Math Club orientation week questions

[Happypig375]

Here are the math club orientation week questions.

1. What mathematical formulation is appropriate for each? How would a math student approach it?
2. What are the appropriate steps and solution to each question?
3. What are the appropriate AngouriMath nodes that accurately represent such a mathematical formulation?
4. What generalized implementation is needed for AngouriMath to solve the mathematical formulation with the requirements of the original question, with adequate performance and memory usage? How about any special forms that can receive better speed and memory usage?
5. Does the above implementation differ from your method? Why?
6. Implement as necessary to solve these. It is expected that all operations, like solve, parse, stringize and latexize, work well with any nodes added. Remember to submit a pull request to CSharpMath too.

[Rafael-SOWNet]

Read from the four sheets; every "today" line below is measured on master at e2476ac5 with the expression given, not inferred. Answers first, then what the library does with each, then the implementation plan in the generalised form question 4 asks for.

Period I

1. Σ_{k≥0} 3^{k+2}(k²+k+1)/(k+3)! — Shift j = k+3 and write the polynomial in falling factorials: Σ_{j≥0} 3^j (j²−5j+7)/j! = e³(9+3−15+7) = 4e³, minus the three terms j = 0, 1, 2 (7, 9, 9/2), divided by 3: (8e³ − 41)/6 ≈ 19.947. Nodes: Summationf over Powf, Factorialf, a polynomial; the +oo bound. Today: sum(3^(k+2)*(k^2+k+1)/(k+3)!, k, 0, +oo).Simplify() is left as written; the finite sum to 30 evaluates to an exact rational agreeing to 27 digits. Gap: infinite sums of the form Σ p(k)·cʲ/(k+a)!, which are all e^c after the index shift and a falling-factorial basis. That is a general rule, not this sum.

2. ∫_1^∞ {x}/⌊x⌋! dx — On [n, n+1) the integrand is t/n!, so the integral is Σ_{n≥1} 1/(2·n!) = (e − 1)/2. Nodes: Floorf exists; the fractional part is x − floor(x); the improper Integralf. Today: integral((x - floor(x))/floor(x)!, x, 1, +oo) is left as written; integral(t/n!, t, 0, 1) answers 1/2 / n! provided not n! = 0; sum(1/(2*n!), n, 1, +oo) is left as written. Gaps: an integral over a range where the integrand is piecewise-constant in ⌊x⌋ splits into a sum over unit intervals (general: any integrand that is a function of x − ⌊x⌋ and ⌊x⌋), and the same exponential-series rule as question 1.

3. The tent map composed five times — g has 32 linear pieces of slope ±32 over [0,1], so the graph's length is Σ √(1 + 32²)·(width) = √1025 = 5√41. Nodes: Piecewise, composition by substitution, Derivativef. Today: substituting f into itself four times and differentiating gives the right derivative, ±32 everywhere, but as a five-deep nest of piecewise whose conditions are themselves piecewise — correct and unreadable; integral(sqrt(1 + 32^2), x, 0, 1) answers 5 * sqrt(41). Gap: flattening a composed piecewise into disjoint intervals (conditions x <= 1/2 under 2x become x <= 1/4), after which the arc length is one integral of a constant per piece.

4. Integers 1..1000 of the form a² − b² — a² − b² = (a−b)(a+b) with both factors of the same parity, so exactly the integers not ≡ 2 (mod 4): 750. Nodes: none — this is a counting argument. Today: x^2 - y^2 = 6 solves for x over the complex numbers, { ±sqrt(y^2 + 6) }, which is right and not the question. Not a CAS operation; a solver over the integers with a count would be the general form, and that is #212's territory.

5. ∫_{3/4}^{4/3} (2x²+x+1)/(x³+x²+x+1) dx — Partial fractions: 1/(x+1) + x/(x²+1), so ln(x+1) + ½ln(x²+1), which at the bounds is 2·ln(4/3) = ln(16/9). Today: answered, ln(4/3) + ln(16/9) / 2 in 194 ms, the indefinite integral in 3 ms. Gap, cosmetic: ln(16/9)/2 is ln(4/3) for a positive rational under a rational power, and the two terms then collect; a rule ln(p^q)/q = ln(p) for decidably positive p.

