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The Banach space ${\ell }^1$ consists of absolutely summable scalar sequences, with norm $\Vert x{\Vert }_1={\sum }_n|x_n|$. Its continuous dual is ${\ell }^{\infty }$, under the pairing ${\sum }_n{x}_n{y}_n$. Finitely supported sequences are norm dense, which identifies a linear functional from its values on coordinate vectors. The space of sequences converging to zero inside this dual is a norming subspace of infinite codimension.

For example, $x=(2^{-1},2^{-2},\dots )$ belongs to the absolutely summable sequence space and has norm one, by the geometric series formula. Merely tending to zero is insufficient: $(1/n)$ tends to zero but its series of absolute values diverges. Thus this sequence space records total absolute size, rather than just the eventual size of individual coordinates. Its completeness ensures that a Cauchy sequence in this total-error norm has an absolutely summable limit.^[ref]

For a discrete-time Markov chain and a set $A$ of absorbing states, its absorption time is the hitting time $T_A=\inf \{m\ge 0:X_m\in A\}$. It is zero when the initial state already lies in $A$, and infinite on a sample path that never enters $A$. Finiteness almost surely need not imply a finite expected value. For a finite-state model, a uniform positive probability of absorption within a fixed number of steps gives a geometric series bound on the survival probabilities and proves that the mean is finite. The mean absorption time of a lazy symmetric random walk is an explicit example.

For real $z>0$, defineThe integrand is positive and at most $e^{-t}$, so $0<F(z)<1$ and the improper integral converges. Substituting $s=z+t$ also gives $F(z)=z{e}^z{E}_1(z)$ in terms of the exponential integral; the original representation makes positivity and the large-$z$ behavior transparent.

Apply the exact finite geometric series identity inside the integral, without assuming that an infinite series can be integrated term by term:The Gamma integral gives ${\int }_0^{\infty }{e}^{-t}{t}^{k}dt=k!$. Therefore, with $N$ retained terms,This remainder formula proves the asymptotic expansion at every fixed order: $R_N=O(z^{-N})=o(z^{-N+1})$ as $z\to +\infty$. Its sign also gives alternating bounds: an odd $N$ overestimates $F(z)$, and an even $N$ underestimates it.

At $z=10$, an independent evaluation of the integral gives $F(10)\approx 0.9156333394$. Three retained terms giveFive retained terms give $S_5(10)=0.9164$ with error bounded by $5!/10^5=0.0012$. The actual errors are about $0.004367$ and $0.000767$, respectively. The error bound is conservative, but established without relying on the numerical integral evaluation.

Nevertheless, the formal infinite series diverges at every fixed $z>0$. The magnitudes $T_k=k!/z^k$ satisfy $T_{k+1}/T_k=(k+1)/z$, so they eventually increase without bound. In particular, the terms fail the necessary zero-term condition for a convergent series. The finite remainder identity remains valid after convergence has failed; this is precisely what makes the example useful.

Aggregate analysis bounds the total cost of a whole sequence directly, then divides by its length to obtain an amortized bound. Summing the copied capacities as a geometric series in dynamic array doubling is an example.

An analytic continuation extends a holomorphic function beyond its original domain without changing its existing values. Formally, if $U\subset V$ are domains and $f$ is holomorphic on $U$, a holomorphic function $F$ on $V$ with $F{|}_U=f$ is an analytic continuation. More locally, holomorphic functions on overlapping open sets continue one another when they agree on the overlap, allowing a chain of patches along a path. The identity theorem for holomorphic functions forces agreement throughout a connected overlap once they agree on a nonempty open set in it. It also makes an extension to a fixed connected domain unique. ^[ref]

The geometric series illustrates why a formula can reach farther than the power series that first represents it. For $z\ne 1$, its partial sum isWhen $|z|<1$, $z^{N+1}\to 0$, so the series converges to $f(z)=1/(1-z)$. The rational function $F(z)=1/(1-z)$ is holomorphic on $\mathbb{C}\setminus \{1\}$ and agrees there with the sum on the unit disc. It therefore continues $f$ to every point except its simple pole at $1$. In particular,Thus analytic continuation gives a value at $2$ without making the original geometric series converge there. Its radius of convergence remains $1$: on $|z|\ge 1$ the terms $z^n$ do not even tend to zero.

