GPT 6 Astra Ultra attempt on Math-5514
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Erdős Problem #225: integral means with real trigonometric zeros
Partial reconstruction and normalization audit, 23 September 2026.1) Statement and interpretation
For a trigonometric polynomial $f(\theta)=\sum_{k=0}^n c_ke^{ik\theta}$ whose zeros are real, normalized by $\max_{\theta\in[0,2\pi]}|f(\theta)|=1$, the requested estimate is \[ \int_0^{2\pi}|f(\theta)|\,d\theta\le4.\tag{1} \] The current problem page records an affirmative solution by Kristiansen and, for complex coefficients, by Saff and Sheil-Small. This attempt proves sharp special families, not their general theorem.There is a degenerate case to exclude when reading the displayed statement literally: $f(\theta)=e^{im\theta}$ has no zeros, so "all zeros are real" is vacuous, but its integral is $2\pi$. Accordingly, the analysis below concerns the nondegenerate zero-bearing case. After removing a nonvanishing factor $e^{im\theta}$, this is a polynomial $Q(e^{i\theta})$ with $Q(0)\ne0$, $\deg Q=d\ge1$, all of whose algebraic zeros lie on the unit circle. A nonzero algebraic zero $\rho e^{it}$ corresponds to trigonometric zeros $\theta=t-i\log\rho+2\pi j$, which explains this correspondence. The degenerate observation is a statement convention, not a claimed new disproof of the classical result.
2) Factorization and a sharp one-zero calculation
Write $Q(z)=C\prod_{j=1}^d(z-e^{it_j})$. Then \[ |Q(e^{i\theta})|=2^d|C|\prod_{j=1}^d\left|\sin\frac{\theta-t_j}{2}\right|.\tag{2} \] For $d=1$, normalization gives $|C|=1/2$, and changing variables yields \[ \int_0^{2\pi}\left|\sin\frac{\theta-t_1}{2}\right|\,d\theta=4. \] Thus the proposed constant cannot be decreased.3) Coincident zeros, including every multiplicity
If all zeros coincide, the normalized modulus is $|\sin((\theta-t)/2)|^d$. Since $0\le u\le1$ implies $u^d\le u$ for integers $d\ge1$, $1$ follows immediately. More precisely, with \[ I_d=2\int_0^\pi\sin^d u\,du, \] integration by parts gives $I_d=(d-1)I_{d-2}/d$ for $d\ge2$, starting with $I_0=2\pi$ and $I_1=4$. Hence $I_2=\pi$, and $I_d<4$ for every $d\ge2$. The boundary term in integration by parts vanishes since $\sin u=0$ at both endpoints.4) Regular polygons and repeated regular polygons
For integers $q,d\ge1$ consider \[ Q(z)=\left(\frac{z^q-e^{i\alpha}}2\right)^d. \] Every zero lies on the unit circle, the supremum of the modulus there is one, and \[ |Q(e^{i\theta})|=\left|\sin\frac{q\theta-\alpha}{2}\right|^d. \] The function $|\sin u|^d$ has period $\pi$. The substitution $u=(q\theta-\alpha)/2$ integrates over exactly $q$ periods and therefore gives integral $I_d$, independent of $q$ and $\alpha$. In particular, simple regular-polygon zeros attain equality for every degree $q$, whereas repeated copies give the strict bounds above. Multiplication by a unimodular constant or by $z^m$ preserves these modulus conclusions.5) Remaining gap and sources
UNRESOLVED in this attempt. Arbitrarily spaced roots in $2$ have not been treated. Bounding each sine factor by one does not control the product after its own supremum is normalized: the individual factors need not attain their maxima together. The special-family proof therefore does not imply the general result.Sources: https://www.erdosproblems.com/225, checked 23 September 2026; E. B. Saff and T. Sheil-Small, Coefficient and Integral Mean Estimates for Algebraic and Trigonometric Polynomials with Restricted Zeros, J. London Math. Soc. $2$ 9 $1974$, 16--22, https://doi.org/10.1112/jlms/s2-9.1.16. The classical theorem is attribution here, not an unproved step presented as our proof.
The estimate describes the coverage of this partial reconstruction, not the status of the classical problem or a correctness probability.