Covered by the official notes: onboard acceleration from rearward photon emission (PDF pp. 7; handwritten pp. 2).^[ref]

For $u,w\in X$, the forcing terms cancel, givingThe scalar function $s\mapsto s^7$ is strictly increasing on all of $\mathbb{R}$, so the second integrand is nonnegative. For the first one, use the quartic gradient monotonicity inequality. Put $d=\xi -\eta$ and $m=(\xi +\eta )/2$ for real vectors $\xi ,\eta$. Expanding givesConsequentlyThe last strict inequality uses the Poincaré inequality: if the weak gradient of $u-w\in W_0^{1,4}(\Omega )$ vanishes, then $u=w$. This proves that $A$ is a strictly monotone operator.

For the spaces in part (b)(i) and fixed $f\in Y$, define $A:X\to X^{\ast }$ byFor each $u$, this is a linear functional in $v$, and the estimates in part (b)(i) showHence $A(u)$ belongs to the continuous dual space $X^{\ast }$, and $A$ sends bounded sets to bounded sets. The weak formulation is exactlyThe forcing term is included in $A$ so that its right side is zero, as requested.

Choose the real zero-trace Sobolev spaceThe Poincaré inequality makes this an equivalent Sobolev norm. Since the dimension is five, the Sobolev conjugate exponent is $4^{\ast }=5\cdot 4/(5-4)=20$. The Sobolev embedding theorem gives $X\hookrightarrow L^{20}(\Omega )$ continuously. Boundedness of $\Omega$ and Lp inclusion on a finite measure space then give $X\hookrightarrow L^8(\Omega )$ continuously, say $\Vert u{\Vert }_8\le C_8\Vert u{\Vert }_X$.

For $f\in Y$, the weak formulation isThe zero Dirichlet boundary condition is imposed through the Sobolev trace. Hölder's inequality verifies every term:Thus $|\nabla u{|}^2\nabla u\in L^{4/3}(\Omega ;{\mathbb{R}}^5)$, while $u^7\in L^{8/7}(\Omega )$, and all three pairings are finite bounded linear functionals in $v$. The choice $Y=L^{8/7}(\Omega )$ is sufficient; it is not intended to be the largest possible space of data.

Let $P=\{w\in L^2(\Omega ):w\ge 0\text{ almost everywhere}\}$. For $g\in L^2(\Omega )$, define $Gg=z\in H_0^1(\Omega )$ byThe Lax-Milgram theorem guarantees a unique $z$: the bilinear form on the left is bounded and coercive in the norm $\Vert \nabla \cdot {\Vert }_2$, and the right side is a bounded linear functional by the Poincaré inequality. Testing with $z$ givesDefine $T:P\to L^2(\Omega )$ by $Tw=G(1+\sqrt{w})$. This actually maps into $P$. Indeed the weak maximum principle says that a weak solution $z\in H_0^1(\Omega )$ of $-\Delta z=g\ge 0$ is nonnegative almost everywhere. Here $1+\sqrt{w}\ge 0$ and belongs to $L^2(\Omega )$. Directly, taking the negative part $z^{-}=\max \{-z,0\}$ as a test function gives $-\Vert \nabla z^{-}{\Vert }_2^2={\int }_{\Omega }g{z}^{-}dx\ge 0$, so $z^{-}=0$ by the Poincaré inequality.

The required fixed point formulation isEvery such fixed point belongs automatically to $H_0^1(\Omega )$ and satisfies part (b)(i). Conversely, uniqueness in the Lax-Milgram theorem makes every nonnegative weak solution a fixed point of $T$.

Use the real zero-trace Sobolev space $H_0^1(\Omega )$ with norm $\Vert \nabla u{\Vert }_2$. The Poincaré inequality on the bounded smooth domain gives $\Vert u{\Vert }_2\le C_P\Vert \nabla u{\Vert }_2$, making this an equivalent Sobolev norm. The weak formulation is to find $u\in H_0^1(\Omega )$ with $u\ge 0$ almost everywhere andThe zero Dirichlet boundary condition is encoded in the Sobolev trace. Formally, the identity follows by multiplying the partial differential equation by a test function and using integration by parts; it then extends to all $v\in H_0^1(\Omega )$ by density.

To check all terms, boundedness of $\Omega$ gives finite Lebesgue measure $m=|\Omega |$. The Cauchy-Schwarz inequality gives $u\in L^1(\Omega )$ andThus the right side is a bounded linear functional of $v$, since its absolute value is at most $\Vert 1+\sqrt{u}{\Vert }_2{C}_P\Vert \nabla v{\Vert }_2$. The left side is finite by the Cauchy-Schwarz inequality. These estimates work for every stated $n\ge 2$.

