Rumour at the pond

POND NEWS. At midnight a small frog croaked a rumour: "the pond will dry up". It was not true. Every frog that heard it told two new frogs in the following hour.

Midnight: a frog croaks a false rumour. Each frog that hears it tells $2$ new frogs in the next hour.

Rumour $R$: each frog that knows it tells $2$ new ones per hour.

The count is simple. In hour one, two new frogs heard it; in hour two, four; in hour three, eight; and it keeps doubling. After eight hours 511 frogs knew, and after nine hours 1023 frogs, which is more than the whole pond of 1000.

New frogs in hour $k$: $2^k$. Hours $1,2,3$: $2,4,8$. After $n$ hours, ${\sum }_{k=0}^n{2}^k=2^{n+1}-1$ frogs know it. For $n=8$ that is $511$; for $n=9$ it is $1023>1000$, the whole pond.

${\sum }_{k=0}^8{2}^k=511<1000$, ${\sum }_{k=0}^9{2}^k=1023>1000$.

Then the oldest frog told the truth: the pond is fed by a spring. But the truth was told by one frog to one frog each hour. After nine hours only ten frogs knew the truth.

The truth: one frog tells one frog per hour. After $9$ hours, $1+9=10$ frogs know it, and $10<1000$.

Truth $T$: $1+9=10<1000$.

Lesson of the day: a rumour spreads by doubling, a correction only by counting. Make sure before you croak.

A rumour doubles ($2^n$), a correction only counts ($n$). For every $n\ge 1$ we have $2^n>n$.

$r(n)=2^n$, $c(n)=n$; $\forall n\ge 1,r(n)>c(n)$.