In the Black-Scholes model the terminal stock price conditional on current price $x$ is $xM$ for a positive lognormal multiplier $M$. Differentiation under the risk-neutral pricing expectation gives the displayed option gamma. Its sign follows that of $f^{\prime \prime }$, so convexity and concavity of the terminal payoff are preserved. An analogous first-derivative identity preserves monotonicity. With appropriate growth conditions, Gaussian integration by parts also gives $V_{\sigma }=\sigma (T-t)x^2{V}_{xx}$, fixing the sign of option vega for these payoff classes.

A smooth value function satisfying the Black-Scholes equation admits a self-financing portfolio with the displayed delta hedge. An arbitrary decomposition $p=xg+Bh$ need not be self-financing. For example $p=0$, $g=1$, $h=-x/B$ solves the value equation but has portfolio gains $dS-\rho Sdt$, which are not zero when volatility is positive. For specified holdings, self-financing is equivalent to the value equation together with $g=p_x$.

If the claim holder receives a cash-flow rate $k(S_t,t)$ before maturity, the self-financing hedge matches the claim's cum-dividend gain $dp+kdt$. Its stock holding is the option delta $p_x$ and its bank-account value is $p-x{p}_x$. Comparing Itô formula with portfolio gains gives the displayed equation, with terminal value equal to the terminal payoff. The positive source term $k$ represents dividends paid by the claim; dividends on the underlying stock would instead alter the stock's pricing drift. Risk-neutral pricing includes both the discounted terminal payoff and the integral of discounted interim payments.

When the underlying stock pays dividends at rate $\theta S_t$, its gain is $d{S}_t+\theta S_{t}dt$. Under the risk-neutral measure its ex-dividend drift is $(\rho -\theta )S_t$. A sufficiently smooth terminal-claim value obeys $p_t+(\rho -\theta )x{p}_x+{\sigma }^2{x}^2{p}_{xx}/2-\rho p=0$. These are dividends on the stock, not a source term paid by the claim itself.

In the Black-Scholes equation with continuous stock dividends, separation of variables gives $g^{\prime }=(r-\lambda )g$ and ${\sigma }^2{S}^2{f}^{\prime \prime }/2+(r-q)S{f}^{\prime }-\lambda f=0$. Substituting $f=S^{\alpha }$ gives the displayed eigenvalue. A stationary solution uses the two roots of ${\sigma }^2\alpha (\alpha -1)/2+(r-q)\alpha -r=0$. For positive volatility and $r>0$, their product is negative, so they are distinct real roots.

Encode $u|0⟩+v|1⟩$ as $u|000⟩+v|111⟩$. Four controlled-NOT gates copy the displayed two data parities into initially zero quantum ancillas. No logical amplitude is measured. The error syndromes $00,01,10,11$ correspond respectively to no bit flip and bit flips on data qubits $1,2,3$. Conditional Toffoli gates, with temporary Pauli X gates implementing zero-valued controls, undo the indicated error. Inverse encoding yields the original logical state and a separate syndrome state. A phase flip has zero bit-flip syndrome and remains as a logical phase error; this three-qubit code is not a general single-qubit-error code.

Assume that each physical qubit undergoes an independent quantum bit-flip channel with probability $p$ during one storage interval, and that encoding, error syndrome extraction, recovery and decoding are all ideal. Use the three-qubit bit-flip syndrome table to apply the indicated minimum-weight recovery. Zero or one bit flips are corrected. Two bit flips have the same error syndrome as the complementary one-site error: for example, $X_1{X}_2$ gives $01$, so recovery $X_3$ leaves $\overline{X}=X_1{X}_2{X}_3$. Three bit flips give $00$ and also leave $\overline{X}$. This exchanges $|000⟩$ and $|111⟩$, and therefore acts as a logical Pauli X gate after decoding.

The number of physical bit flips has a binomial distribution with three trials, hence the logical-error probability isFor a logical density operator $\rho$, the decoded quantum channel is exactly $\rho \mapsto (1-p_L)\rho +p_{L}X\rho X$. Here logical failure means a nontrivial logical Pauli X gate error; a particular input such as $|+⟩$ can be invariant under that error. At $p=0.1$, exactly two flips have probability $3(0.1{)}^2(0.9)=0.027$ and three flips have probability $(0.1{)}^3=0.001$, giving $p_L=0.028$. The block succeeds with probability $0.972$ compared with $0.9$ for an unencoded qubit. At $p=0.01$, $p_L=0.000298$, illustrating the quadratic suppression of rare independent bit flips.