6. Maximum area under a projectile's path — Range 2v² sinθ cosθ / g; area (2/3) v⁴ sin³θ cosθ / g², maximal where tan²θ = 3, so θ = π/3 and the area is √3·v⁴ / (8g²). Today: the integral answers 2/3 * cos(t) * sin(t)^3 * v^4 / g^2 (115 ms, with the right provided), and substituting π/3 gives sqrt(3) * v^4 / (8 * g^2). Gap: an extremum over an interval — derivative, its zeros in the interval, the endpoints, a comparison — is not an operation; Solve needs a statement, so the caller assembles it. A MathS.Maximum(expr, var, interval) is the general form.

7. Expected number of uniform factors until the product is ≤ 1/3 — −ln s_i are unit exponentials, the count of them summing to less than ln 3 is Poisson(ln 3), and the stopping index is one more: 1 + ln 3 ≈ 2.099. Not a CAS operation: no distribution or expectation node exists, and none is proposed here.

8. a_{n+1} = (a+b)/2, b_{n+1} = 2ab/(a+b) — a_n b_n is invariant, = 9, and the arithmetic and harmonic means squeeze to the geometric one: both limits are 3. Nodes: none — a recurrence has no node. The general form is a Sequence/recurrence node with a limit rule for contracting means, which is real work and not this question's alone.

Period II

1. I_m = ∫_0^{2π} Π_{k≤m} sin(kx) dx — Under x → x+π the product picks up (−1)^{1+2+…+m} = (−1)^{m(m+1)/2}, so for m ≡ 1, 2 (mod 4) the integral is its own negative, 0; for m ≡ 0, 3 (mod 4) the constant term of the expanded product is a positive count and the integral is nonzero. Sum of such m ≤ 100: 2575. Today: m = 3 answers 0 (84 ms); m = 4 is left unevaluated after 4.2 s. Gap, twofold: the product-to-sum expansion of sin(x)…sin(mx) over a period is a computation that should be fast for any fixed m, and the symbolic-m question is a parity theorem no computation gives — the answer for the question is reasoning, which the library can check for each m if the fixed-m integral is quick.

2. P(⌈y/x⌉ is a perfect square) — P(⌈y/x⌉ = n) is 1/2 for n = 1 and 1/(2(n−1)) − 1/(2n) for n ≥ 2; summed over n = k²: 1/2 + ½(3/4 − (π²/6 − 1)) = 11/8 − π²/12 ≈ 0.553. Not a CAS operation (a probability over a region); the two series inside it are Σ 1/(k²−1) and Σ 1/k², which is the telescoping/ζ(2) gap in the same family as I.1.

3. Σ_{k=0}^N C(N,k) cos(kπ/3) — Re (1 + e^{iπ/3})^N = Re (√3 e^{iπ/6})^N = 3^{N/2} cos(Nπ/6). Today: the symbolic sum is left as written; N = 6 evaluates to −27, which is 27·cos(π), correct. Gap: binomial sums, Σ C(N,k) x^k = (1+x)^N and its real part with x = e^{iθ} — one rule, the binomial theorem read backwards, which also covers Σ C(N,k) sin(kθ).

4. ∫_0^{π/2} arctan(11 sin x)/sin x dx — With I(a) = ∫ arctan(a sin x)/sin x, I'(a) = ∫ dx/(1 + a² sin²x) = π/(2√(1+a²)), so I(11) = (π/2) asinh(11) = (π/2) ln(11 + √122) ≈ 4.859. Today: both integrals are left as written. Gap: differentiation under the integral sign as a strategy — differentiate in a parameter, integrate the derivative, integrate back — which is general and expensive; and the inner one, ∫ dx/(1 + a² sin²x) over a quarter period, is a tangent half-angle substitution the integrator should have on its own.

5. Maximum arc length of the trajectory — L(θ) = (v²/g)[sinθ + cos²θ · ln(secθ + tanθ)], dL/dθ = (2v²/g) cosθ [1 − sinθ ln(secθ+tanθ)], zero where f(θ) = 1 for the sheet's f, so θ* = f⁻¹(1) ≈ 56.5°, and there ln(secθ*+tanθ*) = 1/sinθ*, giving L_max = v² / (g · sin(f⁻¹(1))). Today: the arc-length integral is computed (with sqrt(cos(t)^2) and sqrt(1/cos(t)^2) left unsimplified, which under 0 ≤ θ ≤ π/2 are cos t and 1/cos t), but the maximisation needs the extremum operation of I.6 plus a named inverse of a non-elementary function, which no node expresses. The honest generalisation is I.6's, with the equation f(θ) = 1 left as a condition when it has no closed form.

