Original public practice · A-1 · October 2026

Build a connected analysis argument.

Ten multipart questions bring together limits, series, calculus and function spaces. The form has 100 practice points.

This is original supported practice, not an official university paper, a calibrated grade prediction or a complete examination of every analysis topic. It samples the declared outcomes below. Further source reconciliation and independent learner evidence remain necessary.

Attempt, review and revise

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For an optional rehearsal allow 210 minutes: 60 for sequences and series, 60 for continuity and calculus, and 90 for uniform limits and spaces. These are author estimates, not measured learner timings. Breaks, separate sessions and accommodations are welcome; no timer is enforced.

Use exact reasoning. State the domain, quantifiers and hypotheses of every theorem. Results in linked lessons may be cited unless the question asks for their proof. A plot, finite computation or theorem name cannot replace a requested argument. Other valid proofs can earn the same criteria.

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1. Sequences and series

30 practice points.

Question 1: Transport a supremum and recover its approximations

10 points

Let A be a nonempty bounded subset of [0,∞), let s=sup A, and set φ(t)=t/(1+t) for t≥0. Let B={φ(a):a∈A}. Use the least-upper-bound property and explicit algebraic estimates; do not merely cite continuity of φ. Suggested time: 20 minutes.

(a) · 3 points

Prove sup B=s/(1+s). Prove also that B has a largest element if and only if A does, identifying the corresponding elements. Include s=0.

(b) · 3 points

Choose aₙ∈A with s−1/n<aₙ≤s and define cₙ=max(a₁,…,aₙ). Prove cₙ is a member of A, increases to s, and φ(cₙ) increases to sup B with error less than 1/n. Explain why this argument does not assume that A is closed or that s∈A.

(c) · 4 points

Suppose zₙ∈B and 0≤sup B−zₙ≤1/n. Put dₙ=zₙ/(1−zₙ). Prove dₙ∈A and 0≤s−dₙ≤(1+s)²/n. Is there one finite constant C such that |u−v|≤C|φ(u)−φ(v)| for all nonnegative u,v? Prove your answer with an explicit family.

Review question 1: worked argument and scoring criteria

(a)

For 0≤a≤s, φ(s)−φ(a)=(s−a)/[(1+s)(1+a)]≥0, so U=φ(s) is an upper bound. If s>0, for every ε>0 the supremum property supplies a∈A with a>s−ε; hence 0≤U−φ(a)≤s−a<ε. Thus no smaller number than U is an upper bound. If s=0, nonemptiness and A⊆[0,s] give A={0} and B={0}. The same difference identity is strictly positive when a<s, so φ is strictly increasing. Therefore A has maximum s exactly when B contains U, in which case its maximum is φ(s). Conversely, a maximum φ(a) of B forces every original element at most a by strict increase, so a=s.

1 point: Derives the exact difference identity and proves U is an upper bound.
1 point: Proves leastness using the supremum approximation property and handles s=0.
1 point: Proves both directions of maximum attainment using strict monotonicity.

(b)

The supremum approximation property supplies each aₙ; for s=0 choose zero. A finite maximum is one of the finitely many selected elements, so cₙ∈A even when A is not closed under other operations. Adding a new candidate cannot decrease a finite maximum. Also s−1/n<aₙ≤cₙ≤s, so 0≤s−cₙ<1/n, proving convergence to s. Strict monotonicity makes φ(cₙ) nondecreasing and the difference identity gives 0≤U−φ(cₙ)=(s−cₙ)/[(1+s)(1+cₙ)]<1/n. Neither limit need be a member of its set; all selections are made from A using leastness of its upper bound, not closedness or attainment.

1 point: Justifies the selections and membership of the finite maximum in A.
1 point: Proves monotonicity and the explicit error bound for cₙ.
1 point: Transfers the bound to φ(cₙ) and separates convergence from closedness/attainment.

(c)

Every zₙ=φ(a) for some a∈A and zₙ≤U<1. Solving the defining equation gives dₙ=a, proving membership and the denominator restriction. Since s=U/(1−U), subtraction gives s−dₙ=(U−zₙ)/[(1−U)(1−zₙ)]. Because zₙ≤U, the denominator is at least (1−U)²=1/(1+s)², proving the bound. A universal inverse Lipschitz constant on the entire nonnegative half-line does not exist. Take u=m,v=2m for positive integers m. Their original difference is m and their transformed difference is m/[(1+m)(1+2m)]. The ratio of the two differences is (1+m)(1+2m), which grows without bound. Thus a bound depending on a fixed upper range, such as s, is essential; the forward transformation compresses large values without making its global inverse uniformly Lipschitz.

1 point: Proves zₙ<1 and recovers an element of A by the inverse formula.
1 point: Derives the exact inverse difference identity.
1 point: Obtains the stated range-dependent error bound.
1 point: Provides an explicit family with unbounded inverse difference ratio, refuting a universal C.