We first prove the compact-domain Schauder fixed-point theorem. Let $K$ be a nonempty compact convex set in a Banach space $X$, and let $T:K\to K$ be a continuous map. The Brouwer fixed-point theorem states that every continuous map of a finite-dimensional closed unit ball into itself has a fixed point. By the homeomorphism proved in part (a)(ii), it applies equally to any nonempty finite-dimensional compact convex set, including a singleton.

For $\epsilon >0$, use the supplied Schauder approximation lemma to obtain a finite-dimensional linear subspace $L_{\epsilon }$ and a continuous map $T_{\epsilon }:K\to K\cap L_{\epsilon }$ with $\Vert Tx-T_{\epsilon }x\Vert \le \epsilon$. The set $K_{\epsilon }=K\cap L_{\epsilon }$ is nonempty because it contains $T_{\epsilon }(K)$. It is a compact convex set, since a finite-dimensional linear subspace is closed. Restricting $T_{\epsilon }$ to $K_{\epsilon }$ and applying the Brouwer fixed-point theorem gives $x_{\epsilon }=T_{\epsilon }{x}_{\epsilon }$, and thusTake ${\epsilon }_j=1/j$. The compactness of $K$ yields a convergent subsequence $x_{{\epsilon }_{j_m}}\to x_{\ast }\in K$. The continuity of $T$ and the preceding estimate giveFor completeness, this also proves the version on a nonempty closed bounded convex set $C$ when $T(C)$ is relatively compact. Let $K=\overline{\mathrm{conv}}T(C)\subset C$. The closed convex hull of a compact set in a Banach space is compact. To verify the fact needed here, cover $T(C)$ by finitely many balls of radius $\delta$. Replacing every term of a convex combination by a centre changes it by at most $\delta$ in norm. The convex hull of those finitely many centres is compact in a finite-dimensional linear subspace, so it has a finite net of radius $\delta$. Thus $\mathrm{conv}T(C)$ is totally bounded. Its closure is complete and totally bounded, hence compact. Now $T(K)\subset T(C)\subset K$, so the compact-domain result supplies a fixed point in $C$.

Let $E=\mathrm{aff}K$ be the affine hull and put $k=\dim E\le n$. If $k=0$, $K$ is a singleton and is homeomorphic to ${\mathbb{B}}^0$. Suppose $k>0$. Choose $k+1$ affinely independent points of $K$ spanning $E$. Their simplex is contained in $K$ because $K$ is a convex set, and its barycentre $a$ has a neighbourhood in $E$ contained in that simplex. Identifying $E-a$ with ${\mathbb{R}}^k$, put $D=K-a$. For some $0<r\le R<\infty$,Here the inner inclusion follows from that neighbourhood, and the outer one follows from the compactness of $K$.

Use the Minkowski functionalThe inclusions imply $|z|/R\le p(z)\le |z|/r$. The fact that $D$ is a convex set implies $p(z+w)\le p(z)+p(w)$: if $z\in sD$ and $w\in tD$, then $(z+w)/(s+t)$ is a convex combination of points of $D$. Also $p(tz)=tp(z)$ for $t\ge 0$. ConsequentlyThus $p$ is a continuous function. Since $D$ is closed and contains zero, $D=\{p\le 1\}$: the forward inclusion is immediate, and for the reverse inclusion one takes a limit of $z/t_j\in D$ with $t_j\to p(z)\le 1$, using positive homogeneity and the fact that $D$ is a convex set. For $z\ne 0$, $p(z)>0$, so this limiting argument has no division by zero.

Define a radial continuous map byEach ray meets $D$ in exactly $\{s\theta :0\le s\le 1/p(\theta )\}$. Hence $h$ is a bijection from ${\mathbb{B}}^k$ to $K$. Away from zero its continuity follows from that of $p$, and at zero $|h(t\theta )-a|\le Rt$. A continuous map that is a bijection from a compact space to a Hausdorff space is a homeomorphism. ThereforeUsing the affine hull is essential: $K$ can have empty interior in the ambient ${\mathbb{R}}^n$.

For Banach spaces $X,Y$, a nonlinear compact operator $T:X\to Y$ is a continuous map that sends every bounded set to a relatively compact subset of $Y$: the closure of its image is a compact set. For a continuous map, the image condition is equivalent to requiring that every bounded sequence $(x_j)$ has a subsequence along which $T(x_j)$ converges in the norm topology of $Y$. For a linear operator, this is the usual compact operator definition; the continuity requirement is included explicitly here because the subsequent applications are nonlinear.

