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For $X=m$, the first $m-1$ digits must all differ from $3$ and the next digit must be $3$. Independence givesThus $X$ has the geometric distribution counting trials up to and including the first success, with success probability $1/3$. For completeness, its mean follows from the tail-sum formula:The sum of the geometric series also proves finiteness of this expectation.

For the first recurrence relation, the master theorem compares $n^2$ with $n^{{\log }_{2}6}$. Since $2<{\log }_{2}6$, the recursive leaves dominate and the answer is $\Theta (n^{{\log }_{2}6})$. More explicitly, the work at depth $j$ is $n^2(6/4{)}^j$, a growing geometric series that stops at depth ${\log }_{2}n-O(1)$, where its sum has that order.

For the second recurrence relation, each step subtracts two and adds one until reaching a base case. For integral $n>4$ there are $\lceil (n-4)/2\rceil$ steps, so $T(n)=1+\lceil (n-4)/2\rceil$ and the answer is $\Theta (n)$.

The third set of printed requirements is inconsistent. The specified base values and its recurrence relation give $T(n)=1$ identically, as strong induction immediately verifies: $4\cdot 1-4\cdot 1+1=1$. In particular $T(8)=1$, while the extra condition demands $T(8)>2T(4)=2$. Thus no function satisfies all the printed conditions. Dropping that extra inequality gives $\Theta (1)$.

The likely intended growth calculation can nevertheless be made precise, without treating it as a solution of the contradictory data. On dyadic arguments put $U_j=T(2^j)$. Its equation has the general formbecause $U_j-1$ has characteristic equation $(r-2{)}^2=0$. The printed values $U_1=U_2=1$ force $A=B=0$. With suitably different initial values, the condition $U_j>2U_{j-1}$ for arbitrarily large $j$ instead requires $B{2}^j-1>0$, hence $B>0$ and $U_j=\Theta (j{2}^j)$, or $T(n)=\Theta (n\log n)$. That last order describes corrected initial data, not the paper as printed. All three bullets belong to this single marked part.

Let $q=1-p$. By the definition of a probability generating function and the geometric series,The series converges absolutely when $|qs|<1$, equivalently $|s|<1/(1-p)$.

A nonempty binary tree in which every level, including the deepest, contains all its possible vertices. Equivalently, it is a full binary tree whose leaves are all at the same graph distance from the root of a tree.^[ref]

Every perfect binary tree is both a full binary tree and a complete binary tree; the five-vertex example above shows that the converse fails even when both properties hold. These definitions fix the terminology used here.

For height of a binary tree $h\ge 0$, the level sizes are $1,2,4,\dots ,2^h$, giving $n=2^{h+1}-1$ by a geometric series. There are $2^h$ leaves. The seven-vertex example has level sizes $1,2,4$ and height of a binary tree $2$; the single-vertex binary tree is already perfect, with $h=0$. More generally, joining two perfect binary trees of equal height of a binary tree under a fresh root of a tree gives the next such shape. Equal heights are essential: fullness alone allows unequal leaf distances. A perfect binary tree realizes the maximum number of vertices for its height of a binary tree, making the exponential capacity behind short lookup graph paths visible.

// Target: probability-and-statistics.bigb

If $f$ is integrable and $2\pi$-periodic and $n$ is a positive integer, then ${\int }_0^{2\pi }f(nx)e^{imx}dx=0$ for every integer $m$ with $0<|m|<n$. Split the integral into $n$ equal intervals and substitute $y=nx-2\pi j$ to obtainThe root of unity sum vanishes by the finite geometric series formula. The zero mode is $\int f$, so a mean-zero dilated function is orthogonal to all trigonometric polynomials of degree at most $n-1$.

Planck's law gives the spectral radiance of an ideal blackbody in thermal equilibrium at absolute temperature $T>0$. It predicts how much power is carried in each small frequency or wavelength interval, per projected area and per solid angle. The whole curve is determined by $T$: no material-specific parameter enters this ideal Planck function.