6. Interval with exactly x multiples of a and y multiples of b — A feasibility question over integers with a case analysis on gcd(a, b); not a CAS operation and I am not going to hand it a formula I have not proved.

7. Expected resistance after N random series/parallel steps — The distribution of R_n is invariant under R ↔ 1/R (both operations swap under it), so E[1/(1+R)] = 1/2 and E[R_n] = 3/4 + ½E[R_{n−1}] from E[R_0] = 1: E[R_N] = 3/2 − 2^{−(N+1)}. Check: N = 1 gives 5/4, the mean of 2 and 1/2. Not a CAS operation.

8. y² | x³+1 and x² | y³+1 in naturals — (1,1), (2,3), (3,2). Today: (x^3 + 1) mod y^2 = 0 solves to the conditional set { x : (x^3 + 1) mod y^2 = 0 }, which is the question restated and, per #1036, the right non-answer. Not a CAS operation without a Diophantine solver.

Question 5 — does the implementation differ from the method?

Yes, in one consistent way: a student reaches each answer by a transformation — shift an index, split at the integers, compose the map, take a real part, differentiate a parameter — and then a routine computation. The library has the routine computations (this sheet's rational integral, the projectile area, the fixed-N binomial sum, the fixed-m sine product for small m) and lacks the transformations, which is what the gaps above name. The seven questions that are counting, probability, recurrence or Diophantine are not CAS operations, and I would not add nodes for them: a Sequence node (I.8) is the only one with a general shape, and it is its own proposal.

Question 4 — what to implement, generalised, in order

1. Exponential-type series — Σ_{k≥0} p(k) c^k / (k+a)! for a polynomial p, whole a, any c: shift, falling-factorial basis, e^c. Covers I.1 and the Σ 1/n! of I.2; extends the closed-form summation of A polynomial summand is summed in closed form (#717) #1152. Special form: p = 1, a = 0 is e^c directly.
2. Binomial sums — Σ_{k=0}^N C(N,k) x^k = (1+x)^N, with x = e^{iθ} taking the real and imaginary parts: II.3, and every Σ C(N,k) cos/sin(kθ).
3. Integrals over floor and fractional part — split [m, +oo) into unit intervals where the integrand depends on ⌊x⌋ and x − ⌊x⌋; I.2.
4. Extremum over an interval — derivative zeros within the interval, endpoints, comparison, with the zeros left as conditions where they have no closed form: I.6 and II.5. This one wants an API (MathS.Maximum/Minimum), so it is a proposal for you before it is a PR.
5. Small ones: ln(p^q)/q → ln p for decidably positive rational p (I.5's presentation); sqrt(cos(t)^2) under an interval condition (II.5); the sine-product integral's speed (II.1, 4.2 s at m = 4).
6. Differentiation under the integral sign (II.4) last: general, expensive, and the inner quarter-period integral is worth having first.

None of these adds a node, so nothing here needs a CSharpMath change; I will note on that repository if item 4's API ends up wanting a printed form.

I will start with 1 and 2, each as its own pull request with the sheet's questions as tests and the general form measured on more than the sheet.

[Happypig375]

It is not correct to leave some mathematical notation out "by scope" alone. We may name ourselves a "computer algebra system" by convention (it is a term that people know) but all the calculus here are already way past algebra. I'm sure sympy doesn't do provided as we do. Yet, it would match how we do math better. If the expectation operator and variables with distributions let statistics be represented as nodes too, we should be able to formulate them as well.

If you want a term that defines what we do, it should be more like #746 - "Math OS", rather than "CAS". We are here to solve math, not to become "just another CAS". It would be reasonable to collect mathematical symbols and see what we can do with all of them. The above questions should have the closest mathematical symbols expressing them instead of English words - try to fit them as nodes. Otherwise, we might have something like provided (do we have a better notation for provided at all?).

Think about this, and see if your agentic instructions need an update.

[Rafael-SOWNet]

Taken, and written into AGENTS.md as its own rule under "no answer is the floor, not the target": "not a CAS operation" is not a reason either — a question is not out of scope because its notation is statistical, combinatorial, number-theoretic or about a sequence; the definition behind the notation is what to fit as a node, and where the reduction is a theorem rather than a computation, the node still states the question and the answer stays unevaluated rather than absent. On provided: it is an Iverson-style restriction and reads fine; the notation I would keep is the one the parser already has, with piecewise for the case-by-case form.