Targeted lesson returns:

Question 2: Repair convergence tests based on local changes

10 points

All sequences below are real and indexed by positive integers. Distinguish control of one successive difference from control of every sufficiently late finite tail. Suggested time: 20 minutes.

(a) · 3 points

Prove that convergence of aₙ implies aₙ₊₁−aₙ→0. Disprove the converse using an explicit sequence, and prove directly that your example is not Cauchy by exhibiting late pairs with a fixed positive separation.

(b) · 3 points

Now assume |aₙ₊₁−aₙ|≤C/[n(n+1)] for every n≥1, with C≥0. Prove convergence and the tail error bound |L−aₙ|≤C/n. Show that the coefficient of C/n cannot be reduced uniformly over all such sequences.

(c) · 4 points

A proposed weaker repair assumes a_(2^k)→L and aₙ₊₁−aₙ→0. Test it using a₁=0 and, for 2ᵏ≤n≤2ᵏ⁺¹ with k≥1, aₙ=min(n−2ᵏ,2ᵏ⁺¹−n)/2ᵏ⁻¹. Then prove the correct equivalence: assuming a_(2^k)→L, the whole sequence tends to L if and only if max over 2ᵏ≤n≤2ᵏ⁺¹ of |aₙ−a_(2^k)| tends to zero.

Review question 2: worked argument and scoring criteria

(a)

If aₙ→L, then |aₙ₊₁−aₙ|≤|aₙ₊₁−L|+|aₙ−L|→0. For the converse take aₙ=Σ from j=1 to n of 1/j. Its successive difference is 1/(n+1), which tends to zero. But a₂ₙ−aₙ=Σ from j=n+1 to 2n of 1/j≥n/(2n)=1/2. For every proposed Cauchy cutoff choose n beyond it; both indices n,2n lie beyond the cutoff and their difference is at least 1/2, violating, for example, tolerance 1/4. Hence the sequence does not converge.

1 point: Proves the necessary condition with a shared limiting value or two-error estimate.
1 point: Gives a sequence with vanishing successive differences.
1 point: Establishes failure of the full Cauchy quantifiers through arbitrarily late pairs.

(b)

For m>n, |aₘ−aₙ|≤Σ from j=n to m−1 of C/[j(j+1)]=C(1/n−1/m)≤C/n. If C>0, choose n beyond C/ε to obtain the Cauchy criterion; if C=0 the sequence is constant. Real completeness gives L. Let m tend to infinity in the finite-pair inequality to get |L−aₙ|≤C/n. For any C>0 the sequence aₙ=C(1−1/n) has differences exactly C/[n(n+1)], limit C and error exactly C/n. Thus no fixed multiplier smaller than 1 works for every admissible sequence. The guarantee comes from the sum of all later step bounds, not from the fact that each step is small.

1 point: Telescopes the finite sum of step bounds and proves the Cauchy condition, including C=0.
1 point: Invokes completeness and derives the error bound by a justified limit.
1 point: Exhibits and checks an equality case proving uniform sharpness.

(c)

At shared block endpoints both formulas give zero, so the sequence is well-defined. Every dyadic sample a_(2^k) is zero. Inside block k the successive slopes have magnitude 2^(1−k), and the same vanishing bound controls steps adjacent to block endpoints. Hence successive differences tend to zero. Yet at the integer midpoint n=3·2^(k−1) the value is 1 for every k, so there is no convergence to zero or any other limit. For the valid criterion, if aₙ→L, then all n in a sufficiently late dyadic block, including its left endpoint, are within ε/2 of L, so the maximum difference is less than ε. Conversely, if the block maxima tend to zero and the dyadic samples tend to L, choose one block cutoff after which both errors are less than ε/2. Every sufficiently large n belongs to such a block, and |aₙ−L|≤|aₙ−a_(2^k)|+|a_(2^k)−L|<ε. This controls all intervening terms rather than only a subsequence and individual increments.

1 point: Checks endpoint consistency, zero dyadic samples and vanishing successive differences.
1 point: Computes the recurring midpoint value 1 and disproves the proposed repair.
1 point: Proves necessity of the block-oscillation condition with one tail cutoff.
1 point: Proves sufficiency by bounding every late index through its dyadic block.

Targeted lesson returns:

Question 3: Construct a bijective rearrangement and audit every partial sum

10 points

Let Σaₙ be a convergent real series that is not absolutely convergent; allow infinitely many zero terms. Fix an arbitrary finite real number T. A rearrangement means a bijection of the positive integer indices, not merely a choice of infinitely many terms. Suggested time: 25 minutes.

(a) · 3 points

Prove that the strictly positive terms have divergent total sum and the magnitudes of the strictly negative terms also have divergent total sum. Identify the term-size fact needed to control crossings later. Do not infer either sign divergence just from the existence of infinitely many terms of that sign.