Writing $\nu$ for frequency in hertz and $\lambda$ for wavelength in metres, the two forms of the Planck function areHere $h$ is the Planck constant, $k_B$ the Boltzmann constant, and $c$ the speed of light in vacuum. Their exact values in the International System of Units are $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, and $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$. ^[ref]

The units matter: $B_{\nu }$ is measured in $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\mathrm{H}{\mathrm{z}}^{-1}$, whereas $B_{\lambda }$ is measured in $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}{\mathrm{m}}^{-1}$. Both are spectral radiance, rather than radiant exitance or energy density. To report $B_{\lambda }$ per micrometre, multiply its per-metre value by $10^{-6}$; a micrometre-wide interval is $10^{-6}$ metres wide.

For a computed example, at $T=6000\mathrm{K}$ and $\lambda =0.500\mu \mathrm{m}$, the exponent is $hc/(\lambda k_{B}T)\approx 4.79592$. Substitution into Planck's law gives $B_{\lambda }\approx 3.17569\times 10^{13}\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}{\mathrm{m}}^{-1}$, or $3.17569\times 10^7\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$ per micrometre. At the same wavelength, the illustrative temperatures $3000\mathrm{K}$ and $4500\mathrm{K}$ give $2.60268\times 10^5$ and $6.37793\times 10^6$ in the latter units. These are ideal predictions for stipulated temperatures.

The key change of variables is $\nu =c/\lambda$. Corresponding positive intervals carry the same power, so $B_{\lambda }d\lambda =B_{\nu }|d\nu |$ andEqual-width wavelength intervals have unequal widths in frequency; the factor $c/{\lambda }^2$ compensates for this. Merely replacing the horizontal-axis labels of $B_{\nu }$ by $c/\nu$ does not produce $B_{\lambda }$. This explains the frequency and wavelength peaks of Planck's law.

A short derivation uses the Boltzmann distribution for a single electromagnetic mode with thermal energies $nh\nu$, $n=0,1,\dots$. Put $q=e^{-h\nu /(k_{B}T)}$. The geometric series gives $Z={\sum }_{n\ge 0}{q}^n=(1-q{)}^{-1}$ and mean thermal energy $h\nu q/(1-q)$. The electromagnetic mode density in a box, including both independent polarization of an electromagnetic wave states of a photon, is $8\pi {\nu }^2/c^3$ per volume per unit frequency. Multiplication gives spectral energy density $u_{\nu }=8\pi h{\nu }^3/[c^3(e^{h\nu /(k_{B}T)}-1)]$; isotropy gives $B_{\nu }=c{u}_{\nu }/(4\pi )$. This counts thermal excitation energy, excluding zero-point energy. ^[ref]

Planck's law connects quantum mechanics to a measurable radiation curve. Its small-$h\nu /(k_{B}T)$ limit is the Rayleigh-Jeans law, $B_{\nu }\approx 2k_{B}T{\nu }^2/c^2$; its large-$h\nu /(k_{B}T)$ limit is the Wien approximation, $B_{\nu }\approx (2h{\nu }^3/c^2)e^{-h\nu /(k_{B}T)}$. The suppression at high frequency makes the total energy density finite. Integrating the curve gives the Stefan–Boltzmann law, while locating its peak gives Wien's displacement law. Thus one formula predicts both the spectral shape and the total emission.

For $0\le r<1$, this Poisson kernel for the unit disk has the uniformly convergent Fourier seriesThe identity follows by taking the real part of the geometric series $1+2{\sum }_{n\ge 1}(r{e}^{i\theta }{)}^n$. For integrable $2\pi$-periodic boundary data $f$, its Poisson extension iswhere $a_n,b_n$ are the usual Fourier coefficients. Each term is a harmonic function in the open disk. For continuous boundary data, the extension solves the Dirichlet problem with those boundary values.

For $0<r<1$, the geometric series gives $P_r(t)=1+2{\sum }_{n\ge 1}{r}^n\cos nt$. Hence this positive even kernel has normalized integral one, and normalized convolution with it multiplies the $n$th Fourier coefficient by $r^{|n|}$. Its mass outside any fixed neighbourhood of zero tends to zero as $r\uparrow 1$. If an integrable function has finite one-sided limits $A,B$ at zero, evenness gives half the mass to each side and the convolution tends to $(A+B)/2$.