The pole at $1$ limits the original Taylor series, although most points of its convergence circle are regular for $F$. Recentring at $-1$ gives another genuinely convergent power series,At $z=-3/2$ this yields $\frac{1}{2}{\sum }_{n\ge 0}(-1/4{)}^n=2/5$, even though the original powers of $-3/2$ grow. This is the useful idea: overlapping local descriptions can represent one holomorphic function on a larger region.

Uniqueness on a connected overlap does not imply that every path leads to the same terminal germ of a holomorphic function. Continuing a complex logarithm around $0$ changes its value by $2\pi i$, giving monodromy; the return patch can carry a different branch of a multivalued function. If two patches have a disconnected overlap, agreement on one component imposes no agreement on the others. ^[ref] Continuation using the rational function $1/(1-z)$ is single-valued: any path avoiding $1$ gives the same terminal germ of a holomorphic function, even if it winds around the pole.

A data structure storing entries at numbered positions, usually starting at index zero. The fixed-capacity representation considered here puts equal-sized slots consecutively in computer memory. The number $n$ of occupied slots can be smaller than the capacity; unused slots provide room for later entries. An array of references stores consecutive references, while the objects they identify can be elsewhere. ^[ref]

Let the first slot have abstract address $b$ and each slot occupy $s$ address units. Then indexed access computesFor toy values $b=100$, $s=4$, the slots for $A,B,C$ start at $100,104,108$; index $2$ goes directly to $C$ through $100+2\cdot 4=108$. These are illustrative slot addresses, not observed allocation addresses. Under the word-RAM (random-access machine) model for time complexity, word-sized arithmetic operations and any addressed computer memory access have constant cost, and the addresses and entries fit in words. Thus one array access takes $O(1)$ time in Big O notation, independently of the index. This is the access assumption needed by binary search. ^[ref]

Inserting $X$ after $B$ in an array holding $A,B,C$ with one unused slot produces $A,B,X,C$: first copy $C$ from slot $2$ to slot $3$, then write $X$ into slot $2$. Writing $X$ first would lose $C$ unless it had been saved separately. More generally, insert at index $i$ by copying occupied slots from right to left, then fill the gap; with spare capacity this moves $n-i$ entries and takes $O(1+n-i)$ time in Big O notation, so insertion near the middle has $O(n)$ time complexity. Removing an entry closes the gap by shifting later entries left. ^[ref]

A fixed-capacity array cannot accept another entry when full. A resizable representation can allocate a larger array and copy the entries. Doubling capacity when full gives $O(1)$ append cost by amortized analysis over a sequence starting empty, but a single append that copies $n$ entries still has $O(n)$ worst-case time complexity. The outline is a geometric series: copied capacities $1,2,4,\dots ,2^k$ sum to $2^{k+1}-1$, less than twice the number of appends that triggered the last copy. Middle insertion still pays for its shifts. ^[ref]

An asymptotic expansion describes a function near a specified limit by successively smaller scales. Each fixed finite truncation has a remainder smaller than its last retained scale. The central question is how the approximation improves as the parameter approaches the limit; convergence of the entire infinite series at a fixed parameter is a separate question. For the inverse-power scale $1,z^{-1},z^{-2},\dots$,means that, for every fixed nonnegative integer $m$,The little o notation makes the accuracy requirement precise. With expansions to all orders, this is also equivalent to an $O(z^{-m-1})$ remainder after the $m$th term; the implied bound may depend on $m$.^[ref]

For a quick convergent example, the finite geometric series identity givesFor $z>1$, both the asymptotic expansion as $z\to +\infty$ and the ordinary convergent geometric series describe this function. The exact remainder displays the required extra inverse power.

The more surprising example under a divergent factorial series can approximate a convergent integral has a convergent integral but a divergent series at every fixed positive $z$:A few terms can estimate $F(10)$ well, even though adding sufficiently many terms eventually destroys the approximation. Optimal truncation explains where to stop.

This makes asymptotic expansions useful when an exact expression is a complicated integral or differential equation solution, but a parameter is very large or very small. The leading term reveals the scale; corrections reveal which effects matter next. The asymptotic sequence must be specified: integer powers, fractional powers, logarithms, and exponential functions can encode different kinds of behavior. A good finite approximation includes a remainder estimate or an independently checked range of use.

For the stationary causal AR(1) with $|\varphi |<1$, matching terms in its infinite moving-average representation gives, for $h\ge 0$,Symmetry of the autocovariance extends this to negative lags. Dividing by $\gamma (0)$ gives the autocorrelation $\rho (h)={\varphi }^{|h|}$. The geometric series explains both the enlarged stationary variance and the diminishing memory: all past disturbances contribute to the present, but their squared weights have a finite sum. At $\varphi =0$, the process reduces to shifted white noise and all nonzero-lag autocorrelations vanish. See Bartlett, lecture 4: AR(1) representation and mean-square convergence.