The compact-domain version of the Schauder fixed-point theorem saysA useful equivalent version replaces $K$ by a nonempty closed bounded convex set $C$ and assumes $T:C\to C$ is a continuous map with relatively compact image. This latter form will be used for the partial differential equation. Both forms assert existence of a fixed point, without asserting uniqueness.

The radial retraction of a punctured ball gives a retraction: project away from the removed interior point until reaching the unit sphere. Write $a=x_0$, $v=x-a\ne 0$, and setThe equation $|a+tv{|}^2=1$ is the quadratic equationIts constant term is negative, so it has exactly one positive root, namely $\tau (x)$. Since $|a+v|=|x|\le 1$, that positive root satisfies $\tau (x)\ge 1$. Thus $r(x)\in \partial {\mathbb{B}}^n$ for every $x\ne a$ in the closed unit ball.

The displayed expression is a continuous map on the punctured closed unit ball: its denominator is nonzero there, and its square root has strictly positive argument. If $|x|=1$, $t=1$ is the unique positive root, so $r(x)=x$. These are exactly the requirements for a retraction. The missing interior point is essential: the formula need not extend continuously to $a$, in agreement with the no-retraction theorem.

A retraction from a topological space $X$ onto a subset $Y$ is a continuous map $r:X\to Y$ satisfying $r(y)=y$ for every $y\in Y$. Equivalently, $r\circ \iota ={\mathrm{id}}_Y$, where $\iota :Y\hookrightarrow X$ is the inclusion.

The no-retraction theorem for the finite-dimensional closed unit ball saysIndeed, if $r$ were such a retraction, $F(x)=-r(x)$ would be a continuous map from ${\mathbb{B}}^n$ into itself. The Brouwer fixed-point theorem would give $x=F(x)=-r(x)$. Since $|r(x)|=1$, $x$ lies on the boundary, where $r(x)=x$. Thus $x=-x$, contradicting $|x|=1$. Here the Brouwer fixed-point theorem asserts that every continuous map of a finite-dimensional closed unit ball into itself has a fixed point.

The native paper prints $n\ge 1$ while specifying $x_0$ alone. This does not determine $x_1$. We use the intended convention $x_{n+1}=\sin x_n$ for $n\ge 0$; even with an independently chosen real $x_1$, the same proof gives the requested convergence.

For the sine fixed-point iteration, after one step $x_1\in [-1,1]$. On this interval, the sine has the same sign as its argument, andFor example, $t-\sin t={\int }_0^t(1-\cos s)ds>0$, since the integrand is positive for $0<s\le 1$. By oddness of the sine, $a_n=|x_n|$ consequently satisfies $a_{n+1}=\sin a_n\le a_n$ for $n\ge 1$. This monotone sequence is bounded below by zero, so it has a limit $a\in [0,1]$. The continuity of the sine gives $a=\sin a$, and the strict inequality above forces $a=0$.

Hence $x_n\to 0$, and $0$ is a fixed point of the sine. It is the only real fixed point: every such point has absolute value at most one, where the preceding strict inequality applies. ThusThe Banach fixed-point theorem does not directly apply to the sine on $[-1,1]$: the ratios $|\sin t-\sin 0|/|t-0|$ tend to one as $t\to 0$, so no uniform contraction mapping constant smaller than one exists there.

The Banach fixed-point theorem states that a contraction mapping $T:X\to X$ of a nonempty complete metric space $(X,d)$ has a unique fixed point, and iterations of a map $x_{n+1}=T(x_n)$ converge to it from every $x_0\in X$. Precisely, assume a single constant $q\in [0,1)$ satisfies $d(Tx,Ty)\le qd(x,y)$ for every $x,y\in X$.

Induction gives $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$. The triangle inequality and the geometric series therefore give, for $m>n$,Thus $(x_n)$ is a Cauchy sequence. By completeness it converges to some $x_{\ast }\in X$. The contraction mapping is a continuous map, so $T{x}_{\ast }={\lim }_{n}T{x}_n={\lim }_n{x}_{n+1}=x_{\ast }$. If $y_{\ast }$ is another fixed point, then $d(x_{\ast },y_{\ast })\le qd(x_{\ast },y_{\ast })$, forcing $x_{\ast }=y_{\ast }$. Letting $m\to \infty$ in the estimate also gives the contraction iteration error bound:If $q=0$, $T$ is constant and the sequence reaches its unique fixed point after the first step; this also removes any ambiguity about $0^0$ in the displayed estimate at $n=0$.