For $|z|<1$, $a\ge 0$ and integer $N\ge 0$, truncating the defining polylogarithm power series after $N$ terms has remainderThe inequality uses monotonicity of $k^a$ and then the geometric series. The bound applies to negative real $z$ as well as positive real $z$, and gives a rigorous absolute truncation tolerance independently of cancellation. It does not bound floating-point rounding error.

For a compact convex set $K\subset \mathbb{C}$ and $\zeta \notin K$, strict separation gives a unit vector $u$ with $\mathrm{Re}(\bar{u}\zeta )>{\max }_K\mathrm{Re}(\bar{u}z)$. Taking $c=-Ru$ for large $R$ makes $q={\max }_K|z-c|/|\zeta -c|<1$. The polynomialhas uniform error at most $q^{N+1}/(|\zeta -c|(1-q))$. This elementary geometric series argument proves exterior-pole approximation without a general complex approximation theorem.

Every fixed power is eventually smaller than an exponential function with base $a>1$: for a fixed nonnegative integer $k$, $n^k/a^n\to 0$. To prove this, the ratio of successive terms is $(1+1/n{)}^k/a\to 1/a<1$, so after finitely many terms the ratio is bounded by a fixed $q<1$. A geometric series bound then proves the limit. In particular $n^k=O(a^n)$ in big O notation.

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.

Scale the offered load and capacity together. Reversing the denominator of the Erlang loss formula givesFor each fixed $j$, the summand tends to $(C/\nu {)}^j$. If $\nu >C$, the dominated convergence theorem applies with the summable geometric series bound $(C/\nu {)}^j$. Otherwise each fixed finite partial sum has limiting value at least its number of terms, so the reciprocal diverges. Taking reciprocals proves the formula, including the critical case $\nu =C$.

A selection algorithm that uses array partitioning around a pivot and recursively searches only the side containing the desired order statistic. If there are $s$ entries before the placed pivot, an earlier rank is unchanged and a later rank is reduced by $s+1$. Segment length strictly decreases, proving algorithm termination. The same reduction can update local bounds in a loop, using constant auxiliary space.

Balanced reductions give linear work by a geometric series, while repeatedly discarding only a constant number of entries can give quadratic work. For distinct entries and independent uniform random pivots, a pivot in the middle half has probability at least one half and reduces the retained length to at most three quarters, up to rounding. The expected number of attempts for each such reduction is bounded, so summing expected work over geometric size scales gives $O(n)$ expected time. This random-pivot guarantee is not a worst-case guarantee for a deterministic median-of-three heuristic. With duplicates and a strict two-way partition, random pivot choice alone can also fail to shrink an all-equal segment; a three-way partition treats equal keys together.

An encoding for radix sort must make the intended total order of sort keys agree with the lexicographic order of its fixed-width numerical digit tuples. The positional notation for bounded nonnegative integers has this property. At the most significant differing position $m$, a one-unit numerical digit difference contributes $R^m$, while all lower positions together contribute at mostThis geometric series shows why lower numerical digits cannot reverse that comparison. Leading zero padding works here because it preserves place values and the represented integer.

For signed integers in a known integer interval $L\le x\le U$, encode $e(x)=x-L$ and choose $d$ with $U-L<R^d$. This order-preserving function is also injective: $x<y$ exactly when $e(x)<e(y)$. For $-8\le x\le 7$, the four-bit encodings of $-8,-3,-1,0,4,7$ are $0000,0101,0111,1000,1100,1111$ after adding eight. Restore the original signed sort keys when displaying the complete sorting records; the encoding need not replace their data. Sorting the unsigned values of raw two's complement bit patterns would put nonnegative integers before negative ones, so their representation alone is not this ordered encoding.

For variable-length strings, first specify the alphabet order and whether a proper prefix precedes its extensions. As an original construction, use $a<b$, let every string have length at most three, encode $a$ as one and $b$ as two, and append zeros to width three. Zero is reserved for an end marker strictly below every real symbol. The strings $\epsilon ,a,aa,ab,b$ encode as $000,100,110,120,200$ in radix three, which preserves their lexicographic order. A differing real symbol decides the comparison; at a prefix boundary, the shorter string's zero comes before the extension's one or two. Left padding would instead encode $b$ as $002$ and $aa$ as $011$, reversing $aa<b$. Padding raw bytes with zero also fails if zero is an allowed data symbol: distinct strings can then receive the same encoding. A reserved end marker and an explicit symbol ordering are essential; byte order alone does not establish a desired human text order.