(b) · 3 points

Specify an algorithm that rearranges the series to T: use positive and negative terms in their original relative orders and account for every zero index. Prove each phase is finite and the resulting index map is a bijection. Explain how the start is handled when T<0.

(c) · 4 points

Prove convergence of every partial sum of your rearrangement, repairing the claim “the crossing sums converge, so nothing else is needed.” For an exact check let a₃ₖ₋₂=1/k, a₃ₖ₋₁=−1/k and a₃ₖ=0 for k≥1. Verify conditional convergence in the original order. For T=5/4 list the original indices used in the first complete cycle of your algorithm, including its inserted zero, and their partial sums.

Review question 3: worked argument and scoring criteria

(a)

Put pₙ=max(aₙ,0), qₙ=max(−aₙ,0). For original prefixes write S_N=Σaₙ and A_N=Σ|aₙ|. Then Σpₙ=(A_N+S_N)/2 and Σqₙ=(A_N−S_N)/2. The nonnegative partial sums A_N tend to infinity because their series diverges, whereas S_N converges and is bounded. Both displayed sign sums therefore tend to infinity. Removing zero terms preserves their values, so each strictly nonzero sign list is infinite with divergent total magnitude. Also aₙ=Sₙ−Sₙ₋₁→0. This vanishing of individual terms, rather than merely sign divergence, will force late crossing errors to vanish.

1 point: Introduces positive/negative parts and derives both finite-prefix identities.
1 point: Uses bounded original partial sums and divergent absolute sums to prove both sign divergences.
1 point: Proves aₙ→0 and distinguishes this needed fact from divergence of sign sums.

(b)

Start with sum zero. Take at least one next unused positive term and continue until the sum is strictly above T; then take at least one next unused negative term and continue until strictly below T. After each complete positive/negative cycle insert the next unused zero term if there is one. Repeat. Every phase ends: finitely many earlier removals leave a divergent remaining positive or negative magnitude sum. No index is repeated since each list is consumed from its front. Every cycle advances both nonzero lists at least once, so each fixed position in either list is eventually used. Each fixed zero-list position is likewise used after finitely many cycles. Thus all original indices occur exactly once, and infinitely many finite cycles define a bijection of the positive integers. If T<0, the first upward phase can finish after its mandatory first term even though its starting sum was already above T. Its error need not satisfy the later overshoot estimate; it is a finite exceptional prefix. All subsequent upward phases start below T.

1 point: Gives an unambiguous strict-crossing rule that consumes nonzero terms and schedules zero terms.
1 point: Proves finite phase termination from the remaining divergent sign sums.
1 point: Proves injectivity/exhaustion and handles the arbitrary-target initial phase.

(c)

After the finite initial exception, an upward phase starts below T and first ends strictly above it. Its final overshoot is at most the last positive term, since the preceding sum was at most T. A downward phase has undershoot at most the magnitude of its final negative term. The sign lists advance in original order without bound, so these final terms tend to zero by part a. Both sequences of crossing sums tend to T. All sums inside a phase lie monotonically between its starting and ending crossings. Given ε>0, after sufficiently many cycles both crossings are within ε of T, which controls every intermediate partial sum as well. Inserted zeros change nothing. For the displayed series each completed triple sums to zero, and the only nonzero intermediate triple sum is 1/k→0; hence the original sum is zero. Its absolute sums contain 2Σ1/k and diverge. For target 5/4 the first cycle uses indices 1,4,2,3, corresponding to terms 1,1/2,−1,0 and partial sums 1,3/2,1/2,1/2. The positive phase crosses upward at 3/2 and the negative phase crosses downward at 1/2. This finite trace illustrates the algorithm but cannot replace the preceding all-indices proof.

1 point: Bounds both crossing errors by late individual term magnitudes and proves those bounds vanish.
1 point: Controls every intermediate partial sum and inserted zero, not just crossing subsequences.
1 point: Proves the supplied original series converges conditionally, including intermediate triple sums.
1 point: Gives the correct first-cycle indices and partial sums and limits the evidential role of the trace.

Targeted lesson returns:

2. Continuity and calculus

30 practice points.

Question 4: Recover an endpoint from derivative information

10 points

Let f:(0,1]→R be continuous and differentiable on (0,1), and suppose lim as x→0+ of f′(x)=L∈R. You may use scalar completeness and the ordinary mean value theorem, but must prove the extension claims below. Suggested time: 18 minutes.

(a) · 3 points

Prove that there is a finite A with f(x)→A as x→0+. Construct A from a sequence, then prove that the same limit holds for all sufficiently small positive inputs, not only that sequence.

(b) · 3 points

Define F(0)=A and F(x)=f(x) for x>0. Prove that the right derivative F′₊(0) exists and equals L. In particular, for each ε>0 prove a bound |F(x)−F(0)−Lx|≤εx for all sufficiently small x>0, explaining why MVT is now legitimate on [0,x].