Factoring the improvement givesThus one level reduces logical-error probability for $0<p<1/2$, is equal at $p=0,1/2,1$, and makes it worse for $1/2<p<1$ under this same decoder. The suppression arises because a logical error requires at least two physical bit flips. With independent disjoint inner blocks, the same noise model and ideal correction at every level, a concatenated quantum error-correcting code has $p_{\ell +1}=3p_{\ell }^2-2p_{\ell }^3$, with $p_0=p$. At $p_0=0.1$, two levels use nine data qubits and give $p_2=3(0.028{)}^2-2(0.028{)}^3=0.002308096$.

This calculation describes ideal correction of stored information with the specified quantum bit-flip channel. Repeated correction in time additionally requires a fresh specified noise model for each interval; imperfect quantum gates, faulty measurements and correlated errors change the calculation. The timed bit flip during repetition-code syndrome extraction and exact encoded-state failure with a faulty parity check show concretely why this storage calculation alone is not a fault-tolerance result.

Measure the commuting stabilizer generators $g_1=Z_1{Z}_2$ and $g_2=Z_2{Z}_3$ in that order. Use error syndrome bit $0$ for eigenvalue $+1$ and bit $1$ for eigenvalue $-1$. For $X_j$ the Pauli X gate on physical qubit $j$, every row holds for arbitrary normalized $\alpha ,\beta$. This ordered parity-check convention follows ^[ref]:

Each pair of computational basis words within a row has the same two parities. Consequently the projective measurement reveals the row without distinguishing its two amplitudes. For example, $|{\psi }_L⟩=(\sqrt{3}/2)|000⟩+(i/2)|111⟩$ becomes $(\sqrt{3}/2)|010⟩+(i/2)|101⟩$ under $X_2$. Both checks give $-1$ with probability one, and applying $X_2$ restores both coefficients exactly, including the relative phase $i$.

One implementation uses two quantum ancillas initialized to $|00⟩$. Two controlled-NOT gates from data qubits 1 and 2 to the first quantum ancilla write their parity; two more from data qubits 2 and 3 to the second quantum ancilla write the other parity. Measuring only the quantum ancillas in the computational basis gives $(s_1,s_2)$. Measuring the individual data qubits instead would reveal the encoded computational basis value and destroy the unknown quantum superposition. The existing coherent syndrome extraction for the three-qubit repetition code uses the alternative ordered checks $(Z_2{Z}_3,Z_1{Z}_3)$, so its bit labels differ from this table.

The recovery also handles coherent errors. Put $E_0=I,E_1=X_1,E_2=X_2,E_3=X_3$. Their images of the code are mutually orthogonal, giving $P{E}_a^{†}{E}_{b}P={\delta }_{ab}P$, the Knill--Laflamme condition for this family. For any linear combination $E={\sum }_{a=0}^3{c}_a{E}_a$, its component in sector $s(E_a)$ is $c_a{E}_a|{\psi }_L⟩$. Measuring that error syndrome has probability $|c_a{|}^2/{\sum }_b|c_b{|}^2$ on the normalized erroneous state, and recovery $E_a$ returns $|{\psi }_L⟩$. More generally one recovery quantum channel corrects every noise quantum channel whose Kraus operators belong to this span. The assumption concerns the whole jointly correctable family; it does not extend to the span of arbitrary single-site Pauli errors. ^[ref].

A fully computed coherent example is the unitary operator $U=\exp (-i\theta X_2)=\cos \theta I-i\sin \theta X_2$. At $\theta =\pi /6$ it givesThe error syndromes $00$ and $11$ occur with probabilities $3/4$ and $1/4$, independently of $\alpha ,\beta$. Doing nothing for $00$ and applying $X_2$ for $11$ restores the same logical qubit in both branches; the scalar $-i$ in the second branch is an irrelevant global phase. This is an example of linearity of coherent quantum error correction, rather than an assumption that the original error was a classical random bit flip.