The reciprocal chart makes infinity an ordinary coordinate origin. Its domain is the Riemann sphere minus the point $z=0$; $w(\infty )=0$ and $w(z)=1/z$ elsewhere. A neighbourhood $U_R$ of infinity becomes the open disc $|w|<1/R$. In sphere coordinates this is exactly the capFor $R=2$, the exterior $|z|>2$ together with infinity becomes $Z>3/5$ and $|w|<1/2$. The boundary itself is excluded from each open set.

The point $z=2+i$ lies in this neighbourhood because $|z|=\sqrt{5}>2$. Its reciprocal coordinate is $w=(2-i)/5$, with $|w|=1/\sqrt{5}<1/2$, and its sphere height is $Z=2/3>3/5$. The negative imaginary part of $w$ is the axis reversal required by the holomorphic stereographic atlas of the sphere.

For a complex-valued holomorphic function $f$ defined outside a large disc, having a removable singularity at infinity means that $g(w)=f(1/w)$ extends to a holomorphic function at $w=0$. This definition reduces questions at infinity to familiar questions at a finite point.^[ref] For example,The geometric series gives a Taylor series at the new origin, so $f(\infty )=1$ and $g^{\prime }(0)=-5$. The coefficient $-5$ is the derivative in the specified $w$ coordinate. In contrast, $f(z)=z^2$ becomes $g(w)=w^{-2}$, displaying a pole of order two at infinity when the target is the complex plane. As a holomorphic map to the Riemann sphere, it uses the target reciprocal coordinate $1/f(1/w)=w^2$, which is regular at zero.

The positive tail after term $n$ in the reciprocal-factorial series for Euler's number satisfies $1/(n+1)!<e-{\sum }_{i=0}^{n}1/i!\le (n+2)/((n+1)(n+1)!)$. The upper bound replaces each subsequent denominator factor by $n+2$ and sums the resulting geometric series; the lower bound retains only the first omitted term. This supplies an explicit error certificate and, with Stirling's formula, a term count of order $d/\log d$ for $d$ decimal digits. The arithmetic evaluation also has its own rounding error, distinct from this truncation bound.

For $y_{tt}=c^2{y}_{xx}$ on $0<x<L$, initially at rest with zero displacement, apply $y(0,t)=a$ for $t>0$ and keep $y(L,t)=0$. A geometric series in the Laplace transform givesHere $H$ is the Heaviside step function, $L,c>0$, and only finitely many terms contribute at any finite time. The second step in each pair is a sign-reversed reflection from the fixed endpoint. The formula solves the wave equation classically away from the fronts and distributionally across them; the midpoint is a square wave of period $2L/c$.

For $n\ge 3$ cyclically ordered planar vertices, let $e_j=v_{j+1}-v_j$, with indices modulo $n$. They describe a convex regular polygon exactly when $e_0\ne 0$ and there is a single choice of sign for which every successive edge vector is obtained by a rotation through $2\pi /n$ in that orientation. Necessity follows from equal sides and equal exterior angles. Conversely, the edge vectors have the form $R^j{e}_0$. Their sum is zero because the corresponding geometric series runs through the $n$ distinct roots of unity. Summing its partial series shows that the vertices are a translated and scaled copy of those roots, so they form a convex regular polygon. A larger turn such as $4\pi /5$ can instead trace a star, which is why merely checking a constant turning angle is insufficient.

For an Elitzur-Vaidman bomb tester using a rotation and its inverse, let $p$ be the probability of entering the absorbing path and $q=1-p$. Conditional on a live absorber, one trial has explosion probability $p$, successful safe identification probability $pq$, and inconclusive probability $q^2$. The inconclusive projective measurement resets the input, so summing a geometric series gives the displayed success probability after at most $N$ trials. Its limit is $q/(1+q)$, approaching one half as $p$ tends to zero through positive values. Exactly zero coupling produces no information.