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Statement

Theorem 7 (p. 9). Let $p(\varepsilon)$ be the length of the longest descending wave in the sequence $a_n=c^n$, where $c=1+\varepsilon$. Then there exist constants $A$ and $B$ such that

$$A/\sqrt{\varepsilon}\le p(\varepsilon)\le B/\sqrt{\varepsilon}.$$

The statement places no restriction on $\varepsilon$. The proof's lower bound is given for $\varepsilon<0.6$, where it takes $A=0.787$, and its upper-bound argument ends by saying that the length is less than, approximately, twice $1/\sqrt{\varepsilon}$ (p. 10).

Source. Brown, T. C., Erdős, P. and Freedman, A. R., Quasi-progressions and descending waves, J. Combin. Theory Ser. A 53 (1990), no. 1, 81--95, doi:10.1016/0097-3165(90)90021-N, read in the authors' copy identified on the source card, whose pages are numbered 1 to 13: the statement on p. 9, the proof on pp. 9--10.

Read depth. Claims checked: the statement was read clause by clause on the print's page. The proof was read but not checked step by step; as noted above, its upper bound is argued only approximately in the print. Nothing here is independently reviewed.

Proof pointer

Section 4, pp. 9--10. Lower bound: the $t+2$ terms $1,c^{t+1},c^{(t+1)+t},\ldots,c^{(t+1)+t+\cdots+1}$ form a descending wave exactly when $c^t\le1+\sqrt{\varepsilon/c}$, which allows $t$ of order $1/\sqrt{\varepsilon}$. Upper bound: in a descending wave $c^{r_1},c^{r_2},\ldots$ the exponent gaps cannot increase, and once the ratio of consecutive terms is close to 1 the exponent gaps are bounded by about $R=1/\sqrt{\varepsilon}$, so the wave has length about at most $2R$.

Dependencies

None outside the paper.

Bears on

No problem page.