Several zeros of a function can lie in one initial root bracket: the bisection method seeks one, rather than enumerating them. For instance, the polynomial $(x-1)(x-2)(x-3)$ has opposite endpoint signs on $[0,4]$, but its first midpoint evaluation returns only the polynomial root $2$. The initial zero of a function need not be unique for the bisection error bound to apply to a retained zero of a function.

A zero of a function without a sign change cannot be located by the strict sign-change test alone. The polynomial $(x-1{)}^2$ touches zero at $1$ but is positive on either side. On $[0,3]$ both endpoint values are positive: although the closed interval contains a polynomial root, it fails the usual starting test for the bisection method. An endpoint or evaluated midpoint zero of a function can still be recognized directly.

Opposite signs certify a root bracket only with the required continuity. For $f(x)=1/x$, the values at $-1$ and $1$ have opposite signs, but $f$ is undefined at the midpoint $0$ and has no zero of a function. Hence $[-1,1]$ is invalid input for the bisection method; the intermediate value theorem does not apply across the gap.

For a requested absolute tolerance $\epsilon >0$, stop when the retained root bracket has width at most $2\epsilon$, and report its midpoint. The bisection error bound then guarantees $|m_k-x_{\ast }|\le \epsilon$ in exact arithmetic. Using the ceiling function and the base-two logarithm, the required number of halvings isThe zero allows stopping before any update when the initial root bracket is already sufficiently narrow. For the polynomial $f(x)=x^2-2$ on $[1,2]$ with $\epsilon =10^{-3}$, $k_{\min }=\lceil {\log }_{2}500\rceil =9$. After nine halvings,Thus the reported midpoint meets the tolerance; its actual error from the square root $\sqrt{2}$ is approximately $0.000825500126905$. After eight halvings the bound is $1/512>10^{-3}$, so eight do not give this worst-case guarantee.

Cache the endpoint values. Each update of the bisection method needs just one new function evaluation, at the old midpoint. With no early zero of a function, $k$ updates therefore cost $k+2$ function evaluations, including the two initial endpoints. Returning the new midpoint requires no additional function evaluation; requesting its value would add one. The nine-update example needs eleven evaluations for the guarantee.

A small function value $|f(m_k)|$ alone does not guarantee a small position error: rescaling $f$ changes those values without changing its zeros of a function. In a practical implementation, use an explicit width test and an iteration limit. With floating-point arithmetic, also stop if the computed midpoint equals either endpoint: the root bracket can no longer shrink, so repeating the update would loop forever. Report this precision limit separately if the requested tolerance remains unmet. Reliable sign evaluations are needed to preserve the root bracket; compare signs directly rather than multiplying endpoint values, which can suffer overflow or underflow in floating-point arithmetic.

For a rounded midpoint $m$ still inside a valid root bracket, the geometric error bound is $\max \{m-a_k,b_k-m\}$; it need not equal $W_k/2$ because of roundoff error. A rigorous computed enclosure also needs conservative arithmetic for that bound and reliable endpoint signs. The exact-arithmetic bisection error bound should not be claimed unchanged after precision prevents further halving.

For practical tolerance conventions, SciPy's bisect combines absolute and relative tolerances through $|x-x_0|\le \mathrm{xtol}+\mathrm{rtol}|x_0|$, and provides a maximum iteration count and convergence status. These differ from the purely absolute width test above.^[ref]

In exact arithmetic, after $k$ completed halvings the retained root bracket has width $W_k=b_k-a_k$ and midpoint $m_k$, whereThe absolute value bound holds for every zero of a function $x_{\ast }$ in the current root bracket, including the one approached by continued bisection method updates. A zero of a function discarded with an earlier half need not satisfy later bounds. Before update $k\ge 1$, the evaluated candidate is $m_{k-1}$, whose corresponding guarantee is $(b-a)/2^k$. After update $k$, the new midpoint $m_k$ has the smaller guarantee $(b-a)/2^{k+1}$, even without evaluating $f(m_k)$.

The guaranteed error halves at every step; the actual absolute value error need not decrease at every step or have successive ratio $1/2$. In the square root example it increases from $|m_6-\sqrt{2}|\approx 0.000151062$ to $|m_7-\sqrt{2}|\approx 0.003755188$, while both errors remain below their guarantees. An exact midpoint zero of a function instead gives zero error and immediate termination. For this example $\sqrt{2}$ is an irrational number, so no finite exact halving can produce it as a midpoint.