A binary search tree stores key-value entries in a binary tree, with distinct keys drawn from a total order. At every vertex, all keys in its left subtree must be smaller than its own key, and all keys in its right subtree must be larger. Thus the entries represent a partial function from keys to values: a key has at most one associated value and may be absent. ^[ref]

Lookup starts at the root of a tree. Compare the target with the current key: equality returns its value, a smaller target continues into the left subtree, and a larger target continues into the right subtree. Reaching an empty subtree reports absence. The ordering rule justifies discarding the other subtree at every step; this is the algorithm correctness argument for lookup. It applies to the entire subtree, even if that subtree contains many vertices. ^[ref]

In the seven-key example, search for $5$. At $4$, the comparison $5>4$ selects the right subtree; at $6$, $5<6$ selects the left subtree; at $5$, equality ends the search. The visited keys are $4\to 6\to 5$, three key comparisons. The right-child chain stores exactly the same keys and satisfies the same binary search tree rule, but lookup of $5$ visits $1\to 2\to 3\to 4\to 5$, five comparisons. Searching for absent key $8$ visits $4,6,7$ in the first binary search tree and all seven keys in the chain before reaching an empty right subtree.

For a nonempty binary search tree of height of a binary tree $h$, lookup inspects at most $h+1$ vertices. Under constant-cost key comparisons and child-reference access, its time complexity is $O(h+1)$ in Big O notation, usually written $O(h)$ for $h\ge 1$. With $n$ entries, $h\le n-1$, giving $O(n)$ worst-case lookup. If the shape is kept balanced so that $h=O(\log n)$, lookup takes $O(\log n)$ for $n\ge 2$. The empty and one-entry cases take constant work. The worst-case linear bound also appears in Sedgewick and Wayne's ordinary, unbalanced implementation. ^[ref]

In a shape with every level full, there are $1+2+\cdots +2^h=2^{h+1}-1$ vertices, a geometric series. This explains how a small height of a binary tree can accommodate many entries. The binary search tree ordering rule alone does not guarantee this shape or logarithmic time complexity. Breadth-first search also gains no logarithmic key-lookup guarantee merely from having a binary tree: in the displayed balanced example, its left-to-right level order is $4,2,6,1,3,5,7$, so it inspects six keys before finding $5$, and an absent target requires all seven. Directed lookup uses the total order to choose one branch.

The ordering requirement includes every ancestor on the graph path from the root of a tree. Consider the smaller binary tree with root key $4$, left child $2$, right child $6$, and right child $3$ under $2$. Key $3$ must satisfy both $2<3$, from its immediate parent in a rooted tree, and $3<4$, from its position in the root's left subtree. Replacing $3$ by $5$ preserves all immediate parent-child comparisons, including $2<5$, but violates $5<4$. Lookup for $5$ would incorrectly go from $4$ to $6$ and then to an empty left subtree, overlooking the stored key.

An inorder traversal visits a left subtree, then its own vertex, then a right subtree. On the valid small example its output is $2,3,4,6$; on the invalid example it is $2,5,4,6$, exposing the descent from $5$ to $4$. Both seven-key binary search trees yield $1,2,3,4,5,6,7$. The theorem is that an inorder traversal of a finite binary search tree with distinct keys lists every key once in strictly increasing total order. For a short proof outline, use mathematical induction on the number of vertices: each smaller subtree produces its own sorted list, and the full ordering rule places every left key before the current key and every right key after it. Concatenating these three parts preserves strict order. Sedgewick and Wayne's keys method uses this left-current-right order. ^[ref]

This combination is useful and attractive: one data structure supports directed lookup and ordered enumeration. An inorder traversal costs $O(n)$ time complexity for $n$ entries, since it visits every vertex; balance accelerates lookup but cannot eliminate the work of outputting all keys.

