MATH40004 2026 Full Paper H

QUESTION 1 — COMPLETE WORKED SOLUTION **(a)** If \(f\) has the required derivatives near \(a\), then \[ f(x)=\sum_{k=0}^{4}\frac{f^{(k)}(a)}{k!}(x-a)^k+R_4(x), \] where, for example under a \(C^5\) hypothesis, \[ R_4(x)=\frac1{4!}\int_a^x(x-t)^4f^{(5)}(t)\,dt \] (or the equivalent Lagrange/Peano local remainder). **(b)** Put \(h=x-1\). The fourth-order expansion is \[ \log(1+h)=h-\frac{h^2}{2}+\frac{h^3}{3}-\frac{h^4}{4}+O(h^5),\quad |h|<1. \] The stated neighbourhood/radius must be checked before substituting a new point. **(c)** In the three regimes listed in the question, the limits are \[ +\infty,\ 1,\ 0. \] The quotient has a removable singularity and is real analytic after setting g(1)=1. **(d)** Near 0, \(g(x)/(1+x^2)\sim-\log x\), which is integrable; near 1 the point is removable; at infinity it is \(O(\log x/x^3)\). The ordinary improper integral converges absolutely. Completeness and checks. State the Taylor theorem and remainder before using it, compute coefficients at the printed centre, and state the interval of convergence. Each limit must be labelled endpoint or internal and justified by an expansion, comparison or a correctly applicable L'Hopital step. Define the removable value before discussing derivatives of the extension. For an improper integral, split at every singular endpoint or internal point and at infinity; give a local equivalent or an explicit dominating function on each piece. Common errors: differentiating an undefined quotient, using one global comparison across different singularities, or confusing a formal series with its interval of convergence. SCORING AND EQUIVALENT METHODS. Each labelled subpart is split into method/conditions marks and conclusion/verification marks in the published marking scheme. Algebraically equivalent formulae, a different valid theorem route, or a different correct drawing order receive the same credit. A consistent arithmetic slip is followed through; it does not erase earlier correct mathematics.

QUESTION 2 — COMPLETE WORKED SOLUTION **(a)** Conservation of volume gives \[ 3\dot h=-2\sqrt h-h. \] The coefficient multiplying \(h\) has dimensions area/time. With \(u=\sqrt h\), \[ dt=-\frac{6\,du}{2+u}, \] so the actual two-loss emptying time is \[ T=6 \log\left(1+\frac{2}{2}\right). \] **(b)** After multiplying the porous loss by \(\lambda\), \[ 3\dot h=-2\sqrt h-\lambda h,\qquad T_\lambda=\frac{6}{\lambda} \log\left(1+\frac{\lambda\sqrt{4}}{2}\right). \] As \(\lambda\to0\), \[ T_\lambda\longrightarrow T_0 =\frac{6\sqrt{4}}{2} =6. \] Since \(\log(1+x)<x\) for \(x>0\), \(T_\lambda<T_0\): the added porous loss makes the tank empty sooner. **(c)** For the signal of height \(A=2\), \[ a_0=2,\qquad a_n=0,\qquad b_n=\frac{2}{\pi n}\bigl(1-(-1)^n\bigr). \] At a jump the series equals the midpoint \(A/2\). At \(x=\pi/2\), \[ \frac\pi4=1-\frac13+\frac15-\frac17+\cdots. \] Completeness and checks. Start with accumulation equals inflow minus outflow, so every outward volume rate has a negative sign. Convert V to h before separating, preserve the initial condition, and check every coefficient against length cubed per time. The emptying time is the first time the non-negative height reaches zero; the formula is not continued physically afterwards. When a loss parameter tends to zero all other physical data remain fixed. For the Fourier part, state the coefficient convention, compute the integrals, use the midpoint value at jumps, and evaluate only at a point whose convergence value is known. Common errors: losing the area factor, changing two parameters in a limiting comparison, or substituting the one-sided value at a jump. SCORING AND EQUIVALENT METHODS. Each labelled subpart is split into method/conditions marks and conclusion/verification marks in the published marking scheme. Algebraically equivalent formulae, a different valid theorem route, or a different correct drawing order receive the same credit. A consistent arithmetic slip is followed through; it does not erase earlier correct mathematics.

