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Statement

Inequality (3) (p. 185). Let z1,…,znz_1,\ldots,z_n be complex numbers with

1=z1≥∣z2∣≥⋯≥∣zn∣,(1)1=z_1\ge\lvert z_2\rvert\ge\cdots\ge\lvert z_n\rvert, \tag{1}

and put sk=∑m=1nzmks_k=\sum_{m=1}^n z_m^k and s=max⁡1≤k≤n∣sk∣s=\max_{1\le k\le n}\lvert s_k\rvert, the paper's (2). Then

s>1/6.(3)s>1/6. \tag{3}

The bound holds for every nn and every choice of z1,…,znz_1,\ldots,z_n subject to (1). The paper presents it, in its words (p. 185), "Without any suggestion that this is a precise value"; it makes no claim that 1/61/6 is best possible.

Context (p. 185). Turán posed the problem of a positive lower bound for ss valid for all choices subject to (1). The paper records Turán's bound (log⁡2)/∑m=1nm−1(\log 2)/\sum_{m=1}^n m^{-1}, de Bruijn's improvement to Clog⁡log⁡n/log⁡nC\log\log n/\log n for some C>0C>0 and sufficiently large nn, and Uchiyama's proof that CC could be taken arbitrarily close to 11; (3) verifies the conjecture that ss has a positive lower bound independent of nn.

Source. F. V. Atkinson, On sums of powers of complex numbers, Acta Math. Acad. Sci. Hungar. 12 (1961), no. 1--2, 185--188, DOI 10.1007/BF02066680: (1), (2) and (3) on p. 185, the proof on pp. 185--188, (13) on p. 187 and the concluding deduction on p. 188. The edition read is identified on the source card.

Read depth. Claims checked: (1), (2), (3) and the inequality (13) with its hypothesis s<1/4s<1/4 were read clause by clause on the page images. The proof was read for its structure, not checked line by line. Nothing here is independently reviewed.

Proof pointer

Sections 2 and 3, pp. 185--188. With $g(\theta)=-\sum_{m=1}^n m^{-1}s_m e^{mi\theta}$, the exponential eg(θ)e^{g(\theta)} equals ∏r=1n(1−zreiθ)\prod_{r=1}^n(1-z_re^{i\theta}) plus a power series in eiθe^{i\theta} starting at the exponent n+1n+1 (equations (4), (5)); since z1=1z_1=1, the product vanishes at θ=0\theta=0. Reading the tail coefficients as Fourier coefficients and integrating by parts gives the identity (8),

1=(2πi)−1∫−ππg′(θ)eg(θ)−g(0)h(θ) dθ,h(θ)=∑m=n+1∞m−1e−miθ.1=(2\pi i)^{-1}\int_{-\pi}^{\pi}g'(\theta)e^{g(\theta)-g(0)}h(\theta)\,d\theta, \qquad h(\theta)=\sum_{m=n+1}^{\infty}m^{-1}e^{-mi\theta}.

Schwarz's inequality, Parseval's equality for g′g' (giving at most 2πns22\pi ns^2), and separate bounds for the second factor on ∣θ∣≤π/n\lvert\theta\rvert\le\pi/n and π/n≤∣θ∣≤π\pi/n\le\lvert\theta\rvert\le\pi, the latter assuming s<1/4s<1/4, give (13) (p. 187):

1<s2e2πs{1+e4s(1−4s)−1}.(13)1<s^2e^{2\pi s}\{1+e^{4s}(1-4s)^{-1}\}. \tag{13}

The right side is independent of nn and increasing on 0<s<1/40<s<1/4, and (13) fails at s=1/6s=1/6 (p. 188), so s>1/6s>1/6. Not reconstructed further here.

Dependencies

None beyond classical analysis: the expansion (4), which the paper takes from Uchiyama (Acta Math. Acad. Sci. Hungar. 9 (1958), 275--278), Schwarz's inequality and Parseval's equality.

Bears on

  • Problem 519: the problem asks whether max⁡1≤k≤n∣∑izik∣\max_{1\le k\le n}\lvert\sum_i z_i^k\rvert exceeds an absolute constant c>0c>0 for all complex z1,…,znz_1,\ldots,z_n with z1=1z_1=1. Inequality (3) gives c=1/6c=1/6 under the paper's condition (1), which adds that every ∣zm∣\lvert z_m\rvert is at most 11. The problem's hypothesis reduces to (1) (an observation of this page, not of the paper): dividing every zmz_m by one of largest modulus M≥1M\ge1 divides ∣sk∣\lvert s_k\rvert by MkM^k, and after relabeling the quotients satisfy (1), so the bound 1/61/6 holds under the problem's hypothesis too. The problem's claim page for this paper records the result.