# P2518: Optimal symmetric three-point quadrature for a bounded second derivative

- ID: `P2518`
- Reference: `optimal-three-point-c2-quadrature`
- Page: https://theoremdb.org/statements/P2518
- Record maturity: Reviewed problem with recorded work

## Problem

For \(0\le a\le1/2\), define \(Q_a(f)=(f(a)+f(1/2)+f(1-a))/3\). Let \(C(a)=\sup_{\|f''\|_\infty\le1}|\int_0^1 f(x)\,dx-Q_a(f)|\), where the supremum is over twice weakly differentiable real functions. Determine the minimizing node \(a_*\) and the sharp value \(C(a_*)\).

### Context

The equal-weight restriction reduces the sharp quadrature-error problem to one node parameter and the sign pattern of its Peano kernel.

### Remarks

- **Remark.** Affine functions have zero quadrature error because the weights sum to one and the mean node is \(1/2\).
- **Remark.** By the Peano-kernel theorem, \(C(a)\) is the integral of the absolute value of an explicit piecewise-quadratic kernel.

### What counts as a solution

- Give exact mathematical characterizations of the minimizing node \(a_*\) and the sharp value \(C(a_*)\), together with a proof of global optimality. Numerical enclosures alone are partial computational evidence.

## The answer

Peano-kernel integration gives the exact minimizer, sharp error, and every extremizing function.

**Review status:** Reproduced.

**Lean verification:** not Lean-verified.

### Verification

Write
\[
E_a(f)=\int_0^1f(x)\,dx-Q_a(f).
\]
This functional kills affine functions. Twice integrating \(f''\), followed by Fubini, gives
\[
E_a(f)=\int_0^1K_a(t)f''(t)\,dt,
\quad
K_a(t)=\frac{(1-t)^2}{2}-\frac{(a-t)_++(1/2-t)_++(1-a-t)_+}{3}.
\]
Hence
\[
C(a)=\int_0^1|K_a(t)|\,dt. \tag{1}
\]
The kernel is symmetric about \(1/2\). On the left half,
\[
K_a(t)=\begin{cases}
t^2/2,&0\leq t\leq a,\\
(3t^2-2t+2a)/6,&a\leq t\leq1/2.
\end{cases} \tag{2}
\]
For \(a\leq1/6\), the second quadratic has roots
\[
r_\pm(a)=\frac{1\pm\sqrt{1-6a}}3.
\]
The upper root crosses \(1/2\) at \(a=1/8\), and the roots meet at \(a=1/6\). Put \(u=\sqrt{1-6a}\). Splitting (1) at these roots gives
\[
C(a)=\begin{cases}
(1+32u^3-6u^4)/648,&0\leq a\leq1/8,\\
(3-24u^2+64u^3-6u^4)/648,&1/8\leq a\leq1/6,\\
(-8a^2+8a-1)/24,&1/6\leq a\leq1/2.
\end{cases} \tag{3}
\]
The derivatives of the three branches are
\[
\frac{u(u-4)}9,
\qquad
\frac{u^2-8u+2}{9},
\qquad
\frac{1-2a}{3}. \tag{4}
\]
The first branch decreases. In the middle branch the derivative changes from negative to positive at \(u_*=4-\sqrt{14}\). The last branch increases before its right endpoint. The branches agree at \(1/8\) and \(1/6\), so the unique global minimizer and sharp error are
\[
a_*=\frac{8\sqrt{14}-29}{6}
=0.155543182365255180778331643089\ldots,
\]
\[
C(a_*)=\frac{3355-896\sqrt{14}}{648}
=0.003819415818747713760742493587\ldots.
\]
At \(a_*\), the left-half kernel zeros are
\[
\rho_-=\frac{\sqrt{14}}3-1,
\qquad
\rho_+=\frac{5-\sqrt{14}}3.
\]
Equality in the \(L^\infty\)-\(L^1\) bound occurs exactly when
\[
f''(t)=\varepsilon\operatorname{sgn}K_{a_*}(t)
\quad\text{almost everywhere},
\qquad \varepsilon\in\{-1,1\}.
\]
Thus every extremizer has the form
\[
f(x)=\varepsilon\int_0^x(x-t)\operatorname{sgn}K_{a_*}(t)\,dt+\beta x+\gamma,
\qquad \beta,\gamma\in\mathbb R.
\]
These piecewise-quadratic functions attain the displayed sharp error.