For a brief proof outline, each retained half has half the previous width and contains a zero of a function by the intermediate value theorem. Every point in the retained root bracket is at most half its width from its midpoint. If the process does not stop at a zero of a function, the nested interval theorem gives one common point as the widths tend to zero; continuity and the opposite endpoint signs force its function value to be zero. Uniqueness in the initial root bracket is unnecessary.

For a biregular graph with $m$ left vertices of degree $r$ and $n$ right vertices of degree $s$, let $X$ be the indicator vector of a uniformly selected independent set. Use base-two information entropy. Shearer inequality gives $H(X_U)\le r^{-1}{\sum }_{w\in W}H(X_{N(w)})$. Conditional on $X_U$, each right vertex with an empty selected neighbourhood is an independent fair binary choice, and the other right vertices are forbidden. Writing $q_w=\mathbb{P}(X_{N(w)}=0)$ gives $H(X)\le r^{-1}{\sum }_w[H(X_{N(w)})+r{q}_w]$. For the $2^s$ neighbourhood patterns, assign weight $2^r$ to the zero pattern and weight one to all others. Jensen's inequality implies $H(Y)+r{q}_w\le {\log }_2(2^r+2^s-1)$. Since $n/r=m/s$, the bound follows from $H(X)={\log }_{2}i(G)$. If $s$ divides $m$, the disjoint union of $m/s$ copies of the complete bipartite graph $K_{s,r}$ attains equality: each component has $2^s+2^r-1$ independent sets.

For an attractive central force of magnitude $c/r^5$, a circle of radius $R$ through the force center has polar equation $r=2R\cos \theta$. With specific angular momentum $h$, substitution into the Binet equation gives $c=8m{h}^2{R}^2$. Consequently $R$ is inversely proportional to the magnitude of the angular momentum, and the angular-time integral gives $T=2\pi R^2/|h|\propto R^3$. This is the finite travel time along the punctured circle from one approach to the singular center to the next; the force law alone does not specify continuation through the collision.

Modulo scalar phases, the n-qubit Pauli group is the additive space ${\mathbb{F}}_2^{2n}$, with label $(a,b)$ for $X^a{Z}^b$. Multiplication adds labels, and commutation contributes the sign $(-1{)}^{\omega }$. The displayed alternating form is nondegenerate. An isotropic subspace $L$ satisfies $L\subseteq L^{\perp }$, hence $\dim L\le n$; it labels commuting stabilizers.

A binary digit is a zero or one in the base-two positional representation of an integer. The sum $s_2(n)={\sum }_j{b}_j$ determines the Thue–Morse sequence and is the Hamming weight of the binary representation.

The code consisting of all even-weight binary words. For $n\ge 1$ its dimension is $n-1$, and its weight enumerator is $[(t+s{)}^n+(t-s{)}^n]/2$, by selecting the even terms of the binomial theorem. The length-one code contains only zero.

A linear subspace $W\subseteq {\mathbb{F}}_2^n$ is self-orthogonal if $u\cdot v=0$ for every $u,v\in W$. The standard bilinear form is nondegenerate, whence $\dim W+\dim W^{\perp }=n$ and $\dim W\le \lfloor n/2\rfloor$. The condition with $u=v$ means every vector has even Hamming weight. Such a binary linear code gives the linear algebra behind the Eventown theorem.

A binary linear code is doubly even when the Hamming weight of every codeword is divisible by four.

// Target: algebra.bigb

For a binary linear code $C$, the linear functional $\sigma (x)={\sum }_j{x}_j$ records Hamming weight modulo two. Its kernel of a linear map is the even-weight subcode. If $\sigma {|}_C$ vanishes, this is all of $C$; otherwise rank-nullity theorem gives dimension $\dim C-1$. This also proves closure under addition without treating ordinary integer weight as a linear function.

A length-$n$, dimension-$k$ binary linear code has full support when no coordinate vanishes on every codeword, equivalently when the union of the supports of a vector of all codewords covers every coordinate. Every coordinate is then a nonzero linear functional on the message vector space ${\mathbb{F}}_2^k$. Translation by a vector on which it is one pairs the zero and one fibers, so exactly $2^{k-1}$ codewords contribute one in that coordinate. Summing these contributions proves the formula. If there are $r$ nonzero columns in a full-rank generator matrix, the same argument gives $r{2}^{k-1}$ instead.