// Destination: computer-science.bigb

Black-body radiation is the electromagnetic radiation in thermal equilibrium with an ideal blackbody. It can be described as a photon gas: a cavity whose walls are all at the same temperature contains radiation with the universal Planck's law spectrum. A small opening in such a cavity approximates a blackbody emitter because incoming radiation undergoes repeated absorption inside. The radiation's equilibrium spectrum depends on temperature, rather than the wall material. ^[ref]

There are three useful quantities to keep distinct: directional spectral radiance $B_{\nu }$ or $B_{\lambda }$, total energy density $u$ in the cavity, and outward radiant exitance $j^{\ast }$ from its opening. Planck's law supplies the first; integrals over frequency and directions supply the others. For isotropy, every direction has the same $B_{\nu }$, so the full $4\pi$ solid angle givesHere $\hslash =h/(2\pi )$ is the reduced Planck constant, and the radiation constant is $a_{\mathrm{rad}}\approx 7.565733250\times 10^{-16}\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-4}$. The explicit Boltzmann constant makes this an SI formula; adopting natural units with $k_B=1$ recovers $u={\pi }^2{T}^4/(15{\hslash }^3{c}^3)$.

For a short integral outline, substitute $x=h\nu /(k_{B}T)$ and useThe expansion is a geometric series with nonnegative terms; repeated integration by parts evaluates each integral, and $\zeta (4)={\pi }^4/90$ is a value of the Riemann zeta function. This gives the coefficient above.

The photon gas also exerts radiation pressure, $p=u/3$. The factor $1/3$ comes from the angular average of the squared directional component of photon momentum, $⟨{\cos }^2\theta ⟩=1/3$. The thermodynamic derivation of blackbody radiation pressure gives the same result. In contrast, outgoing power involves one projected-direction factor over an outward hemisphere, yielding radiant exitanceThis is the Stefan–Boltzmann law with unit emissivity; $\sigma$ is the Stefan-Boltzmann constant. The factor $\pi$ follows from Lambert's cosine law, rather than the $4\pi$ full-sphere factor used for energy density.

For example, an ideal $6000\mathrm{K}$ cavity has $u\approx 0.980519\mathrm{J}{\mathrm{m}}^{-3}$, radiation pressure $p\approx 0.326840\mathrm{Pa}$, and outward radiant exitance $j^{\ast }\approx 7.34881\times 10^7\mathrm{W}{\mathrm{m}}^{-2}$. A stipulated $3000\mathrm{K}$ cavity has one-sixteenth each value because all three scale as $T^4$. These describe stored energy, mechanical pressure, and outgoing power per area, respectively. In a closed equilibrium cavity, equal opposing radiation balances the directional power flow even though $u$ and $p$ are nonzero.

The usefulness of black-body radiation is this common reference: Planck's law, Wien's displacement law, and the Stefan–Boltzmann law consistently connect spectral measurements, temperature, and total emission. The three illustrative curves in Planck's law are ideal equilibrium spectra; applying them to a material requires considering its emissivity and whether thermal equilibrium is a suitable approximation.

For $\mathrm{Re}s>1$, expansion of $(e^x-1{)}^{-1}$ as a geometric series and termwise integration give

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.

Expand each factor as a geometric series indexed by a nonnegative matrix. The RSK correspondence preserves its row and column contents, changing the sum to pairs of Semistandard Young tableaux of the same shape and proving the displayed formal identity. Its kernel is also ${\sum }_{\lambda }{h}_{\lambda }(x)m_{\lambda }(y)={\sum }_{\lambda }{p}_{\lambda }(x)p_{\lambda }(y)/z_{\lambda }$. These are dual-basis kernels for the Hall inner product of symmetric functions, so the Schur expansion proves orthonormality. Applying the symmetric-function involution in one alphabet gives the dual identity ${\prod }_{i,j}(1+x_i{y}_j)={\sum }_{\lambda }{s}_{\lambda }(x)s_{{\lambda }^{\prime }}(y)$.

If $\mathbb{E}|X|<\infty$, the characteristic function $\varphi (u)=\mathbb{E}{e}^{iuX}$ satisfies ${\varphi }^{\prime }(0)=i\mathbb{E}X$. The bound $|(e^{iuX}-1)/u|\le |X|$ and the dominated convergence theorem justify differentiation at zero. For mean zero, $\psi (u)=|1-\varphi (u)|/|u|$ extends continuously by $\psi (0)=0$, and $|1-\varphi (u{)}^n|\le n|u|\psi (u)$ by a finite geometric series. This supplies a direct characteristic-function proof of the weak law of large numbers.

// Target: martingale.bigb

For $N$ bonds of length $a$, fixed forward angle $\theta$ and uniform independent torsions, conditional averaging gives $⟨{\mathbf{a}}_i\cdot {\mathbf{a}}_j⟩=a^2{\alpha }^{|j-i|}$. Consequently $⟨R^2⟩=N{a}^2+2a^2{\sum }_{k=1}^{N-1}(N-k){\alpha }^k$. Dividing by the independent-step result $N{a}^2$ and summing the geometric series gives the displayed $C_N$, for $\alpha \ne 1$. For fixed $|\alpha |<1$, $C_N\to C_{\infty }=(1+\alpha )/(1-\alpha )$.