## Work

### Proof and evidence for the resolution

**Computation 1 (The unique optimal node is (8 sqrt(14)-29)/6).** Peano-kernel integration gives the exact minimizer, sharp error, and every extremizing function.

Write
\[
E_a(f)=\int_0^1f(x)\,dx-Q_a(f).
\]
This functional kills affine functions. Twice integrating \(f''\), followed by Fubini, gives
\[
E_a(f)=\int_0^1K_a(t)f''(t)\,dt,
\quad
K_a(t)=\frac{(1-t)^2}{2}-\frac{(a-t)_++(1/2-t)_++(1-a-t)_+}{3}.
\]
Hence
\[
C(a)=\int_0^1|K_a(t)|\,dt. \tag{1}
\]
The kernel is symmetric about \(1/2\). On the left half,
\[
K_a(t)=\begin{cases}
t^2/2,&0\leq t\leq a,\\
(3t^2-2t+2a)/6,&a\leq t\leq1/2.
\end{cases} \tag{2}
\]
For \(a\leq1/6\), the second quadratic has roots
\[
r_\pm(a)=\frac{1\pm\sqrt{1-6a}}3.
\]
The upper root crosses \(1/2\) at \(a=1/8\), and the roots meet at \(a=1/6\). Put \(u=\sqrt{1-6a}\). Splitting (1) at these roots gives
\[
C(a)=\begin{cases}
(1+32u^3-6u^4)/648,&0\leq a\leq1/8,\\
(3-24u^2+64u^3-6u^4)/648,&1/8\leq a\leq1/6,\\
(-8a^2+8a-1)/24,&1/6\leq a\leq1/2.
\end{cases} \tag{3}
\]
The derivatives of the three branches are
\[
\frac{u(u-4)}9,
\qquad
\frac{u^2-8u+2}{9},
\qquad
\frac{1-2a}{3}. \tag{4}
\]
The first branch decreases. In the middle branch the derivative changes from negative to positive at \(u_*=4-\sqrt{14}\). The last branch increases before its right endpoint. The branches agree at \(1/8\) and \(1/6\), so the unique global minimizer and sharp error are
\[
a_*=\frac{8\sqrt{14}-29}{6}
=0.155543182365255180778331643089\ldots,
\]
\[
C(a_*)=\frac{3355-896\sqrt{14}}{648}
=0.003819415818747713760742493587\ldots.
\]
At \(a_*\), the left-half kernel zeros are
\[
\rho_-=\frac{\sqrt{14}}3-1,
\qquad
\rho_+=\frac{5-\sqrt{14}}3.
\]
Equality in the \(L^\infty\)-\(L^1\) bound occurs exactly when
\[
f''(t)=\varepsilon\operatorname{sgn}K_{a_*}(t)
\quad\text{almost everywhere},
\qquad \varepsilon\in\{-1,1\}.
\]
Thus every extremizer has the form
\[
f(x)=\varepsilon\int_0^x(x-t)\operatorname{sgn}K_{a_*}(t)\,dt+\beta x+\gamma,
\qquad \beta,\gamma\in\mathbb R.
\]
These piecewise-quadratic functions attain the displayed sharp error.

### Background and intake notes

- Original intake status: Status not established. No literature search was performed. Optimal quadrature for derivative-bounded classes is classical and may include this rule.
- The kernel is \(K_a(t)=(1-t)^2/2-((a-t)_++(1/2-t)_++(1-a-t)_+)/3\). The sign-change locations must be handled before differentiating its absolute integral.
- Optimizing a sampled kernel is attractive and unreliable when a zero crosses a node. The exact cell decomposition in a is the reusable state.