QUESTION 3 — COMPLETE WORKED SOLUTION **(a)** For \(A=\begin{pmatrix}1&-4\\1&-1\end{pmatrix}\), direct calculation gives \[ A^2=-3I,\ \lambda=\pm i\sqrt3,\quad X_h=[I\cos(\sqrt3t)+(A/\sqrt3)\sin(\sqrt3t)]C; the origin is a neutrally stable centre. \] **(b)** The forcing is \((e^t,2)^T\). The scalar rate is not an eigenvalue in this case, so no resonance multiplier is required. Solving \[ (I-A)p=\binom10,\qquad Aq=-\binom0{2} \] gives \[ p=\binom{1/2}{1/4},\qquad q=\binom{-8/3}{-2/3}. \] Thus \(X=X_h+pe^t+q\). When the exponential rate is zero, the two constant particular terms may of course be combined. **(c)** With \[ M=2x^{2}y,\qquad N=x^{3}+6xy^{2}, \] the original cross-partials are unequal. On \(x>0\) or \(x<0\), the integrating factor \(\mu=x^{-1}\) gives \[ \widetilde M=2x^{1}y,\qquad \widetilde N=x^{2}+6y^{2}, \] whose cross-partials agree. Integrating yields \[ \Phi(x,y)=x^2y+2y^3=C. \] Completeness and checks. Show the characteristic polynomial, eigenvalues and enough eigenvector work to reconstruct a real fundamental matrix. Its determinant or direct differentiation verifies independence. Stability uses the real parts of eigenvalues, while rotation orientation comes from one explicit vector-field evaluation. In the forced problem, test kI-A before choosing an exponential ansatz and add the homogeneous solution afterwards. For the differential form, identify M and N, compare cross-partials on the stated domain, derive the integrating factor, integrate one component and differentiate the recovered potential as a final check. Common errors: classifying from 'complex' alone or accepting an integrating factor without verification. SCORING AND EQUIVALENT METHODS. Each labelled subpart is split into method/conditions marks and conclusion/verification marks in the published marking scheme. Algebraically equivalent formulae, a different valid theorem route, or a different correct drawing order receive the same credit. A consistent arithmetic slip is followed through; it does not erase earlier correct mathematics.

QUESTION 4 — COMPLETE WORKED SOLUTION **(a)** The vector field factors as \[ F_r(y)=\frac{y(1r+(r-1)y)}{1+y}. \] The singular state is \(y=-1\). The equilibria are \(y=0\) and, for \(r\ne1\), \[ y_*=\frac{1r}{1-r}. \] Their linearisation derivatives are \(F_r'(0)=r\) and \(F_r'(y_*)=r(r-1)\). Thus \(0\) is stable for \(r<0\) and unstable for \(r>0\); \(y_*\) is stable for \(0<r<1\) and unstable for \(r<0\) or \(r>1\). The branches exchange stability in a transcritical bifurcation at \(r=0\). At \(r=1\) the second branch escapes through infinity. The singular line must remain a barrier on the diagram. **(b)** With \[ \widehat u(\xi)=\int_{\mathbb R}u(x)e^{-i\xi x}\,dx, \] \[ \mathcal F(u'')=-\xi^2\widehat u,\qquad \mathcal F(2xu)=2i\widehat u', \] so \[ -\xi^2\widehat u-2i\widehat u'=0,\qquad \widehat u=C\exp\left(\frac{i\xi^3}{6}\right). \] Pairing positive and negative frequencies gives the real oscillatory integral \[ u(x)=\frac C\pi\int_0^\infty \cos\left(\xi x+\frac{\xi^3}{6}\right)\,d\xi, \] equivalently a scaled Airy \(Ai\) solution selected by decay. **(c)** At \((0,1,0)\), \[ F_x=1,\qquad F_y=0,\qquad F_z=-1\ne0. \] Hence the implicit-function theorem applies, \[ z_x=1,\qquad z_y=0,\qquad z(x,y)=x+ o\!\left(\sqrt{x^2+(y-1)^2}\right). \] Completeness and checks. Solve F(y;r)=0 without multiplying through a denominator until the singular state has been recorded separately. Split the parameter axis at every collision, pole or change of sign. Stability must be shown by F_y or a phase-line sign table and encoded consistently: solid for stable, dashed for unstable, and a distinct guide for a singular barrier. Under the declared Fourier convention, transform differentiation and multiplication by x with their correct signs, solve the resulting first-order frequency equation, and justify a real oscillatory inverse representation. For the implicit part verify F_z≠0 at the base point, calculate the partial derivatives, and display the full first-order polynomial. Common errors: joining branches through a pole or invoking the implicit theorem before checking its non-degeneracy condition. SCORING AND EQUIVALENT METHODS. Each labelled subpart is split into method/conditions marks and conclusion/verification marks in the published marking scheme. Algebraically equivalent formulae, a different valid theorem route, or a different correct drawing order receive the same credit. A consistent arithmetic slip is followed through; it does not erase earlier correct mathematics.