Let $k$ count parallel bonds. Then $N-k$ bonds are antiparallel and the energy isThe periodic boundary conditions imposeHence $N-k$ must be even. For odd $N$, $k$ is odd, so its minimum is one: one bond must be parallel. All other bonds alternate. Choosing that bond in $N$ ways and then choosing either value of $S_1$ determines every Ising spin, givingThis is an odd antiferromagnetic Ising ring: geometrical frustration forces one defect, whose location is unrestricted. The ground-state degeneracy includes both its possible locations and global spin inversion symmetry. The formula also includes the self-bond case $N=1$ under the stated Hamiltonian convention.

For $J>0$ each bond minimizes its energy when its two Ising spins are opposite, giving bond energy $-J$. An even ring allows every bond to do this simultaneously, including the closing bond required by the periodic boundary conditions. Its two ground states areThe two configurations are related by spin inversion symmetry. There is no geometrical frustration.

HARKing means presenting a hypothesis generated after inspecting results as though it had been specified beforehand. For example, after trying several possible factor structures in the lecturer data, a researcher might claim that the best-fitting structure was their original prediction. The problem is the misrepresented evidential history, not the act of discovering a pattern.

Such undisclosed selection makes apparent confirmation too persuasive because the reported test does not represent the whole search that produced the hypothesis. Label new ideas as exploratory, record prior plans through preregistration, disclose the search and evaluate the resulting prediction in independent data. Registered reports and transparent reporting can reduce incentives for that misrepresentation while preserving legitimate exploratory work.^[ref]

Preregistration creates a time-stamped record of research questions and planned methods before the relevant results are examined. Specify the design, recruitment and stopping rules, outcome definitions, exclusions, principal tests, model choices and any conditional decision rules. For existing data, disclose prior access and what was already known. A sufficiently precise record helps readers distinguish prior predictions from choices prompted by the results.^[ref]

It does not prohibit useful changes or new discoveries: report departures, their timing and reasons, and distinguish exploratory analysis from confirmatory analysis. A vague or selectively followed registration does not solve selective reporting, and preregistration cannot repair a poor measure or inadequate design of experiments. A registered report adds pre-result peer review and conditional publication acceptance, which are not implied by registration alone.

Bootstrapping estimates sampling uncertainty by repeating an analysis on resampled data. In the ordinary nonparametric bootstrap, draw observations with replacement from the empirical sample, retaining its size, and recalculate the statistic many times. The resulting distribution can estimate a standard error, assess bias or form a justified bootstrap confidence interval. It is useful when an analytic sampling distribution is difficult to obtain.

The resampling unit must match the design: keep paired measurements together, and resample independent clusters when observations are clustered. Resampling individual ratings as if they were independent lecturers would misrepresent uncertainty. The ordinary bootstrap assumes the sample and chosen resampling scheme adequately represent the relevant data-generating variation; it does not repair selection bias, guarantee accurate inference from a tiny sample, or succeed for every statistic. Specify the interval method and computational settings, since a simple percentile interval is not universally appropriate.

Use factor analysis when several observed measurements may reflect fewer unobserved dimensions, such as related questionnaire items about planning, prioritization and record keeping. A common-factor model separates shared variation represented by latent factors from item-specific variation. Factor loadings describe how measurements relate to those latent factors; communalities summarize their shared-variance contributions.

Exploratory factor analysis investigates a structure not yet fixed; confirmatory factor analysis evaluates prespecified measurement restrictions, preferably in independent data. Extraction, factor count and factor rotation affect interpretation, so report them and justify the sample and measurement assumptions. A statistically convenient factor solution does not alone establish construct validity or demonstrate the existence of a literal psychological mechanism. The distinction from principal component analysis is the common-versus-total-variation objective, and from cluster analysis the modeling of measurement relationships rather than the grouping of cases.

Use logistic regression when the response variable is binary, for example whether a lecturer missed a deadline. It relates the event's log odds to one or more predictors, producing fitted probabilities between zero and one. Exponentiating a coefficient gives an odds ratio for a specified predictor change, conditional on the other model terms. An odds ratio is not generally a ratio of probabilities; the corresponding probability change depends on the starting probability and other predictors.

Report the outcome coding, estimated associations with confidence intervals, and relevant model checks. Sparse events, complete separation and dependent observations can make ordinary estimates or inferential procedures unreliable; repeated observations may need a generalized linear mixed model. For prediction, evaluate calibration and discrimination on data not used to tune the model. An adjusted association in an observational study does not automatically identify a causal effect.