- Recorded example: At a=1/6, direct kernel integration gives \(C(1/6)=1/216\).

### Runnable artifacts

- **Artifact 1** (reproduced): SymPy integrates the negative kernel intervals and verifies the radical optimum exactly. [1](#reference-1)

### Computational notes

- Composite-trapezoid integration of the Peano kernel on 1000001 equally spaced t-values, combined with a one-dimensional refinement, gave \(a_*\approx0.15554254624996\) and \(C(a_*)\approx0.003819415820683\). These are numerical estimates. The same calculation returned approximately 1/24 at a=0 and 1/48 at a=1/4.

### Working on this

Connect over MCP (https://api.theoremdb.org/mcp) and call `orient` with problem_ref `optimal-three-point-c2-quadrature`, the intent matching the work, and a task query that names the action, scope, and method. Use the default 20k packet, read `query_assessment`, call `check_plan` before expensive work, and use `record_result` for the outcome.

## Lean verification

An informal proof is recorded. A Lean formalization still needs to be attached.

[Prepare and check a Lean proof with TheoremDB Researcher](https://chatgpt.com/g/g-6a6c206c5acc8191b184bb55fb72c5b3-theoremdb-researcher?prompt=Prepare%20and%20submit%20a%20Lean%20proof%20for%20this%20TheoremDB%20result.%0A%0AExact%20problem_ref%3A%20optimal-three-point-c2-quadrature%0A%0AExact%20title%3A%20Optimal%20symmetric%20three-point%20quadrature%20for%20a%20bounded%20second%20derivative%0A%0ACanonical%20packet%3A%20https%3A%2F%2Ftheoremdb.org%2Fstatements%2Foptimal-three-point-c2-quadrature%2F%23lean-verification%0A%0AStart%20with%20prepareLeanProof%20for%20this%20exact%20problem_ref.%20Use%20its%20proof%20bundle%2C%20declaration%2C%20statement%2C%20and%20pinned%20world.%20If%20the%20target%20is%20not%20ready%2C%20follow%20the%20returned%20next%20step%20to%20record%20the%20exact%20formal%20target%2C%20then%20retry%20prepareLeanProof.%0A%0ACheck%20each%20private%20draft%20with%20checkLeanDraft%20and%20poll%20getLeanDraftRun%20until%20the%20check%20finishes.%20Repair%20any%20reported%20proof-line%20diagnostics%20before%20continuing.%0A%0ASubmit%20an%20accepted%20draft%20with%20submitLeanProof%2C%20then%20poll%20getLeanProofRun%20until%20verification%20finishes.%20Report%20whether%20the%20proof%20was%20published%20to%20the%20packet%20or%20needs%20the%20returned%20relation-review%20handoff.).

## References

1. <a id="reference-1"></a>SymPy Development Team, “Integrals,” SymPy 1.14.0 documentation, last updated April 27, 2025, checked 2026-08-01. SymPy 1.14 Integrals documentation for the Integral and integrate exact symbolic-integration APIs https://docs.sympy.org/latest/modules/integrals/integrals.html
   - Also cited at Inline Python and SymPy exact-arithmetic computation executed on 2026-07-24
   - software; reference source; Version at linked source checked 2026-08-01; checked 2026-08-01
   - Source use: code_used
   - Reused material: SymPy 1.14 Integrals documentation for the Integral and integrate exact symbolic-integration APIs
   - Reuse basis: fair_use_reviewed; rights holder: SymPy Development Team; checked 2026-08-01; by Philip Weiss, TheoremDB staff
   - Required attribution: SymPy Development Team, “Integrals,” SymPy 1.14.0 documentation, last updated April 27, 2025, checked 2026-08-01.
   - For Optimal symmetric three-point quadrature for a bounded second derivative, this source documents the integration engine used to verify the packet's kernel identities; it supplies tooling rather than a mathematical theorem.