(c) · 4 points

Now replace the derivative-limit hypothesis by boundedness of f′ near 0. Determine which of (a) and (b) still follows. Analyze f(x)=x sin(ln x), including a derivative bound and two explicit input sequences that demonstrate the relevant failure. Distinguish a one-sided derivative from differentiability on a neighborhood of 0.

Review question 4: worked argument and scoring criteria

(a)

The derivative limit gives δ∈(0,1) and M=|L|+1 with |f′(t)|≤M on (0,δ). For 0<x,y<δ, MVT on the interval between them gives |f(x)−f(y)|≤M|x−y|. Consequently f(1/n) is Cauchy once 1/n<δ, and scalar completeness supplies a finite limit A. Fix any x∈(0,δ) and let n→∞ in |f(x)−f(1/n)|≤M|x−1/n| to obtain |f(x)−A|≤Mx. This establishes f(x)→A for unrestricted positive inputs. The finitely many early sequence terms do not affect the Cauchy property.

1 point: Obtains a common derivative bound on a punctured interval and the two-input MVT estimate.
1 point: Constructs A by scalar completeness of the sampled sequence.
1 point: Passes the estimate to A and proves the full right-hand limit, not merely sampled convergence.

(b)

Part (a) makes F continuous at 0; it is continuous on [0,x] and differentiable on (0,x). MVT therefore gives (F(x)−F(0))/x=f′(c_x) for some 0<c_x<x. Choose δ_ε>0 so |f′(t)−L|<ε whenever 0<t<δ_ε. For 0<x<δ_ε the same estimate holds at c_x, giving |F(x)−F(0)−Lx|<εx, and hence the requested weak bound. Dividing by x proves the right derivative. No continuity of f′ away from the endpoint is assumed or needed.

1 point: Checks endpoint continuity and interior differentiability before applying MVT.
1 point: Uses 0<c_x<x to transfer the derivative-limit cutoff to every small input.
1 point: Derives the error bound and concludes the right derivative is L.

(c)

The proof of (a) used only a derivative bound, so a finite continuous extension still exists. For f(x)=x sin(ln x), |f(x)|≤x gives extension value 0, and f′(x)=sin(ln x)+cos(ln x) has magnitude at most 2. But F(x)/x=sin(ln x). Along x_n=exp(π/2−2πn) and y_n=exp(−π/2−2πn), n≥1, these quotients are 1 and −1 respectively, and both positive sequences tend to 0. Thus even the right derivative need not exist. A right derivative concerns only inputs in the given domain; it does not assert differentiability of a function defined on an open neighborhood across 0. Under the original hypothesis one could separately choose a compatible left extension, but no left-side values are specified here.

1 point: Retains the continuous-extension conclusion under bounded derivative and explains why.
1 point: Computes the derivative and verifies a valid bound and extension value.
1 point: Supplies two explicit sequences tending to zero with distinct difference-quotient limits.
1 point: States the failure of right differentiability and distinguishes it from a two-sided domain claim.

Targeted lesson returns:

Question 5: What can a quadrature certificate establish?

10 points

Let a<b and let f:[a,b]→R be continuous, differentiable on (a,b), with |f′(x)|≤M for a known M≥0. For a partition a=x₀<…<x_N=b write h_i=x_i−x_(i−1), c_i=(x_(i−1)+x_i)/2 and Q=Σ h_i f(c_i). All integrals are Riemann integrals. Suggested time: 18 minutes.

(a) · 3 points

Starting from MVT, prove |∫ₐᵇ f−Q|≤(M/4)Σ h_i²≤M(b−a) max_i h_i/4. Do not quote a quadrature error theorem. Explain why this proof also handles M=0.

(b) · 3 points

Assume additionally f≥0. Prove ∫ₐᵇ f=0 implies f≡0, treating a positive endpoint value as well as an interior one. For [a,b]=[0,3], M=2 and uniform N-cell partitions, give an integer N sufficient to certify error strictly below 1/100 for every admissible f. Is this a proof of the smallest N for a particular unknown f?

(c) · 4 points

For a fixed uniform N-cell partition of [0,1], construct a nonzero nonnegative continuously differentiable f with |f′|≤1 and Q=0. Compute its integral exactly. Explain why a zero sampled sum does not trigger the rigidity theorem, and why removing continuity also invalidates the theorem’s every-point conclusion.

Review question 5: worked argument and scoring criteria

(a)

1 point: Derives the pointwise cell estimate from MVT with endpoint hypotheses checked.
1 point: Integrates the distance to the midpoint to obtain h_i²/4 and sums errors.
1 point: Proves the mesh bound and handles M=0 without division.

(b)

If f(x₀)>0, continuity supplies a relative neighborhood on which f≥f(x₀)/2. That neighborhood contains a subinterval of positive length inside [a,b], including when x₀ is an endpoint. Integral order and nonnegativity elsewhere force ∫f>0, contradiction. For the numerical certificate the bound is M(b−a)²/(4N)=9/(2N). Strict error below 1/100 follows from N>450, so N=451 is sufficient. This is a uniform sufficient certificate, not the smallest partition count needed for a particular function: constant functions, for example, have zero midpoint error for every N.

