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Source: released examination bundle, Part A 2007 11-1.pdf, pages 121–124. All nine questions are covered. No mark allocations are printed.

Source: released examination bundle, Part A 2007 11-1.pdf, pages 190–194. All nine questions are covered. No mark allocations are printed.

Source: released examination bundle, Part A 2007 11-1.pdf, pages 259–264. The printed title is Algebra, Analysis and Differential Equations I. The paper is cross-listed for mathematics and mathematics and statistics and contains no printed mark allocations.

This paper was printed as Algebra, Analysis and Differential Equations I and was shared by mathematics and mathematics and statistics. Source: released examination archive, member FOI-20210708-1 - Data/4301 (1).pdf, pages 1–4. The examination took place on 18 June 2012. No mark allocations are printed. The solutions cover all nine questions.

Source: Mathematics Part A release, member FOI-20210708-1 - Data/4301.pdf, dated 17 June 2013. All 35 source leaf parts are answered below.

The department's solution-release policy excludes ordinary odd years; bounded exact-paper checks found no matching advertisement of full or selected worked answers. No parts are skipped.

Source: Mathematics Part A release, member FOI-20210708-1 - Data/4322.pdf, dated 26 June 2014. All thirteen source leaf parts are answered below.

Bounded exact-paper checks found no matching advertisement of worked answers. The department's release policy establishes even-year releases back at least to 2016 without specifically confirming this paper; the answer archive requires Oxford SSO. No source parts are skipped.

Source: Mathematics Part A release, member FOI-20210708-1 - Data/a13755w1.pdf. The complete two-page paper is dated 25 June 2015. All fourteen source leaf parts are answered below; the examination statements remain in the private archive.

The department's solution-release policy does not advertise ordinary odd-year solutions. An exact-paper check found no matching advertisement of full or selected worked answers for this paper, so no source parts are skipped.

Source release: Oxford examination archive, member FOI-20210708-1 - Data/4320.pdf. The cover dates this examination to 18 June 2014, 2.30–5.30pm. The six questions have 36 leaf parts, answered in source order below.

Source release: Oxford examination archive, member FOI-20210708-1 - Data/a13753w1.pdf. The cover dates this examination to 17 June 2015, 9.30am–12.30pm.

All six questions and their 38 leaf subparts are answered below. The unnumbered criterion at the start of 2(a) has its own solution before the four field cases.

Every parameter-free definable set is preserved setwise by every automorphism of a first-order structure. For atomic formulas, preservation follows from the definition of a structure automorphism. Logical connectives preserve that equivalence, and a bijection transports existential and universal witnesses in both directions. Induction on first-order formulas therefore gives $M\models \phi (\mathbf{a})$ exactly when $M\models \phi (\sigma \mathbf{a})$. Applied to a defining formula, this proves the displayed equality. With parameters from $B$, restrict to automorphisms fixing $B$ pointwise. An automorphism moving a proposed singleton or taking a proposed function-graph point outside the graph is consequently a direct obstruction to definability.

For $ϵy^{\prime \prime }-y^{\prime }+e^y=0$ with $y(0)=A<0$ and $y(1)=0$, the outer solution is $-\log (e^{-A}-x)$. The right-end boundary layer has width $ϵ$ and leading correction $\log (e^{-A}-1)e^{(x-1)/ϵ}$. This follows by setting $X=(x-1)/ϵ$, solving $Y^{\prime \prime }-Y^{\prime }=0$, and matching its constant part. The leading correction changes sign at $A=-\log 2$.

A fixed order in which Boolean variables may be tested along each branch of an ordered binary decision diagram. Branches may skip a test, but cannot reverse the order. The order can change the number of distinct remaining Boolean cofactors and therefore the size of the representation.

An ordered binary decision diagram with shared equal terminal nodes, no decision node having identical zero and one children, and no two decision nodes having the same tested Boolean variable and the same ordered pair of children. For a fixed variable ordering of a binary decision diagram, it is unique up to labelled graph isomorphism: each nonconstant remaining Boolean cofactor has a uniquely determined first essential Boolean variable, and its two children are recursively determined by its zero and one Boolean cofactors. This uniqueness does not compare different orderings.

Given a bounded-complete one-consistent arithmetic first-order theory with a false theorem, enumerate first-order sentences and successively choose one-consistent one-sentence extensions. Every stage has Sigma-1 soundness, and a false Sigma-1 theorem in the union would already have a formal proof at one stage. The union is therefore one-consistent and complete while retaining the false theorem. It cannot be omega-consistent, since a complete bounded-complete omega-consistent first-order theory is sound.

A bounded-complete complete omega-consistent arithmetic first-order theory is correct on every arithmetic first-order sentence. Completeness and first-order consistency make its theoremhood obey the Boolean truth tables. Omega-consistency ensures that every proved existential sentence has a provable standard numeral instance, since otherwise completeness refutes all such instances. Induction on logical formulas now identifies theoremhood with standard truth.

Name every integer and add a new constant asserted to be prime and unequal to every named integer. Every finite part is satisfied by an ordinary prime outside a finite list, using Euclid's theorem. Compactness theorem with the elementary diagram gives an elementary extension with the desired prime. A countable elementary substructure retaining that prime cannot be isomorphic to the integer ring, since any unital ring isomorphism fixes every standard integer.

Polarizing the displayed Lie algebra identity gives $[x,[y,z]]+[y,[x,z]]=0$. Antisymmetry shows that the three cyclic terms in the Jacobi identity are all equal. Their sum is zero, so three times each term is zero. Away from characteristic three they therefore vanish, giving nilpotency class at most two. No division by two occurs, so this argument also applies in characteristic two for alternating Lie brackets.

If $S$ is the natural cubic spline interpolating data $y_i$ at fixed distinct knots $x_i$, then $aS(x)+b+cx$ is the natural cubic spline interpolating $a{y}_i+b+c{x}_i$.

The transformed function is piecewise cubic, preserves continuity of the first two derivatives, and has endpoint second derivatives $a{S}^{\prime \prime }=0$. It takes exactly the transformed nodal values. Uniqueness of the natural spline therefore proves the claim. This permits affine changes in data or the addition of a linear trend without solving a new spline system.