}_2)$: its formal group law has multiplication series $nt+O(t^2)$ with unit linear coefficient, whose inverse formal power series has integral coefficients and converges on $2{\mathbb{Z}}_2$. This is the prime-to-residue-characteristic multiplication on a formal group.

Thus the odd-primary torsion subgroup injects into the order-five quotient. It also realizes that quotient: choose a lift $P$ of any reduced point, so $[5]P\in E_1$. There is a unique $Q\in E_1$ with $[5]Q=[5]P$. Then $P-Q$ is killed by five and has the prescribed reduction. This proves the prime-to-p torsion lifts uniquely at good reduction directly here. Therefore, for an odd prime $p$,The five-primary subgroup consists of the unique five-torsion lifts of the five reduced points. These lifts need not be rational over $\mathbb{Q}$; the preceding answer about rational torsion points of an elliptic curve does not contradict their existence over the p-adic field.

Work near the identity $O$ of a generalized Weierstrass equation of an elliptic curve. On the projective chart where $Y\ne 0$, use the local parameter $t=-x/y$ and the auxiliary coordinate $w=-1/y$. Substitution givesThe coefficient of $w$ in the linearized equation at $(0,0)$ is one. Consequently, successive coefficient comparison determines a unique formal power series $w(t)$, without dividing by any coefficient of the equation. Its initial terms areRecover the affine coordinates as Laurent series $x=t/w(t)$ and $y=-1/w(t)$; in particular $x=t^{-2}+O(t^{-1})$ and $y=-t^{-3}+O(t^{-2})$.

The elliptic curve group law is a regular morphism near $(O,O)$. Applying the local parameter to the sum and passing to completed local rings producesOne can construct the series without any nonunit division. PutThe divided difference is an integral formal power series, including its tangent line specialization at $T_1=T_2$. The line $w=\lambda t+\nu$ meets the cubic in the two chosen points and a third point $R$. Substitution in the equation shows that its cubic and quadratic coefficients in $t$ are $-A,-B$, whereSince $\lambda$ has total degree at least two and $\nu$ at least three, $A$ is a unit. The sum of the roots therefore givesThe last expression reflects the third intersection according to the chord-and-tangent group law. Both denominators have constant term one. This proves integrality of the universal construction and also provides an explicit coefficient algorithm. Expanding it, the resulting formal group of an elliptic curve beginsThe integral coefficient recursion and the unit-denominator addition calculation work universally over $\mathbb{Z}[a_1,a_2,a_3,a_4,a_6]$; thus these coefficients specialize in every characteristic of a field. This is the initial coefficients of the elliptic formal group at infinity construction, rather than a definition requiring division by the elliptic-curve discriminant.

The identity, commutativity and associativity of the elliptic curve group law give the defining identities of a one-dimensional commutative formal group law:Reflection in the generalized Weierstrass equation of an elliptic curve gives its formal inverse explicitly:Successive addition gives the multiplication series $[n{]}_F(T)=nT+O(T^2)$.

In characteristic zero, an especially useful description comes from the invariant differential on an elliptic curve:Integrating with zero constant term constructs the formal logarithmTranslation invariance of the invariant differential on an elliptic curve implies $L(F(X,Y))=L(X)+L(Y)$. Thus the formal logarithm converts the formal group law into the formal additive group in characteristic zero. Expanding this identity to degree three also recovers the displayed coefficients of $F$; in positive characteristic the integral addition construction remains valid although the logarithmic denominators may not exist.

The product defining the modular discriminant converges locally uniformly for $|q|<1$ and has integral Fourier coefficients. Its logarithmic derivative isTherefore $D\Delta =E_2\Delta$. Comparing the coefficient of $q^n$ gives $n\tau (n)=\tau (n)-24{\sum }_{j=1}^{n-1}{\sigma }_1(j)\tau (n-j)$, orThis also fixes the initial value $\tau (1)=1$.

For the Ramanujan tau congruence modulo five, first justify the link between this product and $E_4,E_6$. Question 1 constructs ${\Delta }_0=(E_4^3-E_6^2)/1728\in S_{12}$ with first coefficient one. Its Serre derivative is in $S_{14}$, which vanishes by the valence formula for the modular group: a nonzero such cusp form would contribute at least $1+1/2>14/12$, since it vanishes at infinity and at $i$. Hence $D{\Delta }_0=E_2{\Delta }_0$. The product and ${\Delta }_0$ have the same differential equation and leading coefficient. Equivalently their quotient has derivative zero near $q=0$ and limit one; the identity theorem givesAlso $D_6{E}_6\in M_8=\mathbb{C}{E}_4^2$. Its constant coefficient is $-1/2$, so the Serre derivative gives $2D{E}_6=E_2{E}_6-E_4^2$.

Work now in the formal power series ring ${\mathbb{F}}_5[[q]]$. The integral expansions give $E_4\equiv 1$ and $E_2\equiv E_6(\mathrm{mod}5)$, because $d^5\equiv d(\mathrm{mod}5)$, $-24\equiv -504\equiv 1$, and $240\equiv 0$. The two identities reduce toThus $3\Delta \equiv -2D{E}_6\equiv 3D{E}_6$, and division by the nonzero element $3$ gives $\Delta \equiv D{E}_6$. Since the coefficient of $q^n$ in $D{E}_6$ is $-504n{\sigma }_5(n)$, we conclude