Here are the seven I wrote off, each as the notation it actually is, what the node reduces to, and what it needs. I have not built any of these yet; a new node touches the parser, the three printers, the C++ export and CSharpMath, so I would rather have your yes on the spelling first, and then build them in the order below.

Distributions and expectation (I.7, II.2, II.7).

• X ~ Uniform(0, 1) — a statement node, distributed as, with the distribution as a value: Uniform(a, b) to start, Exponential(rate), Poisson(rate) as they are needed.
• E[f(X) | X ~ D] and P[statement | X ~ D] — binders like sum and integral: E[f(X)] for X ~ Uniform(0, 1) reduces to integral(f(x), x, 0, 1), and P[A] is E[1_A] with the indicator a piecewise(1 provided A, 0). Two independent variables reduce to a double integral, which the library takes as nested integrals. So II.2, P[ceil(y / x) is a square], is integral(integral(piecewise(1 provided ceil(y/x) is a square, 0), y, 0, 1), x, 0, 1) — and that needs is a square as a predicate over the naturals (below) and the floor/ceil split of An integral across a jump is split at the jumps, never taken through an antiderivative #1215 in two variables.
• I.7 is a stopping time, N = min{ n : s_1 ⋯ s_n <= 1/3 }, and E[N] = sum(P[N > n], n, 0, +oo): the sequence of i.i.d. samples wants an indexed family s_n ~ Uniform(0, 1), and the reduction goes through -ln s_i ~ Exponential(1) and a Poisson count — a theorem the node can name but not compute. II.7 is a Markov chain; its answer 3/2 - 2^-(N+1) comes from a symmetry argument, and the node would state E[R_N] and leave it.

Sequences and recurrences (I.8, and I.7's family).

• a_(n+1) = (a_n + b_n) / 2, a_0 = 1 — a Sequence node: a name, a recurrence, initial values, spelled with the subscript the parser reads for a_n already. limit(a_n, n, +oo) over it. Reduction: a linear recurrence with constant coefficients to a closed form (that is computation); a coupled non-linear one like the arithmetic–harmonic mean to its invariant and fixed point (that is a theorem: a_n b_n is constant, both converge, so both tend to sqrt(a_0 b_0)). The node states it either way.

Divisibility, counting, Diophantine solutions (I.4, II.6, II.8).

• a | b — a statement node, divides, equal by definition to b mod a = 0 and printed as the bar; it already reduces.
• #{ n in [1; 1000] : exists a b : n = a^2 - b^2 } — cardinality of a set builder with a quantifier, which is Quantifiers #225 (Quantifiers, accepted) plus a card/# node. I.4's answer, 750, is the theorem "a^2 - b^2 takes every value not ≡ 2 (mod 4)"; the node states the count, and a bounded enumeration can check it for a concrete range, which is what a student does first.
• II.8, solve(y^2 | x^3 + 1 and x^2 | y^3 + 1, (x, y) in NN^2) — the statement is expressible today with mod; the solver has no Diophantine method, and the honest answer stays the conditional set until one exists. II.6 is the same shape with the interval's ends as the unknowns.

Extremum over a set (I.6, II.5).

• max(f(t), t in [0; pi/2]) and argmax — binders that reduce to the derivative's zeros in the set plus its endpoints, compared; item 4 of the earlier plan. II.5 additionally needs the inverse of a named function, f^(-1)(1) for the sheet's f, which is Lambdas & functions as entities #495 (functions as entities, accepted) and an inverse that stays symbolic where no closed form exists.

Order I would build them: a | b and card/# (smallest, and they fit existing machinery), max/min/argmax over a set (needed by two of the sheet's questions), then ~ with Uniform and E/P reducing to integrals, then Sequence. Each gets its parse/print/LaTeX and a CSharpMath rendering in the same change. Tell me which spellings you want changed and I will start with the first two.

[Happypig375]

The spellings themselves look fine to me. If | will mean divides as a boolean statement, then we'd need to keep (| |) for absolute values (though abs( ) can also be used?)

and you'd need to disambiguate that from the | in E[f(X) | X ~ D] and P[statement | X ~ D] too. Maybe the expectation and probability operators override the meaning of | in the scope, and a new pair of parentheses restore the meaning of |?