1 point: Uses continuity and positive interval length to prove rigidity, including endpoints.
1 point: Computes the bound 9/(2N) and a valid strict-threshold integer such as 451.
1 point: Distinguishes a sufficient all-function certificate from a function-specific minimum.

(c)

Take f(x)=[1+cos(2πNx)]/(2πN). It is nonnegative and continuously differentiable, with derivative −sin(2πNx) of magnitude at most 1. At each midpoint c_i=(i−1/2)/N the cosine is cos(2πi−π)=−1, so Q=0. Its integral is 1/(2πN)>0 because the cosine integrates to zero over N complete cycles. Thus zero sampled sum is not zero true integral; the error certificate gives only an interval containing the true integral. If continuity is removed, the function equal to 1 at one point and 0 elsewhere is bounded, nonnegative and Riemann integrable with integral 0, but does not vanish everywhere. Isolating that point in cells of arbitrarily small total length proves its integrability and zero integral.

1 point: Constructs a nonzero nonnegative function satisfying the derivative bound.
1 point: Checks every midpoint value and evaluates the positive integral exactly.
1 point: Separates sampled zero from the exact integral hypothesis.
1 point: Gives and justifies the discontinuous one-point counterexample.

Targeted lesson returns:

Question 6: Derive a growth-ratio theorem and audit its boundary

10 points

The quotient rule differentiates a function; a derivative-ratio limit theorem is a different assertion. In this question derive the needed limit theorem rather than quoting L’Hôpital. You may use Rolle’s theorem and ordinary limit laws. Suggested time: 20 minutes.

(a) · 3 points

Let f,g be continuous on [u,v], differentiable on (u,v), with u<v and g′>0 there. Derive Cauchy’s mean value identity: (f(v)−f(u))/(g(v)−g(u))=f′(ξ)/g′(ξ) for some ξ∈(u,v). Check both denominators and construct the auxiliary function used in Rolle’s theorem.

(b) · 3 points

Now suppose f,g are differentiable on (a,∞), g′>0, g(x)→+∞ and f′(x)/g′(x)→L∈R. Prove f(x)/g(x)→L without assuming any limit for f itself. Keep the chosen tail anchor fixed while x tends to infinity; do not claim that an unspecified intermediate point must tend to infinity.

(c) · 4 points

Apply (b) to any differentiable f on (1,∞) satisfying |f′(x)−4x³|≤x², proving f(x)/x⁴→1 independently of its starting value. Then test whether the condition g(x)→+∞ can simply be deleted: on (0,∞), take g(x)=2−e^(−x) and f(x)=g(x)+3. Compute the derivative-ratio and function-ratio limits and identify exactly the term in your proof that no longer vanishes.

Review question 6: worked argument and scoring criteria

(a)

Ordinary MVT, itself a consequence of Rolle, gives g(v)−g(u)=g′(η)(v−u)>0. Define H(t)=(f(v)−f(u))(g(t)−g(u))−(g(v)−g(u))(f(t)−f(u)). It is continuous on the closed interval, differentiable inside and zero at both endpoints. Rolle supplies ξ with 0=H′(ξ)=(f(v)−f(u))g′(ξ)−(g(v)−g(u))f′(ξ). Both g(v)−g(u) and g′(ξ) are positive, so division yields the identity.

1 point: Proves the increment denominator is positive.
1 point: Constructs a valid auxiliary function and checks Rolle’s hypotheses.
1 point: Differentiates, checks the derivative denominator and derives the ratio identity.

(b)

Given ε>0, choose a fixed c>a so |f′(t)/g′(t)−L|<ε/2 for every t>c. For every x>c, part (a) on [c,x] bounds |(f(x)−f(c))/(g(x)−g(c))−L|<ε/2, since the intermediate point is greater than the already chosen c. Therefore |f(x)−L g(x)|≤(ε/2)(g(x)−g(c))+|f(c)−L g(c)|. Eventually g(x)>0. Divide by g(x); the ratio (g(x)−g(c))/g(x) tends to 1 and the fixed constant divided by g(x) tends to 0. More explicitly, make the first ratio less than 3/2 and the second term less than ε/4 to obtain |f(x)/g(x)−L|<ε. No behavior of f alone was assumed. The fixed-anchor argument controls every later x and avoids an unjustified claim about the intermediate point.

1 point: Chooses a fixed tail anchor and controls every derivative ratio between it and x.
1 point: Derives the increment estimate including the fixed additive constant.
1 point: Divides only once g is positive and completes a quantified limit argument without an assumption on f.

