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. £
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