In mathematics , the Prouhet–Thue–Morse constant , named for Eugène Prouhet [fr ] , Axel Thue , and Marston Morse , is the number—denoted by τ —whose binary expansion 0.01101001100101101001011001101001... is given by the Prouhet–Thue–Morse sequence . That is,
τ τ -->
=
∑ ∑ -->
n
=
0
∞ ∞ -->
t
n
2
n
+
1
=
0.412454033640
… … -->
{\displaystyle \tau =\sum _{n=0}^{\infty }{\frac {t_{n}}{2^{n+1}}}=0.412454033640\ldots }
where tn n th
Other representations
The Prouhet–Thue–Morse constant can also be expressed, without using tn [ 1]
τ τ -->
=
1
4
[
2
− − -->
∏ ∏ -->
n
=
0
∞ ∞ -->
(
1
− − -->
1
2
2
n
)
]
{\displaystyle \tau ={\frac {1}{4}}\left[2-\prod _{n=0}^{\infty }\left(1-{\frac {1}{2^{2^{n}}}}\right)\right]}
This formula is obtained by substituting x = 1/2 into generating series for tn
F
(
x
)
=
∑ ∑ -->
n
=
0
∞ ∞ -->
(
− − -->
1
)
t
n
x
n
=
∏ ∏ -->
n
=
0
∞ ∞ -->
(
1
− − -->
x
2
n
)
{\displaystyle F(x)=\sum _{n=0}^{\infty }(-1)^{t_{n}}x^{n}=\prod _{n=0}^{\infty }(1-x^{2^{n}})}
The continued fraction expansion of the constant is [0; 2, 2, 2, 1, 4, 3, 5, 2, 1, 4, 2, 1, 5, 44, 1, 4, 1, 2, 4, 1, …] (sequence A014572 OEIS )
Yann Bugeaud and Martine Queffélec showed that infinitely many partial quotients of this continued fraction are 4 or 5, and infinitely many partial quotients are greater than or equal to 50.[ 2]
Transcendence
The Prouhet–Thue–Morse constant was shown to be transcendental by Kurt Mahler in 1929.[ 3]
He also showed that the number
∑ ∑ -->
i
=
0
∞ ∞ -->
t
n
α α -->
n
{\displaystyle \sum _{i=0}^{\infty }t_{n}\,\alpha ^{n}}
is also transcendental for any algebraic number α, where 0 < |α | < 1.
Yann Bugaeud proved that the Prouhet–Thue–Morse constant has an irrationality measure of 2.[ 4]
Appearances
The Prouhet–Thue–Morse constant appears in probability . If a language L over {0, 1} is chosen at random, by flipping a fair coin to decide whether each word w is in L , the probability that it contains at least one word for each possible length is [ 5]
p
=
∏ ∏ -->
n
=
0
∞ ∞ -->
(
1
− − -->
1
2
2
n
)
=
∑ ∑ -->
n
=
0
∞ ∞ -->
(
− − -->
1
)
t
n
2
n
+
1
=
2
− − -->
4
τ τ -->
=
0.35018386544
… … -->
{\displaystyle p=\prod _{n=0}^{\infty }\left(1-{\frac {1}{2^{2^{n}}}}\right)=\sum _{n=0}^{\infty }{\frac {(-1)^{t_{n}}}{2^{n+1}}}=2-4\tau =0.35018386544\ldots }
See also
Notes
References
Allouche, Jean-Paul; Shallit, Jeffrey (2003). Automatic Sequences: Theory, Applications, Generalizations . Cambridge University Press . ISBN 978-0-521-82332-6 Zbl 1086.11015 . Pytheas Fogg, N. (2002). Berthé, Valérie ; Ferenczi, Sébastien; Mauduit, Christian; Siegel, Anne (eds.). Substitutions in dynamics, arithmetics and combinatorics . Lecture Notes in Mathematics. Vol. 1794. Berlin: Springer-Verlag . ISBN 3-540-44141-7 Zbl 1014.11015 .
External links
![{\displaystyle \tau ={\frac {1}{4}}\left[2-\prod _{n=0}^{\infty }\left(1-{\frac {1}{2^{2^{n}}}}\right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2e8057422b4c954b51cdc2e975d308ef61ce41fa)