(c)

For g(x)=x⁴, g′(x)=4x³>0 and g→∞. The hypothesis gives |f′(x)/g′(x)−1|≤1/(4x), so the derivative ratio tends to 1 and part (b) gives f(x)/x⁴→1. Any additive starting constant disappears in this ratio; no separate limit for f was required. In the changed example, g′=e^(−x)>0 and f′/g′=1 identically, but g→2 and f→5, so f/g→5/2 rather than 1. In the proof with L=1 the anchor constant f(c)−g(c)=3 divided by g(x) tends to 3/2, not 0. The example violates the unbounded-denominator hypothesis, not the proved theorem.

1 point: Obtains the derivative-ratio bound 1/(4x) and verifies the comparison denominator hypotheses.
1 point: Concludes the growth ratio for every admissible f, explaining starting-value independence.
1 point: Computes both changed-case limits correctly.
1 point: Identifies the nonvanishing anchor term and the exact missing hypothesis.

Targeted lesson returns:

3. Uniform limits and spaces

40 practice points.

Question 7: Control an entire accumulation curve

10 points

Let g_n be continuous real functions on [0,2] converging uniformly to g. Define A_n(x)=∫₀ˣ(x−t)g_n(t)dt for 0≤x≤2. You may use the fundamental theorem of calculus for continuous integrands and the usual Riemann integral inequalities. Do not replace a uniform claim by checking one fixed x.

(a) · 3 points

Prove g is continuous. Define A by the same integral with g. If δ_n=||g_n−g||∞, obtain a bound for ||A_n−A||∞ and prove uniform convergence.

(b) · 4 points

Without assuming a rule for differentiating integrals with parameter-dependent integrands, prove A_n′(x)=∫₀ˣg_n(t)dt and A_n″(x)=g_n(x) in the interior. Prove A_n′→A′ uniformly and identify A″. Give a joint bound for ||A_n−A||∞+||A_n′−A′||∞.

(c) · 3 points

Can uniform convergence of A_n and A_n′ to zero force uniform convergence of g_n to zero? Test g_n(t)=sin(nt). Derive A_n and its first derivative explicitly and justify the conclusion.

Review question 7: worked argument and scoring criteria

(a)

1 point: Proves continuity with a common uniform approximation and a three-error argument.
1 point: Derives the x-dependent bound δ_n x²/2.
1 point: Takes the supremum to obtain 2δ_n and uniform convergence.

(b)

Write A_n(x)=x∫₀ˣg_n(t)dt−∫₀ˣt g_n(t)dt. Product differentiation and FTC give A_n′=∫₀ˣg_n+xg_n−xg_n=∫₀ˣg_n. FTC once more gives A_n″=g_n. The same argument applies to continuous g and A. Thus |A_n′(x)−A′(x)|≤xδ_n≤2δ_n, proving uniform derivative convergence. Also A″=g on (0,2). Adding the two established bounds gives at most 4δ_n; no general unsupported interchange theorem is needed.

1 point: Rewrites the parameter integral as two ordinary accumulation functions.
1 point: Differentiates with explicit cancellation and identifies both derivatives.
1 point: Proves the uniform derivative bound 2δ_n.
1 point: Identifies A″=g and the joint error bound 4δ_n.

(c)

Direct integration gives A_n′(x)=(1−cos(nx))/n and A_n(x)=x/n−sin(nx)/n². Their supremum norms are bounded by 2/n and 2/n+1/n², respectively, so both tend uniformly to zero. However for every integer n≥1 the point t=π/(2n) lies in [0,2] and g_n(t)=1; hence ||g_n||∞=1. The converse fails: convergence of a function and its first derivative does not control its second derivative.

1 point: Obtains both explicit formulas with their zero initial values.
1 point: Gives uniform vanishing bounds for A_n and A_n′.
1 point: Uses an in-domain point attaining g_n=1 to refute the converse.

Targeted lesson returns:

Question 8: Extend convergence from dense readings

10 points

Let D be a dense subset of [0,1]. Continuous functions f_n:[0,1]→R share a finite Lipschitz constant L≥0: |f_n(x)−f_n(y)|≤L|x−y| for every n,x,y. Suppose f_n(d) converges as n→∞ for each d∈D. Values at other points are not assumed to converge. You may use scalar completeness and compactness of [0,1].

(a) · 4 points

Prove f_n is uniformly Cauchy. Your proof must choose finitely many centers in D and track the three errors. Treat L=0 without dividing by zero.

(b) · 3 points

Construct the uniform limit f and prove it is L-Lipschitz. Show it is the unique continuous function with the prescribed limits on D.

(c) · 3 points

Show the common Lipschitz bound is essential even if convergence is known at every point: analyze h_n(x)=max(1−3n|x−1/n|,0) on [0,1]. State the pointwise limit, a Lipschitz constant and the uniform error.