The angle is measured between forward vectors; a chemical internal bond angle is supplementary. At $\alpha =1$, the finite chain is a rigid rod with $C_N=N$, so taking the infinite-chain limit before the rigid limit would be invalid. The Kuhn length of a freely rotating chain is asymptotically $a{C}_{\infty }$.

For a complex semisimple Lie algebra and an integer $k\ge 0$, the Weyl character formula and Weyl denominator formula giveApply the dilation $e^{\mu }\mapsto e^{(k+1)\mu }$ to the denominator identity to obtain the numerator. Cancellation leaves a finite geometric series for each positive root of a root system. Evaluating the resulting polynomial at all formal exponentials equal to one gives the dimension, without taking an undefined quotient at the identity.

A Chebyshev series expands a function in Chebyshev polynomials $T_j$. If $|c_j|\le M{q}^j$ for $0<q<1$, then $|T_j(x)|\le 1$ on $[-1,1]$ gives uniform convergence and the tail bound $\Vert f-{\sum }_{j=0}^d{c}_j{T}_j{\Vert }_{\infty }\le M{q}^{d+1}/(1-q)$ by the geometric series.

For noninteracting particles in a three-dimensional isotropic quantum harmonic oscillator, measure the one-particle ground energy as zero. The single-particle partition function is $Z_1=(1-e^{-\beta \hslash \omega }{)}^{-3}$, from the product of three geometric series. In the dilute Maxwell–Boltzmann statistics regime, $\ln \Xi =e^{\beta \mu }{Z}_1$, so the grand potential is the displayed expression and $N=e^{\beta \mu }{Z}_1$. For $k_{B}T\gg \hslash \omega$,The same approximation follows from integrating the density of states $g(E)\simeq E^2/[2(\hslash \omega {)}^3]$ against $e^{-\beta E}$. Extrapolating the classical formula until the fugacity is one gives the scale of Bose condensation in a three-dimensional harmonic trap, but the quantitative transition temperature requires the Bose occupation law rather than Maxwell-Boltzmann statistics.

Choose $N$ equal-amplitude, monochromatic odd components with orders $q_j=q_0+2j$, $j=0,\dots ,N-1$, and odd $q_0$. Fix the complex-field conventionHere $I$ is a normalized light intensity after averaging the fast carrier oscillations. Setting every ${\varphi }_j=0$ and summing a geometric series givesAt the removable singularities take the continuous limit. Maxima repeat every $T/2=\pi /\omega$; the first zeros on either side of a maximum lie at $t=\pm T/(2N)$. The zero-to-zero width is $T/N$, which is not the full width at half maximum.

Choose orders $13,15,17,19,21$ and driving wavelength $\lambda =800\mathrm{nm}$. Using $T=\lambda /c$ with the speed of light $c$ gives $T=2.66851\mathrm{fs}$ and burst spacing $1.33426\mathrm{fs}$. Solving the exact half-maximum equation gives a full width at half maximum of approximately $240.6$ attoseconds. These are values of the specified five-component model, not measurements from the 2001 experiment.

A deliberately chosen spectral phase ${\varphi }_q=q\omega t_0$ translates the whole attosecond pulse train by $t_0$ without changing its shape. Adding $\kappa (q-17{)}^2/2$ makes neighboring phase increments unequal: the same component intensities then produce a different temporal profile. For the choice $t_0=0.20\mathrm{fs}$ and $\kappa =0.08$ radians per squared harmonic order, the adjacent-component delay estimates arenamely $0.0981$, $0.1660$, $0.2340$ and $0.3019\mathrm{fs}$ for $q=13,15,17,19$. This finite difference estimates the group delay at the pair's midpoint. It is not a single common arrival time.

The infinite monochromatic sum is periodic forever and has infinite total energy. A real attosecond pulse train has a finite envelope and finite component linewidths. This calculation isolates coherent addition; it cannot describe an isolated burst or establish its contrast against satellites. As a further check, averaging $I(t)$ over one half-period gives $N|a{|}^2$ for any constant component spectral phases, because the cross terms integrate to zero by Fourier orthogonality. Spectral phase redistributes the light intensity in time while leaving that average fixed.