this claim almost certainly holds.

With the chosen initial data (the first point at re

uration trajectory enters the wedge in the direction,

vertical side. In this case, the number of reflections

modification of formula (1.3), namely

N(k) =

7T

- 1 .

arctan (lO-^)

This fact is established by the same unfolding method

For now, denote 10~~fc by x. This x is a very sma

one expects arctan x to be very close to x. More preci

1 V

(1.5) 0

arctan x x

x for x 0.

Exercise 1.5. Prove (1.5) using the Taylor expansion

The first k digits of the number

1 - ri0*7rl - 1 =

|_10*TTJ

coincide with the first k + 1 decimal digits of TT. The

follows from the fact that 10fc7r is not an integer; \

function, the greatest integer not greater than y.

We will be done if we show that

(1.6)

By (1.5),

(1.7)

arctan x

7T

arctan x

7T

h

7TX

X

The number TTX = 0.0. ..031415... has k — 1 zeros a

dot. Therefore the left- and the right-hand sides in

only if there is a string oi k — 1 nines following the firs

the decimal expansion of n. We do not know whethe

ever occurs, but this is extremely unlikely for large

one does not have such a string, then both inequaliti

equalities, (1.6) holds, and the claim follows. £