Review question 8: worked argument and scoring criteria

(a)

Fix ε>0 and choose δ=ε/[6(L+1)]. Density makes the δ-balls centered at points of D cover [0,1]; compactness supplies finitely many centers d₁,…,d_r. Each numerical sequence f_n(d_i) is Cauchy, so choose one N making |f_n(d_i)−f_m(d_i)|<ε/3 for all centers and n,m≥N. For any x choose a center within δ. Then |f_n(x)−f_m(x)|≤Lδ+ε/3+Lδ<2ε/3<ε. The finite maximum of center cutoffs is legitimate. The choice δ uses L+1, so L=0 is included.

1 point: Obtains a finite net with centers in the specified dense set.
1 point: Chooses a single cutoff from finitely many Cauchy numerical sequences.
1 point: Controls both spatial errors and the center error uniformly.
1 point: Handles L=0 and avoids a maximum over infinitely many cutoffs.

(b)

The uniform Cauchy estimate makes every f_n(x) scalar Cauchy, defining f(x). Fixing n,x and passing m→∞ in a common tail estimate proves uniform convergence, using ε/2 first to accommodate a non-strict limiting inequality. Passing to the limit in |f_n(x)−f_n(y)|≤L|x−y| gives the same inequality for f. If another continuous h agrees with these limits on D, choose points d_j∈D tending to any x. Continuity gives h(x)=lim h(d_j)=lim f(d_j)=f(x).

1 point: Constructs the pointwise candidate and proves uniform convergence using one cutoff.
1 point: Passes the common Lipschitz inequality to the limit.
1 point: Proves uniqueness by density and continuity.

(c)

The support is contained in [2/(3n),4/(3n)]∩[0,1]. At x=0 all values are zero; every fixed x>0 is outside that support for sufficiently large n. Thus the pointwise limit is zero. Each h_n is Lipschitz with constant 3n, since absolute value and truncation at zero preserve that bound. At x=1/n the value is 1, so ||h_n||∞=1. The constants are not uniformly bounded, and uniform convergence fails despite a continuous pointwise limit.

1 point: Proves pointwise convergence including the endpoint zero.
1 point: Justifies the individual constant 3n and its lack of a common bound.
1 point: Computes the persistent peak and diagnoses the failed uniform conclusion.

Targeted lesson returns:

Question 9: A weighted constraint prevents escaping mass

10 points

In real ℓ² define K={x:Σ from k=1 to infinity of k²x_k²≤1}. The sum is the supremum of its finite nonnegative partial sums. You may use scalar Bolzano–Weierstrass and the norm/completeness results for ℓ², but not a general compactness theorem for weighted sequence sets.

(a) · 3 points

Prove every x∈K has ||x||₂≤1 and that truncation after J coordinates has error at most 1/(J+1). Explain why the bound is uniform over K.

(b) · 4 points

Prove every sequence of members of K has a subsequence converging in ℓ² to a member of K. Construct the subsequence and verify membership of the candidate before invoking a norm difference.

(c) · 3 points

Compare the unweighted closed unit ball of ℓ² with K. Also decide whether K is compact in the stronger weighted distance d_w(x,y)=(Σk²(x_k−y_k)²)^(1/2), defined on K. Supply explicit sequences rather than relying on dimension slogans.

Review question 9: worked argument and scoring criteria

(a)

Because k²≥1, Σx_k²≤Σk²x_k²≤1. For k>J one has k²≥(J+1)²; therefore Σ_(k>J)x_k²≤(J+1)^(−2)Σ_(k>J)k²x_k²≤(J+1)^(−2). Taking square roots gives the tail bound. It depends only on J, not on which member x was chosen, so it controls all sequences in K simultaneously.

1 point: Bounds the full ℓ² norm by the weighted sum.
1 point: Uses k≥J+1 on the omitted tail before taking square roots.
1 point: States the resulting uniform bound 1/(J+1) with its quantifiers.

(b)

Every kth coordinate lies in [−1/k,1/k]. Successive applications of scalar Bolzano–Weierstrass and a strictly increasing diagonal selection give a subsequence converging at every coordinate, to x_k. For fixed J, finite-sum continuity gives Σ_(k≤J)k²x_k²≤1. Taking the supremum over J shows x∈K, hence x∈ℓ². For ε>0 choose J with 2/(J+1)<ε/2. The subsequence and its limit both have tails bounded by 1/(J+1), while convergence of the fixed first J coordinates makes the prefix norm difference <ε/2 eventually. The triangle inequality yields total error <ε. Thus K is sequentially compact in ℓ²; in a metric space this also implies compactness if that established equivalence is invoked.

1 point: Constructs a genuine increasing diagonal subsequence from bounded coordinates.
1 point: Proves x∈K by finite weighted sums before taking their supremum.
1 point: Controls both tails by a single J independent of subsequence index.
1 point: Combines finite-coordinate convergence with the tails to prove ℓ² convergence.

(c)

The unit vectors e_n belong to the unweighted unit ball and have pairwise ℓ² distance √2, so no subsequence can converge; that ball is not compact. In K use v^(n)=e_n/n. Each weighted sum is 1, so v^(n)∈K. Their ordinary ℓ² norms tend to zero, consistently with part (b), but for different m,n the weighted distance is √2. Thus K is not compact in d_w. The weighted distance is finite on K by the triangle inequality for the sequences (kx_k); compactness depends on the chosen metric.

1 point: Uses unit vectors and fixed separation to refute unweighted-ball compactness.
1 point: Checks membership of e_n/n and computes its weighted pairwise distance.
1 point: States the distinct metric conclusions and justifies the weighted distance domain.

Targeted lesson returns:

Question 10: Construct and compare a nonlinear solution

10 points

Consider y′(t)=t+y(t)² with y(0)=1/2 for t≥0. For h=1/5 let X={u∈C([0,h]):||u−1/2||∞≤1/2} and Tu(t)=1/2+∫₀ᵗ(s+u(s)²)ds. You may use the contraction theorem on a nonempty complete metric space, but must verify its hypotheses. No global existence assertion is requested.

(a) · 4 points

Prove T has exactly one fixed point in X by checking completeness, the self-map property and a strict contraction constant. Explain why this fixed point is a classical IVP solution and why every classical solution staying in X is a fixed point.

(b) · 3 points

For any v∈X suppose ||Tv−v||∞≤η. Bound its distance to the exact fixed point. Apply your bound to the constant v(t)=1/2 by calculating its exact residual on [0,1/5]. Distinguish an error upper bound from an exact error.

(c) · 3 points

Suppose two classical solutions with initial value 1/2 both exist on [0,T], where T is finite, and both satisfy |y(t)|≤B for a finite B≥0. Prove they agree on [0,T] without requiring 2BT<1. Explain why this comparison does not itself prove existence for every T.

Review question 10: worked argument and scoring criteria

(a)

X is a nonempty closed ball in complete C([0,h]) with the supremum norm. For u∈X, 0≤u≤1, so |s+u²|≤6/5 and ||Tu−1/2||∞≤(1/5)(6/5)=6/25≤1/2. The integrand is continuous, so T maps X into X. For u,v∈X, |u²−v²|≤2|u−v|, giving ||Tu−Tv||∞≤(2/5)||u−v||∞. Banach gives a unique fixed point. FTC makes its derivative t+y² and its initial value 1/2. Conversely, integrating any classical solution with that initial value gives its fixed-point equation; the range restriction places it in X. Endpoint derivatives are interpreted one-sidedly.

1 point: Identifies a nonempty complete closed function ball.
1 point: Checks continuity and the self-map bound 6/25≤1/2.
1 point: Derives the contraction factor 2/5 and applies Banach.
1 point: Proves both fixed-point/IVP directions and states the range restriction.

(b)

For the fixed point y, ||v−y||∞≤||v−Tv||∞+||Tv−Ty||∞≤η+(2/5)||v−y||∞, so ||v−y||∞≤5η/3. For v=1/2, Tv−v=t²/2+t/4 is nonnegative increasing on [0,1/5], and its maximum is 1/50+1/20=7/100. Hence ||v−y||∞≤7/60. This is a certified upper bound derived from a contraction estimate; it is not an exact formula for the actual error.

1 point: Derives the residual error estimate 5η/3 from the fixed-point equation.
1 point: Computes the exact residual supremum 7/100.
1 point: Obtains 7/60 and correctly distinguishes bound from exact error.

(c)

For two solutions u,v, the state difference in the vector field is bounded by |u²−v²|≤2B|u−v|. Choose segment length ℓ=1/[2(2B+1)], so 2Bℓ<1. Subdivide [0,T] into finitely many segments of length at most ℓ. If u=v at a segment start, integration from that start gives supsegment|u−v|≤2Bℓ supsegment|u−v|, forcing equality throughout the segment. Starting at zero propagates equality across the finite partition, including a shorter final segment. The proof compares already existing bounded solutions; it does not ensure such solutions or a common B exist for arbitrarily long times.

1 point: Derives the common state-Lipschitz bound 2B.
1 point: Uses a finite short-segment partition and propagates equality from shared endpoints.
1 point: Separates uniqueness of existing solutions from global existence.

Targeted lesson returns:

Turn missing steps into new practice

Award only points supported by the written argument. Allow valid later reasoning based on a carried-forward value unless the criterion requires the correct earlier result. Do not award a correct-result point for an incorrect result. Record missing hypotheses, invalid limit interchanges and quantifier errors separately from arithmetic mistakes.

The total is out of 100 with no pass threshold or university-grade conversion. A high score on this public sample does not establish full-course readiness. Keep your first attempt, use the targeted returns, explain the repair, then attempt a fresh problem. Ask a reviewer to inspect the complete proof and choose an unfamiliar case.

For sustained writing and revision, use the analysis manuscript project. It supplies a separate investigation and does not replace this cumulative practice or certify the entire course.

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